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I have solved the Killing vector equations for a 2-sphere and got the following answer. $A,B,C$ are three integration constants as expected.

$$\xi_{\theta}=A \sin{\phi}+B\cos{\phi}$$ $$\xi_{\phi}=\cos{\theta} \sin{\theta}(A \cos{\phi}-B\sin\phi)+c \sin^2{\theta}$$

From this, how do I find the basis elements in terms of the tangent vectors $\frac{\partial}{\partial \phi}$ and $\frac{\partial}{\partial \theta}$? One obvious elements if $\frac{\partial}{\partial \phi}$ itself, as the metric is independent of $\phi$. Another method IMO could be to derive it from the angular momentum generators $x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}$ etc. Is this correct? Would changing the coordinates of these generators give the right answer?

Lastly, How do I do this for any general metric? Is there a standard procedure to find the killing vector fields, as a basis.

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Got it for the 2 sphere.

$\xi=\xi^{\theta}\partial_{\theta}+\xi^{\phi}\partial_{\phi}=-A L_x + B L_y + C L_z$

A, B, C are the same integration constants as in the question and $L_x, L_y, L_z$ are the angular momentum operators written in the spherical polar coordinates.

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    $\begingroup$ In general, the maximal number of Killing vectors (isometries) in N-dimensional space is $\frac{N(N+1)}{2}$. The 2-sphere has 3 killing vectors, so it is maximally symmetric. The isometries of the 2-sphere, are the generators of the 3 rotations about axis $x, y, z$. $\endgroup$
    – Trimok
    Jun 10, 2013 at 8:33
  • $\begingroup$ @Trimok: Thanks! How can I prove the maximum number of killing vectors if $n(n+1)/2$? It doesnt seem obvious to me. $\endgroup$
    – user7757
    Jun 10, 2013 at 8:51
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    $\begingroup$ First, for a standard Euclidean or Lorentzian space, you have $n$ translations and $\frac{n(n -1)}{2}$ "rotations" or "rotations+boosts", so $\frac{n(n + 1)}{2}$ isometries (these space are maximally symmetric), . Then, for a curved manifold, the idea is that the tangent space, at some point, is a vector space, and is some copy of an Euclidean or Lorentzian space (whose inner product is given by the value of the metric at that point). So the number of Killing vectors corresponding to the infinitesimal isometries could not excess $\frac{n(n + 1)}{2}$ $\endgroup$
    – Trimok
    Jun 10, 2013 at 9:28
  • $\begingroup$ How does a sphere have isometries due to translations? Shouldn't it be just rotations $\frac{n(n-1)}{2}$? $\endgroup$
    – user7757
    Jun 11, 2013 at 9:39
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    $\begingroup$ The tangent space of the n-sphere $S^n$ is isomorphic to $R^n$, so the maximum of isometries is $\frac{n(n + 1)}{2}$. This maximum is effectively reached because $S^n$ is maximally symmetric. On the other hand, you can consider the n-sphere $S^n$ as a version of $R^{n+1}$ with norm 1. So, $S^n$ has the rotation isometries of $R^{n+1}$, but not the translations isometries of $R^{n+1}$. So the number of isometries of $S^n$ is $$\frac{(n + 1) (n + 1 - 1)}{2} = \frac{n(n + 1)}{2}$$ $\endgroup$
    – Trimok
    Jun 11, 2013 at 16:55

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