Complex normal coordinates in Kähler manifolds

Question

Let $(M, g, J, \omega)$ be a Kähler manifold. That is, $(M, J)$ is a complex manifold, $g$ is a Hermitian metric on $M$ and $$\omega (X, Y) = g(JX, Y)$$ is a closed two form.

As a Riemannian manifold $(M, g)$, for each $x\in M$, one can find a geodesic normal coordinates around each $x$. In the case of Kähler metric, one actually have more:

Proposition: (Complex Normal Coordinates on Kähler manifolds) For each $x\in M$, there is a local holomorphic coordinates around $x$ so that the metric $g = g_{i\bar j}$ satisfies $$g_{i\bar j}(x) = \delta_{ij}, \ \ d g_{i\bar j} (x) = 0, \ \ \ \frac{\partial^2 g_{i\bar j}}{\partial z_k \partial z_l} (x) = 0.$$

While the first two conditions are similar to what we have for geodesic normal coordinates in Riemannian geometry, there is no corresponding analogue for the last condition. Also, even in a Kähler manifold, the geodesic normal coordinates might not be holomorphic.

I am looking for a proof of this proposition.

Answer 1

The result is local, we may assume that $p =0 \in U$, where $U \subset \mathbb C^n$ is an open set. By a complex linear transformation, we may also assume $h_{i\bar j}(0) = \delta_{ij}$. Consider the holomorphic mapping $\phi : B_\epsilon \to U$, given by $$z_i=\phi (w)_i = w_i + A^i_{mn} w_mw_n + B^i_{pqr} w_pw_qw_r,$$

here $A^i_{mn}, B^i_{pqr}$ are constants to be chosen, and repeated indices means sum over. We also assume that $A^i_{mn}, B^i_{pqr}$ are fully symmetric with respect to the lower indices. Since $d\phi_0 = Id$, $\phi$ is a biholomorphism onto its image (restricting to smaller sets when necessary). Let $g = \phi^* h$ be the pullback metric. Then using the new cordinates $(w_1, \cdots, w_n)$,

\begin{align}\tag{1} g_{\alpha \bar \beta} &= \frac{\partial z_i}{\partial w^\alpha}\overline{\frac{\partial z_j}{\partial w^\beta}}h_{i\bar j} \end{align} and $$\tag{2} \frac{\partial z_i}{\partial w_\alpha} = \delta_{i\alpha} + 2A^i_{\alpha m} w_m + 3B^i_{\alpha pq } w_pw_q$$ Then we have \begin{align}g_{\alpha\bar\beta} &= h_{\alpha\bar\beta} +2A^i_{\alpha m} h_{i\bar\beta}w_n + 2\overline{A^j_{\beta n}} h_{\alpha \bar j} \bar w_n \\ &\ \ \ + 3B^i_{\alpha pq } w_pw_q h_{i\bar\beta} + 3\overline{B^j_{\beta rs} } \bar w_r \bar w_s h_{\alpha \bar j} + 4 A^i_{\alpha m} \overline{A^j_{\beta n} } h_{i\bar j}w_m \bar w_n\\ &\ \ \ + O(|w|^3). \end{align} From here it's obvious that $g_{\alpha\bar\beta}(0) = \delta_{\alpha\beta}$. Also, we have \begin{align} \partial_\gamma g_{\alpha\bar\beta} (0) &= \partial_\gamma h_{\alpha\bar\beta} (0) + 2A^\beta_{\alpha\gamma}, \\ \bar\partial_\gamma g_{\alpha\bar\beta} (0) &= \bar\partial_\gamma h_{\alpha\bar\beta} (0) + 2\overline{A^\alpha_{\beta\gamma}}, \\ \partial_\eta \partial _\gamma g_{\alpha\bar\beta} (0) &= \partial_\eta \partial _\gamma h_{\alpha\bar\beta} (0) + 6B^\beta_{\alpha\gamma\eta}. \end{align}

Now we choose $A^\beta_{\alpha\gamma} = -\frac 12 \partial_\gamma h_{\alpha\bar\beta}$. First of all, $A$ so chosen is really symmetric in the lower indice, since $$\tag{3}\partial_\gamma h_{\alpha\bar\beta} =\partial_\alpha h_{\gamma \bar\beta}$$ when $h$ is a Kähler metric (see here). Next, since $h$ is Hermitian, $$\bar \partial_\gamma h_{\alpha\bar\beta} (0) = \overline{\partial_\gamma \overline{h_{\alpha\bar\beta}(0)}} = \overline{\partial_\gamma h_{\beta\bar\alpha}(0)} = -\frac{1}{2} \overline{A^\alpha_{\beta\gamma}}, $$ where in the last equality we used the definition of $A$. Thus we have $dg_{i\bar j}(0) = 0$.

Lastly, we choose $B^\beta_{\alpha\gamma\delta} = - \partial_\eta \partial _\gamma h_{\alpha\bar\eta} (0)$. Again by (3), $\partial_\eta \partial _\gamma h_{\alpha\bar\eta}$ is symmetric in $\alpha, \gamma, \eta$. Thus $B$ is again well defined, and this finishes the proof.

In this book, they state the following proposition:

Proposition 1.6: (Normal coordinates in Kähler case) Let $M$ be a Kähler manifold with a real analytic Kähler metric. Given $x\in M$, there exist local complex coordinates $(z_1, \cdots, z_n)$ unique modulo unitary linear transformations such that $g_{i\bar j}(x) = \delta_{ij}$, $dg_{i\bar j}(x)= 0$ and $$ \frac{\partial ^l g_{i\bar j}}{\partial z_{i_1} \cdots \partial z_{i_k}} (x) = 0$$ for all $l\ge 0$ and $i_1 + \cdots + i_k = l$, and this also holds for its conjugate.

They also suggest a reference, p.286, claiming that there's an elegant proof.

Answer 2

Although the question is somewhat old, I still write here a beautiful proof due to J.P. Demailly in his online book Complex analytic and differential geometry located on page 298.

Theorem A: Let $\omega$ be a closed smooth positive definite $(1,1)$ form on a complex manifold. Let $\omega$ be Kaehler, then it is necessary and sufficient that to every $x_0\in X$ corresponds a holomorphic coordinate system $(z_1,\ldots, z_n$) centered at $x_0$ such that $$\omega=i \sum \omega_{lm}dz_l\wedge d\overline z_m,\,\,\,\, \omega_{lm}=\delta_{lm}+O(|z|^2).$$

Some explanations： For the fixed point $x_0$, after translation, one may assume $x_0=0$. We then hope to achieve that $h_{ij}(0)=\delta_{ij}$. To this end, suppose that the old coordinate systems are $\left\{x_i\right\}$. Set $\left<\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\right>=g_{ij}$. Then we use constant linear transformation $x_i=x_i^\prime c_{ij}$ such that $$C^{T}g_{ij}(0)\overline C=\delta_{ij}=\left<\frac{\partial}{\partial x^\prime_i},\frac{\partial}{\partial x^\prime_j}\right>(0).$$ The final goal of Theorem A is obtained by using the Taylor expansion method to eliminate the linear term.

The above theorem corresponds the first two conditions in the question (also pointed out by Arctic Char in the comment: this theorem also proved by Griffiths-Harris), however, the coordinates are not so elaborate. Thus, to make the last condition hold, we need the following theorem. Demailly also write down the coefficients of the order 2 term explicitly, they are just the opposite numbers of the coefficients of the Chern curvature tensor!

Theorem B: If $\omega$ is Kaehler, we can choose the coordinate system $\left\{{z_i}\right\}$ such that $$\omega_{lm}=\delta_{lm}-c_{jklm}z_j\overline z_{k}+O(|z|^3),$$ where $(c_{jklm})$ are the coefficients of the Chern curvature tensor $$\Theta(T_X)_{x_0}=\sum_{j,k,l,m}c_{jklm}\,\,dz_j\wedge d\overline z_{k}\otimes (\frac{\partial}{\partial z_l})^*\otimes \frac{\partial}{\partial z_m}$$ associated to $(T_X,\omega)$ at $x_0$.

We now come to prove Theorem B.

Proof: Suppose that Theorem A holds, as a result one then can find holomorphic coordinates $\left\{w_i\right\}$ adapted to Theorem A such that $$\widetilde{\omega_{lm}}=\delta_{lm}+O(|w|^2)=\delta_{lm}+a_{jklm}w_j\overline w_k+a^\prime_{jklm}w_j w_k+a^{\prime\prime}_{jklm}\overline w_j\overline w_k+O(|w|^3).$$ Since $\widetilde{\omega_{lm}}=\overline{\widetilde{\omega_{ml}}}$, we get $$a_{jklm}=\overline a_{kjml},\,\,\,\,\,\,a^\prime_{jklm}=\overline a^{\prime\prime}_{jkml}.$$ It is easy to see also that $a^\prime_{jklm}=a^\prime_{kjlm}$. The Kaehler condition implies that $$\frac{\partial \widetilde{\omega_{lm}}}{\partial w_j}=\frac{\partial \widetilde{\omega_{jm}}}{\partial w_l},$$ therefore $$a_{jklm}=a_{lkjm},\,\,\,\,\,\,a^\prime_{jklm}=a^\prime_{lkjm}.$$ Particularly, $a^\prime_{jklm}$ is invariant under all permutations of $j,k,l$.

We want to find a change of coordinates ($b_{jklm}$ is to be determined) $$w_m=z_m+\frac{1}{3} b_{jklm}z_jz_kz_l.$$ such that $$\omega_{lm}=\delta_{lm}-c_{jklm}z_j\overline z_{k}+O(|z|^3).$$ Thus, by letting $b_{jklm}=-a^\prime_{jklm}$, we know that it is invariant under all permutations of $j,k,l$, one then obtains $$dw_m=dz_m-a^\prime_{jklm}z_j z_kdz_l.$$ Then,

$$\frac {2}{\sqrt{-1}}\omega=dz_m\wedge d\overline z_m+a_{jklm}z_j\overline z_kdz_l\wedge d\overline z_m+O(|z|^3).$$

It is now easy to compute the Chern curvature tensor $\Theta(TX)_{x_0}$ in terms of the coefficients. Indeed $$\partial<\frac{\partial}{\partial z_l},\frac{\partial}{\partial z_m}>=\left\{D^\prime\frac{\partial}{\partial z_l},\frac{\partial}{\partial z_m}\right\}=a_{jklm}\overline z_k dz_j+O(|z|^2),$$ $$\Theta(T_X)\cdot\frac{\partial}{\partial z_l}=\overline{\partial}D^\prime(\frac{\partial}{\partial z_l})=-a_{jklm}dz_j\wedge d\overline z_k\otimes\frac{\partial}{\partial z_m}+O(|z|).$$

We 're done.