First Cohomology Group

Question

Is it true that the first cohomology group of a differentiable manifold with finite fundamental group is trivial? If so, could you explain why? Thanks very much

Answer 1

It depends on what you mean by "cohomology group". If you mean singular cohomology with integer coefficients, note that for homology with integer coefficients there is an isomorphism $H_1(X;\Bbb Z) \cong \pi_1^{ab}(X) = \pi_1(X)/[\pi_1(X),\pi_1(X)]$. In particular, if $\pi_1$ is abelian, $H_1(X;\Bbb Z) \cong \pi_1(X)$. Now the universal coefficient theorem provides an isomorphism $H^1(X;\Bbb Z) \cong \text{Hom}(H_1(X;\Bbb Z),\Bbb Z) = \text{Hom}(\pi_1(X),\Bbb Z)$ (the last equality because $\Bbb Z$ is abelian). The Ext term vanishes because $H_0$ is always free. Because $H_1(X;\Bbb Z)$ (or $\pi_1$, if you like) is finite, all homomorphisms to the integers are trivial, so $H^1(X;\Bbb Z) = 0$.

You might also mean the de Rham cohomology of $X$. The de Rham theorem says that $H_{dR}^1(X) \cong H^1(X;\Bbb R)$, the singular cohomology of $X$ with coefficients in $\Bbb R$; the universal coefficient theorem, then, implies that this is isomorphic to $\text{Hom}(H_1(X;\Bbb Z),\Bbb R).$ The same argument as before shows that this is trivial.

There is also a more direct argument showing that $H_{dR}^1(X)$ is trivial; pick a closed 1-form $\omega$ on $X$. Let $\tilde X$ be the universal cover of $X$, and $f: \tilde X \to X$ the universal covering map. Then $\tilde \omega := f^*\omega$ is closed, and necessarily exact Suppose $h: \tilde X \to \Bbb R$ has $d\tilde h = \tilde \omega$. Recall that $G:=\text{Deck}(\tilde X,X) \cong \pi_1(X)$; consider $$h'(x) = \frac{1}{|G|}\sum_{g \in G} \tilde h(g(x)).$$ Now show that $h'$ descends to a real-valued function $h$ on $X$ with $dh = \omega$.

(An identical argument actually shows, more generally, that if $f: X \to Y$ is a finite covering of smooth manifolds, the induced map on cohomology $H_{dR}^*(Y) \to H_{dR}^*(X)$ is an injection. This is far from true for singular cohomology with integer coefficients!)

Answer 2

If you mean the de Rham cohomology, then it is simply because the first de Rham cohomology group is isomorphic to the group of homomorphisms from the fundamental group to $\mathbb R$.