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I'm looking for a proof (that I can understand) of the following fact: If $K$ and $G$ are Lie groups, and $K$ is compact, then nearby homomorphisms $K\to G$ are conjugate.

That is, if $\mathrm{Hom}(K,G)$ is the set of Lie group homomorphisms, endowed with a suitable topology (I'd like to say compact-open), then the orbits of the conjugation-by-$G$ action on it are open. (Note that there are obvious conterexamples if $K$ is not compact.)

This is referred to in The space of Lie group homomorphismsThe space of Lie group homomorphisms. A reference is given there to Connor-Floyd, Differentiable Periodic Maps, Ch. VIII, Lemma 38.1.

Following the reference, we see that Connor and Floyd derive this as an easy consequence of a theorem from Montgomery-Zippin, Topological Transformation Groups, p. 216. That is, by thinking about the graph $K\to K\times G$, of a homomorphism, the statement can be deduced from the following:

  • If $K\subseteq G$ is a compact subgroup, then there exists a neighborhood $U$ of $K$ in $G$ such that for any subgroup $H\subset U$, there exists $g\in G$ such that $gHg^{-1}\subseteq K$. (I.e., all subgroups "close" to a compact subgroup are conjugate to a subgroup of it.)

Montgomery-Zippin's proof is an exercise involving geodesics in symmetric spaces, which is opaque to me and will probably always remain so. (They have statements such as: "there exists a neighborhood $U$ of $x$ such that for any geodesic in $U$, for any points $a,b,c$ in that order along the geodesic, $d(x,b)< \mathrm{max}(d(x,a),d(x,c))$" (quoting from memory, don't take it literally). I'm just a simple algebraic topologist, and sort of thing goes right over my head.)

Can anyone describe a more modern proof, or give a reference? I'm imagining such a proof will be an exercise involving the exponential map. (In fact, it seems easy to prove that any subgroup "close" to the identity is trivial in just this way.)

I'm looking for a proof (that I can understand) of the following fact: If $K$ and $G$ are Lie groups, and $K$ is compact, then nearby homomorphisms $K\to G$ are conjugate.

That is, if $\mathrm{Hom}(K,G)$ is the set of Lie group homomorphisms, endowed with a suitable topology (I'd like to say compact-open), then the orbits of the conjugation-by-$G$ action on it are open. (Note that there are obvious conterexamples if $K$ is not compact.)

This is referred to in The space of Lie group homomorphisms. A reference is given there to Connor-Floyd, Differentiable Periodic Maps, Ch. VIII, Lemma 38.1.

Following the reference, we see that Connor and Floyd derive this as an easy consequence of a theorem from Montgomery-Zippin, Topological Transformation Groups, p. 216. That is, by thinking about the graph $K\to K\times G$, of a homomorphism, the statement can be deduced from the following:

  • If $K\subseteq G$ is a compact subgroup, then there exists a neighborhood $U$ of $K$ in $G$ such that for any subgroup $H\subset U$, there exists $g\in G$ such that $gHg^{-1}\subseteq K$. (I.e., all subgroups "close" to a compact subgroup are conjugate to a subgroup of it.)

Montgomery-Zippin's proof is an exercise involving geodesics in symmetric spaces, which is opaque to me and will probably always remain so. (They have statements such as: "there exists a neighborhood $U$ of $x$ such that for any geodesic in $U$, for any points $a,b,c$ in that order along the geodesic, $d(x,b)< \mathrm{max}(d(x,a),d(x,c))$" (quoting from memory, don't take it literally). I'm just a simple algebraic topologist, and sort of thing goes right over my head.)

Can anyone describe a more modern proof, or give a reference? I'm imagining such a proof will be an exercise involving the exponential map. (In fact, it seems easy to prove that any subgroup "close" to the identity is trivial in just this way.)

I'm looking for a proof (that I can understand) of the following fact: If $K$ and $G$ are Lie groups, and $K$ is compact, then nearby homomorphisms $K\to G$ are conjugate.

That is, if $\mathrm{Hom}(K,G)$ is the set of Lie group homomorphisms, endowed with a suitable topology (I'd like to say compact-open), then the orbits of the conjugation-by-$G$ action on it are open. (Note that there are obvious conterexamples if $K$ is not compact.)

This is referred to in The space of Lie group homomorphisms. A reference is given there to Connor-Floyd, Differentiable Periodic Maps, Ch. VIII, Lemma 38.1.

Following the reference, we see that Connor and Floyd derive this as an easy consequence of a theorem from Montgomery-Zippin, Topological Transformation Groups, p. 216. That is, by thinking about the graph $K\to K\times G$, of a homomorphism, the statement can be deduced from the following:

  • If $K\subseteq G$ is a compact subgroup, then there exists a neighborhood $U$ of $K$ in $G$ such that for any subgroup $H\subset U$, there exists $g\in G$ such that $gHg^{-1}\subseteq K$. (I.e., all subgroups "close" to a compact subgroup are conjugate to a subgroup of it.)

Montgomery-Zippin's proof is an exercise involving geodesics in symmetric spaces, which is opaque to me and will probably always remain so. (They have statements such as: "there exists a neighborhood $U$ of $x$ such that for any geodesic in $U$, for any points $a,b,c$ in that order along the geodesic, $d(x,b)< \mathrm{max}(d(x,a),d(x,c))$" (quoting from memory, don't take it literally). I'm just a simple algebraic topologist, and sort of thing goes right over my head.)

Can anyone describe a more modern proof, or give a reference? I'm imagining such a proof will be an exercise involving the exponential map. (In fact, it seems easy to prove that any subgroup "close" to the identity is trivial in just this way.)

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Charles Rezk
Charles Rezk
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Nearby homomorphisms from compact Lie groups are conjugate

I'm looking for a proof (that I can understand) of the following fact: If $K$ and $G$ are Lie groups, and $K$ is compact, then nearby homomorphisms $K\to G$ are conjugate.

That is, if $\mathrm{Hom}(K,G)$ is the set of Lie group homomorphisms, endowed with a suitable topology (I'd like to say compact-open), then the orbits of the conjugation-by-$G$ action on it are open. (Note that there are obvious conterexamples if $K$ is not compact.)

This is referred to in The space of Lie group homomorphisms. A reference is given there to Connor-Floyd, Differentiable Periodic Maps, Ch. VIII, Lemma 38.1.

Following the reference, we see that Connor and Floyd derive this as an easy consequence of a theorem from Montgomery-Zippin, Topological Transformation Groups, p. 216. That is, by thinking about the graph $K\to K\times G$, of a homomorphism, the statement can be deduced from the following:

  • If $K\subseteq G$ is a compact subgroup, then there exists a neighborhood $U$ of $K$ in $G$ such that for any subgroup $H\subset U$, there exists $g\in G$ such that $gHg^{-1}\subseteq K$. (I.e., all subgroups "close" to a compact subgroup are conjugate to a subgroup of it.)

Montgomery-Zippin's proof is an exercise involving geodesics in symmetric spaces, which is opaque to me and will probably always remain so. (They have statements such as: "there exists a neighborhood $U$ of $x$ such that for any geodesic in $U$, for any points $a,b,c$ in that order along the geodesic, $d(x,b)< \mathrm{max}(d(x,a),d(x,c))$" (quoting from memory, don't take it literally). I'm just a simple algebraic topologist, and sort of thing goes right over my head.)

Can anyone describe a more modern proof, or give a reference? I'm imagining such a proof will be an exercise involving the exponential map. (In fact, it seems easy to prove that any subgroup "close" to the identity is trivial in just this way.)