Let $G$ be a reductive group over $\mathbb{C}$ and $H\subseteq G$ a reductive subgroup. Let $\rho$ be a faithful irreducible finite dimensional representation of $G$ over $\mathbb{C}$. Assume that $\rho|_H$ is reducible. Then is it always the case that the centralizer $Z_G(H)$ is strictly larger than the center $Z(G)$?
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1$\begingroup$ Probably it would have been fine to edit your original question, where after a counterexample you now replaced the assumption that $H$ is a proper subgroup, with the assumption that $G$ is irreducible but not $H$ (in the original form $G$ was $\mathrm{SO}_n$ acting on $\mathbf{C}^n$ and the counterexample was the image of an odd-dimensional rep of $\mathrm{SL}_2$). $\endgroup$– YCorCommented Dec 20, 2022 at 16:47
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1$\begingroup$ And then a variation works here. Take dimension 4, $G=\mathrm{SO}_4$, and $H$ the image of the 3-dim irreducible of $\mathrm{SL}_2$ (i.e., $H$ is the upper left $\mathrm{SO}_3$). The centralizer of $H$ in $\mathrm{GL}_4$ consists of those diagonal$(a,a,a,b)$ matrices, and hence in $\mathrm{SO}(4)$ it consists of $\{\pm 1\}$, and this is precisely the center of $\mathrm{SO}(4)$. $\endgroup$– YCorCommented Dec 20, 2022 at 16:58
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4$\begingroup$ @Windi: Hmm... 24 questions but only 1 accepted answer? You must be very difficult to please. $\endgroup$– Alex M.Commented Dec 20, 2022 at 17:32
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1$\begingroup$ @Windi as Alex M. suggests, you should accept answers of questions whenever the answers are correct. Accepting correct answers is useful to everybody in this site and shows consideration to people who're trying to help you. Please take this into account. $\endgroup$– YCorCommented Dec 21, 2022 at 6:43
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1$\begingroup$ @LSpice It won't usually. I should have taken $G$ to be simple: if $H<G$ then $L(H)$ is always a submodule of $L(G)$. The question asked for the restriction to be reducible though, not irreducible. I just chose the Lie algebra because proper subgroups are guaranteed to yield reducible restrictions. $\endgroup$– David A. CravenCommented Dec 21, 2022 at 21:41
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1 Answer
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$\DeclareMathOperator\GL{GL}$No. Take $H=\GL_2$ embedded diagonally into $G=\GL_2\times \GL_2$ and take $\rho$ equal to $\mathbb C^2 \otimes (\mathbb C^2)^*$ with the natural action of $\GL_2$ on $\mathbb C^2$.
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1$\begingroup$ I think you mean "take $\rho$ to be the natural action of $G = \operatorname{GL}_2 \times \operatorname{GL}_2$ on $\mathbb C^2 \otimes (\mathbb C^2)^*$" (which then specifies how $H = \operatorname{GL}_2$ acts). To think about it another way, $\rho$ is the action of $G$ on $\operatorname{Lie}(H)$ by simultaneous left and right multiplication $(h_1, h_2)\cdot X = h_1 X h_2^{-1}$, and the restriction to $H$ is then the adjoint action, which is reducible. $\endgroup$– LSpiceCommented Dec 20, 2022 at 18:05
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