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
Differential Geometry – First Cohomology Group
algebraic-topologydifferential-geometryhomology-cohomology
Related Solutions
Real projective space $\mathbb{RP}^n$ has fundamental group $\mathbb Z/2\mathbb Z$ for $n\geq 2$. This the quotient of the sphere $S^n$ by the antipodal action $x\sim -x$. In fact $S^n$ is a $2$-sheeted universal cover, which implies by covering space theory that its fundamental group is of order $2$.
The claim as stated is false. What is true is that if $U \subset \mathbb{R}^3$ is open, then $H_1(U;\mathbb{Z})=0$ if and only if $H^1_{dR}(U)=0$. Basically, the error lies in statement $2.$
So let's first talk about why the claim is false. It is known that there exists an embedding $K$ of the disk $D^2$ in $S^3$ such that $S^3-K$ is not simply connected. However, every embedding of any disk of any dimension in any sphere of any dimension is such that the complement is acyclic, i.e. has vanishing (reduced) singular homology. (You can see this in the chapter about Jordan's curve theorem in Bredon's book Topology and Geometry.) We can embed $S^3-K$ in $\mathbb{R}^3$ by picking a stereographic projection based on an element of $K$, and this gives us our open set that serves as a counterexample.
However, the claim that $H_1(U;\mathbb{Z})$ is torsion-free is true, and with that the result I mentioned follows, as you seem to observe in the question. If I recall correctly, a proof goes as follows: embed $U$ again in the sphere and let $C$ be the complement. By Alexander duality (also in Bredon's book), $$H_1(U;\mathbb{Z})\simeq \check{H}^1(C;\mathbb{Z}). $$ Now let $X$ be any space. By the universal coefficients theorem, $H^1(X;\mathbb{Z}) \simeq \mathrm{Hom}(H_1(X);\mathbb{Z})$, since the $\mathrm{Ext}$ part vanishes as $H_0$ is free abelian. On the other hand, $\mathrm{Hom}(H_1(X);\mathbb{Z})$ is torsion-free. Thus, $H^1(X;\mathbb{Z})$ is torsion-free. It follows that $\check{H}^1(C;\mathbb{Z})$ is a direct limit of torsion-free abelian groups, and therefore is also torsion-free. (If I recall correctly, one way to see this is by noting that over $\mathbb{Z}$ this is equivalent to being flat, and flatness is preserved by direct limits.) So $H_1(U;\mathbb{Z})$ is torsion-free.
Some further points:
- In the comments, Moishe Kohan mentions that the wikipedia article first assumes that $U$ is an open subset of $\mathbb{R}^3$ and then switches mid-proof to being an open subset of $\mathbb{R}^2$. In $\mathbb{R}^2$, it is true that $H_1=0 \iff \pi_1=0$, and thus by making minor adaptations it is true that $\pi_1=0 \iff H^1_{dR}=0$. This is due to the fact that an open subset of $\mathbb{R}^2$ must have a free fundamental group, and thus $H_1$ is the free abelian group with the same number of generators of the $\pi_1$ by Hurewicz's theorem. This, of course, does not exempt the sloppiness on the wikipedia article.
- The result that $H_1=0 \iff H^1_{dR} =0$ does not hold for general open subsets of Euclidean spaces. The simplest example is obtained by embedding $\mathbb{R}P^2$ in some $\mathbb{R}^n$ and taking a tubular neighbourhood of it. This will have trivial $H^1_{dR}$, but $H_1=\mathbb{Z}_2$. This also shows that the result is not true as soon as $n=4$, since $\mathbb{R}P^2$ embeds in $\mathbb{R}^4$.
Best Answer
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!)