# Killing vectors for a given Kahler potential

1. Jul 5, 2012

### L0r3n20

Hi all!
It's already a couple of days that I'm trying to solve this problem. I've been given the following Kahler potential
$$\mathcal{K} = - Log\left[ i \left(s-\bar{s}\right)\left(t-\bar{t}\right)\left(u-\bar{u}\right)\right]$$
and I have to compute the Killing vectors of such a manifold. I've tried to compute them using
$$\nabla_a \bar{\xi_b} + \bar{\nabla}_b \xi_a =0$$
but then I can't solve the PDE system if not trivially.
Any suggestion? =)

2. Jul 5, 2012

### L0r3n20

By the way I forgot to tell you I've tried using Cartan's calculus
$$\mathcal{L}\left(g\right) =\iota_{\xi}\left(\mathrm{d}g\right) + \mathrm{d}\left(\iota_{\xi}g\right)$$
but I wasn't able to write the metric in a "covariant" way. It should be something like (I know this formula is wrong it's just to give you the taste)
$$-\frac{1}{\left(x_i-\bar{x}_j\right)\delta^{ij}}\delta_{ij}$$
since I have a diagonal metric...

3. Jul 6, 2012

### Ben Niehoff

This formula is not applicable, because g is not a differential form. However, you can use the product rule:

\begin{align*}\mathcal{L}_\xi g &= \mathcal{L}_\xi (g_{ij} \, dx^i \otimes dx^j) \\ &= (\mathcal{L}_\xi g_{ij}) \, dx^i \otimes dx^j + g_{ij} \, (\mathcal{L}_\xi dx^i) \otimes dx^j + g_{ij} \, dx^i \otimes (\mathcal{L}_\xi dx^j) \end{align*}
What do you mean?

Might want to work on that. It appears that you don't understand how the index gymnastics works. You have too many i, j indices.

4. Jul 8, 2012

### L0r3n20

$$g_{i \bar{j}} \mathrm{d} z^i \wedge \mathrm{d} \bar{z}^{\bar{j}} = -\frac{1}{\left( s-\bar{s}\right)} \mathrm{d} s \wedge \mathrm{d} \bar{s} -\frac{1}{\left( t-\bar{t}\right)} \mathrm{d} t \wedge \mathrm{d} \bar{t} -\frac{1}{\left( u-\bar{u}\right)} \mathrm{d} u \wedge \mathrm{d} \bar{u}$$ from which I'm not able to extract $$g_{i \bar{j}}$$