1
$\begingroup$

$\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\S{S}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PSU{PSU}\newcommand{\irr}{\mathrm{irr}}\DeclareMathOperator\T{T}\DeclareMathOperator\A{A}\DeclareMathOperator\N{N}$Is the $6$-dimensional $ (2,0) $ irrep of $ \SU(3) $ maximal in $ \SU(6) $?

I started wondering this the other day when I tried to write down the maximal subgroups of $ \SU(6) $, if you are curious that list of subgroups can be found exactly in the original MSE question .

It might also be of interest to you that, in a similar situation, the irreducible $ SO(3) $ subgroup is maximal in $ SO(5) $ see Maximal Closed Subgroups of $ SO(5) $.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ If you are interested in the general question of classifying the maximal subgroups of compact Lie groups, you might want to look at mathoverflow.net/questions/60315/… and the references that it cites. $\endgroup$
    – Robert Bryant
    Commented May 23 at 11:15
  • $\begingroup$ @RobertBryant Yes there are three types of maximal subgroups $ M $ of a compact (simple) Lie group $ G $. Type NS for "not simple" where the identity component of $ M $ is positive dimensional but not simple, type S for "simple" where the identity component of $ M $ is simple and type $ F $ for "finite" where $ M $ is a finite subgroup of $ G $. The type NS maximal subgroups are explicitly classified in arxiv.org/abs/math/0605784 Type $ S $ are exactly the normalizers of the groups listed as (ii) and (iii) in prop 2.3 of link.springer.com/article/10.1007/s00209-019-02324-7 $\endgroup$
    – Ian Gershon Teixeira
    Commented May 23 at 12:16
  • $\begingroup$ continued... Type S groups listed under (ii) are just the natural $Sp(n/2)$ and $SO(n)$ subgroups of $SU(n)$ which is fine and explicit enough, but (iii) just says "a compact simple Lie group acting irreducibly on V...not isomorphic to a classical group on V." The reference points out that some subgroups of this type are maximal but some are not however it is not clear how to tell the difference. This question is about exactly such a case, the irreducible $SU(3) \subset SU(6) $, which turns out to be maximal. By contrast the irreducible $SO(3) \subset SO(7)$ is not maximal because of $G_2$. $\endgroup$
    – Ian Gershon Teixeira
    Commented May 23 at 12:30

1 Answer 1

Reset to default
10
$\begingroup$

The Lie algebra of $SU(6)$ splits as the Lie algebra of $SU(3)$ plus a $27$-dimensional irreducible representation. (This follows from Pieri's rule: The adjoint representation is formed from the tensor product of the dual representation and the original representation, modulo the trivial subrepresentation. The dual has highest weight $(0,0,-2)$ so by Pieri's rule the tensor product is a sum of irreps with highest weights $(2,0,-2), (1,0,-1),(0,0,0)$. The $(0,0,0)$ is the trivial and the $(1,0,-1)$ is the adjoint so the remainder is irreducible.)

Thus the Lie algebra of any proper subgroup containing $SU(3)$ must be the Lie algebra of $SU(3)$, so any proper subgroup containing $SU(3)$ is the normalizer of $SU(3)$. But the centralizer of $SU(3)$ is just the scalars which are contained in the image of $SU(3)$ and the outer automorphism group of $SU(3)$ consists of a single element of order $2$ which does not fix this representation, so the centralizer is $SU(3)$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Just to make sure I'm following, so you are tensoring the $ (2,0) $ irrep of $ SU(3) $ with its dual the $(0,2) $ irrep of $ SU(3) $ and then find that you get a direct sum of the $ (2,2), (1,1) $ and $ (0,0) $ irrep. So the action by conjugation of this $ SU(3) $ subgroup of $ SU(6) $ on the Lie algebra of $ SU(6) $ gives a direct sum of a $ (1,1) $ irrep of $ SU(3) $, which is the adjoint rep for the Lie algebra of $ SU(3) $, with a $ (2,2) $ irrep of $ SU(3) $. $\endgroup$
    – Ian Gershon Teixeira
    Commented May 23 at 0:00
  • $\begingroup$ And this argument works in general for the case $ H $ is a connected irreducible subgroup of $ SU(n) $ and tensoring the natural rep of $ H $ with the dual of the natural rep of $ H $ gives a direct sum with 1 trivial factor one adjoint rep of $ H $ and one other irrep. For example this could be used to prove that $ Sp(2) $ (or rather its normalizer) is a maximal subgroup of $ SU(4) $. Is that a correct read of your argument? $\endgroup$
    – Ian Gershon Teixeira
    Commented May 23 at 0:06
  • $\begingroup$ @IanGershonTeixeira That all sounds right to me. $\endgroup$
    – Will Sawin
    Commented May 23 at 0:57
  • $\begingroup$ ok slight addendum, if $ I_6 $ denotes the $ 6 \times 6 $ identity matrix then the irreducible $ SU(3) $ subgroup has the scalar matrix $ \zeta_3 I_6 $ but it doesn't have any central elements of order $ 6 $ so it does not contain the scalar matrix $ \zeta_6 I_6 $, so I guess to be maximal technically we would have to add that $ \zeta_6 I_6 $ in. So I agree with you that the normalizer of $ SU(3) $ is maximal and that the normalizer of $ SU(3) $ equals the centralizer, but I think the full centralizer does include the $ \zeta_6 I_6 $ which is not in $ SU(3) $. $\endgroup$
    – Ian Gershon Teixeira
    Commented May 23 at 12:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .