2
$\begingroup$

Let $X$ be a Riemannian surfaces with a divisor $D$ and let $E$ be a holomorphic complex vector bundle of rank r on $X$.

1) The Riemann-Roch theorem is used to give an estimate of the dimension of the vector space of the holomorphic sections of $E$, i.e

$ \operatorname{dim}(H^{0}(X,E))-\operatorname{dim}(H^{1}(X,E))=\deg(E)-rk(E)(1-g(X))$

where g(X) is the genus of $X$.

Here my question: let $D$ be as above, is it possible to write a version of the above formula that gives informations about the dimension of the vector space of meromorphic sections of $E$ with pole in $D$? (this is possible for line bundle thanks to the the correspondence line bundles---divisors, for this reason I'm expecetd something involve the determinant bundle of $E$)

$\endgroup$
  • $\begingroup$ Sorry, I don't know how to answer but I feel confused by the statement : do you mean that $X$ is a Riemann surface ? Or an algebraic surface ? Because, you added the tag "Riemann surfaces" but on a curve any divisor $D$ is a NCD. $\endgroup$ – user171326 Jul 3 '17 at 9:53
  • $\begingroup$ right, now is edited $\endgroup$ – Cepu Jul 3 '17 at 12:59
1
$\begingroup$

In the case of a line bundle $L$, to consider sections with a pole at a point $P$, is the same as considering the line bundle $L(P)$. The same Riemann-Roch formula applies. If $E$ is a vector bundle, then a pole at $P$ may be in "several different directions". These will correspond to sections of a bundle $E'$ lying between $E$ and $E(P)$. In contrast to the line bundle case, this $E'$ is not unique. However, if you allow just one pole, its degree will be one greater, and the same Riemann-Roch formula as before will apply. Note that Riemann-Roch does not give you the dimension of the space of sections, only its Euler characteristic, i.e. the difference with the first cohomology group.

$\endgroup$

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.