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Let $G$ a compact semisimple Lie group, $H$ a subgroup of $G$. Is it always possible to find an irreducible representation $R$ of $G$ such that the stabilizer of an $x\in R$ is "locally isomorphic" to $H$? I am only interested in the case when $H$ is a continuous subgroup, and "locally isomorphic" means has the same Lie algebra.

If the result is not true, what would be the simplest counterexample? What about weaker results when $R$ is not required to be irreducible, or when stabilizers of a finite set and not just of a single element are allowed?

Let me give an example of what I have in mind. I am only interested in the case when both $G$ and $H$ are finite dimensional compact Lie groups. As a representative example let's take $G=SU(3)$ and two subgroups $H_1=SU(2)$, $H_2=SU(2)\times U(1)$ (all groups over $\mathbb{C}$). I can realize $H_1$ as the stabilizer of $x=(0,0,1)^t$ in the fundamental representation of $SU(3)$. I can also realize $H_2$ as the stabilizer of $x=diag(1,1,-2)$ in the adjoint representation of $SU(3)$. Can I always do this? What would be an algorithm to construct the representation given the set of generators of $G$ which generate $H$?

Update: the answers below show that the answer is yes, if you allow reducible representations (which I don't mind). There remains a problem of how to construct $R$ concretely and simply, given the list of generators for $H$ inside $G$. Such constructions can be also extracted from the proofs below, however they do not look elementary (for me).

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  • $\begingroup$ "locally isomorphic" is clearly not a good terminology for what you mean. I don't know what you mean by "continuous subgroup" but clearly a necessary condition is that $H$ is closed. So you could rephrase your question as: "suppose $H$ is a closed connected subgroup of $G$; does there exist an irreducible representation and an $x$ whose stabilizer $G_x$ admits $H$ as unit component". $\endgroup$
    – YCor
    Commented Nov 27, 2013 at 17:20
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    $\begingroup$ Since $G$ is compact, $G$ is naturally the group of $\mathbf{R}$-points of a real linear algebraic group, and any closed subgroup is Zariski closed. By (Borel, Linear Algebraic Groups, Springer GTM 126: Theorem 5.1 page 89), $H$ is the stabilizer of a line in some representation. By compactness, some closed subgroup $H'$ of index at most 2 of $H$ acts trivially on the line, and hence $H'$ is the stabilizer of a point. Francois' answer seems to indicate you don't even have to go to index 2. But indeed both arguments do not necessarily provide irreducible representations. $\endgroup$
    – YCor
    Commented Nov 27, 2013 at 17:32
  • $\begingroup$ Thanks. However my question is both more elementary and more practical in nature. Is it really necessary to use algebraic geometry to answer it? Can't I use basic Cartan theory of roots and weights to construct such a representation explicitly? $\endgroup$
    – Slava Rychkov
    Commented Nov 27, 2013 at 18:18
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    $\begingroup$ The proof (from algebraic geometry) is practical. You start with $A$, the $\mathbf{R}$-algebra of real-valued polynomial functions on $G$; let $I$ be its ideal of functions vanishing on $H$. There exists a finite dimensional (say $d$-dimensional) $G$-invariant subspace $V$ of $A$ such that $I$ is generated by $W=V\cap I$. Then $H$ is precisely the stabilizer of $W$. This is close to what you require: to get $H$ as stabilizer of a line instead of a subspace, consider the exterior power $\Lambda^dV$. Then $H$ is the stabilizer of the line $\Lambda^dW$. $\endgroup$
    – YCor
    Commented Nov 27, 2013 at 18:36
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    $\begingroup$ I don't know... at the first place you need a way to encode a way to input a pair $(G,H)$, i.e., you need a natural way to have at your disposal an enumeration all compact semisimple groups (this is ok) and, say, their closed subgroups up to conjugacy. Then there are several steps to achieve... (compute generators for the ideal, compute generators for the $G$-submodule they generate, convert the output into standard representation-theoretic invariants...) I'm not competent enough to go ahead (and I'm not sure this is the best approach). $\endgroup$
    – YCor
    Commented Nov 27, 2013 at 19:11

2 Answers 2

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You need $H$ to be closed. The Mostow-Palais theorem (1,2,3) then gives what you want -- with "equal" in place of "locally isomorphic", but with a possibly reducible representation. I'm not aware of conditions ensuring that the representation can be chosen irreducible.

(1): http://en.wikipedia.org/wiki/Mostow-Palais_theorem

(2): http://books.google.com/books?id=oCO0xOzNLhAC&pg=PA373

(3): http://books.google.com/books?id=yqbocEpFdyQC&pg=PA104

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  • $\begingroup$ Could you be a bit more specific how to get the result I need from the Mostow-Palais theorem? I've looked at the Mostow paper and the page from the book that you mention, but I don't see how to apply these results to the group and its subgroup simultaneously ensuring the stabilizer condition. $\endgroup$
    – Slava Rychkov
    Commented Nov 27, 2013 at 18:26
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    $\begingroup$ That's because Google will not show the next page, which says: COROLLARY 2. Let $H$ be a closed subgroup of $G$. There exists an analytic linear representation of $G$ on a finite dimensional vector space $E$ and a point $v\in E$ with fixer $H$. $\endgroup$
    – Francois Ziegler
    Commented Nov 27, 2013 at 18:31
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    $\begingroup$ (Proof:) Apply Cor. 1 to the canonical operation of $G$ on the compact manifold $G/H$. This gives an analytic linear representation $\rho : G\to GL(E)$ and an embedding $\varphi : G/H \to E$ such that $\varphi(gx) = \rho(g)\varphi(x)$, $g \in G$, $x \in G/H$. Let $\bar e\in G/H$ be the class of $e \in G$, and $v = \varphi(\bar e)$ its image. For all $g\in G$, we have $\rho(g)v = v \Leftrightarrow \varphi(g\bar e) = \varphi(\bar e) \Leftrightarrow g\bar e = \bar e \Leftrightarrow g \in H$. $\endgroup$
    – Francois Ziegler
    Commented Nov 27, 2013 at 18:36
  • $\begingroup$ Thanks now I see how it works. It would be nice to know an efficient and explicit way to find $E$ in concrete situations. $\endgroup$
    – Slava Rychkov
    Commented Nov 27, 2013 at 18:52
  • $\begingroup$ I know this is false in case $G$ is not compact, but can we still say something in that case? Like further conditions that $H$ needs to satisfy? $\endgroup$
    – Nicolò Cavalleri
    Commented Oct 27, 2022 at 19:37
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Though it's not so readily available online, there is an elementary textbook treatment of the basic theory here in Representations of Compact Lie Groups by Brocker and tom Dieck (GTM 98, Springer, 1985).

1) As they note early in the book, the isotropy group $H$ of a point $v$ must be closed in an arbitrary topological group $G$ acting continuously: here $H$ is the inverse image of $v$ under the (continuous) orbit map $g \mapsto g \cdot v$.

2) As a corollary of the Peter-Weyl theorem (using representative functions), they derive easily in III, (4.6) for any compact Lie group $G$ (not necessarily semisimple): Every closed subgroup $H$ of $G$ appears as the isotropy group of an element of some $G$-module.

3) Unfortunately, since all of this theory is somewhat abstract, it doesn't seem to shed light on your question about finding an irreducible representation. However, you do have complete reducibility here of all (necessariy finite-dimensional) representations, which are usually studied in the essentially equivalent (complexified) Lie algebra setting. So if there is a counterexample it would probably be best located there. It seems hard to compute directly with group elements and group representations, but for example Willem de Graaf has worked extensively with computer algorithms for both real and complex Lie algebras.

P.S. Though Dan Mostow (who just turned 90) was on my thesis committee, I don't think you need to get into his more general results involving group actions on manifolds.

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