### 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

- It’s readily decidable, on purely mathematical grounds.
- 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

in a suitable noncompact real form of $E_8$, such that one finds 3 copies of the representation

in the decomposition of the 248 of $E_8$. Here $SL(2,\mathbb{C})= Spin(3,1)_0$ 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*.

- $E_{8(8)} \supset Spin(16)$ as a maximal compact subgroup
^{1}. In $E_{8(8)}$, the 248 decomposes as $248 = 120 +\mathbf{128}$ - $E_{8(-24)}\supset SU(2)\times E_7$ as a maximal compact subgroup. In $E_{8(-24)}$, the 248 decomposes as $248 = (3,1) +(1,133) +\mathbf{(2,56)}$

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 $F_4\times G_2\subset E_8$, 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 $F_4\times G_2$ has rank 6, its commutant inside of $E_8$ is discrete. The 248 decomposes as(3)$248 = (1,14) + (52,1) + (26,7)$
- The group $G$ that we are trying to embed also has rank 6, so — if it’s possible — the embedding in $F_4\times G_2$ is essentially unique.
- There are
*two*noncompact real forms of $F_4$- $F_{4(-20)}\supset Spin(9)$ as a maximal compact subgroup. $\begin{aligned} 52 &= 36 +\mathbf{16}\\ 26 &= 1+9+16 \end{aligned}$
- $F_{4(4)}\supset SU(2)\times Sp(3)$ as a maximal compact subgroup. $\begin{aligned} 52 &= (3,1)+(1,21) +\mathbf{(2,14)}\\ 26 &= (1,14)+ (2,6) \end{aligned}$

It turns out that $F_{4(-20)}\times G_2 \subset E_{8(8)}$ and $F_{4(4)}\times G_2 \subset E_{8(-24)}$. We can see how this works by decomposing under the common (compact) subgroup. For $E_{8(8)}$, we have $Spin(16)\supset Spin(9)\times G_2$ and $\begin{aligned} 120 &= (1,14) + (36,1) +{\color{red} (1,7) +(9,7)}\\ \mathbf{128} &= \mathbf{(16,1)} +{\color{red} \mathbf{(16,7)}} \end{aligned}$ and for $E_{8(-24)}$, we have $SU(2)\times E_7\supset SU(2)\times Sp(3)\times G_2$ and $\begin{aligned} (3,1) &= (3,1,1)\\ (1,133) &= (1,1,14) + (1,21,1) + {\color{red} (1,14,7)}\\ \mathbf{(2,56)} &= \mathbf{(2,14,1)} +{\color{red} \mathbf{(2,6,7)}} \end{aligned}$ where, in each case, I’ve indicated the additional generators (the ones not in $F_4\times G_2$) in red.

So that was, actually, no help. The only thing to do is to try to embed $G$ in $F_{4(-20)}\times G_2$ and in $F_{4(4)}\times G_2$ 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* $F_{4(-20)}\times G_2$ or $F_{4(4)}\times G_2$. 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 $F_4\times G_2$, 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, $E_{8(8)}$, the *best* you can do is obtain 2 copies of $R$. To see how that goes, embed the maximal compact subgroup

of $G$ in an $SU(2)\times SU(5)$ subgroup of $Spin(16)$, such that^{2}
$\begin{aligned}
120 &= (1, 24) + (3, 1) + (1, 10+\overline{10}) +2(1, 5+\overline{5}) + 5(1, 1) + 2(2, 5+\overline{5}) + 4(2, 1)\\
128 &= (3, 1) + (1, 1) + 2(2, \overline{5}+10) +2(2, 5+\overline{10}) +2(2, 1)
\end{aligned}$
In addition to the (24-dimensional) adjoint of $SU(5)$, and the (6-dimensional) adjoint of $SL(2,\mathbb{C})$, the 248 contains

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 exercise^{3} for the reader to repeat the analysis for $E_{8(-24)}$.

#### 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 $F_4\times G_2$. 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 $F_{4(-20)} \supset Spin(8,1) \supset Spin(3,1)\times Spin(5)\supset SL(2,\mathbb{C})\times SU(2)\times U(1)$ and $F_{4(4)} \supset Spin(5,4) \supset Spin(3,1)\times Spin(2,3)\supset SL(2,\mathbb{C})\times SU(2)\times U(1)$ In the latter case, one obtains $\begin{aligned} 26 = 1 + 9 + 16 &= (\mathbf{1},1)_0 + (\mathbf{4},1)_0 + (\mathbf{1},3)_0 + (\mathbf{1},1)_2 + (\mathbf{1},1)_{-2}\\ &\quad + (\mathbf{2},2)_1 + (\mathbf{2},2)_{-1} + (\overline{\mathbf{2}},2)_1 + (\overline{\mathbf{2}},2)_{-1}\\ 52 = 36 + 16 &= (\mathbf{Adj},1)_0 + (\mathbf{1},3)_0 + (\mathbf{1},1)_0 + (\mathbf{1},3)_2 + (\mathbf{1},3)_{-2} + (\mathbf{4},3)_0 + (\mathbf{4},1)_2 + (\mathbf{4},1)_{-2}\\ &\quad + (\mathbf{2},2)_1 + (\mathbf{2},2)_{-1} + (\overline{\mathbf{2}},2)_1 + (\overline{\mathbf{2}},2)_{-1} \end{aligned}$ In the former case, there are two distinct embeddings of $SU(2)\times U(1)\subset Spin(5)$. 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 $\begin{aligned} 26 &= 2(\mathbf{1},1)_0 + (\mathbf{4},1)_0 + (\mathbf{1},2)_1 + (\mathbf{1},2)_{-1}\\ &\quad + (\mathbf{2},2)_0 + (\mathbf{2},1)_1 + (\mathbf{2},1)_{-1} + (\overline{\mathbf{2}},2)_0 + (\overline{\mathbf{2}},1)_1 + (\overline{\mathbf{2}},1)_{-1}\\ 52 &= (\mathbf{Adj},1)_0 + (\mathbf{1},3)_0 + (\mathbf{1},1)_0 + (\mathbf{4},1)_0 + (\mathbf{1},1)_2 + (\mathbf{1},1)_{-2} + (\mathbf{1},2)_1 + (\mathbf{1},2)_{-1} + (\mathbf{4},2)_1 + (\mathbf{4},2)_{-1}\\ &\quad + (\mathbf{2},2)_0 + (\mathbf{2},1)_1 + (\mathbf{2},1)_{-1} + (\overline{\mathbf{2}},2)_0 + (\overline{\mathbf{2}},1)_1 + (\overline{\mathbf{2}},1)_{-1} \end{aligned}$ Putting these, together with the embedding of $SU(3)\subset G_2$, $\begin{aligned} 7 &= 1+3+\overline{3}\\ 14 &= 8+3+\overline{3} \end{aligned}$ 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 $SL(2,\mathbb{C})$ 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 $SL(2,\mathbb{C})$, we’re embedding $Spin(4)=SU(2)_L\times SU(2)_R$. Consider the left-handed “fermions” (the $(2,1)$ representation of $Spin(4)$), which transform as electroweak doublets. If they lie in a*generation*, then they transform as $3_{1/6} + 1_{-1/2}$ under $SU(3)\times U(1)_Y$. 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 $SL(2,\mathbb{C})$ which are *not* related by “Wick rotation” to embeddings of $Spin(4)$ 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 $Spin(8,8)$ in the case of $E_{8(8)}$ or $Spin(12,4)$, in the case of $E_{8(-24)}$. 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

where
$\mathfrak{R} = (3,2)_{1/6}+(\overline{3},1)_{-2/3}+(\overline{3},1)_{1/3}+(1,2)_{-1/2} +(1,1)_{1}$
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

The reason why (6) *cannot* occur is very simple. Since we are talking about the spinor representation of $Spin(16-4k,4k)$, 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 $(1,8,1)_0$ 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} $E_{8(8)}$ 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
$16 = (1,5) + (1,\overline{5}) + 2(2,1) + 2(1,1)$

## 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.