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This question makes sense for any topological group $G$, but I'd particularly like to know the answer for $G$ a compact, connected Lie group. $G$ acts on itself by conjugation. One has the equivariant ...

Let $n$ be an even number. Let $X$ be a $n$-dimensional complex projective manifold with $H^{2m+1}(X,\mathbb{Z})=0$, for all $0\leq m\leq n-1$. $H^{2m}(X,\mathbb{Z})$ is a free $\mathbb{Z}$-module ...