# connections between Grothendieck's and Serre's duality

Hi, I would like to show that if $f: X \rightarrow Y=Spec \, \mathbb{C}$, where $X$ is a nonsingular complex projective variety, is the projection to a point, then the complex $f^! \mathcal{O}_Y$, appearing in Grothendieck's duality, is the dualizing sheaf for $X$. Let $F$ be a coherent sheaf on $X$. Starting from $Hom_{\mathcal{O}_X}(F, f^! \mathcal{O}_Y) \simeq Hom_{{O}_Y}(Rf_* F, \mathcal{O}_Y)$ applying the cohomology functor $H^i$ we obtain $Ext^i(F, f^! \mathcal{O}_Y) \simeq Ext^i(Rf_* F, \mathcal{O}_Y).$ Using Yoneda's Formula, the right term becomes $Hom^i_{D(Y)}(Rf_* F, \mathcal{O}_Y) \cong Hom_{D(Y)}(Rf_* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$ where, for the last isomorphism, I use Theorem 8.5 from Hartshorne'a Algebraic Geometry, p.251. Now, by Corollary 5.5 pag.151 from Hartshorne, and remembering that $\Gamma(Y, \mathcal{O}_Y)= \mathbb{C}$, the last term is equal to $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}(X,F)'$. Now, we have $Hom^i_{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$ and, shifting by $(-n+1-i)$, $Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$. So, $f^!(\mathcal{O}_Y)[-n+1]$ is a dualizing sheaf for $X$. But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf $\omega$. Thus, we must have $f^!\mathcal{O}_Y[-n+1] \cong \omega$, then $f^! \mathcal{O}_Y = \omega[n-1]$.

I should obtain $f^! \mathcal{O}_Y = \omega[n]$... what's wrong? Thank you

• I'm too lazy to check through the whole thing, but certainly your identification $Rf_*F= \widetilde{H^1(X,F)}$ isn't right. – Donu Arapura Jun 13 '12 at 11:26
• emmy, how do you define the dualizing sheaf for $X$? Do you just want to show it satisfies the usual Serre duality? – Karl Schwede Jun 13 '12 at 13:38
• If you just want to find the mistake in the argument above, then you could try specialising it to the case where $X$ is a point and $n=0$ :-) because I reckon that even in this degenerate case $\omega$ is non-zero (I guess it's the structure sheaf). – Kevin Buzzard Jun 13 '12 at 21:26
• emmy, that's right, but how are you getting $H^1$? You should get $\widetilde{R \Gamma(X, F) }$. – Karl Schwede Jun 13 '12 at 23:06
• emmy, no, you definitely don't get that. $R\Gamma(X,F)$ is a complex (an object in the derived category). You get $h^1(R \Gamma(X, F)) = H^1(X,F)$. In particular, one of the cohomologies of that complex is what you want. – Karl Schwede Jun 14 '12 at 12:06

The proof that if $f \colon X \to Y$ is proper and smooth then $f^!\mathcal{O}_Y \cong \Omega^n_F[n]$ ($n=$ dim. rel.$(f)$) assuming only the existence of $f^!$ follows from theorem 3 in
The proof relies on the "fundamental local isomorphism" and several formal properties of $f^!$. Hope it helps.