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We say that a Riemannian metric $g$ on a complex manifold $(X,I)$ is Hermitian if $$ g(x,y)=g(Ix,Iy) $$ for any $x,y\in \Gamma(X,TX)$. Here we consider $X$ as a real even dimensional manifold with complex structure $I$.

How can one show that $g$ is locally of the form $$ g=\sum_{i,j}g_{i,\overline{j}}dz_{i} \otimes d\overline{z}_{j} $$ where $z_1,\dots$ are local complex coordinate of $X$.

I am confused with two definition of complex structure; one given by $I\in \Gamma(X,End(TX))$ and the other given by local coordinate.

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Sorry, but I don't understand. A Riemannian metric is real-valued, whereas a Hermitian metric is complex-valued. How can the latter be a particular case of the former? What i think is true is that a Riemannian metric invariant under $J$ is the real part of a Hermitian metric. Moreover, being a Hermitian a section of $TX\otimes (TX)^*$, the formula you wrote is pretty obvious... –  wisefool Dec 13 '12 at 19:03
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1 Answer

up vote 2 down vote accepted

K., this stuff confuses me too, but here's how I understand it. Consider $X$ as a real, $2n$ dimensional manifold equipped with a (Riemannian) metric $g$, and a complex stucture $J$. You are correct in saying it is called Hermitian if $g(JX,JY) = g(X,Y)$, but this does not make it a hermitian inner product in the usual sense.

Now for any $p\in X$ $J_p: T_pX \rightarrow T_pX$ satisfies $J_p^2=-Id$, and so we may always choose real coordinates $x_1,y_1,\ldots,x_n,y_n$ on $X$ such that $J(\frac{d}{dx}) = \frac{d}{dy}$ and $J(\frac{d}{dy}) = -\frac{d}{dx}$. So here's how you get $J$ in local coordinates.

If we extend $J_p$ to the complexification of $T_pX$ it will have two eigenvalues $i$ and $-i$ and $T_pX\otimes\mathbb{C}$ will split into two eigenspaces: $T_pX\otimes\mathbb{C}=T^{'}_pX\oplus T^{''}_pX$. $T^{'}_pX$, the $i$ eigenspace, is the holomorphic tangent space and is spanned by vectors $\frac{d}{dz} = \frac{d}{dx}-i\frac{d}{dy}$ while $T^{''}_pX$ is spanned by $\frac{d}{d\bar{z}_i} = \frac{d}{dx}+i\frac{d}{dy}$. Note that if $\xi \in T_p^{'}X$ $\bar{\xi}\in T_p^{''}$.

We can extend $g$ by complex linearity to be defined on $T_pX\otimes\mathbb{C}$. In coordinates we could write this as $g = \sum g_{ij}dz_i\otimes dz_j + \sum g_{\bar{i}\bar{j}}\bar{dz}_i\otimes\bar{dz}_j + \sum g_{\bar{i}j} \bar{dz_i}\otimes dz_j$. Observe that $g(J\frac{d}{dz_i},J\frac{d}{dz_j}) = g(i\frac{d}{dz_i},i\frac{d}{dz_j}) = -g(\frac{d}{dz_i},\frac{d}{dz_j})$. But using, the fact that $g$ is Hermitian we also have $g(J\frac{d}{dz_i},J\frac{d}{dz_j})= g(\frac{d}{dz_i},\frac{d}{dz_j})$, hence $g_{ij} = 0$, and similarly $g_{\bar{i}\bar{j}} = 0$ so we have $g = \sum g_{\bar{i}j} \bar{dz_i}\otimes dz_j$ as required. $g$ now creates a Hermitian inner product on each holomorphic tangent space $h(\xi,\zeta) = g(\xi,\bar{\zeta})$

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Daniel, thank you for the detailed answer. This makes my understanding clearer. –  M. K. Feb 4 '13 at 7:51
You're most welcome! –  Daniel Mckenzie Feb 7 '13 at 13:16
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