# Locus where a vector bundle is null

I am reading a paper in differential geometry, Hitchin's Langlands duality and G2 spectral curves (see the end of page 8 in the arxiv version), where $f: E \rightarrow F$ is a morphism of holomorphic vector bundles on a Riemann surface, and the kernel bundle $K$ is considered. A standard argument shows that $K$ is a vector bundle too. In the paper, the nature of the map ensures also that $K$ is not a zero bundle.

Anyway, the author considers a proper divisor $D$ on the Riemann surface "where $K$ is null". Does it make any sense?

(I study algebraic geometry so maybe there is some kind of "differential" approach I am missing here, I don't know...)

• Are these vector bundles morphisms, or sheaf homomorphisms? – Tabes Bridges Dec 18 '17 at 5:45
• It is for sure a morphism of $O_X$-modules where X is the Riemann surface. Don't know about the locally constant rank. Thanks for the clarification. Anyway, the argument about the kernel being locally free works also in this context as far as I know for Riemann surfaces – Ramac Dec 18 '17 at 11:44
• What's the reference to the paper? – j.c. Dec 20 '17 at 15:12
• @TabesBridges: morphisms ov vector bundles are exactly the same thing as morphisms between (the corresponding locally free) sheaves. What changes is the notion of injective and surjective morphism: for bundles the definitions involve fibers, while for sheaves the definitions involve the stalks. – Qfwfq Dec 21 '17 at 0:18
• @j.c. Hitchin's Langlands duality and G2 spectral curves, end of page 8 in the arxiv version – Ramac Dec 21 '17 at 14:36

Mistery solved. The bundle $E$ is endowed with a bilinear, non-degenerate, symmetric form. In this context, "null" means that $K$ is isotropic at certain points.