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.

  • $\begingroup$ 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... $\endgroup$
    – wisefool
    Commented Dec 13, 2012 at 19:03

2 Answers 2


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 $J(\frac{\partial}{\partial x}) = \frac{\partial}{\partial y}$ and $J(\frac{\partial}{\partial y}) = -\frac{\partial}{\partial x}$. 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: $T_pX\otimes\mathbb{C}=T^{'}_pX\oplus T^{''}_pX$. The $i$ eigenspace, $T^{'}_pX$, is the holomorphic tangent space and is spanned by vectors $\frac{\partial}{\partial z_i} = \frac{\partial}{\partial x_i}-i\frac{\partial}{\partial y_i}$ while $T^{''}_pX$ is spanned by $\frac{\partial}{\partial\bar{z}_i} = \frac{\partial}{\partial x_i}+i\frac{\partial}{\partial y_i}$. 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 dual to the basis $\{\frac{\partial}{\partial z_i}, \frac{\partial}{\partial \bar{z_j}}\}$ we can write this as:

$$g = \sum g_{ij}dz_i\otimes dz_j + \sum g_{\bar{i}\bar{j}}\bar{dz}_i\otimes\bar{dz}_j + \sum g_{\bar{i}j} \bar{dz_i}\otimes dz_j + \sum g_{j\bar{i}} dz_j\otimes\bar{dz_i}$$

As in Javier's comment, where $g_{ij} = g(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j})$ and so on.

Observe that

$$ g(J\frac{\partial}{\partial z_i},J\frac{\partial}{\partial z_j}) = g(i\frac{\partial}{\partial z_i},i\frac{\partial}{\partial z_j}) = -g(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j})= -g_{ij}$$

Using the fact that $g$ is Hermitian however we also have $$g(J\frac{\partial}{\partial z_i},J\frac{\partial}{\partial z_j})= g(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j}) = g_{ij}$$

Hence $g_{ij} = 0$. Similarly $g_{\bar{i}\bar{j}} = 0$. Finally observe that:

$$ g_{\bar{i}j} = g(\frac{\partial}{\partial \bar{z_i}}, \frac{\partial}{\partial z_j}) = g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) + g(\frac{\partial}{\partial y_i}, \frac{\partial}{\partial y_j}) + i\left(g(\frac{\partial}{\partial y_i},\frac{\partial}{\partial x_j}) - g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j})\right) $$ But, using the fact that $g$ is Hermitian again, as well as the definition of $\frac{\partial}{\partial y_i}$ and $\frac{\partial}{\partial x_i}$: $$ g(\frac{\partial}{\partial y_i},\frac{\partial}{\partial x_j}) = g(J\frac{\partial}{\partial x_i},-J\frac{\partial}{\partial y_j}) = - g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j}) $$ We also have $$ g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) = g(\frac{\partial}{\partial y_i},\frac{\partial}{\partial y_j}) $$

and so:

$$g_{\bar{i}j} = 2g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) -2ig(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j})$$

Using the same argument (i.e. using the fact that $g$ is Hermitian as well as the relationship between $\frac{\partial}{\partial x_i}$ and $\frac{\partial}{\partial y_i}$) we can show that: $$g_{j\bar{i}} = 2g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) -2ig(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j}) = g_{\bar{i}j}$$ We can also show that

$$g_{i\bar{j}} = \overline{g_{\bar{i}j}} $$

So we have $g = 2\sum g_{\bar{i}j} (\bar{dz_i}\otimes dz_j+dz_j\otimes\bar{dz_i})$ as required.

A word of caution; $g$ as defined above is a Hermitian metric in the sense of Griffiths and Harris. That is, a "positive definite Hermitian inner product: $$ (\ ,\ )_{p}: T^{'}_{p}M\otimes\overline{T^{'}_{p}M} \to \mathbb{C}$$ on the holomorphic tangent space at $p$ for each $p\in M$ depending smoothly on $p$" (pg. 27). I find it clearer to think of

$$h_p: T^{'}_{p}M\otimes T^{'}_{p}M \to \mathbb{C}$$ $$h_p(\xi,\zeta) = g(\xi,\overline{\zeta})$$ as the Hermitian metric on $T^{'}_{p}M$ as this more obviously (to me at least) a Hermitian inner product on each tangent space.

I tried to find a good reference for this construction; I think the best would be either Huybrecht's Complex Geometry pg. 28-31 or Gross, Huybechts and Joyce's "Calabi-Yau Manifolds and Related Geometries" (look at the beginning of the chapter written by Joyce).

  • $\begingroup$ Daniel, thank you for the detailed answer. This makes my understanding clearer. $\endgroup$
    – M. K.
    Commented Feb 4, 2013 at 7:51
  • $\begingroup$ You're most welcome! $\endgroup$ Commented Feb 7, 2013 at 13:16
  • $\begingroup$ What confuses me is why do you have only 3 terms in the decomposition of g, rather than 4. This would lead to a necessity of having one more condition that is $g(X,Y)=\overline{g(Y,X)}$ . $\endgroup$
    – Edvard
    Commented Aug 20, 2014 at 16:02
  • $\begingroup$ It is very unclear for me, how do you get $\overline{g_{\bar{i}j}}=g_{i\bar{j}}$. In particular, why do you consider $g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial y^j})$ to be real? $\endgroup$
    – Edvard
    Commented Jan 18, 2015 at 11:11
  • $\begingroup$ Well, $g$ was initially a (real-valued) Riemannian metric on $T_{p}(X)$ and $\frac{\partial}{\partial x^{i}}$, $\frac{\partial}{\partial y^{i}}$ are real tangent vectors (i.e. in $T_{p}X$),hence $g(\frac{\partial}{\partial x^{i}},\frac{\partial}{\partial y^{i}})$ is a real number. When we extend $g$ from $T_{p}X$ to $T_{p}X\otimes\mathbb{C}$ this remains true. $\endgroup$ Commented Jan 19, 2015 at 15:45

$∑g_{i\bar j}dz^i⊗d\bar z^j+∑g_{\bar j i}d\bar z^j⊗dz^i=$ $∑g_{i\bar j}(dz^i⊗d\bar z^j+d\bar z^j⊗dz^i)$

I think it should be this. I only write it because the notation is missleading.


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