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Let $X$ and $Y$ be compact manifolds. $p:X \rightarrow Y$ is a covering. Generally it is not true that

$$ H^* (X , \mathbb{Q} ) = H^* (Y , \mathbb{Q} )$$

For instance, if $Y$ is a sphere with $y$ handles (let $y > 1$), then $X$ is a sphere with $x$ handles. Moreover if it is not trivial covering then $x > y$.

But if $X$ and $Y$ are Lie groups and $p$ is homomorphism, then $ H^* (X , \mathbb{Q} ) = H^* (Y , \mathbb{Q} )$. It seems a bit like a miracle.

To prove it you have to consider Chevalley-Eilenberg complex. It counts the cohomology of Lie group. On the other hand it is formulated purely in terms of Lie algebra.

Question: I want a purely topological condition on $p$, $X$ and $Y$ to keep isomorphism $$ H^* (X , \mathbb{Q} ) = H^* (Y , \mathbb{Q} )$$.

For instance, what about condition "$Y$ has trivial tangent bundle"?

Peter Cotowsky
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    That parallelizability condition is not sufficient. Every orientable 3-manifold is parallelizable, but there is a double cover $S^2 \times S^1 \to \Bbb{RP}^3 # \Bbb{RP}^3$; the former has $b_1 = 1$, and the latter has $b_1 = 0$. (As a side note, your induced map on cohomology is always injective.) –  May 20 '15 at 23:52
  • The triviality of a tangent bunde is a topological condition? There exist pairs of homeomorphic non diffeomorphic smooth manifolds with non isomorphic tangeent bundles, at least http://math.stackexchange.com/questions/469992/tangent-bundles-of-exotic-manifolds – Mariano Suárez-Álvarez May 21 '15 at 04:56
  • @MarianoSuárez-Alvarez: When one says 'topological invariant' one might frequently mean 'differential topological invariant', not strictly 'homeomorphism invariant'. I don't see any particular reason to object to this use of the terminology; I find it perfectly clear in context, e.g. here. –  May 21 '15 at 07:31
  • Well it is a rather inconvenient terminology when it is inconvenient. As I mentioned, there are homeomorphic smooth manifolds with non-isomorphic tangent bundles, so it is quite conceivable that there exist pairs of manifolds which are homeomorphic and of which exactly one has trivial tangent bundle. In that case, «has trivial tangent bundle» could only very weirdly be called a topological condition. – Mariano Suárez-Álvarez May 21 '15 at 10:13
  • Having tangent bundle with zero Pontriagin classes, for example, is a property of smooth manifolds which is a topological condition, and that is a significant theorem — and its significance, of course, is, precisely that there is a non-trivial content in claiming that a condition on the tangent bundle is topological. – Mariano Suárez-Álvarez May 21 '15 at 10:15
  • @Mariano: Certainly I agree that there are diffeomorphism invaraints of smooth manifolds that are not homeomorphism invariants. I just mean that I often find far less use of the phrase 'purely topological' to mean 'homeomorphism invariant', but more often find use from 'of or relating to topology', rather than, say, geometry (i.e. 'supports a flat metric' is a diffeomorphism invariant, but I would not call it a 'topological condition'!) I think when one uses it this way - and claim that this is how it was used here - the flavor of the condition one is looking for is clear. –  May 21 '15 at 14:53

1 Answers1

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If $X \to Y$ is a finite Galois cover with Galois group $G$ then it's known that the induced map $H^{\bullet}(Y, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q})$ is injective and induces an isomorphism

$$H^{\bullet}(Y, \mathbb{Q}) \cong H^{\bullet}(X, \mathbb{Q})^G.$$

(This has nothing to do with $X$ and $Y$ being compact or manifolds.) So here the answer is that the natural map is an isomorphism iff the action of $G$ on $H^{\bullet}(X, \mathbb{Q})$ is trivial. (This has nothing to do with the tangent bundle of $X$ or $Y$.)

If $p : X \to Y$ is a finite covering map of connected Lie groups (both "finite" and "connected" are essential to what follows; I leave it to you to find counterexamples if either is dropped), then it is Galois with Galois group $G = \text{ker}(p)$, and the covering map is given by quotienting by $G$. The action of $G$ on $H^{\bullet}(X, \mathbb{Q})$ is trivial because it factors through the action of $X$, which is trivial because $X$ is connected. No need to pass to Lie algebras.

So you can expect the action of $G$ to be trivial if it factors through the action of a connected Lie group; for example, you might know that $G$ acts by isometries of some Riemannian metric on $X$ with respect to which the isometry group of $X$ is a connected Lie group. Other than that, all bets are off.

Edit: Note also that if $X, Y$ are compact manifolds then they have Euler characteristics which satisfy $\chi(X) = n \chi(Y)$ where $n$ is the degree of the cover, so a necessary condition in this case (if $n \neq 1$) is that $\chi(X) = \chi(Y) = 0$, which is of course always satisfied for compact Lie groups so no problem there. The way this is reflected in the above argument is that if $G$ is a nontrivial group acting freely on $X$ but trivially on the cohomology of $X$, then by the Lefschetz fixed point theorem $\chi(X) = 0$.

Qiaochu Yuan
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