# When is the cohomology cross product square nonzero?

Let $X$ be a finite CW complex, and $A$ an abelian group. Given a nonzero class in singular cohomology $$x\in H^k(X;A),$$ when is its cross product square $$x\times x\in H^{2k}(X\times X;A\otimes A)$$ nonzero?

Remarks: The tensor product is over $\mathbb{Z}$. By the Künneth theorem $x\times x$ is always nonzero if $A$ is a field, or more generally if $A$ is a domain and $H^\ast(X)$ is flat over $A$. Examples where $x\times x=0$ arise in the study of Lusternik-Schnirelmann category, in particular in spaces $X$ for which $\operatorname{cat}(X\times X)<2\operatorname{cat}(X)$.

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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. – Ralph Jun 20 '12 at 17:36
I mean of course $Tor_1^\mathbb{Z}(A,A)=0$. – Ralph Jun 20 '12 at 19:17
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 $2\in \mathbb{Z}/4$. – Greg Friedman Jun 21 '12 at 22:53
@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). – Mark Grant Jun 22 '12 at 9:50
@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). – Mark Grant Jun 22 '12 at 9:54