Killing field in Minkowski space-time

Question

If we look at the killing equation for a vector field $X$ in $\mathbb{R}^{(p,q)}$ (or on an open subset thereof) in coordinates with constant diagonal pseudo-metric we get:

$$X_{\mu,\nu}+X_{\nu,\mu}=0 \tag{1}$$

In the case $\mu=\nu$ it is clear that this implies $X^\mu$ does not depend on the coordinate $q^\mu$. In the book I am reading it is remarked that the equation also implies that $X^\mu$ in general can be put in the form:

$$X^\mu=c^\mu+\omega^\mu_\nu q^\nu$$

Where (1) implies for $\omega$:

$$\omega^Tg+g\omega=0$$

where $g$ is the metric.

My question is, why does the Killing equation imply that $X$ is linear in the chosen coordinates?

Answer 1

Here is one method:

1. The Killing equation $$({\cal L}_X\eta)_{\mu\nu}~=~0\tag{1}$$ for a constant metric $$\eta_{\mu\nu}={\rm const}\tag{2}$$
   reads $$ X_{\mu,\nu}+X_{\nu,\mu}~=~0,\tag{3}$$ as OP correctly mentions.

2. On the other hand consider a coordinate transformation $$ x^{\mu}~~\longrightarrow~~ x^{\prime \nu}~=~f^{\nu}(x), \tag{4}$$ that preserves the metric (2), i.e. $$ \eta_{\mu\nu}~=~\frac{\partial x^{\prime\kappa}}{\partial x^{\mu}}\eta_{\kappa\lambda}\frac{\partial x^{\prime\lambda}}{\partial x^{\nu}}. \tag{5}$$ Note that for an infinitesimal coordinate transformation of the form $$ \delta x^{\mu}~=~x^{\mu \prime}-x^{\mu}~=~ \varepsilon X^{\mu}, \tag{6}$$ eq. (5) becomes exactly the same eq. (3)!

3. The solutions (4) to eq. (5) is proven to be affine transformations in e.g. this Phys.SE post using various methods.

Answer 2

This is essentially a consequence of the connection on $\mathbb{R}^n$ being flat.

One can give the $\omega$ in $X = c + \omega x$ explicitly as $\omega_{\mu\nu} = \partial_\mu X_\nu$, and the $c$ as $c = X - \omega x$.

For a Killing field $X$, one has that $\nabla_\mu\nabla_\nu X_\rho = {R^\sigma}_{\mu\nu\rho}X_\sigma$, where $R$ is the Riemann tensor, but $R=0$ for flat connections, so $\partial_\mu\omega_{\nu\sigma} = \partial_\mu\partial_\nu X_\sigma = 0$, i.e. $\omega$ is constant.

Evaluating $\partial_\mu c_\nu$, one also finds that $c$ is constant if $\omega$ is antisymmetric, and $\partial_\mu X_\nu$ is antisymmetric by virtue of the Killing equation.

Altogether, this gives $X= c+\omega x$ for $c,\omega$ constant and $\omega$ antisymmetric.

Answer 3

Sorry to reply 8 years later but if you take the killing equation, differentiate wrt an arbitrary third coordinate and amtisymmetrise, you get a pair of equations which yields the result you want.