Timeline for Wanted: Odd-dimensional integral cohomology class whose square is nonzero
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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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 |