Rigid representations of triangle groups

TL;DR

This paper generalizes a conjecture about how finite groups of Lie type can be generated, using geometric analysis of representation varieties and tangent spaces.

Contribution

It introduces a new geometric approach to understanding generation properties of finite groups of Lie type, extending previous conjectures.

Findings

Proves a generalized conjecture on group generation

Uses geometric properties of representation varieties

Analyzes tangent spaces to derive results

Abstract

We prove a generalization of a conjecture of C. Marion on generation properties of finite groups of Lie type, by considering geometric properties of an appropriate representation variety and associated tangent spaces.

Equations

$$T(a_1,a_2,\dots ,a_n)\stackrel{\text{def}}{=}⟨x_1,\dots ,x_n\mid x_1^{a_1}=\dots =x_n^{a_n}=x_1{x}_2\dots x_n=1⟩.$$

$$\mathrm{Hom}(T(a_1,\dots ,a_n),G)\cap (C_1\times \dots \times C_n).$$

$$d{f}_{\rho }(X)=c_1^{\prime }\cdot \alpha ({\gamma }_1)+\cdots +c_n^{\prime }\cdot \alpha ({\gamma }_n)=c_1^{\prime }\cdot X_1+\cdots +c_n^{\prime }\cdot X_n.$$

$$d(f\circ h)$$

Full text

The first author is supported by the Alexander von Humboldt Foundation.

1. Introduction

Let $G$ be a linear algebraic group over an algebraically closed field $k$ of characteristic $p\geq 0$. Given $d\in\mathbb{N}$, let $j_{d}$ be the largest dimension of a conjugacy class of elements of $G$ of order $d$. For $a,b$ and $c$ positive integers, let $T(a,b,c)$ be the triangle group given by $T(a,b,c)=\langle x,y,z\mid x^{a}=y^{b}=z^{c}=xyz=1\rangle$. In this note we prove a conjecture of Marion (cf. [8]).

Theorem 1.1**.**

Suppose that $G$ is simple and defined over a finite field $\mathbb{F}_{q}$, and suppose $j_{a}+j_{b}+j_{c}=2\dim(G)$. Then there are only finitely many $r\in\mathbb{N}$ such that the finite group of Lie type $G(q^{r})$ is a quotient of $T(a,b,c)$.

In fact our results are more general. For integers $a_{1},a_{2},\ldots,a_{n}\in\mathbb{N}$ let

[TABLE]

Recall also that if an algebraic group $G$ is defined over a finite field $\mathbb{F}_{q}$, then a Steinberg endomorphism of $G$ is a homomorphism $F\colon G\to G$ of algebraic groups such that some power of $F$ is the Frobenius endomorphism corresponding to an $\mathbb{F}_{q^{t}}$-structure of $G$ $(t\geq 1)$. In particular, the fixed-point subgroup $G^{F}$ is finite whenever $F$ is a Steinberg endomorphism. Let $\mathcal{S}_{G}$ be the set of isomorphism classes of groups of the form $G^{F}$, as $F$ varies over Steinberg endomorphisms of $G$.

Theorem 1.2**.**

Suppose that $G$ is connected and reductive, and let $a_{1},a_{2},\ldots,a_{n}\in\mathbb{N}$ such that $\sum_{i=1}^{n}j_{a_{i}}=2(\dim G-\dim Z(G))$. Then only finitely many members of $\mathcal{S}_{G}$ are a quotient of the group $T(a_{1},\ldots,a_{n})$.

Remark 1.3*.*

Theorem 1.1 was proved ‘generically’ in [5, Theorem 1.7], under the hypothesis that the defining characteristic of $G$ is coprime to the integers $a$, $b$, $c$, and to the determinant of the Cartan matrix of $G$. Using this, the authors show that a certain cohomology group vanishes, which implies that a certain map of tangent spaces is surjective. Our results follow Theorem 2.1 below, which is in a similar vein but with weaker hypotheses. Indeed, one can show that under the restriction on the characteristic given in [5], each appropriate generating set for a finite group $H=G(q^{r})$ will satisfy the hypothesis of Theorem 2.1 below.

2. Representation varieties

We start with a few preliminaries. For a linear algebraic group $G$ we write $\mathfrak{g}$ for the Lie algebra $\operatorname{Lie}(G)$ considered as the adjoint $G$-module, and $g\cdot X$ for $\operatorname{Ad}_{g}(X)$ if $g\in G$ and $X\in\mathfrak{g}$. Given $H\leq G$, we set $D_{H}(\mathfrak{g})=\{X-h\cdot X\mid h\in H\}$ and let $\mathfrak{g}_{H}=\mathfrak{g}/D_{H}(\mathfrak{g})$, the space of coinvariants of $H$ on $\mathfrak{g}$. The Zariski closure of $H$ is denoted $\overline{H}$.

Let $\Gamma$ be a finitely generated group. Define $\operatorname{Hom}(\Gamma,G)$ to be the set of homomorphisms from $\Gamma$ to $G$. It is well-known that $\operatorname{Hom}(\Gamma,G)$ is an affine algebraic variety over $k$, called the representation variety. To see this concretely, let $\gamma_{1},\ldots,\gamma_{n}$ be a set of generators for $\Gamma$. Then the map $\rho\mapsto(\rho(\gamma_{1}),\ldots,\rho(\gamma_{n}))$ embeds $\operatorname{Hom}(\Gamma,G)$ into $G^{n}$ as the closed subvariety of $n$-tuples satisfying the defining relations of $\Gamma$. In particular, if $F_{n}$ denotes the free group on $n$ generators then we may (and shall) identify $\operatorname{Hom}(F_{n},G)$ with $G^{n}$ via a choice of free generators. If $\gamma\in\Gamma$ then we have a morphism of varieties $\operatorname{ev}_{\gamma}\colon\operatorname{Hom}(\Gamma,G)\to G$ given by $\operatorname{ev}_{\gamma}(\rho)=\rho(\gamma)$.

If $C_{1},\ldots,C_{n}$ are conjugacy classes of $G$, let $\mathbf{C}=C_{1}\times\cdots\times C_{n}$. Define $f\colon\mathbf{C}\to G$ by $f(c_{1},\ldots,c_{n})=c_{1}\ldots c_{n}$ and set $\mathbf{C}_{0}=\{(c_{1},\ldots,c_{n})\in\mathbf{C}\mid c_{1}\ldots c_{n}=1\}=f^{-1}(1)$. The group $G$ acts on $G^{n}$ by conjugation, and this action stabilises both $\operatorname{Hom}(\Gamma,G)$ and $\mathbf{C}_{0}$.

An element $\rho=(g_{1}\ldots,g_{n})\in\operatorname{Hom}(\Gamma,G)$ determines an $n$-tuple of classes $\mathbf{C}=C_{1}\times\cdots\times C_{n}$ (where $g_{i}\in C_{i}$ for each $i$), and we may regard $\mathbf{C}$ as a (quasi-affine) subvariety of $G^{n}$. If $\rho\in\mathbf{C}_{0}$ then we say that $\rho$ is rigid if the orbit $G\cdot\rho$ is an open subset of $\mathbf{C}_{0}$. It is immediate that $G\cdot\rho$ is open if and only if $G^{\circ}\cdot\rho$ is open. Note that if the generators of $\Gamma$ multiply to $1$, then $\rho\in\mathbf{C}_{0}$ for all $\rho$.

Recall that a subgroup $H$ of a connected reductive group $G$ is said to be $G$-irreducible ($G$-ir) if $H$ is not contained in any proper parabolic subgroup of $G$; in this case, if $H$ is closed then $H^{\circ}$ is reductive and the centraliser $C_{G}(H)$ is finite. For a subset $X\leq\operatorname{Hom}(\Gamma,G)$, write $X_{\rm ir}$ for the set of homomorphisms $\rho\in X$ such that $\overline{\rho(\Gamma)}$ is $G$-ir.

Theorem 2.1**.**

*Suppose that $G$ is semisimple and that there exists $\rho=(c_{1},\ldots,c_{n})\in(\mathbf{C}_{0})_{\rm ir}$, such that the closed subgroup $H=\overline{\langle c_{1},\ldots,c_{n}\rangle}$ satisfies $\mathfrak{g}_{H}=0$. Then the following are equivalent.

(a) $\rho$ is rigid.

(b) $\dim(C_{1})+\cdots+\dim(C_{n})=2\dim(G)$.

(c) $\dim(C_{1})+\cdots+\dim(C_{n})\leq 2\dim(G)$.*

We postpone the proof of Theorem 2.1 to the next section, and now use it to prove Theorem 1.2 and hence Marion’s conjecture (Theorem 1.1) as follows. We first reduce to the case that $Z(G)$ is finite, i.e. $G$ is semisimple. We then replace $G$ by its simply-connected cover. Neither of these operations changes the quantity $\dim G-\dim Z(G)$, nor the dimension of the conjugacy class of a fixed element. In order to make use of these reductions, we prove a slightly stronger result than Theorem 1.2 for simply-connected semisimple groups.

For our first lemma, note that if $H$ is a closed, characteristic subgroup of a connected reductive group $G$, then a Frobenius morphism of $G$ induces a Frobenius morphism of $G/H$, and if $H$ is connected then $G^{F}/H^{F}\to(G/H)^{F}$ is an isomorphism [7, Proposition 23.2].

Lemma 2.2**.**

Let $G$ be connected and reductive, let $Z=Z(G)^{\circ}$ be the identity component of the centre of $G$, and write $\underline{G}=G/Z$. If infinitely many distinct members of $\mathcal{S}_{G}$ are images of some $T(a_{1},\ldots,a_{n})$, then so are infinitely many members of $\mathcal{S}_{\underline{G}}$.

Proof.

We claim that if $F$ is a Frobenius morphism, then the order of $Z^{F}$ is bounded in terms of $G$ and $(a_{1},\ldots,a_{n})$ but independently of $F$; this implies that an infinite sequence of distinct groups of the form $G^{F}$ gives rise to an infinite sequence of distinct groups of the form $\underline{G}^{F}$.

The group $[G,G]\cap Z\leq Z([G,G])$ has finite order bounded by the Lie type of $[G,G]$. The quotient by this subgroup sends $Z$ onto a connected central subgroup of $G/([G,G]\cap Z)$, whose fixed points under the induced Frobenius morphism contain the image of $Z^{F}$. Thus it suffices to prove the claim assuming that $[G,G]\cap Z=1$, so that $G=[G,G]\times Z$. But in this case, if $G^{F}$ is a quotient of $T(a_{1},\ldots,a_{n})$ then so is $Z^{F}$, and $|Z^{F}|$ is then bounded by the size of the abelianization of $T(a_{1},\ldots,a_{n})$, which is bounded in turn by $a_{1}a_{2}\ldots a_{n}$. ∎

Thus in the setting of Theorem 1.2 it suffices to prove that only finitely many members of $\mathcal{S}_{\underline{G}}$ are images of $T(a_{1},\ldots,a_{n})$. Note that the dimension of a conjugacy class of $G$ is equal to the dimension of its image in $\underline{G}$.

Next, let $\widetilde{G}$ be the simply-connected cover of $\underline{G}$, with canonical projection $\pi\colon\widetilde{G}\to\underline{G}$. Then $\pi$ gives rise to an obvious map $\widetilde{G}^{n}\to\underline{G}^{n}$ which we also denote by $\pi$, abusing notation. We identify $\operatorname{Hom}(T(a_{1},a_{2},\ldots,a_{n}),\underline{G})$ with the set of $n$-tuples $\mathbf{h}=(h_{1},h_{2},\ldots,h_{n})\in\underline{G}^{n}$ such that $h_{i}^{a_{i}}=1$ for each $i$ and $h_{1}h_{2}\ldots h_{n}=1$. Then $\pi^{-1}(\operatorname{Hom}(T(a_{1},a_{2},\ldots,a_{n}),\underline{G}))$ is a set of $n$-tuples in $\widetilde{G}^{n}$ of elements whose product lies in $Z(\widetilde{G})$. We identify these with $(n+1)$-tuples $(\widetilde{h_{1}},\widetilde{h_{2}},\ldots,\widetilde{h_{n}},\widetilde{h_{n+1}})\in\widetilde{G}^{n+1}$ such that $\widetilde{h_{1}}\widetilde{h_{2}}\ldots\widetilde{h_{n}}\widetilde{h_{n+1}}=1$ and $\widetilde{h_{n+1}}\in Z(\widetilde{G})$. Since $Z(\widetilde{G})$ is finite, the set of such $(n+1)$-tuples occurring this way is contained in a finite union of subsets of the form $\operatorname{Hom}(T(\widetilde{a_{1}},\widetilde{a_{1}},\ldots,\widetilde{a_{n}},\widetilde{a_{n+1}}),\widetilde{G})$ for integers $\widetilde{a_{i}}$ such that $\widetilde{a_{n+1}}\leq|Z(\widetilde{G})|$ and $a_{i}\leq\widetilde{a_{i}}\leq|Z(\widetilde{G})|a_{i}$ for $1\leq i\leq n$. Moreover the dimension of the conjugacy class $\widetilde{G}\cdot h_{i}$ equals that of $\underline{G}\cdot h_{i}$ for $1\leq i\leq n$, and is [math] for $i=n+1$.

The Steinberg endomorphism $F$ is induced from a Steinberg endomorphism of $\widetilde{G}$ (cf. [7, Proposition 22.7]), and $\pi^{-1}(\underline{G}^{F})$ contains (but may be strictly larger than) the subgroup $\widetilde{G}^{F}$. The image $\pi(\widetilde{G}^{F})$ has index in $\underline{G}^{F}$ equal to $Z(\widetilde{G}^{F})$ (see [7, §24.1]), in particular this index is bounded by $|Z(\widetilde{G})|$, independently of $F$.

So suppose infinitely many members of $\mathcal{S}_{G}$ are images of $T(a_{1},\ldots,a_{n})$. Taking the image in $\underline{G}$ and then lifting to $\widetilde{G}$, by the above discussion we obtain infinitely many distinct subgroups of $\widetilde{G}$ which are generated by an $(n+1)$-tuple of elements in $\operatorname{Hom}(T(\widetilde{a_{1}},\ldots,\widetilde{a_{n}},\widetilde{a_{n+1}}),\widetilde{G})\cap(C_{1}\times\ldots\times C_{n},C_{n+1})$, where $C_{i}$ are conjugacy classes of $\widetilde{G}$ with $\sum_{i=1}^{n+1}\dim C_{i}\leq\sum_{i=1}^{n}j_{a_{i}}$. Each of the subgroups occurring this way contains a subgroup of the form $\widetilde{G}^{F}$, with index bounded independently of $F$.

It therefore suffices show the following.

Lemma 2.3**.**

Let $G$ be a simply-connected semisimple group and let $\mathcal{S}$ be a collection of subgroups of $G$, such that each member of $\mathcal{S}$ contains a subgroup of the form $G^{F}$ for a Steinberg endomorphism $F$ of $G$, and such that for any infinite sequence of pairwise distinct members of $\mathcal{S}$, the corresponding sequence of subgroups $G^{F}$ has unbounded order.

If $C_{1},\ldots,C_{n}$ are conjugacy classes of $G$ such that $\sum\dim C_{i}\leq 2\dim G$, then only finitely many members of $\mathcal{S}$ are generated by elements in an $n$-tuple in the set

[TABLE]

Proof.

We claim that all but finitely many subgroups $H\in\mathcal{S}$ are $G$-irreducible and satisfy $\mathfrak{g}_{H}=0$, which allows us to apply Theorem 2.1. By the hypothesis on $\mathcal{S}$, it suffices to prove this for the subgroups $G^{F}$ themselves, for if $G^{F}\leq H$ and $G^{F}$ is $G$-irreducible and has no coinvariants on $\mathfrak{g}$, then this also holds for $H$.

The hypothesis on $G$ implies that $G$ is the direct product of its simple factors, and its Lie algebra $\mathfrak{g}$ is the direct sum of the Lie algebras of these factors. Moreover the Lie algebra of a simply-connected simple group has no coinvariants (see for instance [3, Table 1]). Thus $\mathfrak{g}_{G}=0$. Next, let $F$ be a Steinberg endomorphism of $G$, and consider the image of $G^{F}$ under projection to some direct factor $G_{1}$ of $G$. From the classification of Steinberg endomorphisms of semisimple groups (cf. [2, Corollary 1.5.16]) it follows that this image is a finite group of Lie type, in fact it is equal to $G_{1}^{F^{m}}$ for some $m\geq 1$ such that $F^{m}(G_{1})=G_{1}$. Thus if $\mathfrak{g}_{G^{F}}\neq 0$ for infinitely many pairwise non-isomorphic subgroups $G^{F}$, then there exists a simple factor of $G$ (say $G_{1}$ without loss of generality) and an infinite collection of subgroups $G_{1}^{F_{1}}$, $G_{1}^{F_{2}}$, $\ldots$ for Steinberg endomorphisms $F_{i}$ of $G_{1}$, such that $G_{1}^{F_{i}}$ has coinvariants on the Lie algebra $\mathfrak{g}_{1}$ of $G_{1}$ for all $i$. On the other hand, for all but finitely many $F_{i}$, the $G_{1}$-module composition factors of $\mathfrak{g}_{1}$ all remain irreducible upon restriction to $G_{1}^{F_{i}}$ [4, Theorem 2.11], with non-isomorphic factors remaining non-isomorphic. Then results relating cohomology of $G_{1}^{F_{i}}$ to that of $G_{1}$ [1, Theorem 7.1], combined with results relating group cohomology to submodule structure [6, Proposition 1.4], tell us that for all but finitely many values of $i$, each $G_{1}^{F_{i}}$-submodule of $\mathfrak{g}_{1}$ is a $G_{1}$-module. In particular, this implies that $G$ has coinvariants on its Lie algebra; a contradiction.

Note also that, again with finitely many exceptions, the group $G_{1}^{F_{i}}$ is quasi-simple and does not map non-trivially into a proper Levi subgroup of $G_{1}$; it follows that with finitely many exceptions, a group of Lie type $G^{F}$ is $G$-irreducible.

We have therefore shown that, with finitely many exceptions, the subgroups $H\in\mathcal{S}$ are $G$-irreducible and satisfy $\mathfrak{g}_{H}=0$. Let $\mathcal{S}^{\prime}$ denote the subset of $\mathcal{S}$ satisfying these conditions. If $H\in\mathcal{S}^{\prime}$ and $\rho=(h_{1},\ldots,h_{n})\in\operatorname{Hom}(T(a_{1},a_{2},\ldots,a_{n}),G)$ such that $\rho(T(a_{1},a_{2},\ldots,a_{n}))=H$, we may regard $\rho$ as an element of $(\mathbf{C}_{0})_{\rm ir}$ where we choose $C_{i}=G\cdot h_{i}$ for $i=1,\ldots,n$. By Theorem 2.1, $\rho$ is rigid. Hence if $H_{1}$, $H_{2}\in\mathcal{S}^{\prime}$ and $\rho_{1}$, $\rho_{2}\in\operatorname{Hom}(T(a_{1},a_{2},\ldots,a_{n}),G)$ have respective images $H_{1}$ and $H_{2}$, then $G\cdot\rho_{1}$ and $G\cdot\rho_{2}$ are nonempty disjoint open subsets of $\mathbf{C}_{0}$, hence cannot both be contained in the same irreducible component. But the variety $\mathbf{C}_{0}$ has only finitely many irreducible components, and the desired result follows. ∎

3. Tangent spaces and 1-cocycles

In this section we prove Theorem 2.1. Let $H$ be a closed subgroup of $G$, and recall that $D_{H}(\mathfrak{g})=\{X-h\cdot X\mid h\in H\}$.

Lemma 3.1**.**

Suppose that $H=\overline{\langle h_{1},\ldots,h_{n}\rangle}$ for some $h_{1},\ldots,h_{n}\in H$. Then $D_{H}(\mathfrak{g})$ is the subspace spanned by $\{X-h_{i}\cdot X\mid X\in\mathfrak{g},1\leq i\leq n\}$.

Proof.

Let $V$ be the subspace $\{X-h_{i}\cdot X\mid X\in\mathfrak{g},1\leq i\leq n\}$. Let $m\in\mathbb{N}$. Suppose $X-h\cdot X\in V$ for all $X\in\mathfrak{g}$ and all $h\in H$ such that $h$ is a word in the $h_{i}$ of length at most $m$. Let $h$ be a word in the $h_{i}$ of length $m+1$ and let $X\in\mathfrak{g}$. We can write $h=h_{i}h^{\prime}$ for some $1\leq i\leq m$ and some $h^{\prime}$ that is a word of length $m$ in the $h_{i}$. Then $X-h\cdot X=X-h_{i}h^{\prime}\cdot X=X-h^{\prime}\cdot X+h^{\prime}\cdot X-h_{i}h^{\prime}\cdot X=X-h^{\prime}\cdot X+h^{\prime}\cdot X-h_{i}\cdot(h^{\prime}\cdot X)\in V$. It follows by induction on $m$ that $X-h\cdot X\in V$ for every $h\in\langle h_{1},\ldots,h_{n}\rangle$. The result now follows. ∎

We recall some facts from the theory of 1-cocycles and representation varieties, see for instance [9]. Let $\rho\in\operatorname{Hom}(\Gamma,G)$. The group $\Gamma$ acts on $\mathfrak{g}$ by $\gamma\cdot X=\operatorname{Ad}_{\rho(\gamma)}(X)$; we denote the resulting $\Gamma$-module by $\mathfrak{g}(\rho)$. Recall that a 1-cocycle is a function $\alpha\colon\Gamma\to\mathfrak{g}$ such that $\alpha(\gamma\gamma^{\prime})=\alpha(\gamma)+\gamma\cdot\alpha(\gamma^{\prime})$. We define $Z^{1}(\Gamma,\mathfrak{g}(\rho))$ to be the set of 1-cocycles; this is a vector space over $k$ with respect to the obvious pointwise operations. There is a unique linear map $A_{\rho}$ from the tangent space $T_{\rho}(\operatorname{Hom}(\Gamma,G))$ to $Z^{1}(\Gamma,\mathfrak{g}(\rho))$ such that if $X=(X_{1},\ldots,X_{n})\in G^{n}$ then $A_{\rho}(X)(\gamma_{i})=dR_{g_{i}}^{-1}(X_{i})$ for $1\leq i\leq n$, where $g_{i}\overset{\textup{def}}{=}\rho(\gamma_{i})$ and $R_{g}\colon G\to G$ is multiplication by $g$. Moreover, if $\gamma\in\Gamma$, $X\in T_{\rho}(\operatorname{Hom}(\Gamma,G))$ and $\alpha=A_{\rho}(X)$ then $d(\operatorname{ev}_{\gamma})(X)=dR_{g}(\alpha(\gamma))$.

Now consider the special case $\Gamma=F_{n}=\langle\gamma_{1},\ldots,\gamma_{n}\rangle$. Then for conjugacy classes $C_{1},\ldots,C_{n}$, with $\mathbf{C}=C_{1}\times\ldots\times C_{n}$ as before, the product map $f\colon C_{1}\times\cdots\times C_{n}\to G$ coincides with $\operatorname{ev}_{\gamma}$, where $\gamma=\gamma_{1}\ldots\gamma_{n}$. Let $c\in\mathbf{C}_{0}=f^{-1}(\{1\})$. Set $c_{1}^{\prime}=1$ and $c_{i}^{\prime}=c_{1}\ldots c_{i-1}$ for $1\leq i\leq n$. For any $X\in T_{\rho}(\operatorname{Hom}(\Gamma,G))$,

[TABLE]

Define $h\colon G^{n}\to\mathbf{C}$ by $h(g_{1},\ldots,g_{n})=(g_{1}c_{1}g_{1}^{-1},\ldots,g_{n}c_{n}g_{n}^{-1})$. Let $(Y_{1},\ldots,Y_{n})\in\mathfrak{g}^{n}$. It is easily seen that $dh_{1}(Y_{1},\ldots,Y_{n})=(dR_{c_{1}}(Y_{1}-c_{1}\cdot Y_{1}),\ldots,dR_{c_{n}}(Y_{n}-c_{n}\cdot Y_{n}))$. We see that the derivative of $f\circ h$ is given by

[TABLE]

Proof of Theorem 2.1.

Let $V\subseteq\mathfrak{g}$ be the image of $df$. Let $H$ be the subgroup of $G$ generated by $c_{1},\ldots,c_{n}$. Then $H$ is also the subgroup generated by $c_{1}^{\prime}c_{1}(c_{1}^{\prime})^{-1},\ldots,c_{n}^{\prime}c_{n}(c_{n}^{\prime})^{-1}$. It follows from Lemma 3.1 and the previous displayed equation that $V=D_{H}(\mathfrak{g})$. By hypothesis, $\mathfrak{g}_{H}=0$, so ${\rm im}(df)=D_{H}(\mathfrak{g})=\mathfrak{g}$. This shows that $df$ has rank equal to $\dim(G)$. Since $\rho$ is a smooth point of $\mathbf{C}$, it follows that $df$ has rank equal to $\dim(G)$ at all points in an open neighbourhood of $\rho$. We deduce that ${\rm im}(f)$ is dense in $G$ and $f$ is smooth at $\rho$; in particular, any irreducible component of $\mathbf{C}_{0}$ that contains $\rho$ has dimension $\dim(\mathbf{C})-\dim(G)$.

Let $\mathbf{D}$ be an irreducible component of $\mathbf{C}_{0}$ that contains $G^{\circ}\cdot\rho$. Then $\dim(G^{\circ}\cdot\rho)=\dim(G)$ as $\rho$ belongs to $(\mathbf{C}_{0})_{\rm ir}$, so $\dim(\mathbf{D})\geq\dim(G)$. But $\dim(\mathbf{D})=\dim(\mathbf{C})-\dim(G)$ from the previous paragraph, so $\dim(\mathbf{C})\geq 2\dim(G)$, with equality if and only if $\dim(G)=\dim(\mathbf{D})$, which is the case if and only if $\rho$ is rigid. The result follows. ∎