Let $X$ be a complex compact manifold. I want to understand the holomorphic tangent vector field on $X$. I know a smooth vector field can define an infinitesimal diffeomorphism of $X$ (more precisely, a one-parameter diffeomorphism group). Similarly, if we consider a holomorphic tangent vector field, will it determine to a infinitesimal holomorphic equivalence of $X$ (a one-parameter holomorphic equivalence group)?
Holomorphic tangent vector field determines a infinitesimal holomorphic equivalence
algebraic-geometrycomplex-geometrymanifolds
Related Solutions
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.
Since you ask for additional resources for dealing with holomorphic tangent fields specifially, I encourage you to have a look at the Bochner--Weitzenböck formulas you asked about on MO the other day. These are often used to show that there are no non-zero holomorphic vector fields on a manifold (a fun exercise is to prove this by using the Kähler--Einstein metric on a projective manifold with ample canonical bundle -- try Ballmann or Zheng's books if you need help on this).
It’s possible to view the Poincaré metric on the unit disk as the curvature form of a Hermitian metric on the trivial line bundle over the disk. In this trivialization the metric is identified with a smooth function on the disk. This function must be real-valued to be a Hermitian metric (because a Hermitian inner product on a one-dimensional complex vector space is just a positive real number), so it cannot be holomorphic.
More generally, if a metric $h$ on a holomorphic vector bundle is holomorphic it follows that it is flat: Pick a local holomorphic frame for the bundle and let $H$ be the matrix of $h$ in that frame. Then the entries of $H$ are holomorphic functions. However, as $H$ is Hermitian we have $H = \overline H^t$, so the conjugates of the entries of $H$ are holomorphic functions as well. It follows that the entries of $H$ are constant, so its Chern curvature tensor vanishes. Now, there are holomorphic vector bundles that do not admit flat metrics (like the tangent bundle of the projective space), so those bundles do not admit holomorphic Hermitian metrics.
Best Answer
Yes, the flow will give a 1-parameter group of biholomorphisms. See chapter 3 of S. KOBAYASHI, Transformations Groups in Differential Geometry.