The killing vectors of flat space in polar coordinates

Question

The metric of flat three-dimensional space is written in usual Cartesian coordinates

  ds^2 = dx^2 + dy^2 + dz^2

There are 3 killing vectors (1,0,0) , (0,1,0) and (0,0,1) that corresponds to three translational symmetries of flat space.

But when the polar coordinates are used

ds^2 = dr^2 + r^2 d theta^2 + r^2sin^2 theta d phi^2

The killing vectors of this polar metric is (0,0,1) in polar coordinates but (-y,x,0) in Cartesian coordinates

Can someone please explain how do we get (-y,x,0) ???

Answer 1

There are several ways to approach the computation. One typical method is to push forward the coordinate vector field $\partial_\phi$ by the coordinate transformation formula, i.e., the identity map $\operatorname{id}$ written in its coordinate representation: \begin{align*} x &= r \sin \theta \cos \phi \\ y &= r \sin \theta \sin \phi \\ z &= r \cos \theta . \end{align*}

Computing the relevant components, $\frac{\partial x^i}{\partial \phi}$, of the matrix representation of the tangent map (i.e., the Jacobian matrix), gives $$[T \operatorname{id}] = \pmatrix{ * & * & -r \sin \theta \sin \phi \\ * & * & r \sin \theta \cos \phi \\ * & * & 0 } . $$ So, the matrix representation of $\partial_\phi$ with respect to the Cartesian coordinate frame is $$[T \operatorname{id}] [\partial_\phi] = \pmatrix{ * & * & -r \sin \theta \sin \phi \\ * & * & r \sin \theta \cos \phi \\ * & * & 0 } \pmatrix{0\\0\\1} = \pmatrix{ -r \sin \theta \sin \phi \\ r \sin \theta \cos \phi \\ 0 } .$$ Thus, as an explicit linear combination of the Cartesian coordinate vector fields, $\partial_\phi$ is $$\partial_\phi = (- r \sin \theta \sin \phi) \partial_x + (r \sin \theta \cos \phi) \partial_y + (0) \partial_z = \boxed{-y \partial_x + x \partial_y} .$$