Connected, compact, orientable smooth manifold with finite fundamental group

Question

The problem is prove that a connected, compact, orientable smooth manifold $M$ with finite fundamental group has $H^{1}_{dR}(M)=0$. The hint is to apply

Suppose $\tilde M$ and $M$ are smooth $n$-manifolds and $\pi:\tilde M\to M$ is a smooth $k$-sheeted covering map. If $\tilde M$ and $M$ are oriented and $\pi$ is orientation-preserving, show that $\int_{\tilde M}\pi^*\omega=k\int_{M}\omega$ for any compactly supported $n$-form $\omega$ on $M$.

but i don't know how this proves the result!

Any tips?

Answer 1

I don't see immediately how to use a hint specifically about top dimensional forms to deal with something related to $1$-forms (unless it was intended for you to deal with $1$-dimensional submanifolds $N$ of $M$, but you would have to prove something like: if $\int_N \omega = 0$ for every submanifold $N$, then $\omega$ is exact. This can be a bit of a hassle), so I'll allow myself to keep the core idea of that hint but to adapt it to the case of $1$-forms, while also trying to keep elementary. From here on, $\omega$ is a closed $1$-form in $M$.

Given any path $c: I \to M$ and a fixed point $p \in \pi^{-1}(c(0))$, you can lift it uniquely to the universal cover as a path $\widetilde{c}$ starting from such point $p$. It is trivial, by definition, that $\int_c \omega=\int_{\widetilde{c}} \pi^*\omega$. But the problem is that since we are aiming to gather conclusions about fundamental groups and loops, we want to handle loops, and the lift of a loop $c$ need not be a loop (of course, otherwise everything would be simply connected!).

But since there are finitely many points in $\pi^{-1}(c(0))$, we can concatenate the loop $c$ and start at where the lifted path ended each time, until some loop $l$ is formed. So, we get that $n(l)\int_c \omega = \int_l \pi^* \omega$, where we don't even know what $n(l)$ is (only that it is a non-negative integer), but it doesn't matter: since $l$ is a loop and the universal cover is simply connected, this implies that the right side is zero, and thus $\int_c \omega=0$.

But this holds for every $c$. Thus, $\omega$ is exact. Note that this also disregards orientability altogether, which is not necessary for the result to hold.

There are plenty of other ways to see this. The link provided by Arnaud in the comments of his answer is one of them. For yet another way, we can argue algebraically, since this is immediate from the Hurewicz theorem, the characterization of finite abelian groups, the universal coefficients theorem and the fact that deRham cohomology is isomorphic to real-valued singular cohomology.

Answer 2

I honestly don't know what I was thinking when I wrote that hint -- it doesn't seem to be useful at all.

A more appropriate hint would have been something like this: "If $\omega$ is a closed $1$-form on $M$, let $\widetilde\omega$ be the pullback of $\omega$ to the universal cover of $M$, and let $\tilde f$ be a potential function for $\omega$. Define $f\colon M\to \mathbb R$ by letting $f(x)$ be the average value of $\tilde f$ over the preimages of $x$, and show that $f$ is a smooth potential function for $\omega$."

This answer by @TedShifrin gives a little bit more detail about how to carry out that argument.

There's a different proof of this result in the second edition of my Introduction to Smooth Manifolds -- see Corollary 17.18.

Answer 3

Let $\tilde{M}$ be the universal cover of $M$. As you know, it can be made into a smooth manifold such that the projection $\pi : \tilde{M} \to M$ is smooth. Moreover since $\pi_1(M)$ is finite, the projection $\pi$ is a finite, $k$-sheeted covering, where $k = \# \pi_1(M)$.

Now let's use your hint and Poincaré duality. Suppose that $H^1(M) \neq 0$. Choose some nonzero class $\alpha \in H^1(M)$. Then by Poincaré duality, there exists a form $\beta \in H^{n-1}(M)$ such that $\int_M \alpha \beta = 1$. Call the form $\omega = \alpha \beta$. Thanks to your hint, you know that $\int_{\tilde{M}} \pi^*(\omega) = k \int_M \omega = k \neq 0$.

However, $\pi^*(\omega) = \pi^*(\alpha) \wedge \pi^*(\beta)$. But $\tilde{M}$ is simply connected, therefore $H^1(\tilde{M}) = 0$, and therefore $\pi^*(\alpha) \in H^1(\tilde{M})$ is necessarily zero. This is in contradiction with $\int_{\tilde{M}} \pi^*(\alpha) \wedge \pi^*(\beta) = k \neq 0$. Therefore the initial hypothesis is wrong and $H^1(M) = 0$.

Answer 4

(Ar.... this one assumes the existence of harmonic forms, so it is again an overkill) Let $[\omega] \in H^1_{dR} (M)$. We need to show that $[\omega] = 0$. Using exercise 15-9 (let's say we give $M$ a Riemannian metric), one can choose $\omega$ so that $d^*\omega = 0$. Then $*\omega$ is also closed. Now

$$\| \omega \|^2 = \int_M \omega \wedge *\omega = \frac{1}{k} \int_{\tilde M} \pi^* (\omega \wedge *\omega) = \frac 1k\int_{\tilde M} \pi^* \omega \wedge \pi^*(*\omega)$$

Since $\tilde M$ is simply connected and $\pi^*\omega$ is closed, $\pi^*\omega = df$ for some function $f : \tilde M \to \mathbb R$. Together with $d(\pi^* *\omega) = \pi^* (d*\omega) = 0$,

$$\int_{\tilde M} \pi^* \omega \wedge \pi^*(*\omega) = \int_{\tilde M}df\wedge \pi^*(*\omega) = \int_{\tilde M}d( f\pi^*(*\omega)) = 0$$

by Stokes theorem. Thus $\|\omega\|^2 = 0$ and so $\omega = 0$. Thus $H^1_{dR}(M) = \{0\}$.

Answer 5

Hint: The fact that the covering space in your hint is $k$-sheeted (so has finite fibre) and $M$ has finite fundamental group should ring a bell.