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I am taking a Kaehler geometry course this semester. The book we use is Tian's Canonical Metrics in Kaehler Geometry. I got a little confused about the calculation there in.

For example, $\mathbb{C}$ is a Kaehler 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 Kaehler form $\omega_g$ for $\mathbb{C}$ defined on previous page.

But here is my calculation:

$$ g\left(\frac{\partial }{\partial z},\frac{\partial }{\partial \bar{z}}\right)= \frac14 g\left(\frac{\partial }{\partial x}-\sqrt{-1}\frac{\partial }{\partial y}, \frac{\partial }{\partial x}+\sqrt{-1}\frac{\partial }{\partial y}\right )\\=\frac14 \left (g\left( \frac{\partial }{\partial x}, \frac{\partial }{\partial x}\right) + g\left( \frac{\partial }{\partial y}, \frac{\partial }{\partial y}\right)\right)=\frac12$$

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Hmm, your calculation looks correct to me. Could it be an error in the text? –  anon Feb 28 '12 at 3:21
    
The text should also be correct, cause I can use the Kaehler form $\omega_g$ to calculate the length of $\frac{\partial}{\partial z}$ –  Sun Feb 28 '12 at 3:44
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