Killing vector and one-form [closed]

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?

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?

closed as unclear what you're asking by ACuriousMind♦, Gert, user36790, Sebastian Riese, yuggibJan 7 '16 at 14:09

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• 1. Please include all relevant information into the question. What is $b$, what is $\sigma$, why is $V$ timelike, what manifold are we even on? 2. If $V$ is an ordinary vector field, $V^2 = -4\lvert b \rvert^2$, $V = \partial_t$ and $V_t = -4\lvert b\rvert^2$ do not make any sense if $V_t$ is meant to be the temporal component of $V$. By definition, $V_t = 1$ if $V = \partial_t$. – ACuriousMind Jan 5 '16 at 1:34
• @ACuriousMind - If $V = \partial_t$ then $V^t = 1$ not $V_t$. – Prahar Jan 5 '16 at 1:48
• @Prahar you're absolutely right since we're talking about dual maps. Maybe it is a typo from the authors or something deeper than that :S. – Beyond-formulas Jan 5 '16 at 1:53
• @Prahar: Ah, yes, sorry. Beyond-formulas: It's a bit weird in the first place to claim to be able to write a vector field as a 1-form, since 1-form are the duals to vector fields. But if they mean that the dual form to $V$ can be written as such, then the claim follows directly from $V_t = -4|b|^2$: $V^\flat = V_\mu\mathrm{d}^\mu = V_t \mathrm{d}t + V_i\mathrm{d}x^i = V_t (\mathrm{d}t + \frac{V_i}{V_t}\mathrm{d}x^i)$ and defining $\sigma = \frac{V_i}{V_t}\mathrm{d}x^i$. – ACuriousMind Jan 5 '16 at 1:57

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.

• Thanks for your answer, but I'm afraid this doesn't answer my question. Maybe what ACuriousMnid mentioned above solved my issue. I can't see if your answer is adding anything to ACuriousMind's answer. Is it? – Beyond-formulas Jan 5 '16 at 2:42
• @Beyond-formulas Well, he made several comments from the mathematical point of view and, I think, he did not answered your questions directly. My answer is direct and simple, also it does not contradict his :) But everything he said is right. – Yuri Jan 5 '16 at 2:47
• $V=\partial_t$ means $V^t=1$, not $V_t=1$! To get the object with lower index you need to use the metric $V_t=g_{t\alpha}V^\alpha$. – Yuri Jan 5 '16 at 2:57
• @PhilosophicalPhysics $V=\partial_t$ means that $V$ is a Killing vector along $t$, which means that the metric is independent of $t$. In general, if the Killing vector has the form $V^{y}=const$ it means that the metric is independent of $y$. – Yuri Jan 5 '16 at 4:14
• @PhilosophicalPhysics I will give you an example, when I write $X=A dt+ B dz$ it means nothing but $X_t=A, X_z=B$. It is just the compact way to write vectors, instead of using smth like $X=(A,B)$. – Yuri Jan 5 '16 at 4:32