Musings (original) (raw)

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

MathML-enabled post (click for more details).

We would like to find an embedding of

(1)G=SL(2,ℂ)×SU(3)×SU(2)×U(1)G = SL(2,\mathbb{C})\times SU(3)\times SU(2)\times U(1)

in a suitable noncompact real form of E 8E_8, such that one finds 3 copies of the representation

(2)R= 2⊗[(3,2) 1/6+(3¯,1) −2/3+(3¯,1) 1/3+(1,2) −1/2+(1,1) 1] + 2¯⊗[(3¯,2) −1/6+(3,1) 2/3+(3,1) −1/3+(1,2) 1/2+(1,1) −1]\begin{aligned} R = &\mathbf{2}\otimes [(3,2)_{1/6}+(\overline{3},1)_{-2/3}+(\overline{3},1)_{1/3}+(1,2)_{-1/2} +(1,1)_{1}]\\ +&\overline{\mathbf{2}}\otimes [(\overline{3},2)_{-1/6}+(3,1)_{2/3}+(3,1)_{-1/3}+(1,2)_{1/2} +(1,1)_{-1}] \end{aligned}

in the decomposition of the 248 of E 8E_8. Here SL(2,ℂ)=Spin(3,1) 0SL(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 8E_8 are we talking about? There are two.

where, in both cases, I’ve indicated the noncompact generators in bold.

Garret never deigns to tell us which real form of E 8E_8 he is using. But he does say that the embedding of GG in E 8E_8 is supposed to proceed via the subgroup F 4×G 2⊂E 8F_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

It turns out that F 4(−20)×G 2⊂E 8(8)F_{4(-20)}\times G_2 \subset E_{8(8)} and F 4(4)×G 2⊂E 8(−24)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)E_{8(8)}, we have Spin(16)⊃Spin(9)×G 2Spin(16)\supset Spin(9)\times G_2 and120 =(1,14)+(36,1)+(1,7)+(9,7) 128 =(16,1)+(16,7)\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)E_{8(-24)}, we have SU(2)×E 7⊃SU(2)×Sp(3)×G 2SU(2)\times E_7\supset SU(2)\times Sp(3)\times G_2 and(3,1) =(3,1,1) (1,133) =(1,1,14)+(1,21,1)+(1,14,7) (2,56) =(2,14,1)+(2,6,7)\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×G 2F_4\times G_2) in red.

So that was, actually, no help. The only thing to do is to try to embed GG in F 4(−20)×G 2F_{4(-20)}\times G_2 and in F 4(4)×G 2F_{4(4)}\times G_2 and see what happens.

To make a long story short, GG is not embeddable (actually, it is embeddable; there are just no suitable embeddings) as a subgroup of either F 4(−20)×G 2F_{4(-20)}\times G_2 or F 4(4)×G 2F_{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 GG in F 4×G 2F_4\times G_2, let’s just find some embedding of GG in E 8E_8. Clearly, that’s possible to do in quite a number of ways. Demanding that the representation RR appear in the decomposition of the 248 is, however, highly restrictive.

For the split real form, E 8(8)E_{8(8)}, the best you can do is obtain 2 copies of RR. To see how that goes, embed the maximal compact subgroup

(4)G 0=SU(2) MM×SU(3)×SU(2)×U(1)G_0 = SU(2)_{\text{MM}}\times SU(3)\times SU(2)\times U(1)

of GG in an SU(2)×SU(5)SU(2)\times SU(5) subgroup of Spin(16)Spin(16), such that2 120 =(1,24)+(3,1)+(1,10+10¯)+2(1,5+5¯)+5(1,1)+2(2,5+5¯)+4(2,1) 128 =(3,1)+(1,1)+2(2,5¯+10)+2(2,5+10¯)+2(2,1)\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)SU(5), and the (6-dimensional) adjoint of SL(2,ℂ)SL(2,\mathbb{C}), the 248 contains

(5)“fermions”: 2[(2,5¯+10)+(2¯,5+10¯)]+(2,5+5¯)+(2¯,5+5¯)+3[(2,1)+(2¯,1)] “bosons”: 2(1,5+5¯)+(1,10+10¯)+6(1,1)\begin{aligned} \text{“fermions”:} & 2[(\mathbf{2}, \overline{5}+10) + (\overline{\mathbf{2}}, 5+\overline{10})] + (\mathbf{2}, 5+\overline{5}) +(\overline{\mathbf{2}}, 5+\overline{5}) + 3[(\mathbf{2},1)+(\overline{\mathbf{2}}, 1)]\\ \text{“bosons”:} & 2(1, 5+\overline{5}) + (1, 10+\overline{10}) +6(1, 1) \end{aligned}

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 RR. 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(−24)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 GG does not embed in F 4×G 2F_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 RR). I simply didn’t bother considering such embeddings, when I was preparing this post. For the record, thoughF 4(−20)⊃Spin(8,1)⊃Spin(3,1)×Spin(5)⊃SL(2,ℂ)×SU(2)×U(1) 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)andF 4(4)⊃Spin(5,4)⊃Spin(3,1)×Spin(2,3)⊃SL(2,ℂ)×SU(2)×U(1) 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 26=1+9+16 =(1,1) 0+(4,1) 0+(1,3) 0+(1,1) 2+(1,1) −2 +(2,2) 1+(2,2) −1+(2¯,2) 1+(2¯,2) −1 52=36+16 =(Adj,1) 0+(1,3) 0+(1,1) 0+(1,3) 2+(1,3) −2+(4,3) 0+(4,1) 2+(4,1) −2 +(2,2) 1+(2,2) −1+(2¯,2) 1+(2¯,2) −1\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)×U(1)⊂Spin(5)SU(2)\times U(1)\subset Spin(5). For the one under which 4=2 1+2 −14 = 2_1 +2_{-1}, one obtains the same result as above. For the one under which 4=2 0+1 1+1 −14= 2_0 + 1_1 + 1_{-1}, one obtains26 =2(1,1) 0+(4,1) 0+(1,2) 1+(1,2) −1 +(2,2) 0+(2,1) 1+(2,1) −1+(2¯,2) 0+(2¯,1) 1+(2¯,1) −1 52 =(Adj,1) 0+(1,3) 0+(1,1) 0+(4,1) 0+(1,1) 2+(1,1) −2+(1,2) 1+(1,2) −1+(4,2) 1+(4,2) −1 +(2,2) 0+(2,1) 1+(2,1) −1+(2¯,2) 0+(2¯,1) 1+(2¯,1) −1\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)⊂G 2SU(3)\subset G_2,7 =1+3+3¯ 14 =8+3+3¯\begin{aligned} 7 &= 1+3+\overline{3}\\ 14 &= 8+3+\overline{3} \end{aligned}into (3), one obtains a completely nonchiral representation of GG.

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 GG with two generations. To do that, I had optimistically assumed that there is an embedding of SL(2,ℂ)SL(2,\mathbb{C}) in a suitable noncompact real form of A 4A_4, such that the 55 decomposes as 5=1+2+25=1+2+2. This is incorrect. It is easy to show that only 5=1+2+2¯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 GG in either noncompact real form of E 8E_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 8E_8), the “Euclidean” case (associated to the compact real form). Instead of SL(2,ℂ)SL(2,\mathbb{C}), we’re embedding Spin(4)=SU(2) L×SU(2) RSpin(4)=SU(2)_L\times SU(2)_R. Consider the left-handed “fermions” (the (2,1)(2,1) representation of Spin(4)Spin(4)), which transform as electroweak doublets. If they lie in a generation, then they transform as 3 1/6+1 −1/23_{1/6} + 1_{-1/2} under SU(3)×U(1) YSU(3)\times U(1)_Y. If they lie in an anti-generation, then they transform as 3¯ −1/6+1 1/2\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,ℂ)SL(2,\mathbb{C}) which are not related by “Wick rotation” to embeddings of Spin(4)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 8e_8. The “bosons” correspond to the subalgebra, on which the involution acts as +1+1; the “fermions” correspond to generators on which the involution acts as −1-1. 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 8e_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 −1-1 eigenvalues is 128128. The “bosonic” subalgebra is a certain real form of d 8d_8, and the 128128 is the spinor representation.

We’re interested in embedding GG in the group generated by the “bosonic” subalgebra, which is Spin(8,8)Spin(8,8) in the case of E 8(8)E_{8(8)} or Spin(12,4)Spin(12,4), in the case of E 8(−24)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

(6)128=?2(2,ℜ+(1,1) 0)+2(2¯,ℜ¯+(1,1) 0)128 \stackrel{?}{=} 2 \left(\mathbf{2},\mathfrak{R}+(1,1)_0\right) + 2 \left(\overline{\mathbf{2}},\overline{\mathfrak{R}}+(1,1)_0\right)

whereℜ=(3,2) 1/6+(3¯,1) −2/3+(3¯,1) 1/3+(1,2) −1/2+(1,1) 1\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+1 on 56 generators and as −1-1 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=(2,ℜ+(1,1) 0+ℜ¯+(1,1) 0)+(2¯,ℜ+(1,1) 0+ℜ¯+(1,1) 0)128 = \left(\mathbf{2},\mathfrak{R} + (1,1)_0 + \overline{\mathfrak{R}} + (1,1)_0\right) + \left(\overline{\mathbf{2}},\mathfrak{R} + (1,1)_0 + \overline{\mathfrak{R}} + (1,1)_0\right)

The reason why (6) cannot occur is very simple. Since we are talking about the spinor representation of Spin(16−4k,4k)Spin(16-4k,4k), we should have∧ 2128⊃120 \wedge^2 128 \supset 120In particular, we should find the adjoint representation of GG in the decomposition of the antisymmetric square. This does not happen for (6); in particular, you won’t find the (1,8,1) 0(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 8E_8 you use, or how you attempt to embed GG.

Quod Erat Demonstratum. Merry Christmas, y’all!

2 Start with the decomposition of the fundamental representation16=(1,5)+(1,5¯)+2(2,1)+2(1,1) 16 = (1,5) + (1,\overline{5}) + 2(2,1) + 2(1,1)

3 As before, once you specify how the 22 of SU(2)SU(2) and the 5656 of E 7E_7 decompose under G 0G_0, everything else is determined. And there just aren’t that many possibilities…