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Given a compact simple Lie group $G$ of dimension $n$ acting irreducibly on some vector space $\mathbb{C}^r$ or $\mathbb{R}^r$, I wish to deduce the (real) dimension of the orbit space $G\cdot v$, for $v$ a generic vector in the irrep.

For certain groups and fundamental reps, counting is easy, but I would like to find a more systematic approach. For example, $SO(n)$ in the fundamental representation acts naturally on the $n-1$-sphere, and being simple the orbit is the whole sphere and so the orbit space is $n-1$ dimensional, also seen as the fact the normalization of a vector is preserved.

For $SU(n)$ fundamental reps it's similar, we have a $2n$ dimensional space, but the norm of a vector is fixed, and so the orbit of a generic vector in $\mathbb{C}^n$ is $2n-1$ real dimensional.

For other groups and reps its not clear if there is a more systematic way to proceed, for generic irreps and groups, such as $Spin(n)$, $Sp(n)$, or even say the $5$ dimensional rep of $SO(3)$?

Is there a nice way to show this? My first thought was merely the rank of a generic transformation in some irrep, but the image of the transform will span the whole space, just not be the whole space itself.

Edit: the accepted answer discusses stabilizers and not orbits; see Callum's answer as to why these are equivalent.

Craig
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  • In the case you speak of - of $SO(n)$ acting on $\mathbb{R}^n$ with vector $v:=e_1$ and orbit $SO(n)v \cong SO(n)/SO(n)_V \cong SO(n)/SO(n-1) \cong S^{n-1}$ - in this case the vector $v$ is not a generic vector - do you have a comment? Why do you use the notion "generic vector" when the example you are studying is not? – hm2020 Feb 17 '25 at 09:22
  • My misuse of language is twofold. Firstly I'm a physicist and so I try my best (imperfectly) to translate the problems I encounter into a language I know the mathematicians will find palatable. Secondly the spirit of the question is really something more projective. If the orbit of a vector is an $n-1$ sphere, than the stabilizer group is the same stabilizer subgroup for any scalar multiple of the vector. – Craig Feb 17 '25 at 15:47
  • What I really want to know (in part) is if there are vectors for which the stabilizer subgroup I find would be truly different, i.e. non conjugate. For $SO(n)$ this is the same as making sure the orbit covers a sphere, I.e. if there are any orbit types which are not diffeomorphic. I asked a question like this earlier to no avail, so I tried finding a different point of view and it worked, as Moishe Kohan realized what I was after and found a near perfect reference. – Craig Feb 17 '25 at 15:49
  • In my example with $G:=SL(V)$ and $Sym^2(V)$, the two vectors $v:=e_1^2, w:=e_1e_2$ give different stabilizer groups: $G_v=P$ and $G_w=N_G(T)$. You must look for a similar example for $SO(n)$. – hm2020 Feb 17 '25 at 16:03
  • Note: for $G:=SL(V)$ it follows the quotient $G/G_v$ is projective iff $P_v$ is parabolic, which is iff $v$ is a highest weight vector. – hm2020 Feb 17 '25 at 16:05
  • Oh I see now. I assumed you were talking about the complexified groups. For noncompact real forms of groups we will have distinct orbits through various 'little groups' and so these would need to be classified explicitly (physics terminology of spacelike, light like, timelike etc) But for compact or complexified group actions this does not seem to happen, every vector seemingly can be mapped to any other to get orbits up to diffeo. – Craig Feb 17 '25 at 16:20
  • the finite dimensional irreducible $SO(n)$-modules are classified, and if $V$ is such an irreducible module with $dim(V)>> dim(SO(n)):=d$ this cannot happen for dimension reasons - there can be no dense open suset $U \subseteq V$ with $dim(U)=d$. – hm2020 Feb 17 '25 at 16:22

3 Answers3

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First of all, this problem is reduces to the one for representations of simple complex Lie groups (by taking the complexification). In the complex case, there will be more orbit types though. The description of representations with trivial stabilizers of generic orbits is in table at the end of the book

Popov, V. L.; Vinberg, Eh. B., Invariant theory, Algebraic geometry. IV: Linear algebraic groups, invariant theory, Encycl. Math. Sci. 55, 123-278 (1994); translation from Itogi Nauki Tekh., Ser. Sovrem. Probl. Mat., Fundam. Napravleniya 55, 137-309 (1989). ZBL0789.14008.

For every simple group $G$ there are only finitely many irreducible representations $(V,G)$ where stabilizers of generic vectors are nontrivial. Finiteness of stabilizers of generic vectors is equivalent to the inequality $\dim V > \dim G$ (see Theorem 7.9 in their book). The exceptions are listed in the table as well as stabilizers of generic vectors.

In case you care, in the case of general algebraically closed fields the problem is also solved, in a series of recent papers by Garibaldi and Guralnick.

Moishe Kohan
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  • @Moshie Kohan - If $G:=SO(n)$ over the real numbers and if $V$ is a finite dimensional irreducible (real) $G$-module with highest weight vector $v$ (these are completely classified), do they classify the stablilizer $G_v$? If yes, what is the group $G_v$? – hm2020 Feb 15 '25 at 11:09
  • @hm2020: They classify everything, but the table is huge, goes over several pages. As I said, once dimension is large enough (roughly speaking, twice $\dim G$ times the rank), the stabilizer is trivial. – Moishe Kohan Feb 15 '25 at 11:12
  • do they prove this for groups such as $SL(n), GL(n)$ as well? – hm2020 Feb 15 '25 at 11:15
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    @hm2020: The case of $GL(n)$ is a bit tricky because it's reductive and I am not sure, but for $SL(n)$ - yes. As I wrote, they solved the problem for all simple groups (over $\mathbb C$). But most proofs are in the references to multiple papers. – Moishe Kohan Feb 15 '25 at 11:18
  • This is a problem. The example in my answer holds for any $Sym^d(V)$ where $dim(V)=2$. If you choose a highest weight vector $v_d \in Sym^d(V)$ it will (for all $d \geq 1$) have non trivial stablilizer $P_d \subseteq SL(V)$ and $SL(V)/P_d \cong \mathbb{P}^1$ in the $d$-uple embedding in $\mathbb{P}(Sym^d(V)^)$. There is a closed immersion $\phi_d: SL(V)/P_d \rightarrow \mathbb{P}(Sym^d(V)^)$. – hm2020 Feb 15 '25 at 11:21
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    @hm2020: This means that the highest weight vector is not generic. I was writing about generic orbits. As for highest weight vectors, I would have to take a closer look at their book, but you can do the same. – Moishe Kohan Feb 15 '25 at 11:26
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    This table is great! The book will surely prove a worthwhile reference. Thank you. – Craig Feb 16 '25 at 06:00
  • @Craig: You are welcome. – Moishe Kohan Feb 16 '25 at 08:59
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By the orbit-stabiliser theorem we have an bijection between the orbit $G\cdot v$ and $G/G_v$ where $G_v$ is the stabiliser. Under certain conditions such as the orbit being open, the orbit being closed or the action being proper (see here) this bijection is a diffeomorphism. In your case, any continuous action of a compact group on a smooth manifold is proper so this is immediate.

Now we simply compute $\dim G \cdot v = \dim G/G_v = \dim G - \dim G_v$. Thus we just need to find the dimension of the stabiliser of $v$.

Callum
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    Aha, this question arose precisely because I am actually interested in the dimension of stabilizer subgroups for various irreps. So the question is just as hard. I'm looking for something somewhat systematic, it is rather cumbersome to do this explicitly for every irrep, for each group. – Craig Feb 13 '25 at 23:14
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    @Craig Stabilisers are probably easier to work with than orbits in general but finding generic facts about all irreps seems quite hard here. Already if you just consider the action of $SU(n)$ on $\bigwedge^k \mathbb{C}^n$ there are many different size orbits. Maybe there are some results about principal orbits here but I'm not familiar with them. – Callum Feb 13 '25 at 23:25
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Note: If $k$ is a field of characteristic zero and if $V:=k\{e_1,e_2\}$ and if $G:=SL(V)$ it follows $W:=Sym^2(V):=k\{e^2_1,e_1e_2,e^2_2\}$ and $v:=e^2_1$ is a "highest weight vector". The orbit $Gv \subsetneq P(W^*)$ is isomorphic to $\mathbb{P}^1$. The representation $Sym^2(V)$ is irreducible. There is an isomorphism $G/P \cong \mathbb{P}^1$ where $P$ is the subgroup stablilizing $v$. When $v$ is a highest weight vector in a finite dimensional irreducible module, it follows the stablilizer $P$ is a parabolic subgroup and these are classified when $G$ is reductive (I believe this holds for $SL(W), GL(W)$ for $W:=k^n$ but I do not have a precise reference). You get an isomorphism $Gv \cong G/P$ and there are dimension formulas for $P$. A closed subgroup $P \subseteq G$ (if $G:=SL(V)$) is parabolic iff there is a flag of subspaces $W_i \subseteq W$ with $dim_k(W_i):=n_i$ and $\sum_i n_i =n$

$$ (0)= W_1 \subsetneq W_2 \subsetneq \cdots \subsetneq W_l = W $$

where $P$ is the subgroup stablilizing the flag $W_i$.

For compact groups such as $SO(n)$ I have not seen similar results but Fulton/Harris book has some information for $SL(n), GL(n)$. In the book they discuss the general case as well.

In the above example: If $w:=e_1e_2$ you may check that $G_w=N_G(T)$ is the normalizer of the diagonal torus $T \subseteq G$. Hence $G/G_w \cong G/N_G(T)$ and $dim(G/G_w)=2$.

hm2020
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    This example has a noncompact group but the question is about compact groups. In fact all $G/P$ for $G$ complex, $P$ parabolic can aldo be realised as a homogeneous space for the compact real form of $G$ instead but this is not obvious. – Callum Feb 14 '25 at 11:23
  • @Callum - I believe the result is well known in the case when $G:=SL(W), GL(W)$ and when $v$ is a highest weight vector in a finite dimensional irreducible representation. Hence one may look for similar results for other reductive groups such as $SO(n)$. For other vectors I believe there are no general classifications or formulas. – hm2020 Feb 14 '25 at 14:55