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 $ℤ_{2}$ -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: ${\underset{︸}{A}}_{\in Ω^{(1, even)} (Y, g)} + {\underset{︸}{Ψ}}_{\in Ω^{(0, odd)} (Y, g)} (1)$ where $g$ is some $ℤ_{2}$ -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 $ℤ_{2}$ -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 $ℤ_{2}$ -graded Lie algebra, can be read as a Quillen superconnection on a (necessarily) $ℤ_{2}$ -graded vector bundle $E \to X$ over spacetime $X$ , which is associated to a principal $G$ -bundle by some representation of $G$ on $ℤ_{2}$ -graded vector spaces:

$A : = d + A + Ψ$ $A : (Ω^{•} (X, E))_{even, odd} \to (Ω^{•} (X, E))_{odd, even}$ $\forall ω \in Ω^{1} (X) : [A, ω] = d ω .$

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 $ℤ_{2}$ -graded Lie algebras and super Lie algebras. As follows:

there are precisely two different symmetric braided monoidal structures on the category of $ℤ_{2}$ -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 $ℤ_{2} Vect$ . The other one is $S Vect$ .

A $ℤ_{2}$ -graded Lie algebra is a Lie algebra internal to $ℤ_{2} Vect$ . A super Lie algebra is a Lie algebra in $S Vect$ .

$ℤ_{2}$ -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 $A$ on a $ℤ_{2}$ -graded $e_{8}$ associated vector bundle and (3.2) is the corresponding Quillen curvature $F_{A} = A^{2} .$

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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146 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 $ℤ_{2}$ -grading you like. In particular with the trivial $ℤ_{2}$ -grading that assigns even degree to every element.

But if you do it for the “natural” $ℤ_{2}$ -grading of $e_{8}$ that Lisi considers, then a Quillen superconnection with respect to that $ℤ_{2}$ -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 $ψ^{2}$ !

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.)

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 $ψ^{2}$ 0-form piece.

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

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,” $Ψ$ , which are Lie algebra valued Grassmann numbers. A new effective Lagrangian for this theory, using standard BRST, is

(5)

L_{eff} = B_{g} D Ψ + F_{H} * F_{H}

in which

(6)

D Ψ = (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 ( $Ψ Ψ$ ) 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 $Ψ$ . This is the analogue to supersymmetry. However, we have chosen a BRST action that takes $C$ to zero. So all we have is $Ψ$ and $H$ (with no transformation between these two).

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

(8)

A' = H + Ψ

and compute its generalized curvature,

(9)

F' = (d H + H H) + (d Ψ + H Ψ + Ψ H) + (Ψ Ψ)

from which the effective Lagrangian can be built. In this construction, the $Ψ Ψ$ 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 $Ψ$ ’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 $ℤ_{2}$ gradings on the Lie algebra.

Indeed, the main reason why Lisi’s idea fails is that Marcel Berger classified the possible $ℤ_{2}$ -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 $e 8$ , 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 $E 8$ ’s Lie algebra you’re referring to is:

(1)

e 8 = h + k = (so (8) + so (8)) + (8_{+} \times 8_{+} + 8_{-} \times 8_{-} + 8_{v} \times 8_{v})

In this decomposition, the $H = SO (8) + SO (8)$ is a symmetric subgroup of $E 8$ , so there’s no problem with building the superconnection, $A + Ψ$ , with $A$ a 1-form field valued in $h$ , and $Ψ$ 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 $ℤ_{2}$ -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 $ℤ_{2}$ -grading comes from an embedding of $SL (2, ℂ)$ 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 $e 8$ does not break up as:

(1)

e 8 = h + k = (so (8) + so (8)) + (8_{+} \times 8_{+} + 8_{-} \times 8_{-} + 8_{v} \times 8_{v})

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 $ℤ_{2}$ grading of the $e_{8}$ Lie algebra under which “ $k$ ” is odd, much less a $ℤ_{2}$ grading of the $e_{8}$ Lie algebra which comes from an embedding $SL (2, ℂ) \mapsto E_{8}$ (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 $ℤ_{2}$ -graded vector space, then why bother with this one? Why not choose any old $ℤ_{2}$ -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 $ψ ψ$ 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 $ψ ψ$ 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 $ℤ_{2}$ -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 $ℤ_{2}$ -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 $ψ ψ$ 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 $e 8$ 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 $ψ ψ$ ) is not completely ad hoc; it comes directly from application of the BRST technique to an (ad hoc) action for an $e 8$ 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 $E 8$ 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 $E 8$ 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 $E 8$ 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 $SL (2, ℂ)$ 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)

e 8 = h + (k 1 + k 2 + k 3) = (so (8) + so (8)) + (8_{+} \times 8_{+} + 8_{-} \times 8_{-} + 8_{v} \times 8_{v})

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

(2)

[k 1, k 2] \subset k 3

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 $k 1$ ; 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 $k 2$ and $k 3$ , but in such away that they had the quantum numbers of $k 1$ , 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 $ψ$ which Garrett suppressed.

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

At the moment I don’t really understand how to interpret this. If we want the $ψ$ 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 $ψ$ s are odd sections of a super vectorbundle. While then we’d have $ψ^{a} ψ^{b} = - ψ^{b} ψ^{a}$ , it wouldn’t imply that the bilinear term in the $ψ$ 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 $ψ ψ$ 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 $ψ ψ$ 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 $e 8$ . (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 $ψ ψ$ 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 $ℤ_{2}$ 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 $ψ$ alone is already an odd endomorphism (such as if the $ψ$ 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 $ψ$ -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 $ψ ψ$ term isn’t in the action, we don’t strictly need to use a $ℤ_{2}$ grading of the algebra

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

That’s why you can’t use any old $ℤ_{2}$ grading but, rather, one induced from an embedding of $Spin (3,1) ≃ SL (2, ℂ)$ 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 $SL (2, ℂ)$ 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 $SL (2, ℂ) \times SU (3) \times SU (2) \times U (1) ↪ D_{4} \times D_{4} \subset E_{8}$ . None of them agree with the quantum numbers you claim to have found.

Am I correct in thinking a $ℤ_{2}$ grading requires $[k, k] \in h$ , or is this not necessary?

Yes, it is necessary. But, then, Marcel Berger classified all such $ℤ_{2}$ 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 $ℤ_{2}$ grading of the Lie algebra involved.

An embedding of $Spin (3,1)$ in $E_{8}$ furnishes a $ℤ_{2}$ 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 $ℤ_{2}$ 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 $ψ ψ$ term vanishes because $ψ$ is “Grassmann valued”. But we really have several $ψ$ s, the term is rather a linear combination of $ψ^{a} ψ^{b}$ . This does not vanish, in ge

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