# What is a Holomorphic Vector Field?

On a smooth manifold $M$, a smooth vector field is an element of $\Gamma(M, TM)$ which is the space of all smooth sections of the bundle $TM \to M$.

If $M$ is a complex manifold, then we have the holomorphic tangent space $T^{1,0}M$. We can form the space $\Gamma(M, T^{1,0}M)$ of smooth sections, but locally, an element can be written as $$f_1\frac{\partial}{\partial z^1} + \dots + f_n\frac{\partial}{\partial z^n}$$ where $n = \mathrm{dim}_{\mathbb{C}}M$ and the functions $f_1, \dots, f_n$ are smooth complex-valued functions, they are not necessarily holomorphic. This makes me think that these vector fields shouldn't be called holomorphic, but maybe I'm wrong.

What is the definition of a holomorphic vector field on a complex manifold?

Any additional resources dealing with such vector fields would also be appreciated.

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"$f_1\frac{\partial}{\partial z^1} + \dots + f_n\frac{\partial}{\partial z^n}$ where $n = \mathrm{dim}_{\mathbb{C}}M$ and the functions $f_1, \dots, f_n$ are smooth complex-valued functions" Why don't you just replace "smooth" by "holomorphic"? – Makoto Kato Sep 18 '12 at 8:15
I'm not sure whether this is what is meant, and if it is, what the correct global description is (i.e. sections of the bundle $T^{1,0}M \to M$ such that...). – Michael Albanese Sep 18 '12 at 8:19
The holomorphic tangent bundle has the canonical complex structure. Its holomorphic sections are holomorphic vector fields. The local representation of a holomorphic vector field is of the above form. – Makoto Kato Sep 18 '12 at 8:30
We can define holomorphic sections of any holomorphic vector bundle in the same way as we define holomorphic functions: if $E \to X$ is a bundle, then $\overline \partial$ acts on $E$ (define it locally as usual and observe that the resulting operator glues b/c the transition functions are holomorphic). Then a holomorphic section $\sigma$ of $E$ is a smooth section such that $\overline \partial \sigma = 0$. In particular, this entails that your local functions $f_j$ are holomorphic. – Gunnar Þór Magnússon Sep 18 '12 at 8:32
Would either/both of you be willing to write your comment in an answer so that I may accept it? – Michael Albanese Sep 18 '12 at 8:35

We can define holomorphic sections of any holomorphic vector bundle in the same way as we define holomorphic functions. Let $X$ be a complex manifold and let $E \to X$ be a holomorphic vector bundle over $X$. We can extend the $\overline\partial$ to act on sections of $E$: Let $E_U \to U \times \mathbb C^r$ be a local trivialization and $(e_1, \dots, e_r)$ be a local holomorphic frame of $E$. If $\sigma = \sum_j s_j e_j$ is a section of $E$ over $U$, then we set $$\overline\partial \sigma := \sum_j \overline \partial s_j \otimes e_j.$$ If $E_V \to V \times \mathbb C^r$ is another trivialization, then we write $g(z,\lambda) = (z, g(z) \lambda)$ for the induced transition function. These are holomorphic, so $g(z)$ is a $r \times r$ matrix of holomorphic functions. If we write $\sigma_U$ and $\sigma_V$ for the representations of the section $\sigma$ in the frames over $U$ and $V$, then $\sigma_U = g \sigma_V$. It follows that $$\overline \partial \sigma_U = g \overline \partial \sigma_V$$ because $g$ is holomorphic, so the $\overline \partial$ operator glues to define an operator on the space of sections of $E$.
We now define holomorphic sections of $E$ to be smooth sections $\sigma$ such that $\overline \partial \sigma = 0$. If we pick a local holomorphic frame $(e_1, \dots, e_r)$ and write $\sigma = \sum_j s_j e_j$ as before, then this entails that $\sigma$ is holomorphic if and only if all the functions $s_j$ are holomorphic.
We could of course have defined holomorphic sections as being those sections that satisfy that the "coordinate functions" $s_j$ are holomorphic in any local holomorphic frame. Since the transition functions of $E$ are holomorphic, this is well defined. This is basically the same as what we did here.