2
$\begingroup$

...for $M$ a compact, connected, orientable manifold without boundary.

This is part of a question in a past paper I'm working on: the full thing is at https://www1.maths.ox.ac.uk/system/files/attachments/C3_1_2015.pdf, in question 2. I've tried finding a low-dimensional counterexample: from what I remember, the statement holds for tori of any genus (ruling out a 2-dimensional example). I then tried to find a 3-fold with perhaps a torsion $H_1$, so that $H^2$, isomorphic to $ H_1^{free} \oplus H_0^{tor}$ by the universal coefficient theorem and Poincaré duality, would be zero; and then $H^1 \cup H^2$ would definitely be zero. But I can't find an example, so I'm starting to think it's true...?

This is all $\mathbb{Z}$ cohomology, by the way! Huge thanks in advance to anyone that can find a simple counterexample or a proof at the level of a first course in algebraic topology.

$\endgroup$
2
$\begingroup$

Consider $\mathbb{RP}^3$, the real projective space. It is a compact, connected, orientable manifold of dimension $3$. Its cohomology is given by: $$H^p(\mathbb{RP}^3;\mathbb{Z}) = \begin{cases} \mathbb{Z} & p = 0 \\ 0 & p = 1 \\ \mathbb{Z}/2\mathbb{Z} & p = 2 \\ \mathbb{Z} & p = 3 \\ 0 & p > 3 \end{cases}$$ Then for the nonzero element $\alpha \in H^2(\mathbb{RP}^3;\mathbb{Z})$, there is no element in $H^1$ which multiplies with $\alpha$ to a nonzero element – simply because there's nothing in $H^1$.

As you guessed the problem is indeed torsion: if we consider (co)homology with coefficients in a field $\Bbbk$ instead, let $0 \neq \alpha \in H^k(M; \Bbbk)$. By Poincaré duality, $H^k(M;\Bbbk) \cong H_{n-k}(M;\Bbbk)$, and by the UCT $H_{n-k}(M;\mathbb{Q}) \cong H^{n-k}(M;\Bbbk)$ ($M$ is compact thus $H_{n-k}(M;\Bbbk)$ is a finite-dimensional vector space). Let $\beta \in H^{n-k}(M;\Bbbk)$ correspond to the Poincaré dual of $\alpha$, then $\alpha \cup \beta$ is the fundamental class of $M$ (and hence nonzero), pretty much by definition of Poincaré duality and the UCT isomorphism.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That's great, thanks a lot! $\endgroup$ – Latimer Leviosa Apr 6 '16 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.