Automorphism group of complex manifold The (holomorphic) automorphism group  of the unit disc is $SU(1,1)$ which is not a complex Lie group!
So why (intuitively) the (holomorphic) automorphism group of a complex manifold  wouldn't have
to be a complex Lie group even though if it is finite dimensional?
 A: First of all, the automorphism group of a complex manifold $M$ is not always finite dimensional, the standard example is ${\mathbb C}^2$, its automorphism group contains the infinite-dimensional subgroups of "shears"
$$
(z,w)\mapsto (z, h(z)+w),
$$
where $h$ is an arbitrary entire function on ${\mathbb C}$. However, if $M$ is compact then indeed, $Aut(M)$ is naturally a real Lie group. Typically, though, the automorphism group is trivial.
Suppose you impose further assumptions on $M$, say, that it is a simply-connected Hermitian symmetric space (HSS). Two extreme cases of HSS's are:

*

*Hermitian symmetric spaces of noncompact type. For instance, the open unit ball in ${\mathbb C}^n$.


*Compact HSS. For instance, complex projective spaces.
As it turns out, in the second case, the automorphism group is naturally a complex Lie group, meaning that it has the structure of a complex Lie group $G$ such that the action map
$$
G\times M\to M 
$$
is holomorphic. Briefly, the space of holomorphic vector fields on $M$ is finite-dimensional (since $M$ is compact) and is naturally a complex vector space (that much is indeed clear and is true regardless of compactness, etc.). Each holomorphic vector field is integrable (again, by compactness), hence, you conclude that the identity component of $G$ is naturally a complex Lie group.
On the other hand, in the former case, the automorphism group is never naturally a complex Lie group. Here is a reason (hopefully, intuitive enough). As it turns out, each HSS of noncompact type is biholomorphic to a bounded domain $D$ in ${\mathbb C}^n$ and the stabilizer of a point $x\in D$ is a compact connected nontrivial Lie subgroup $K< G$. Now, if $G$ were to admit a complex structure such that its action on $D$ is holomorphic, then $K$ would be also a complex Lie subgroup of $G$. But then, for each point $y\in D$ the $K$-orbit
$$
Ky=\{k(y): k\in K\}
$$
would be a compact complex submanifold of $D$, hence, of ${\mathbb C}^n$. Thus, $Ky$ would be a single point (namely, $y$). It would then follow that $K$ is a trivial subgroup, which is a contradiction.
