# Killing vectors for SO(3) (rotational) symmetry

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.

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Is this better suited for Maths SE? –  ramanujan_dirac Sep 22 '12 at 21:41
If you know the metric and use the natural connection, you solve $\mathcal{L}_X g_{ab}=X_{a;b}-X_{b;a}=0$. These are the vector fields that if you follow them they leave the metric unchanged. –  kηives Sep 22 '12 at 23:22

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.

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Sir, thanks for the reply. Unfortunately, I don't know anything about the Maurer-Cartan form. I am aware of the so called killing equation $V^λ ∂_λ g_{μν} + ∂_μ V^λ g_{λν} + ∂_ν V^λ g_{μλ}$ which comes from $\nabla_\mu V_\nu + \nabla_\nu V_\mu=0$.Can one solve this problem using the above mentioned killing equation rather than the method proposed above. –  ramanujan_dirac Sep 23 '12 at 10:04
@ramanujan_dirac: Actually, for the construction described above, one does not need to know any property of the Maurer-Cartan form except its evaluation by the insertion of the Euler parameter formula of $g$ into its definition. This form has many interseting properties and applications, please see for further reading Shlomo Sternberg lectures: math.harvard.edu/~shlomo/docs/lie_algebras.pdf. –  David Bar Moshe Sep 23 '12 at 10:43
cont. For the computation of the Killing vectors according to the given Wikipedia page, one needs in advance to construct the invariant metric. Furthermore, given the invariant metric, the Killing vector components satisfy differential equations which are harder to solve than the algebraic equations in the method described in the answer. –  David Bar Moshe Sep 23 '12 at 10:44