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Aug 29, 2012 at 13:57 vote accept Mark Grant
Aug 29, 2012 at 11:49 answer added Mark Grant timeline score: 9
Aug 21, 2012 at 20:54 comment added John Klein Let $n$ be odd. Then it cannot happen for $(n-1)$-connected, closed $(2n)$-manfolds, since such a manifold $M$ is up to homotopy a finite wedge of $n$-spheres with a $(2n)$-cell attached. The attaching map $\alpha\: S^{2n-1} \to \vee S^n$ is has the property that the composite with each projection $\vee S^n \to S^n$ is a map $S^{2n-1} \to S^n$ and such a map has finite order in homotopy. The projections generate the cohomology and this will imply that the square of all $n$-dimensional classes will vanish.
Aug 21, 2012 at 19:22 comment added Mark Grant @Ralph: I would like to accept your example, if you'd care to leave it as an answer.
Aug 21, 2012 at 8:06 comment added Ralph @Mark: Concerning your Künneth-Tor question: I think it has rather to do with properties of the connecting hom. /Bockstein hom. than with Tor: $c=\delta(xy)$ where $x,y$ are the one-dim. generators of the mod-2 cohomology.
Aug 21, 2012 at 7:42 comment added Ralph As the classifying space of $\mathbb{Z}/2$ is a real projective space, my comment above leads to the simple example $X=\mathbb{R}P^4 \times \mathbb{R}P^4$ (maybe $\mathbb{R}P^3 \times \mathbb{R}P^3$ or $\mathbb{R}P^4 \times \mathbb{R}P^2$ will also do - I didn't check).
Aug 21, 2012 at 1:46 comment added Mark Grant @Will: Yes, $n=6$ should suffice (long exact sequence of the pair $(X,X^{(n)})$ shows that the cohomology of $X$ injects into that of $X^{(n)}$ up to dimension $n$).
Aug 21, 2012 at 0:57 comment added Mark Grant @Ralph: This is interesting. There is a short exact Kunneth sequence for $X$, and $c$ comes from the Tor-term. It's cup square is an expression involving the torsion classes which gave birth to it, namely $a$ and $b$. Is this symptomatic of a more general phenomenon?
Aug 21, 2012 at 0:34 comment added Will Sawin In particular $n=6$, no?
Aug 20, 2012 at 22:54 comment added Ralph You can take $X=B(\mathbb{Z}/2 \times \mathbb{Z}/2)$ where $H^\ast(X;\mathbb{Z})=\mathbb{Z}[a_2,b_2,c_3]/(c^2-ab^2-a^2b,2a,2b,2c)$. If you want a finite dimensional space, you can take the $n$-skeleton of $X$ for n sufficiently large.
Aug 20, 2012 at 22:18 comment added Tom Goodwillie It can't happen for one-dimensional classes. You can see this either by universal example (second cohomology of circle is trivial) or by an explicit construction. There are examples in $H^3$, but I don't have an easy one.
Aug 20, 2012 at 22:01 history asked Mark Grant CC BY-SA 3.0