Let $X$ be a a compact oriented manifold of dimension $n$. Assume that its (co)homologies have no torsion. Then Poincaré duality says that $$ H^{k}(X,\mathbb{Z})\cong H_{n-k}(X,\mathbb{Z}) $$ holds for $1 \le k \le n$. Does this mean that for any primitive $\alpha\in H^{k}(X,\mathbb{Z})$, there exists $\beta \in H^{n-k}(X,\mathbb{Z})$ such that the cup pairing $\alpha\cup\beta=1$? Here we use the identification $H^{n}(X,\mathbb{Z})\cong \mathbb{Z}$ via the orientation of $X$.

Note that we say that $\alpha$ is primitive if it is not a multiple of any other element, i.e. if $\alpha=k\gamma$ for some $k\in \mathbb{Z}$ and $\gamma \in H^{k}(X,\mathbb{Z})$, then $k=\pm1$.


1 Answer 1


Yes, more is true. Given any primitive element $\alpha \in H^k$, you can extend it to a basis $\{\alpha_1,\cdots,\alpha_l\}$ of $H^k$ where $\alpha = \alpha_1$. Then there exists a basis $\{ \beta_1, \cdots, \beta_l\}$ of $H^{n-k}$ so that

$$ \alpha_i \cup \beta_j = \delta_{ij}$$

of course you need to assume your manifold is connected for this to make sense.

There's a proof of this in the textbook Characteristic Classes by Milnor and Stasheff.


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