Let $\ \mathrm{Hom}(H,G)\ $ be the space of Lie group homomorphisms between compact connected Lie groups $H$, $G$. What is known about homology (or homotopy) groups of $\mathrm{Hom}(H,G)$?

UPDATE: $G$ acts on $\mathrm{Hom}(H,G)$ by conjugation, and the orbits are
a) connected, as $G$ is connected,
b) closed, as easily follows from compactness of $G$, and
c) open for it is an classical result that any nearby representations of $H$ into $G$ are conjugate; this uses compactness of $H$, and a proof can be found in the book by Conner-Floyd, "Differentiable periodic maps", Chapter VIII, Lemma 38.1.

Thus $\mathrm{Hom}(H,G)$ is the disjoint union of the $G$-orbits, and the $G$-orbit that contains a representation $r$ is homeomorphic to $G/Z_G(r(H))$, where $Z_G(r(H))$ is the centralizer of $r(H)$ in $G$.

What I do not yet understand is how to see whether $\mathrm{Hom}(H,G)$ has infinitely many connected components.

UPDATE: the topology on $\mathrm{Hom}(H,G)$ is that of uniform convergence.

  • 1
    $\begingroup$ What's your topology on $\mathrm{Hom}(H,G)$? It is most likely discrete in many cases.... $G = U(1)$ and $H = U(1)$ being the simplest example. One can begin with the case where $G = U(n)$, then it's actually about representations of $H$, which is, as we know, discrete and indexed by the root lattice. $\endgroup$ – Bo Peng Mar 30 '10 at 16:59
  • $\begingroup$ Bo Peng, I just realized that nearby homomorphisms of $H$ into $G$ are conjugate in $G$, which is what you probably mean when talking about discreteness. Thanks! $\endgroup$ – Igor Belegradek Mar 30 '10 at 17:35
  • $\begingroup$ Here I deleted the comment in which I was stating that the topology on $\mathrm{Hom}(H,G)$ is that of pointwise convergence; this is incorrect, and it should be uniform convergence topology. Hats off to Tom Church for pointing this out. $\endgroup$ – Igor Belegradek Mar 30 '10 at 22:01
  • $\begingroup$ I deleted my answer below, because half of it was wrong, and the other half no longer applies after the update to the question. Thanks to Charles Rezk for pointing out my mistake. $\endgroup$ – Tom Church Mar 30 '10 at 22:33

The general criterion is:

Given connected compact Lie groups $G,H$, $\mathrm{Hom}(G,H)$ has finitely many components if and only if $H$ is semisimple or $G=1$.

Proof. If $G=1$ there is nothing to prove.

Assume that H is not semisimple and $G\neq 1$. Then there exists a 1-dimensional torus $T$ in $H$ and a closed normal subgroup $N$ of codimension 1 such that $TN=H$. Define $T=T\cap N$, so that $Z$ is finite and define $T'=T/Z$. Then $\mathrm{Hom}(T',G)$ can be identified to the set of elements in $\mathfrak{g}$ whose exponential is 1; we have already seen that it has infinitely many components.

We have a continuous restriction map $\mathrm{Hom}(H,G)\to \mathrm{Hom}(T,G)$. Its image contains $\mathrm{Hom}(T',G)$ (identified to those homomorphisms trivial on $N$), and $\mathrm{Hom}(T',G)$ is clopen in $\mathrm{Hom}(T,G)$ (indeed, let $Z$ have exponent $n$; then 1 is an isolated point in the set $K$ of elements $g\in G$ such that $g^n=1$ and $\mathrm{Hom}(T',G)$ is the fiber of $(1,\dots,1)$ for the mapping $\mathrm{Hom}(T,G)\to K^Z$ mapping $f$ to $z\mapsto f(z)$). Thus $\mathrm{Hom}(H,G)$ has infinitely many components (the proof even shows that it has a continuous map onto an infinite discrete set).

Conversely, suppose $H$ semisimple.

Consider the map $L:\mathrm{Hom}(H,G)\to\mathcal{L}(\mathfrak{h},\mathfrak{g})$ mapping $f$ to its tangent map between Lie algebras ($\mathcal{L}(\mathfrak{h},\mathfrak{g})$ denoting the whole space of linear maps). Then $L$ is injective. Since $L(f)$ is locally conjugate to $f$ by the exponential map, $L$ is continuous, and actually is a homeomorphism to its image (because if $L(f_n)$ tends to $L(f)$, then in a given compact neighborhood of 1 we have $f_n$ tending to $f$ uniformly, and this implies that $f_n$ tends to $f$ uniformly on all of $H$).

If $H$ is simply connected, then the image of $f$ is equal to the set of Lie algebra homomorphisms $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$, which is Zariski closed, and hence has finitely many components in the ordinary topology. In general (still assuming $H$ semisimple), let $\tilde{H}$ be its universal covering and $Z$ the (finite) kernel of $\tilde{H}\to H$. Recalling that the exponential is surjective, write $Z=\exp(Z')$ for some finite subset $Z'$ of $\mathfrak{h}$; then the image of $L$ is the set of elements $u$ in $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ such that for all $z\in Z'$ we have $u(\exp(z))=1$. Let $\mathfrak{g}_1$ be the set of elements $x$ in $\mathfrak{g}$ such that $\exp(x)=1$: then we have the restatement: the image of $L$ is $$\{f\in \mathrm{Hom}(\mathfrak{h},\mathfrak{g}): f(Z')\subset \mathfrak{g}_1\}.$$

Fix a faithful continuous linear representation of $G$ (in $\mathrm{SO}(n)$ for some $n$), so that we have a faithful representation of $\mathfrak{g}$ (by antisymmetric matrices): then $\mathfrak{g}_1$ is the set of elements in $\mathfrak{g}$ whose eigenvalues are in $2i\pi\mathbf{Z}$. Note that as soon as $G\neq 1$, it has infinitely many connected components (since all these eigenvalues can be achieved, using a 1-parameter subgroup whose image in $G$ is a 1-dimensional torus).

Nevertheless, observe that (still assuming $H$ compact semisimple) $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ is a compact subset of the vector space $\mathcal{L}(\mathfrak{h},\mathfrak{g})$. Indeed, $\mathfrak{h}$ admits a basis $(e_j)$ such that each $e_j$ belongs to some subalgebra $\mathfrak{h}_j$ isomorphic to $\mathfrak{so}(3)$, such that some 2-dimensional complex representation of $\mathfrak{h}_i$ maps $e_j$ to the diagonal matrix $(i,-i)$ (where $i^2=-1$). It follows from representation theory of $\mathfrak{sl}_2$ that for every $n$-dimensional complex representation $\rho$ of $\mathfrak{h}_j$ (and hence of $\mathfrak{h}$) maps $e_j$ to an element with eigenvalues in $\{-i(n-1),\dots,i(n-1)\}$. Thus the trace of $-\rho(e_j)^2$ is $\le (n-1)^2n$. Since $-\mathrm{trace}(XY)$ is a definite positive symmetric bilinear form on antisymmetric matrices, this shows that $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ is bounded, hence compact.

In particular, there exists $n_0$ such that eigenvalues of $f(z)$ for all $z\in Z'$ and $f\in\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ are in the interval $[-2i\pi n_0,2i\pi n_0]\subset i\mathbf{R}$. Thus the image by $L$ of $\mathrm{Hom}(H,G)$ is the set of $f\in\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ such that for every $z\in Z'$ we have $\prod_{k=-n_0}^{n_0} (f(z)-2i\pi k)=0$. This is a Zariski-closed subset, and hence has finitely many connected components in the ordinary topology, and hence so does the homeomorphic space $\mathrm{Hom}(H,G)$.

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  • $\begingroup$ Thank you! I now cannot recall why I asked the question. I will have to study this more carefuly to see how everything works. $\endgroup$ – Igor Belegradek May 1 '15 at 17:03

In general there are infinitely many connected components. Look at the case when both groups are the one dimensional torus, then every $k\in\mathbb Z$ gives a homomorphism by $z\mapsto z^k$. So the set is isomorphic to $\mathbb Z$ in this case. But this phenomenon has to do this central tori. In general, the groups are products of simple Lie-groups and central tori. If both groups are simple, they have only nontrivial homomorphisms, if they are isomorphic, so it boils down to determining isomorphisms, these induce isomorphisms on the Lie-algebra which preserve the Killing form. The latter is definite, hence its isometry group is compact, which should suffice to make the set of conjugacy classes finite.

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  • $\begingroup$ Thanks for the input. Perhaps there is no "easy to state" way to enumerate components of $\mathrm{Hom}(H,G)$. One extreme is when $G$, $H$ are tori; then everything boils down to $\mathrm{Hom}(T^k, S^1)$, i.e. to enumerating codimension 1 subtori in $H=T^k$. At the other extreme if $H$, $G$ are isomorphic and simple, then mod conjugation we are looking at $\mathrm{Out}(G)$, which I think is finite. $\endgroup$ – Igor Belegradek Apr 1 '10 at 12:33

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