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$\newcommand{\O}{\mathcal{O}}$ Let $X$ be a smooth projective curve and $D$ and effective divisor on it. The normal bundle of $D$ is defined as $$ \O_D(D)\; = \; \O_x(D)\;\otimes_{\O_X}\, \O_D$$ where $\O_D$ is just the restriction of the stricture sheaf of $X$ to the support of $D$. In the literature I found the claim that, as a special case of Serre's duality, the dual space to $H^0(X, \O_D(D))$ is $$ H^0(X, \O_D(D))^* \cong H^0(X,K\otimes\O_D), $$ where $K$ is the canonical sheaf.

Now, I'm not an expert of Serre's duality, but I was expecting the dual of that space to be something like $$ H^0(X,(K-D)\otimes\O_D). $$ Could you please explain the reason why the above is the right answer and mine is the wrong one?

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The adjunction formula says that the dualizing sheaf $\omega_D$ on $D$ is $K\otimes O_D(D)$. For any invertible sheaf $L$ on $D$, we have $H^0(D, L)^*=H^0(D, \omega_D\otimes L^{-1})$. Now apply this to $L=O_D(D)$. –  Cantlog Oct 20 '13 at 9:03
    
Thanks! I found it in Hartshorne II - 8.20 (page 182) –  Abramo Oct 20 '13 at 9:12
    
Ah, then please write and accept your answer. –  Cantlog Oct 20 '13 at 9:13
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Dear Abramo and @Cantlog, beware that Hartshorne's result only works in the special case that $D$ is smooth. –  Georges Elencwajg Oct 20 '13 at 10:08
    
Dear Georges, @GeorgesElencwajg: too bad. The adjunction formula holds in general for any effective Cartier divisor. But I can't find it in Fulton's "Intersection Theory". Maybe in SGA 6 ? –  Cantlog Oct 20 '13 at 17:10

1 Answer 1

up vote 4 down vote accepted

From Proposition 8.20 of Hartshorne (chapter II, page 182), we have a formula for the canonical sheaf $\omega_D$ of $D$: $$ \omega_D = \omega_X \otimes \O_D(D). $$ So applying Serre's duality for $D$ we find $$ H^0(\O_D(D))^* \cong H^0(\omega_D \otimes \O_D(D)^{-1}) \cong H^0(\omega_X \otimes\O_D(D)\otimes \O_D(D)^{-1}) \cong H^0(\omega_X \otimes \O_D). $$

UPDATE:

To make the notation easier let's use the following shorthands: $$H^0(D) \leadsto H^0(X,\mathcal{O}_X(D)) \quad\text{ and }\quad H^0(D_D) \leadsto H^0(X,\mathcal{O}_X(D)\otimes \mathcal{O}_D)$$ I'd like to point out that once one proves (like in the original book of Serre Algebraic Groups and Class Fields) that for every divisor $D$ we have $$ H^1(D)^{\vee} \cong H^0(K-D), $$ then the above duality between $H^0(D_D)$ and $H^0(K_D)$ follows by abstract nonsense, without using the adjunction formula. Indeed from the short exact sequences of invertible sheaves $\newcommand{\SES}[3]{ 0 \to #1 \to #2 \to #3 \to 0 }$ $$ \SES{\O_X}{\O_X(D)}{\O_X(D)\otimes \O_D} $$ and $$ \SES{K(-D)}{K}{K\otimes \O_D} $$ we get the commutative diagram with exact rows $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 & \;\;\ra{} \;& k & \;\ra{} & H^0(D) & \;\;\ra{} & H^0(D_D) & \;\;\ra{} & H^1(\O_X) & \;\;\ra{} & H^1(D) & \;\;\ra{} & 0 \\ & & \da{\;f_0} & & \da{\;f_1} & & \da{\;f_2} & & \da{\;f_3} & & \da{\;f_4} & & \\ 0 & \;\;\ra{} \;& k & \;\ra{} & H^1(K-D)^{\vee} & \;\;\ra{} & H^0(K_D)^{\vee} & \;\;\ra{} & H^0(K)^{\vee} & \;\;\ra{} & H^0(K-D)^{\vee} & \;\;\ra{} & 0 \\\end{array} $$ where $f_0$, $f_1$, $f_3$ and $f_4$ are the duality isomorphisms. Using the 5 lemma we can now deduce that $f_2$ is an isomorphism as well.

Remark: we are just using the hypothesis of Serre duality from Serre's book, so that $X$ is a smooth projective curve over an algebraically closed field.

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