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Why is there a dual bundle in Serre duality?


Let $\mathcal E$ be a vector bundle over complex manifold $X$, without any metric anywhere, then one has a pairing $$(\Omega^{0,q} \otimes \mathcal E) \otimes (\Omega^{n,n-q} \otimes \mathcal E^*) \to \Omega^{n,n} \cong K_X,$$ which naturally induces a mapping $$\Omega^{0,q} \otimes \mathcal E \to (\Omega^{n,n-q})^* \otimes \mathcal E \otimes K_X.$$ Now choosing Riemann metric on $X$, and by some reason not using the fact $K_X$ is trivial, one gets a mapping $$\Omega^{0,q} \otimes \mathcal E \to \Omega^{n,n-q} \otimes \mathcal E \otimes K_X.$$

Its restriction to harmonic forms gives isomorphism $$\mathcal H^q(\mathcal E) \cong \mathcal H^{n-q}(\mathcal E \otimes K_X).$$


But everywhere there is a star over the second $\mathcal E$. What have I missed?

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There are some errors throughout. We start with a compact $n$-dimensional complex manifold $X$. First of all, $K_X \cong \Omega^n$ is the sheaf of holomorphic $n$-forms. Second, Serre duality actually gives us duality, not isomorphism, between $H^q(\Omega^p(\mathcal E))$ and $H^{n-q}(\Omega^{n-p}(\mathcal E^*))$. Via the Dolbeault isomorphism, this is a non-degenerate pairing $$H^{p,q}(\mathcal E) \otimes H^{n-p,n-q}(\mathcal E^*) \to H^{n,n}(X)\cong\mathbb C.$$The Dolbeault cohomology, in turn, can be represented by harmonic $(p,q)$-forms, once you introduce hermitian metrics.

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