E8 Quillen Superconnection

May 10, 2008

E₈ Quillen Superconnection

Posted by Urs Schreiber

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A remark on the nature of Quillen superconnections with values in $\mathbb{Z}_2$ -graded Lie algebras, such as $e_8$ .

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Quick remark on the background for those unfamiliar with Yang-Mills (super) gauge theory

For decades physicicts have been playing around with symbols, trying to identify the mathematical structure that underlies the mess they read off their particle accelerators.

The biggest general success has been the identification that all forces in nature can be identified with the mathematical structure of a connection on a fiber bundle. Known as “Yang-Mills theory”, this is the heart of the standard model of particle physics today.

Apart from forces, there is matter. It turns out that while forces are described by connections on fiber bundles, these fiber bundles are spin bundles and matter turns out to be encoded by sections of these Spin bundles. The picture thus obtained is called the standard model.

While nice, this insight makes some people think that one should look for mathematical structures further unifying this picture. There is an obvious guess: extend the structure Lie group (called the gauge Lie group in this context) of your fiber bundles to a super Lie group. That will make the connection have odd graded components which may possibly be identified with the spin sections we had before. Hence it would allow to identify forces and matter alike with the mathematical structure given by a connection, distinguished just by the super degree of the components of that connection.

And in fact, in supergravity theories it works just like this: the gravitational force itself is a connection with value in the Poincaré Lie algebra $iso(n,m)$ , and the “matter” piece in this context, called the gravitino, is simply the odd part obtained when extending this to a super Poincaré connection with values in $s iso(n,m)$ .

But that also makes it clear that, unlike the matter observed in the standard model, gravitinos are spinor-valued 1-forms, not 0-forms.

What structure?

In physics people often feel free to manipulate their symbols first, and worry about finding the mathematical interpretation of these manipulations later. One needs to be careful with this, but lots of interesting structure was found this way over the centuries, beginning with Newton’s “fluxions”.

In arXiv:0711.0770 the proposal is – implicitly – not to worry about the above problem – that a superconnection is locally given by a super 1-form – and just formally add to an ordinary connection 1-form an odd 0-form: $\underbrace{A }_{\in \Omega^{(1,even)}(Y,g)} + \underbrace{\Psi }_{\in \Omega^{(0,odd)}(Y,g)} \;\;\;\; (1)$ where $g$ is some $\mathbb{Z}_2$ -graded Lie algebra, for instance $e_8$ I, II.

The proposal that this might be the key to making big progress with understanding the standard model of particle physics had created an unprecedented amount of attention and lots of criticism.

Much of that criticism revolved around the interpretation of the formal sum above. While arXiv:0711.0770 offers an interpretation, this was perceived as unviable.

Quillen superconnections on $\mathbb{Z}_2$ -graded bundles

Since I have never seen it mentioned in any of these discussions, all I want to point out here is that (1), for $g$ a $\mathbb{Z}_2$ -graded Lie algebra, can be read as a Quillen superconnection on a (necessarily) $\mathbb{Z}_2$ -graded vector bundle $E \to X$ over spacetime $X$ , which is associated to a principal $G$ -bundle by some representation of $G$ on $\mathbb{Z}_2$ -graded vector spaces:

$\mathbf{A} := d + A + \Psi$ $\mathbf{A} : (\Omega^\bullet(X,E))_{even,odd} \to (\Omega^\bullet(X,E))_{odd,even}$ $\forall \omega \in \Omega^1(X) : [\mathbf{A},\omega] = d\omega \,.$

Notice two things:

1) Quillen superconnections are different from other notions of superconnections. In particular, Quillen superconnections do not come from a path-lifting property and are not related to an ordinary notion of parallel transport. For a discussion of Quillen superconnections and also of super parallel transport I can recommend

Florin Dumitrscu,
Superconnections and Parallel Transport
(pdf).

2) It is crucial to distinguish here between $\mathbb{Z}_2$ -graded Lie algebras and super Lie algebras. As follows:

there are precisely two different symmetric braided monoidal structures on the category of $\mathbb{Z}_2$ -graded vector spaces: the trivial one and the one which introduces a sign when two odd vectors are exchanged in a tensor product.

The symmetric monoidal category with the trivial braiding here is $\mathbb{Z}_2 Vect$ . The other one is $S Vect$ .

A $\mathbb{Z}_2$ -graded Lie algebra is a Lie algebra internal to $\mathbb{Z}_2 Vect$ . A super Lie algebra is a Lie algebra in $S Vect$ .

$\mathbb{Z}_2$ -graded Lie algebras such as $e_8$ are not super Lie algebras: while they look alike a lot (especially $e_8$ does look a lot alike a super $so(16)$ Lie algebra) they are different.

Main point, summarized.

Whatever the physical viability of the proposal of arXiv:0711.0770, the expression in equation (3.1) on p. 23 is to be interpreted as a Quillen superconnection $\mathbf{A}$ on a $\mathbb{Z}_2$ -graded $e_8$ associated vector bundle and (3.2) is the corresponding Quillen curvature $F_{\mathbf{A}} = \mathbf{A}^2 \,.$

So if one wants to examine the possibility of describing particle physics with this approach, the mathematical structure to determine would seem to be something like “Quillen Yang-Mills theory”.

Posted at May 10, 2008 8:20 AM UTC

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148 Comments & 3 Trackbacks

Re: E8 Quillen Superconnection

Wow, this sounds cool Urs.

Posted by: Bruce Bartlett on May 10, 2008 11:06 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

with values in Z₂-graded Lie algebras, such as e₈.

Every Lie algebra admits Z₂-gradings. Just consider some Z-grading, e.g. the ones corresponding to a simple root, modulo 2.

Posted by: Thomas Larsson on May 10, 2008 12:01 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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You can play this game with any Lie algebra with any $\mathbb{Z}_2$ -grading you like. In particular with the trivial $\mathbb{Z}_2$ -grading that assigns even degree to every element.

But if you do it for the “natural” $\mathbb{Z}_2$ -grading of $e_8$ that Lisi considers, then a Quillen superconnection with respect to that $\mathbb{Z}_2$ -grading is essentially what Lisi considers.

Posted by: Urs Schreiber on May 10, 2008 3:46 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Yes, but every Lie algebra (at least simple, finite-dim) admits a non-trivial Z_2-grading, like this:

Every root is a sum of simple roots, with coefficients +1, 0, -1. The even (odd) root sublattice consists of roots where the sum of coefficients is even (odd). The CSA belongs to the even part.

Hm. I wonder if you get a non-isomorphic Z_2-grading by choosing different simple roots.

Posted by: Thomas Larsson on May 12, 2008 6:29 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Yes.

(That’s basically all I wanted to say, but I’ll elaborate, and ask two questions.)

I referred to this construction as a “superconnection” a few times in the paper (including in the abstract), having come across Quillen’s work. But I tried to do it in a way that would make it clear I was not using supersymmetry, since, as you say, it’s different.

The way I got there, and so the way I still think about a superconnection, is as the “generalized connection” that shows up in BRST. Often, in the physics literature, this is related to the “Russian formula,” which is just the supercurvature. I don’t know why people don’t like this construction–probably because they’re afraid of ghosts. But, all I care about is whether the BRST generalized connection is, in fact, a kind of Quillen superconnection. Urs, could you confirm this is true, so I’m not the only one saying it?

The second question has to do with the use of super Lie algebras vs Lie algebras. Basically, I don’t think the choice is physically discernible, because the Lie brackets between the “matter” parts of the superconnection don’t appear in the physical action. And this is where the sign flip occurs. So, for the purpose of matching the standard model, it shouldn’t matter whether we start with a super Lie algebra or a Lie algebra–at least until we talk about including this quadratic matter term and going beyond the standard model. Does this make sense?

Thanks Urs!

-Garrett

Posted by: Garrett on May 10, 2008 4:55 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Garrett wrote:

Yes.

I was half and half expecting that after posting I would just be told: “Yes, sure, what you just said is what we were talking about all along.”

But there are different notions of superconnections. And there seemed to be puzzlement of how to think of the 0-form part of your construction. There may still be puzzlement about this for the physical interpretation, but at least I thought it might be helpful to point out that it fits at least into the existing concept of Quillen superconnections.

And, you know, it’s not quite a “Yes” even if we agree on that: you mention “Grassmann” parts which I can’t quite identify from the point of view I have described it, nor do I see the need for them. That may (and probably is) just a matter of terminology, there are a couple of signs here that go by different names.

This is important: the curvature of the Quillen connection that I mentioned actually differs from curvature by one term: in general it contains the 0-form term $\psi^2$ !

You say in your article that this vanishes due to the Grassmann nature of $\psi$ . But in the setup I described $\psi$ is not a Grassmann thing, but just a 0-form with values in odd endomorphisms of a $\mathbb{Z}_2$ -graded vector space. (More on that below.)

But apart from that, I am glad we agree on this.

What I would like to discourage you frm, though, is to use the motivation via BRST. Even if technically correct (I am not sure yet I see exactly what you have in mind) it is at best misleading: in BRST formalism the ghosts are, by the very construction, not physical. So it is sub-optimal to claim that that your supposedly physical fermions are BRST ghosts.

Not every group one runs into is a gauge group. Not every grading one sees is a ghost.

And please remind me what you mean by the “BRST generalized connection”. Sorry, I guess I should know what you mean, but I am not sure that i do. I remember you gave pointers somewhere, but I can’t find it right now. You don’t just mean the BRST operator itself, do you?

I don’t think the choice is physically discernible

Maybe, I am not sure yet.

because the Lie brackets between the “matter” parts of the superconnection don’t appear in the physical action.

As I remarked above, at least the way I tried to make sense of the formula, they do appear, since the Quillen curvature has a $\psi^2$ 0-form piece.

Maybe that drops out again when you feed that curvature into your BV action and extremize? I don’t know.

Personally, I’d feel like fixing very carefully the true mathematical meaning of the starting point of your discussion before asking questions about physical observables. You should feel free, as you do, to speed ahead and see if the train will be running in the right direction, but I’d right now be more interested in checking that it runs on steadfast tracks from the beginning.

So: I’d be happy if we could first make absolutely clear what the precise nature of the “superconnection” is that we are considering.

Posted by: Urs Schreiber on May 10, 2008 9:46 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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I was half and half expecting that after posting I would just be told: “Yes, sure, what you just said is what we were talking about all along.”

Well, that would have been disingenuous, which is why I elaborated on my “yes.” From my perspective, I was playing with this construction, and noted the similarity to Quillen’s superconnection (enough to refer to it as a superconnection). But, because the literature I encountered on superconnections didn’t seem entirely coherent, I wasn’t confident enough in my understanding of the correspondence to use it, and used the BRST construction instead.

This is why I’m very interested in your post–I’m hoping to learn more about superconnections.

But there are different notions of superconnections. And there seemed to be puzzlement of how to think of the 0-form part of your construction. There may still be puzzlement about this for the physical interpretation, but at least I thought it might be helpful to point out that it fits at least into the existing concept of Quillen superconnections.

Yes, I’d very much like to understand this in more detail.

And, you know, it’s not quite a “Yes” even if we agree on that: you mention “Grassmann” parts which I can’t quite identify from the point of view I have described it, nor do I see the need for them. That may (and probably is) just a matter of terminology, there are a couple of signs here that go by different names.

I consider it to be a matter of terminology, but the different terminology lends itself to different constructions–and I would like to find the most natural construction.

In BRST, the algebraically odd valued part of the connection 1-form is replaced with Grassmann fields valued in the odd part, which can be formally added to the algebraically even valued 1-forms to make a generalized connection–what we’re calling the superconnection. There is a rather Byzantine (old and new) literature devoted to developing a more geometric description of this construction. (I believe a frequent commenter, Jim Stasheff, wrote a related paper.) The most successful idea seems to be to consider these “Grassmann” parts as 1-forms in the space of connections. This may or may not jive with the description you provided of these 1-forms being “orthogonal” to the space of the even valued 1-forms (I’m oversimplifying). I’d really like to understand the geometry of a Quillen superconnection better, since using BRST doesn’t make me so happy either.

This is important: the curvature of the Quillen connection that I mentioned actually differs from curvature by one term: in general it contains the 0-form term ψ 2!

You say in your article that this vanishes due to the Grassmann nature of ψ. But in the setup I described ψ is not a Grassmann thing, but just a 0-form with values in odd endomorphisms of a ℤ 2-graded vector space. (More on that below.)

I agree that this quadratic term is in the supercurvature. I end up throwing this term out, explicitly, by not contracting it with a corresponding part of a $B$ in the action–which is built by hand to agree with the standard model action. I also think doing this by hand is unsatisfactory, and would like to find better mathematical justification for why the action is what it is.

But apart from that, I am glad we agree on this.

What I would like to discourage you frm, though, is to use the motivation via BRST. Even if technically correct (I am not sure yet I see exactly what you have in mind) it is at best misleading: in BRST formalism the ghosts are, by the very construction, not physical. So it is sub-optimal to claim that that your supposedly physical fermions are BRST ghosts.

Not every group one runs into is a gauge group. Not every grading one sees is a ghost.

I’m quite amenable to dropping the BRST interpretation. It is simply the way I got here, and the way I thought it would be most understandable (if controversial) to other high energy physicists.

The reason I would be so willing to drop BRST is because the use of Grassmann fields seems geometrically unnatural. It took me quite a while to come to a good geometric understanding and appreciation of what a principal bundle connection is, with its curvature described by a natural geometric object: the Frölicher-Nijenhuis bracket of the connection with itself.

As far as I know, there is no similarly nice, uncomplicated geometric construction for the BRST generalized connection, though many people have tried various more or less promising approaches. So I’m not particularly attached to it.

I will be very happy if I can learn about a natural geometric construction that exists for Quillen’s superconnection and its curvature, though I suspect it may take me a while.

I’m also encouraged that, thanks to John’s TWF262, I now have a more geometric understanding of what a Lie superalgebra is, in terms of a Killing superalgebra. It seems I’m getting more “super,” despite initial resistance. If I end up understanding Quillen superconnections in terms of D-branes, I’ll be my own worst enemy. ;)

And please remind me what you mean by the “BRST generalized connection”. Sorry, I guess I should know what you mean, but I am not sure that i do. I remember you gave pointers somewhere, but I can’t find it right now. You don’t just mean the BRST operator itself, do you?

OK, no, not the BRST operator. Although there are other references, I first learned this from van Holten’s review,

http://arxiv.org/abs/hep-th?0201124

especially page 70. I will paraphrase–but I have to start a new comment, since this one has gotten too long.

Posted by: Garrett on May 11, 2008 4:01 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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(continuation of an unusually long comment)

Following van Holten, we start with a principal bundle connection,

(1)

A = H + C

in which $H$ and $C$ are Lie algebra valued 1-forms inhabiting the “even” and “odd” parts of some Lie algebra. Then we can construct (by hand) a modified BF Lagrangian that is invariant under arbitrary gauge transformation of the $C$ part,

(2)

L = B F + B_H*B_H

in which $B = B_H + B_C$ is a Lagrange multiplier field that lives in both ( $H$ and $C$ ) parts of the algebra. If the Lie group corresponding to $H$ is a symmetric subgroup, then the curvature of $A$ also breaks into even and odd parts,

(3)

F = d A + A A

(4)

= (d H + H H + C C) + (d C + H C + C H)

(The $CC$ term is the source of confusion, and it will go away in the next step.) Using the BRST gauge fixing technique, we can replace the $C$ part of the connection with “ghosts,” $\Psi$ , which are Lie algebra valued Grassmann numbers. A new effective Lagrangian for this theory, using standard BRST, is

(5)

L_eff = B_g D \Psi + F_H*F_H

in which

(6)

D \Psi = (d C + H C + C H)

is the covariant derivative, $B_g$ are BRST anti-ghosts, and the $H$ part of the curvature is

(7)

F_H = d H + H H

The CC term goes away because $C$ has been replaced by ghosts, and there is no ghost-ghost ( $\Psi \Psi$ ) term in the effective Lagrangian from BRST. This should answer the question of what is going on.

The BRST transformation, involving the BRST operator, relates $C$ and $\Psi$ . This is the analogue to supersymmetry. However, we have chosen a BRST action that takes $C$ to zero. So all we have is $\Psi$ and $H$ (with no transformation between these two).

Now, in the next step, we formally add $H$ and $\Psi$ in an extended connection (also called a “generalized connection”),

(8)

A' = H + \Psi

and compute its generalized curvature,

(9)

F' = (d H + H H) + (d \Psi + H \Psi + \Psi H) + (\Psi \Psi)

from which the effective Lagrangian can be built. In this construction, the $\Psi \Psi$ is “ghost grade 2,” and doesn’t appear in the effective Lagrangian because $B_g$ is only ghost grade -1. So I just drop this term.

Since this quadratic term is getting dropped, it doesn’t matter what its sign is, and it shouldn’t matter whether this part of the algebra is super.

As far as I can tell, this kind of BRST extended connection is a kind of Quillen superconnection. Is this true?

I see you discussing this with Jacques Distler now, who claims this would only be a kind of “Schreiber superconnection.” If he’s right, that’s also OK with me, since I drop the quadratic term anyway. It would just mean having to start with a supergroup to have a Quillen connection. For physics, a supergroup that resembles $E_8$ .

As I remarked above, at least the way I tried to make sense of the formula, they do appear, since the Quillen curvature has a ψ 2 0-form piece.

Maybe that drops out again when you feed that curvature into your BV action and extremize? I don’t know.

Yes, that’s exactly what I’m doing.

Personally, I’d feel like fixing very carefully the true mathematical meaning of the starting point of your discussion before asking questions about physical observables. You should feel free, as you do, to speed ahead and see if the train will be running in the right direction, but I’d right now be more interested in checking that it runs on steadfast tracks from the beginning.

I agree with you. But, remember, I got here by building up from the standard model, and I’m as conservative as possible in what mathematical machinery I’m willing to add.

It was extremely valuable, for me, to work out in detail the natural, geometric description of a principal bundle connection. (Getting my description on steadfast tracks.) The BRST formulation is some standard machinery necessary in QFT, but it’s very frustrating for me that it doesn’t have a nice, geometric construction.

When I first showed this BRST interpretation of fermions to John Baez, he wrinkled his nose and said he preferred to think of that as the odd part of $Z_2$ graded 1-form. That’s fine with me, and even better if I can come to understand a nice, geometric description of a Quillen connection.

So: I’d be happy if we could first make absolutely clear what the precise nature of the “superconnection” is that we are considering.

Me too. Is there a review paper on Quillen superconnections, suitable for physicists?

Posted by: Garrett on May 11, 2008 4:13 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Errata: Equation (6) should be all $\Psi$ ’s, instead of $C$ ’s.

Posted by: Garrett on May 11, 2008 4:23 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Ah, these lecture notes by Shlomo Sternberg look good as an introduction. But there still isn’t a natural geometric description–I’ll have to dig more.

Posted by: Garrett on May 11, 2008 4:40 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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I find the van Holten review a bit problematic, at least for our purposes here. It’s not just the he consistently writes “principle bundle” instead of “principal bundle”. :-)

Posted by: Urs Schreiber on May 12, 2008 9:01 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Heh, yes, van Holten does possess strong moral fiber. But his is the most practical introduction to BRST I know of that includes the generalized connection.

There is this more mathematically oriented review of BRST:
http://arxiv.org/abs/hep-th/9408003
but I don’t think they cover generalized connections.

This is a good, mathematically oriented paper:
http://arxiv.org/abs/hep-th/9705123
and it introduces the generalized connection on p51.

This very recent paper is entirely devoted to the BRST extended connection:
http://arxiv.org/abs/0710.5698

And this paper explicitly discusses it as a superconnection:
http://www.atlantis-press.com/php/download_paper.php?id=364

There is also this recent paper by some of the pioneers in this field:
http://arxiv.org/abs/hep-th/9611168
It strikes a nice balance of mathematical accuracy and practicality.

These are some of the more recent papers discussing the BRST generalized connection, but it is a very rich area and the extensive literature stretches back to the original papers on BRST, and the “Russian formula” for the supercurvature. The mathematical foundation of this construction is varied, but certainly firm.

And as much as I would be amused to refer to this as a “Schreiber superconnection,” its existence has a long and rich history.

But, much of this mathematics is rather convoluted. If you know of a more natural, geometric description of this kind of superconnection, built from geometric parts rather than built up from BRST pieces, I’d be interested to learn about it.

Posted by: Garrett on May 13, 2008 1:21 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Garrett,
In what sense is a formula on p. 51 (thanks for a precise ref as well as all the others)a **generalized** connection?

Urs etal,
For a sizeable chunk of your audience, graded Lie algebra means with appropriate signs, both for the `anti-symm’ and for the Jacobi. In homotopy theory, these objects (without the name) go back to the 50s, cf.
MR0091473 (19,974g)
Uehara, Hiroshi; Massey, W. S.
The Jacobi identity for Whitehead products. Algebraic geometry and topology. A symposium in honor of S. Lefschetz, pp. 361–377. Princeton University Press, Princeton, N. J., 1957.

If you do wnat to talk about a Lie algebra whihc respects some additional grading, how about calling it a Lie algebra with grading. Do you really see any use for such a gadget?

In the graded world, e.g. of graded differentials, it is the graded commutator which is called for.

jim

Posted by: jim stasheff on May 13, 2008 2:12 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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In what sense is a formula on p. 51 (thanks for a precise ref as well as all the others)a **generalized** connection?

Hi Jim,

I am merely repeating what others have called it. Roughly, a connection is usually a Lie algebra valued 1-form over a base manifold, so my guess is this has been called a “generalized” connection because it’s a formal addition of Lie algebra valued 1-forms and Lie algebra valued Grassmann numbers.

I suppose Quillen’s superconnection would also qualify as a generalized connection in a similar sense. But then we’re going to have to talk about specific generalized connections, which is silly. So maybe we’re better off calling this a “Schreiber superconnection” after all. ;)

There are quite a few different geometric interpretations of what the Grassmann part of this connection means. And I should also include this nice description in terms of jet bundles, which you’re probably familiar with:

http://arxiv.org/abs/hep-th/9712157

But, as I’ve said, I’m not completely happy with any of these, and would prefer to have a more elementary geometric understanding of this object.

Posted by: Garrett on May 13, 2008 4:02 PM | Permalink | Reply to this

ℤ2 gradings

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If you do wnat to talk about a Lie algebra whihc respects some additional grading, how about calling it a Lie algebra with grading. Do you really see any use for such a gadget?

One place where these are used is in the theory of Symmetric Spaces. Their classification comes down to classifying the possible $\mathbb{Z}_2$ gradings on the Lie algebra.

Indeed, the main reason why Lisi’s idea fails is that Marcel Berger classified the possible $\mathbb{Z}_2$ -gradings of $e_8$ , and the one desired by Lisi doesn’t exist. (Well, OK, there are other problems, too …)

Posted by: Jacques Distler on May 13, 2008 4:46 PM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

Maybe it is clear for him trat the SM can’t work with that. I guess what he is trying to find now it is universe like ours, with a broken SM using e8, as a toy model.

Garrett, can you confirm this?

Posted by: Daniel de França MTd2 on May 13, 2008 5:14 PM | Permalink | Reply to this

Re: ℤ2 gradings

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Hi Daniel,

I’ve said all along that the second and third fermion generations don’t seem to have the correct quantum numbers under the current decomposition. This is the main deficiency of the current model. So, this discrepancy needs to be understood, or we need to use a different decomposition of $e8$ , or we need to use a different group. Regardless of how this problem is solved, it’s fairly interesting (to me anyway) that all fields of the standard model can be embedded in one superconnection. And if we can figure out a nice geometric description of this superconnection, all the better.

Posted by: Garrett on May 13, 2008 6:02 PM | Permalink | Reply to this

Re: ℤ2 gradings

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Hi Jacques,

I’m amenable to using other groups or supergroups. But the particular decomposition of $E8$ ’s Lie algebra you’re referring to is:

(1)

e8 = h + k = \big( so(8) + so(8) \big) + \big( 8_+ \times 8_+ + 8_- \times 8_- + 8_v \times 8_v \big)

In this decomposition, the $H=SO(8)+SO(8)$ is a symmetric subgroup of $E8$ , so there’s no problem with building the superconnection, $A+\Psi$ , with $A$ a 1-form field valued in $h$ , and $\Psi$ a Grassmann field valued in $k$ . This is a fairly standard construction, as you can see if you look at any of the references I provided in my previous comment.

Posted by: Garrett on May 13, 2008 5:54 PM | Permalink | Reply to this

Re: ℤ2 gradings

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No, Garrett, that’s wrong.

There is no $\mathbb{Z}_2$ -grading of $e_8$ with 192 odd generators and 56 even generators.

You keep insisting that there is (and that, moreover, this is “standard”). But there isn’t.

(What you actually require for your construction is something even more stringent, namely that the $\mathbb{Z}_2$ -grading comes from an embedding of $SL(2,\mathbb{C})$ in $E_8$ . But let’s leave that aside for the moment.)

Posted by: Jacques Distler on May 13, 2008 7:01 PM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

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So, Jacques, are you claiming that $e8$ does not break up as:

(1)

e8 = h + k = \big( so(8) + so(8) \big) + \big( 8_+ \times 8_+ + 8_- \times 8_- + 8_v \times 8_v \big)

Or that the Lie brackets between these parts are not:

(2)

[h,h] \in h

(3)

[h,k] \in k

(4)

[k,k] \in h+k

Or is something else causing you to say this is “wrong?” I suspect it’s the last line that’s upsetting you. In the BRST construction, the action is built in such a way that this $[k,k]$ part is irrelevant–it does not appear because it is Grassmann grade two. I’m not making this up–it’s in the references I cited.

This term is, however, in the supercurvature; but when this is used to build the action, there is no corresponding part of the $B$ field in the $BF$ action to contract with this Grassmann grade two part.

Posted by: Garrett on May 13, 2008 9:10 PM | Permalink | Reply to this

Re: ℤ2 gradings

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I am making the statement that there is no $\mathbb{Z}_2$ grading of the $e_8$ Lie algebra under which “ $k$ ” is odd, much less a $\mathbb{Z}_2$ grading of the $e_8$ Lie algebra which comes from an embedding $SL(2,\mathbb{C})\mapsto E_8$ (which is required if you want to interpret the generators in $k$ as corresponding to fermions).

If you want to say that the Lie algebra structure of $e_8$ is “irrelevant,” and all you have is a $\mathbb{Z}_2$ -graded vector space, then why bother with this one? Why not choose any old $\mathbb{Z}_2$ -graded vector space, and dispense with the pretense that it’s also a Lie algebra?

Posted by: Jacques Distler on May 13, 2008 9:24 PM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

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If you want to say that the Lie algebra structure of $e_8$ is “irrelevant”

I think he is saying that the $[k,k]$ -bracket of $e_8$ is irrelevant, or supposed to be, because the would-be quadratic $\psi \psi$ contribution in the curvature does not, for one reason or another, appear in the action.

In other words, he is using the Lie algebra $h \subset e_8$ as his gauge Lie algebra and the rep of $h$ on $k \subset e_8$ given by the bracket in $e_8$ , but is not making use of the bracket $[k,k]$ .

I am not sure I already see why the $\psi \psi$ term should drop out in the end.

But on the other hand, as far as just the model building goes it seems to me that Garrett could actually drop the superconnection point of view and just say: consider an $h = so(8) \oplus so(8)$ gauge theory with fermions that happen to live in the rep $k$ coming from the Lie bracket of $e_8$ .

The superconnection point of view would possibly provide a more “unified” picture, but maybe it is good to disentangle different somehwat independent ideas for the time being.

Posted by: Urs Schreiber on May 14, 2008 6:15 AM | Permalink | Reply to this

Re: ℤ2 gradings

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In other words, he is using the Lie algebra $h\subset e_8$ as his gauge Lie algebra and the rep of $h$ on $k\subset e_8$ given by the bracket in $e_8$ , but is not making use of the bracket $[k,k]$ .

In other words, he is using some $\mathbb{Z}_2$ -graded vector space, which happens to be a representation of $h$ .

“[N]ot making use of the bracket $[k,k]$ ” is just a euphemism for saying that the Lie algebra structure is incompatible with the grading he would like to impose.

My point is, that once you abandon the requirement that $h\oplus k$ is a Lie algebra, I don’t see the rationale for choosing this particular $\mathbb{Z}_2$ -graded vector space, which happens to be a representation of $h$ , as opposed to some other.

Posted by: Jacques Distler on May 14, 2008 6:36 AM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

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Urs’ description above is correct. The $\psi \psi$ term is not in the action, making the $[k,k]$ bracket irrelevant. This action is built by hand. There is currently no good justification for it, other than trying to cook up an action that matches the standard model. And this action is what controls how $e8$ breaks up into $h$ and $k$ . I agree this is unsatisfactory, and this is the deficiency Lee tried to remedy with his paper.

However, the fermionic part of the action (including the disappearing $\psi \psi$ ) is not completely ad hoc; it comes directly from application of the BRST technique to an (ad hoc) action for an $e8$ principal bundle connection, which involves the complete Lie algebra. This is much more satisfying than just starting with an $h$ gauge theory and a $k$ rep. It still leaves the question of why the initial action is what it is, but the resulting decomposition into $h$ and $k$ parts, and the reunion of the corresponding 1-form and Grassmann fields into a superconnection, follows a direct application of BRST.

The reason (for me at least) why this approach is so nice is that the initial physical geometry–an $E8$ principal bundle and connection–is completely natural. By this is meant that it involves only maps between vector fields describing diffeomorphisms of manifolds. And, also, it’s very pretty.

I agree that a justification needs to be found for why nature chooses the standard model action for this principal bundle connection, but that’s for future work.

Posted by: Garrett on May 14, 2008 9:25 AM | Permalink | Reply to this

Re: ℤ2 gradings

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it comes directly from application of the BRST technique to an (ad hoc) action for an e8 principal bundle connection, which involves the complete Lie algebra.

I don’t understand that statement.

The complete Lie algebra is incompatible with the grading you wish to assign. I don’t understand how you could possibly have obtained something with your grading by applying “the BRST technique” (whatever that means) to the $e_8$ Lie algebra.

You really need to make a choice: if your starting point is $e_8$ , then you have to choose a grading compatible with $e_8$ (in fact, as I keep emphasizing, the restrictions on your choice of grading are even more stringent).

Otherwise, you might as well acknowledge that what you’re doing is choosing $h$ and a representation $k$ and be done with it.

The reason (for me at least) why this approach is so nice is that the initial physical geometry–an E8 principal bundle and connection

You never, ever have an $E_8$ principal bundle. We are discussing whether you even have a “Schreiber superconnection” associated to $e_8$ . It certainly appears that you do not.

Posted by: Jacques Distler on May 14, 2008 1:54 PM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

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What I am doing is described explicitly in this previous comment, which begins “Following van Holten, we start with a principal bundle connection…” The $H$ is valued in $h$ , and the $C$ is valued in $k$ ; together these are parts of an $E8$ principal bundle connection, with the decomposition as I’ve described. This technique is straightforward, I’ve described it explicitly, and I’ve provided many references.

An important question is, “Does this $E8$ model match up with known physics?” I know you don’t think it does, and I certainly acknowledge some problems with it, but discussing that is probably outside the scope of Urs’ post.

A question relevant to Urs’ post is “Does this match up with the mathematics of Quillen’s superconnection?” I believe we are making progress on answering this in the negative. Quillen’s superconnection (from what I’m gathering) involves the formal addition of 1-forms and real fields valued in a Lie superalgebra, whereas the superconnection I’m using involves the formal addition of 1-forms and Grassmann fields valued in a Lie algebra. Also, Quillen’s superconnection appears to require that the supergroup decomposition into $h + k$ satisfy

(1)

[h,h] \in h

(2)

[h,k] \in k

(3)

[k,k] \in h

whereas what I’m doing requires only that the first two of these hold. This, in my opinion, is useful information for this discussion.

Posted by: Garrett on May 14, 2008 4:09 PM | Permalink | Reply to this

Re: ℤ2 gradings

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A question relevant to Urs’ post is “Does this match up with the mathematics of Quillen’s superconnection?” I believe we are making progress on answering this in the negative.

Not only is it incompatible with the Quillen superconnection, it is incompatible with the Schreiber superconnection as well (which is the more relevant statement).

Moreover, it is incompatible with the grading coming from an embedding of $SL(2,\mathbb{C})$ in $E_8$ .

This last is most damning of all, because that is how fermions are supposed to arise in your model.

Posted by: Jacques Distler on May 14, 2008 5:10 PM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

Hi Garrett,

So, what happens when you try to get a “Schreiber connection using” the Z2 grading from that spinorial embending of e8? What is the difference from that? Could you write down the equations?

I would love to hear the difference, because I am terribly confused.

Posted by: Daniel de França MTd2 on May 15, 2008 5:35 AM | Permalink | Reply to this

Re: ℤ2 gradings

I meant, what it the difference between this result, and the formula from Quillen connection and Schreiber connection?

Posted by: Daniel de França MTd2 on May 15, 2008 5:46 AM | Permalink | Reply to this

Re: ℤ2 gradings

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Hi Daniel, The problem is that the decomposition of e8 I employed is not quite a $Z_2$ grading. Specifically, the decomposition is

(1)

e8 = h + (k1+k2+k3) = \big( so(8) + so(8) \big) + \big( 8_+ \times 8_+ + 8_- \times 8_- + 8_v \times 8_v \big)

and the Lie bracket between what I’ve called fermion “generations”,

(2)

[k1,k2] \subset k3

breaks the $Z_2$ grading. This description of generations is dubious anyway, and gives Jacques apoplexy–justifiably. However, I think we could build a Schreiber superconnection valued in $h$ and $k1$ ; but then we’d still be left wondering about the second and third generations of fermions. It’s wishful thinking, but if the geometry of this superconnection led to the natural reappearance of $k2$ and $k3$ , but in such away that they had the quantum numbers of $k1$ , that would be spiffy.

Quillen’s superconnection is no longer on the table, because it involves the formal addition of 1-forms and real numbers valued in a superalgebra, not 1-forms and Grassmann numbers valued in a Lie algebra, as we need for the Standard Model.

Posted by: Garrett on May 16, 2008 2:49 AM | Permalink | Reply to this

Re: ℤ2 gradings

“The problem is that the decomposition of e8 I employed is not quite a Z 2 grading.”

Distler said below that by embending this decompositon, e8 in general, in a spinorial representation, you would automaticaly get a Z2. So, it is really a Z2 grading.

I don’t know if this is what he meant, but that’s what I understood.

Posted by: Daniel de França MTd2 on May 16, 2008 5:47 AM | Permalink | Reply to this

Re: ℤ2 gradings

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It seems that “Schreiber superconnection” is closer to what Garrett Lisi had in mind than “Quillen superconnection”.

The difference here is hard to tell, because the choices differ only in the term in the curvature which is bilinear in the “spinorial” part $\psi$ which Garrett suppressed.

Garrett argues that $\psi$ is “Grassmann valued” and that this implies that the quadratic term in $\psi$ vanishes.

At the moment I don’t really understand how to interpret this. If we want the $\psi$ s to be not only valued in the odd-graded part of the Lie algebra, but also be odd functions themselves, then the way to make that precise which I could come up with is to say that the $\psi$ s are odd sections of a super vectorbundle. While then we’d have $\psi^a \psi^b = - \psi^b \psi^a$ , it wouldn’t imply that the bilinear term in the $\psi$ s vanishes.

Garrett says, I think, that he has some BRST-inspired construction in mind, which this is supposed to be modeled after. So far I had a look at the part in van Holten’s review that he pointed us to, but couldn’t quite find a superconnection being discussed there. I still need to look at some of the other references.

Posted by: Urs Schreiber on May 15, 2008 6:46 AM | Permalink | Reply to this

Re: ℤ2 gradings

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The $\psi \psi$ term does not vanish from the curvature. It simply is not used when constructing the action. This is a typical action one gets through BRST.

Since the $\psi \psi$ term isn’t in the action, we don’t strictly need to use a $Z_2$ grading of the algebra we are breaking up into $h$ and $k$ , since $[k, k]$ does not have to be in $h$ for this model to work – it can also return a part in $k$ , as it does in the current decomposition of $e8$ . (This addresses Jacques’ concerns.)

Am I correct in thinking a $Z_2$ grading requires $[k,k]\in h$ , or is this not necessary? And, is it sensible to call it a “superconnection” if we only have

(1)

[h,h] \in h

(2)

[h,k] \in k

(3)

[k,k] \in h+k

and not necessarily $[k,k] \in h$ ?

Posted by: Garrett on May 15, 2008 5:14 PM | Permalink | Reply to this

Re: ℤ2 gradings

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The $\psi \psi$ term does not vanish from the curvature.

Okay, that’s good.

It simply is not used when constructing the action.

Not used by decree? Or drops out by some natural mechanism?

Am I correct in thinking a $\mathbb{Z}_2$ grading requires $[k,k] \in h$ ,

Yes. I would write $[k,k] \subset h$ .

In general, let $K$ be an abelian group. Then a $K$ -graded Lie algebra is a $K$ -graded vector space $g$ , i.e. a direct sum $g = \otimes_{k \in K} g_k$ of vector spaces labeled by elements in $k$ equipped with a Lie bracket that satisfies $[g_{k_1}, g_{k_2}] \subset g_{k_1 k_2} \,.$

And, is it sensible to call it a “superconnection” if we only have

Depends a bit on what “it” is here. If “it” is either of the two constructions that Jacques spells out, then the answer is no.

I suggested a fix in some other comment: if multiplying by $\psi$ alone is already an odd endomorphism (such as if the $\psi$ s odd functions on a supermanifold, what Jacques called fermionic fields below his equation (5)) then one could simply ignore any grading on the Lie algebra, hence regard all actions of the Lie algebra element on the representation space as even, and then the $\psi$ -prefactor would make the last term an odd endomorphism of some vector bundle and would lead back to the notion of Quillen superconnection once again.

Posted by: Urs Schreiber on May 15, 2008 7:06 PM | Permalink | Reply to this

Re: ℤ2 gradings

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Not used by decree? Or drops out by some natural mechanism?

The quadratic term just isn’t used in the BRST framework. It doesn’t appear. I described this in detail in a previous comment. This isn’t a problem in the BRST framework, but the quadratic term is going to be more of a problem if we use a Schreiber superconnection.

It would be very interesting to describe the Standard Model using this superconnection–instead of using a “BRST generalized connection” as I have. This description would work well, and might lead to some interesting physics. It would be very similar to what I’ve done, but the Lie algebra (or superalgebra) would not be E8

Posted by: Garrett on May 16, 2008 2:16 AM | Permalink | Reply to this

Re: ℤ2 gradings

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Since the $\psi\psi$ term isn’t in the action, we don’t strictly need to use a $\mathbb{Z}_2$ grading of the algebra

Sure you do. It’s called fermion number (mod 2). $\mathbb{Z}_2$ odd generators are supposed to correspond to fermions and $\mathbb{Z}_2$ even generators are supposed to correspond to bosons.

That’s why you can’t use any old $\mathbb{Z}_2$ grading but, rather, one induced from an embedding of $Spin(3,1)\simeq SL(2,\mathbb{C})$ in $E_8$ . That way, fermions transform in spinorial representation of $Spin(3,1)$ (and bosons in tensorial ones).

we are breaking up into $h$ and $k$ , since $[k,k]$ does not have to be in $h$ for this model to work

Sure it does. The product of two fermions is a boson (the product of two spinorial representations of $SL(2,\mathbb{C})$ is tensorial)

– it can also return a part in k,

Not if you want something physically (or mathematically) sensible.

as it does in the current decomposition of $e_8$ . (This addresses Jacques’ concerns.)

No it doesn’t, and in fact, the group-theoretical analysis in your paper is just plain wrong. I classified all embeddings of $SL(2,\mathbb{C})\times SU(3)\times SU(2)\times U(1) \hookrightarrow D_4\times D_4 \subset E_8$ . None of them agree with the quantum numbers you claim to have found.

Am I correct in thinking a $\mathbb{Z}_2$ grading requires $[k,k]\in h$ , or is this not necessary?

Yes, it is necessary. But, then, Marcel Berger classified all such $\mathbb{Z}_2$ gradings. None of them satisfied your requirements.

Posted by: Jacques Distler on May 15, 2008 7:24 PM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

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I’ve described what I’m doing explicitly–using the BRST framework–in a previous comment. This does not require a $Z_2$ grading of the Lie algebra involved. I’ve provided a collection of references, which is the tip of a large iceberg of literature stretching back thirty years, and involves a well established mathematical framework essential to successful constructions in Quantum Field Theory.

Using Urs’ superconnection, on the other hand, does require a $Z_2$ grading of the algebra. In that context your statements are correct, and the Lie group matching the standard model would not be E8. What would it be? Interesting question, I think.

Posted by: Garrett on May 16, 2008 2:30 AM | Permalink | Reply to this

Re: ℤ2 gradings

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This does not require a $\mathbb{Z}_2$ grading of the Lie algebra involved.

An embedding of $Spin(3,1)$ in $E_8$ furnishes a $\mathbb{Z}_2$ grading of the Lie algebra (corresponding to fermions and bosons).

You do claim to embed $Spin(3,1)\times SU(3)\times SU(2)\times U(1)$ in $E_8$ , do you not?

If so, then you have defined a $\mathbb{Z}_2$ grading of the Lie algebra.

If not, then it’s pretty hard to understand what you do claim to have done in your paper.

In that context your statements are correct, …

“That context” is your context.

Posted by: Jacques Distler on May 16, 2008 2:47 AM | Permalink | PGP Sig | Reply to this

Re: ℤ2 gradings

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I’ve described what I’m doing explicitly–using the BRST framework–in a previous comment.

Okay, so the problem is: I haven’t really understood what you say there, yet, it seems. Here are some points that confuse me:

1) I don’t see that van Holten discusses superconnections.

2) In the previous comment you seem to say at one point that the $\psi \psi$ term vanishes because $\psi$ is “Grassmann valued”. But we really have several $\psi$ s, the term is rather a linear combination of $\psi^a \psi^b$ . This does not vanish, in general, even if the $\psi$ s anticommute.

3) But a few sentences later you say “so I simply drop this term”. I am confused about what exactly you mean.

4) In general, I am not clear about what you mean when you explain your construction in terms of BRST. It seems you have some analogy in mind, a “BRST inspired” process?. Maybe instead of referring back to this supposed analogy, it would be helpful if you could give a first principles description of what you mean you are doing. My impression is that I am not the only one not following this part.

5) Finally, let me repeat a comment I made before:

originally I had tried to to interpret your term $A + \psi$ as a Quillen superconnection by assuming $\psi$ to be an even field with values in an odd vector space endomorphism. Alternatively, it seems we’d rather want it to be an odd field with values in an even vector space endomorphism. That would again make it an odd endomorphism, in total, of a suitable super vectorbundle and you’d have an honest Quillen superconnection. And you could forget about the $\mathbb{Z}_2$ -grading. (Well, at least for this purpose. Maybe you need it elswhere.)

Have to run now. Will give some talks today, so probably won’t be able to reply until much later.

Posted by: Urs Schreiber on May 16, 2008 9:41 AM | Permalink | Reply to this

Re: ℤ2 gradings

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1) I don’t see that van Holten discusses superconnections.

On page 70, equation (4.49) is what we called

(1)

A' = H + \Psi

It is the formal addition of 1-forms, $H = dx^i H_i^B T_B$ , and “ghosts,” $\Psi = \Psi^B T_B$ , with anti-commuting (Grassmann) coefficients, valued in the $h$ and $k$ parts of a Lie algebra, respectively. If this $h$ and $k$ correspond to a $\mathbb{Z}_2$ grading of the Lie algebra, then this $A'$ is a Shreiber superconnection. Whether they correspond to a $\mathbb{Z}_2$ grading or not, it is referred to as a “generalized connection” in the BRST literature.

There are many different possible geometric interpretations of what this $\Psi$ is. The most common interpretation is that it is a 1-form on the space of connections.

2) In the previous comment you seem to say at one point that the $\psi \psi$ term vanishes because $\psi$ is “Grassmann valued”.

No, I have never said this. Someone else claimed I said this, but I didn’t.

3) But a few sentences later you say “so I simply drop this term”. I am confused about what exactly you mean.

This term simply is not used in the action when one follows the BRST formulation.

4) In general, I am not clear about what you mean when you explain your construction in terms of BRST. It seems you have some analogy in mind, a “BRST inspired” process?

I’m really doing standard BRST. I’ve laid out the steps explicitly in a previous comment. I’d be happy to elaborate on any of these steps in more detail if you like, or provide page numbers in van Holten or elsewhere for each one.

(5)…Alternatively, it seems we’d rather want it to be an odd field with values in an even vector space endomorphism. That would again make it an odd endomorphism, in total, of a suitable super vectorbundle and you’d have an honest Quillen superconnection. And you could forget about the $\mathbb{Z}_2$ -grading. …

I thought a

\mathbb{Z}_2

grading of the algebra was necessary to build a Quillen (or Schreiber) superconnection?

Posted by: Garrett on May 17, 2008 4:19 AM | Permalink | Reply to this

Re: ℤ2 gradings

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1) I don’t see that van Holten discusses superconnections.

On page 70, equation (4.49) is what we called $A' = H + \Psi$

It is the formal addition of 1-forms, $H = d x^i H_i^B T_B$ , and “ghosts,” $\Psi = \Psi^B T_B$

Now I see what you mean. I am not sure we can interpret this construction as a superconnection.

That expression that equation (4.49) is about is, technically, the way van Holten set it up, the (ordinary) Ehresmann connection on the total space of a trivial $G$ -bundle over $M$ .

The point is that if decomposed into its horizontal part $A$ and vertical part $C$ as in (4.31), one finds that the vertical part behaves as a ghost for the horizontal 1-form field on $M$ .

I like to say this differently: the ghosts for gauge theory are the 1-forms on the space of $g$ -valued 1-forms. This is described in 9.3.1 starting on p. 87.

But in the picture of von Holten’s equation (4.49), we see just one single ordinary $g$ -valued 1-form on the total space of a $G$ -bundle, to be interpreted as an Ehresmann connection 1-form. One can think of its horizontal/vertical decomposition as yielding the algebr of the gauge ghosts acting on the connection field.

But I don’t really see how to get from there to the point of view of superconnections.

And, would it matter much for what we are discussing?

Posted by: Urs Schreiber on May 17, 2008 8:03 AM | Permalink | Reply to this

Re: ℤ2 gradings

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This $\Psi$ (which van Holten calls $c$ ) is a Lie algebra valued Grassmann field over the base. The geometric interpretation of it as a 1-form on the total space is just one possibility, among several. On the bottom of page 68, van Holten says:

Clearly, the above system of equations are in one-to-one correspondence with the BRST transformations of the Yang-Mills fields, described by the Lie-algebra valued one-form $A=dx^\mu A_\mu^a T_a$ , and the ghosts described by the Lie-algebra valued grassmann variable $c=c^a T_a$

This is typical of how these geometric interpretations are built. But the construction of the generalized connection,

(1)

A' = H + \Psi

is common through much of the BRST literature.

I am not sure we can interpret this construction as a superconnection.

This is a good question. Certainly if $H$ is valued in $h$ , and $\Psi$ is valued in $k$ , and this is a $\mathbb{Z}_2$ grading of the Lie algebra, then this is a Schreiber superconnection over the base. So, a BRST generalized connection can be a superconnection under certain specific conditions. But, if we have

(2)

[h,h] \subset h

(3)

[h,k] \subset k

but we don’t have

(4)

[k,k] \subset h

what should we call it? It seems very close to your Schreiber superconnection, but without the $\mathbb{Z}_2$ grading it might be best to refer to it as a generalized connection.

And, more interesting than establishing terminology, is there a nice way to interpret this object geometrically, if we consider abandonning BRST and starting with this object a priori?

Posted by: Garrett on May 17, 2008 3:10 PM | Permalink | Reply to this

Re: ℤ2 gradings

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Hi Garrett,

okay, I see now what you mean concerning van Holten.

Essentially, the point similar to the one I made when saying that the odd 0-form part in the Quillen superconnection can be read as a 1-form in an orthogonal direction.

Okay, good. Sorry that it took me so long to see that this is just what you mean. I think that the reference to BRST kept irritating me, since it seemed to indicate that supposedly more is going on. But the point here is not really about BRST, but about 1-forms on direct product spaces.

And, more interesting than establishing terminology, is there a nice way to interpret this object geometrically, if we consider abandonning BRST and starting with this object a priori?

Yes. I made this comment a couple of times already:

Forget the $\mathbb{Z}_2$ -grading. In my original post I interpretd $\psi$ as an even function with values in an odd endomorphism. But apparently we’d rather want it the other way around: an odd function with values in an even endomorphism.

Switch from $e_8$ to $u(1)$ for a moment, to see the point more clearly. Let $X$ be a supermanifold of dimension $n|1$ , for definiteness. This means simply that, locally, in addition to the $n$ different standard coordinate functions, there is one odd coordinate function, i.e. that locally over a small patch $U$ the algebra of functions is $C^\infty(X)|_U \simeq C^\infty(U) \otimes \wedge^\bullet \langle \theta \rangle$ where $\theta$ is a generator in degree 1.

Consider a line bundle over this superspace. Locally the collection of sections of this line bundle coincides precisely with the above function algebra.

Now put a Quillen superconnection on that super line bundle by choosing some connection 1-form locally of the form $A = \sum_i f^i d g^i$ for $\{f^i, g_i\}$ even elements of the super function algebra $C^\infty(X)$ and a 0-form locally of the form $\Psi = \theta h$ for $h$ some even function.

That’s nothing but what you am, I suppose, calling a Grassmann valued function. Then $A + \Psi$ is a Quillen superconnection on our super line bundle over the supermanifold $X$ : $\Psi$ simply acts by multiplication, and since it is an odd function, such multiplication is an odd endomorphism of the bundle.

In this particular case actually $\Psi \Psi = 0 \,,$ but that’s only because I started by assuming that the superdimension of $X$ is $n|1$ . We can play the same game with superdimensions $n|m$ . But maybe it would help your approach not to do that.

So take now $g$ to be any Lie algebra. Might be $e_8$ or might not for the following purpose. Don’t worry about gradings on $g$ .

Pick an ordinary representation $V$ of $g$ . $V$ is some vector space of dimension $r$ .

Take $X$ again to be a supermanifold of dimension $n|1$ , now with a rank $r|0$ vector bundle $E$ over it. By definition, the sections of this vector bundle are locally the same as $r$ copies of the function algebra:

$\Gamma(E)|_U \simeq (C^\infty(U) \otimes \wedge^\bullet \langle \theta\rangle)^r \,.$

Now you can take $A$ to be any $r \times r$ matrix of even 1-forms and $\Psi$ any $r\times r$ matrix of odd functions and regard $A + \Psi$ as a Quillen superconnection on $E$ . Choosing a basis $\{T_a\}$ of the algebra of $r \times r$ -matrices, you can write this in components as $A + \Psi = A^a T_a + \Psi^a T_a$ for $\{A^a\}$ a collection of even 1-forms and $\{\Psi^a\}$ a collection of odd functions (ordinary functions times $\theta$ ).

My impression is that this is what you had in mind all along, and that the issue of $\mathbb{Z}_2$ -gradings on $g$ has been a red herring (as far as the interpretation of the term $A + \Psi$ goes.)

And in fact, set up this way we have $\Psi \Psi = 0$ . Now take the $T^a$ to be not arbitrary matrices, but only those appearing in the chosen rep of your Lie algebra $g$ . Then there you go: $A + \Psi$ is a Quillen superconnection on a super vector bundle on a supermanifold $X$ of dimension $n|1$ which carries a rep of $g$ . And it is probably just what you had in mind anyway, all along.

If so, I’d suggest to stop mentioning BRST and start saying: Quillen superconnection on a rank $r|0$ supervectorbundle over a supermanifold $X$ of dimension $n|1$ :-)

Posted by: Urs Schreiber on May 17, 2008 6:38 PM | Permalink | Reply to this

Re: ℤ2 gradings

Garrett,

did you ever think of using the Monster Group?

Posted by: Daniel de França MTd2 on May 14, 2008 4:55 AM | Permalink | Reply to this

Re: ℤ2 gradings

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Hi everybody,

I feel a bit bad for having started this discussion and now not participating as actively as I feel I should. But behind the scenes a bunch of things need my attention. I’ll come back as sson as possible, with a little luck tomorrow.

But one quick remark here:

Garrett writes:

This is a fairly standard construction, as you can see if you look at any of the references I provided in my previous comment.

I started looking at some of the articles you linked to, but haven’t looked yet at all of them and not in full detail.

But notice for instance that the superconnection defined in (3.2) of

Abramov, Piivapuu, Geometric approach to BRST-symmetry and $\mathbb{Z}_N$ -generalization of superconnection, http://www.atlantis-press.com/php/download_paper.php?id=364

is, if I am reading them correctly, the Quillen superconnection, not the “Schreiber superconnection”.

When you write $A + \psi$ you should, at some point, say very precisely how you want to think of these symbols.

I suppose another way to interpret what you want to interpret is to forget the $\mathbb{Z}_2$ -grading on $e_8$ but consider the base to be a supermanifold and $\psi$ an odd function (locally). Then $\psi$ would act by odd endomorphisms on the superbundle, if we interpret the latter as something like the free (locally) module over the superfunctions generated by a rep of $e_8$ . Then we’d be back to Quillen superconnections once again.

In any case, some concrete specification of what precisely $A + \psi$ is supposed to mean is necessary – as the discussion we are having clearly shows ;-)

Posted by: Urs Schreiber on May 13, 2008 7:15 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Hi Urs,

Once in a while, you mention our “sleeping baby”, i.e. the discrete geometry stuff we worked on together, and since that is about the only thing I am even remotely capable of commenting on, I just thought I’d point out the eerie similarity between this Quillen curvature and the lattice Yang-Mills curvature (Equation (5.8)) we (you) worked out.

One of the reasons I lurk around this blog is that I’m convinced that it all relates back to the discrete stuff somehow (e.g. perhaps what we did was some simplified version of calculus on n-categories). I’m just not smart enough to draw the connection myself. Hopefully, you find the similarities between the Quillen curvature and lattice Yang-Mills curvature compelling.

Best regards

PS: Here is our paper for anyone who hasn’t seen it

Discrete Differential Geometry on Causal Graphs E.F and U.S.

“It is maybe remarkable that the non-commutativity of 0-forms and 1-forms on the lattice drastically simplifies the notion of gauge covariant derivative to a simple algebraic product of the gauge holonomy 1-form with itself. The discrete gauge theory is in this sense conceptually actually simpler than the continuum theory. In order to illustrate the relevant mechanism in more detail let us restrict attention to a single plaquette as illustrated in figure 9 and work out the value of [Edit: the lattice Yang-Mills curvature] $H^2$ on that plaquette in full detail.”

Posted by: Eric on May 10, 2008 5:05 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Hi Eric,

did you have a look at the entry on Quillen superconnections I kept pointing to?

There my main point is to discuss what a Quillen superconnection with 0-form and 1-form parts (which is the situation we are considering here) amounts to in terms of intergrated parallel transport.

A simple argument shows that the odd 0-form part can be interpreted as parallel transport along a finite edge with is “perpendicular” to the manifold that the connection seems to live on. See my diagrams there.

This observation was one of the main hints which I tried to afterwards connect to a theory of “supercategories” where “odd” directions correspond to finite edges. It does work to some extent, for instance that concept of supercategory reproduces a couple of structures that people thinkig about super D-branes have come across (see Lazaroiu on G-Flows on Categories), as well as relating to the concept of odd tangent spaces (see More on tangent categories) but nevertheless I still feel I am missing something.

Yesterday I had a big discussion about the supersymmetry proof of the index theorem and especially the Bismut formula for the pushforward in K-theory, which does crucially involve Quillen connections. When one reads that circle of phenomena in the light of what I just said one can see essentially that idea come true, that the path integral in quantum mechanics is just push-forward in the corresponding differential cohomology theory: the Chern character of the Bismut push-forward of a vector bundle $E$ with connection on some compact Riemannian spin space $X$ is $\mathrm{tr} \exp(i \mathbf{A}^2)$ where $\mathbf{A}$ is a Quillen connection on the point (!) whose odd 0-form part is the Dirac operator $D_E$ acting on the space of sections of the former spin bundle, which is now the fiber over the point. Hence that Chern character is $\mathrm{tr} \exp(i (D_E)2)$ and hence nothing but the partition function of the charged particle that coupled to the original vector bundle, and the Quillen 0-form curvature is the Hamiltonian. But this describes parallel transport (quantum propagation) then along the “hidden” direction of the Quillen connection in the above sense.

So something is going on here which looks interesting. I want to understand this better. But it’s not entirely clear to me, still.

Posted by: Urs Schreiber on May 10, 2008 9:28 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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I’m a little puzzled at your statement that you can define a Quillen superconnection for any $\mathbb{Z}_2$ -graded Lie algebra (as opposed to a Lie superalgebra). I thought I had a passably good understanding of Quillen’s theory of superconnections. But all of the references on the subject that I’ve read, including the Dumitrscu reference above, use the fact that if $V= V^0\oplus V^1$ is a $\mathbb{Z}_2$ graded vector space, then $end(V)$ is naturally a Lie superalgebra. Thus, for instance, the curvature of Quillen superconnection on a $\mathbb{Z}_2$ graded vector bundle, $E\to X$ , is an even element of $\Omega^\bullet(X,end(E))$ . And they use the fact that ${\Omega^\bullet(X,end(E))}^{\text{even}}$ is endowed with the structure of a Lie algebra, because $end(V)$ is a Lie superalgebra.
Even if we were interested in $\mathbb{Z}_2$ graded Lie algebras, the main point of my discussion was that $e_8$ does not possess a $\mathbb{Z}_2$ grading suitable to Lisi’s purposes.

Quite apart from its application to Lisi’s “theory”, I would like to hear an explication of your notion of Quillen superconnections for $\mathbb{Z}_2$ graded Lie algebras.

Posted by: Jacques Distler on May 10, 2008 8:13 PM | Permalink | PGP Sig | Reply to this

Quillen Superconnection versus

I’m confused, but it’s partly a language problem. I had long thought a Lie super algebra was the same as a Z_2 graded Lie algebra. If not, which is which? To some of us, a Z_2 graded Lie algebra has a bracket of degree 0 which is GRADED skew symmetric.
I suppose one could consider a Lie algebra with a Z_2 grading so that the bracket is the sum of a bracket on each piece, but I’ve never seen that used - unless that is super?

As for end(V), it in turn is Z_2 graded with end_0 of degree 0 and end_1 of degree 1 mod 2. Now you can DEFINE the bracket as the commutator OR the graded commutator -
whihc do you want?

Posted by: jim stasheff on May 10, 2008 9:33 PM | Permalink | Reply to this

Re: Quillen Superconnection versus

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I had long thought a Lie super algebra was the same as a $\mathbb{Z}_2$ graded Lie algebra.

Hi Jim,

remember, you and me sorted this out just a few weeks ago in another thread?

By $\mathbb{Z}_2$ -graded Lie algebra I mean: ordinary Lie algebra and the brackets happen to respect a grading.

By super Lie algebra I mean something that may have symmetric pieces in its bracket.

Posted by: Urs Schreiber on May 10, 2008 10:00 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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I’m a little puzzled at your statement that you can define a Quillen superconnection for any $\mathbb{Z}_2$ -graded Lie algebra

My argument was supposed to be this: a Quillen superconnection is a structure which can be put on any $\mathbb{Z}_2$ -graded vector bundle (as in Dumitrescu section 2.5). Given any $\mathbb{Z}_2$ -graded Lie algebra $g$ with Lie group $G$ and a linear $\mathbb{Z}_2$ -graded representation, we get from any $G$ -bundle such a $\mathbb{Z}_2$ -graded vector bundle on whose fibers the even/odd elements of $g$ act as even/odd endomorphisms.

That’s just what I mean. Let me know if you think I am mixed up here.

the main point of my discussion was that $e_8$ does not possess a $\mathbb{Z}_2$ grading suitable to Lisi’s purposes.

Ah, thanks for pointing that out. I didn’t remember that bit. Will reread your discussion. I had thought what I had to say here was entirely independent of the representation theoretic issues. But maybe it’s not.

Posted by: Urs Schreiber on May 10, 2008 9:53 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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To review:

$\wedge^\bullet T^*$ is $\mathbb{Z}$ -graded vector space with a graded-commutative product (here and below, we’ll only need to keep track of the $\mathbb{Z}/2$ grading). Let $\mathfrak{g}$ be a Lie superalgebra, that is, a $\mathbb{Z}/2$ graded vector space, with a graded-anticommutative product (which, moreover, satisfies an associativity relation called the super-Jacobi identity).

Then $\wedge^\bullet T^*\otimes \mathfrak{g}$ is, again, a $\mathbb{Z}/2$ graded vector space with graded-anticommutative product (satisfying a super-Jacobi identity). That is, it is a Lie superalgebra. Its even part, ${\left(\wedge^\bullet T^*\otimes \mathfrak{g}\right)}^{\text{even}}= \wedge^{\text{even}} T^*\otimes \mathfrak{g}^0 \oplus \wedge^{\text{odd}} T^*\otimes \mathfrak{g}^1$ is a Lie algebra. Globalizing, we have the same statement for $\Omega^\bullet(X,W)$ .

In particular, the curvature of the Quillen superconnection $\mathbb{F}\in {\left(\Omega^\bullet(X,end(E))\right)}^{\text{even}}$ is Lie-algebra valued — and hence Quillen can go through his construction of generalized Chern characters.

Note that this would not have worked if $\mathfrak{g}$ was a $\mathbb{Z}/2$ graded Lie algebra instead of a Lie superalgebra.

As Jim noted, in either case, $\mathfrak{g}$ is a $\mathbb{Z}/2$ graded vector space. But in the former case, the product is anticommutative, whereas in the latter case, the product is graded-anticommutative.

That’s the reason why the entire literature on Quillen superconnections (or, at least, that part of it that I am aware of) makes the latter choice.

If you want to make the former choice, you ought to choose a new name (“the Schreiber superconnection” ?) and work out the theory of this new kind of superconnection.

Posted by: Jacques Distler on May 10, 2008 11:41 PM | Permalink | PGP Sig | Reply to this

Re: E8 Quillen Superconnection

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You seem to be complaining that the inclusion of $\mathbb{Z}_2$ -graded vector spaces $\Omega^\bullet(X,g) \hookrightarrow \Omega^\bullet(X,end(E))$ induced from a $\mathbb{Z}_2$ -graded rep of the $\mathbb{Z}_2$ -graded $g$ on the fibers of $E$ is not a homomorphism of graded or super Lie algebras if we regard $end(E)$ with its super Lie algebra structure (as we have to, yes) but take $g$ to be $\mathbb{Z}_2$ -graded instead of super. True, it’s not.

But it’s still true that this inclusion sends every element $\psi + \nabla \in \Omega^0(X,g_{odd}) + covariant derivative acting through g_{even}$ to an element $\hat \psi + \hat \nabla \in \Omega^0(X,end(E)_{odd}) + even covariant derivative$ hence to a Quillen connection.

For clarity: locally (on trivial bundles) it sends every $\psi + A \in \Omega^0(X,g_{odd}) + \Omega^1(X,g_{even})$ to an element $\hat \psi + \hat A \in \Omega^0(X,end(E)_{odd}) + \Omega^1(X,end(E)_{even}) \,.$

For that it is irrelevant that $g$ has any Lie algebra structure at all. What matters is that it acts by odd and even endomorphisms on $E$ .

Posted by: Urs Schreiber on May 11, 2008 9:10 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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For that it is irrelevant that $g$ has any Lie algebra structure at all. What matters is that it acts by odd and even endomorphisms on $E$ .

And that’s where we disagree.

The fact that $\mathfrak{g}$ is a Lie (super)algebra is important both to Quillen, who is interested in recovering the equivalences that lead to K-theory (ultimately, also its differential version) from the isomorphisms generated by $\mathfrak{g}$ , and to physicists (for whom, for instance, the $\Omega^0(X,\mathfrak{g}^1)$ are the tachyons in some D-brane configuration). In physical terms, $\mathfrak{g}$ is the gauge algebra of this setup.

And, yes, we actually care about its algebra structure.

Posted by: Jacques Distler on May 11, 2008 2:25 PM | Permalink | PGP Sig | Reply to this

Re: E8 Quillen Superconnection

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I don’t see the disagreement. I gave a way to construct a Quillen superconnection. That way involved using a non-super but graded Lie algebra. It amounts to saying:

from any $\mathbb{Z}_2$ -graded rep $V$ of a $\mathbb{Z}_2$ -graded Lie algebra $g$ we get a super Lie algebra $end(V)$ and a canonical injection of $\mathbb{Z}_2$ -graded vector space $g \hookrightarrow end(V)$ which allows us to read any odd $g$ -valued form as an odd $end(V)$ -valued form.

For applying the Quillen formalism you can forget that the superconnection obtained this way arose via $g$ .

It’s still true that this way the expression (3.1) in Lisi’s article can be read as a Quillen superconnection on the $\mathbb{Z}_2$ -graded bundle obtained from a rep of $g$ , and that the curvature givesn in (3.2) is the corresponding Quillen curvature (except that (3.2) is missing the $\psi^2$ term).

It is just the observation that Lisi adds forms with values in odd and even endomorphisms, and that this can be read as a Quillen superconnection.

Whether or not this is physically viable is another question. For instance whether or not $g$ here still deserves to be addressed as the “gauge Lie algebra” is a question.

Posted by: Urs Schreiber on May 11, 2008 3:20 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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I don’t see the disagreement. I gave a way to construct a Quillen superconnection. That way involved using a non-super but graded Lie algebra.

No, it involved using a graded vector space. You didn’t used the Lie algebra structure of $\mathfrak{g}$ at all.

For Quillen’s construction, it is important that isomorphism classes of such data are generated by $\mathfrak{g}$ gauge transformations. Which, under the present circumstances, requires that $\mathfrak{g}$ be a Lie superalgebra.

It is just the observation that Lisi adds forms with values in odd and even endomorphisms, and that this can be read as a Quillen superconnection.

Perhaps it can be read as some kind of superconnection. But “Quillen” is probably the wrong name to attach to it

Because it’s missing the $\psi^2$ term in the “curvature.”
Because it involves $\mathbb{Z}_2$ -graded Lie algebras, rather than Lie superalgebras.
Because, if you want to read Lisi’s paper literally, it involves a $\mathbb{Z}_2$ -graded vector space, rather than a Lie algebra (since the grading, he purports to use, is incompatible with the Lie algebra structure of $e_8$ ).
Because …

If you want to call what Lisi is doing a “Quillen superconnection” because it kinda, sorta looks vaguely like Quillen’s construction, you’re welcome to do so. But, rather than putting Lisi’s ideas on a firm mathematical foundation, I think doing so just sows more confusion.

Posted by: Jacques Distler on May 11, 2008 5:39 PM | Permalink | PGP Sig | Reply to this

Re: E8 Quillen Superconnection

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I wrote:

As Jim noted, in either case, $\mathfrak{g}$ is a $\mathbb{Z}/2$ graded vector space. But in the former case, the product is anticommutative, whereas in the latter case, the product is graded-anticommutative.

Just to be clear, in both cases, the product respects the grading (e.g., the product of two odd guys is even, etc.).

Posted by: Jacques Distler on May 11, 2008 1:27 AM | Permalink | PGP Sig | Reply to this

Re: E8 Quillen Superconnection

I don’t know if I sound reasonable, but… Why don’t you both, Urs and Jacques, try to develop your arguments from first principles of this theory by actually writing down the equations and explicitly developing them, and posting them here?

This is a subject in which the details are obvious for both of you, but you seem to diverge on either the derivation of Quillen superconnection or its definition.

Posted by: Daniel de França MTd2 on May 11, 2008 10:13 PM | Permalink | Reply to this

Read the post Superconnections for Dummies
Weblog: Musings
Excerpt: A brief review of Quillen superconnections, and an introduction to the "Schreiber" superconnection.
Tracked: May 12, 2008 4:01 AM

Re: E8 Quillen Superconnection

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Though Jacques has already alluded to them, I want to say a bit more about the relation between symmetric spaces and $\mathbb{Z}/2$ -graded Lie algebras — that is, not Lie superalgebras, but ordinary Lie algebras $\mathbf{g}$ equipped with a splitting into an ‘even’ subspace $\mathbf{h}$ and an ‘odd’ subspace $\mathbf{p}$ :

$\mathbf{g} = \mathbf{h} \oplus \mathbf{p}$

satisfying:

$[\mathbf{h}, \mathbf{h}] \subseteq \mathbf{h}$

$[\mathbf{h}, \mathbf{p}] \subseteq \mathbf{p}$

$[\mathbf{p}, \mathbf{p}] \subseteq \mathbf{h}$

Note that $\mathbf{h}$ is a Lie subalgebra but $\mathbf{p}$ is not. If we form the corresponding simply-connected groups $G$ and $H$ , we get a homogeneous space $G/H$ of a special sort, called a ‘symmetric space’.

Very loosely, a symmetric space where the view at any point in any direction looks the same as the view at the same point in the opposite direction. In physics jargon, there’s parity symmetry at each point.

This idea is easiest to make precise in the case where $G$ is compact and we pick a left- and right-invariant Riemannian metric on it. Then $G/H$ is a Riemannian symmetric space. In other words, $G/H$ gets a metric with a special property: for each point $x \in G/H$ , there’s a map

$f : G/H \to G/H$

which preserves the metric and the point $x$ , and acts as $-1$ on the tangent space at $x$ . We get this map directly from the $\mathbb{Z}/2$ grading on $\mathbf{g}$ .

The classic example is $G = SO(n)$ , $H = SO(n-1)$ , where $G/H$ is the $n$ -sphere. A more funky example is $G = E_8$ (the compact real form), $H = Spin(16)/(\mathbb{Z}/2)$ , and $G/H$ a certain $2 \times 8^2 = 128$ -dimensional Riemannian manifold that Boris Rosenfeld calls the ‘octooctonionic projective plane’. It’s not a projective plane in the axiomatic sense, and I wish I understood it better! Each tangent space looks like the chiral spinor representation of $E_8$ .

Besides this $E_8$ example, there are examples corresponding to $E_7$ , $E_6$ and $F_4$ . I wrote about these and some vague possible relations to fermions and bosons in particle physics back in week253. There’s a lot of nice math there — including relations to the theory of quandles — but I really don’t see how to apply it to particle physics. That’s why I’m so interested in this discussion, and so sad at the difficulty in clear communication.

Derek Wise worked out an interesting generalization of MacDowell–Mansouri gravity for any $\mathbb{Z}/2$ -graded Lie algebra with an invariant inner product. It’s geometrically very nice — but I don’t see any relation to fermions and bosons, alas.

Posted by: John Baez on May 17, 2008 1:33 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Jacques wrote:

There is no $\mathbb{Z}_2$ -grading of $e_8$ with 192 odd generators and 56 even generators.

Garrett almost wrote:

So, Jacques, are you claiming that $e_8$ does not break up as:

$e_8 = h \oplus k$

with

$h = so(8) \oplus so(8)$

and

$k = 8_+ \otimes 8_+ \oplus 8_- \otimes 8_- \oplus 8_v \otimes 8_v$

Or that the Lie brackets between these parts are not:

$[h,h] \in h$ $[h,k] \in k$ $[k,k] \in h+k ?$

I doubt he’s questioning these facts, because they’re true! It’s pretty clear he’s claiming something else: there’s no way to make $e_8$ into a $\mathbb{Z}_2$ -graded Lie algebra with 192 odd generators and 56 even generators. And he’s right about that.

In other words: there’s no way to write $e_8$ as a direct sum of a 192-dimensional ‘even’ subspace $h$ and a 56-dimemsional ‘odd’ subspace $k$ where the brackets obey

$[h,h] \in h$ $[h,k] \in k$ $[k,k] \in h$

Sure, you can get

$[h,h] \in h$ $[h,k] \in k$ $[k,k] \in h+k$

but that’s not giving you a $\mathbb{Z}_2$ -graded Lie algebra, since the bracket of two odd generators may not be even!

This may or may not matter to your ideas, Garrett. But let’s be clear about the usual definition: in a $\mathbb{Z}_2$ -graded Lie algebra,

the bracket of even guys is even,
the bracket of an even guy and an odd guy is odd,
the bracket of odd guys is even.

Your decomposition of $e_8$ does not obey the last property.

On the other hand, there’s a very nice $\mathbb{Z}_2$ -grading on $e_8$ where the even part is 120-dimensional and the odd part is 128-dimensional. You know about this, surely, but I just recalled it in my previous post in this thread.

Whoops! I guess I’m a bit late in catching up on this big discussion. After writing the above, I see you wrote:

The problem is that the decomposition of $e_8$ I employed is not quite a $\mathbb{Z}_2$ grading.

So, I guess we all agree on that now. I’ll still post this comment, since the discussion is quite convoluted and some readers might appreciate a little extra clarification.

Posted by: John Baez on May 17, 2008 2:59 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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Hi John, Now that you’ve written a nice summary, clarifying things, this gives me the opportunity to elaborate on my “not quite” and add some confusion. If we use this decomposition of $e8$ :

(1)

h = so(8) \oplus so(8)

(2)

k=k1 \oplus k2 \oplus k3

(3)

k1 = 8_+ \otimes 8_+

(4)

k2 = 8_- \otimes 8_-

(5)

k3 = 8_v \otimes 8_v

Then we can build a Shreiber superconnection valued in $h$ and $k1$ , since

(6)

h \oplus k1 = so(16)

is a $\mathbb{Z}_2$ grading. So, this subgroup of $e8$ is amenable to this construction.

The $k2$ and $k3$ parts are related to $k1$ by triality (which is why we’re able to call that $so(16)$ ). So, that’s why I said “sort of.” I’m not sure exactly what to make of this, since $k2$ and $k3$ come in via this hand waving. But that’s about the same place I was in the paper, with the second and third generations not having the correct quantum numbers, but being related to $k1$ by $e8$ triality. So the Schreiber superconnection may still have a role here.

Or, as I suggested in an earlier comment, it would be interesting to fit the Standard Model into a Schreiber superconnection for a Lie algebra other than $e8$ .

Posted by: Garrett on May 17, 2008 4:42 AM | Permalink | Reply to this

Triality

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since $h\oplus k_1 = so(16)$ is a $\mathbb{Z}_2$ grading. So, this subgroup of $e_8$ is amenable to this construction.

The $k_2$ and $k_3$ parts are related to $k_1$ by triality …

Umh. No.

There are 2 problems with this:

We are interested in noncompact real forms (precisely which ones are listed here) of $D_4\times D_4$ . While the compact real form of $D_4$ has a triality symmetry, the noncompact real forms do not. In particular, $d_8 = h \oplus k_3$ . In the cases of interest, $k_1$ and $k_2$ are complex, and complex conjugates of each other. There is no triality symmetry relating them to $k_3$ . And there’s no $\mathbb{Z}_2$ grading of the sort you claim.
In any case, we are not interested in any old $\mathbb{Z}_2$ grading, but rather one induced from an embedding of $Spin(3,1)$ (so that the generators which transform as spinor representations of $Spin(3,1)$ are odd, whereas those in tensor representations are even.) Again, the possible embeddings of $Spin(3,1)\times SU(3)\times SU(2)\times U(1)\subset D_4\times D_4\subset E_8$ are known. They lead to a $\mathbb{Z}_2$ grading of the sort that John likes — with $k_1\oplus k_2$ odd — 128 odd generators.

Posted by: Jacques Distler on May 17, 2008 5:48 AM | Permalink | PGP Sig | Reply to this

Summary

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At Urs’s request, I am posting the following summary of this discussion. Personally, I think it’s flogging a dead horse, but Urs thinks it would be worthwhile summarising all the main points again.

This is my summary of Lisi’s programme (at least, as best I have been able to understand it).

Choose an embedding of Spin(3,1)×SU(3)×SU(2)×U(1)Spin(3,1)\times SU(3)\times SU(2)\times U(1) in (noncompact) E 8E_8. The generators of the Lie algebra, e 8e_8, then transform as some representation of this subgroup.
- In particular, the action of the center of $SL(2,\mathbb{C})\simeq Spin(3,1)$ gives a $\mathbb{Z}_2$ grading on $e_8$ . The generators of $e_8$ which transform in spinorial representations of $Spin(3,1)$ are “odd”; the generators which transform in tensorial representations are “even.”
- There is a similar $\mathbb{Z}_2$ grading on the fields of any QFT: fermions are “odd” and bosons are “even.” The spin-statistics theorem requires that these be the same grading.
- While Lisi say that he doesn’t want to envoke a $\mathbb{Z}_2$ grading, one is clearly physically required, and mathematically provided by the aforementioned embedding of $Spin(3,1)$ . He might as well say that he doesn’t want to speak in prose.
Use this ℤ 2\mathbb{Z}_2 grading to build a Schreiber superconnection. The bosonic fields transform as 1-forms with values in various tensor representations of Spin(3,1)Spin(3,1); the fermionic fields transform as 0-forms in spinor representations of Spin(3,1)Spin(3,1).
- Lisi says that he isn’t using a Schreiber superconnection. Instead he’s doing ‘standard’ BRST. I can’t make head or tails of his usage of the term “BRST.” In the end, to each generator of $e_8$ , he associates either a bosonic or a fermionic field. Spin-statistics dictates that he do this in a fashion compatible with the $\mathbb{Z}_2$ grading. Which is to say that his fields comprise a Schreiber superconnection. Protestations to the contrary he, again, seems to be speaking in prose.
Use this Schreiber superconnection to build an action.
Quantize that action.
Try to extract some quasi-realistic physics from it.

Unfortunately, the construction falls down at step 1.

Lisi wants there to be 192 odd generators, with respect to some embedding of $Spin(3,1)$ . This, of course, is impossible.
Moreover, in his paper, Lisi embeds $Spin(3,1)\times SU(3)\times SU(2)\times U(1)$ via a $D_4\times D_4$ subgroup of $E_8$ . I classified all such embedding. They all lead (via the above prescription) to a non-chiral fermion spectrum. The closest one can come to the Standard Model spectrum of fermions is to get 1 generation and 1 anti-generation.
This, in fact, is completely general. Any embedding $SL(2,\mathbb{C})\times SU(3)\times SU(2)\times U(1)\hookrightarrow E_8$ yields a nonchiral spectrum of fermions, with — at best — a generation and an anti-generation of Standard Model particles.

None of these statements is particularly hard to prove. In fact, once you know that there’s no $\mathbb{Z}_2$ grading of $e_8$ with more than 128 odd generators, you know that it’s impossible to accommodate 3 generations. The best you could get is 2, but even that proves not to be possible.

That said, there is something kinda cool about the elements of the construction:

An embedding of $Spin(d-1,1)$ in $G$ gives a $\mathbb{Z}_2$ grading on $\mathfrak{g}$ .
Using the corresponding Schreiber superconnection, one naturally gets a theory with fermions, corresponding to the odd generators of $\mathfrak{g}$ , transforming as spinors $Spin(d-1,1)$ .

It would be mildly interesting to see what sort of actions one could build with this construction.

Unfortunately, Lisi seems to be hung up on finding a “Theory of Everything,” so it doesn’t look like he, himself, is going to take up this obvious question.

Posted by: Jacques Distler on May 22, 2008 1:57 PM | Permalink | PGP Sig | Reply to this

Re: Summary

Hi Distler,

Reading your post, and the one made by [Garrett right above](http://golem.ph.utexas.edu/category/2008/05/e8_quillen_superconnection.html#c016743), it seems the idea is to use the SM and the Anti-SM from the non chiral embeding as components,related by triality, in a certain superconetion, . And in the end, recover the SM.

So, what you said is no possible [here](http://golem.ph.utexas.edu/~distler/blog/archives/001532.html#c013815), on number 4, it is perhaps what Garrett is trying to solve using the superconnection. Anyway, I don’t know if in this case the SM and its anti version are embeded in a compact D4xD4 to allow triality…

Posted by: Daniel de França MTd2 on May 23, 2008 3:50 AM | Permalink | Reply to this

Re: Summary

I tried to embed the hyperlinks in the works just like as I saw in “PGP Sig”, and it didn’t work…

Posted by: Daniel de França MTd2 on May 23, 2008 3:54 AM | Permalink | Reply to this

Re: Summary

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Hi Jacques,

The program presented in the paper is something weirder than using a $\mathbb{Z}_2$ grading of $E_8$ , so I understand why this may cause confusion. If you insist on interpreting what’s in the paper in terms of a $\mathbb{Z}_2$ grading of something, there is a way of looking at it as a $\mathbb{Z}_2$ grading of $D_8$ induced by a $D_4 \times D_4$ subgroup. This gives only one generation, as the “odd” part of this decomposition. This $D_8$ can be related by triality to two other $D_8$ ’s in $E_8$ , all sharing the same $D_4 \times D_4$ subgroup, with the “odd” parts of the three different $D_8$ ’s disjoint in $E_8$ . The union of these three odd parts are not the odd part of a $\mathbb{Z}_2$ grading of $E_8$ . And these three different odd parts do not have the same quantum numbers with respect to the shared $D_4 \times D_4$ subgroup. Nevertheless, I consider this construction interesting – even if you do not.

However, I agree with you that it’s interesting to consider constructing the standard model using a Schreiber superconnection, using a more honest $\mathbb{Z}_2$ grading of some algebra other than $E_8$ .

Posted by: Garrett on May 23, 2008 12:42 PM | Permalink | Reply to this

Re: Summary

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Hi Garrett,

this argument about triality, do I understand correctly that it says the following:

1) There is a way to choose a $\mathbb{Z}_2$ -grading on $e_8$ such that the model obtained from it describes one single generation of the standard model.

(?)

2) And there is not just one such way, but actually three isomorphic such ways to find a 1-generation standard model in an $e_8$ superconnection.

(?)

Is that what you are saying?

I’d like to concentrate on the statement for the case of one single generation a bit:

is there a $\mathbb{Z}_2$ -grading on $e_8$ such that the “Schreiber superconnection” built from it following the programme that Jacques summarized has gravity, the standard model gauge fields and one single generation of fermions sitting in it?

My understanding is that this is your main point, that this is true. But I am just checking again.

Posted by: Urs Schreiber on May 23, 2008 1:15 PM | Permalink | Reply to this

Triality, Trishmality

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This argument about triality, do I understand correctly that it says the following:

I feel that I am require to endlessly repeat well-known facts. There is NO triality symmetry for the noncompact real forms of $D_4$ that Garrett is using.

The compact real form of $D_4$ has three 8-dimensional representations, which are all real.
There is an outer automorphism of the Lie algebra which permutes these representations.

This is not true for the noncompact real forms, needed by Garrett. The $8_v$ representation is still real, but the two eight dimensional spinor representations are complex and are complex conjugates of each other. There is no automorphism that permutes these three representations. (There is an involution that exhanges the two spinor reps, but that’s not what Garrett wants.)

Is that clear? Can we stop discussing the nonexistent triality symmetry, now?

2) And there is not just one such way, but actually three isomorphic such ways to find a 1-generation standard model in an $e_8$ superconnection.

This, again, is a mistake in Garrett’s paper. He claims to find a complete Standard Model generation in the $(8_s,8_s)$ , leaving the $(8_v,8_v)$ and $(8_{s'},8_{s'})$ “free” to be interpreted in some exotic fashion.

That is not correct for any possible embedding of $Spin(3,1)\times SU(3)\times SU(2)\times U(1)$ in $D_4\times D_4\subset E_8$ . There are a finite number of inequivalent such embeddings, and you can study them all. There are only two cases

$Spin(7,1)\times Spin(5,3)$ and
$Spin(7,1)\times Spin(1,7)$

in which the $(8_v,8_v)\oplus (8_s,8_s)\oplus(8_{s'},8_{s'})$ contains any copies of the Standard Model representation. In both cases, you find that a Standard Model generation does not fit into a single $(8,8)$ , as claimed by Garrett. Rather, it sits partly in both the $(8_s,8_s)$ and the $(8_{s'},8_{s'})$ . But, along with it, sits an anti-generation (the same representation(s) of $SU(3)\times SU(2)\times U(1)$ paired with the complex conjugate spinor representation(s) of $Spin(3,1)$ ).

So one doesn’t even have the claimed decomposition of $(8_v,8_v)\oplus (8_s,8_s)\oplus(8_{s'},8_{s'})$ , which the triality symmetry (if it existed, which it doesn’t) would be invoked to save.

Before we start throwing around high falutin’ concepts, like the “Schreiber superconnection”, can we at least get the basic group representation theory right?

I apologize for the somewhat overheated character of this comment. Please believe me that it is considerably toned down from the first draft.

Posted by: Jacques Distler on May 23, 2008 3:36 PM | Permalink | PGP Sig | Reply to this

Compact real D4

When he talked about relating the elements of D4 by triality, I was sure he wanted to make the anti generation vanish by using the compact real form in the superconnection. The output would be somewho related to a non compact D4xD4 :S

I’m sorry.

Posted by: Daniel de França MTd2 on May 23, 2008 4:44 PM | Permalink | Reply to this

Re: Summary

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Hi Urs,

Apologies for the long delay in responding–I got distracted. Your understanding is correct.

Specifically, for (1), $e8$ can be broken into even and odd $d8$ and $16_{S+}$ parts; with gravity and the standard model bosons fitting in $d8$ and one generation fitting in $16_{S+}$ . As far as I can tell, Jacques used to disagree with this claim of fitting the first generation, but now agrees that the first generation can fit, but in a slightly different way–but still in the same $\mathbb{Z}_2$ grading.

For, (2), this is what I have been saying. I’ve been attempting to put a generation in $8_{S+} \times 8_{S+}$ , which is isomorphic by triality (for compact $e8$ ) to $8_{S-} \times 8_{S-}$ and $8_{V} \times 8_{V}$ –which I’ve speculated (tentatively) may relate to the second and third generations. Jacques is objecting strongly to this–apparently because he thinks the transformation I’m using, which is a triality rotation for compact $e8$ , isn’t a symmetry of non-compact $e8$ . Since my speculation on the second and third generations is so tentative anyway, I didn’t feel the need to argue the point.

Posted by: Garrett on July 12, 2008 6:45 PM | Permalink | Reply to this

Re: Summary

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As far as I can tell, Jacques used to disagree with this claim of fitting the first generation, but now agrees that the first generation can fit, but in a slightly different way–but still in the same $\mathbb{Z}_2$ grading.

I am not sure what you think I am now “agreeing to”.

I classified all possible embeddings, of $d_4\times d_4$ , in my post seven months ago.
I explained that your alleged decomposition of $e_8$ under such an embedding was incorrect, and showed that the correct decomposition leads to one generation plus an anti-generation (or to decompositions in which all “fermions” are in $SU(2)$ doublets).

In particular, when you say

For, (2), this is what I have been saying. I’ve been attempting to put a generation in $8_{S+}\times 8_{S+}$ …

you are repeating a statement which you know to be incorrect. Half of a generation and half of an anti-generation fit into $8_{S+}\times 8_{S+}$ .

Jacques is objecting strongly to this–apparently because he thinks the transformation I’m using, which is a triality rotation for compact $e8$ , isn’t a symmetry of non-compact $e8$ .

“Thinks” is a funny way of characterizing a simple mathematical fact.

Posted by: Jacques Distler on July 13, 2008 6:36 PM | Permalink | PGP Sig | Reply to this

Re: Summary

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I am not sure what you think I am now “agreeing to”.

Hi Jacques,

To clarify: In a previous post, to my understanding, you argued that the standard model (plus gravity) gauge group and one generation of fermions could not be fit in E8. Now I think you would agree that this can be done, but in a slightly different way than I claimed (incorrectly) to do it in the paper. Is that correct?

showed that the correct decomposition leads to one generation plus an anti-generation

I don’t see where you said that in your post. You seemed to be saying that this model couldn’t work for even one generation.

By the way, I’m quite grateful to you, Gregg Zuckerman, David Vogan, and Jeff Adams, for figuring out that so(7,1)+so(8) is not a subalgebra of either noncompact real form of e8, but that so(7,1)+so(1,7) and so(5,3)+so(7,1) are.

Posted by: Garrett on July 16, 2008 10:09 PM | Permalink | Reply to this

Re: Summary

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The Standard Model is a chiral theory. It has 3 generations and no anti-generations.

I showed in this post that, after correcting the various mistakes in your paper, you obtained one generation plus one anti-generation — a completely nonchiral spectrum.

I don’t see where you said that in your post.

I gave the decomposition of all of the relevant representations (in particular, of the $8_s\times 8_s$ ). Look at the material between (2) and (3).

It’s plain, from those decompositions that half a generation and half an anti-generation come from the $8_s\times 8_s$ , with the other halves coming from the $\overline{8}_s\times \overline{8}_s$

I then went on to show that this is the best you could possibly do.

This is the maximum number of fermions one can obtain from an embedding in $E_8$ .
Any such embedding is necessarily nonchiral.

Obtaining a chiral Standard Model spectrum (even one generation, let alone three) is impossible.

(And, no, the complex form of $e_8$ does not make things better.)

By the way, I’m quite grateful to you, Gregg Zuckerman, David Vogan, and Jeff Adams, for figuring out that so(7,1)+so(8) is not a subalgebra of either noncompact real form of e8, but that so(7,1)+so(1,7) and so(5,3)+so(7,1) are.

You’re welcome. (Though I don’t think the observation that leads to that conclusion — namely, that the irreducible spinor representations of so(15,1) and so(7,9) are complex — is particularly profound.)

Posted by: Jacques Distler on July 16, 2008 11:41 PM | Permalink | PGP Sig | Reply to this

Re: Summary

Hmm, this seems to confuse the question. I’m trying to find your position on Urs’ question (1). Let me try to rephrase it, apart from issues with chiral symmetry breaking: Do you agree that the gravitational and standard model gauge group and one generation of fermions fit in the split real form of E8, with other stuff left over? And that the bosons fit in the even Z2 part while the fermions fit in the odd part?

Posted by: Garrett on July 17, 2008 6:26 AM | Permalink | Reply to this

Re: Summary

The answer to Urs’s question (1) is a resounding no!

And, in return, could you please explain what you mean when you say:

apart from issues with chiral symmetry breaking …

What do you mean by “chiral symmetry breaking” in this context?

Surely, you are not still labouring under the misapprehension that one can turn a non-chiral theory (such as this one) into a chiral one (like the 1-generation Standard Model), via some “symmetry-breaking” process.

Posted by: Jacques Distler on July 17, 2008 6:56 AM | Permalink | PGP Sig | Reply to this

Re: Summary

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OK, I must have misunderstood. When you said

In both cases, you find that a Standard Model generation does not fit into a single (8,8), as claimed by Garrett. Rather, it sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

you did not mean that a Standard Model generation sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ ?

Posted by: Garrett on July 17, 2008 7:19 AM | Permalink | Reply to this

Re: Summary

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The fermions in your model consist of a Standard Model generation and a Standard Model anti-generation, each of which sits partly in the $(8_s,8_s)$ and partly in the $(8_{s'},8_{s'})$ .

That your model has a non-chiral spectrum (net number of generations = 0) is not an accident. You can tinker with it till the cows come home; there is no way to get a nonzero net number of generations (let alone a net of 3 generations).

Posted by: Jacques Distler on July 17, 2008 7:34 AM | Permalink | PGP Sig | Reply to this

Re: Summary

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If it’s true that

there’s a Standard Model generation and a Standard Model anti-generation, each of which sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

then it’s true that

a Standard Model generation sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

yes?

Posted by: Garrett on July 17, 2008 7:58 AM | Permalink | Reply to this

Re: Summary

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Garrett,

I have proven, in complete generality, for any embedding in $e_8$ , that the net number of generations is zero. Not one, not two, and most certainly not three.

then it’s true that … yes?

I have no idea what this lawyerly splitting of hairs is supposed to achieve.

Probably, it is bound up in the phrase

apart from issues with chiral symmetry breaking …

that I asked you about previously. Maybe you could explain what you meant by that, and we could call it a night.

Posted by: Jacques Distler on July 17, 2008 8:17 AM | Permalink | PGP Sig | Reply to this

Re: Summary

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I’d rather not go off on a tangent. I’m trying to understand our disagreement in as direct a way as possible. My reasoning is very basic, starting with your statements:

It is true that (A)

There’s a Standard Model generation and a Standard Model anti-generation, each of which sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

yes? And so it is true that (B)

A Standard Model generation sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

yes? And it is true that (C) $(8_s,8_s)+(8_{s′},8_{s′})$ is the odd part of a $\mathbb{Z}_2$ grading of e8, yes?

Putting together (B) and (C), it is true that (D)

A Standard Model generation sits in the odd part of a $\mathbb{Z}_2$ grading of e8.

yes? I agree that there is extra stuff (which you’re calling an anti-generation) in this odd part. And that may be what’s confusing this discussion. But it is true that (E)

There is a $\mathbb{Z}_2$ grading of e8 such that the gravitational and standard model bosons sit in the even part, and a generation of fermions sits in the odd part, with extra stuff in both parts.

yes? If you say “that’s all true,” then I’m satisfied. (Even if you say “that’s all true, but …”) If you say one of (A) through (E) is false, then I’m confused, but interested to hear why.

Posted by: Garrett on July 17, 2008 10:11 AM | Permalink | Reply to this

Re: Summary

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I agree that there is extra stuff (which you’re calling an anti-generation) in this odd part.

It’s not some random “extra stuff”, that I’m (perversely) calling an anti-generation.

If a generation transforms as

(1)

(\mathbf{2},R) \oplus (\overline{\mathbf{2}},\overline{R})

under $Spin(3,1)\times\text{SM}$ , then an anti-generation transforms as

(2)

(\mathbf{2},\overline{R}) \oplus (\overline{\mathbf{2}},R)

I have no idea why you think you can get away with calling (1) a generation, and then proceeding to ignore (2) as just some random junk.

In a (perhaps vain) attempt to carry the discussion forward, let me explain the crucial physical difference between a chiral theory, like the Standard Model (which contains three copies of (1) and no copies of (2)) and a nonchiral theory, like your “model”, which contains an equal number of copies of each.

In the former case, the fermions are necessarily massless, until the electroweak gauge symmetry is broken. In the latter case, a generation and an anti-generation can pick up a (large) gauge-invariant mass.

(For mathematicians, this is the statement that there’s a nondegenerate, invariant pairing between the representations (1) and (2), but there’s no such pairing of (1) with itself.)

Thus, in your model, there are – generically – no fermions at all in the low energy spectrum (having all picked up large (presumably Planck-scale) gauge-invariant masses).

This is completely unlike the Standard Model.

Previously, you were trying to maintain that the problems with your model would be fixed up by choosing a different embedding in $E_8$ . No such alternate embedding is possible.

If you say “that’s all true,” then I’m satisfied.

Maybe you could explain why you find this decomposition (that I wrote down in my post seven months ago) satisfactory.

Posted by: Jacques Distler on July 17, 2008 2:30 PM | Permalink | PGP Sig | Reply to this

Re: Summary

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I’d be happy to discuss these things. But first (I risk being sphexish here) do you agree that the following statement is true?

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the gravitational and standard model bosons sit in the even part, and a generation of fermions sits in the odd part, with extra stuff in both parts.

Posted by: Garrett on July 17, 2008 3:56 PM | Permalink | Reply to this

Re: Summary

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Seven months ago, I classified all possible embedding of $Spin(3,1)\times SU(3)\times SU(2)\times U(1) \subset D_4\times D_4 \subset E_8$ (for both noncompact real forms of $E_8$ ) and wrote down the resulting spectrum of fermions.

Do you agree that I did that correctly?

If so, then I am not sure what you wish me to add to the results presented there.

If not, then we should discuss our points of disagreement.

Posted by: Jacques Distler on July 17, 2008 4:24 PM | Permalink | PGP Sig | Reply to this

Re: Summary

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Once again, I’d be happy to discuss these things, but first (I increasingly risk being sphexish here) do you agree that the following statement is true?

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the gravitational and standard model bosons sit in the even part, and a generation of fermions sits in the odd part, with extra stuff in both parts.

Posted by: Garrett on July 17, 2008 4:50 PM | Permalink | Reply to this

Re: Summary

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The $\mathbb{Z}_2$ even part contains the Standard Model gauge bosons (along with other junk). The $\mathbb{Z}_2$ odd part consists of a generation and an anti-generation.

Satisfied?

If you’re not satisfied, then what, in the above statement, do you disagree with?

If you can’t find anything in the above statement to disagree with, then let’s move on.

Now, is the classification of all such embeddings, in my post from seven month ago, correct, or isn’t it?

Is the proof that this is the best that one can do — for any embedding of $Spin(3,1)\times SU(3)\times SU(2)\times U(1) \subset E_8$ (for either noncompact real form of $E_8$ ) correct, or isn’t it?

I expect that you’ve discussed this extensively with Gregg Zuckerman, David Vogan, and Jeff Adams. What do they have to say?

Posted by: Jacques Distler on July 17, 2008 5:20 PM | Permalink | PGP Sig | Reply to this

Re: Summary

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For some reason you don’t like to include so(3,1) now? Strange, but OK, so you agree with the following then, yes?

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the standard model bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

After we agree to this, then we can talk about other stuff, but I want to make sure we agree to this first. I think this is true, and I want to make sure you also think it’s true, otherwise we have nowhere to work from. So, that is true?

Posted by: Garrett on July 17, 2008 7:43 PM | Permalink | Reply to this

Re: Summary

I have made very precisely clear what is correct. And you haven’t seen fit to disagree with my statement. (If you do disagree with it, I will be happy to have a substantive discussion of the points of our disagreement.)

If you wish to postpone, indefinitely, a discussion of substantive matters, we can keep playing games.

I, at least, have laid my cards on the table, and made some precise mathematical statements about what is and is not possible in your “E8” theory.

Where do you stand on the questions posed above? Inquiring minds, like Christine, want to know.

Posted by: Jacques Distler on July 17, 2008 8:02 PM | Permalink | PGP Sig | Reply to this

Re: Summary

Do I understand correctly that the “other stuff” that sits in the odd part is considered important for Distler (it is the “anti-generation” which for him is one of the points that would make the whole approach doomed to be incorrect), whereas for Lisi the “other stuff” – whatever it is – can be worked out, eventually avoiding a possible invalidation? Is this the point of tension?

Posted by: Christine Dantas on July 17, 2008 8:34 PM | Permalink | Reply to this

Re: Summary

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I am trying to work constructively, by starting with something we both agree with. Remember, this discussion we’re having is about Urs’ question (1); basically, whether a generation fits in e8 or not. You have said (A)

There’s a Standard Model generation and a Standard Model anti-generation, each of which sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

Which implies that (B)

A Standard Model generation sits partly in both the $(8_s,8_s)$ and the $(8_{s′},8_{s′})$ .

And now you have said (F)

The $\mathbb{Z}_2$ even part contains the Standard Model gauge bosons (along with other junk). The $\mathbb{Z}_2$ odd part consists of a generation and an anti-generation.

And so it should be very direct for you agree that it is true that (G)

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the standard model gauge bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

yes?

I would rather not argue where we disagree, but prefer that we agree on a statement, (G), that we both think is true.

I am being very stubborn about this because I think this point has been misunderstood in the past by others (including me), who thought you were saying that one can’t even get one generation to fit in e8.

I’d rather not take this discussion away from Urs’ question (1) yet. Please don’t confound the issue by rephrasing it, or taking us off on (interesting) tangents.

Just please answer, is (G) true, as written?

Posted by: Garrett on July 17, 2008 8:46 PM | Permalink | Reply to this

Re: Summary

a generation of fermions (along with other stuff) sits in the odd part.

Not some random “other stuff”, an anti-generation. This is physically important, for the reasons stated above.

Do you actually disagree with the statement that this “other stuff” is an anti-generation? If so, then we can discuss it. If not, then I have a hard time crediting your statement that

I am trying to work constructively…

At issue is whether the E8 theory can ever be chiral (let alone have three net generations). I have argued that it cannot. These arguments are not new. They were stated very clearly, and very precisely, seven months ago.

What I see here is a studious attempt to avoid discussing this central issue (or any other issue of substance). You are not, by any stretch of the imagination, being “constructive.”

If you want to be constructive, you could start by answering these questions.

Posted by: Jacques Distler on July 17, 2008 9:19 PM | Permalink | PGP Sig | Reply to this

Re: Summary

I will try to say what I’ve understood from Garrett, or what he wants to say:

I think he agrees with your calculation, but even so wants to use the junk anti generation, even if it provides a massless spectrum for the SM. From what I understood, instead, he wants to figure out a way to give mass to the particles by coupling it to a gravity field. It seems, according to him, that his worst problem it is that he wants to break the coupling with the gravitational field, so that a mass can be provided to that massless spectrum, but such thing is not possible unless he breaks the symmetry by hand.

I guess that’s sums up what comes up to me when I try to understand what he tries to argue.

Posted by: D on July 17, 2008 9:57 PM | Permalink | Reply to this

Re: Summary

Sorry, my name was not printed above…

Posted by: Daniel de França MTd2 on July 17, 2008 10:28 PM | Permalink | Reply to this

Re: Summary

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You keep suggesting interesting tangents, but I would like this settled first; then we can discuss whatever you like. You have persuaded many people (intentionally or unintentionally) that one generation of Standard Model fermions does not fit in e8. I would like to clear this up, here and now, before we move on. So, this statement:

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the gravitational so(3,1) and standard model gauge bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

Is this true? A simple “yes” would be fine, and then we could discuss whatever you’d like. Even a “Yes, but…” would be fine. But I need to see the “yes” here before we move on. It seems obviously true, based on what you’ve said already, but I want it settled conclusively with your “yes.”

Perhaps, as a service to your many readers, you might then add it as clarification to your post about it on Musings, which I think caused some confusion. But, really, this is all I’m after. Having one generation fit in e8 this way is important to me, for making it worthwhile to spend my time on.

After you agree, I’m perfectly happy to discuss the impossibility of fitting three generations in e8. But I need to see your agreement, via a “yes,” to the above statement. Please don’t suggest tangents, rephrase it, or bring up any side considerations to modify it. A “yes,” followed by qualifying statements about anti-generations would be fine.

Posted by: Garrett on July 17, 2008 10:18 PM | Permalink | Reply to this

Re: Summary

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In your paper, you stated

That you embedded one generation of SM fermions in the $(8_s,8_s)$ of $D_4\times D_4\subset E_8$ .
That the $(8_s,8_s)$ is related by triality to the $(8_{s'},8_{s'})$ and the $(8_v,8_v)$ .

You have repeated these statements on numerous occasions, most recently this past Saturday.

Both of these statements are false, as was pointed out to you seven months ago.

My post stated very clearly that the correct fermion spectrum for an embedding via $D_4\times D_4$ , is a generation plus an anti-generation¹. And I have repeated that statement, ad nauseum, here.

Do you agree that what I wrote in my post (and have repeated endlessly here) is correct? If so, then I do not see what possible purpose could be served by further belabouring this point.

If not, then please state what it is that you disagree with, so that we can proceed to the heart of the matter.

If you feel that, somehow, it is necessary for me to assent to the obvious fact that the phrase “a generation plus an anti-generation” contains the phrase “a generation”, before we can proceed, then so be it.

Is that really the intellectual level at which you wish to hold this discussion?

Having descended to that level of mindless triviality, can we please discuss something substantive?

¹ To be completely accurate, this is for two of the five possible embeddings of a

D_4\times D_4\subset E_8

. For the other three, all fermions are in

SU(2)

doublets.

Posted by: Jacques Distler on July 17, 2008 11:02 PM | Permalink | PGP Sig | Reply to this

Re: Summary

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These tangents are fascinating, and I look forward to discussing them with you in the future (though they might not be appropriate here on Urs’ blog). And it’s curious that you would want to change this into an argument about phrases, when it is about mathematics, physics, and basic logic. But, I must resist this great temptation to stray from what is now surely sphexishness on my part, and repeat this statement again:

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the gravitational so(3,1) and standard model gauge bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

Do you agree that this is true? A “yes” will do nicely as a response, and we can then get to these other interesting questions. A “no” response would be surprising. Further attempts to introduce tangents, rephrasing of the statement, or otherwise deflect this discussion will lead only to our mutual frustration or amusement (depending on temperament) and what is now my predictable behavior of repeating this statement and entreating you to directly affirm or refute it. So, what do you say, is the above statement true? Surely you know it is true. Will you provide a “yes, that is true, but the model can’t work because…”?

Posted by: Garrett on July 18, 2008 12:51 AM | Permalink | Reply to this

Re: Summary

What a silly game. Reminds me of the time I got J.D. to denounce a certain fanatical string theory colleague over at cosmicvariance :)

So the necessity of combining the SM generation with an anti-generation is considered a *tangent* here? Not the central point? Wow. All those people who concerned themselves with anomaly cancellation in the past shouldn’t have bothered – just take your chiral gauge theory (SM or whatever you have), stick it together with the corresponding anti-fermion theory, and presto you get a nice vector-like theory which is automatically anomaly-free. No need to have those annoying anomaly cancellation conditions restricting your fermion content any more. Yipee!

But to be a little less frivolous for a minute, it is apparently possible to get back the SM generation by itself after combining it with the anti-generation. In a paper by Eichten and Preskill (NPB, 1986) it was suggested how, starting from the vector theory consisting of a chiral fermion theory (e.g., an SM generation) plus its chiral anti-fermion theory, one can introduce non-chiral interaction terms such that the excitations of the chiral anti-fermion part get heavy masses and effectively decouple from the chiral fermion part (which remains light). This was proposed as a possible way to regularize chiral gauge theories, by doing it via a vector theory which is easy to regularize (e.g., on the lattice). Here the chiral anti-fermion fields are fictitious and just introduced as a trick, similar in spirit to Pauli-Villars regularization.

Presumably one can play this game in the present situation, i.e., introduce interactions a la Eichten-Preskill to make the SM anti-generation excitations heavy and decoupled from the SM generation itself. But then the anti-generation excitations are not fictitious but really out there, so you get all that extra heavy junk besides the SM generation itself. So in conclusion I would say it may be doable in principle, but looks really contrived and not at all interesting/appealing as a model for physics beyond the SM. And that’s even before taking into account that you only have one of the 3 SM generations.

Posted by: amused on July 18, 2008 6:55 AM | Permalink | Reply to this

Eichten-Preskill

You’re talking about this paper? That was about lattice fermions, and the fermion doubling problem for chiral gauge theories on the lattice. As I recall, they proposed a variant of the Wilson mechanism, which would work for chiral gauge theories. The “anti-generations” (which got a gauge-invariant “Wilson” mass with the doubler modes) were not fundamental fields, but rather, were composite fermions, bound by some funky interactions that they introduced in the lattice action.

Did their scheme actually work, in the end?

In any case, I don’t see how that would apply to the continuum theories discussed here. Here, we apparently have fundamental fields, transforming as anti-generations, no doubler modes, and no scope for writing down the funky lattice interactions considered by E&P.

Relevant or not, it’s cool to be reminded of that paper. Takes me back to my graduate school days …

Posted by: Jacques Distler on July 18, 2008 7:24 AM | Permalink | PGP Sig | Reply to this

Re: Eichten-Preskill

Although I gave Eichten-Preskill as the reference, the work I actually had in mind was more recent: This paper by Poppitz and Shang. For gauge theories with vector-like fermion rep they discuss how to split into “light” and “mirror” chiral reps, with the idea that the mirror part should become heavy and decouple due to funky interactions. They consider this in the lattice setting, but without the issues of doublers, Nielsen-Ninomiya etc since they use the recent overlap/Ginsparg-Wilson formulation which isn’t afflicted with those problems.

Basically the motivation is this: Formulation of lattice chiral gauge theories using overlap/GW is still incomplete since no one has showed anomaly cancellation except when the gauge group is U(1). So Poppitz-Shang want to get around this by starting with vectorlike lattice theory in overlap/GW framework, which is on a completely fine footing, and then get to the chiral theory by decoupling the mirror fermions as mentioned. If I remeber rightly, they show that a necesary condition for being able to do this is that the usual (continuum) anomaly cancelation condition must hold. Didn’t show that it is a sufficient condition though. A nice aspect of their approach is that it can apparently be investigated numerically via lattice simulations.

That’s not my game though, so I didn’t study the paper in much detail. Their focus seems to be mainly on strongly interacting chiral gauge theories, so I don’t know about applicability of the ideas and techniques to the SM… but I didn’t get the impression that there was anything uniquely “latticey” about the constructions… So if Garrett wants to try to keep his boat afloat this might offer one possibility.

Poppitz-Shang described their approach as implementing the idea of Eichten-Preskill, which was why I refered to that paper in previous comment. Embarrassingly, I hadn’t actually read that paper though… (Just looked at it now and it seems to be quite different from Poppitz-Shang.)

Posted by: amused on July 18, 2008 9:40 AM | Permalink | Reply to this

Poppitz-Shang

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Poppitz-Shang, eh? Boy, you really do make me work for my my money, don’t you?

I haven’t read the paper in any detail, but I think I disagree with you: the effect they are discussing is very latticey.

In the continuum, you can’t give the anti-generation a mass, without breaking either $Spin(3,1)$ or $G$ -gauge invariance. (In Lisi’s case, $Spin(3,1)$ is also a gauge symmetry, which gives yet another reason not to break it.)

On the lattice, there’s no $Spin(3,1)$ (or $Spin(4)$ ) symmetry, and you can — at least, in principle — generate the requisite mass term, without breaking the $G$ gauge-symmetry.

Posted by: Jacques Distler on July 18, 2008 1:50 PM | Permalink | PGP Sig | Reply to this

Re: Poppitz-Shang

There is no possibility of a usual (Dirac) mass term for the chiral mirror fermions in any case - in this regard the situation for overlap/GW lattice formulation is the same as for the continuum one. Nothing to do with Spin(4) being restricted to the lattice hypercube transformations as far as I can tell. (And that restriction together with gauge invariance is still enough to prove all the needed things like renormalizability.)

From my (very limited) understanding of Poppitz-Shang, the way the decoupling of the mirror fermions is supposed to happen is as follows: The mirror fermions are coupled together via funky interaction terms which don’t break any of the theory’s symmetries. This is supposed to make the mirror fermions bind together in composite states (a bit like quarks binding together to form baryons perhaps). And apparently these composite states can be massive (unlike the original mirror fermion fields). Hence they decouple from the low energy dynamics of the light (non-mirror) chiral fermions when the masses are large.

However, from what P-S write, it seems the only senario where this situation has been shown (at some level) to arise is in the strong bare lattice coupling limit of the lattice gauge theory. The interest of P-S (and E-P as well it seems) is in asymptotically free chiral gauge theories. For these, a continuum limit can be taken in the limit of vanishing bare lattice coupling. The expectation/hope seems to be that the decoupling of mirror fermions which happens at strong bare lattice coupling will persist in the weak coupling regime where the continuum limit is approached. (I guess this is similar to how confinement in QCD, which can be proved at strong lattice coupling, is expected - and at this point pretty much demonstrated - to persist at weak lattice coupling where the continuum limit is approached.)

The SM is of course not asymptotically free, so the preceding mirror decoupling expectations don’t apply in that case. But I haven’t seen any statement that such decoupling couldn’t also occur in that case… Is there anything known about this? It would be nice if someone has (or could) prove a no-go theorem stating that the mirror decoupling stuff described above can never happen in weakly coupled chiral gauge theories theories like the SM.

How about writing a blog post about the Poppitz-Shang/Eichten-Preskill papers sometime :)

Posted by: amused on July 18, 2008 3:28 PM | Permalink | Reply to this

Re: Poppitz-Shang

This is supposed to make the mirror fermions bind together in composite states (a bit like quarks binding together to form baryons perhaps). And apparently these composite states can be massive (unlike the original mirror fermion fields).

The obvious constraint, here, are the ‘Hooft Anomaly Matching Conditions.

In the case of the strongly-coupled Standard Model (the Abbott-Farhi Model), the solution to these constraints is well-known. The spectrum of massless composites is … (wait for it) … the same as that of the conventional, weakly-coupled Standard Model.

How about writing a blog post about the Poppitz-Shang/Eichten-Preskill papers sometime :)

Could be a plan. It’s been a while since I wrote a post about Lattice Gauge theory.

Posted by: Jacques Distler on July 18, 2008 4:13 PM | Permalink | PGP Sig | Reply to this

Re: Poppitz-Shang

Thanks, that’s interesting, I didn’t know that. Actually, if you could write a general post on this topic, or maybe a series of posts, it could be really useful. There seems to be a lore about this stuff which is known to insiders but which isn’t to be found in textbooks (that I’ve seen) and which is hard to dig out of the literature by oneself.

Posted by: amused on July 18, 2008 4:51 PM | Permalink | Reply to this

Re: Summary

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Since I started the discussion, let me just try to summarize the latest exchange for the potential benefit of whoever is following this.

Garrett keeps asking for a direct confirmation of this statement:

There is a $\mathbb{Z}_2$ grading of the split real form of $e_8$ such that the gravitational $so(3,1)$ and standard model gauge bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

Do you agree that this is true?

From the discussion it seems clear that the answer is Yes.

The disagreement seems to be about how to phrase this such that it is not misleading, either way or other, given that we are in the end not just talking about representation theory by itself but about its use in a gauge theory.

To get a viable field theory model, Garrett will have to invoke some trick or other to argue that the “other stuff” plays no real role.

Now, part of the “other stuff” is not just random other stuff, but an anti-copy of some of the stuff one would wish to keep. So the question is if there is any trick to be invoked that removes even such mirror-image kind other stuff.

Mr. “amused” mentioned one potential idea. Maybe it can be employed here, maybe not.

If nothing works, then the statement “there fits one generation in $e_8$ plus other stuff” is, while true, somewhat misleading from the point of view of gauge theory, bcause the presence of the other stuff will make the first part of the statement useless (for the purposes discussed here).

But maybe something does work. I don’t know. But as far as I can tell it is true that one generation of the standard model is sitting in $e_8$ in the way quoted above. Next to some other stuff. :-)

Posted by: Urs Schreiber on July 18, 2008 3:41 PM | Permalink | Reply to this

Re: Summary

No, the above summary is not correct.
Here’s a mathematical analog: an index theorem relates the sum of the critical indices of a vector field on a manifold to its Euler character.
So the winding numbers of a vector field on the surface of a (genus one) torus necessarily sum to zero.
Consider a vector field with two zeroes, of index +1 and -1. Let’s see what Salviati and Simplicio say

Distler: The net winding number is zero
Lisi: The winding number is one plus junk

In the physical situation, the standard model requires non-zero *net* winding number (fermion chirality). To claim otherwise represents a fundamental misunderstanding regarding what is the standard model of particle physics, and how fermion masses are generated.

Posted by: Summerizer on July 18, 2008 4:44 PM | Permalink | Reply to this

Re: Summary

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No, the above summary is not correct.

[…]

In the physical situation,

That’s why I made a distinction between the mathematical fact and its physical application.

There is enough room in the odd part of $e_8$ to fit in one generation. Anyone disagrees with that?

Posted by: Urs Schreiber on July 18, 2008 5:03 PM | Permalink | Reply to this

Re: Summary

> There is enough room in the odd part of e8 to fit in one generation.
> Anyone disagrees with that?

Yes, particle theorists would disagree since you are confusing the colloquial and technical notions of “fit”. Technically, a generation does not “fit” into a representation if it necessarily carries along an anti-generation.
This is not a question of string theory versus loop quantum gravity.
No particle theorist active during the past three decades would claim that a model with a generation and anti-generation constitutes a “one generation model”. Any phenomenologist or grand unified model builder would be absolutely flabbergasted that the discussion here is even taking place.
As evident from the discussion above, the rhetorical claim of a “fit” is an attempt to confuse rather than clarify the physics.

A miraculous mechanism to achieve a low energy chiral spectrum from Garrett’s non-chiral particle content is about as likely at this point as a mechanism to give his theory sensible interacting dynamics.
Reality is tough to handle.

Posted by: Summerizer on July 19, 2008 1:48 AM | Permalink | Reply to this

Re: Summary

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the rhetorical claim of a “fit” is an attempt to confuse

(Except for the imputation of malice) this is what I said.

There are other problems with Garrett’s model, too, by the way, some of which we had discussed in this thread.

Posted by: Urs Schreiber on July 19, 2008 11:53 AM | Permalink | Reply to this

Re: Summary

..just to make that clear for mathematicians: why does the presence of some ‘extra structure’ in some representation which allows a non-degenerate pairing between the acclaimed (existing) generation of fermions and the ‘anti-generation’ make the former structure ‘useless’ for the ‘purposes of gauge theory’? Which pre-assumptions on the physical interpretation of mathematical structures imply that ‘non-chiral’ theories cannot be used to give predictions concerning the masses of fermions? I suppose, here as elsewhere, there are a number of ‘normative discoursive assumptions’ between physicists involved which are simply ignored by Garrett, which is the actual reason for that disagreement.

I would love to see a mathematical theorem expressing (to be clear: translating) Distler’s concerns.

Posted by: a.k. on July 19, 2008 9:21 PM | Permalink | Reply to this

Re: Summary

Pardon my ignorance - I guess I don’t speak even prose or at least not Z_2 prose:
The bosonic fields transform as 1-forms with values in various tensor representations of Spin(3,1); the fermionic fields transform as 0-forms in spinor representations of Spin(3,1).

I thought bosons commuted and fermions anti-commuted??

Posted by: jim stasheff on May 23, 2008 1:43 PM | Permalink | Reply to this

Super duper

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If you write a connection, locally, as a Lie algebra-valued 1-form, $A = dx^\mu A_\mu^a(x) T_a$ then $A_\mu^a(x)$ is an ordinary (commuting) bosonic field. $\psi^\alpha(x)$ is a fermionic (anticommuting) field.

If you want a mathematical theory of such gadgets, you can think about families of supermanifolds, where the “base” supermanifold has infinite odd dimensionality. Odd functions on such a supermanifold are what physicists call fermionic fields.

As much as I like the theory of supermanifolds, I have never found this a particularly useful point of view.

Posted by: Jacques Distler on May 23, 2008 4:24 PM | Permalink | PGP Sig | Reply to this

Re: Super duper

Thanks.
I have trouble thinking of the coefficient functions A^a_\mu(x) as the field. I’m a disciple of Rainich who taught that e.g. a vector was a thing in itself and not a tuple of coordinates.

Posted by: jim stasheff on May 24, 2008 1:59 PM | Permalink | Reply to this

Proof from the Infinite Book; Re: Super duper

Naive question: how in Cat or n-Cat theory do we define even and odd infinite dimensionalities?

I seem to half remember a joke or puzzle about whether the middle page of a book with an odd infinite number of pages had one or two sides…

Posted by: Jonathan Vos Post on May 24, 2008 8:40 PM | Permalink | Reply to this

Read the post HIM Trimester on Geometry and Physics, Week 4
Weblog: The n-Category Café
Excerpt: Talk in Stanford on nonabelian differential cohomology.
Tracked: May 29, 2008 11:22 PM

Re: E8 Quillen Superconnection

I do not wish to interrupt the interesting exchange, so you may ignore my comment if that is the case.

If I understand it correctly, it is agreeded on both parts that Lisi’s model as a whole results in a non-chiral spectrum (net number of generations = 0). Furthermore, Distler appears to have shown that there are no decompositions of E8 allowing the inclusion of the 3 SM generations. (Does Lisi agree with the latter?)

So, I was wondering – is it really all there is to be concerning the use of E8 (or any other group, for what is worth)?

I mean, on speculative grounds, is it possible that simply using the group “as it is” is not the whole story, but actually one could gain more room for analysis or insight by seeing the group from a different “perspective”?

What I have in mind here comes from something I was reading superficially about, groups of polynomial growth and the work of Gromov. Does E8 have any relation to such groups? If so, would it be possible to prove whether the “net # gen = 0” feature shown by Distler for the E8 is preserved (or not) when considering related groups of polynomial growth (if that is possible at all), in which the group is “seen from infinity”?

This is an innocent question from a layman, so I apologize in advance if this speculation has no meaning.

Posted by: Christine Dantas on July 17, 2008 7:06 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Jacques says that the odd part of E8
contains
one generation of fermion particles
and
one anti-generation of fermion particles.

Could the anti-generation of fermion particles
be physically interpreted as representing
one generation of fermion anti-particles?

If so, Garrett’s model might look better as a one-generation model,
and then the other two generations would have to be not-fundamental, but somehow constructed from the fundamental stuff,
perhaps by having the second and third generations being composites of the first.

Tony Smith

Posted by: Tony Smith on July 18, 2008 11:48 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Could the anti-generation of fermion particles be physically interpreted as representing one generation of fermion anti-particles?

No.

Those have already been accounted for. (And, besides, the quantum numbers would be wrong.)

Posted by: Jacques Distler on July 19, 2008 12:18 AM | Permalink | PGP Sig | Reply to this

Re: E8 Quillen Superconnection

I think Tony wants to unaccount for them then match the particles and antiparticles with the two half-spinors. That kind of only requires E6 but Tony I know goes up to E8 by having Dirac gammas too. Tony can still get Garrett’s two D4s doing what he is doing. Having the two D4s is nice.

Posted by: John G on July 19, 2008 7:07 AM | Permalink | Reply to this

Circling the Wagons

Urs wrote:

Except for the imputation of malice…

The only imputations of malice that I have seen were from champions of Lisi’s, and were directed at those (myself, in particular) who took up the thankless task of pointing out the problems with Lisi’s idea.

While, in knowledgeable circles, the pendulum may have swung in the other direction, the recent New Yorker article makes clear that such imputations of malice are alive and well elsewhere.

Posted by: Jacques Distler on July 19, 2008 6:49 PM | Permalink | PGP Sig | Reply to this

Re: Circling the Wagons

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Hi Jacques,

what I meant was that Garrett is surely not making a deliberate malicious “attempt to confuse” # any one on purpose.

He is clearly trying to fiddle around to save his model – as one does. So here he is trying to make sure that he has at least that one generation sitting there, hoping that he can get rid of the unwanted stuff somewhow, but glad that at least that one generation is there (where it seemed for a while that not even that might be true). Let’s grant him that and advise him that even so the model is still in big trouble. I think we had seen that this is not even the only trouble.

It’s a pity that the insane media attention is turning this into what it is turning into including all the emotional side-effects on all sides, because otherwise we had a better chance of having just an interesting discussion about a failed idea (I like to think of theoretical physics as a place where one can discuss with gain about failed ideas) which however contains aspects that, with good advice from experts such as most notably you, could be followed up fruitfully. I am very fond of the summary and constructive advice you gave here. My advice to Garrett would be: follow that.

I should maybe make that post of yours a blog entry, since it will be buried under the comments here…

Posted by: Urs Schreiber on July 19, 2008 7:19 PM | Permalink | Reply to this

Re: Circling the Wagons

Hi Jacques,

This is getting rather personal, and I can’t speak for colleagues, but I hold your abilities and character in high regard. You have shown a tendency to avoid answering questions directly if you think it will hurt your position, and you’ve attempted to paint things in the worst possible light. But you haven’t lied outright as far as I’ve seen. Sadly, you do seem to view me and my “champions” with contempt — considering us “chumps” — but as long as you speak your mind honestly, I can’t really hold that against you. And you’ve done an impressive job analyzing the mathematics involved with this E8 theory, even catching a mistake I made (thinking so(7,1)+so(8) was in a real form of e8), and helpfully correcting it. So I have no malice towards you, and actually look forward to meeting you in person soon.

Posted by: Garrett on July 19, 2008 10:51 PM | Permalink | Reply to this

Re: Circling the Wagons

Garrett,

I am quite sure we would have a smashing time, and look forward to meeting you in person, someday.

However, I don’t see that there is anything to be gained by injecting personalities into this discussion.

And you’ve done an impressive job analyzing the mathematics involved with this E8 theory, even catching a mistake I made (thinking so(7,1)+so(8) was in a real form of e8), and helpfully correcting it.

Is that really the only flaw you’re willing to admit to?

It is, I will agree, trivial, and easily corrected. But it is entirely peripheral to the much more serious problems discussed in my posts, and reviewed here.

That, even seven months later, you won’t even acknowledge their existence (and seemed to be surprised that the spectrum of fermions in your model is non-chiral, consisting of a generation plus an anti-generation) does not speak well of you.

But, then, I said I was going to avoid injecting personalities into this discussion. So let me confine myself to giving you one piece of advice.

As a general rule, if you are trying to do “X”, and someone comes along and claims to prove that “doing X is impossible”, the circumspect thing to do is study that proof very carefully.

Either you find that the proof is correct, in which case further effort to do “X” would be wasted. Or you discover a loophole in the proof, which provides you the crucial clue as to how to succeed at doing “X”.

Either way, you save yourself much wasted effort, and not a small amount of public embarrassment.

Posted by: Jacques Distler on July 19, 2008 11:32 PM | Permalink | PGP Sig | Reply to this

Re: Circling the Wagons

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Jacques,

I am quite sure we would have a smashing time, and look forward to meeting you in person, someday.

Great. Will you agree to have lunch with me at Perimeter on September 8th, before your talk?

It is, I will agree, trivial, and easily corrected. But it is entirely peripheral to the much more serious problems discussed in my posts, and reviewed here. That, even seven months later, you won’t even acknowledge their existence (and seemed to be surprised that the spectrum of fermions in your model is non-chiral, consisting of a generation plus an anti-generation) does not speak well of you.

If it is true (and you agree with me, Urs, and, err, yourself) that

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the gravitational so(3,1) and standard model gauge bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

then this raises the question of what to make of the other stuff in e8. If we do the straightforward thing of starting with the Z_2 graded, e8 valued superconnection, then this doesn’t work for matching the standard model. Not only is there a 64 block of “anti-fermions” in the $128_{S+}$ odd part of the superconnection, but there’s another 64 block of “1-form fermions” in the d8 part. Also, their quantum numbers aren’t correct, with respect to the d4’s. This is why I said (and emphasized) in the paper I wrote, that:

When considered as independent ﬁelds with E8 quantum numbers,…, the second and third generation of ﬁelds do not have correct charges and spins.

My point of view is that it is interesting that the standard model (with one generation at least) can be formulated using a superconnection, and it is also interesting that the algebra of this superconnection fits nicely in e8, with two blocks of 64 remaining. I don’t think the standard model can be modeled using a Z2 graded, e8 valued superconnection – I suspect it would have to be something weirder, making use of these two 64 blocks in a different way. I am afraid this is pretty vague speculation at this point, motivated by the triality symmetry of d4 and e8. But I have been emphasizing this specific problem from the very beginning, doing the best I can to have it appear in every description in the media.

As to the future… I intend to pursue the research program you suggested and Urs agreed with, as well as researching other ideas. I will attempt to include you in this, to the degree you are willing.

Posted by: Garrett on July 20, 2008 1:43 AM | Permalink | Reply to this

Re: Circling the Wagons

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Will you agree to have lunch with me at Perimeter on September 8th, before your talk?

Sure.

but there’s another 64 block of “1-form fermions” in the d8 part.

Umh, no. Those are bosons There are never more that 128 fermions in any $\mathbb{Z}_2$ grading of $e_8$ .

(Which is why I wasted considerable time trying to find a two-generation variant of your model. To my great embarrassment, I wrote that I thought I had found one, in the first version of my first post on your model. As it turns out, I was wrong. And, more fundamentally, I had been wasting my time, because the fermion spectrum, for any embedding in $e_8$ , is always non-chiral.)

That said, it is true that the bosonic sector of your model is full of all sorts of phenomenologically unacceptable crap (colour-triplets of various spins, being the most notable example).

Perhaps these can be given masses and drop out of the low energy spectrum. I haven’t thought about it, but it seems like a potentially solvable problem, unlike the fundamentally non-chiral nature of your model.

I don’t think the standard model can be modeled using a $\mathbb{Z}_2$ graded, $e_8$ valued superconnection.

On that, I think we can agree …

Posted by: Jacques Distler on July 20, 2008 2:32 AM | Permalink | PGP Sig | Reply to this

Re: Circling the Wagons

Jacques Distler said, about Garrett Lisi’s E8 model:
“… the spectrum of fermions … is non-chiral, consisting of a generation plus an anti-generation …”.

That may be true for superstring theories at tree-level. For instance, in his book String Theory (volume II, Cambridge 1998), Joseph Polchinski says:
“… At tree level the only way generations and antigenerations could become massive is in pairs … This leaves the Euler number unchanged. …”.
So, at tree level in superstring theory you cannot make the antigenerations unobservably massive without making the generations also unobservably massive.

However, even in superstring theory context (and for all I know Garrett Lisi’s connnection may be constructed somehow differently from the relevant constructions of superstring theory), Polchinski goes on to say:
“… One might think that the argument above about generations and antigenerations becoming massive in pairs would exclude any Euler number changing conifold transition … However … study of nonperturbative dynamics in field and string theory …[shows]… that at a nontrivial fixed point … one can have phase transitions that cannot be described by any classical Lagrangian …”.

So, even in the context of superstring theory, is it not possible that there might exist some nonperturbative process that would leave Garrett Lisi’s one generation OK and move the antigeneration to an unobservably massive state ?

Tony Smith

Posted by: Tony Smith on July 20, 2008 4:26 AM | Permalink | Reply to this

Re: Circling the Wagons

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That may be true for superstring theories at tree-level.

This has nothing whatsoever to do with superstring theory. It is basic quantum field theory.

… [stuff about conifold transitions] …

The conifold transitions have quite prosaic quantum field-theoretic interpretations in Type-II theory, where there is $N=2$ supersymmetry, and the field theories undergoing the transition are completely non-chiral.

When embedded in heterotic string theory, the transitions are decidedly much more “stringy” in nature.

More generally, it is true that different branches of the moduli space of certain quantum field theories (with different local lagrangian descriptions on each branch) can be separated by multicritical points, with no local Lagrangian description.

Lisi has written down a local Lagrangian (well, kinda, sorta…) which, alas, does not describe the physics that he wants. You are free to speculate about the existence of other branches (with different local Lagrangian descriptions), which do have the desired physics.

Posted by: Jacques Distler on July 20, 2008 5:39 AM | Permalink | Reply to this

Re: Circling the Wagons

You’re missing the point, Tony. What happens is that generations and anti-generations pair up to become massive as Polchinski says. It is thus the net number of generations that is relevant for the low energy theory because that’s what remains massless. String theory compactifications, as Polchinski discusses earlier, do produce a chiral spectrum with the net number of generations given by half the Euler character.

Posted by: Aaron Bergman on July 20, 2008 6:05 AM | Permalink | Reply to this

Re: Circling the Wagons

Half the Euler character being 3 for the moduli of the 6 punctured sphere, the real points of which are tiled by the 3d Stasheff associahedron, labelled by chorded hexagons …..

Posted by: Kea on July 20, 2008 6:30 AM | Permalink | Reply to this

Re: Circling the Wagons

Hi Garret, Hi Distler,

“Jacques,
I am quite sure we would have a smashing time, and look forward to meeting you in person, someday.
Great. Will you agree to have lunch with me at Perimeter on September 8th, before your talk?”

So, did you arrive at any conclusion during yesterday’s lunch you had together?

Posted by: Daniel de França MTd2 on September 9, 2008 7:04 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

This is a fascinating discussion, but completely unreadable due to the absurdly deep nesting of comments, over 20 levels deep at some points, with (I think) a mixture of top and bottom posting!

Sorry if it sounds cheeky of a non-regular to ask on someone else’s blog - it probably is - but with all due respect might there be any chance of people trimming “replies before last” where appropriate, so bystanders have a better chance of actually learning something?

Posted by: John R Ramsden on July 22, 2008 1:28 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

To my taste, I only read comments chronologically (not threaded). You have this option by clicking the button at the lower right side of the page.

Posted by: Christine Dantas on July 22, 2008 12:05 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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John Ramsden wrote

[This discussion is] completely unreadable due to the absurdly deep nesting of comments #

I know it can be hard to read. Just in case, for those not aware of it, I’ll mention that little button “view chronologically” at the bottom of each page, which makes the comments arrange in the linear order in which they came in, suppressing the nesting.

might there be any chance of people trimming “replies before last” where appropriate #

Not sure I understand what you are asking for. (?)

I think it is good practice to always verbatim quote the part that one is directly replying to in one’s own comment, using blockquote.

Even more robust as far as later readability goes is to give the quote and end it with a hyperlinked “#” that links to the comment which the quote is taken from.

so bystanders have a better chance of actually learning something? #

In the best of all worlds, somebody, such as one of the blog-hosts, would write an executive summary of what has happened, after it has happened.

For the first part of the discussion we had here, such a summary was posted by Jacques Distler, here.

Posted by: Urs Schreiber on July 22, 2008 12:17 PM | Permalink | Reply to this

citations

Urs wrote:

Even more robust as far as later readability goes is to give the quote and end it with a hyperlinked “#” that links to the comment which the quote is taken from.

Except that you don’t have to make the hyperlink “by hand.” It’s generated automatigically (by a Javascript) when you use the cite attribute, as in

<blockquote cite="#c017916"><p>Even more robust
as far as...</p></blockquote>

where the value of the cite attribute is a URL (in the case of comments, here, the “Permalink” URL to that comment).

Posted by: Jacques Distler on July 22, 2008 2:45 PM | Permalink | PGP Sig | Reply to this

Re: citations

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Ah, thanks. Didn’t know that.

Posted by: Urs Schreiber on July 22, 2008 2:57 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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John Ramsden wrote:

[This discussion is] completely unreadable due to the absurdly deep nesting of comments

Yeah, it’s annoying. When I’m posting comments on big threads I avoid clicking on ‘Reply to this’ in the comment I’m replying to. Instead, I click on ‘Reply to this’ in a location that makes my reply appear below the comment I’m replying to, but no more deeply nested… just equally deep.

I also religiously quote the text I’m replying to.

And, as already mentioned, there’s that handy ‘view chronologically’ button.

Also, you can skim through about 20 grumpy exchanges on this thread once you realize that Lisi keeps asking if some representation of the Standard Model gauge group is sitting inside the adjoint rep of $E_8$ , and Distler keeps saying “(yeah, but) in fact this representation plus something else is sitting inside there, which makes this useless for physics” — without ever saying out loud the “yeah, but” part that’s what Lisi wanted to hear. Instead of saying “yeah, but”, Distler keeps saying “no, in fact”.

Posted by: John Baez on July 22, 2008 3:13 PM | Permalink | Reply to this

Obfuscation

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Sigh. And here I thought we had put this one to bed.

Let’s be very clear on why it is that Lisi finds

The $\mathbb{Z}_2$ even part contains the Standard Model gauge bosons (along with other junk). The $\mathbb{Z}_2$ odd part consists of a generation and an anti-generation.

unacceptable, and insists on the vaguer

There is a $\mathbb{Z}_2$ grading of the split real form of $e8$ such that the standard model bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

Why? Because

He wishes to maintain the pretense that the “ $\mathbb{Z}_2$ odd part” is 192-dimensional, rather than 128-dimensional.
He wishes to assert that a model with zero net generations is a “one-generation model”.

The only way to achieve these goals is to attempt to obfuscate the fermion content. “A generation of fermions (along with other stuff)” is vague enough to be compatible with both these goals. “A generation and an anti-generation” is not.

Sorry, but that really does border on dishonesty, and I am not going to be a party to it, even passively.

The $\mathbb{Z}_2$ odd part consists of a generation and an anti-generation.

is both correct and more importantly, is not deliberately misleading.

I stand by that characterization, unless someone can give me a cogent reason why some other characterization would be superior. “Lisi prefers it.” is, for the reasons just explained, not a cogent reason.

Posted by: Jacques Distler on July 22, 2008 7:53 PM | Permalink | PGP Sig | Reply to this

Re: Obfuscation

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Jacques,

You have falsely stated what my wishes are, and used these claims to suggest I’m bordering on dishonesty…nice trick.

I have not tried to use a $\mathbb{Z}_2$ grading of $E8$ as a physical model. It is disingenuous to say that I have, or to speculate that I wish to, just so you can argue over and over again why it won’t work.

Posted by: Garrett on July 22, 2008 10:19 PM | Permalink | Reply to this

Re: Obfuscation

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I have not tried to use a $\mathbb{Z}_2$ grading of $E8$ as a physical model.

Already addressed that. I gather you would prefer the phrase “fermion content” instead of “ $\mathbb{Z}_2$ odd part”? If so do a global replace:

s/ $\mathbb{Z}_2$  odd part/fermion content/g

on my previous comment. Regardless of how one phrases it, you do still insist that your model has 192 (rather than 128) fermions, and that it is a one-generation, rather than a zero-generation model.

It’s clear that the vague statement

The fermion content consists of a generation (along with other stuff).

is compatible with those assertions, whereas the precise statement

The fermion content consists of a generation and an anti-generation.

is not.

If that’s not why you insist on my assenting to the former statement (and reject the latter), then why?

Posted by: Jacques Distler on July 22, 2008 11:32 PM | Permalink | PGP Sig | Reply to this

Clarification

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If that’s not why you insist on my assenting to the former statement (and reject the latter), then why?

Ah, thank you for asking. Your latter statement is irrelevant because it is a straw man. Let me try to make it clear what I’m doing. In a previous paper, I described how gravity and the standard model (with only one generation) could be expressed using a superconnection valued in part of a Clifford algebra. Then, in this more recent paper, now under discussion, I basically said “Holy crap! This algebra sits in E8! How cool is that!” And I was so excited that I did make some mistakes, which you helped fix, but this fact held true. I also thought it worth noting that there were these two other blocks of 64 elements in E8, which might relate to the second and third generations – and I emphasized that this was encouraging, but merely speculation. This is why my only claim is, and remains, this statement which Urs, John, and I have agreed that you agree is true:

There is a $\mathbb{Z}_2$ grading of the split real form of e8 such that the gravitational so(3,1) and standard model gauge bosons (along with other stuff) sit in the even part, and a generation of fermions (along with other stuff) sits in the odd part.

It is curious that you refuse to agree that you agree this is true – a cause of much amusement. Surely, as a champion of truth, you might remedy this now by agreeing this is true? (Eek, the return of the Sphex to its hole.)

I am not suggesting that this $\mathbb{Z}_2$ grading of E8 be used for a physical model – that is a straw man of your creation. I am mentioning this $\mathbb{Z}_2$ grading of E8 because Urs brought it up in a question, which I answered affirmatively via the above statement. But, given that the above statement is true, I am interested in considering what related constructions might give us a physical model, even if this $\mathbb{Z}_2$ grading of E8 does not. But I’d rather not engage in discussing these possible constructions without your explicit assent that our starting point, above, is true, as a sign of good faith.

Posted by: Garrett on July 23, 2008 1:04 AM | Permalink | Reply to this

Re: Clarification

Derek Wise cited your paper hereMacDowell–Mansouri technique, (p.4), as an example of application of . This paper was also a theme of a blog post made by John Baez. I guess Derek Wise could give us a nice insight about your ideas.

Posted by: Daniel de França MTd2 on July 23, 2008 1:37 AM | Permalink | Reply to this

Adieu

I am sorry, Garrett, a model¹ whose fermion content consists of a generation plus an anti-generation is not a “one-generation model”. It is a “zero-generation model”.

Clearly, you either

disagree that the fermion content of your E8 model consists of a generation plus an anti-generation (certainly, you never mentioned this highly salient fact in your paper)

or you

feel no compunction about attempting to obfuscate that fact, in the interest of maintaining the fiction that it is a “one-generation model”.

Unfortunately, I am unwilling to play along with that deception.

Perhaps Kea is right that nothing further can be gained by this discussion.

¹ I haven’t looked at your previous paper, but if it contains the assertion that the

(8_{s} {,8}_{s})

D_{4} \times D_{4}

contains a generation (as opposed to half a generation and half an anti-generation), then it’s wrong, too. That part of the analysis, of course, has nothing to do with

E_{8}

Posted by: Jacques Distler on July 23, 2008 4:31 AM | Permalink | PGP Sig | Reply to this

Re: Clarification

“Then, in this more recent paper, now under discussion, I basically said “Holy crap! This algebra sits in E8! How cool is that!”“

Not only that, but you also went around telling journalists that your model predicted new stuff potentially detectable at the LHC. The color-triplets in the bosonic sector of your model being one example of such stuff. What about the anti-generation that comes along with the SM fermion generation in your model? Should that not be detectable as well? If your model is correct, or a step towards a correct theory of particle physics, why haven’t the anti-generation fermions been seen already in experiments along with the fermions of the SM generation to which they correspond? Why do experiments only see left-handed neutrinos and not right-handed ones?

Clearly, to salvage this proposal of yours, you need some mechanism to give heavy masses to the new stuff, including the anti-generation. As Jacques mentioned, it might be possible to do this for the bosonic stuff, but for the fermionic anti-generation it is very unlikely that a workable mechanism exists (although as far as I can see it hasn’t been definitively ruled out yet either). But from the way you keep replying to Distler it seems that you just don’t get (or don’t want to get) this essential point.

Posted by: amused on July 23, 2008 5:20 AM | Permalink | Reply to this

Re: Clarification

Hmmm, just occured to me that Aaron’s remark above,

“generations and anti-generations pair up to become massive as Polchinski says. It is thus the net number of generations that is relevant for the low energy theory because that’s what remains massless.”

is probably a perturbation theory statement. So it should hold for weakly coupled gauge theories such as the SM, but not necessarily for strongly coupled ones. I guess this explains how the above statement can avoid being in contradiction with the expected possibility to obtain chiral fermions from vector fermions through decoupling of “mirror fermions” in the Poppitz-Shang setting I mentioned earlier, since that setting applies for strongly coupled gauge theories. So, yeah, it seems that there is no possibility to get a *weakly coupled* chiral gauge theory like the SM from a vector theory.

Posted by: amused on July 23, 2008 4:23 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

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John Baez writes:

Also, you can skim through about 20 grumpy exchanges on this thread once you realize that Lisi keeps asking if some representation of the Standard Model gauge group is sitting inside the adjoint rep of E8, and Distler keeps saying “(yeah, but) in fact this representation plus something else is sitting inside there, which makes this useless for physics” — without ever saying out loud the “yeah, but” part that’s what Lisi wanted to hear. Instead of saying “yeah, but”, Distler keeps saying “no, in fact”.

This gets my vote for most succinct recap. Although, sadly, the Sphex wasn’t mentioned.

Posted by: Garrett on July 22, 2008 6:26 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

I think it’s less sphex and more Fudd. “Be vewwy vewwy quiet – I’m hunting ‘yeah, but’s”.

Posted by: John Armstrong on July 22, 2008 6:59 PM | Permalink | Reply to this

Re: E8 Quillen Superconnection

I suspect a lot of people are tired of this exchange. Distler is obviously correct and I don’t see why this cannot be acknowledged so everybody can stop procrastinating and get back to work.

Posted by: Kea on July 23, 2008 12:01 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Distler has said at times that Lisi is not even using E8 and one interpretation for that would be that Lisi’s two D4s are coming after a symmetry break of E8/D8. This would fit with 128 fermions rather than 192. Lisi has mentioned that he has not given up on a Kaluza-Klein-like model and going from D8 to two D4s intuitively seems like a Kaluxa-Klein-like symmetry break. It’s just way too early to be dotting every i and crossing every t in general. Tony Smith’s E8 model (the only one I’m really familiar with) does not have a generation number as a quantum number so I’m quite biased against Lisi needing one, especially given Aaron Bergman’s comment about it being related to string theory compactification, is there some general need for it? To me triality really makes sense as an emergent spacetime-particle-antiparticle thing.

Posted by: John G on July 23, 2008 12:46 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

Jacques Distler said that “the (8s,8s) of D4xD4” contains … half a generation and half an anti-generation …”.

amused asked about Garrett Lisi’s E8 model “… why have not the anti-generation fermions …[such as]… right-handed neutrinos … been seen already in experiments along with the fermions of the SM generation to which they correspond? …”.

Here is a suggestion:

The D4xD4 (with its antigeneration of fermions) is part of D8.

If you look at these D8 rerpreesentations:

the 120-dim adjoint - denoted by D8adj

the 128-dim +half-spinor - denoted by D8s+

the 128-dim -half-spinor - denoted by D8s-

and
if you make the (admittedly unconventional, but it seems to me to be possibly workable) physical interpretations:

D8adj as gauge bosons plus more bosonic stuff (possibly spacetime vectors)

D8s+ as one generation of fermion particles and antiparticles

Ds- as one antigeneration of fermion particles and antiparticles

THEN

If you try to form a Lie algebra from

D8adj + Ds+ + Ds-

it does not work,

but

if you try to form a Lie algebra from

D8adj + Ds+

you succeed and get E8
with the 64+64 = 128-dim Ds+ representing one generation of fermion particles (one 64 of Ds+) and one generation of fermion antiparticles (the other 64 of Ds+).

So, maybe the math structure of Lie algebras is telling you that there is no physical D8s- antigeneration of fermions,
and
that one generation of D8s+ fermions lives inside E8.

If you do that,
then you have to deal with the Atiyah-Singer index giving the net number of generations,
which
is an issue conventionally formulated in terms of the Euler index of the compact manifold (6-dim) used to reduce 10-dim spacetime to physical 4-dim.

For an E8 model, you could see spacetime as 8-dim reduced to a Kaluza-Klein
M4 x CP2
and look at the index structure of the CP2.

CP2 has no spin structure and the index formula gives you -1/8 which is not an integer for generation number,
so
follow Hawking and Pope (Phys. Lett. 73B (1978) 42-44) and Chakraborty and Parthasarathy (Class. Quantum Grav. 7 (1990) 1217-1224)
and define a generalized spin structure.
You see from those papers that CP2 can be given a series of such generalized spin structures, the first two being:

index = 1 for 1 generation (the E8 prior to dimensional reduction)

index = 3 for 3 generations (the E8 model after dimensional reduction induces the second and third generations to emerge as effective composites of the first).

Tony Smith

PS - If you were to insist on starting with a 10-sim spacetime, you could still reduce it using the compact CP2 with generalized spin structure,
leaving an 6-dim conformal spacetime
that naturally gives you 4-dim spacetime,
using conformal correspondences related to quaternions, twistors, etc.

Posted by: Tony Smith on July 23, 2008 4:44 PM | Permalink | Reply to this

Read the post My Dinner with Garrett
Weblog: Musings
Excerpt: By popular demand, another post on this stuff.
Tracked: September 15, 2008 4:49 PM

Re: E8 Quillen Superconnection

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There is no “Theory of Everything” inside E8

Jacques Distler, Skip Garibaldi (Submitted on 16 May 2009)

We analyze certain subgroups of real and complex forms of the Lie group E8, and deduce that any “Theory of Everything” obtained by embedding the gauge groups of gravity and the Standard Model into a real or complex form of E8 lacks certain representation-theoretic properties required by physical reality. The arguments themselves amount to representation theory of Lie algebras along the lines of Dynkin’s classic papers and are written for mathematicians.

Garrett has made a little comment and a question about this article:

“Coin:” “Your understanding looks good to me, as a summary. But it gets more interesting. “The e8 TOE” in this paper refers to the idea of starting with E8 and seeing how its elements transform as representation spaces under the gravitational and standard model gauge groups as subgroups. A generation of standard model fermions necessarily transforms as a complex (i.e. chiral) representation space of the gravitational and standard model Lie groups. So, Distler and Garibaldi have this ToE2 requirement on page 1: the relevant representation space in E8 needs to be complex. And then they go on to prove that this representation space in E8 is always non-complex, so that ToE2 fails. Well, that may be true. But here’s the interesting question:”

“Are Distler and Garibaldi claiming that one cannot find a (necessarily complex) representation space in E8 that transforms under the gravitational and standard model subgroups as one generation of fermions?”

//////////

Posted by: Daniel de França MTd2 on May 20, 2009 12:59 AM | Permalink | Reply to this

Re: E8 Quillen Superconnection

I am quoting Garrett from several places in the above thread:

“”The e8 TOE” in this paper refers to the idea of starting with E8 and seeing how its elements transform as representation spaces under the gravitational and standard model gauge groups as subgroups. A generation of standard model fermions necessarily transforms as a complex (i.e. chiral) representation space of the gravitational and standard model Lie groups. So, Distler and Garibaldi have this ToE2 requirement on page 1: the relevant representation space in E8 needs to be complex. And then they go on to prove that this representation space in E8 is always non-complex, so that ToE2 fails. Well, that may be true. But here’s the interesting question:”

“Distler and Garibaldi say in their conclusion that “it is impossible to obtain even the one-generation Standard Model in this fashion.” If by this statement they are implying that one cannot find a representation space in E8 that transforms under the gravitational and standard model subgroups as one generation of fermions, then they are being blatantly dishonest. Since they are almost but not saying that, their paper is merely misleading.”

“Their ToE2 requirement is a straw man setup. For a generation of standard model fermions to be in E8, in a complex representation space of the gravitational and standard model gauge groups, it is not necessary for their complex conjugates not to be in E8. Distler and Garibaldi demand this, via ToE2, and then prove that it cannot be satisfied. But this ToE2 is not required in order to embed a standard model generation in E8 as I have done, so proving ToE2 can’t be satisfied does not invalidate my work.

“ToE2 requires that a generation of standard model fermions (in a complex representation space) not be a subspace of a non-complex representation space in E8. But this requirement is superfluous, because a complex representation space can be a subspace of a non-complex one. As an example, if you have so(6) acting on a real 6 representation space, then there is a su(3) subalgebra of the so(6) that acts on a 3 and on a bar{3} subspace in the 6. Distler would have you believe that, according to ToE2, it is impossible to obtain the complex 3 representation space “inside” the real 6 representation space because the bar{3} is their too. That is the ridiculous argument he uses to conclude that “it is impossible to obtain even the one-generation Standard Model [inside E8] in this fashion.” It is extremely misleading, if not an outright lie.”

“Distler’s use of chirality is nonstandard – it usually refers to how the weak force interacts with only left chiral fermions – so I prefer to speak of complex and non-complex representations, which I think is what Distler is calling chiral and non-chiral. But, as I’ve explained above, it is possible to find a “chiral” representation space as a subspace of a “non-chiral one.” The standard model algebra I’m working with is the usual algebra of the gravitational so(1,3) and standard model s(u(2)xu(3)) acting on the 64 dimensional representation space of one generation of fermions. This is the algebra that I find embedded in E8.”

“What I do next is look at how a triality automorphism maps this algebra into other parts of E8, and see if I can figure out how to relate this to the other two fermion generations. It is Distler’s red herring to look directly at how the rest of E8 transforms under the gravitational and standard model subalgebras, as if that is what I’m doing, because I’m not. What I’m doing is to embed one generation of the standard model and gravity in E8 and then see where I can go from there. To say that there’s a conjugate represenation space to the generation of fermions in E8 that makes that generation “not there” is silly in that context – it depends on false assumptions about how E8 is being broken up and interpreted in my work.”

Posted by: Daniel de França MTd2 on May 21, 2009 8:01 PM | Permalink | Reply to this

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May 10, 2008

Posted by Urs Schreiber

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