Let $V$ be a vector space and $H$ a unipotent subgroup of the group of affine transformations of $V$. Let $\Gamma$ be a lattice in $H$ (discrete and cocompact) that acts properly discontinuously on $V$ and such that $\Gamma\backslash V$ is compact. Then $\Gamma\backslash H$ and $\Gamma\backslash V$ are Eilenberg-MacLane spaces. If $\widetilde{f}:H\rightarrow V$ is the orbit map, $\widetilde{f}(h)=h(0)$, why is the induced map $f: \Gamma\backslash H\rightarrow \Gamma\backslash V$ a homotopy equivalence?

I came across this in the article by Fried, Goldman and Hirsch "Affine manifolds with nilpotent holonomy". The statement is written without proof, I guess it is something well known for the experts. Can somebody help me understand why this is true?


1 Answer 1


It is a theorem that for a group $G$, all Eilenberg–Mac Lane spaces for $G$ are homotopy equivalent. (Section 1.B in Hatcher's Algebraic Topology is my reference for Eilenberg–Mac Lane spaces; this fact is proved there.)

In fact, something stronger is true. If $X$ and $Y$ are Eilenberg–Mac Lane spaces for $G$ and $x\in X$ and $y\in Y$ are points, any isomorphism $\pi_1(X,x)\to \pi_1(Y,y)$ is induced by a a based map $h\colon(X,x)\to(Y,y)$ that is a homotopy equivalence and unique up to a homotopy fixing $x$. (This is a restatement of 1.B.9 in Hatcher.)

So, having shown that $\Gamma\backslash H$ and $\Gamma\backslash V$ are Eilenberg–Mac Lane spaces, all it remains to show is that $f$ induces an isomorphism on fundamental groups. I'd prefer to leave that step to you, although I'm happy to say more about it if you'd like. This completes the proof because the proposition I quoted guarantees that the inverse isomorphism is induced by a map that is a homotopy inverse for $f$.

Since we agree that both spaces are Eilenberg–Mac Lane spaces, let's show that $f$ induces an isomorphism on fundamental groups. Firstly, we know that the action of $\Gamma$ on $H$ is a covering space action, so given a lift of the basepoint, say $1_H \in H$ and its image in $\Gamma\backslash H$, elements of the fundamental group of $\Gamma\backslash H$ are in one-to-one correspondence between homotopy classes of paths from $1_H$ to $g.1_H = g \in H$ for $g \in \Gamma$. Fixing $g$, let $\gamma_g \colon [0,1] \to H$ be such a path. Because $\tilde f\colon H \to V$ is continuous, $\tilde f\circ \gamma_g$ is a path from $\tilde f(\gamma_g(0))=\tilde f(1_H) = 0\in V$ to $\tilde f(\gamma_g(1)) = \tilde f(g) = g(0)$. But this homotopy class of paths is exactly those that correspond to $g$ as an element of the fundamental group of $\Gamma\backslash V$! Is this helpful?

  • $\begingroup$ Thanks, I was not precise enough in my question. I am aware of the fact that it is enough to prove that the map induced on fundamental groups is an isomorphism (the groups are, of course, isomorphic), but that is precisely the step that I cannot seem to do. I would be glad if you could help me with this. $\endgroup$
    – User1407
    Commented Aug 5, 2019 at 17:06
  • $\begingroup$ @User1407 I tried to add a little about that bit $\endgroup$ Commented Aug 6, 2019 at 2:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .