# Killing spinor equation

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The Supersymmetry transformation is:

$$\delta \psi_\mu^i=(\partial_\mu +1/4 \gamma^{ab}\omega_{\mu ab})\epsilon^i -1/8\sqrt{2}\kappa \gamma^{ab}F_{ab}\epsilon^{ij} \gamma_\mu \epsilon_j$$ For the time part, we substitute $\mu$ by $t$ so equation (2) is:

$$\delta \psi_{tA} = \partial_t \epsilon _A + \frac{1}{4} \gamma^{ab} \omega_{tab} \epsilon_A - \frac{1}{8} \sqrt{2} \kappa F_{ab} \gamma^{ab} \gamma_t \epsilon_{AB} \epsilon^B =0$$

Automatically since the spin connections I found to be:

$\omega^{0i} = e^U \partial_iU e^0$ and $\omega ^{ij} = -dx^i\partial_jU+dx^j\partial_iU$

Implies that the ${a, b}$ indices in equation (2) should either be ${0,i}$ or ${i, j}$ $$\delta \psi_{tA} = \partial_t \epsilon _A + \frac{1}{4} \gamma^{0i} \omega_{t0i} \epsilon_A - \frac{1}{8} \sqrt{2} \kappa F_{0i} \gamma^{0i} \gamma_t \epsilon_{AB} \epsilon^B + \frac{1}{4} \gamma^{ij} \omega_{tij} \epsilon_A - \frac{1}{8} \sqrt{2} \kappa F_{ij} \gamma^{ij} \gamma_t \epsilon_{AB} \epsilon^B =0$$

From here, I am having difficulty in proceeding. Some papers like http://arxiv.org/abs/hep-th/0608139 see (4.3.27) separate the $\gamma^{0i}$ into $\gamma^{0}\gamma^{i}$.

This is the first time I encounter such problem, I have no idea how to proceed from here in order to find 2 conditions for ERN BH.

This post imported from StackExchange Physics at 2014-12-03 15:54 (UTC), posted by SE-user beyondtheory

@Sean if I continued from where I stopped, this would be: $$\partial_t \epsilon_A + \frac{1}{4} \gamma^{0i} e^U \partial_iUe^0 \epsilon_A + \frac{1}{4} \gamma^{ij}(-dx^i \partial_jU+dx^j \partial_iU )\epsilon_A -\frac{1}{8} \sqrt{2} \kappa (-\partial_i A_t) \gamma^{0i} \gamma_t \epsilon_{AB} \epsilon^B = 0$$ But the final answer written in the book (actually this is not a homework but an exercise I was following) is: $$\delta \psi_{tA} = \partial_t \epsilon _A +1/2 e^{2U} \partial_i U\gamma^i \gamma^0 \epsilon_A -1/4 \sqrt{2}\kappa e^u \partial_i A_t \gamma^i \epsilon_{AB} \epsilon^b =0$$
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