Musings

Despite all the talk, in his paper, about $F_4\times G_2$, I eventually got the following clear statement from Lisi: $G$ is embedded in a $Spin(7,1)\times Spin(8)$ subgroup¹ of $E_{8(8)}$, and a generation (a copy of $R$) fits in the $(8_v,8_v)$. Now, the spinor representation of $Spin(7,1)$ is complex. So, if there were such an embedding of $G$ in $E_{8(8)}$, then it’s possible that this would lead to a chiral “fermion” representation (in particular, to a copy of $R$).

But that’s an “If pigs could fly …” sort of statement. There is no such embedding of $Spin(7,1)\times Spin(8)$ and I was too generous (or credulous or whatever) in assenting that there was. If there were, then it would sit inside a $Spin(15,1)$ or $Spin(9,7)$ subgroup, neither of which exists.

In fact, one does not get a chiral fermion representation for any embedding based on a subgroup $D_4\times D_4\subset E_8$. Since the group theory might be of moderate interest to someone, let’s go through it.

There are two possible embeddings² of $D_8\subset E_8$, compatible with the real structure $E_{8(8)}$

$$\begin{aligned} & Spin(16)\\ & Spin(8,8) \end{aligned}$$ and one $$Spin(12,4)$$ compatible³ with the real structure $E_{8(-24)}$.

Correspondingly, there are six possible embeddings of $D_4\times D_4\subset D_8\subset E_8$, compatible with the real structure $E_{8(8)}$: $$\begin{gathered} Spin(8,0)\times Spin(8,0)\\ Spin(8,0)\times Spin(0,8),\, Spin(7,1)\times Spin(1,7),\, Spin(6,2)\times Spin(2,6),\, Spin(5,3)\times Spin(3,5),\, Spin(4,4)\times Spin(4,4) \end{gathered}$$

and three compatible with the real structure $E_{8(-24)}$:

$$Spin(8,0)\times Spin(4,4),\, Spin(7,1)\times Spin(5,3),\, Spin(6,2)\times Spin(6,2)$$

Lisi’s choice, $Spin(7,1)\times Spin(8)$, is not on either list.

From the above nine, demanding that $G$ be embedded as a subgroup narrows the choices down to five:

Finding the representation $R$ in the decomposition of the 248 narrows the choices down to two (one appropriate to $E_{8(-24)}$, and one appropriate to $E_{8(8)}$):

$$Spin(7,1)\times Spin(5,3),\, Spin(7,1)\times Spin(1,7)$$

The spinor representations of $Spin(7,1)$ and $Spin(5,3)$ are complex. The 248 decomposes as

$$248 = (28,1) + (1,28) + (8_v,8_v) + (8_s, 8_s) + (\overline{8}_s,\overline{8}_s)$$

Under the decomposition

we have $$\begin{aligned} 8_v &= 3_{-2} + \overline{3}_{2} + 1_0 + 1_0\\ 8_s &= 3_1 + \overline{3}_{-1} + 1_{-3} + 1_3\\ \overline{8}_s &= 3_1 + \overline{3}_{-1} + 1_{-3} + 1_3\\ 28 &= 8_0 + 1_0 + 1_0 + 3_{-2} + 3_{-2} + 3_4 + \overline{3}_2 + \overline{3}_2 + \overline{3}_{-4} \end{aligned}$$ and under $$Spin(1,7)\, \text{or}\, Spin(5,3) \supset SL(2,\mathbb{C})\times SU(2) \times SU(2)_B$$ we have $$\begin{aligned} 8_v &= (\mathbf{4}, 1, 1) + (1, 2, 2)\\ 8_s &= (\mathbf{2}, 2, 1) + (\overline{\mathbf{2}}, 1, 2)\\ \overline{8}_s &= (\overline{\mathbf{2}}, 2, 1) + (\mathbf{2}, 1, 2)\\ 28 &= (\mathbf{Adj}, 1, 1) + (1, 3, 1) + (1, 1, 3) + (\mathbf{4}, 2, 2) \end{aligned}$$

Identifying $U(1)_Y = \tfrac{1}{6} U(1)_a + (T_3)_B$, where $(T_3)_B$ is the Cartan generator of $SU(2)_B$, we find

a completely nonchiral representation.

I leave it as an exercise for the reader to work out the remaining cases in (1). The spinor representations of $Spin(6,2)$ are pseudoreal, while those of $Spin(8)$ and $Spin(4,4)$ are real, neither of which lead to a “matter content” remotely resembling that of the Standard Model (in particular, all the “fermions” are in $SU(2)$ doublets).

And, no, I don’t intend to comment on the rest of Smolin’s paper. I’ll leave that to Sean or Bee or Steinn. Why should I have to do all the hard work around here?

Appendix: Pati-Salam

Lee Smolin complained that I failed to use the phrase “Pati-Salam” in this post. The reason I didn’t is that anyone familiar with the Pati-Salam model could easily fill in the necessary step. Anyone unfamiliar with it would not find its invocation the least bit helpful. It would, instead, serve only to clutter the notation.

But, to keep Lee happy (and to correct a small typo), note that the $SU(3)\times U(1)_a$ subgroup in (2) actually sits inside an $SU(4)\subset SO(7,1)$: $$\begin{aligned} 4 &= 3_1 + 1_{-3}\\ \overline{4} &= \overline{3}_{-1} + 1_3\\ 6 &= 3_{-2} +\overline{3}_{2}\\ 15 &= 8_0 +1_0 + 3_4 +\overline{3}_{-4} \end{aligned}$$ Rather than $G$, above, one can talk, instead, of $$G_{PS} = SL(2,\mathbb{C}) \times SU(4)\times SU(2)\times SU(2)_B$$ and a “generation” is $$R_{PS} = (\mathbf{2},\mathfr{R}_{PS}) + (\overline{\mathbf{2}}, \overline{\mathfr{R}}_{PS})$$ where $$\mathfr{R}_{PS} = (4, 2, 1)+ (\overline{4}, 1, 2)$$ Note that $\mathfr{R}_{PS}$ is a complex representation of $SU(4)\times SU(2)\times SU(2)_B$, so this theory is chiral (which is no surprise, since $\mathfr{R}_{PS}$ consists of a Standard Model generation plus an extra $SU(3)\times SU(2)\times U(1)_Y$ singlet).

Nothing about the analysis changes, except that one can write (3) as

$$(8_s, 8_s) + (\overline{8}_s, \overline{8}_s) = (\mathbf{2},\mathfr{R}_{PS} + \overline{\mathfr{R}}_{PS}) + (\overline{\mathbf{2}},\mathfr{R}_{PS} + \overline{\mathfr{R}}_{PS})$$

which, unlike the desired result, is every bit as non-chiral as before.

Oh, and I don’t see why the “Euclidean” (compact) case, where one replaces $SL(2,\mathbb{C})$ by $Spin(4)=SU(2)_L\times SU(2)_R$, with $\mathbf{2}\to (2,1)$ and $\overline{\mathbf{2}}\to (1,2)$, is supposed to be any better. Nothing changes, except that you need to use the notion of “chiral” appropriate to Euclidean signature.