$\mathbb C^2\setminus\{p_1,p_2\}$ is not a homogeneous space

Question

Why the complex manifold $\mathbb C^2\setminus\{p_1,p_2\}$ is not homogeneous where $p_1,p_2\in \mathbb C^2$ are distinct? I mean who to show that there is no complex Lie group acts holomorphically and transitively on $\mathbb C^2\setminus\{p_1,p_2\}$?

Answer 1

Edit: I've goofed and missed/misremembered an important closed assumption in the definition of rationally elliptic spaces. I think this might still be salvage-able, but it is not correct as written.

Simply-connected homogeneous spaces are rationally elliptic. Rationally elliptic spaces have positive Euler characteristic. Since Euler characteristic is a homotopy invariant, it suffices to consider the Euler characteristic of a wedge of two $S^3$s, as this is homotopy-equivalent to $\mathbb{C}^2\setminus \{p_1,p_2\}$. But this space has an Euler characteristic of $-1$. So $\mathbb{C}^2\setminus \{p_1,p_2\}$ cannot be homogeneous.

Answer 2

Let us write $M=\mathbb C^2\setminus \{p_1,p_2\}$.
Using a suitable invertible affine transformation we can assume that $p_1=(0,0)$ and $p_2=(1,0).$
a) The $2$- dimensional Lie group $G$ of invertible matrices of the form $\begin {pmatrix} 1&b\\0&a\end {pmatrix} \\$ $(a,b\in \mathbb C, a\neq 0)$ then acts on the manifold $M$ because it acts on $\mathbb C^2$ while fixing $p_1$ and $p_2$.
That action on $M$ is transitive on the open subset $M_0=\{(z,w)\in \mathbb C^2\vert w\neq 0\}\subset M$ but unfortunately fixes all the points $(z,0)$.
b) To remedy this defect we introduce the group $H=(\mathbb C,+)$ acting on $M$ through $$t\cdot (z,w)=(z,w+tz(z-1))$$ This allows us to send biholomorphically any point of $M$ on the $z$-axis to a point not on the $z$-axis. c) Composing automorphisms of types a) and b) we see that the biholomorphic automorphisms of $M$ act transitively on $M$.
However I don't know how to combine the actions of $G$ and $H$ to an action of a complex Lie group of automorphisms on $M$.