Timeline for Examples of non-simply connected manifolds with trivial H^1
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| Jul 8, 2011 at 10:22 | comment | added | Francesco Polizzi | 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 $H^1(X, Z)=0$). | |
| Jul 8, 2011 at 9:59 | comment | added | Craig Westerland | 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). | |
| Jul 8, 2011 at 7:09 | comment | added | Francesco Polizzi | 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. | |
| Jul 8, 2011 at 6:59 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Jul 8, 2011 at 6:52 | comment | added | Francesco Polizzi | 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 complex-analytic 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. | |
| Jul 7, 2011 at 23:24 | comment | added | Craig Westerland | 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... | |
| Jul 7, 2011 at 12:16 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |