homework and exercises - Killing vector and one-form

p. 21 in this paper (http://arxiv.org/abs/0704.0247)

$V$ is Killing vector, where $V^2 = −4b\bar{b}$, which means it is timelike Killing vector.

The authors say:

From $V^2 = −4|b|^2$ and $V = ∂_t$ as a vector we get $V_t = −4|b|^2$,

My question here is how did the authors set $V_t$ equal to this value?

They add,

From $V^2 = −4|b|^2$ and $V = ∂_t$ as a vector we get $V_t = −4|b|^2$, so that $V = −4|b|^2(dt+σ)$ as a one-form, with $σ_t = 0$.

Why did they assume that?

Answer

My question here is how did the authors set $V_t$ equal to this value?

On page 21 the authors say: "Let us choose coordinates $(t, z, x_i)$ such that $V = \partial_t$ and $i = 1, 2$."

So they chose the coordinates such that $V=\partial_t$, which means $V^t=1$. Note that the other components of $V^\mu$ are zeros. Next we have $$V^2=V_\mu V^\mu=-4|b|^2=V_t V^t+V_{x_1}V^{x_1}+V_{x_2}V^{x_2}+V_{z}V^{z}=V_t*1,$$ from which you find $V_t=-4|b|^2$. Here we used $V^{x_i}=V^z=0$.

Though what I don't get is their requirement that σt be equal to zero and why did they place a dt next to it. Why did they assume that?

$\sigma$ is a general one form on coordinates $x^i, z$, which means $\sigma=\sigma_{1}dx^1+\sigma_2 dx^2+\sigma_3 dz$, note that later they use the gauge freedom to set $\sigma_z=0$. They chose the coordinates to fix $V_t$ and the rest it is the most general one form on $x^i$. For example, the most general one form on coordinates $t,x^1, x^2,z$ is $\alpha=\alpha_0 dt+\alpha_1 dx^1+\alpha_2 dx^2+\alpha_3 dz$. Compare it to their expression for $V$ (after (4.7)) and you will see that they chose only the first components, the rest is arbitrary.