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?
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4$\begingroup$ Such families do not exist as each finite group has only finitely many conjugacy classes of representations into a fixed target Lie group. $\endgroup$– MishaCommented Feb 11, 2015 at 17:02
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$\begingroup$ @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. $\endgroup$– Paul TaylorCommented Feb 11, 2015 at 18:08
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1$\begingroup$ @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. $\endgroup$– Ben Webster ♦Commented Feb 11, 2015 at 19:01
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$\begingroup$ @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. $\endgroup$– AnonCommented Feb 12, 2015 at 4:46
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1$\begingroup$ 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? $\endgroup$– Paul TaylorCommented Feb 12, 2015 at 9:58
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2 Answers
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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.
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4$\begingroup$ 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. $\endgroup$– YCorCommented Feb 12, 2015 at 21:05
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$\begingroup$ 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 $\endgroup$ Commented Jun 8, 2015 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 $r$.
answered Feb 12, 2015 at 23:02