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I am reading a paper that states the following theorem without proof:

Poincaré duality in middle dimension: Let $M$ be a connected oriented manifold of even dimension $2d$. Then the cup product induces a non-degenerate bilinear form

$$H^n(M ; \mathbb{R}) \times H^n(M ; \mathbb{R}) \to H^{2n}(M ; \mathbb{R})$$ which is symmetric if $n$ is even and skew-symmetric if $n$ is odd.

I have been looking through Hatcher for a possible proof or more description of the theorem.

I found that the cup product pairing is non-degenerate in the case where coefficients are in a field or $\mathbb{Z}$ and torsion is factored out.

In the case of $\mathbb{R}$ our cohomology groups are torsion-free, and so at least the cup product pairing will be non-singular.

However, the theorem as stated in the paper refers to the bilinear form induced by the cup product and not just the cup product pairing.

Why is the bilinear form non-singular for the cup product?

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    $\begingroup$ A bilinear form is just a pairing of a space with itself. $\endgroup$
    – anomaly
    Apr 22, 2016 at 23:16
  • $\begingroup$ The cup product pairing was defined as $\alpha \smile \alpha [M]$ where $[M] \in H^{2n}(M; R)$ represents the fundamental class. It seems that the above theorem is slightly more general than that. $\endgroup$
    – user7090
    Apr 22, 2016 at 23:20
  • $\begingroup$ Actually, I guess since the fundamental class generators $H^{2n}(M ; R)$ the cup product pairing as defined in Hatcher is thus defined for all elements of $H^{2n}(M ;R)$. $\endgroup$
    – user7090
    Apr 22, 2016 at 23:22

1 Answer 1

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There's a little abuse of notation. First, $M$ has to be compact for all this to make sense; it's probably written somewhere before in the paper. When $M$ is compact, orientable, and connected, $H^{2d}(M;\mathbb{R})$ is isomorphic to $\mathbb{R}$, and the choice of an orientation basically amounts to the choice of an isomorphism $H^{2d}(M;\mathbb{R}) \cong \mathbb{R}$, given by $\alpha \mapsto \langle \alpha, [M] \rangle$.

You can then compose the cup product and this isomorphism to get a bilinear form (a bilinear form is nothing more than a bilinear map $V \times V \to \mathbb{R}$): $$H^d(M;\mathbb{R}) \times H^d(M;\mathbb{R}) \xrightarrow{\smile} H^{2d}(M;\mathbb{R}) \xrightarrow{\langle -, [M] \rangle} \mathbb{R}.$$ To say that the cup-product is non-degenerate amounts exactly to say that this bilinear form is non-degenerate, pretty much by definition. There's no funny business going on, just an identification of $H^{2d}(M;\mathbb{R})$ with $\mathbb{R}$.

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