Symmetry of metric

A scalar field has a symmetry or isometry under a coordinate transformation if

$\varphi ({\tilde{x}}^{\mu }=x_0^{\mu })=\varphi (x^{\mu }=x_0^{\mu })$

Let us now consider an infinitesimal transformation

${\tilde{x}}^{\mu }=x^{\mu }+ϵX^{\mu }\text{for some small }ϵ\text{ and vector field }{X}^{\mu }$

Then, in general

$\varphi (x^{\mu })=\varphi ({\tilde{x}}^{\mu }(x^{\rho }))=\varphi ({\tilde{x}}^{\mu }=x^{\mu }+ϵX^{\mu })=\varphi ({\tilde{x}}^{\mu }=x^{\mu })+ϵX^{\lambda }{\partial }_{\lambda }\varphi ({\tilde{x}}^{\mu }=x^{\mu })+...$

So, to first order in ε, for φ to be invariant, the derivative must vanish in some direction.

Let us generalise this to the metric. The metric has a symmetry under a coordinate transformation if

${\tilde{g}}_{\mu \nu }({\tilde{x}}^{\mu }=x_0^{\mu })=g_{\mu \nu }(x^{\mu }=x_0^{\mu })$

Recall

$g_{\mu \nu }(x^{\mu })=(\frac{\partial {\tilde{x}}^{\rho }}{\partial x^{\mu }})(\frac{\partial {\tilde{x}}^{\sigma }}{\partial x^{\nu }}){\tilde{g}}_{\rho \sigma }({\tilde{x}}^{\mu })$

Again, consider an infinitesimal transformation and follow the steps to obtain

${\tilde{g}}_{\mu \nu }({\tilde{x}}^{\mu }=x_0^{\mu })=g_{\mu \nu }(x^{\mu }=x_0^{\mu })⟹{\partial }_{\mu }{X}^{\rho }{g}_{\rho \sigma }+{\partial }_{\nu }{X}^{\sigma }{g}_{\mu \sigma }+X^{\lambda }{\partial }_{\lambda }{g}_{\mu \nu }=0$

To make it coordinate independent, let us change the partial derivatives to covariant derivatives and obtain theKillings-Equation

${\nabla }_{\mu }{X}^{\nu }+{\nabla }_{\nu }{X}^{\mu }=0$ and a vector satisfying it is called theKilling-Vector.

By Noether’s Theorem, there must be a conserved quantity associated to geodesics for a symmetry of the metric: The scalar

$K=W_{\mu }{V}^{\mu }$ is constant along the geodesic

$\frac{dK}{ds}=V^{\alpha }{\nabla }_{\alpha }(W_{\mu }{V}^{\mu })=\frac{1}{2}{V}^{\alpha }{V}^{\mu }({\nabla }_{\alpha }{W}_{\mu }+{\nabla }_{\mu }{W}_{\alpha })=0$