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 CWcomplex, 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 nonsimply connected manifolds with trivial first homology group?

6$\begingroup$ You've almost answered your own question. You can realise any finitely presented group as the fundamental group of a closed 4manifold. You take your 2dimensional CW complex, embed it in R^5 (dim 5 is necessary so you can separate all the 2cells) and then take the boundary of a tubular neighbourhood (and perhaps smooth off any corners). The result is a closed 4manifold with the same fundamental group as the CW complex, unless I'm mistaken. $\endgroup$– Joel FineJul 7, 2011 at 14:27

$\begingroup$ @Joel: why is the fundamental group of the boundary of a regular neighbourhood the same as that of the original complex? $\endgroup$– Neil StricklandJul 7, 2011 at 19:19

2$\begingroup$ To explain Joel Fine's answer. By transversality you can make a loop in the regular neighborhood miss the 2complex and hence a smaller regular neighborhood. Then the loop is homotopic into the boundary. If a loop in the boundary is null homotopic in the regular neighborhood the disk it bounds in the regular neighbor can similarly be made to miss the 2complex and hence is homotopic into the boundary. Thus the two complex and the boundary of the regular neighborhood have the same fundamental group. – Tom Mrowka 11 hours ago $\endgroup$– Tom MrowkaJul 8, 2011 at 9:48

$\begingroup$ 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. $\endgroup$– user16335Jul 9, 2011 at 20:18

$\begingroup$ Some of these answers have $H_1=0$ also. $\endgroup$– J.C. OttemJul 9, 2011 at 20:47
4 Answers

1$\begingroup$ The "easiest" family of examples in this class is provided by ±1surgery on any nontrivial knot (as pointed out in the page you linked): the first homology group is trivial by MayerVietoris, and it's easy to write down a presentation for the fundamental group of the manifold using SeifertVan Kampen. $\endgroup$ Jul 7, 2011 at 18:39
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 torsionfree cocompact arithmetic subgroup of $PU(2,1)$).
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 $S^1\times B^{n1}\to P_k$, remove the interior, and replace it with $B^2\times S^{n3}$. (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.

$\begingroup$ I thought you can achieve this for a manifold without boundary by just doubling what you get above. But I have to admit, the details of why this works eludes me right now. $\endgroup$ Jul 7, 2011 at 12:43

1$\begingroup$ @Donu. Indeed you can double the manifold and preserve the fundamental group if its built from handles of small index. From the presentation build handle one handles corresponding to the generators and 2handles corresponding to the relations. Doubling will add n2 and n1 handles to arrive at a closed manifold. If n>5 adding these doesn't change the fundamental group. $\endgroup$ Jul 8, 2011 at 11:36
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 fixedpoint 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 $H^1(X, \mathbb{Z})=0$.

2$\begingroup$ This is a little misleading; the vanishing of $H^1$ here is an artifact of the universal coefficient theorem. If $\pi_1 = \mathbb{Z} / 2 \mathbb{Z}$ then $H_1$ is also $\mathbb{Z} / 2 \mathbb{Z}$. But $H^1 = Hom(H_1, Z)$ is 0 because $H_1$ is torsion; a fact which makes its presence felt in $H^2$ through an Ext term. The same trick will hold for any space with torsion $H_1$; e.g., projective spaces, lens spaces, products of these... $\endgroup$ Jul 7, 2011 at 23:24

1$\begingroup$ Why misleading? For any complex projective surface the rank of $H^1(X, Z)$ is equal to $q/2$, where $q$ is the irregularity, by the Hodge decomposition and the Dolbeault isomorphism. Moreover $H^1(X, Z)$ has no torsion by the exponential sequence. Since $q=0$ for an Enriques surface, we are done. This is a complexanalytic argument, and no universal coefficient theorem is needed. Of course you can see the vanishing of $H^1$ in the way you said, but it is a matter of taste. $\endgroup$ Jul 8, 2011 at 6:52

1$\begingroup$ And, strictly speaking, the argument about the torsion is no really necessary. The vanishing of $q(X)=H^1(X,\mathcal{O}_X)$ and the inclusion given by the exponential sequence are sufficient to conclude. These arguments of course apply only for complex projective varieties, in the case of other spaces with torsion $\pi_1$ one must use the universal coefficient theorem. $\endgroup$ Jul 8, 2011 at 7:09

$\begingroup$ I agree that it is a matter of taste! Your machinery undoubtedly makes you as happy as mine does me. I'm just saying that the way that the question is phrased (although, admittedly, not the title of the question), Mari is asking for manifolds with trivial first homology (not cohomology). $\endgroup$ Jul 8, 2011 at 9:59

$\begingroup$ Needless to say, I appreciated your comment. And yes, I answered the question in the title. The question about homology seems to me also quite easy. For any finitely presented group $G$ it is easy to construct a (smooth) compact $4$manifold $X$ with $G$ as its fundamental group. Now take as $G$ a finite perfect group, for instance $A_5$. Then the abelianization of $G$ is trivial, hence $H_1(X, Z)=0$ (and consequently, also $H^1(X, Z)=0$). $\endgroup$ Jul 8, 2011 at 10:22