I am reading a paper in differential geometry, [Hitchin's Langlands duality and G2 spectral curves](https://arxiv.org/abs/math/0611524) (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](https://mathoverflow.net/questions/168521/holomorphic-vector-bundles-over-a-riemann-surface-does-not-satisfy-mathbfab2) 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...)



Locus where a vector bundle is null
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.


