most recent 30 from http://mathoverflow.net 2013-05-24T21:56:31Z http://mathoverflow.net/feeds/user/22668 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38098/geometrization-for-3-manifolds-that-contain-two-sided-projective-planes/93185#93185 jonathan 2012-04-05T03:38:22Z 2012-04-05T03:38:22Z

<p>If $N$ is a non-orientable closed 3-manifold then $H_3(N;Q)=0$. Since $\chi(N)=0$ the first rational Betti number must be strictly positive. Therefore there is an essential map $f:N\to{S^1}$, and so $N$ contains an essential 2-sided surface. Haken-ness then follows easily.</p>

http://mathoverflow.net/questions/71543/embedding-the-product-of-three-circles-in-the-4-sphere/93056#93056 jonathan 2012-04-04T00:42:48Z 2012-04-04T00:42:48Z

<p>No embedding of a product $M=T_g\times{S^1}$ in $S^4$ can have one complementary component $X$ with $H_1(X)=0$. For otherwise, the other component $Y$ would have $H_2(Y)=0$. But then the inclusions of $M$ and of a wedge of $2g+1$ circles into $Y$ would induce isomorphisms on the lower central series quotients of the fundamental groups, by an old theorem of Stallings. This cannot be so, as $\pi_1(M)$ has a central factor, whereas the free group $F(2g+1)$ does not.</p>

http://mathoverflow.net/questions/18661/pd3-groups-and-pd4-complexes/93052#93052 jonathan 2012-04-03T23:48:42Z 2012-04-03T23:48:42Z

<p>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.</p>