Is every simply connected Kähler manifold homogeneous?

Question

Does every connected simply connected Kähler manifold admit a transitive action by a Lie group?

Answer 1

I am not a complex geometer, but if you can believe Wikipedia, then the (a?) $K3$ surface is not homogeneous for any action of a (finite dimensional) Lie group. I will sketch the most elementary proof I know below.

First, topologically, $K3$ is a closed simply connected $4$-manifold with second Betti number $b_2(K3) = 22$.

Let us suppose for a contradiction that there is some Lie group $G$ which acts transitively on $K3$. By modding out by the ineffective kernel, we may assume the action is effective. Because $K3$ is connected, the identity component of $G$ acts transitively, so we may as well assume $G$ is connected. Further, because $K3$ is compact and simply connected, a theorem of Montgomery implies that the maximal compact subgroup of $G$ acts transitively. In particular, we may assume $G$ is compact. Now this implies we can find a Riemannian metric on $K3$ for which the $G$ action is isometric.

Let $H$ denote the isotropy group at some point $p\in K3$. Since $H$ fixes $p$, it acts (effectively) on the set of unit vectors in $T_p K3$, so we may view $H\subseteq O(4)$. Thus, $H$ has rank at most $2$.

Also, we get a long exact sequence $\rightarrow...\pi_2(H)\rightarrow \pi_2(G)\rightarrow \pi_2(K3)\rightarrow \pi_1(H)\rightarrow \pi_1(G)\rightarrow 0$.

Since $\pi_2$ of any Lie group vanishes, $\pi_2(K3)$ must inject into $\pi_1(H)$. But, from the Hurewicz theorem, $\pi_2(K3)\cong H_2(K3)\cong \mathbb{Z}^{22}$, so $\mathbb{Z}^{22}\subseteq \pi_1(H)$.

This implies that $H$ has a finite cover of the form $H_0\times T^k$ with $k\geq 22$ (and $H_0$ simply connected). In particular, the rank of $H$ is at least $22$.

Since we already know the rank is at most $2$, we have a contradiction.

(In particular, this proof shows that a simply connected closed $4$-manifold which is homogeneous must have $b_2 \leq 2$.)