How Do You Compute Killing Vectors for a Given Metric?

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To compute the Killing vectors for the metric x²dx² + xdy², the Killing equation must be applied: Lχg_ab = g_ad∂_bX^d + g_bd∂_aX^d + X^e∂_e g_ab = 0. The process begins by calculating the partial derivatives of the metric components, which are essential for substituting into the Killing equation. This involves evaluating derivatives like ∂_0 g_{00} = 2x and ∂_1 g_{11} = 2xy, among others. After substituting these derivatives into the Killing equation, one can solve for the components of the Killing vectors, X^0 and X^1. Understanding how to manipulate the indices and derivatives is crucial for starting the computations effectively.
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I want to find all the killing vectors of the metric ##x²dx² + xdy²##. We could guess somethings by intuition and check it, but i decided to use the equation itself. Unfortunatelly, i realized that i am not sure how to manipulate the equation

$$L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} =^{?} 0$$

When i wrote this equation, i was not sure how the index should work here, so i guessed that we need to check it for all possible combinations of ab. ##(a,b) = [(0,0),(0,1),(1,0),(1,1)]##$. So, seeing this way, we need to find ##\vec X = (X^{0},X^{1})##? THe answer is something in partial derivatives. How would i got that just with the equation? I would appreciate if you give a tip of how to really start the computations, because as you can see, i am stuck right at the start.
 
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The Killing equation for a metric $g_{ab}$ is given by$$L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} = 0.$$In this case, you are trying to find the Killing vectors $\vec{X}=(X^0,X^1)$ for the metric $$g_{ab}=\begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}.$$ To solve for $\vec{X}$, begin by computing the partial derivatives of the metric components with respect to each coordinate:$$\partial_0 g_{00} = \partial_0 (x^2) = 2x,$$$$\partial_1 g_{00} = \partial_1 (x^2) = 0,$$$$\partial_0 g_{01} = \partial_0 (0)=0,$$$$\partial_1 g_{01} = \partial_1 (0) = 0,$$$$\partial_0 g_{11} = \partial_0 (xy^2) = y^2,$$$$\partial_1 g_{11} = \partial_1 (xy^2) = 2xy.$$Now, substitute these partial derivatives into the Killing equation and solve for $\vec{X}$:\begin{align*}L_\chi g_{ab} &= g_{ad}\partial_b X^d + g_{bd}\partial_a X^d + X^e \partial_e g_{ab} \\&= \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\begin{bmatrix}\partial_0 X^0 & \partial_1 X^0 \\ \partial_0 X^1 & \partial_1 X^1\end{bmatrix} + \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\begin{bmatrix}\partial_0 X^0 & \partial_1 X^0 \\ \partial
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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