Examples of non-simply connected manifolds with trivial H^1 - MathOverflow

Examples of non-simply connected manifolds with trivial H^1

2011-07-07T11:39:29Z

It is known that, if a topological space is simply connected,its first homology group vanishes. The converse is not true, since for every presentation of a (say, finite) perfect group G we can construct a CW-complex, via generators and relations, having G as a fundamental group. Are there such examples in the class of topological or differentiable manifolds? In other words, does there exist a non-simply connected manifolds with trivial first homology group?

Answer by Matthew Kahle for Examples of non-simply connected manifolds with trivial H^1

2011-07-07T12:06:37Z

The classical examples are homology spheres.

Answer by Francesco Polizzi for Examples of non-simply connected manifolds with trivial H^1

2011-07-07T12:16:56Z

Enriques surfaces.

These are complex projective surfaces (hence, real $4$-manifolds) with $p_g(X)=q(X)=0$, obtained by taking the quotient of a $K3$ surface (which is simply connected) by a fixed-point free involution.

So, if $X$ is such a surface we have $\pi_1(X)=\mathbb{Z}/ 2 \mathbb{Z}$.

On the other hand, for any compact complex surface $X$, the first cohomology group $H^1(X, \mathbb{Z})$ injects into $H^1(X, \mathcal{O}_X)= \mathbb{C}^{b^1(X)}$ by the standard exponential sequence of sheaves

$$0 \to \mathbb{Z} \to \mathcal{O}_X \stackrel{\textrm{exp}}{\longrightarrow} \mathcal{O}_X^* \to 0$$

(in particular, it follows that $H^1(X, \mathbb{Z})$ has no torsion).

Since for an Enriques surface $X$ we have $b^1(X)=\frac{q(X)}{2}=0$, we have $H^1(X, \mathbb{Z})=0$.

Answer by Neil Strickland for Examples of non-simply connected manifolds with trivial H^1

2011-07-07T12:26:42Z

If you are willing to use manifolds with boundary then the question is easy. For any finitely presented group $G$ you can build a finite simplicial complex $X$ with $\pi_1(X)=G$, then embed $X$ in a simplex and let $M$ be a regular neighbourhood of $X$ in the second barycentric subdivision; this will be a manifold with boundary homotopy equivalent to $X$.

If you want to restrict to smooth closed manifolds then the problem is harder, but I think that the answer is the same. Fix a sufficiently large number $n$ (I think $5$ will do) and let $P_k$ be the connected sum of $k$ copies of the $n$-torus. By a small exercise with the van Kampen theorem, $\pi_1(P_k)$ is the free product of $k$ copies of $\mathbb{Z}^n$. Thus, for any finitely presented $G$ there is an epimorphism $\pi_1(P_k)\to G$ for some $k$, with finitely generated kernel. Each generator of the kernel can be represented by a map $u:S^1\to P_k$, which we can assume to be an embedding by a transversality argument. If the normal bundle to $u$ is trivial then we can thicken it to an embedding $S^1\times B^{n-1}\to P_k$, remove the interior, and replace it with $B^2\times S^{n-3}$. (In other words, we perform surgery on $u$). This gives a new manifold, and using van Kampen again we see that the new $\pi_1$ is obtained from the old one by killing $u$. After repeating this process for each generator we get a smooth closed manifold with $\pi_1=G$.

I am not sure what to do if the normal bundle of $u$ is nontrivial, but I doubt that this is a serious problem. I also think that I have seen a more efficient construction in the literature, but I do not remember it at the moment.

Answer by J.C. Ottem for Examples of non-simply connected manifolds with trivial H^1

2011-07-07T14:12:19Z

Fake Projective planes

These are smooth complex projective surfaces with the same betti numbers as $\mathbb{CP}^2$, but with infinite fundamental group $\pi_1(X)$ (in fact it is isomorphic to a torsion-free cocompact arithmetic subgroup of $PU(2,1)$).

Answer by Luciano Mari for Examples of non-simply connected manifolds with trivial H^1

2011-07-09T20:18:27Z

Ops, I have just seen the mistake in the title. Sorry, in fact I mean H_1 and not H^1... . I would like to thank you all for your quick answers and useful examples.