Exercise 3.14 Killing vectors on two-sphere

Question

Consider the three Killing vectors of the two-sphere, (3.188). Show that their commutators satisfy the following algebra:
\begin{align}
\left[R,S\right]=T&\phantom {10000}(1)\nonumber\\
\left[S,T\right]=R&\phantom {10000}(2)\nonumber\\
\left[T,R\right]=S&\phantom {10000}(3)\nonumber
\end{align}

Answer

The three Killing vectors at 3.188 were
\begin{align}
R&=\partial_\phi&\phantom {10000}(4)\nonumber\\
S&=\cos{\phi}\partial_\theta-\cot{\theta}\sin{\phi}\partial_\phi&\phantom {10000}(5)\nonumber\\
T&=-\sin{\phi}\partial_\theta-\cot{\theta}\cos{\phi}\partial_\phi&\phantom {10000}(6)\nonumber
\end{align}It is easy to prove the algebra as long as we remember that the ##\partial_\theta,\partial_\phi## are just the basis vectors and Carroll's 2.23 that ##\left[X,Y\right]^\mu=X^\lambda\partial_\lambda Y^\mu-Y^\lambda\partial_\lambda X^\mu## which we proved back in exercise 2.04. So first we could write (4),(5),(6) as ##R=\left(0,1\right),\ S=\left(\cos{\phi},-\cot{\theta}\sin{\phi}\right),\ T=\left(-\sin{\phi},-\cot{\theta}\cos{\phi}\right)##.

What is more difficult to understand is what does this neat algebra mean? A common way to show a commutator (which Carroll uses when introducing the Riemann tensor) is thus

The commutator ##\left[R,S\right]## is the difference between doing ##R## then ##S## and ##S## then ##R##. So the algebra that the Killing vectors satisfy corresponds to this geometry:

##R## has always been shown as a unit vector in increasing ##\phi## direction, ##T,S## are more flexible. I'm not sure how this helps.

The proof is given at Ex 3.14 Killing vectors on two-sphere.pdf (2 pages)