# Twisting holomorphic vector bundles and Euler characteristics

Given a holomorphic vector bundle $$\mathcal{V}$$ over a compact complex manifold $$M$$, it seems that even if $$\mathcal{V}$$ is non-trivial, then it can still have trivial Euler characteristic, that it, $$\sum_{k=0}^{\mathrm{dim}(M)} (-1)^k H^{(0,k)}(\mathcal{V}) = 0.$$ Is it true that for a positive line bundle $$\mathcal{L}$$, one can always find a large enough $$n$$, such that the twisted vector bundle $$\mathcal{V} \otimes \mathcal{L}^{\otimes n}$$ will have non-trivial Euler characteristic?

• If $M$ is projective, then for $n$ large enough one has that $\mathcal V\otimes \mathcal L^{\otimes n}$ is globally generated and that $H^q(M, V\otimes \mathcal L^{\otimes n})=0$ for $q>0$. Therefore, the holomorphic Euler characteristic of $V\otimes \mathcal L^{\otimes n}$ is positive. Oct 10, 2019 at 12:57
• ... and since $M$ carries a positive line bundle (by assumption), it is in fact projective.
– Ben
Oct 15, 2019 at 17:49