I am reading a paper$^1$ by Manton and Gibbons on the dynamics of BPS monopoles. In this, they write the Atiyah-Hitchin metric for a two-monopole system. The first part is for the one monopole moduli manifold, and other terms for a 4-dimensional hyper kahler surface which is $SO(3)$ symmetric parameterized by the euler angles. He obtains two sets of $SO(3)$ killing vectors. What is the systematic way to obtain these two various sets? What are the equations involved?
$$\xi^R_i=\cot{\theta}\cos{\psi}\frac{\partial}{\partial{\psi}}-\sin{\psi}\frac{\partial}{\partial{\theta}}+\frac{cos{\psi}}{\sin{\theta}}\frac{\partial}{\partial{\phi}}$$
$$\xi^R_2=-\cot{\theta}\sin{\psi}\frac{\partial}{\partial{\psi}}+\cos{\psi}\frac{\partial}{\partial{\theta}}+\frac{sin{\psi}}{\sin{\theta}}\frac{\partial}{\partial{\psi}}$$
$$\xi^R_3=\frac{\partial}{\partial{\psi}}$$
and the other set by
$$\xi^L_1=\cot{\theta}\cos{\phi}\frac{\partial}{\partial{\phi}}+\sin{\phi}\frac{\partial}{\partial{\theta}}-\frac{\cos{\phi}}{\sin{\theta}}\frac{\partial}{\partial{\psi}}$$
$$\xi^L_2=-\cot{\theta}\sin{\phi}\frac{\partial}{\partial{\phi}}-\cos{\phi}\frac{\partial}{\partial{\theta}}-\frac{\sin{\psi}}{\sin{\theta}}\frac{\partial}{\partial{\psi}}$$
$$\xi^L_3=-\frac{\partial}{\partial{\phi}}$$
References:
$^1$ G.W. Gibbons and N.S. Manton, Classical and quantum dynamics of BPS monopoles, Nucl. Phys. B274 (1986) 183.
Best Answer
The specific form of the Killing vectors depends on the parameterization of the group element, from the notation (and the results), one can deduce that Euler angle parameterization has been used:
$ g = exp(i\sigma_3 \psi) exp(i \sigma_1 \theta) exp(i \sigma_3 \phi)$
where the sigmas are the generators of rotations with respect to Cartesian axes in the three dimensional representation.
Also the two sets of killing vectors correspond to the left and right action of $SO(3)$ on itself which preserve the invariant metric. I'll describe to you the case of the left action for example.
The basic equation definining the lect killing vectors is
$ K_A^L g = \sigma_A g$ (for the right action $K_A^R g = g \sigma_A$, I'll skip from now the superscript understanding it is a left action).
$K_A$ is a differential operator:
$K_A = K_A^{\phi} \frac{\partial}{\partial \phi} +K_A^{\theta} \frac{\partial}{\partial \theta} +K_A^{\psi} \frac{\partial}{\partial \psi}$
For convenience we shall call $x^1 = \phi$, $x^2 = \theta$, $x^3 = \psi$,
So our task is to compute $K_A^j$
In order to do that,we remember that Maurer-Cartan one form $g^{-1} dg$ is a Lie algebra valued one form, i.e.,
$m = g^{-1} dg = i a_j^A \sigma_A dx^j$ (With summation convention)
Thus, the first task to be done is to explicitely compute the coefficients $ a_j^A$, this is done by computing the derivatives in the given parameterization.
If we contract this form with a killing vector, we obtain:
$<K_A, m> = i K_A^j a_j^B \sigma_B = g^{-1} K_A g =g^{-1} \sigma_A g$
Using the orthogonality relations
$tr(\sigma_A \sigma_B) = \delta_{AB}$
We obtain:
$K_A^j a_j^B = tr(\sigma_B g^{-1} \sigma_A g)$
Thus by solving this system of linear equations or equivalently, inverting the matrix A we get the formula for the Killing vectors components:
$K_A^j = (a^{-1})_B^j tr(\sigma_B g^{-1} \sigma_A g)$
in summary, one needs to compute the coefficient matrix of the Maurer-Cartan form and invert it and to compute the traces required by the last equation, then compute the Killing vector components by matrix multiplication.