Consider a complex manifold $X$ and $Aut(X)$ the Lie group of biholomorphisms of $X$. I want to calculate the Lie algebra of this Lie groups so as to derive a result analogue in the holomorphic realm to that for real smooth manifolds, namely

$$

\text{Lie}(\text{Diff}(M)) \simeq \mathfrak{X}(M)

$$

i.e. that infinitesimal diffeomorphisms are given by smooth vector fields. In the real case, one would take a 1-parameter family of diffeomorphisms $f_t$ close to the identity and argue that at each point, taking the derivative gives a tangent vector $\gamma'_p(0) = \frac{d}{dt}|_{t=0} f_t(p)$, and after some calculations, deduce that this induces a smooth field of vectors over $M$.

For the holomorphic case, I suspect that the infinitesimal biholomorphisms of a complex manifold correspond to **holomorphic vector fields**, namely holomorphic sections of the complex vector bundle $T^{1,0}X \rightarrow X$ endowed with the natural holomorphic structure coming from $X$.

However, I don't know how to prove that the derivative $\gamma'_p(0)$ above is in a natural way a (1,0)-vector in $T\underline{X}_\mathbb{C}=T^{1,0}X \oplus T^{0,1}X$ (where $\underline{X}$ is the smooth manifold underlying $X$).

## Best Answer

I think that your intuition is correct, but the question is a bit subtle. FIrst of all, there are topological obstructions for the group of biholomorphisms to be a complex Lie group, as pointed out in the comment by Andrew Hwang. One simple case in which such obstructions vanish (and to be honest, the only one I ever studied in any detail) is when the manifold is compact.

So, assume for the moment we have a compact manifold $X$, together with a complex structure $J$. I will denote by $\operatorname{Aut}(X,J)$ the group of biholomorphisms. A $1$-parameter subgroup, say $\{f_t\mid t\in\mathbb{R}\}\subset\operatorname{Aut}(X,J)$ gives you a vector field $v$ on $X$ by setting $$v_p:=\partial_{t=0}f_t(p).$$ Notice that this is just a real vector field, i.e. a section of $TX\to X$. As $f_t^*J=J$, we see however that $v$ satisfies \begin{equation}\label{realhol} \tag{$\star$} \mathcal{L}_vJ=0 \end{equation} and for any $v$ that satisfies \eqref{realhol}, its flow $f_t$ will be a subgroup of $\operatorname{Aut}(X,J)$ - the flow will be complete, as $X$ is compact. I guess we can already see that this might be a problem for noncompact manifolds.

Sections of $TX$ that satisfy \eqref{realhol} are usually called

real-holomorphic vector fields. From this point of view then, the lie algebra of $\operatorname{Aut}(X,J)$ is the algebra of real-holomorphic vector fields. Notice however that:if $v$ is real-holomorphic, also $Jv$ is;

$v$ is real-holomorphic if and only if $v^{1,0}$ is holomorphic, meaning that it is a holomorphic section of the holomorphic vector bundle $T^{1,0}X$.

So we can identify the Lie algebra of $\operatorname{Aut}(X,J)$ either with holomorphic or real-holomorphic vector fields, and it has a complex structure coming either from remark $1$ above, if you like to use real-holomorphic vector fields, or just from multiplication by $i$, if you prefer holomorphic vector fields. However, we are still missing an useful tool, as one might wonder what the complex analogue of the flow of a real-holomorphic vector field is.

So, let $v^{1,0}$ be a holomorphic vector field, and consider the real-holomorphic vector fields $v$ and $Jv$. notice that \begin{equation} v=2\,\Re(v^{1,0})$\mbox{ and }Jv=-2\,\Im(v^{1,0}) \end{equation} and notice also that by \eqref{realhol}, $[v,Jv]=0$. So the flows of $v$ and $Jv$ commute; call them $f^v_t$ and $f^{Jv}_s$, and consider the composition \begin{equation} \Phi_{z}:=f^v_t\circ f^{Jv}_s,\ \mbox{ for }z=t+\operatorname{i}s \end{equation} Then $\Phi_z$ is a subgroup of $\operatorname{Aut}(X,J)$, and it can be checked that for any $p\in M$, $$v^{1,0}_p=\partial_{z=0}\Phi_z(p).$$ In other words, $\Phi_z$ is the "complex flow" of the holomorphic vector field $v^{1,0}$.