Consider a Kähler manifold of complex dimension $n$ with Kähler metric $\omega$ and Riemann metric $g$ which we complex linearly extend to $h$. My ultimate goal is to prove that $\omega^n/n!$ is the volume form of $g$. My idea of proving this was the following:
- It is enough to prove it locally at some $p\in M$
- Consider local coordinates $z_1,\dots,z_n$ around $p$
- Up to performing a linear coordinate change, we can assume that $h_p\left(\frac{\partial}{\partial z_i},\frac{\partial}{\partial \overline{z_j}}\right)=\frac{\delta_{ij}}{2}$ (Kronecker delta)
- Then $\omega_p=\sum_{i}dx_i\wedge dy_i$ and $\omega_p^n=n!dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n$.
The volume form of $g$ at $p$ is given by $\sqrt{\det G(p)}\,dx_1\wedge\cdots\wedge dx_n$, where $G(p)$ is the positive definite matrix corresponding to $g$ at $p$. All that is left to show is thus that $\det G(p)=1$. I believe that $G(p)$ is the identity, that is,
$$g_p\left(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\right)=\delta_{ij},\quad g_p\left(\frac{\partial}{\partial y_i},\frac{\partial}{\partial y_j}\right)=\delta_{ij},\quad g_p\left(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j}\right)=0.$$
Is this true? If so, what would be a hint to prove it?
Edit: Another idea I just had is that maybe $G(p)$ is not necessarily the identity, but a diagonal block matrix, whose blocks on the diagonal are $2\times 2$ matrices of determinant $1$. This seems like it could follow from the assumption of bullet point 3.
Best Answer
You need a formula relating $\omega$ and $g$, which is of course
$$ g(X, Y) = \omega (X, JY), $$ where $J$ is the usual complex structure
$$ J \left( \frac{\partial }{\partial x^i} \right)= \frac{\partial }{\partial y^i}, \ \ J \left( \frac{\partial }{\partial y^i}\right) = - \frac{\partial }{\partial x^i}, \ \ \forall i=1, \cdots , n. $$
Then you can show that $g$ is the identity. For example,
$$g \left( \frac{\partial }{\partial x^i}, \frac{\partial }{\partial x^j}\right) = \omega \left( \frac{\partial }{\partial x^i}, J\frac{\partial }{\partial x^j}\right) = \omega \left( \frac{\partial }{\partial x^i}, \frac{\partial }{\partial y^j}\right) = \delta_{ij}$$
(We use, of course, that $\omega = \sum_i dx^i \wedge dy^i$ in the last equality).