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Suppose we have a continuous family of finite subgroups of a compact Lie group G. All the subgroups are necessarily isomorphic. Alternately, we can say we have a continuous family of homomorphisms from a finite group K to G. Can we say that the images of all homomorphisms in this family land in the same conjugacy class in G. Can I have a reference or a proof?

Anon
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    Such families do not exist as each finite group has only finitely many conjugacy classes of representations into a fixed target Lie group. – Misha Feb 11 '15 at 17:02
  • @Misha, In $SO(3)$ for any integer $n$, consider the family of subgroups of rotations by multiples of $2\pi/n$, indexed by the axis of rotation. I agree that some thought needs to be put into what a "continuous family of finite subgroups" is, but this is a meaningful question. – Paul Taylor Feb 11 '15 at 18:08
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    @PaulTaylor Which Misha answered. The space of representations of a finite group into a Lie group is a discrete union of conjugacy classes, so any continuous map from a connected topological space lands inside one conjugacy class. – Ben Webster Feb 11 '15 at 19:01
  • @Ben-Webster, I am interested in what Ben Webster says - 'The space of representations of a finite group into a Lie group is a discrete union of conjugacy classes'. Is there a proof or a reference? I edited the question to make it clearer. – Anon Feb 12 '15 at 04:46
  • it's still a bit unclear because you haven't defined "continuous family". Here's a precise statement which is hidden behind Misha's answer: Let $I$ be an interval with 0, $F$ a finite group, $G$ a compact Lie group. Let $(\pi_i)_{i\in I}$ be a continuous family of homomorphisms $F\to G$ (that is, $(i,h)\mapsto\pi_i(h)$ is continuous). Then there exists a continuous map $i\mapsto g_i$ from I to G such that $\pi_i(h)=g_i\pi_0(h)g^{−1}_i$ for all $h\in F$. Besides, up to conjugation in G, there are finitely many homomorphisms $F\to G$. – YCor Feb 12 '15 at 07:47
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    The notion of "continuous family" does have a clear meaning, so long as we know the relevant topologies. We need that on the set of finite, or better compact, subspaces of a compact Hausdorff space. This is provided by the Hausdorff metric or Vietoris topology. For the conjecture to be true, (the indexing space of) the family must also be connected. However, without worrying about unfamiliar topologies, a simpler result would suffice to answer the Question: given $g:[0,1]\to G$ with $g(t)^n=id$, are $g(0)$ & $g(1)$ conjugate? @YCor's Comment seems to solve this, so why not make it an Answer? – Paul Taylor Feb 12 '15 at 09:58
  • In a general setting "continuous family" is not clear. You want a list of ambiguities? 1) Family indexed by what? a topological space? an interval? 2) does continuous mean locally injective? 3) Do you specify the group $F$ and look at representation, or do you consider the space of closed subgroups of $G$? 4) in the rep case, do you work in $Hom(F,G)$ or its quotient by conjugation, or its GIT quotient? In the case $F$ finite and $G$ compact Lie groups, all points of view are equivalent but it sort of comes after asking the basic questions. – YCor Feb 12 '15 at 10:07
  • @PaulTaylor: I voted to move to stackexchange where I'd be happy to post an answer (but I'd need to find a reference for the finiteness result). – YCor Feb 12 '15 at 10:12
  • @YCor, why not just post a full answer here? – Paul Taylor Feb 12 '15 at 10:14
  • Because it's off-topic here. – YCor Feb 12 '15 at 10:17
  • It's a (fairly) clearly stated mathematical question asked on a site for such things, so it is not "off topic". – Paul Taylor Feb 12 '15 at 11:10
  • This question is pretty similar to (actually, is a special case of) this one. Some good answers are given there, though I'm not satisfied that I know a clean proof. – Charles Rezk Feb 12 '15 at 20:54

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I do not know a reference to this off-hand, but here is a proof. Consider $Hom(K, G)$, where $K$ is a finitely-generated group. It is a real-algebraic set. Its Zariski tangent space at each point $r\in Hom(K,G)$ is isomorphic to the space of cocycles $Z^1(K, Ad\circ r)$, where $Ad$ is the adjoint representation of $G$. It is well-known (see e.g. Brown's book "Cohomology of groups") that $H^1(K, V)=0$ for each finite group $K$ and each ${\mathbb R}K$-module $V$. Therefore, $Z^1(K, Ad\circ r)= B^1(K, Ar\circ r)$ for each $r: K\to G$, $K$ is finite. It therefore, follows that each $r\in Hom(K,G)$ is locally rigid, i.e., all nearby representations are conjugate to it. (You can find details for this type of arguments e.g. in Raghunathan's book "Discrete subgroups of Lie groups", the argument itself was first enunciated by A.Weil in "Remarks on cohomology of groups"< Annals of Math, 1964.) By compactness of $Hom(K,G)$ (I am now using the classical topology), the entire space $Hom(K,G)$ is a disjoint union of finitely many $G$-orbits of representations. qed

There is an alternative argument to this one, it uses ultralimits and is technically more complicated: Consider $Hom(K, G({\mathbb C}))$. If $Hom(K, G)$ contains infinitely many conjugacy classes, then the character variety $X(K)=Hom(K, G({\mathbb C}))//G({\mathbb C})$ is noncompact and, hence, there exists a divergent sequences of representations of the finite group $K$ into $G({\mathbb C})$. By taking a suitable ultralimit, one obtains an isometric action of $K$ on an affine building which does not have a fixed point. This contradicts Cartan-Tits fixed-point theorem.

Misha
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    Here's a reference. The precise result is: Let $G$ be a compact Lie group. Let $F$ be a finite group. Then the space $Hom(F,G)$ can be viewed as the set of real points of some algebraic variety. Then the result is that all real points in each component are conjugate under $G_0$. Reference: D.H. Lee, T.S. Wu. On conjugacy of homomorphisms of topological groups. Illinois J. Math. 13 1969 694–699. – YCor Feb 12 '15 at 21:05
  • There is a more general discussion of the conjugacy of nearby fibres (beyond the compact case) in the following article: D. Coppersmith, Deformations of Lie subgroups, Trans. Amer. Math. Soc. 233 (1977), 355-366. link – Alexander Alldridge Jun 08 '15 at 21:41
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Here is Lemma 38.1 in Chapter VIII of "Differentiable periodic maps" by Conner and Floyd: if $r: H\to G$ is a Lie group homomorphism and $H$ is compact, then any Lie group homomorphism $H\to G$ that is sufficiently close to $r$ is conjugate to $r$.

Igor Belegradek
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