xml version='1.0' encoding='iso-8859-1'?> A Little Group Theory ... | Musings

## November 21, 2007

### A Little Group Theory …

I really wasn’t going to post about Garrett Lisi’s paper. Preparing a post like this requires work and, in this case, the effort expended would be vastly incommensurate with any benefit to be gained.

So I gritted my teeth through a series of credulous posts in the Physics blogosphere and the ensuing media frenzy. (Yes, Virginia, science reporters do read blogs. And if you think something is worth posting about, there’s a good chance — especially if it has the phrase “Theory of Everything” in the title — they will conclude that it’s worth writing about, too.) But, finally, it was Sean Carroll’s post that pushed me over the edge. Unlike the others, Sean freely admitted that he hadn’t actually read Lisi’s paper, but decided it was OK to post about it anyway.

So here goes.

I’m not going to talk about spin-statistics, or the Coleman-Mandula Theorem, or any of the Physics issues that could render Garrett’s idea a non-starter. Instead, I will confine myself to a narrow question in group representation theory. This has the advantage that

1. It’s readily decidable, on purely mathematical grounds.
2. Since it involves the starting point of Garrett’s analysis, a negative answer would render all of the other questions moot.

We would like to find an embedding of

(1)$G=\mathrm{SL}\left(2,ℂ\right)×\mathrm{SU}\left(3\right)×\mathrm{SU}\left(2\right)×U\left(1\right)$

in a suitable noncompact real form of ${E}_{8}$, such that one finds 3 copies of the representation

(2)$\begin{array}{rl}R=& 2\otimes \left[\left(3,2{\right)}_{1/6}+\left(\overline{3},1{\right)}_{-2/3}+\left(\overline{3},1{\right)}_{1/3}+\left(1,2{\right)}_{-1/2}+\left(1,1{\right)}_{1}\right]\\ +& \overline{2}\otimes \left[\left(\overline{3},2{\right)}_{-1/6}+\left(3,1{\right)}_{2/3}+\left(3,1{\right)}_{-1/3}+\left(1,2{\right)}_{1/2}+\left(1,1{\right)}_{-1}\right]\end{array}$

in the decomposition of the 248 of ${E}_{8}$. Here $\mathrm{SL}\left(2,ℂ\right)=\mathrm{Spin}\left(3,1{\right)}_{0}$ is the connected part of the Lorentz Group, the “gauge group” in the MacDowell-Mansouri formulation of gravity.

Now, the first question you might ask is, which noncompact real form of ${E}_{8}$ are we talking about? There are two.

• ${E}_{8\left(8\right)}\supset \mathrm{Spin}\left(16\right)$ as a maximal compact subgroup1. In ${E}_{8\left(8\right)}$, the 248 decomposes as $248=120+128$
• ${E}_{8\left(-24\right)}\supset \mathrm{SU}\left(2\right)×{E}_{7}$ as a maximal compact subgroup. In ${E}_{8\left(-24\right)}$, the 248 decomposes as $248=\left(3,1\right)+\left(1,133\right)+\left(2,56\right)$
where, in both cases, I’ve indicated the noncompact generators in bold.

Garret never deigns to tell us which real form of ${E}_{8}$ he is using. But he does say that the embedding of $G$ in ${E}_{8}$ is supposed to proceed via the subgroup ${F}_{4}×{G}_{2}\subset {E}_{8}$, and he devotes page after mind-numbing page to describing the details of that embedding. Does this provide a clue?

Let us note 3 facts

• Despite the fact the ${F}_{4}×{G}_{2}$ has rank 6, its commutant inside of ${E}_{8}$ is discrete. The 248 decomposes as
(3)$248=\left(1,14\right)+\left(52,1\right)+\left(26,7\right)$
• The group $G$ that we are trying to embed also has rank 6, so — if it’s possible — the embedding in ${F}_{4}×{G}_{2}$ is essentially unique.
• There are two noncompact real forms of ${F}_{4}$
• ${F}_{4\left(-20\right)}\supset \mathrm{Spin}\left(9\right)$ as a maximal compact subgroup. $\begin{array}{rl}52& =36+16\\ 26& =1+9+16\end{array}$
• ${F}_{4\left(4\right)}\supset \mathrm{SU}\left(2\right)×\mathrm{Sp}\left(3\right)$ as a maximal compact subgroup. $\begin{array}{rl}52& =\left(3,1\right)+\left(1,21\right)+\left(2,14\right)\\ 26& =\left(1,14\right)+\left(2,6\right)\end{array}$

It turns out that ${F}_{4\left(-20\right)}×{G}_{2}\subset {E}_{8\left(8\right)}$ and ${F}_{4\left(4\right)}×{G}_{2}\subset {E}_{8\left(-24\right)}$. We can see how this works by decomposing under the common (compact) subgroup. For ${E}_{8\left(8\right)}$, we have $\mathrm{Spin}\left(16\right)\supset \mathrm{Spin}\left(9\right)×{G}_{2}$ and $\begin{array}{rl}120& =\left(1,14\right)+\left(36,1\right)+\left(1,7\right)+\left(9,7\right)\\ 128& =\left(16,1\right)+\left(16,7\right)\end{array}$ and for ${E}_{8\left(-24\right)}$, we have $\mathrm{SU}\left(2\right)×{E}_{7}\supset \mathrm{SU}\left(2\right)×\mathrm{Sp}\left(3\right)×{G}_{2}$ and $\begin{array}{rl}\left(3,1\right)& =\left(3,1,1\right)\\ \left(1,133\right)& =\left(1,1,14\right)+\left(1,21,1\right)+\left(1,14,7\right)\\ \left(2,56\right)& =\left(2,14,1\right)+\left(2,6,7\right)\end{array}$ where, in each case, I’ve indicated the additional generators (the ones not in ${F}_{4}×{G}_{2}$) in red.

So that was, actually, no help. The only thing to do is to try to embed $G$ in ${F}_{4\left(-20\right)}×{G}_{2}$ and in ${F}_{4\left(4\right)}×{G}_{2}$ and see what happens.

To make a long story short, $G$ is not embeddable (actually, it is embeddable; there are just no suitable embeddings) as a subgroup of either ${F}_{4\left(-20\right)}×{G}_{2}$ or ${F}_{4\left(4\right)}×{G}_{2}$. This is not surprising. As I said, since the ranks are equal, there’s no “wiggle-room” in choosing an embedding.

A pessimist would probably pack it in, at this point. But let’s try to give Garrett the benefit of the doubt and relax our assumptions a bit.

Rather than attempting to embed $G$ in ${F}_{4}×{G}_{2}$, let’s just find some embedding of $G$ in ${E}_{8}$. Clearly, that’s possible to do in quite a number of ways. Demanding that the representation $R$ appear in the decomposition of the 248 is, however, highly restrictive.

For the split real form, ${E}_{8\left(8\right)}$, the best you can do is obtain 2 copies of $R$. To see how that goes, embed the maximal compact subgroup

(4)${G}_{0}=\mathrm{SU}\left(2{\right)}_{\text{MM}}×\mathrm{SU}\left(3\right)×\mathrm{SU}\left(2\right)×U\left(1\right)$

of $G$ in an $\mathrm{SU}\left(2\right)×\mathrm{SU}\left(5\right)$ subgroup of $\mathrm{Spin}\left(16\right)$, such that2 $\begin{array}{rl}120& =\left(1,24\right)+\left(3,1\right)+\left(1,10+\overline{10}\right)+2\left(1,5+\overline{5}\right)+5\left(1,1\right)+2\left(2,5+\overline{5}\right)+4\left(2,1\right)\\ 128& =\left(3,1\right)+\left(1,1\right)+2\left(2,\overline{5}+10\right)+2\left(2,5+\overline{10}\right)+2\left(2,1\right)\end{array}$ In addition to the (24-dimensional) adjoint of $\mathrm{SU}\left(5\right)$, and the (6-dimensional) adjoint of $\mathrm{SL}\left(2,ℂ\right)$, the 248 contains

(5)$\begin{array}{rl}\text{“fermions”:}& 2\left[\left(2,\overline{5}+10\right)+\left(\overline{2},5+\overline{10}\right)\right]+\left(2,5+\overline{5}\right)+\left(\overline{2},5+\overline{5}\right)+3\left[\left(2,1\right)+\left(\overline{2},1\right)\right]\\ \text{“bosons”:}& 2\left(1,5+\overline{5}\right)+\left(1,10+\overline{10}\right)+6\left(1,1\right)\end{array}$

I’ve put scare quotes around “fermions” and “bosons”, for reasons that are obvious to anyone who has taken more than a passing glance at Garrett’s paper. No matter. We have failed to find an embedding of three copies of $R$. The best we were able to do was embed 2 copies.

I leave it as an exercise3 for the reader to repeat the analysis for ${E}_{8\left(-24\right)}$.

#### Update (11/23/2007):

Those who think I have been too harsh in condemning the Physics blogosphere as an intellectual wasteland can probably point to Wikipedia as being measurably worse. If the article doesn’t make your head explode, try reading the Talk page.

#### Update (11/29/2007):

David Vogan, from MIT, wrote me to point out that I was too fast in saying that $G$ does not embed in ${F}_{4}×{G}_{2}$. It is possible to find such an embedding, but it necessarily leads to a completely nonchiral “fermion” representation (and hence contains no copies of $R$). I simply didn’t bother considering such embeddings, when I was preparing this post. For the record, though ${F}_{4\left(-20\right)}\supset \mathrm{Spin}\left(8,1\right)\supset \mathrm{Spin}\left(3,1\right)×\mathrm{Spin}\left(5\right)\supset \mathrm{SL}\left(2,ℂ\right)×\mathrm{SU}\left(2\right)×U\left(1\right)$ and ${F}_{4\left(4\right)}\supset \mathrm{Spin}\left(5,4\right)\supset \mathrm{Spin}\left(3,1\right)×\mathrm{Spin}\left(2,3\right)\supset \mathrm{SL}\left(2,ℂ\right)×\mathrm{SU}\left(2\right)×U\left(1\right)$ In the latter case, one obtains $\begin{array}{rl}26=1+9+16& =\left(1,1{\right)}_{0}+\left(4,1{\right)}_{0}+\left(1,3{\right)}_{0}+\left(1,1{\right)}_{2}+\left(1,1{\right)}_{-2}\\ & \phantom{\rule{1em}{0ex}}+\left(2,2{\right)}_{1}+\left(2,2{\right)}_{-1}+\left(\overline{2},2{\right)}_{1}+\left(\overline{2},2{\right)}_{-1}\\ 52=36+16& =\left(\mathrm{Adj},1{\right)}_{0}+\left(1,3{\right)}_{0}+\left(1,1{\right)}_{0}+\left(1,3{\right)}_{2}+\left(1,3{\right)}_{-2}+\left(4,3{\right)}_{0}+\left(4,1{\right)}_{2}+\left(4,1{\right)}_{-2}\\ & \phantom{\rule{1em}{0ex}}+\left(2,2{\right)}_{1}+\left(2,2{\right)}_{-1}+\left(\overline{2},2{\right)}_{1}+\left(\overline{2},2{\right)}_{-1}\end{array}$ In the former case, there are two distinct embeddings of $\mathrm{SU}\left(2\right)×U\left(1\right)\subset \mathrm{Spin}\left(5\right)$. For the one under which $4={2}_{1}+{2}_{-1}$, one obtains the same result as above. For the one under which $4={2}_{0}+{1}_{1}+{1}_{-1}$, one obtains $\begin{array}{rl}26& =2\left(1,1{\right)}_{0}+\left(4,1{\right)}_{0}+\left(1,2{\right)}_{1}+\left(1,2{\right)}_{-1}\\ & \phantom{\rule{1em}{0ex}}+\left(2,2{\right)}_{0}+\left(2,1{\right)}_{1}+\left(2,1{\right)}_{-1}+\left(\overline{2},2{\right)}_{0}+\left(\overline{2},1{\right)}_{1}+\left(\overline{2},1{\right)}_{-1}\\ 52& =\left(\mathrm{Adj},1{\right)}_{0}+\left(1,3{\right)}_{0}+\left(1,1{\right)}_{0}+\left(4,1{\right)}_{0}+\left(1,1{\right)}_{2}+\left(1,1{\right)}_{-2}+\left(1,2{\right)}_{1}+\left(1,2{\right)}_{-1}+\left(4,2{\right)}_{1}+\left(4,2{\right)}_{-1}\\ & \phantom{\rule{1em}{0ex}}+\left(2,2{\right)}_{0}+\left(2,1{\right)}_{1}+\left(2,1{\right)}_{-1}+\left(\overline{2},2{\right)}_{0}+\left(\overline{2},1{\right)}_{1}+\left(\overline{2},1{\right)}_{-1}\end{array}$ Putting these, together with the embedding of $\mathrm{SU}\left(3\right)\subset {G}_{2}$, $\begin{array}{rl}7& =1+3+\overline{3}\\ 14& =8+3+\overline{3}\end{array}$ into (3), one obtains a completely nonchiral representation of $G$.

#### Update (12/10/2007):

For more, along these lines, see here.

#### Correction (12/11/2007):

Above, I asserted that I had found an embedding of $G$ with two generations. To do that, I had optimistically assumed that there is an embedding of $\mathrm{SL}\left(2,ℂ\right)$ in a suitable noncompact real form of ${A}_{4}$, such that the $5$ decomposes as $5=1+2+2$. This is incorrect. It is easy to show that only $5=1+2+\overline{2}$ arises. Thus, instead of two generations, one obtains a generation and an anti-generation. That is, the spectrum of “fermions” is, again, completely non-chiral. I believe (but haven’t proven) that this is a completely general result: for any embedding of $G$ in either noncompact real form of ${E}_{8}$, the spectrum of “fermions” is always nonchiral. Let’s have a contest, among you, dear readers, to see who can come up with a proof of this statement.

I apologize if I’d gotten anyone’s hopes up, with the above example. Not only can one never hope to get 3 generations out of this “Theory of Everything”; it appears that one can’t even get one generation.

#### Update (12/16/2007):

This post is still receiving a huge number of hits, but no one has taken me up on my challenge above. So let me give the easy part of the proof. Consider, instead of the Minkowskian case (associated to some noncompact real form of ${E}_{8}$), the “Euclidean” case (associated to the compact real form). Instead of $\mathrm{SL}\left(2,ℂ\right)$, we’re embedding $\mathrm{Spin}\left(4\right)=\mathrm{SU}\left(2{\right)}_{L}×\mathrm{SU}\left(2{\right)}_{R}$. Consider the left-handed “fermions” (the $\left(2,1\right)$ representation of $\mathrm{Spin}\left(4\right)$), which transform as electroweak doublets. If they lie in a generation, then they transform as ${3}_{1/6}+{1}_{-1/2}$ under $\mathrm{SU}\left(3\right)×U\left(1{\right)}_{Y}$. If they lie in an anti-generation, then they transform as ${\overline{3}}_{-1/6}+{1}_{1/2}$. But the 248 is real, ergo the number of generations and anti-generations must be equal, and the theory is non-chiral. QED.

That much was trivial. The gnarly bit is to work out what happens for embeddings of $\mathrm{SL}\left(2,ℂ\right)$ which are not related by “Wick rotation” to embeddings of $\mathrm{Spin}\left(4\right)$ in the compact real form.

#### Final Update (Christmas Edition)

Still no responses to my challenge. I suppose that the overlap between the set of people who know some group theory and those who are (still) interested in giving Lisi’s “Theory of Everything” a passing thought is empty.

But, since it’s Christmas, I guess it’s time to give the answer.

First, I will prove the assertion above, that there can be at most 2 generations in the decomposition of the 248. Then I will proceed to show that even that is impossible.

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 $-1$. Note that we are not replacing commutators by anti-commutators for the “fermions.” While that would make physical sense, it would correspond to an “${e}_{8}$ Lie superalgebra.” Victor Kač classified simple Lie superalgebras, and this isn’t one of them. Nope, the “fermions” will have commutators, just like the “bosons.”

We would like an involution which maximizes the number of “fermions.” Marcel Berger classified such involutions, and the maximum number of $-1$ eigenvalues is $128$. The “bosonic” subalgebra is a certain real form of ${d}_{8}$, and the $128$ is the spinor representation.

We’re interested in embedding $G$ in the group generated by the “bosonic” subalgebra, which is $\mathrm{Spin}\left(8,8\right)$ in the case of ${E}_{8\left(8\right)}$ or $\mathrm{Spin}\left(12,4\right)$, in the case of ${E}_{8\left(-24\right)}$. And we’d like to count the number of generations we can find among the “fermions.” With a maximum of 128 fermions, we can, at best find

(6)$128\stackrel{?}{=}2\left(2,\Re +\left(1,1{\right)}_{0}\right)+2\left(\overline{2},\overline{\Re }+\left(1,1{\right)}_{0}\right)$

where $\Re =\left(3,2{\right)}_{1/6}+\left(\overline{3},1{\right)}_{-2/3}+\left(\overline{3},1{\right)}_{1/3}+\left(1,2{\right)}_{-1/2}+\left(1,1{\right)}_{1}$ That is, we can, at best, find two generations.

Lisi claimed to have found an involution which acted as $+1$ on 56 generators and as $-1$ on 192 generators. This, by Berger’s classification, is impossible.

In the first version of this post, I mistakenly asserted that I had found a realization of (6). This was wrong, and I had to sheepishly retract the statement. Instead, it — and Lisi’s embedding (after one corrects various mistakes in his paper) — is nonchiral

(7)$128=\left(2,\Re +\left(1,1{\right)}_{0}+\overline{\Re }+\left(1,1{\right)}_{0}\right)+\left(\overline{2},\Re +\left(1,1{\right)}_{0}+\overline{\Re }+\left(1,1{\right)}_{0}\right)$

The reason why (6) cannot occur is very simple. Since we are talking about the spinor representation of $\mathrm{Spin}\left(16-4k,4k\right)$, we should have ${\wedge }^{2}128\supset 120$ In particular, we should find the adjoint representation of $G$ in the decomposition of the antisymmetric square. This does not happen for (6); in particular, you won’t find the $\left(1,8,1{\right)}_{0}$ in the decomposition of the antisymmetric square of (6). But it does happen for (7). So (6) can never occur. It doesn’t matter which noncompact real form of ${E}_{8}$ you use, or how you attempt to embed $G$.

Quod Erat Demonstratum. Merry Christmas, y’all!

1 ${E}_{8\left(8\right)}$ is the split real form of ${E}_{8}$, which also engendered a slew of blog posts.

2 Start with the decomposition of the fundamental representation $16=\left(1,5\right)+\left(1,\overline{5}\right)+2\left(2,1\right)+2\left(1,1\right)$

3 As before, once you specify how the $2$ of $\mathrm{SU}\left(2\right)$ and the $56$ of ${E}_{7}$ decompose under ${G}_{0}$, everything else is determined. And there just aren’t that many possibilities…

Posted by distler at November 21, 2007 11:50 PM

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### Re: A Little Group Theory …

Thank you Jacques for taking the pains going through and put an end to this nonsense!

It is just incredible how the blogosphere distorts things, fueled by the dearest wish to outsmart string physicists and “kick string theory in the back” (Lisi). All this hype and media engineering would be worth a news story on its own.

Boy, those have just no idea of what it takes to do sensible science and create a “better” model inspite hundreds of physicists have tried for decades, and more often than not in such a naive way.

Posted by: Moveon on November 22, 2007 4:26 AM | Permalink | Reply to this

### Re: A Little Group Theory …

“Garrett never deigns to tell us which real form of E8 he is using.”

Actually he does, on page 29: E8(24).

“We would like to find an embedding of G=SL(2,ℂ)×SU(3)×SU(2)×U(1)”

But he doesn’t use SL(2,ℂ), he uses SO(4,1). It’s an LQG thing.

Posted by: mitchell porter on November 22, 2007 8:27 AM | Permalink | Reply to this

### Re: A Little Group Theory …

Actually he does, on page 29: ${E}_{8\left(24\right)}$.

I stand corrected. On the last page of the paper, in the penultimate paragraph, he says

This relationship may also shed light on how and why nature has chosen a non-compact form, E IX, of E8.

It’s a bit of a mystery how (and where) that figures into his analysis.

Suffice to say that I did look at both noncompact forms of ${E}_{8}$, and chose to present the one which came closer to actually succeeding.

But he doesn’t use SL(2,ℂ), he uses SO(4,1). It’s an LQG thing.

No. In his paper, he exclusively talks about the “gravitational so(3,1)” group. SO(4,1) never appears. Needless to say, trying to embed the latter group in ${E}_{8}$ would make the problem even worse.

Posted by: Jacques Distler on November 22, 2007 8:53 AM | Permalink | PGP Sig | Reply to this

### Re: A Little Group Theory …

Yes, I should have said SO(3,1). Though he does talk about SO(4,1).

Posted by: mitchell porter on November 22, 2007 9:17 AM | Permalink | Reply to this

### Re: A Little Group Theory …

Yes, I should have said SO(3,1).

Sigh. As I explained in my post, we’re not interested in $\mathrm{SO}\left(3,1\right)$, we’re interested in its double-cover, $\mathrm{Spin}\left(3,1\right)$. And the connected component of $\mathrm{Spin}\left(3,1\right)$ is $\mathrm{SL}\left(2,ℂ\right)$.

Posted by: Jacques Distler on November 22, 2007 9:42 AM | Permalink | PGP Sig | Reply to this

### Re: A Little Group Theory …

I have the impression that in loop quantum gravity they often don’t use the double cover. But it doesn’t affect your real point, which Garrett has conceded.

Posted by: mitchell porter on November 22, 2007 5:19 PM | Permalink | Reply to this

### Re: A Little Group Theory …

Hello Jacques,
It wasn’t my intent to instigate this media frenzy. I apologize for the pressure it put on you to read and discuss the paper, but at the same time I am happy that you did. Your analysis is entirely correct, and focuses on exactly the issue that I agree needs a better understanding in this theory. What this theory does currently have is an embedding of G and one copy of R in E8. The other two copies of R are related to the first by a triality rotation. Under a strict interpretation of the embedding, as you present here, these other two copies of R do not have the same quantum numbers as the first – so by this interpretation they are not good copies. However, they do have the correct quantum numbers under the triality rotated G. This, admittedly, is unsatisfactory hand waving – and I have discussed this inadequacy clearly in the paper, going so far as to explicitly state it is currently the main problem with the theory. Addressing this inadequacy in the theory is what I will be working on over the coming months, and if I or someone else can’t figure out a good way to solve this problem, the theory won’t work. Nevertheless, I consider it nontrivial that G and one copy of R fit in E8 in such a way that the standard model and GR Lagrangian may be efficiently constructed. For this reason, and the prospect of relating the other two generations to R through E8 triality, I consider this to be a developing theory that is worth my time to work on, as a long shot.
Best,
Garrett

Posted by: Garrett on November 22, 2007 12:07 PM | Permalink | Reply to this

### While you’re here …

Addressing this inadequacy in the theory is what I will be working on over the coming months, and if I or someone else can’t figure out a good way to solve this problem, the theory won’t work.

I can honestly say that I have no idea what you are hoping for.

Still, since you are here, perhaps you can explain something that has eluded me.

Do you claim that $G$ can be embedded as a subgroup of ${F}_{4\left(4\right)}×{G}_{2}$? If so, can you please enlighten me and write down the decomposition of the $\left(26,1\right)$ and of the $\left(1,7\right)$ under $G$?

Thanks.

Posted by: Jacques Distler on November 22, 2007 7:01 PM | Permalink | PGP Sig | Reply to this

### Re: While you’re here …

Sure, you’re a very smart guy, and I’ll probably learn some things here. The $G$ is not embedded in $F4×G2$ (thanks for the benefit of the doubt). The $G$ is embedded in a $D4×D4$ subgroup of $E8$. If we write $G$ in terms of the Lie algebra breakdown, $g=\mathrm{so}\left(3,1\right)+\mathrm{su}\left(2\right)+u\left(1\right)+\mathrm{su}\left(3\right)$ then this is in a $\mathrm{so}\left(7,1\right)+\mathrm{so}\left(8\right)$ of $e8$ via the Pati-Salam, left-right symmetric model, $g\prime =\mathrm{so}\left(3,1\right)+\mathrm{su}\left(2{\right)}_{L}+\mathrm{su}\left(2{\right)}_{R}+\mathrm{su}\left(4\right)$ The $\mathrm{so}\left(3,1\right)+\mathrm{su}\left(2{\right)}_{L}+\mathrm{su}\left(2{\right)}_{R}$ is in $\mathrm{so}\left(7,1\right)$, the $\mathrm{su}\left(4\right)$ is in $\mathrm{so}\left(8\right)$, and a $u\left(1\right)$ from the $\mathrm{su}\left(2{\right)}_{R}$ and a $u\left(1{\right)}_{B-L}$ from the $\mathrm{su}\left(4\right)$ combine to give the weak hypercharge $u\left(1\right)$ in $g$. The $u\left(1{\right)}_{B-L}$ in the $\mathrm{su}\left(4\right)$ is not in $F4$ or $G2$. I wrote the paper by starting with $G2$ because it was pedagogically useful as the simplest example for getting the point across that I was including the fermions in the adjoint.

The $\mathrm{so}\left(7,1\right)+\mathrm{so}\left(8\right)$ acts on the ${8}_{S+}×{8}_{S+}$ as the first generation of fermions. That part works great. The structure of $E8$ suggests that the second and third generations relate to the triality partners of the first, ${8}_{S-}×{8}_{S-}$ and ${8}_{V}×{8}_{V}$, but I don’t understand this relationship yet. As you know, and as I described in the paper, these second and third triality partners cannot literally be the second and third generation particles as the theory is currently constructed – the relationship is merely suggestive, and I suspect something more interesting is going on. I will probably end up using a slightly different (non-triality) assignment of the fermions, and may even end up using a different group for gravity. Or I might not be able to get it to work. I’ve tried to be very clear, both in the paper and to the press, that this idea is still in development. Most physicists seem to understand this theory is work in progress, and treat it accordingly – but thank you for spending the time to elucidate this fact so that others will understand.

Posted by: Garrett on November 23, 2007 11:08 AM | Permalink | Reply to this

### Re: While you’re here …

Garrett, I will take my opportunity:
Congratulations, very appealing theory - there is a certain amount of beauty (and importance) in it regardless if it’s 100% true. Your paper is very readable, and I’m sure that everyone looking here appreciates it very much that you do your science in such an open way. The same amount of gratitude to Jacques.

Posted by: msal on November 23, 2007 11:35 AM | Permalink | Reply to this

### Re: While you’re here …

The G is embedded in a $D4×D4$ subgroup of E8. If we write G in terms of the Lie algebra breakdown, $g=\mathrm{so}\left(3,1\right)+\mathrm{su}\left(2\right)+u\left(1\right)+\mathrm{su}\left(3\right)$ then this is in a $\mathrm{so}\left(7,1\right)+\mathrm{so}\left(8\right)$ of $e8$.

I’m sorry. Now you have me baffled again.

Above, Mitchel Porter pointed me to the passage in your paper where you state that $G$ is embedded in ${E}_{8\left(-24\right)}$. Now you say it’s embedded in $\mathrm{Spin}\left(7,1\right)×\mathrm{Spin}\left(8\right)$. That’s not a subgroup of ${E}_{8\left(-24\right)}$. It is a subgroup of ${E}_{8\left(8\right)}$, which is the case I analyzed in detail in my post.

If you want people to have a fighting chance of understanding what you are talking about, it would be best to pick one story and stick to it.

Moreover, while the spinor representations of $\mathrm{Spin}\left(8\right)$ are real, and related by triality to the vector representation, the spinor representations of $\mathrm{Spin}\left(7,1\right)$ are complex, and unrelated by any symmetry to the vector representation (which is real).

Posted by: Jacques Distler on November 23, 2007 12:23 PM | Permalink | PGP Sig | Reply to this

### Re: While you’re here …

Ah, I knew I’d learn something here. I’ve been building this model from the ground up, and only recently encountered the potential match to $E8$. It sounds like I made a mistake in thinking that $\mathrm{so}\left(7,1\right)+\mathrm{so}\left(8\right)$ is in the Lie algebra of $E\mathrm{IX}$, when in fact it’s in the split real form. Thanks.

Posted by: Garrett on November 23, 2007 1:59 PM | Permalink | Reply to this

### Re: A Little Group Theory …

That was something that caught my eye and confused me when I looked at the paper the first time.

The triality symmetry you take acts like its simply a symmetry from a larger group as opposed to acting like its something internal. It doesn’t seem like the triality im used too.

Further, its not clear to me at all what generates different mass hierarchies between generations. Even imposing it by hand makes it seem like the spontaneous symmetry breaking terms are working from a much bigger group.

What am I missing?

Posted by: Haelfix on November 22, 2007 5:30 PM | Permalink | Reply to this

### Re: A Little Group Theory …

The bottom line is that there isn’t a single “E8 theory” here (I’m just talking classically, since the quantum theory hasn’t been defined). There are three theories, related by triality, and they each only get one generation right.

Posted by: mitchell porter on November 22, 2007 6:50 PM | Permalink | Reply to this

### Re: A Little Group Theory …

Gee, Mitchell, you’re on to it. And I don’t think Garrett was claiming to have understood the generation structure.

Posted by: Kea on November 22, 2007 7:12 PM | Permalink | Reply to this

### Re: A Little Group Theory …

> I don’t think Garrett was claiming to have understood the generation structure

I am not sure at this point what new insight Garrett *is* claiming to have found.
What do you call a knife without blade where the handle is missing?

Posted by: wolfgang on November 22, 2007 8:53 PM | Permalink | Reply to this

### Re: A Little Group Theory …

What do you call a knife without blade where the handle is missing?

An idea of a knife contains the idea of a cutting edge.

Posted by: Kea on November 22, 2007 9:18 PM | Permalink | Reply to this

### Re: A Little Group Theory …

An idea of a knife

Yeah, it’s that element of the infamous “wishful thinking” in hep physics:

with no idea being easily proved right, we are left with estimating which ideas are promising. Agreeing on that is much harder than agreeing on what is right, which can already be pretty hard.

Anyway, one idea I definitely like is that of blogs: I found the above quite informative, and mainly because of it containing both an entry and the comments by Garrett.

Posted by: Urs Schreiber on November 23, 2007 6:24 AM | Permalink | Reply to this

### Generations

Anyway, one idea I definitely like is that of blogs: I found the above quite informative, and mainly because of it containing both an entry and the comments by Garrett.

For Kea’s benefit, I will add one more remark, perhaps elucidating why this “triality” idea is a non-starter.

Let $H$ be the representation $H=\left(1,1,2{\right)}_{-1/2}$ of $G$, and $\overline{H}$ be its complex conjugate. Let $R={R}_{c}+{\overline{R}}_{c}$, where ${R}_{c}$ is the first line of (2) and its complex conjugate, ${\overline{R}}_{c}$, is the second line. There are $G$-invariant trilinear forms

$\begin{array}{rl}{\lambda }_{l},{\lambda }_{d}:& {\wedge }^{2}{R}_{c}\otimes H\to ℂ\\ {\lambda }_{u}:& {\wedge }^{2}{R}_{c}\otimes \overline{H}\to ℂ\end{array}$

Since there are 3 generation of $R$, each of the $\lambda$’s is a $3×3$ matrix in “generation-space”.

These matrices not simultaneously diagonalizable. At best, you can simultaneously diagonalize two of the three (say, ${\lambda }_{l}$ and ${\lambda }_{u}$). The remaining one is ineluctably non-diagonal, and its off-diagonal entries have been measured to be nonzero in the real world.

Posted by: Jacques Distler on November 23, 2007 9:46 AM | Permalink | PGP Sig | Reply to this

### Re: Generations

Jacques, thank you for your interest and informative post on this topic. I suppose that many people are looking into this, and it is a great chance to educate, your post brings its best.

I am also waiting for response from Garrett, but regardless if the embedding you ask for is possible, it looks to me that Garrett just found some embedding in E8 where one generation of fermions agrees with everything beautifully, two others only after triality rotation.

What is your opinion on what Garrett proposed in his paper(p.13): that physical fermions may be linear combinations of triality partners (maybe, for some reason, in experiments we only get quantum numbers’ sets we know, others still being there but hidden…?).

Also, as someone already remarked here, it might be that “triality” transformation comes from some larger group.
Thanks for your great post (and exercises).

Posted by: msal on November 23, 2007 11:19 AM | Permalink | Reply to this

### Re: Generations

Thanks, Jacques. I realise you probably don’t care, but I’m actually not interested in a triality arising from ordinary representation theory, as has been made clear at great length on my blog.

Posted by: Kea on November 23, 2007 9:09 PM | Permalink | Reply to this

### Re: A Little Group Theory …

Jacques Distler wrote:

Those who think I have been too harsh in condemning the Physics blogosphere as an intellectual wasteland can probably point to Wikipedia as being measurably worse.

Wow. That article manages to quote both Motl and Woit, without exploding in a matter-antimatter annihilation. The fact that one can’t learn any physics from it is almost inconsequential by comparison.

Posted by: Blake Stacey on November 23, 2007 10:03 PM | Permalink | Reply to this

### Re: A Little Group Theory …

What is the upside to adopting such a contentious and dismissive tone?

If you’re wrong, you’re the poster boy for the leaden establishment physics community.

If you’re right…you still look like an asshole.

Apparently, between you and Motl, it takes sharp elbows to get anywhere in physics. Which I suppose explains why physics has gone nowhere in 20 years.

Posted by: Molon Labe on November 24, 2007 12:14 AM | Permalink | Reply to this

### Haggard old crone

I really didn’t want to post on this subject. (It took over a week before I finally broke down and wrote this post.)

You wanna know why?

One of the reasons (admittedly, not the only reason) was that I knew that, if I did write this post, I would have to field comments from people who, unironically, use phrases like “the leaden establishment physics community.”

Thanks for reminding me why I should steer clear of subjects like this in the future.

Posted by: Jacques Distler on November 24, 2007 1:40 AM | Permalink | PGP Sig | Reply to this

### Re: Haggard old crone

Thanks for reminding me why I should steer clear of subjects like this in the future.

Dear Jacques,

I actully learnt some E8 representation theory from your post, and also saw why Garrett’s paper doesn’t even work field-contentwise.

When a long paper appears which claims to solve “everything”, and when people like Smolin who are part of the academia endorse it, then the question of “to-read or not-to-read” begins to become a serious one.

Assuming I am not the exceptional (no pun intended) case, this is one of those situations where your scientific journalism saved a lot of people at least some amount of time. So I am not sure that an unqualified “steering clear” of things like this is the right thing to do in the future.

Maybe you should pick and choose?

Warm regards from cold Brussels,
Chethan.

Posted by: chethan on November 26, 2007 5:35 AM | Permalink | Reply to this

### Labour-saving devices

Assuming I am not the exceptional (no pun intended) case, this is one of those situations where your scientific journalism saved a lot of people at least some amount of time.

Thanks for the kind words, Chethan.

Obviously, there’s a great collective saving of labour in sparing the rest of y’all the need to go through the tedious group-theoretical analysis yourselves.

However, I’d much rather spend my time discussing topics where the result is actually interesting — where your reaction to reading my post is

Boy, I’d like to look into that more closely!

rather than

On the other hand, I’ve gotten quite a number of emails, thanking me for this post, for essentially the reasons you have. You’d be surprised at some of the people who, apparently, read this blog.

Under the circumstances, I think I will take consolation in that knowledge.

Posted by: Jacques Distler on November 27, 2007 2:11 AM | Permalink | PGP Sig | Reply to this

### Re: Labour-saving devices

You have my thanks as well (and my gratitude also goes to Chethan for making my point before I could find the words).

Having lurked around and learned at least a little from your discussions of topics “where the result is actually interesting”, I hope you continue doing that for a long time to come!

Posted by: Blake Stacey on November 28, 2007 11:38 AM | Permalink | Reply to this

### Re: A Little Group Theory …

If you want to take a clue as to the generation structure from Koide’s lepton mass formulas, it is hardly a surprise that the three generations do not match each other in quantum numbers. The three generations show up as a real singlet (the tau generation) and a complex pair (the electron and muon).

The symmetry breaking between the electron and muon families is by a rotation by the mysterious angle delta = 0.22222204717. Gerald Rosen uses 2/9.

The formulas (ignoring an overall scale) are:
sqrt(m) = 1 + sqrt(2)cos(delta + phi)
where phi is 0 for the tau (singlet) and +/- 2 pi/3 for the electron and muon.

Posted by: Carl Brannen on November 24, 2007 12:15 AM | Permalink | Reply to this

### I have a theory…

Carl (and the rest of the Anne Elks among you): this is not a forum for airing your pet theories of physics.

If I have to start deleting comments, I will.

Posted by: Jacques Distler on November 24, 2007 12:30 AM | Permalink | PGP Sig | Reply to this

### Re: I have a theory…

Carl (and the rest of the Anne Elks among you): this is not a forum for airing your pet theories of physics. If I have to start deleting comments, I will.

As far as me changing the subject to a pet theory, I am the “C. Brannen” that is mentioned on Lisi’s blog with regard to Koide’s mass formula, and my comments are exactly in line with what you are discussing.

And oh, please do delete my comments. I publish things on the web to get priority, not convince arrogant morons of their errors or to get web traffic. You might consider a similar attitude.

Posted by: Carl Brannen on November 24, 2007 4:33 PM | Permalink | Reply to this

### Re: A Little Group Theory …

For the SO(3,1) the (3,1) is both spacetime and vectors so I would think the vector part of the Triality should be an emergent spacetime (there’s already an emergent gravity why not an emergent spacetime)? For the E6 GUTs of superstrings, is the vector part of the Triality the bosons? Spacetime seems better than bosons even and still leaves only spinors for the fermions.

The reason D4xD4 seems to be needed for the color/electroweak bosons is that Lisi uses for bosons not just the adjoint for the D4 Triality but also uses its Hodge dual. That’s certainly unusual. In an A-D-E series view of E8, the Hodge dual would be up at the E7 and E8 level. That’s above the E6 GUT level. E7 and E8 are up at the level where Smolin started thinking about the Triality in big Jordan Algebras and how to apply that to string theory (yes Virginia, Smolin has worked on string theory).

Posted by: John G on November 24, 2007 10:08 AM | Permalink | Reply to this

### Stringing buzzwords into syntactically-correct sentences

For the SO(3,1) the (3,1) is both spacetime and vectors so I would think the vector part of the Triality should be an emergent spacetime (there’s already an emergent gravity why not an emergent spacetime)? For the E6 GUTs of superstrings, is the vector part of the Triality the bosons? Spacetime seems better than bosons even and still leaves only spinors for the fermions.

The reason D4xD4 seems to be needed for the color/electroweak bosons is that Lisi uses for bosons not just the adjoint for the D4 Triality but also uses its Hodge dual. That’s certainly unusual.

I have no idea what you are talking about.

But, as I explained above, there is no triality for $\mathrm{Spin}\left(7,1\right)$ and, even if there were, it would not help.

(yes Virginia, Smolin has worked on string theory).

Don’t even get me started…

I’m annoyed enough at the apparent intellectual standards of the physics blogosphere.

Posted by: Jacques Distler on November 24, 2007 11:35 AM | Permalink | PGP Sig | Reply to this

### Re: Stringing buzzwords into syntactically-correct sentences

Yes I have no reason to doubt you as far as using Triality for the 3 generations but if one uses just the spinors for fermions that leaves the vectors for spacetime. To make the spacetime complex you actually use D5 as in D5/D4xU(1). The spinors will also be complex at E6 as in E6/D5xU(1). As for signature problems, I’m just an unauthorized messenger, but perhaps this from Tony Smith helps:

“Since the more realistic Minkowski Physical SpaceTime with -+++ Signature has Quaternionic Structure, a useful Clifford path is this Clifford Path that is Quaternionic from Cl(3,5) through Cl(2,4) to Cl(1,3) and from Cl(2,6) through Cl(2,5) and Cl(2,4) to Cl(1,3):

http://www.valdostamuseum.org/hamsmith/ClifQPath.gif

D5 is taken to be SL(2,O) = Spin(1,9), and spinors are R16 + R16 of Cle(1,9) = Cl(1,8), where Cle(p,q) = Cl(p,q-1) and Cle(p,0) = Cl(0,p-1)”

All I’m saying is that perhaps Lisi needs to stay unconventional in some areas (like having gravity in the GUT) and become more conventional in other areas (use only spinors for fermions as in E6 GUTs for strings)

Other areas are perhaps more open to debate, like moving bosons into one D4 instead of D4xD4. This would get everything down into an E6 GUT and leave E7 and E8 open to ideas like this from Horowitz and Susskind (perhaps you prefer them to Smolin, this was the paper Smolin was referencing anyways):

http://xxx.lanl.gov/abs/hep-th/0012037

Posted by: John G on November 26, 2007 1:45 PM | Permalink | Reply to this

### Re: Stringing buzzwords into syntactically-correct sentences

All I’m saying is that perhaps Lisi needs to stay unconventional in some areas (like having gravity in the GUT) and become more conventional in other areas (use only spinors for fermions as in E6 GUTs for strings)

If you want to include the MacDowell-Mansouri $\mathrm{Spin}\left(3,1{\right)}_{0}$, along with the Standard Model gauge group, in ${E}_{8}$, then there is not enough “room” to also include 3 generations of quarks and leptons in the 248. That was what Lisi was aiming for. And I think we are all agreed that it doesn’t work.

If you have some alternative proposal, I invite you to write it up (not here, though).

Posted by: Jacques Distler on November 26, 2007 2:17 PM | Permalink | PGP Sig | Reply to this

### Re: Stringing buzzwords into syntactically-correct sentences

Yes I agree you only get one generation. The other two would have to be some kind of “composite” particles with their own math outside of the E8 GUT. Tony Smith may have written some of this up when the archive was at Los Alamos. Since Smith and his former advisor, David Finkelstein, are both thanked in Lisi’s paper, maybe Finkelstein and Georgia Tech might endorse some newer stuff from Smith. Finkelstein seems just as enthusiastic about Lisi as Smolin is.

Those Georgia Tech people seem good at doing things behind the scenes with Clifford Algebra. I personally have more trouble thinking in Clifford Algebra terms. My previous use of the term Hodge dual could have been appropriate if I was talking two D4-like things in Cl(8) but for two D4s in E8, my use of the term was nonsense (I think).

Posted by: John G on November 27, 2007 9:53 AM | Permalink | Reply to this

### some wiki herald1X ((delete-item))

“In the language of flowers, the thistle (Germ. DISTEL) (like the burr) is an ancient Celtic symbo