Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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. –  Bo Peng Mar 30 '10 at 16:59
    
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! –  Igor Belegradek Mar 30 '10 at 17:35
    
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. –  Igor Belegradek Mar 30 '10 at 22:01
    
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. –  Tom Church Mar 30 '10 at 22:33
add comment

1 Answer

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.

share|improve this answer
    
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. –  Igor Belegradek Apr 1 '10 at 12:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.