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Let $G$ be a finitely generated discrete group, $H\le G$ a subgroup of finite index $d$, and let $\rho : H\rightarrow \operatorname{GL}(n,\mathbb{C})$ be a representation.

Let $\tilde{\rho} := \operatorname{Ind}_H^G\rho$ be the induced representation. I want to say that the Zariski closures of the images of $\rho$ and $\tilde{\rho}$ have isomorphic identity components. In a formula, I want to say: $$\overline{\tilde{\rho}(G)}^0 \cong \overline{\rho(H)}^0$$ as algebraic groups. Is something like this correct? I'm happy to assume that $H$ is normal in $G$. What's a good way to see this?

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  • $\begingroup$ Why would you expect this? $\endgroup$
    – LSpice
    Commented Apr 30, 2023 at 20:58

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Assume that $H$ is normal in $G$.

If $G/H$ acts non-trivially on $\rho$ by conjugation then @SashaP has given a counter-example in the comments to this answer.

$\DeclareMathOperator\GL{GL}$If $G/H$ acts trivially on $\rho$, then your guess is correct. Consider the diagonal map $\Delta : \GL_n \to \GL_{n \cdot d}$, then there exists an element $\alpha \in \GL_{nd}(\mathbb{C})$ such that $$\Delta \circ \rho = \alpha \circ \widetilde{\rho} \rvert_{H} \circ \alpha^{-1}.$$

Since $\Delta$ is a closed map, it induces an isomorphism $\overline{\rho(H)} \cong \overline{\widetilde{\rho}(H)}$. Now it suffices to show that $\overline{\widetilde{\rho}(H)}^\circ \cong \overline{\widetilde{\rho}(G)}^\circ$.

If $G = \bigcup_{i} g_i H$ for some $g_1, \dots, g_d \in G$, then also $$ \widetilde{\rho}(G) \subset \bigcup_{i} \widetilde{\rho}(g_i) \overline{\widetilde{\rho}(H)}, $$ hence $\overline{\widetilde{\rho}(H)} \subset \overline{\widetilde{\rho}(G)}$ is a finite index subgroup. In particular, $\overline{\widetilde{\rho}(H)} \subset \overline{\widetilde{\rho}(G)}$ is open and closed. It follows that there is an isomorphism of identity components $$ \overline{\widetilde{\rho}(H)}^{\circ} \cong \overline{\widetilde{\rho}(G)}^{\circ} $$ as desired.

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    $\begingroup$ $\newcommand\r{\rho}\newcommand\tr{\widetilde\r}$I am a little confused by filling in the details here. $\cong$ probably does not mean an isomorphism of groups, since $\r(H)\cup\dotsb\cup\r(H)$ does not carry a natural group structure. So I think you just mean that there is a finite-index subgroup $\tr(H)$ of $\tr(G)$ that is isomorphic to $\r(H)$. Then it is not clear to me that the cosets of $\tr(H)$ remain disjoint upon passing to the Zariski closure. That is OK, since all you need is that the closure of $\tr(H)$ has finite index in … $\endgroup$
    – LSpice
    Commented May 1, 2023 at 1:00
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    $\begingroup$ …$\newcommand\r{\rho}\newcommand\tr{\widetilde\r}$ the closure of $\tr(G)$; and some finite collection of copies of the closure of $\tr(H)$ covers $\tr(G)$ and is closed, hence covers the closure of $\tr(G)$. So that is fine. But why is the closure of $\tr(H)$ isomorphic to the closure of $\r(H)$? $\endgroup$
    – LSpice
    Commented May 1, 2023 at 1:01
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    $\begingroup$ Yes I agree, I have edited my answer accordingly. $\endgroup$
    – Lambert A'Campo
    Commented May 1, 2023 at 13:49
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    $\begingroup$ Isn't the restriction of $Ind^G_H\rho$ rather isomorphic to the sum of twists of $\rho$ under the conjugation action of $G/H$? I think that situations where these twists are different give many counterexamples to the original statement. E.g if $G=Z/2\ltimes (C^{\times})^2$ with $Z/2$ swapping the factors and $H=(C^{\times})^2$, the character $(t_1,t_2)\mapsto t_1$ of $H$ has $1$-dimensional Zariski closure of the image, while the restriction of the induced representation to $H$ is $(t_1,t_2)\mapsto diag(t_1,t_2)$ and therefore has $2$-dimensional Zariski closure of the image. $\endgroup$
    – SashaP
    Commented May 1, 2023 at 14:17
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    $\begingroup$ @SashaP, re, $(\mathbb C^\times)^2$ isn't finitely generated (right?), but I guess you could just take, say, $G = \mathbb Z/2 \ltimes \langle(\pi, 1), (1, \pi)\rangle$ or something, since we're going to close it up anyway. $\endgroup$
    – LSpice
    Commented May 1, 2023 at 16:58
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Let $V$ be the induced representation and $W$ the $H$-subspace from which it is induced. Certainly, the Zariski closure of the image of $H$ in $GL(V)$ is of finite index in the Zariski closure of the image of $G$ in $GL(V)$. Write $V=W\oplus s_{2}W\oplus\cdots\oplus s_{n}W$, where $1,\ldots,s_{n}$ is a set of coset representatives for $H$ in $G$. Then (assuming it is normal) $H$ acts on $s_{i}W$ by the conjugate representation, and the Zariski closure of the image of $H$ in $GL(V)$ is the Zariski closure of the image of $H$ in $GL(W)$ embedded "diagonally" in $GL(W)\times\cdots\times GL(s_{n}W)\subset GL(V)$. Hence the conclusion.

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  • $\begingroup$ How does this differ from @LambertA'Campo's answer? $\endgroup$
    – LSpice
    Commented May 1, 2023 at 13:51

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