Is the following claim true?:
Let $G$ be an algebraic group such that $G^\circ$ is reductive. Suppose $G$ acts irreducibly on $V$. Is it true that $V$ is decomposed (written as direct sum) into $G^\circ$ components of equal dimension?
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$\begingroup$ I don't think equal dimension implies isotypic in general. Which one do you want? $\endgroup$– S. Carnahan ♦Commented Sep 25, 2014 at 1:31
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$\begingroup$ I mean equal dimension: Decompose in to irreducible components of equal dimension. $\endgroup$– VanyaCommented Sep 25, 2014 at 1:50
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$\begingroup$ It is true: take any irreducible $G^0$ submodule $W$ of the irreducible $G$-module $V$; then $V$ is the sum of $g(W)$ for $g\in G$. Hence we have a surjection from the direct sum of $g(W)$ onto $V$. This means that $V$ is a direct sum of irreps of the same dimension. Of course, isotypical is false. $\endgroup$– VenkataramanaCommented Sep 25, 2014 at 1:55
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3$\begingroup$ Itis easy to see that if $V$ is any (say finite dimensional ) module over a group $H$ which is a sum $\sum W_i$ of irreducible $H$ modules $W_i$, then $V$ is a direct sum of irreducibles. To see this, let $W$ be an $H$ subspace of $V$ which is of the largest dimension, which is a direct sum of irreducibles; then $W\neq 0$ since $W$ contains one of the $W_i$. If some $W_j$ does not lie in $W$, then $W+W_j$ is the direct sum of $W$ and $W_j$, since $W_j$ is irreducible; this contradicts maximality of $W$ and hence $W=V$. $\endgroup$– VenkataramanaCommented Sep 25, 2014 at 2:48
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1$\begingroup$ Take the standard representation $V$ of $G=N(T)$ the normalizer of $T$, the group of diagonals in $GL(V)$. Then $V$ is irreducible for $G$; as a rep of $G^0=T$, $V$ decomposes as a direct sum of one dimensional distinct characters. $\endgroup$– VenkataramanaCommented Sep 25, 2014 at 2:55
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Yes, and even in much more generality. This is a direct application of Clifford's Theorem, which I believe works for any group $G$ acting irreducibly on a finite-dimensional vector space $V$, with a reductive normal subgroup $N$ of finite index (in this case, $N = G^{0}$). Indeed, the proof given by others in the comments is one of the usual module-theoretic proofs of Clifford's Theorem, e.g. the one on wikipedia.
(As to why it's not necessarily isotypic as an $N$-module, asked as a follow-up in the comments. The $N$-irreps appearing can be twisted-equivalent to one another, twisted by an automorphism of $N$ coming from the conjugation action of $G$ on $N$.)
answered Jun 3, 2023 at 23:22
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$\begingroup$ The problem does not require that $V$ be finite dimensional. Also, Clifford theory as usually stated is for representations of abstract groups and their finite-index subgroups. It probably works just fine over an infinite field $k$ with $G(k) \to \pi_0(G)(k)$ surjective because of Zariski density of $G(k)$ in $G$, but I wouldn't be completely confident of that, or that things can't fail for $G$ with "irrational components" or over finite fields. $\endgroup$– LSpiceCommented Jul 4, 2023 at 3:43
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1$\begingroup$ I assumed that "algebraic group" meant G was finite-dimensional and that $G/G^0$ was finite. And then doesn't it follow that any irrep of $G^0$, and hence $G$, is finite-dimensional? But re: other fields, yes, I see there is potentially an issue for non-algebraically closed fields. $\endgroup$ Commented Jul 4, 2023 at 16:28
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1$\begingroup$ Re, also, now that I think about it, as long as the case where $G(k) \to \pi_0(G)(k)$ is surjective works, and as long as every irreducible representation of $G$ decomposes into finitely many components on passage to $k^\text{sep}$, I guess everything is probably fine by Galois descent. (But I haven't thought carefully.) $\endgroup$– LSpiceCommented Jul 4, 2023 at 18:38