Calculating the tangent space of special euclidean group?

Question

Let $SE(n) = \{ \begin{bmatrix} R & v \\ 0 & 1 \end{bmatrix} : R \in SO(n), v \in \mathbb{R}^n , 0 \in \mathbb{R}^n \}$ how does one calculate $T_{I}(SE(n))$ I know how to prove that $T_I(SO(n)) = Anti-symmetric matrices$

Answer 1

You know the tangent space for the SO(n) part --- it's spanned by all matrices of the form $$ E_{ij} - E_{ji} $$ where $1 \le i, j \le n$, and $i \ne j$.

So the only interesting part is the other $n$ dimensions corresponding to the $v$ in your description. Consider, for each vector $v$, the path $$ c_v(t) = \begin{bmatrix} I & tv \\ 0 & 1 \end{bmatrix} $$ Its derivative, at $t = 0$, is a tangent vector at the identity. You can get $n$ independent ones by picking $v = e_1, e_2, \ldots, e_n$; the resulting derivatives will be $$ E_{i, n+1} $$ So that gives you $(n^2 - n)/2$ tangent directions from the SO part, and $n$ more from the $v$ part, for a total of $(n^2 + n)/2$ tangent directions.

Post-Comment-Addition

In summary:

The vectors in the tangent space (which are each matrices) look like $$ \pmatrix{ M & u\\ 0 & 0} $$ where $M$ is skew-symmetric, and $u$ is any element of $\Bbb R^n$. In other words, it looks a whole lot (as a set) like $so(n) \times \Bbb R^n$. What the lie-algebra structure is? That's beyond my skills at this hour.