When discussing Hodge numbers, the identity $h^{p,q} = h^{n-p,n-q}$ is called Serre duality. AFAIU, this identification follows from applying the Hodge $*$ isomorphism. In particular, $H^q(X,\Omega^p_X)$ is dual to $H^{n-q}(X,\Omega^{n-p}_X)$.

If $X$ is now projective, we know that $H^q(X,\Omega_X^p)$ is dual to $H^{n-q}(X, \Omega_X^n \otimes \wedge^p T_X )$. I would like to then deduce the identity of Hodge numbers above by saying that $\Omega_X^{n-p}$ is isomorphic to $\Omega_X^n \otimes \wedge^p T_X$ but I don't know why that should be true.

It would certainly follow if I knew that the wedging map $\Omega^p_X \otimes \Omega^{n-p}_X \to \Omega^n_X$ were an isomorphism.

  • $\begingroup$ Of course it does. $\endgroup$ – Rüdiger Apr 11 '17 at 9:38

The result you're looking for is a special case of the following: "If $\mathcal F$ is a vector bundle of rank $n$, then $\wedge^p \mathcal F \cong (\wedge^{n-p} \mathcal F)^\vee\otimes \wedge^n \mathcal F$." (This is Hartshorne exercise 5.16.)

Note that $(\wedge^{n-p} \mathcal F)^\vee\otimes \wedge^n \mathcal F \cong \mathcal {Hom}_{\mathcal O}(\wedge^{n-p} \mathcal F, \wedge^n \mathcal F)$, so we just need to define an isomorphism from $\wedge^p \mathcal F$ to $\mathcal {Hom}_{\mathcal O}(\wedge^{n-p} \mathcal F, \wedge^n \mathcal F)$. To define this isomorphism, pick a trivialising open cover $\{ U_\alpha \}$. Since $\mathcal F|_{U_\alpha} \cong \mathcal O|_{U_\alpha}^{\oplus n}$, we can write any section of $\mathcal F|_{U_\alpha}$ as an $n$-component vector, whose components are sections of $\mathcal O$. We can then define our isomorphism from $\wedge^p \mathcal F|_{U_\alpha}$ to $\mathcal {Hom}_{\mathcal O|_{U_\alpha}}(\wedge^{n-p} \mathcal F|_{U_\alpha}, \wedge^n \mathcal F|_{U_\alpha})$ in the same way that we would define the natural isomorphism from $\wedge^p V $ to ${\rm Hom}(\wedge^{n-p}V, \wedge^n V)$ if $V$ is an $n$-dimensional vector space. Finally, this isomorphism is "basis independent", which ensures that it glues properly between the different $U_\alpha$'s.


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.