## February 20, 2008

### Kostant on E8

#### Posted by John Baez

At Riverside we recently heard a fascinating talk:

• Bertram Kostant, On Some Mathematics in Garrett Lisi’s ‘E8 Theory of Everything’, U. C. Riverside, February 12th.

Abstract: A physicist, Garrett Lisi, has published a highly controversial, but fascinating, paper purporting to go beyond the Standard Model in that it unifies all 4 forces of nature by using as gauge group the exceptional Lie group $E_8$. My talk, strictly mathematical, will be about an elaboration of the mathematics of $E_8$ which Lisi relies on to construct his theory.

Luckily we had a video camera on hand. So, at the above link you can see streaming or downloadable videos of Kostant’s talk, as well as lecture notes.

Kostant’s talk was quite technical! If it’s too tough, you might prefer the warmup talk I gave earlier that day. But, Kostant described some ideas whose charm is easy to appreciate:

• The dimension of $E_8$ is $248 = 8 \times 31$. There is, in fact, a natural way to chop up $E_8$ into 31 spaces of dimension 8.
• There is a nice way to see the product of two copies of the Standard Model gauge group sitting inside $E_8$.
• The Standard Model gauge group is a subgroup of $SU(5)$. There is also a nice way to see the product of two copies of $SU(5)$ sitting inside $E_8$.
• The dimension of $SU(5) \times SU(5)$ is 48, and $248- 48 = 200$. The adjoint action of $SU(5) \times SU(5)$ on the Lie algebra of $E_8$ thus gives a $200$-dimensional representation, and this is $(5 \otimes 10) \oplus (\overline{5} \otimes \overline{10}) \oplus (10 \otimes 5) \oplus (\overline{10} \otimes \overline{5})$ [Actually this is wrong; see the discussion below.]

Garrett Lisi’s ideas have received serious criticism from Jacques Distler and others. I’ve included links to Lisi’s paper and also Distler’s comments. But, the work Kostant presents here is logically independent — beautiful math, regardless of its possible applications to physics. It makes heavy use of recent work on certain finite subgroups of $E_8$, most notably $GL(2,F_{32})$ and $(\mathbb{Z}/5)^3$. As Kostant said, “$E_8$ is a symphony of twos, threes and fives”.

Posted at February 20, 2008 6:06 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1606

### Re: Kostant on E8

Grr! It seems like the downloadable video website is down again! It was working last night…

Posted by: John Baez on February 20, 2008 7:30 PM | Permalink | Reply to this

### Re: Kostant on E8

Wow, this is great. I had no idea su(5) x su(5) was a subalgebra of E8 – that is very cool. I haven’t made it through the whole talk yet, but thinking out loud… it raises the possibility that one su(5) could be used for the standard model gauge group, and the second su(5) could be split into a su(2) for chiral gravity and an su(3) to relate the three generations. Like I said, just thinking out loud, and I haven’t looked at the details of such a breakdown, but a very nice possibility that may solve the problems of the d4 x d4 model.

Posted by: Garrett on February 20, 2008 7:55 PM | Permalink | Reply to this

### Re: Kostant on E8

I don’t know if it’s relevant for you, but the two factors in the SU(5) x SU(5) in E_8 are different, in the sense that the automorphism that switches them doesn’t extend to one of E_8.

As a group, it’s (SU(5) x SU(5))/Z_5, where the Z_5 is generated by (exp(2 pi i/5), exp(4 pi i/5)). Note that this group still has a central Z_5; indeed, it’s the centralizer of that element of E_8 of order 5.

You can see it on the Dynkin diagram level by extending E_8 to the affine diagram, then removing a vertex next to the trivalent, leaving two A_4 diagrams over. That vertex connects to the end of one A_4, and the middle of the other, which leads to the asymmetric amalgamation picture above.

Posted by: Allen Knutson on February 20, 2008 8:34 PM | Permalink | Reply to this

### Re: Kostant on E8

That’s interesting about the asymmetry of the two copies of $\SU(5)$. Is this just a ‘global’ issue? I.e., is there a Lie algebra automorphism of $e_8$ that switches these two copies of the Lie algebra $su(5)$? I ask because the representation of this $su(5) \oplus su(5)$ on $e_8/su(5) \oplus su(5)$ looks completely symmetrical. Of course, that doesn’t rule out an asymmetry at the Lie algebra level.

Posted by: John Baez on February 22, 2008 9:35 PM | Permalink | Reply to this

### Re: Kostant on E8

I was wrong; the two SU(5)s are indeed conjugate. Here’s the proof Bert showed me at his 80th birthday conference.

As explained above, the (SU(5)xSU(5))/Z_5 arises as the centralizer of a certain element of order 5. [Borel-de Siebenthal] tell us that the conjugacy classes with semisimple centralizer are indexed by the nodes on the affine Dynkin diagram, with the centralizer given by omitting the node, and the order of the conjugacy class in the adjoint group given by the usual label on the node. E_8 is already adjoint, and there is only one vertex labeled 5, so that’s the one.

Let tau be such an element. Then tau^2 is another special element of order 5. Hence there exists a g such that g tau g^-1 = tau^2. Plainly g conjugates C_G(tau) into C_G(tau^2), which is again C_G(tau) since tau,tau^2 generate the same Z_5.

We can distinguish the two factors of C_G(tau) by looking at the projection of tau into them; in one case we get the generator of SU(5)’s center (or we get its inverse, depending on identification), and in the other we get the square (or its inverse).

This rule reverses when we look at the projection of tau^2 instead. Summing up, conjugating by g switches the two factors of C_G(tau).

Posted by: Allen K. on May 25, 2008 3:24 AM | Permalink | Reply to this

### Re: Kostant on E8

The embedding of $a_4\times a_4 \subset e_8$ was the basis of my (embarrassing) failed attempt to find 2 generations in your model.

The more general argument, that it’s impossible to get even 2 generations is independent of any of the details of how the Standard Model is embedded in $E_8$. It relies on Berger’s classification of involutions of $e_8$ and on the fact that the center of $SL(2,\mathbb{C})$ acts as $-1$ on fermions and as $+1$ on bosons (so that any embedding of $SL(2,\mathbb{C})\subset E_8$ provides such an involution).

Posted by: Jacques Distler on February 20, 2008 9:03 PM | Permalink | PGP Sig | Reply to this

### Re: Kostant on E8

So basicaly, you are right, and Konstant is wrong. If so, why is he wrong?

Posted by: Daniel de França MTd2 on February 21, 2008 2:06 PM | Permalink | Reply to this

### Re: Kostant on E8

Huh?

Please elaborate. What statement of Kostant’s (or of mine) do you think is wrong?

Posted by: Jacques Distler on February 21, 2008 2:54 PM | Permalink | PGP Sig | Reply to this

### Re: Kostant on E8

Right in the summary above, according to Baez:

“The Standard Model gauge group is a subgroup of SU(5). There is also a nice way to see the product of two copies of SU(5) sitting inside E 8.”

He got the Standard Model, you didn’t. You say that the 3 generations are part of the standard model, but that Lisi’s theory does not have a way contain that. So, maybe that means either the Standard Model group does not require 3 generations or 3 generations does not imply the Standard Model.

AND, also, it means that one of you is wrong.

Posted by: Daniel de França MTd2 on February 21, 2008 3:35 PM | Permalink | Reply to this

### Re: Kostant on E8

“The Standard Model gauge group is a subgroup of SU(5). There is also a nice way to see the product of two copies of SU(5) sitting inside $E_8$.”

I just said that I had used the (well-known) fact that $E_8$ contains $(SU(5)\times SU(5))/\mathbb{Z}_5$ as a subgroup.

How can that possibly be interpreted as a contradiction with Kostant’s observation that $E_8$ contains $(SU(5)\times SU(5))/\mathbb{Z}_5$ as a subgroup?

There are lots of ways to embed $G= (SU(3)\times SU(2)\times U(1))/\mathbb{Z}_6$ (or even $G\times SL(2,\mathbb{C})$) in (a noncompact real form of) $E_8$.

That’s not the point. The point was to get 3 generations from the rest of the generators of $E_8$.

AND, also, it means that one of you is wrong.

There is a third possibility.

Posted by: Jacques Distler on February 21, 2008 4:06 PM | Permalink | PGP Sig | Reply to this

### Re: Kostant on E8

If you recognize that “There are lots of ways to embed G”, why did you say “The more general argument, that it’s impossible to get even 2 generations is independent of *any* of the details of how the Standard Model is embedded in E 8” when what matter is “The point was to get 3 generations from the rest of the generators of E 8.” Isn’t it a way to just dismiss his article?

It is possible to show that his model is compatible with the Standard Model, yet, you try to use the elements of his theory NOT belonging to it and showing that they do NOT compose the standard model. It sounds like similar to that case in which you would be using gluons to prove that the Standard Model is wrong, since you can’t get photons with a combination of them.

Posted by: Daniel de França MTd2 on February 21, 2008 4:57 PM | Permalink | Reply to this

### Re: Kostant on E8

Daniel wrote:

He [Kostant] got the Standard Model, you [Distler] didn’t.

Kostant didn’t ‘get the Standard Model’ and he never claimed to. Nothing Kostant said disagrees with anything Distler has on his blog.

In particular, both Kostant and Distler agree that $su(5) \times su(5)$ sits inside $e_8$. But this is not ‘getting the Standard Model’.

There are lots of things one might mean by ‘getting the Standard Model’. One can imagine breaking this task down into several steps:

1. Getting the Standard Model gauge group.
2. Getting its correct representation on all the particles of the Standard Model.
3. Getting the Standard Model Lagrangian, which describes how all the particles interact — or something that reduces to this Lagrangian in some limit.

You’d need to do all 3 steps to ‘get the Standard Model’ in the fullest sense of that phrase.

In the sort of theory Lisi is envisaging, there’s no difficulty with step 1: finding the Standard Model gauge group $S(U(3) \times U(2))$ inside $E_8$. Everyone agrees that’s easy.

Lisi never claimed to do step 3. So, let’s not even worry about that!

The controversy, if any, seems to be over step 2. Distler claims to show that one cannot find the Standard Model Lie algebra and its correct representation on all the particles of the Standard Model inside the Lie algebra of $E_8$. (In fact, he claims something much stronger.) I haven’t checked the details of his argument yet. But, the fact that he knows this stuff pretty well, and Lisi isn’t arguing with him about it — nor is Kostant — strongly suggests he’s right.

Posted by: John Baez on February 21, 2008 6:57 PM | Permalink | Reply to this

### Re: Kostant on E8

Thanks John, now I get it. :) I had an issue of language redundancy.

I have some questions, I ask them below. This subthread is too messy.

Posted by: Daniel de França MTd2 on February 21, 2008 9:05 PM | Permalink | Reply to this

### Re: Kostant on E8

John Baez said “… [step 3] Getting the Standard Model Lagrangian, which describes how all the particles interact - or something that reduces to this Lagrangian in some limit.
… Lisi never claimed to do step 3. So, let us not even worry about that!…”.

It seems to me that Garret Lisi did in fact claim to do step 3 = Getting the Standard Model Lagrangian.
In his paper 0711.0770 Garrett Lisi said:
“… an action for everything can be economically expressed as a modified BF theory action over a four dimensional base manifold …
The details of the action, and its agreement with the standard model and general relativity, can be worked out for each sector of the E8 Lie algebra. …
The electroweak part of the action is …[equation on page 27]…
And the so(8) part of the action is …[another equation on page 27]… which includes the action for the gluons …
the massive Dirac action in curved spacetime …[ is another equation on page 27]…
The coframe … in this action contracts with the frame part of the graviweak connection … to give the standard Higgs coupling term …
This action works very well for one generation of fermions. The action for the other two generations should be similar …”.

Garrett Lisi only claims that his action (Lagrangian) is realistic for one generation,
which is consistent with Jacques Distler’s position that E8 has 128 = 64+64 fermionic dimensions and “… instead of two generations [from that 64 + 64],
one obtains a generation and an anti-generation …” to describe one generation of particles and antiparticles.

So it seems that Garrett Lisi’s model, if modified to describe only one generation (the second and third being non-fundamental such as being composite) would be consistent with Jacques Distler’s math ideas.

Tony Smith

PS - As I said in a previous comment here, the remaining objection of Jacques Distler, about chirality or handedness, can be dealt with by considering fermion particles to be fundamentally left-handed and fermion antiparticles to be fundamentally right-handed, with the other handed-nesses arising dynamically as described by Okun in his book Leptons and Quarks.

Posted by: Tony Smith on February 22, 2008 3:33 PM | Permalink | Reply to this

### Re: Kostant on E8

Tony wrote:

Garrett wrote:

The details of the action, and its agreement with the standard model and general relativity, can be worked out for each sector of the $E_8$ Lie algebra ….

Hmm. I hadn’t noticed that. That’s an unfortunate statement. He certainly makes no serious attempt to get the Standard Model Lagrangian in all its detail. Nor does it seem possible (without further feats of genius). For example, his Lagrangian has no place for the $\sim 25$ adjustable constants contained in the Standard Model: particle masses, coupling constants, etc. Nor does he attempt to derive these constants.

So it seems that Garrett Lisi’s model, if modified to describe only one generation (the second and third being non-fundamental such as being composite) would be consistent with Jacques Distler’s math ideas.

If you’re just trying to describe one generation of Standard Model particles, gauge bosons, Higgs and gravity, an $E_8$–based model sure seems to have a lot of extra gunk.

PS - As I said in a previous comment here, the remaining objection of Jacques Distler, about chirality or handedness, can be dealt with by considering fermion particles to be fundamentally left-handed and fermion antiparticles to be fundamentally right-handed, with the other handed-nesses arising dynamically as described by Okun in his book Leptons and Quarks.

I like that book, but I don’t have it, and I forget what that mechanism is supposed to be.

In any event, we’ve now switched to talking about dreams of a possible theory, instead of what Lisi actually did. I’d been talking about what Lisi actually did; he did not ‘get the Standard Model’.

Posted by: John Baez on February 22, 2008 5:47 PM | Permalink | Reply to this

### Re: Kostant on E8

John wrote:

In particular, both Kostant and Distler agree that su(5 )×su(5 ) sits inside e8 . But this is not ‘getting the Standard Model’.

Does anyone have a concrete description
of this a4+a4 subalgebra of e8. Are these
complex or real algebras? (rational?)

PS. My first post here, please excuse any
formatting errors. It’ll get better if
I manage to stay here long enough.

Posted by: rntsai on February 23, 2008 8:32 AM | Permalink | Reply to this

### Re: Kostant on E8

Never mind. I found it. Over the rationals, four (obvious) roots generate
one a4, the second a4 is it’s centralizer
in e8.

Posted by: rntsai on February 24, 2008 6:38 AM | Permalink | Reply to this

### Re: Kostant on E8

Well I think it will take me a few decades to get on to Kostant’s talk, but John’s talk seems to be streaming fine, and pitched at the right level for me :-)

Posted by: Bruce Bartlett on February 20, 2008 11:46 PM | Permalink | Reply to this

### Re: Kostant on E8

Yeps, very nice talk. I didn’t know Lie algebras were related to densely packing spheres! I am also ashamed to admit I learnt about roots and weights the “old-school Lie-algebra-only” way; no-one told me/stressed that the lattices $A_n$, $D_n$, $E_8$ and so on are just the kernel of the exponential map on the maximal torus, which is definitely helpful.

Posted by: Bruce Bartlett on February 21, 2008 10:23 AM | Permalink | Reply to this

### Re: Kostant on E8

I’m glad you liked my talk.

No-one told me/stressed that the lattices $A_n$, $D_n$, $E_8$ and so on are just the kernel of the exponential map on the maximal torus, which is definitely helpful.

Thanks! If you like this approach to classifying simple Lie algebras — thinking a lot about lattices, and bringing Lie groups into the game early instead of avoiding them out of some misguided fanatical devotion to ‘purely algebraic’ methods — I strongly recommend Frank Adams’ book Lectures on Lie Groups. It was only after I learned this approach that I grew to love simple Lie algebras.

Or, if you’re in a rush, try week63week65, which I wrote shortly after I overcame my early hatred of roots and weights.

You can think of my talk as a miniaturized version of those Weeks, with a ridiculous overemphasis on $E_8$

Posted by: John Baez on February 21, 2008 7:09 PM | Permalink | Reply to this

### Re: Kostant on E8

The three pentagons are certainly interesting: reminiscent of one half (upper half plane) of an associahedron.

Posted by: Kea on February 21, 2008 5:04 AM | Permalink | Reply to this

### Re: Kostant on E8

Haven’t had time to view the lecture
Can you clue me in as to page or how far in I’ll find half my beloved

Posted by: jim stasheff on February 21, 2008 12:40 PM | Permalink | Reply to this

### Re: Kostant on E8

Jim wrote:

Can you clue me in as to page or how far in I’ll find half my beloved

I don’t understand this cryptic comment: what’s ‘half your beloved’?

But, if you want to see Kostant’s lecture, or my lecture notes, or my warmup lecture, just click on the tex in blue in my blog entry — or right here!

Posted by: John Baez on February 22, 2008 5:54 PM | Permalink | Reply to this

### Re: Kostant on E8

The clue is where Kea says “one half … of an associahedron”.

Posted by: David Corfield on February 22, 2008 7:47 PM | Permalink | Reply to this

### Re: Kostant on E8

Oh, I see. Unfortunately I have no idea what “three pentagons… half of an associahedron” Kea was talking about. I don’t remember Kostant drawing any pentagons, and I don’t remember drawing any myself.

Posted by: John Baez on February 22, 2008 8:57 PM | Permalink | Reply to this

### Re: Kostant on E8

someone commented on the 3 pentagons in re: half the associahedron K_5

Posted by: jim stasheff on February 23, 2008 1:30 PM | Permalink | Reply to this

### Re: Kostant on E8

Jacques Distler said in a comment above:
“… The more general argument, that it is impossible to get even 2 generations is independent of any of the details of how the Standard Model is embedded in E 8 . …”
and
he has a link to his blog where he gives more detail:
“… What we seek is an involution of the Lie algebra, e 8 .
The bosons correspond to the subalgebra, on which the involution acts as +1 ;
the fermions correspond to generators on which the involution acts as -1 .

the maximum number of -1 eigenvalues is 128 … the 128 is the spinor representation
…”.

So, Jacques Distler is only saying that you have 128 dimensions to play with to make fermions in an E8 model.
In his representation of each generation of fermions,
Jacques Distler (on his blog entry mentioned above where he uses more detailed notation than I am using on this text-type comment)
defines R = (3,2) + (3,1) + (3,1) + (1,2) + 1,1)
and
uses as representation for each generation of particles and antiparticles
(2,R+(1,1)) + (2,R+(1,1))
for a total of
2x(6+3+3+2+1+1) + 2x(6+3+3+2+1+1) = 2x16 + 2x16 = 64
dimensions to represent each generation
so
he notes that 128 = 2 x 64 and says
“… we can, at best, find two generations …”.
However,
he goes on to say that two generations will not work using the 64 + 64 = 128,
because
“… instead of two generations [from that 64 + 64],
one obtains a generation and an anti-generation …”.

So, you can (as I do in my version of E8 physics) use a 64 = 8x8 to describe 8 fundamental first-generation fermion particles
and the other 64 = 8x8 to describe 8 fundamental first-generation fermion antiparticles,
with
the second and third generations being composites of the first.

Jacques Distler raises a further objection about fermion chirality, saying
“… the spectrum of fermions is always nonchiral …”.

However, just as the composite nature of generations 2 and 3 allows construction of a realistic E8 model with one generation of fermion particles and antiparticles,
the chirality (or handed-ness) of fermions is not a problem with my E8 model because
fundamentally all fermion particles are left-handed and all fermion antiparticles are right-handed,
with the opposite handedness emerging dynamically for massive fermions.
Such dynamical emergence of handed-ness is described by L. B. Okun, in his book Leptons and Quarks (North-Holland (2nd printing 1984) page 11) where he said:

“… a particle with spin in the direction opposite to that of its momentum …[is]… said to possess left-handed helicity, or left-handed polarization. A particle is said to possess right-handed helicity, or polarization, if its spin is directed along its momentum. The concept of helicity is not Lorentz invariant if the particle mass is non-zero. The helicity of such a particle depends oupon the motion of the observer’s frame of reference. For example, it will change sign if we try to catch up with the particle at a speed above its velocity. Overtaking a particle is the more difficult, the higher its velocity, so that helicity becomes a better quantum number as velocity increases. It is an exact quantum number for massless particles …
The above space-time structure … means … that at …[ v approaching the speed of light ]… particles have only left-handed helicity, and antparticles only right-handed helicity. …”.

So Distler’s arguments do not discredit E8 physics, but instead show how to construct it as a solid realistic physics model.

Tony Smith

Posted by: Tony Smith on February 21, 2008 4:25 PM | Permalink | Reply to this

### Re: Kostant on E8

So, Konstant’s result was the representation 5 ⊗ 10 ⊕ 5* ⊗ 10* ⊕ 10 ⊗ 5 ⊕ 10* ⊗ 5*, right?

1. Is this a new result? Or is he just pointing out an old idea?

2. So, according to Distler, this representation will never contain the standard model, since no matter how you embed SM group in E(8), it won’t a proper representation for it. Right?

Posted by: Daniel de França MTd2 on February 21, 2008 9:40 PM | Permalink | Reply to this

### Re: Kostant on E8

Daniel wrote:

So, Konstant’s result was the representation 5 ⊗ 10 ⊕ 5* ⊗ 10* ⊕ 10 ⊗ 5 ⊕ 10* ⊗ 5*, right?

I don’t know if that’s new. Experts like Allen Knutson already knew — or could quickly see — that $su(5) \times su(5)$ sits inside $e_8$. As a trained professional, once you know something like this, it’s straightforward to work out the representation of $su(5) \times su(5)$ that this embedding gives.

So, while that result is pretty, I suspect the most novel part of Kostant’s talk is what he spent the most time talking about: the way various interesting groups arise inside $E_8$ as centralizers of finite subgroups, and also how characters of the $(\mathbb{Z}/2)^5$ subgroup allow you to chop $e_8$ into 31 ‘pieces of eight’.

(Yes, I know I’m abusing the original meaning of ‘pieces of eight’.)

Posted by: John Baez on February 22, 2008 6:50 PM | Permalink | Reply to this

### Re: Kostant on E8

I don’t want to interrupt the expert discussion but only thanking John Baez for his continuing exciting and illuminating explanations of Lie groups. I got the twf 63-65, news about E8 (from 3.8.07) and the octonions paper all printed out. When I read them that really gets my heart racing. Add the E8 excitement all over the internet and the insight that Lie groups is an important concept in physics and math anyway, that just makes you want to become a real expert in Lie groups!

Unfortunately, especially for newcomers, the Lie groups/ Lie algebra textbook world is such a disaster. Not only do most of the 1001 Lie groups books out there prefer talking about Haar measure than about Dynkin diagrams and exceptional groups. But also the weights and roots material is often obscured behind all kinds of abstract language.

Also funny, after all the rage about E8, even making the general public take notice, there so little information to find about E8. Only John Baez, a wiki entry and a book by Adams which is out of print.

Posted by: Lutz Gillner on February 21, 2008 10:51 PM | Permalink | Reply to this

### Re: Kostant on E8

Don’t forget the letter written by E8 himself to counter the canard that he is too hard to understand and one should work up through E6 and E7.

Posted by: jim stasheff on February 22, 2008 4:33 AM | Permalink | Reply to this

### Re: Kostant on E8

Lutz wrote:

Unfortunately, especially for newcomers, the Lie groups/ Lie algebra textbook world is such a disaster. Not only do most of the 1001 Lie groups books out there prefer talking about Haar measure than about Dynkin diagrams and exceptional groups. But also the weights and roots material is often obscured behind all kinds of abstract language.

I know what you mean. I’ve actually loved Haar measure and other general concepts in Lie theory since I was a student, but it took me quite a while to penetrate the veil and learn to love the semisimple theory — roots, weights, Cartans, Borels, and all that jazz. Why? At least for me, these turned out to be more beautiful in concrete examples than in abstract generality… but my first introduction to them never properly explained the examples!

What could be cooler than a way of packing spheres in 4 dimensions built by first packing spheres in a cubical lattice as tightly as possible, and then discovering there’s still enough room to slip another copy of this cubical lattice of spheres in the cracks between the first one? And then to discover that when you’ve done this, each sphere touches 24 others: 16 sitting at the vertices of a hypercube, and 8 sitting at the vertices of a hyperoctahedron? And then, to realize that these 24 vertices are the corners of a 4-dimensional Platonic solid, the 24-cell — the sixth and final one, the one that makes there be more Platonic solids in 4 dimensions than any other dimension? And then, to realize that the lattice you’ve got can be thought of as consisting of integer quaternions? And then, to discover that this lattice is the root lattice of the group $Spin(8)$, the double cover of the rotation group in 8 dimensions? And then, to discover that the symmetry of permuting the quaternions $i, j,$ and $k$ gives this lattice and thus $Spin(8)$ a special symmetry called ‘triality’, which no other Spin group has? And then, to discover that triality gives rise to the octonions? And then … well, the wonders never cease!

So, you have to penetrate the veil and see some of this beauty firsthand!

Someday I want to write a book called Exceptional Beauty, which will be a guided tour of all this material.

But for now, the main books I recommend are Adams’ Lectures on Lie Groups and Fulton and Harris’ Representation Theory — A First Course. Start with the pictures and work your way out.

Also, read all of This Week’s Finds!

Posted by: John Baez on February 22, 2008 7:15 PM | Permalink | Reply to this

### Re: Kostant on E8

As someone pointed out on Cosmic Variance, there is a review about a similar subject:

It seems that a similar aproach as the one proposed by Lisi, using the E(8) to find the SM, was studied in the 80’s.

One of the things that called my attention was the citation of articles concerning the phenomenology of the anti-family, that one found by Distler on his blog post.

I don’t have access to the articles cited, so I won’t comment, further.

In case any of you have acceess to them, it would be nice to tell us all if there is anything interesting there…

PS.: If someone could find a way to read them, it would be nice. Of course, if there is something worthwhile there.

Posted by: Daniel de Franï¿½a MTd2 on February 23, 2008 5:56 AM | Permalink | Reply to this

### Re: Kostant on E8

That paper by Adler looks fun, and I urge people interested in grand unified theories to take a peek. As some of you know, this is the Stephen Adler who gained fame for the Adler–Bell–Jackiw anomaly, got a job at the Institute for Advanced Studies, and then spent many years working on quaternionic quantum mechanics — a subject most physicists regard as a lost cause.

(I’ve spent plenty of time on QQM myself, so this isn’t a criticism of Adler: I’m just reporting a sociological fact! But, I part company with Adler when he starts tensoring ‘quaternionic vector spaces’ that are defined to be left modules of the quaternions — I think one really should define them to be bimodules, since the quaternions are a noncommutative ring! My former student Toby Bartels succesfully worked out a fair amount of quaternionic functional analysis after realizing this fact.)

Anyway, the first thing to note is that Adler is studying supersymmetric $E_8$ Yang–Mills theory, so he’s not trying to pack both fermions and bosons into $E_8$ the way Lisi is — and he’s not trying to include gravity, either.

He’s not trying to pack both fermions and bosons into $E_8$or is he? I’m a bit confused: he says “in this $E_8$ model the fermions and gluons [gauge bosons] are in the same supermultiplet, achieving a complete unification of matter fields and force-carrying fields. The point that an $E_8$ unification is automatically supersymmetric was made independently more than twenty years by… [lots of guys].”

Anyway, apart from this remark, he seems to be considering fermions and bosons separately, but both in the adjoint representation. He uses an embedding

$su(3) \times \mathrm{e}_6 \subset \mathrm{e}_8$

together with the usual embedding

$so(10) \times \mathrm{u}(1) \subset \mathrm{e}_6$

to fit the 16-dimensional spinor rep of $so(10)$ (familiar from the $so(10)$ GUT) into the 27-dimensional rep of $\mathrm{e}_6$ (which means lots of extra unseen fermions) and then get three generations of these fermions related by $su(3)$ symmetry.

However, he writes, “Despite the attractive features of automatic supersymmetry and natural inclusion of three families, the reason that $E_8$ has not been further pursued as a unification group is that in addition to three families, it also contains three mirror families. Thus, under $E_8 \supset SU(3) \times E_6$, in addition to three $27$’s there are three $\overline{27}$’s. The presence of mirror families leads to potential phenomenological and theoretical difficulties.”

Another depressing but cute remark is this: he says an asymptotically free $E_8$ gauge theory can’t do symmetry-breaking via a Higgs, because the Dynkin index of the smallest candidate Higgs is already too large — it’s a whopping 3875-dimensional rep.

Posted by: John Baez on February 24, 2008 4:32 AM | Permalink | Reply to this

### Re: Kostant on E8

Ah, he references Bars and Gunaydin, who notably no longer work on naive GUT approaches. Bars works on 2Time physics now, with motivation from twistor theory, and Gunaydin works a lot with Jordan algebras, especially the 3x3 octonion one.

Posted by: Kea on February 24, 2008 4:44 AM | Permalink | Reply to this

### Re: Kostant on E8

quaternionic functional analysis

For the record, the proper url for this is (and should forever be) http://toby.bartels.name/notes#quaternions.

Posted by: Toby Bartels on February 27, 2008 2:50 AM | Permalink | Reply to this

### Re: Kostant on E8

John wrote

• The dimension of $SU(5)\times SU(5)$ is 48, and $248−48=200$. The adjoint action of $SU(5)\times SU(5)$ on the Lie algebra of $E_8$ thus gives a 200-dimensional representation, and this is $(5\otimes 10)\oplus (\overline{5}\otimes \overline{10})\oplus(10\otimes 5)\oplus(\overline{10}\otimes \overline{5})$

Umh. Not quite. That should read

$(5\otimes 10)\oplus (10\otimes \overline{5})\oplus(\overline{5}\otimes \overline{10})\oplus(\overline{10}\otimes 5)$

Posted by: Jacques Distler on February 23, 2008 8:12 AM | Permalink | PGP Sig | Reply to this

### Re: Kostant on E8

Okay — so the asymmetry between the two $SU(5)$’s is manifest at the Lie algebra level. I just watched Kostant’s video again, and this part is right near the end. It’s sort of amusing: he says $\overline{10} \otimes 5$, but he writes $\overline{10} \otimes \overline{5}$ — and that’s what I copied down. It seemed nice and symmetrical!

I wish I were quicker at calculations like this: I would like to see for myself how it works.

Posted by: John Baez on February 23, 2008 8:32 AM | Permalink | Reply to this

### Re: Kostant on E8

I wish I were quicker at calculations like this: I would like to see for myself how it works.

You might find the following helpful. Denoting the four 50-dimensional representation, in the order I wrote them, as $V_i,\, i=0,1,2,3$, the Lie bracket relations are encoded in

\begin{aligned} \wedge^2 V_i &\supset V_{i+1}\\ V_i\otimes V_{i+1} &\supset V_{i-1}\\ V_{0}\otimes V_{2} &\supset (24,1)\\ V_{1}\otimes V_{3} &\supset (1,24)\\ \end{aligned}

where the subscripts are defined $\mod{4}$. Starting with $V_0=(5,10)$, this should be enough for you to generate the rest, and check that you’ve done it correctly.

I should further point out that almost everything Kostant talks about, interesting though it is, applies to the compact real form of $E_8$, not the noncompact real forms relevant to Lisi.

So the Lisi enthusiasts who seem to be hanging around here, breathlessly hoping to learn some tidbit which will “save” Lisi’s theory are going to come away a little disappointed…

Posted by: Jacques Distler on February 23, 2008 8:00 PM | Permalink | PGP Sig | Reply to this

### Re: Kostant on E8

You might find the following helpful…

Hmmm. That’s kinda obscure, even by my standards. Let me try to give an equivalent, but slightly more conceptual description.

$(SU(5)\times SU(5))/\mathbb{Z}_5$ has center $\mathbb{Z}_5$. Let’s decompose (over $\mathbb{C}$!) the 248 according to the action of this $\mathbb{Z}_5$.

(1)$\begin{gathered} W_0 = (24,1)\oplus (1,24),\quad W_1 = (5,10),\\ W_2 = (10,\overline{5}),\quad W_3 = (\overline{10},5),\quad W_4 = (\overline{5},\overline{10}) \end{gathered}$

(Note the change in ordering, relative to what I wrote before!)

The Lie bracket commutes with the $\mathbb{Z}_5$ action, so

(2)$[ W_a, W_b] \subset W_{a+b\pmod{5}}$

And, again, knowing that $W_1=(5,10)$, you can reconstruct the rest of (1) from (2).

Posted by: Jacques Distler on February 24, 2008 5:00 AM | Permalink | PGP Sig | Reply to this

### Re: Kostant on E8

Interesting relation. I was able to verify
it for e8/(a4+a4) : with W0=(24,1)+(1,24),
W1,W2,W3,W4 the other invariant subpaces
as you describe. These 5 subspaces are compatible with a Z5 group.

There’s a similar relation between the subspaces of e8/(d4+d4):
W0=(28,1)+(1,28),W1=64,W2=64,W3=64. These are
compatibe with a Z2xZ2 group. (W0=identity, W1*W1 in W0,W2*W2 in W0,W3*W3 in W0,
W1*W2 in W3,…).

PS. Again excuse the poor formatting.

Posted by: rntsai on February 24, 2008 8:20 AM | Permalink | Reply to this

### Re: Kostant on E8

Okay — so the asymmetry between the two SU(5)’s is manifest at the Lie algebra level.

Since E8 is simply connected, any Lie algebra automorphism extends to a Lie group automorphism, which is the simplest reason that the asymmetry is not a purely group phenomenon.

I’m not at all surprised that Bert would say one thing and write another. Be glad he wrote something at all!

Incidentally I heard today he’s been telling people your cousin is your mother. I’m sure you get that all the time.

Posted by: Allen Knutson on February 29, 2008 6:37 AM | Permalink | Reply to this

### Re: Kostant on E8

Allen wrote:

Incidentally I heard today he’s been telling people your cousin is your mother. I’m sure you get that all the time.

They don’t usually put it quite that way. When they do, I beat them up.

Just to set the record straight: Joan is my dad’s brother’s daughter.

Posted by: John Baez on February 29, 2008 6:57 AM | Permalink | Reply to this

### Re: Kostant on E8

Perhaps you should fix the representation claimed up in the post.

Posted by: Allen Knutson on February 29, 2008 5:56 PM | Permalink | Reply to this

### Re: Kostant on E8

I’ll just add a remark saying it’s wrong, since the ensuing discussion wouldn’t make much sense without that mistake.

I fixed it in the lecture notes.

Posted by: John Baez on February 29, 2008 7:53 PM | Permalink | Reply to this

### Re: Kostant on E8

Dear John Baez:

I cannot download your video lecture. First, I was getting a timeout. More recently, I’m being asked a username and password… I’m not the only one experiencing this problem. Some other person from Nepal reported the same problem.

Kind regards,
Christine

Posted by: Christine Dantas on February 28, 2008 10:54 AM | Permalink | Reply to this

### Re: Kostant on E8

The people at the media center, who run the computers where the downloadable videos are stored, accidentally made it impossible to access the videos without a password. I told them about this and they are trying to fix it.

When I last checked, the streaming videos were working… but perhaps they are not so convenient for you.

Posted by: John Baez on February 29, 2008 1:34 AM | Permalink | Reply to this

### Re: Kostant on E8

Well, when I try the streaming videos, my browser (Safari 2.0.4) simply disapears and the program is shutdown by itself. The same happens with Firefox 2.0.0.6. I’m using Quicktime 7.2 in a mac OS X 10.4.10. Perhaps it is the case for updating those programs to see whether the problem is solved. Alternatively, it may have to do with my bandwidth connection. Anyway, that is why I was trying the downloadable videos. But then…

Thanks,
Christine

Posted by: Christine Dantas on February 29, 2008 11:25 AM | Permalink | Reply to this

### Re: Kostant on E8

Posted by: rntsai on February 29, 2008 5:43 PM | Permalink | Reply to this

### Re: Kostant on E8

I think you need some sort of special “director’s” arrangement with YouTube to download videos longer than 10 minutes. That’s why the Catsters’ videos are all shorter than this time limit.

I don’t know how hard it is to get this arrangement with YouTube.

Posted by: John Baez on February 29, 2008 7:24 PM | Permalink | Reply to this

### Re: Kostant on E8

It used to be the case that some people could get permission to upload videos longer than ten minutes. However, that permission is no longer granted.

Posted by: Tom Leinster on February 29, 2008 7:46 PM | Permalink | Reply to this

### Re: Kostant on E8

As of now, both the streaming and downloadable videos are working for me. The silly “password” business is gone. If they’re still not working for you, it could be a problem with your browser or (for the streaming form) an insufficient rate of data transmission.

Just for comparison purposes, I’m using Firefox 2.0.0.12 on a Windows machine, with QuickTime 7, in Windows XP. I have Firefox set to open .mov files using QuickTime.

Posted by: John Baez on February 29, 2008 8:10 PM | Permalink | Reply to this

### Re: Kostant on E8

Thank you! Download is working. Streaming still makes my browser crash, but that’s ok, since I can dowload the file.

Best wishes,
Christine

Posted by: Christine Dantas on March 2, 2008 2:00 PM | Permalink | Reply to this

### Re: Kostant on E8

I thought his decomposition of e8 into 31 cartan’s was interesting.
It’s actually very closely related to e8/(d4+d4) decomposition :
e8/(d4+d4)= (28,1)+(1,28) + (8v,8v) + (8S+,8S+) + (8S-,8S-)

The last 3 terms can be seen as 24 8-dim spaces invariant under one of the d4’s.
You can verify that these are abelian, so calling them cartan is justified.

The other 7 cartans are inside d4+d4; one of them is the “original” e8 cartan,
there are probably several ways to identify the others. One easy way I found
is to decompose d4 under (a1+a1+a1+a1) : d4 = 3 + 3 + 3 + 3 + 16. The four 3’s
are a1 algebras, taking the positive root of each of the 4 and pairing these
with their counterparts in the second d4 gives one more cartan; the negative
root gives anothers. Breaking the 16 under a1+a1 gives 4 invariant subspaces
,each abelian, which (again after paring with the second d4 counterparts) give
the remaining 4 cartans.

Posted by: rntsai on March 2, 2008 9:25 AM | Permalink | Reply to this

### Re: Kostant on E8

rntsai mentioned Kostant’s “… decomposition of e8 into 31 cartan …” and said:
“… related to e8/(d4+d4) decomposition :
e8/(d4+d4)= (28,1)+(1,28) + (8v,8v) + (8S+,8S+) + (8S-,8S-)
The last 3 terms can be seen as 24 8-dim spaces invariant under one of the d …
The other 7 cartans are inside d4+d4 … there are probably several ways to identify [them]…”.

Another way (other than the one mentioned by rntsai) is to
decompose d4 into a 14-dim G2 plus two 7 spheres S7 + S7, getting
d4 = 14 + 7 + 7

14-dim rank-2 G2 has 7 = 14/2 Cartans
and G2 can be seen as the sum of two 7-dimensional representations.
If each 7 is represented by the 7 imaginary octonions { i,j,k,e,ie,je,ke }
then the 7 Cartans of G2 are the 7 pairs (one from each of the 7 in G2):
i i
j j
k k
e e
ie ie
je je
ke ke

Note that to make an Abelian Cartan, the pairs must match, because only matching pairs close to form an Abelian Cartan (this can be seen by looking at the octonion products).

28-dim rank-4 d4 has 28/4 = 7 Cartans
and d4 looks like G2 plus S7 plus S7
and since G2 decomposes into two 7 representations
d4 decomposes into 7 + 7 + 7 + 7 ( where the first two 7 are from G2 and the other two come from the two S7 )
and
the 7 Cartans of d4 are ( in terms of octonion imaginaries )
i i i i
j j j j
k k k k
e e e e
ie ie ie ie
je je je je
ke ke ke ke

Again, note that all elements of the quadruples must match to get Abelian Cartan structure.

When you look at d4 + d4 to get 8-element Cartans of E8, all 8 elements must again match up to get Abelian Cartan structure, so the 8 Cartans of E8 that come from d4 + d4 look like
i i i i i i i i
j j j j j j j j
k k k k k k k k
e e e e e e e e
ie ie ie ie ie ie ie ie
je je je je je je je je
ke ke ke ke ke ke ke ke

Of course, this octonion structure is also reflected in the
“… 24 8-dim spaces …” described by rntsai as
“… (8v,8v) + (8S+,8S+) + (8S-,8S-) …”

so that all 31 of the 8-dim Cartans of E8 have nice octonionic structure.

Also note that when you make a 240-element E8 root vector diagram of 8 circles each with 30 vertices, 8 of the 248 E8 generators are missing, so that you must leave out one of the 31 Cartan 8-element sets.
Seeing E8 in terms of E8 = 120 + 128 = d4 + d4 + 8x8 + 8x8 + 8x8
it is most natural to see the Cartan as being one of the Cartan sets of 8 coming from the d4 + d4,
but you could see the E8 from other points of view by using other Cartan sets of 8 to determine which of the 248 were the 8 omitted from the root vector diagram.

Tony Smith

Posted by: Tony Smith on March 2, 2008 5:43 PM | Permalink | Reply to this

### Re: Kostant on E8

Sorry for a typo:
I said in my immediately preceding comment
“… the 8 Cartans of E8 that come from d4 + d4 …”
when I should have said
“… the 7 Cartans of E8 that come from d4 + d4 …”.

Tony Smith

Posted by: Tony Smith on March 2, 2008 6:24 PM | Permalink | Reply to this

### Re: Kostant on E8

“When you look at d4 + d4 to get 8-element Cartans of E8, all 8 elements must again match up to get Abelian Cartan structure, so the 8 Cartans of E8 that come from d4 + d4 look like
i i i i i i i i
j j j j j j j j
…”

I’m afraid I don’t see how this construction with octanions is done.
Are you building e8 with 8-tuples of octanions? so iiiiiiii is
{(i,0,…,0),(0,i,…,0),…(0,…,i)}?
what happens with the other combinations?
there are too many to fit into the 248
roots of e8

Posted by: rntsai on March 3, 2008 4:27 AM | Permalink | Reply to this

### Re: Kostant on E8

rntsai asks about my construction of the 7 Cartans of E8 that come from d4+d4,
and here is my current effort at further explanation. If this does not make it clear, then please just feel free to look at it in whatever other way might seem clear to you.

No,
iiiiiiii
is not in my picture
{(i,0, … ,0),(0,i, … ,0), … (0, … ,i)}

I am just looking at d4 + d4
which can be represented as 7+7+7+7 + 7+7+7+7
(from d4 = G2 plus S7 plus S7)
and
then each 7 is represented by the imaginary octonions
i
j
k
e
ie
je
ke

so that the eight 7s of d4+d4 = 7+7+7+7+7+7+7+7
include things like the octuple
i i j k e i k j
but
nice Cartan structure only comes with the ones like
i i i i i i i i
j j j j j j j j
k k k k k k k k
e e e e e e e e
ie ie ie ie ie ie ie ie
je je je je je je je je
ke ke ke ke ke ke ke ke

where all the entries in the octuple are similar
and so you get the seven 8-element E8 Cartan subalgebras related to d4+d4 which, when added to the other 24 from the three 8x8 things, give the full set of 31.

Tony Smith

Posted by: Tony Smith on March 3, 2008 5:25 AM | Permalink | Reply to this

### Re: Kostant on E8

I still don’t see it, but that’s ok. Who knows, maybe someday I’ll warm up to
octonions (note the correct spelling!) and all this becomes obvious.

Back to the topic, finding 7 cartans in d4+d4 is by no means uniques.
There are 7 4-dim abelian subspaces in each d4; each one will commute with all
7 from the other d4, so you can pair them in many ways.

Posted by: rntsai on March 3, 2008 9:56 PM | Permalink | Reply to this

### Re: Kostant on E8

Nice suggestion Rntsai.I am trying to solve it let see where will i reach.

Posted by: Alex Peterson on June 5, 2008 7:05 AM | Permalink | Reply to this

### flavour

To matematicians, let me note that there is a world where SU(5) already appears: the quark model of hadrons. Usually this symmetry (which is not a local gauge one, of course) appears explicitly broken down, via quark masses, to SU(3). Or more precisely it breaks to SU(3)xSU(2), from u,d,s and b,c quarks.

For instance, to classify meson multiplets we can use the $24$ of SU(5) extracted out of the product $5 \otimes \bar 5$. Perhaps there are situations where we should prefer to use two separate SU(5) copies. And then there is the question of diquark spectroscopy, from $5 \otimes 5$. In real practice all these questions become murkier because you need to consider spin, colour, parity… So sometimes larger groups are invoked.

On other hand, perhaps off-topic and -for sure- very speculatively, let me add that I think that these representations also hint a different way to build three generations of particles, if you add colour to the picture: the 24 out of $5 \otimes \bar 5$ contains the same charges and degrees of freedom than the six leptons of the SM, and the 15 out of $5 \otimes 5$ contains the 12 charges and d.o.f. of a given colour of 3 generations of quarks, plus 3 extra exotic chiral objects (which should be ruled out somehow when asking for the electroweak force to enter the game).

Posted by: Alejandro Rivero on August 3, 2008 3:31 AM | Permalink | Reply to this

Post a New Comment