# Lie algebra of the group of biholomorphisms

complex-geometrydifferential-geometry

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 $$X$$).

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 $$$$\label{realhol} \tag{\star} \mathcal{L}_vJ=0$$$$ and for any $$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:

1. if $$v$$ is real-holomorphic, also $$Jv$$ is;

2. $$v$$ 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 $$$$v=2\,\Re(v^{1,0})\mbox{ and }Jv=-2\,\Im(v^{1,0})$$$$ and notice also that by \eqref{realhol}, $$[v,Jv]=0$$. So the flows of $$v$$ and $$Jv$$ commute; call them $$f^v_t$$ and $$f^{Jv}_s$$, and consider the composition $$$$\Phi_{z}:=f^v_t\circ f^{Jv}_s,\ \mbox{ for }z=t+\operatorname{i}s$$$$ Then $$\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}$$.