I am taking a Kähler geometry course this semester. The book we use is Tian's Canonical Metrics in Kähler Geometry. I got a little confused about the calculation there in.
For example, $\mathbb{C}$ is a Kähler manifold with standard Euclidean metric. Let's assume $x, y$ are the standard coordinates for $\mathbb{C}$, such that the Riemannian metric is given by $g\left(\frac{\partial }{\partial x}, \frac{\partial }{\partial x}\right)= g\left(\frac{\partial }{\partial y}, \frac{\partial }{\partial y}\right)= 1,\\ g\left(\frac{\partial }{\partial x}, \frac{\partial }{\partial y}\right)=0 $ Let $z=x+\sqrt{-1}y$, then one has $\frac{\partial }{\partial z} =\frac12\left(\frac{\partial }{\partial x}-\sqrt{-1}\frac{\partial }{\partial y}\right),\ \ \frac{\partial }{\partial \bar{z}} =\frac12\left(\frac{\partial }{\partial x}+\sqrt{-1}\frac{\partial }{\partial y}\right)$
Now in the book the author claims that $g\left(\frac{\partial }{\partial z},\frac{\partial }{\partial \bar{z}} \right)=1$ See the definition of $g_{i\bar{j}}$ on page 3 and the Kähler form $\omega_g$ for $\mathbb{C}$ defined on previous page.
But here is my calculation:
$ g\left(\frac{\partial }{\partial z},\frac{\partial }{\partial \bar{z}}\right)= \frac14 g\left(\frac{\partial }{\partial x}-\sqrt{-1}\frac{\partial }{\partial y}, \frac{\partial }{\partial x}+\sqrt{-1}\frac{\partial }{\partial y}\right )\\=\frac14 \left (g\left( \frac{\partial }{\partial x}, \frac{\partial }{\partial x}\right) + g\left( \frac{\partial }{\partial y}, \frac{\partial }{\partial y}\right)\right)=\frac12$