Is the irreducible $ SU(3) $ subgroup of $ SU(6) $ maximal?

Question

Is the 6 dimensional $ (2,0) $ irrep of $ SU(3) $ maximal in $ SU(6) $?

For those of you who are interested in context, I started wondering this the other day when I tried to write down the maximal subgroups of $ SU(6) $. My guess so far is that the full list of maximal (proper closed) subgroups of $ SU(6) $ is:

Type I (normalizer of maximal connected subgroup) \begin{align*} & U(5) \cong S(U(5) \times U(1)) \\ & S(U(4) \times U(2)) \\ & S(U(3) \times U(3))\rtimes S_2 \\ & 6 \circ_2 Sp(3) \\ & 6 \circ_2 SO(6) \\ & 6 \circ_3 SU(3)_{irr} \end{align*} Type II (finite maximal closed subgroup, for the last 2 groups GAP subscripts are used to label the center and the outer automorphisms when multiple groups of this structure description exist) \begin{align*} & 6.A_7 \\ &6.PSL(3,4).2_1 \\ &6_1.PSU(4,3).2_2 \end{align*} Type III (normalizer of a subgroup which is connected but not maximal connected) \begin{align*} & N(T^6)=S(U(1) \times U(1) \times U(1) \times U(1) \times U(1) \times U(1)) \rtimes S_6\\ &S( U(2) \times U(2) \times U(2) ) \rtimes S_3\\ \end{align*}

Note on notation. $ \rtimes $ means split extension (semidirect product). $ \cdot $ means nonsplit extension. $ \circ $ denotes central product, in most cases here we have $ 6 \circ_2 H $, which is just the group generated by $ H $ and $ \zeta_6I $ but that group is not a direct product since already $ -I \in H $, we get a central product essentially with three $ H $ components. Similar idea for $ 6 \circ_3 SU(3)_{irr} $ having two components.

Here $ N $ denotes normalizer. Recall that a positive dimensional (type I and type III above) maximal subgroup of a simple Lie group equals the full normalizer of its identity component.

https://arxiv.org/pdf/math/0605784.pdf classifies all maximal closed subgroups of $ SU(n) $ whose identity component is not simple (here trivial counts as simple). According to table 5 the maximal closed subgroups of $ SU(4) $ of this type are:

The normalizer of the maximal torus (row 4 table 5, $ \ell=6, p=1 $) $$ N(T)=S(U(1) \times U(1) \times U(1) \times U(1)) \rtimes S_6 $$ and (row 4 of table 5, $ \ell=3, p=2 $) $$ S( U(2) \times U(2) \times U(2) ) \rtimes S_3 $$ As well as (row 1 table 5, $ p=5,q=1 $ ) $$ S(U(5) \times U(1) )\cong U(5) $$ and (row 1 table 5, $ p=4,q=2 $ ) $$ S(U(4) \times U(2) ) $$ and the normalizer of $ S(U(3) \times U(3))= \{\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}:A,B\in U(3),det(A)det(B)=1 \} $ which is a split extension (row 1 table 5 $ p=q=3 $) $$ < S(U(3) \times U(3)),SWAP_{\oplus}> \cong S(U(3) \times U(3)) \rtimes S_2 $$ where the normalizing matrix $ SWAP_{\oplus}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $ swaps the two blocks in the direct sum.

Next, we consider maximal closed subgroups with nontrivial simple connected component.

By dimension, such a subgroup would be isogeneous to $ SU(2),SU(3),Sp(2), G_2, SU(4), SO(7), Sp(3), SU(5), SO(8) $ of dimensions $ 3,8,10,14,21,21,24,28 $ respectively.

Of these the only one with 6d irreps are: 6d irrep of $ SU(2) $, the $ (2,0) $ 6d irrep of $ SU(3) $, fundamental irrep of $ Sp(3) $,

Of these only $$ 6 \circ_2 Sp(2)=<\zeta_6 I,Sp(3)> $$ is maximal subgroup of $ SU(6) $.

Even dimensional irreps of $ SU(2) $ are always symplectic so all $ SU(2) $ subgroups of $ SU(6) $ are contained in a conjugate of $ Sp(3) $. See

Understanding the 4 dimensional irrep of $ SU_2 $

Finally we consider subgroups with trivial connected component. These are finite since $ SU(6) $ is compact. To be maximal they must at least be primitive. For example there is a very large $ 6 \circ_2 2.J_2 $ subgroup of $ SU(6) $ but it is not maximal because it is not even Lie primitive: it is contained in $ 6 \circ_2 Sp(3) $. Also there is an $ A_7 $ subgroup that is not Lie primitive, it is contained in $ SO(6) $ since it is the standard $ A_{n+1} $ subgroup of $ SO(n) $ arising from the deleted permutation representation.

Even Lie primitive subgroups may not be maximal if they are contained in another larger Lie primitive finite subgroup. For example there is a subgroup $ 3.A_7 \subset 6.PSU(4,3) \subset SU(6) $ which is Lie primitive but not maximal.

A maximal finite subgroup which is irreducible in the adjoint representation is always a maximal closed subgroup. This includes the following subgroups The central product $$ 6.A_7 $$ of order $ 6(2,520)=15,120 $ (maximal closed since it is maximal finite and a 2-design) $$ 6.PSL(3,4).2_1 $$ of order $ 6(20,160)2 $ (maximal closed since it is maximal finite and a 3-design). $$ 6_1.PSU(4,3).2_2 $$ of order $ 6(3,265,920)2 $ (maximal closed since it is maximal finite and a 3-design).

For references on designs and maximality see Finite maximal closed subgroups of Lie groups

This is consistent with the fact that a maximal $ 2 $-design group is maximal closed ( all $ 3 $ designs are $ 2 $ designs).

Answer 1

This is adapted from the answer by Will Sawin given here https://mathoverflow.net/a/467269/148524

The irreducible $ SU(3) $ subgroup of $ SU(6) $, corresponding to the $ (2,0) $ irrep of $ SU(3) $, is a projectively maximal subgroup of $ SU(6) $, but the actual maximal subgroup is the one generated by adding in the scalar matrix $ \zeta_6 $ ($ SU(3) $ already has the scalar matrix $ \zeta_3 $ hence the full maximal group has shape $ 6 \circ_3 SU(3) $ where $ \circ $ denotes a central product).

Consider the following argument. The adjoint irrep of $ SU(6) $ irrep is just the natural irrep tensored with its dual, and with a trivial irrep removed, so you get dimension $ 6^2-1=35 $ as expected. Restricting this representation to the $ SU(3) $ subgroup you are tensoring the $ (2,0) $ rep of $ SU(3) $ with its dual the $ (0,2) $, which gives a direct sum of a $ (2,2) $ irrep with a $ (1,1) $ irrep and a $ (0,0) $ irrep, then you remove the trivial irrep to get that the adjoint rep of $ SU(6) $ restricted to this $ SU(3) $ subgroup is a direct sum of the $ (2,2) $ and $ (1,1) $ irreps of $ SU(3) $.

Recall that the irrep $ (p,q) $ of $ SU(3) $ has dimension $ \frac{1}{2} (p+1)(q+1)(p+q+2) $. So we can even verify that $ \frac{1}{2} (3)(3)(6) +\frac{1}{2} (2)(2)(4) = 35 $.

Anyway, the $ (1,1) $ irrep of $ SU(3) $ is the adjoint rep of $ SU(3) $ and any closed subgroup containing $ SU(3) $ would have Lie algebra closed under the $ SU(3) $ conjugation action but we just showed the conjugation action decomposes into $ SU(3) $ irreps as $ (2,2) \oplus (1,1) $ thus the irreducible $ SU(3) $ subgroup of $ SU(6) $ is a maximal connected subgroup and so its normalizer must be a maximal subgroup. The normalizer must equal the centralizer since this copy of $ SU(3) $ is not fixed by the outer automorphism of $ SU(3) $ (I guess the intuition for that is that this irrep has Frobenius-Schur indicator $ 0 $, i.e. the irrep is complex, and the outer automorphism is basically complex conjugation so we shouldn't expect it to fix a complex irrep. Thus $ SU(3) $ irrep is a maximal connected subgroup and its normalizer is therefore a maximal subgroup and since the normalizer is equal to the centralizer since we have ruled out any outer automorphisms then we can conlude that the centralized of $ SU(3) $, which is indeed $ <\zeta_6, SU(3)> $ as remarked above, is a maximal subgroup of $ SU(6) $.