Solve the Killing equation for a vector field in $\mathbb{R}^2$ with the Euclidean metric

Question

I know that the vector field $$X = a_1\partial_1 + a_2\partial_2$$ where $a_1,a_2 : \mathbb{R}^2 \rightarrow \mathbb{R}$ are smooth, is a Killing field on $\mathbb{R}^2$ with the Euclidean metric $dx_1^2 + dx_2^2$.

I have to solve the Killing equation $$\mathcal{L}_X(dx_1^2 + dx_2^2) = 0$$ for $a_1$ and $a_2$.

I know I have to use the definition of Lie derivative and find that is equal to zero but I am struggling a bit with the computation, can someone help me ?

I was thiking about using Cartan's formula, is it a good approach ?

Answer 1

Let $U$,$V$ and $X$ be three vector fields and $g$ be the metric tensor field. Then, \begin{align} \left(L_Xg\right)(U,V) &= X\cdot g(U,V) - g(L_XU,V) - g(U,L_XV) \\ &= g(\nabla_XU,V) + g(U,\nabla_XV) - g(L_XU,V)-g(U,L_XV) \\ &= g(\nabla_XU - L_XU,V) + g(U,\nabla_XV-L_XV) \\ &= g(\nabla_UX,V) + g(U,\nabla_VX) \end{align} Thus, $L_Xg=0$ if and only if for every vector fields $U$ and $V$, $$ g(\nabla_UX,V) + g(U,\nabla_VX) = 0 $$ that is, if and only if $\nabla X : U \mapsto \nabla_UX$ is a skew-symmetric operator.

In the case $g$ is the euclidean metric of $\mathbb{R}^2$, every vector field $U$ is a smooth combination of $\partial_1$ and $\partial_2$, and $$ L_Xg = 0 \iff g(\nabla_{\partial_1}X,\partial_1) = 0,~ g(\nabla_{\partial_2}X,\partial_2)=0 \text{ and } g(\nabla_{\partial_1}X,\partial_2) = -g(\partial_1,\nabla_{\partial_2}X) $$ If $X = a_1\partial_1 + a_2 \partial_2$, recall that $\partial_1$ and $\partial_2$ are parallel for $g$, and: \begin{align} \nabla_{\partial_1}X &= \nabla_{\partial_1}\left(a_1\partial_1 + a_2 \partial_2\right) \\ &=(\partial_1a_1) \partial_1 + (\partial_1a_2)\partial_2 \\ \nabla_{\partial_2}X &= \nabla_{\partial_2}\left(a_1\partial_1 + a_2 \partial_2\right) \\ &=(\partial_2a_1) \partial_1 + (\partial_2a_2)\partial_2 \end{align} Hence, $X$ is a Killing vector field if and only if \begin{align} \partial_1a_1 &= 0, & \partial_2a_2&=0, & \partial_1a_2 &= - \partial_2 a_1 \end{align} I let you continue the calculations.

Important comment Be carefull with the Cartan magic formula. It says that for a differential form $\omega$, $L_X \omega = (d\circ i_X + i_X\circ d)\omega$. A tensor is not, in general, a differential form. A simple reason why this would not make sense is this: how do you define $dg$ when $g$ is a metric tensor?