Intuition for Kähler manifolds?

Question

Define a Kähler manifold to be a complex manifold whose associated (1,1) form is closed.

One can show this condition leads to many interesting properties. For example, the Hodge and Lefschetz decompositions force symmetries of the Hodge numbers.

Despite my best efforts, I still don't really understand what's going on here. One attractive explanation I've heard is that the Kähler condition forces a connection between real potential theory associated with a Riemannian metric and the complex structure, but I'm not sure how exactly this connection works.

What, precisely, is that connection? What is the best way to intuitively understand the Kähler condition? Are any of the alternative definitions more helpful for building intuition? And why does it happen that so many of the manifolds we meet in mathematics are Kähler?

Answer 1

Let's consider a complex manifold $X$ of (complex) dimension $n$, and equip it with some Hermitian metric $h$. It's not hard to see that there is a unique connection $\nabla_{Ch}$ that's compatible with the metric and whose $(0,1)$-part is the $\bar\partial$ operator. This is the Chern connection of $h$.

Now let's view the manifold $X$ as a real manifold $M$ of (real) dimension $2n$. The Hermitian metric is now just a Riemannian one, and thus comes with a unique connection $\nabla_{LC}$ that's torsion-free and compatible with the metric. This is the Levi-Civita connection.

A fairly natural question is when those two connections are the same, that is, when can we expect the complex differential geometry of the manifold to coincide with its Riemannian geometry? After all, we know a lot about Riemannian geometry and it would be nice to exploit that knowledge to study the complex geometry. This happens when the Chern connection has no torsion, as one can work out, and we call the Hermitian metrics that satisfy this condition Kähler metrics.

The link between this definition of a Kähler metric and the exterior form one is as follows: Let $\omega = - \operatorname{Im} h$ be the Kähler form of $h$, and let $\tau$ be the torsion tensor of the Chern connection of $h$. If we take three holomorphic tangent fields $\xi, \nu, \eta$, then $$ \partial\omega(\xi, \nu, \overline \eta) = h(\tau(\xi, \nu), \overline \eta). $$ Thus $\tau = 0$ if and only if $\partial \omega = 0$, which is equivalent to $d \omega = 0$ as the form $\omega$ is real.

I like this definition because it's quite natural from a differential-geometric point of view. It can also motivate Hodge theory, because on the Riemannian side we have the Hodge isomorphism between cohomology groups and harmonic forms on compact manifolds. If we ask what that isomorphism looks like on the complex side on a compact Kähler manifold, we'll eventually invent Hodge theory.

Answer 2

The following may not be entirely satisfying (I don't have time to add in the details), but is too long for a comment.

The Kahler condition fits naturally into the framework of $G$-structures.

Def: Given a Lie subgroup $G \leq \text{GL}_n(\mathbb{R})$.

• A $G$-structure on an $n$-manifold $M$ is a $G$-subbundle $B \subset FM$ of the general frame bundle $FM$ of $M$.
• A $G$-structure $B \subset FM$ is flat iff it admits a section $s \in \Gamma(B)$, denoted say $s(p) = (E_1(p), \ldots, E_n(p))$, such that $[E_i, E_j] = 0$. (That is, $s$ is locally a coordinate frame.)
• A $G$-structure $B \subset FM$ is torsion-free iff it admits a torsion-free connection.

Flat implies torsion-free, but the converse generally fails. It can be shown that "torsion-free" is equivalent to being "flat to first-order" in a certain precise sense.

Example: Let $\text{O}(n) \leq \text{GL}_n(\mathbb{R})$ be embedded in the usual way. An $\text{O}(n)$-structure on $M^n$ is equivalent to a Riemannian metric on $M^n$. The $\text{O}(n)$-structure is flat iff $M$ is locally isometric to $(\mathbb{R}^n, g_{\text{std}})$. The Fundamental Lemma of Riemannian Geometry is exactly the statement that every $\text{O}(n)$-structure is torsion-free.

Example: Let $\text{U}(n) \leq \text{GL}_{2n}(\mathbb{R})$ be embedded in the usual way. A $\text{U}(n)$-structure on $M^{2n}$ is equivalent to a pair $(J,g)$ on $M^{2n}$, where $J$ is an almost-complex structure and $g$ is an almost-Hermitian metric for $J$. The $\text{U}(n)$-structure is flat iff $(M^{2n}, J, g)$ is locally isomorphic to $(\mathbb{C}^n, J_{\text{std}}, g_{\text{std}})$. The $\text{U}(n)$-structure is torsion-free iff it is Kahler.

On a compact manifold $M^n$ with a torsion-free $G$-structure, where $G \leq \text{O}(n)$ is a closed Lie subgroup, one can often (maybe always?) state an analogue of the Hodge and Lefschetz decompositions. At the very least, this can be done when $G$ is a group arising in Berger's List: $\text{U}(n)$, $\text{SU}(n)$, $\text{Sp}(n)$, $\text{Sp}(n) \text{Sp}(1)$, $\text{G}_2$, or $\text{Spin}(7)$.

As to why so many manifolds we meet are Kahler? Every submanifold of $\mathbb{C}^n$ and of $\mathbb{CP}^n$ inherits a Kahler structure, and complex affine/projective varieties come up a lot.