# Is $\bigwedge^{p,0}M$ a holomorphic vector bundle?

I have difficulties in understanding how to show that a vector bundle is holomorphic. For instance, how can I prove that $\bigwedge^{p,0}M$ is a holomorphic vector bundle, where $M$ is a complex manifold ? I think I should determine the transition maps and show that they are holomorphic. However, it is unclear to me how to do this in a correct way.

• What is $\bigwedge^{p,0} M$ when $M$ is simply a holomorphic vector bundle? You should have something like a real structure on $M$. – Ahr Jun 5 '18 at 12:15
• Well I don't understand the meaning of $\bigwedge^{p,0}M$ for $M$ a holomorphic vector bundle over a complex manifold. Could you precise what you mean by that notation? – Ahr Jun 5 '18 at 12:37
• $(M, J)$ is a complex manifold. $\Lambda^{p, 0}M = \Lambda^{1, 0} \wedge \ldots \wedge \Lambda^{1, 0}$ (p-times) , $\Lambda^{1, 0}M = \{\tau \in \Lambda_{\mathbb{C}}^1(M) \mid \tau(Z) = 0, \forall\ Z \in T^{1, 0}M \}$ and $T^{1,0}M = \{X - iJX \mid X \in TM\}$. – Adrian Jun 5 '18 at 12:48
• Ok, I just realized that you mean $\Omega^{p,0} M$ the bundle of complex forms of type $(p,0)$. Then it is just a matter of noting that the exterior power of a holomorphic bundle is holomorphic and that $\Omega^{1,0}M$ is holomorphic, by looking at the transition functions, or by noting that $T^{1,0}M$ is isomorphic as a complex bundle to the holomorphic tangent bundle. – Ahr Jun 5 '18 at 13:07
A complex rank r vector bundle $E$ over $M$ is given by a cocycle $$\{\varphi_{ij}\colon U_i\cap U_j \longrightarrow {\rm GL}(r,\mathbb{C}) \}.$$ $E$ is holomorphic if the $\varphi_{ij}$ are holomorphic. Then we can apply "any linear algebra construction" such as exterior and symmetric powers to $E$ (fiberwise) and the result will be also a holomorphic bundle. For example, the determinant of $E$ will have a cocycle given by $\det(\varphi_{ij})$.