Timeline for When is the cohomology cross product square nonzero?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| S Apr 30 at 13:32 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
fixed broken link to Wikipedia
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| Apr 30 at 12:50 | review | Suggested edits | |||
| S Apr 30 at 13:32 | |||||
| Jun 22, 2012 at 9:54 | comment | added | Mark Grant | @Ralph: Thanks for pointing out this general formulation of Kunneth, which I had forgotten about (worse is that I forgot to look in Spanier for the most general formulation of a result). | |
| Jun 22, 2012 at 9:50 | comment | added | Mark Grant | @Greg Friedman: That's a good point that even $x\otimes x$ may be zero. However, I suppose it might also happen that the cross product fails to be injective (if $\operatorname{Tor}(A,A)$ is non-trivial). | |
| Jun 21, 2012 at 22:53 | comment | added | Greg Friedman | Building on Ralph's answer I would think this question reduces completely to algebra. Using the Kunneth theorem and the cohomology cross product, $x\times x$ should correspond precisely to the image of $x\otimes x\in H^k(X;A)\otimes H^k(X;A)$ under the injective cross product. So the question really becomes, given an element of a group $g\in G$, when is it true that $g\otimes g=0\in G\otimes G$. As Ralph notes, this can't happen if $G$ is torsion-free (or even if $g$ generates an infinite cyclic subgroup?), but I guess it could happen if, for example, we have $2\in \mathbb{Z}/4$. | |
| Jun 20, 2012 at 19:17 | comment | added | Ralph | I mean of course $Tor_1^\mathbb{Z}(A,A)=0$. | |
| Jun 20, 2012 at 17:36 | comment | added | Ralph | According to the Künneth formula [Spanier, 5.3.10], $x \times x \neq 0$ if $Tor_1^{\mathbb{Z}}(A,A) \neq 0$. This holds for example if $A$ is torsion-free. | |
| Jun 20, 2012 at 16:10 | history | asked | Mark Grant | CC BY-SA 3.0 |