Show that $V^{\mu} = (\cos\phi, -\cot\theta \sin\phi)$ is a Killing's vector on the 2-sphere.

Solution 1:

PSA: I am using the mathematics convention for spherical coordinates. I.e, I am using the coordinates $(\theta,\phi)$, in that order, where $\theta$ is the azimuthal angle (coordinate 1) and $\phi$ is the polar angle (coordinate 2).

To show that a vector $\boldsymbol v$ is a Killing vector, you need to show that it satisfies Killing's equation: $$\nabla_i v_j+\nabla_jv_i=0$$ We have the vector $$\boldsymbol v=\begin{bmatrix} -\cot \phi\sin\theta & \cos\theta \end{bmatrix}^\intercal$$ Its corresponding dual components are $$v_i=g_{ij}v^j$$ Recall that on the two sphere (of radius $1$), the metric is $$\mathbf{g} =\begin{bmatrix} \begin{bmatrix} \sin^{2} \phi & 0 \end{bmatrix} & \begin{bmatrix} 0 & 1 \end{bmatrix} \end{bmatrix}$$ You can verify that $$v_1=g_{1j}v^j=g_{1~1}v^1=-\sin^2\phi\cot\phi\sin\theta=-\sin\phi\cos\phi\sin\theta \\v_2=g_{2j}v^j =g_{2~2}v^2=v^2=\cos\theta$$ Hence $$\boldsymbol{v}_\flat=\begin{bmatrix} -\sin\phi\cos\phi\sin\theta & \cos\theta \end{bmatrix}$$ The covariant derivatives are $$\nabla_iv_j=\partial_j v_i-\Gamma^k_{ij}v_k$$ Recall again that the Christoffel symbols on the two sphere (of radius $1$) are $$ \Gamma^1_{1~2}=\Gamma^1_{2~1}=\cot\phi \\ \ \Gamma^2_{1~1}=-\cos\phi\sin\phi$$ With all others zero. Let's return to Killing's equation $$\nabla_i v_j+\nabla_jv_i=0$$ And compute all the components. First $i=j=1$: $$\nabla_1v_1=\partial_1v_1-\Gamma^k_{1~1}v_k \\ =\frac{\partial}{\partial \theta}(-\sin\phi\cos\phi\sin\theta)-\Gamma^{1}_{1~1}v_1-\Gamma^2_{1~1}v_2 \\ =-\sin\phi\cos\phi\cos\theta-(-\cos\phi\sin\phi)\cos\theta=0$$ Now let's do $i=j=2$: $$\nabla_2v_2=\partial_2v_2-\Gamma^k_{2~2}v_k$$ All of the $\Gamma^k_{2~2}$ are zero, and $\partial_2 v_2=\frac{\partial}{\partial\phi}(\cos\theta)=0$ hence $\nabla_2v_2=0$. Finally let's do $i=1,j=2$. Since Killing's equation is symmetric wrt to $i,j$ this covers the $i=2,j=1$ case as well. $$\nabla_1v_2+\nabla_2v_1= \left[\partial_2v_1-\Gamma^k_{1~2}v_k\right]+\left[\partial_1v_2-\Gamma^k_{2~1}v_k\right]$$ It looks like we need to use the product rule to calculate $\partial_2v_1=\frac{\partial}{\partial\phi}(-\sin\phi\cos\phi\sin\theta)$ but you can save yourself a lot of work using the identity $$\sin\phi\cos\phi=\frac{1}{2}\sin(2\phi)$$ So, $$\nabla_1v_2+\nabla_2v_1=\left[-\cos(2\phi)\sin\theta-\cot\phi(-\sin\phi\cos\phi\sin\theta)\right]+\left[(-\sin\theta)-\cot\phi(-\sin\phi\cos\phi\sin\theta)\right]$$ We can simplify this: $$\nabla_1v_2+\nabla_2v_1=2\cot\phi(\sin\phi\cos\phi\sin\theta)-\sin\theta(1+\cos(2\phi)) \\ =2\cos^2\phi\sin\theta-\sin\theta(1+\cos(2\phi)) \\ =\sin\theta(2\cos^2\phi-(1+\cos(2\phi)))=0$$

I'll let you verify the trigonometric identity $2\cos^2\phi=1+\cos(2\phi)$ on your own. Hence we have shown $\boldsymbol{v}$ is a Killing vector field.