# PD3 groups and PD4 complexes

I am interested at the moment in what groups can occur as the fundamental group of a 4-manifold (or more generally, a 4-dimensional CW complex) with prescribed conditions on the intersection form. I have what I am hoping is a basic homotopy theory question:

A (orientable) PD-$n$ group is a group $G$ such that the Eilenberg-Maclane space $K(G,1)$ admits "Poincare duality", i.e. there is an $n$-dimensional integer homology class in $K(G,1)$ (thought of as the "fundamental class") such that cap product with it yields an isomorphism between the corresponding cohomology and homology groups (like for closed oriented manifolds). This is more general than saying that $K(G,1)$ admits the structure of an orientable closed manifold of dimension $n$.

Let $G$ be a PD-3 group. Is there any reason why $G$ cannot be the fundamental group of an orientable PD4 complex $X$ with vanishing second homotopy group, $\pi_2(X)=0$?

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Can you explain what PD-3 and PD-4 mean? –  Andy Putman Mar 18 '10 at 21:31
By the way, you do know that every finitely presentable group is the fundamental group of a 4-manifold, right? Even a symplectic one! –  Andy Putman Mar 18 '10 at 21:35
Thanks for your interest. Yes, I knew that. I have given a rough definition of what a Poincare-duality group is. –  Raphael Mar 18 '10 at 22:55
Great! I'd have known what you meant if you just used the words "Poincare duality group" instead of the abbreviation PD. I think when asking questions in forums like this that attract a broad spectrum of mathematicians, it is best to at least make your terminology googleable. –  Andy Putman Mar 18 '10 at 23:04
What is a Poincare duality complex? I thought a Poincare complex must be aspherical by definition. If so, I do not understand the condition on vanishing of 2nd homotopy group. –  Igor Belegradek Mar 19 '10 at 0:55

$G=Z^3$ is such an example. It is $\pi_1(T^3)$ hence a PD-3 group. If $X$ is a Poincare 4-complex with fund group $Z^3$, then the injective (by Hopf) map on cohomology $H^2(G)\to H^2(X)$ cannot be onto, because its image is lagrangian for the intersection form by naturality of cup products. Dually $H_2(X)\to H_2(G)$ is not injective, and so $\pi_2(X)$ is not zero. Following this kind of idea is what math.GT/0307101 and math.GT/0608103 is based on.

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This is a beautiful answer. Just to expand on your parenthetical, since I was momentarily confused: you can turn $X$ into a $K(G,1)$ by adding 3-cells to kill $\pi_2$, then higher-dimensional cells. This shows that $H_2(X)$ surjects to $H_2(G)$, and in fact that $H_2(G)$ is the quotient of $H_2(X)$ by the image of $\pi_2(X)$. It seems this shows that any PD-3 group that surjects to $\mathbb{Z}$ is an example, since $H^2(G)$ will be nontrivial, and thus can't be all of $H^2(X)$. Perhaps this can be extended to all non-perfect groups, except I'm not sure about duality if $H^2(X)$ is torsion. –  Tom Church Mar 19 '10 at 1:31
I agree. As far as non-perfect groups, the argument doesnt extend verbatim since the intersection form is only non-degenerate after you mod out the torsion. Incidentally, the example I gave is (in a slightly different context) attributed to Kreck. I first saw it in a beautiful article of Hausmann and Weinberger. –  Paul Mar 19 '10 at 2:32
Thanks for your comments! I guess any group G such that $H^2(G;\mathbb{Z})$ contains an infinite cyclic group is not going to work, just because of the naturality of the cup product Paul mentioned and because $H^4(G;\mathbb{Z})=0$ by the assumption that $G$ is a $PD3$ group. I could imagine there being no restriction if $G$ has only 2-torsion in its second cohomology group, but of course this would need a slightly more careful thought. –  Raphael Mar 19 '10 at 6:34

Suppose that $M$ is a closed 4-manifold (or $PD_4$-complex) with fundamental group a $PD_3$-group $G$. Then $M$ cannot be aspherical. Since the homology of the universal cover $\widetilde{M}$ is 0 in degree 1 (it is simply-connected),' in degree 3 (since $H^1(G;\mathbb{Z}[G])=0$, i.e., $G$ has one end) and in degrees greater than 3 (since $G$ is infinite), $\pi_2(M)=H_2(\widetilde{M};\mathbb{Z})$ must be non-zero.

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Suppose that an infinite group $G$ is the fundamental group of a non-aspherical compact $PD(4)$ complex $X$ with $\pi_2(X)=0$. Then $G$ has to have at least 2 ends (since $H_3(\tilde{X})\ne 0$, where $\tilde{X}$ is the universal cover of $X$). Since $PD(3)$ groups are 1-ended, they are never fundamental groups of $PD(4)$ complexes $X$ with $\pi_2(X)=0$.

Furthermore, conjecturally, if $G$ is torsion-free and is isomorphic to the fundamental group of a compact $PD(4)$ complex $X$ with $\pi_2(X)=0$, then $G$ splits a free product of infinite cyclic groups and $PD(4)$ groups.

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