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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

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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
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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
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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
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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

1 Answer 1

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

Verdier, Jean-Louis Base change for twisted inverse image of coherent sheaves. 1969 Algebraic Geometry (Internat. Colloq., Tata Inst. Fund. Res., Bombay, 1968) pp. 393–408 Oxford Univ. Press, London.

The proof relies on the "fundamental local isomorphism" and several formal properties of $f^!$. Hope it helps.

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