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 $E$)

  • $\begingroup$ 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

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.


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