Let $M$ be a compact connected $4$-manifold such that $H_2(M;\mathbb{F}_2)$ is non-zero. Show that the cup product map $H^2(M;\mathbb{F}_2) \times H^2(M;\mathbb{F}_2) \to H^4(M;\mathbb{F}_2)$ given by $(x, x) \mapsto x \cup x$ is surjective.

Now by Poincaré Duality we know that $H^4(M;\mathbb{F}_2) \cong H_0(M;\mathbb{F}_2) = \mathbb{F}_2$. So it suffices to show that there exists an $x \in H^2(M;\mathbb{F}_2)$ which is non-zero such that $x \cup x \neq 0$.

I know that we have the mod-$2$ Poincaré Duality pairing isomorphism $$H^2(M;\mathbb{F}_2) \xrightarrow{\cong} \operatorname{Hom}(H^2(M;\mathbb{F}_2), \mathbb{F}_2)$$ given by $$\alpha \mapsto \langle \alpha, \_\ \rangle : H^2(M;\mathbb{F}_2) \to \mathbb{F}_2$$ where $\langle \alpha, \_\ \rangle : H^2(M;\mathbb{F}_2) \to \mathbb{F}_2$ is in addition given by $$\langle \alpha, \beta \rangle = \Phi(\alpha \cup \beta)[M]$$ where $\Phi : H^4(M;\mathbb{F}_2) \to \operatorname{Hom}(H_4(M;\mathbb{F}_2), \mathbb{F}_2)$ is the surjective map occurring in the universal coefficients theorem and $[M]$ is the fundamental class.

I guess that if all $x \in H^2(M;\mathbb{F}_2)$ has the property that $x \cup x = 0$ then we'd get a contradiction from the mod-$2$ Poincaré Duality pairing isomorphism, but I do not see where the contradiction would occur.

For example if $x \neq 0 \in H^2(M;\mathbb{F}_2)$ and $x \cup x = 0$, then $\langle x, x \rangle = 0$. If I could show that $\langle x, \_ \rangle : H^2(M;\mathbb{F}_2) \to \mathbb{F}_2$ however was the zero morphism then I'd have the needed contradiction, however I don't see why this would be the case.

  • $\begingroup$ if you like, you can use $\smile$ (that its latex is $\smile$ ) instead of $\cup$ like most books! :) $\endgroup$
    – C.F.G
    Commented Sep 17, 2021 at 6:13

2 Answers 2


This is false as stated.

Let $M = S^2\times S^2$. Then $H^\ast(S^2\times S^2; \mathbb{F}_2)\cong \mathbb{F}_2[x,y]/ I$ where $I$ the ideal generated by $x^2$ and $y^2$. In more detail, from Kunneth, $H^\ast(S^2\times S^2)\cong \mathbb{Z}[\alpha,\beta]/J$ where $J$ is the ideal generated by $\alpha^2$ and $\beta^2$, and then by universal coefficients, we just get $x$ and $y$ as the mod $2$ reduction of $\alpha$ and $\beta.$

Now, given any class $ax + by\in H^2(S^2\times S^2;\mathbb{F}_2)$, we have $(ax+by)^2 = a^2 x^2 + 2ab xy + b^2 y^2$. But both $x^2$ and $y^2$ are trivial, and $2 = 0 $ in $\mathbb{F}_2$, so $(ax+by)^2 = 0$.

  • $\begingroup$ Thanks for the answer! Just a small comment, I think you wanted $(ax+by)^2 = a^2x^2+2abxy+b^2y^2$ rather than $(ax+by)^2 \cup (ax+by)$. $\endgroup$ Commented Sep 17, 2021 at 6:42
  • $\begingroup$ @Pertubative: Fixed! Thanks for catching that. $\endgroup$ Commented Sep 17, 2021 at 11:11

This holds if and only if $w_1(M)^2 + w_2(M) = 0$. When M is orientable this holds if and only if M is "spin". This condition on the intersection form is known as being "even" while its negation is known as being "odd".

To see my first claim you need to know about Wu classes and Steenrod squares. On a closed n-manifold there are cohomology classes $\nu_k \in H^k(M;\Bbb Z/2)$ so that $$\nu_k \smile x = \text{Sq}^k x$$ for all cohomology classes of degree n-k. Further, these rate to the Stiefel-Whitney classes via $$w_k(M) = \sum_{i=0}^{\lfloor k/2\rfloor} \text{Sq}^i(\nu_{k-i}).$$ Lastly, $\text{Sq}^k x = x^2$ when $x$ has degree k, $\text{Sq}^k x = 0$ if $x$ has degree larger than k, and $\text{Sq}^0$ is the identity.

All of these facts can be found in the book of Milnor and Stasheff.

Observe that the formula $w_1(M) = \nu_1$ and thus $w_2(M) + w_1(M)^2 = \nu_2$ follow for all manifolds.

Now on a closed 4-manifold, if x is a class of degree 2, $\text{Sq}^2 x = x^2$. Thus one has $\nu_2 \smile x = x^2$ for all x of degree 2 in a closed 4-manifold. It this follows from Poincare Duality that $x^2 = 0$ for all x of degree 2 if, and only if, $\nu_2 = 0$. Thus the first line of this post.


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