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Let $G$ be a compact, simply-connected, and simple Lie group. Let $H$ be a closed subgroup of $G$ that has the same centralizer as the center of $G$. Is there a nice classification of such subgroups?

Edit: Given Misha's example, it seems that classifying such subgroups for all simple groups is hard. So let me pose the following question:

Are there a maximal torus $T$ and $h \in H$ such that $h\in T$ and $h$ is not fixed by any non-trivial element of the Weyl group corresponding to the chosen torus?

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  • $\begingroup$ Do you mean: "Is there a closed proper subgroup $H$ such that $C_G(H) = Z(G)$?" (equivalently, $C_G(H) = C_G(G)$) $\endgroup$
    – S. Carnahan ♦
    Commented Feb 13, 2014 at 2:09
  • $\begingroup$ Yes, that's what I meant. $\endgroup$
    – user46897
    Commented Feb 13, 2014 at 2:20
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    $\begingroup$ $A_5$ in $SU(2)$, assuming you are using the standard notion of simple Lie group. $\endgroup$
    – Misha
    Commented Feb 13, 2014 at 4:36
  • $\begingroup$ @Misha Thanks for your example. I edited my post to two questions which are probably more approachable. $\endgroup$
    – user46897
    Commented Feb 13, 2014 at 21:02
  • $\begingroup$ I guess the second question is obviously wrong. $\endgroup$
    – user46897
    Commented Feb 13, 2014 at 23:58

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For $G=SU(n)$, the answer is given by Schur's lemma: a subgroup $H$ of $G$ satisfies $Z_G(H)=Z(G)$ if and only if it acts irreducibly on $\mathbb{C}^n$, that is, if and only if it is not contained in a subgroup $(U(p)\times U(q))\cap SU(n)$. This gives a rather simple criterion.

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    $\begingroup$ Sure! But I don't how to use this classification to answer questions like 1 and 2 above? $\endgroup$
    – user46897
    Commented Feb 13, 2014 at 21:03

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