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Apr 1, 2010 at 12:33 comment added Igor Belegradek 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.
Apr 1, 2010 at 11:20 history answered user1688 CC BY-SA 2.5