Why Kähler geometry?

I am by no means an expert in complex geometry. This post is an attempt to capture the basics of Kähler geometry from the point of view of a non-expert, and is written for those of you who have limited background in such and wonder what makes Kähler manifolds worth studying. Most of the content here is based on a talk I gave in a student differential grometry seminar in Stony Brook.

Hermitian manifolds

We would like to think of a complex vector space as a pair $(V,J)$ where $V$ is a real vector space and $J :V \to V$ is an automorphism such that $J^2 = - Id$. An $\mathbb{R}$-linear map $L: V \to V$ is $\mathbb{C}$-linear if $LJ=JL$. We would also like to think of a complex manifold as a pair $(M,J)$ where $M$ is a smooth real manifold and $J$ is a global section of the bundle $End(TM)$ such that $(T_pM, J_p)$ is a complex vector space for every $p \in M$, and that $M$ satisfies further integrability condition. (This is the Newlander-Nirenberg theorem.)

A Riemannian metric $\langle \cdot, \cdot \rangle$ on a complex manifold $(M,J)$ is Hermitian if at every point $p \in M$, for every pair of tangent vectors $X, Y \in T_pM$, $\langle J_p X, J_p Y \rangle = \langle X, Y \rangle$. The fundamental form of a Hermitian metric $\langle \cdot,\cdot \rangle$ is a real, nowhere vanishing $2$-form $w$ where $w(X,Y) = \langle JX, Y \rangle$.

It should be easy to see that the fundamental form also defines the metric because we have the formula $\langle X, Y \rangle = w(X, JY)$. Also, it is not difficult to make Hermitian metrics out of Riemannian metrics. For example, for any Riemannian metric $\langle \cdot, \cdot \rangle$, the metric defined by $\langle X,Y\rangle + \langle JX ,JY \rangle$ is automatically Hermitian.

On $\mathbb{C}^n$, it is standard to take $J$ to be multiplication by $i$: $J(z_1, \ldots, z_n) = (iz_1, \ldots, iz_n)$. (Of course, there are many other possible choices of $J$.) The inner product $\langle x, y \rangle = \text{Re}(x^T \bar{y})$ on $\mathbb{C}^n$ is Hermitian.

Kähler manifolds

One of the fundamental concepts in Riemannian geometry is parallel transport. It is natural to ask when is the complex structure $J$ is compatible with such process. It turns out that this is reduced to a mere computation of the exterior derivative $dw$.

Lemma: Let $\langle \cdot,\cdot \rangle$ be a Hermitian metric on a complex manifold $(M,J)$ and let $\nabla$ denote the corresponding Levi-Civita connection. The following are equivalent.

1. $\nabla_Y (JX) = J \nabla_Y X$ for every pair of vector fields $X$ and $Y$;
2. $dw = 0$.

I had a hard time finding an easy proof of this in the literature, so I wrote one up myself here. By the invariant formula, for any triplet of vector fields $X_1$, $X_2,$ and $X_3$, we have

where $S = \{(1,2,3),(2,3,1),(3,1,2)\}$. Since the connection $\nabla$ is torsion-free and compatible with the metric, the expression above becomes

Since the metric is Hermitian, the second and fourth sums become zero:

We are left with

which directly proves the lemma.

A Hermitian metric $\langle \cdot, \cdot \rangle$ and its associated fundamental form $w$ on a complex manifold $(M,J)$ are called Kähler if statement 1 and/or 2 in the lemma above holds. Also, we say that $(M,J)$ is Kähler if it admits a Kähler metric.

For computations, it is convenient to know what Hermitian and Kähler metrics look like in locally. In local coordinates, the fundamental form $w$ can be written as

where $(h_{\alpha \bar{\beta}})_{\alpha, \beta}$ is a positive definite Hermitian matrix. Moreover, it is Kähler precisely when $\partial w = \bar{\partial} w = 0$, or equivalently $\frac{\partial h_{\alpha, \bar{\beta}}}{\partial z_\gamma} = \frac{ \partial h_{\gamma, \bar{\alpha}}}{\partial z_\beta}$ for all $\alpha, \beta,\gamma$.

The Kähler assumption induces some topological constraints. Let $w$ be a Kähler form on a compact $n$-dimensional complex manifold $M$. For $k\geq 2$, the $2k$-form $w^k := w \wedge \ldots \wedge w$ is closed because $dw^k = k dw \wedge w^{k-1} = 0$. By direct computation, one can show that in local coordinates,

and this actually coincides with the volume form! Therefore, $w^n$ (and so every $w^k$) cannot be exact.

Corollary: If $M$ is a compact Kähler manifold of complex dimension $n$, then for all $k = 1,\ldots n$, $H^{2k}(M,\mathbb{R}) \neq 0$.

There are actually further constraints on the cohomology of a compact Kähler manifold arising from Hodge theory. This is a big subject and I’ll leave it to you to explore on your own. :)

Let’s list some easy examples and non-examples of Kähler manifolds.

1. $\mathbb{C}^n$ is Kähler. The flat metric has fundamental form $w_{\mathbb{C}^n} = \frac{i}{2} \sum_j dz_j \wedge d\bar{z}_j$.
2. For any lattice $\Lambda \subset \mathbb{C}^n$, the complex $n$-torus $\mathbb{C}^n/ \Lambda$ is Kähler. The form $w_{\mathbb{C}^n}$ is translation invariant and therefore induces a Kähler form on $\mathbb{C}^n/ \Lambda$.
3. The Fubini-Study metric (which can be obtained by pushing forward the round metric on $S^{2n+1} = \{ z \in \mathbb{C}^{n+1} : \vert z\vert = 1\}$ by the quotient $S^{2n+1} \to \mathbb{P}^n = S^{2n+1}/_{z \sim e^{i\theta} z \text{ for any }\theta}$) $w_{FS}$ on the complex projective space $\mathbb{P}^n$ is Kähler.
4. For $n\geq 2$, the Hopf manifold $M = (\mathbb{C}^n \backslash \{0\})/_{z \sim 2z}$ is a compact complex manifold that is diffeomorphic to $S^{2n-1} \times S^1$. Since $H^2(M,\mathbb{R}) = 0$ (e.g. using the Kunneth formula), $M$ is not Kähler.

Kähler submanifolds

We say that a manifold $M$ is a(n) (embedded complex) submanifold of a complex manifold $(\bar{M}, \bar{J})$ if it is the image of an embedding $\iota: M \to \bar{M}$ satisfying $D\iota \circ \bar{J}\vert_M = \bar{J}\vert_M \circ D\iota$ on $M$, or equivalently $\bar{J}_p T_pM \subset T_pM$ at every point $p \in M$. A submanifold $M \subset \bar{M}$ is naturally equipped with the complex structure $J:= \bar{J}\vert_M$. In the point of view of Riemannian geometry, submanifolds of Kähler manifolds turn out to be quite special.

Theorem: Let $(\bar{M}, \bar{J})$ be a complex manifold with Kähler form $w_{\bar{M}}$ and $\iota: M \to \bar{M}$ be a submanifold.

1. $w_M := \iota^*w_{\bar{M}}$ is a Kähler form on $M$;
2. $M$ is minimal, i.e. it has zero mean curvature.

The first statement should be immediate because $dw_M = \iota^* dw_{\bar{M}} = 0$. The second can be obtained through the concept of the second fundamental form $B$, that the symmetric bilinear form $B$ where for any vector fields $X$ and $Y$ of $M$, $B(X,Y)$ is defined to be the normal component $(\bar{\nabla}_X Y)^N$. The trace of $B$ is the mean curvature of $M$. Since $M$ is also Kähler, $B$ is $\mathbb{C}$-linear:

The fact that $M$ is a submanifold blesses $M$ with the existence of a local orthonormal basis of the tangent bundle of the form $\{e_1, J e_1, \ldots, e_m, Je_m\}$. We finally see that $B$ vanishes everywhere:

Submanifolds of $\mathbb{C}^n$ are often called Stein manifolds, whereas submanifolds of $\mathbb{P}^n$ are often called projective manifolds. Both Stein and projective manifolds are objects of high interest in the field of complex geometry. For one thing, we now know that they are minimal!

References

1: Lawson. B. Minimal Varieties In Real and Complex Geometry. Presses de l’Université de Montréal, 1974.
2: Huybrechts. D. Complex Geometry: An Introduction. Springer, 2005.
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