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My question is how to define the twisting of a sheaf $\mathcal{L}$ by a divisor $D$.

In specific I'm interested in the twisting of the canonical bundle $\omega$ of a non-compact Riemann surface $X$ by a divisor of points. (The points are the missing points of the compactification, in my case they are a finite number.)

My guess on the definition is the following. I take $D=p_1+...+ p_n$. On $X$ Weil and Cartier divisors are the same, so $D$ is also a Cartier divisor and it is well defined the associated sheaf $\mathcal{L}(D)$. Then I "define" the twisted canonical bundle by:

$\omega(D)=\omega \otimes \mathcal{L}(D)$.

Is my "definition" correct? If it is I don't really see its geometric meaning. If it is not can you please point out a reference for it?

Thank you!

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1 Answer 1

up vote 6 down vote accepted

Your guess is correct. The geometric (or function-theoretic) meaning is that sections of $\omega(D)$ are sections of $\omega$ that are allowed to have simple poles along $D$.

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Thank you! A further investigation, can I relate the sheaf cohomology groups of these two sheaves? –  Giovanni De Gaetano May 9 '12 at 11:34
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Dear Giovanni, There is a short exact sequence $$0 \to \omega \to \omega(D) \to \omega(D)_{| D} \to 0.$$ When the ambient variety is a curve, so that $D$ is just a sum of points, the third terms is just a skyscraper sheaf supported at the points of $D$ (its length at $P$ is the multiplicity with which $P$ occurs in $D$). In particular, it has no higher cohomology, and so you get $$0 \to H^0(X,\omega) \to H^0(X,\omega(D)) \to H^0(D,\omega(D)_{|D}) \to H^1(X,\omega) \to H^1(X,\omega(D)) \to 0,$$ where the dimension of $H^0(D,\omega(D)_{|D})$ is just equal to $\deg(D)$. Regards, –  Matt E May 9 '12 at 11:39
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P.S. You can see from this that $H^1(X,\omega(D))$ is a quotient of $H^1(X,\omega)$. If $\deg(D)$ is large enough, then in fact $H^1(X,\omega(D))$ will vanish. So twisting by $D$ tends to subtract dimensions from $H^1$, and add them to $H^0$. The process of twisting is closely related to the theory of the Hilbert polynomial, and to the Riemann-Roch theorem. E.g. by varying $D$, and using the known fact that $\chi(\omega) = g-1$ --- where $\chi$ denotes Euler characteristic --- by taking Euler characteristics in the above exact sequence you obtain the formula ... –  Matt E May 9 '12 at 11:43
    
... $\chi(\omega(D)) = \deg(D) + g - 1,$ which is a form of Riemann-Roch. Regards, –  Matt E May 9 '12 at 11:44
    
Well, I'm sorry I have finished all my chances to give you reputation! Thank you very much again! –  Giovanni De Gaetano May 9 '12 at 11:44

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