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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 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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You could derive this from the fixed point property for affine actions of compact groups, a la the proof of Property T. The point is that lack of local rigidity for representations of a compact group means that $H^1$ of $K$ with coefficients in Lie algebra is nonzero, which is impossible. Note also that geodesic property you are referring to is just the fact that the distance function on a totally normal neighborhood in Riemannian mld is convex. –  Misha Mar 5 '13 at 16:11
    
Check out the short survey staff.science.uu.nl/~Schat001/survey_Lie_algebras.pdf I think it might go in the direction you want. –  Claudio Gorodski Mar 5 '13 at 16:11
    
Claudio: it goes in a nice direction, but I don't see that it gets there. The claim is true for $\mathrm{Hom}(U(1),U(1))$, but false for $\mathrm{Hom}(\mathbb{R}, U(1))$, so I don't see how I can prove it purely from Lie algebra considerations. As Misha suggests, I probably need to know something about $H^1(K,\mathfrak{g})$, not just $H^1(\mathfrak{k},\mathfrak{g})$. What I'm missing is probably really easy. –  Charles Rezk Mar 5 '13 at 19:47
    
That $H^1(K,\mathgfrak{g})$ is easy: indeed a continuous 1-cocycle induces an affine continuous action of $K$ on the finite-dimensional vector space $\mathfrak{g}$, which has a fixed point iff the 1-cocycle is a 1-coboundary. Now for $K$ compact there is a fixed point (integrate along an orbit, or fix a $K$-invariant Euclidean metric and take the circumcenter of an orbit). –  YCor Mar 5 '13 at 21:47
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It looks like you need the vanishing of a kind of nonabelian continuous cohomology: If $K$ acts continuously on $G$ then any continuous 1-cocycle $K\to G$ (crossed homomorphism, $f(xy)=f(x)^yf(y)$) that is close enough to the trivial one is a coboundary, i.e. is determined by an element $a\in G$ ($f(x)=a^xa^{−1}$). –  Tom Goodwillie Mar 6 '13 at 2:52
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3 Answers

Here is a sketch of the proof expanding on my comment.

Suppose that there exists a sequence of continuous homomorphisms $\rho_i: K\to G$ which converges (uniformly) to a representation $\rho: K\to G$. Repeating the arguments from A.Weil "Remarks on cohomology of groups", Annals of Math. (1964), you obtain a cocycle $\zeta\in Z^1(K, {\mathfrak g}_{Ad\circ \rho})$. (Think of rescaling the group $G$, so that in the limit it becomes its Lie algebra ${\mathfrak g}$ and the homomorphism condition for $\rho_i$'s becomes a cocycle condition for $\zeta$.)

Note they Weyl's arguments deal with finitely-generated groups, while your group $K$ is compact and infinitely generated, so you have to work a bit more and use uniformity of convergence to guarantee that the cocycle $\zeta$ is continuous. Now, if you have a continuous $V$-valued cocycle $\zeta$ of a topological group $K$ (where $V$ is a topological vector space), it gives you a continuous affine action of $K$ on $V$ by the formula $g\cdot v= L(g)v + \zeta(g)$, where $L(g)$ is the linear action of $g$ on $V$. In your case, $V$ is the Lie algebra ${\mathfrak g}$ of $G$ and the linear action of $K$ on $V$ is via composition of $Ad$ and $\rho$. Since $K$ is compact, the action on the finite-dimensional space $V$ in question will be isometric for some choice of an inner product, thus, by taking the center of a $K$-orbit on $V$ (with respect to the invariant Euclidean metric), we conclude that the affine action of $K$ has a fixed point. In other words, the cocycle $\zeta$ is a coboundary.

Now, the space of continuous coboundaries $B^1_{c}(K, V)$ is tangent to the orbit of $\rho$ under the group $G$ acting on representations $K\to G$ via conjugation (see Weil's paper). If, again, $K$ were finitely-generated, the space of cocycles would be finite-dimensional and you could take a complementary subspace ${\mathcal H}^1(K, V)\subset Z^1(K, V)$ to $B^1(K, V)$ and postcompose the representations $\rho_i$ with the action of $G$ by conjugation, so that the sequence $\rho_i$ converges to $\rho$ in a conical neighborhood of ${\mathcal H}^1(K, V)$. This would ensure that the cocycle $\zeta$ cannot be a coboundary, thereby giving you a contradiction. In your setting, you maye have to do some analysis to make sure that this argument works (again, using uniformity of convergence).

One possible simplification would be to take a finitely-generated dense subgroup $F$ in $K$ (it always exists: The proof goes back to Hausdorff-Banach-Tarsky paradox) for the last part of the argument and argue with the restriction of the cocycle $\zeta$ and representations $\rho_i$ to $F$, thereby reducing the problem to finite-dimensional one.

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This is very helpful! Though I'm a little confused by the penultimate paragraph, since it seems that by this point we've shown that $Z^1=B^1$. Looking at Weil's paper, it seems that the deformation theory already tells us that $\mathrm{Hom}(K,G)$ is a manifold (assuming $K$ finitely generated), and so we are done once we have $H^1=0$. –  Charles Rezk Mar 7 '13 at 14:22
    
Charles: Yes, you are absolutely right. I lost track of the fact that in your setting $Hom(K, G)$ is a real-analytic variety (as a homomorphism for connected $K$ is determined by the homomorphism of Lie algebras), so things are easier than I thought. Thus, everything reduces to the fact that $H^1_{cont}(K, {\mathfrak g})=0$, as in Weil's paper. There is one issue you need to check though: Uniform convergence of representations implies $C^1$-convergence (in order to use Lie algebras). But, if you use the topology of $C^1$-convergence, this will not be a problem. –  Misha Mar 7 '13 at 23:01
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Here is a proof sketch using cohomological ideas. The argument is in four main steps:

I. General theory of families of Lie algebra homomorphisms.

II. The case of a semisimple $G$.

III. The case of a torus (here is a major gap in my argument)

IV. combining both cases.

A preliminary observation: if $H$ is the full linear group of a complex vector space, then the result is well-known, because up to conjugacy, a homomorphism $G \to H$ is given by its character; and the set of characters is a discrete subspace of the space of all smooth maps $G \to \mathbb{C}$.

I. For arbitrary Lie algebras, $Hom_{Lie -alg} (\mathfrak{g},\mathfrak{h})$ is a real algebraic variety and thus it is locally path- connected. Therefore, nearby homomorphisms can be connected by smooth families of homomorphisms (I am not entirely sure whether this is true, but it seems so).

Now consider a smooth family $f_t$, $t \in \mathbb{R}$, of Lie algebra homomorphisms. We study the problem of finding $h: \mathbb{R} \to H$ such that $f_t (X) = Ad (h(t)) f_0 (X)$ holds for all $t$ and $X \in \mathfrak{g}$. If $f_t$ is the derivative of a smooth family of group homomorphisms $\phi_t$, then $h(t)$ conjugates $\phi_0$ to $\phi_t$ and thus solves the original problem.

Let $F_t$ be the derivative of $f_t$ with respect to $t$. Differentiating the equation $[f_t X,f_t Y]=f_t [X,Y]$ shows that $F_t\in Hom (\mathfrak{g},\mathfrak{h})$ satisfies $F_t ([X,Y])= [F_t (X);Y]-[F_t (Y);X]$. This means that $F_t$ is a $1$-cocycle in the Chevalley-Eilenberg complex for $H^{\ast}(\mathfrak{g};f_t)$. By the cohomology I mean cohomology of $\mathfrak{g}$ with coefficients in $\mathfrak{h}$, viewed as a $\mathfrak{g}$-module via $f_t$.

We can consider the collection of all Chevalley-Eilenberg complexes $C^{\ast} (\mathfrak{g},f_t)$ as a complex of vector bundles on the real line; denote the vector bundles by $C^{\ast}(\mathfrak{g},f)$. The derivatives $F_t$ are a smooth family of $1$-cocycles and $[F_t]$ is a family of cohomology classes, smooth in a certain sense. I say that $[F_t]$ is uniformly trivial if there is a smooth family $H_t$ of $0$-cochains such that $[f_t (X);H_t]=F_t (X)$ for all $t$ and all $X \in \mathfrak{g}$ (this means that $d H_t =F_t$, but in a ''uniform way'').

Suppose that the cohomology class $[F_t]$ is uniformly trivial. Then

$$f_t (X) = \int_{0}^{t} F_s (X) ds = - \int_{0}^{1}[H_s;f_s (X)] ds;$$

in other words $f_t (X)$ solves the ODE $\frac{d}{dt} f_t (X) = - [H_t;f_t(X)]$ with initial value $f_0$. Another solution of the same ODE is $Ad (h(t)) f_0(X)$, where $h(t) \in H$ solves $\frac{d}{dt} h(t)= H_t$. So $f_t$ is conjugate to $f_0$. Vice versa, if $Ad (h(t)) f_0(X)$, then $[F_t]$ is uniformly trivial.

If $f_t$ is the derivative of a group homomorphism $G \to H$ and $G$ is compact, then pointwise triviality ($[F_t]=0$ for each $t$) implies uniform triviality. This is by the preliminary observation, which implies that $d_0:C^0 (\mathfrak{g},f) \to C^1 (\mathfrak{g},f)$ has constant rank and so its image is a vector bundle (pass to the complexification of $\mathfrak{h}$, which is unproblematic as we are only interested in the dimension of the invariant subspace). Thus we can pick a smooth $r: im (d_0) \to C^0 (\mathfrak{g},f)$ with $d_0 r = id$. Choosing $H_t:= r (F_t)$ solves the problem. Thus we arrive at

THEOREM: ''If $f_t: \mathfrak{g} \to \mathfrak{h}$ is a family of homomorphisms of Lie algebras and $H$ a Lie group with Lie algebra $\mathfrak{h}$, then there is a smooth map $h: \mathbb{R} \to H$ with $f_t = Ad (h(t))f_0$ if and only if the obstruction cocycle $[F_t]$ is uniformly trivial.''

ADDENDUM: ''If $G$ is a compact Lie group with Lie algebra $\mathfrak{g}$ and if $f_t$ is the derivative of a smooth family of homomorphisms $G \to H$, then pointwise triviality of $[F_t]$ implies uiform triviality.''

II.

Assume $G$ is semisimple. For each representation $V$ of $\mathfrak{g}$, we have an isomorphism $H^{\ast} (\mathfrak{g};V) \cong H^{\ast} (\mathfrak{g};V^{\mathfrak{g}})$, because of the compactness of $G$. But $H^1 (\mathfrak{g})=0$ since $G$ is semisimple, and so the cohomology class $[F_t]$ is zero, and by the addendum, it is uniformly trivial. Thus by the theorem, nearby homomorphisms are conjugate if $G$ is semisimple.

III.

Assume $G=T$ is a torus (sketch). Let $V$ be the universal cover (equal to $\mathfrak{t}$) and $\Gamma \subset V$ be the kernel; this is a lattice. Smooth families $f_t:\mathfrak{t} \to \mathfrak{h}$ are in bijection with smooth families $\psi_t: V \to H$ and induce families of group homomorphisms $g_t: \Gamma \to H$. As Misha indicates, there is a parallel obstruction theory for such families; with an obstruction in $H^{1}_{group}(\Gamma;\mathfrak{h})$. Consult Weil's paper quoted in Mishas answer.

There is the Van Est isomorphism $H_{Lie}^{\ast} (\mathfrak{t},\mathfrak{h}) \cong H_{smooth} (V,\mathfrak{h})$ to smooth group cohomology and furthermore a restriction $H_{smooth} (V,\mathfrak{h}) \to H^{\ast}_{group}(\Gamma; \mathfrak{h})$; this latter map is an isomorphism. This isomorphism should map the corresponding obstructions onto each other (this is the part of the argument where I do not know the details).

So a family of group homomorphisms $V \to H$ is constant up to conjugacy iff the restriction to the lattice $\Gamma$ is constant up to conjugacy. If the family $V \to H$ is the universal cover of a family $T \to H$, then the restriction to $\Gamma$ is constant; thus $T \to H$ is constant up to conjugacy.

IV.

Consider an arbitrary compact $G$. Without loss of generality, we can pass to a finite cover and thus assume $G=T \times K$, $T$ a torus and $K$ semisimple. Consider a family of group homomorphisms $\phi_t:G \to H$, with Lie algebra maps $f_t$ and obstruction cocycle $F_t$ as above. By the solution of the problem for $T$, the restriction $F_t|_{\mathfrak{t}}$ is uniformly trivial. But by the Künneth formula, the restriction $H^1 (\mathfrak{g} ) \cong H^1 (\mathfrak{k})\oplus H^1 (\mathfrak{t})\to H^1 (\mathfrak{t})$ is an isomorphism. Therefore, $[F_t]$ is trivial and thus uniformly trivial, again by the addendum.

Afterthought: It is probably better to study the whole question in the context of smooth cohomology. A family $\phi_t:G \to H$ should give an obstruction class in $H^{1}_{smooth} (G; \mathfrak{h})$. If $G$ is compact, this space is trivial by invariant integration.

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How's this for another sketch? I don't know if it's "modern". It's a homemade attempt by another simple algebraic topologist.

Let $f:K\to G$ and $f_1:K\to G$ be continuous homomorphisms, where $K$ is a compact group and $G$ is a Lie group. There is a continuous action of $K$ on $G$ given by $f$, which we denote by writing $g^x=f(x)^{-1}gf(x)$. Define $h:K\to G$ by $f_1(x)=f(x)h(x)$. The fact that $f_1$ is another homomorphism means that $$ h(xy)=h(x)^yh(y). $$ Call $h$ a crossed homomorphism if it satisfies this equation. Assume that $f_1$ is close to $f$. This means that $h$ is close to the constant function taking all of $K$ to the identity in $G$. We want $a\in G$ such that $f_1(x)=a^{-1}f(x)a$. This is equivalent to $$ h(x)=a^xa^{-1}.$$ So forget about $f$; the problem is to show that if $K$ acts continuously on $G$ by homomorphisms then any small crossed homomorphism $h:K\to G$ can be expressed in this way for some $a$.

Actually $G$ acts on the set of crossed homomorphisms $h$ as follows: given $a\in G$ and $h$, let $h'(x)=(a^x)^{-1}h(x)a$. So the problem is to show that if $h$ is small then some $a$ takes it to the trivial map.

First look at the case where $G$ is abelian. Switching to additive notation, we have $G$ acting linearly on a finite-dimensional vector space and the problem is to show that if $h$ is such that $$h(xy)=h(x)^y+h(y)$$ then there exists $a$ such that $$h(x)=a^x-a.$$ Let $-a$ be the average $Av_x h(x)$. Then $$ -a=Av_x h(xy)=Av_x (h(x)^y + h(y))=-a^y+h(y).$$ Thus $h(x)=a^x-a$.

Now for the general case: use linear coordinates in $G$ near the identity, writing the group multiplication as $x\ast y=x+y+Q(x,y)$ where $|Q(x,y)|\le |x||y|$ if $|x|$ and $|y|$ are small enough. So we have $K$ acting on a neighborhood of $0$ so as to preserve the multiplication $\ast$, and we have a small $h$ such that $$ h(xy)=h(x)^y\ast h(y)$$ and we want to be able to modify $h$ into $$ h'(x)=(a^x)^{-1}\ast h(x)\ast a$$ so as to make it $0$. Do it in steps. First, again let $a$ be the average of $h$. Now, modulo second-order terms, averaging $ h(xy)=h(x)^y\ast h(y)$ over $x$ again we have $a=a^y+h(y)$. That is, if we modify $h$ by this $a$ then the $h'$ that we get is zero modulo second-order terms. If $h$ is small enough, $|h(x)|\le \epsilon $, then $|h'(x)|\le \epsilon ^2$.

Repeating this infinitely often and taking a limit, surely this gives $0$ as the result of acting on the original $h$ by some $a\in G$, the infinite product of smaller and smaller elements..

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I'm worried about averaging $h(x)=h(x)^y*h(y)$ over $x$. In the local coordinates, $x\mapsto x^y$ isn't linear, and a non-linear transformation might not play well with the integral. –  Charles Rezk Mar 8 '13 at 15:24
    
That is the worrisome part, isn't it? But I believe that the error should just give more second-order terms. –  Tom Goodwillie Mar 8 '13 at 16:47
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