MTW Exercise 9.13. Can the tangent to a curve on the manifold $\mathcal{SO}\left(3\right)$ be a singular matrix?

Question

This is from Misner, Thorne and Wheeler, Exercise 9.13. We have established

$$\mathcal{K}_{3}=\begin{bmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix},$$

and

$$\mathcal{R}_{z}\left(\zeta\right)=\exp\left(\mathcal{K}_{3}\zeta\right)=\begin{bmatrix}\cos\zeta & \sin\zeta & 0\\ -\sin\zeta & \cos\zeta & 0\\ 0 & 0 & 1 \end{bmatrix}.$$

Part (e) of the exercise

Let $\mathcal{C}$ be the curve $\mathcal{P}=\mathcal{R}_{z}\left(t\right)$ through the identity matrix, $\mathcal{C}\left(0\right)=\mathcal{I}\in\mathcal{SO}\left(3\right).$ Show that its tangent, $\left(d\mathcal{C}/dt\right)\left(0\right)=\dot{\mathcal{C}}\left(0\right)$ does not vanish by computing $\dot{\mathcal{C}}\left(0\right)f_{12},$ where $f_{12}$ is the function $f_{12}\left(\mathcal{P}\right)=P_{12},$ whose value is the $12$ matrix element of $\mathcal{P}.$

Every way I can think of to differentiate $\mathcal{R}_{z}\left(\zeta\right)$ leads to the third row and third column being zero.

$$\mathcal{R}_{z}\left(\zeta\right)\equiv\exp\left(\mathcal{K}_{3}\zeta\right)=\sum_{n=0}^{\infty}\frac{\theta^{n}}{n!}\left(\mathcal{K}_{3}\right)^{n}=e^{\mathcal{K}_{3}\zeta}$$

$$\mathcal{R}_{z}^{\prime}\left(\zeta\right)=\mathcal{K}_{3}e^{\mathcal{K}_{3}\zeta}=\begin{bmatrix}-\sin\zeta & \cos\zeta & 0\\ -\cos\zeta & -\sin\zeta & 0\\ 0 & 0 & 0 \end{bmatrix}$$

Is this correct? The exercise doesn't to address the value of $P_{33}.$

I'm not asking if I have the right answer to the question. I'm asking if a matrix which is not an element of $\mathcal{SO}\left(3\right)$ can serve as a tangent to a curve on $\mathcal{SO}\left(3\right).$ That is different from the case of tangent vectors to manifolds (surfaces) in $\mathbb{R}^n$.

Answer 1

The answer to your question is yes. In Lie group theory language, what you are doing is constructing the Lie algebra of a Lie group. So the tangent vector you have created is an element of the Lie algebra $\mathcal{so(3)}$ not an element of the group $\mathcal{SO(3)}$.

In general a manifold does not have a vector space structure, but the tangent space at any point is always a vector space (so like $\mathbb{R}^n$). In that sense $\mathbb{R}^n$ is a special case because the manifold is already a vector space. The group $\mathcal{SO(3)}$ is not a vector space.