What is the killing field on $S^2$ and $SO[3]$?

Question

What is the killing field on $S^2$ and $SO[3]$? I understand the structure on $S^1$, but not sure about how the vectors work on $S^2$. Thanks in advance!

Answer 1

From an intuitive point of view, a Killing field on $S^2$ is an "infinitesimal" rotation about an axis; e.g. $y \partial_x - x \partial_y$ is a rotation about the $z$ axis.

I'll try to explain how we can find the Killing fields once we know the isometry group - you may need to do a little reading on Lie groups to understand everything that's going on here.

The isometry group of $S^2$ is the Lie group $O(3)$ acting by matrix multiplication. The Killing fields on $S^2$ are those vector fields whose flows are isometries; i.e. $X$ such that there is some one-parameter subgroup $\{\phi_t \in O(3):{t\in \mathbb{R}}\}$ with $$X_p = \frac{d}{dt}\bigg|_{t=0}\phi_t \cdot p.$$ The isometries that can be reached by composing the flows of Killing fields are exactly those in $SO(3)$, the connected component containing the identity in $O(3)$ - so we can call $SO(3)$ the "continuous isometry group" of $S^2$.

To study the one-parameter subgroups, we must study the Lie algebra of $O(3)$, which is $$\mathfrak{o}(3) = \{\xi \in M_{3\times 3} : \xi + \xi^T = 0\}$$ and represents the infinitesimal rotations. For any generator $\xi \in \mathfrak{o}(3)$, we have a corresponding Killing field

$$X_p = \frac{d}{dt}\bigg|_{t=0} \exp(t \xi)(p) = d_e A_p(\xi)$$ where $A_p : O(3) \to S^2$ is just the action $\phi \mapsto \phi \cdot p$.

For example, consider the generator given by the matrix

$$ \xi = \left( \begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right)$$ and compute for a point $p=(x,y,z)\in S^2$, noting that since the action is matrix multiplication and therefore is linear, the differential of the action is once again just matrix multiplication:

$$ \begin{align} X_p &= d_e A_p(\xi)\\ &= \xi \cdot p \\ &= \left( \begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \\ &= y \partial_x - x \partial_y, \end{align}$$

the infinitesimal rotation I mentioned in the first sentence.

To find all the Killing fields, just find a basis for $\mathfrak{o}(3)$ (which is 3-dimensional) and repeat the above process on each basis element to get three Killing fields. Since the correspondence between Killing fields and Lie algebra generators is bijective and linear, these will then form a basis for the space of Killing fields on $S^2$.