# Wanted: Odd-dimensional integral cohomology class whose square is nonzero

Does anyone know of a nice simple example of a space $X$ with an odd-dimensional integral cohomology class $a\in H^{2k+1}(X;\mathbb{Z})$ whose square is nonzero?

I once thought that the one-dimensional generator $a\in H^1(K;\mathbb{Z})\cong\mathbb{Z}$ in the cohomology of the Klein bottle had $a^2\in H^2(K;\mathbb{Z})\cong\mathbb{Z}/2$ nonzero, but it appears this is not the case.

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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. –  Tom Goodwillie Aug 20 '12 at 22:18
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. –  Ralph Aug 20 '12 at 22:54
@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? –  Mark Grant Aug 21 '12 at 0:57
@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 $n$). –  Mark Grant Aug 21 '12 at 1:46
@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. –  Ralph Aug 21 '12 at 8:06

## 1 Answer

As Ralph is being modest, I have decided to make his comment into a CW answer.

Recall that the short exact coefficient sequence $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\to 0$ leads to a long exact sequence $$\cdots \to H^\ast(X;\mathbb{Z})\to H^\ast(X;\mathbb{Z})\stackrel{\rho}{\to} H^\ast(X;\mathbb{Z}/2)\stackrel{\beta}{\to} H^{\ast+1}(X;\mathbb{Z})\to\cdots$$ for any space $X$, where $\rho$ denotes reduction mod 2 and $\beta$ is the Bockstein homomorphism. Recall also that $\rho$ is a ring homomorphism, and that $\rho\circ\beta = Sq^1$, the first Steenrod square.

Let $X=\mathbb{R}P^\infty\times\mathbb{R}P^\infty$, and let $x,y\in H^1(X;\mathbb{Z}/2)$ denote the generators over each axis. Note that $c=\beta(xy) \in H^3(X;\mathbb{Z})$ is nonzero, since $$\rho(c) = Sq^1(xy) = Sq^1(x)y + x Sq^1(y) = x^2y + xy^2 \neq 0.$$

Similarly $c^2 \in H^6(X;\mathbb{Z})$ is nonzero, since we have $$\rho(c^2) = \rho(c)^2 = (x^2y+xy^2)^2 = x^4y^2 + x^2 y^4 \neq 0.$$

As Will Sawin noted in his comment, we can get a finite-dimensional example by taking $Y=X^{(6)}$, the $6$-skeleton of $X$ (this is the smallest possible dimension, by Tom Goodwillie's comment). We could also take $Y=\mathbb{R}P^4\times\mathbb{R}P^4$ (but $\mathbb{R}P^3\times\mathbb{R}P^3$ won't work, since its six-dimensional cohomology is torsion-free).

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