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Does $ G_2 $ have an $ A_8 $ subgroup? I think the answer is yes and that moreover this $ A_8 $ subgroup is maximal among the closed subgroups of $ G_2 $.

Some circumstantial evidence is that $ A_8 $ has a degree 7 irrep with Frobenius–Schur indicator 1 so it is a subgroup of $ SO_7 $ and $ G_2 $ is also a subgroup of $ SO_7 $. Also $ A_8 $ has a degree 14 irrep with Frobenius–Schur indicator 1, which could very possibly be the action of $ A_8 $ by the adjoint representation on the Lie algebra of $ G_2 $, which has dimension 14 ( a finite subgroup of a connected Lie group like $ G_2 $ being maximal among the closed subgroups is closely related to acting irreducibly in the adjoint representation).

This is all just conjecture though, I'm not familiar enough with $ G_2 $ to prove any of this. For what it's worth generators for the 7d irrep of $ A_8 $ are given

https://brauer.maths.qmul.ac.uk/Atlas/alt/A8/gap0/A8G1-Zr7B0.g

and generators for the 14d irrep of $ A_8 $ are given

https://brauer.maths.qmul.ac.uk/Atlas/alt/A8/gap0/A8G1-Zr14B0.g

Especially for the $ 7 \times 7 $ generators I would imagine someone out there can "recognize" when matrices are " $ G_2 $ matrices" the same way that one can "check" if a matrix is symplectic.

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2 Answers 2

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$G_2$ is famously the group of symmetries of an antisymmetric trilinear form in its 7 dimensional representation.

The 7 dimensional representation $V$ of $A_8$ does not preserve any antisymmetric trilinear form. Preserving an antisymmetric trilinear form would correspond to an invariant vector in $\Lambda^3(V)$, but $\Lambda^3(V)$ is irreducible.

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  • $\begingroup$ Very interesting! Could you say more about why $ \Lambda^3(V) $ is irreducible for $ V $ the 7d irrep of $ A_8 $? Is it the 35d irrep of $ A_8 $? $\endgroup$ Commented Oct 21, 2022 at 0:38
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    $\begingroup$ Yes. If you take the standard $n-1$ dimensional representation $V$ of $S_n$ then all of its exterior powers $\Lambda^k(V)$ are irreducible, combinatorially these correspond to so-called hook shaped partitions $(n-k, 1^k)$. For $A_n$ the story is similar with the slight exception that if $n = 2k+1$ is odd then $\Lambda^k(V)$ decomposes into two parts. $\endgroup$
    – Nate
    Commented Oct 21, 2022 at 14:46
  • $\begingroup$ +1 Wow that's such a cool fact I'll definitely remember that. I'd upvote it twice if I could! $\endgroup$ Commented Oct 21, 2022 at 15:04
  • $\begingroup$ can this fact be used to conclude that not only is $ A_{n+1} $ a subgroup of $ SO_n(\mathbb{R}) $ but moreover the action of $ A_{n+1} $ on $ \mathfrak{so}_n(\mathbb{R}) $ in the adjoint representation is irreducible for $ n \neq 4 $? (Since the adjoint rep here coincides with $ \Lambda^2 $ the second exterior power of the standard representation) $\endgroup$ Commented Nov 10, 2022 at 22:50
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No

There is no $ A_8 $ subgroup of $ G_2 $. According to 1.4 Corollary 2 of

Basic conjugacy theorems for $ G_2 $

by Greiss https://link.springer.com/content/pdf/10.1007/BF01884298.pdf

All finite subgroups are either subgroups of positive dimensional subgroups like $ SU_3:2 $ or $ SU_2 \otimes SU_2 \cong SO_4 $ or they are from a short list of finite subgroups. These positive dimensional subgroups contain some interesting finite subgroups related to $ A_5,A_6,GL(3,2) $ but nothing related to $ A_8 $. The short list of finite subgroups of $ G_2 $ is given in part (2) as $$ GL(3,2),2^3\cdot GL(3,2),GL(3,2):2\cong SL(2,7), G_2(2)\cong Aut(PSU(3,3)),G_2(2)'\cong PSU(3,3), SL(2,8), PSL(2,13) $$ none of these are $ A_8 $. Indeed $ A_8 $ has order 20,160, and so it appears to be significatly larger than any irreducible finite subgroup of $ G_2 $ (the largest of which $ G_2(2) $ has size 12,096). Thus there is no $ A_8 $ subgroup of $ G_2 $.

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