I am trying to find the Lie algebra for $E(n) = \left\{\begin{bmatrix}1 & 0^t \\
\mathbf{x} & A
\end{bmatrix}: A \in SO(n), \mathbf{x} \in \mathbb{E}^n \right\}$. In particular, I would like to show that $\mathfrak{e}(n) = \left\{\begin{bmatrix}0 & 0^t \\ \mathbf{b} & B \end{bmatrix}:B \in \mathfrak{so}(n),\mathbf{b} \in \mathbb{E}^n \right\}$ using only the definition that a Lie algebra is the tangent space at the identity of the Lie group.
I've managed to show that $\mathfrak{so}(n)$ is the set of skew-symmetric matrices but I'm not sure how to proceed from there.
Thank you in advance.
Best Answer
Proof: suppose that $\gamma(t)$ is a path in the Lie Group with $\gamma(0) = I$. $\gamma$ must have the form $$ \gamma(t) = \pmatrix{1&0\\ \mathbf x(t) & A(t)} $$ It follows that $\gamma'(0)$ has the form $$ \gamma'(0) = \pmatrix{0&0\\ \mathbf x'(t) & A'(0)} $$ which is of the desired form.
On the other hand, take any $B \in so$ and $\mathbf b \in \Bbb R^n$. We can define $A(t)$ in $SO$ such that $A(0) = I$ and $A'(0)=B$, and define $$ \gamma(t) = \pmatrix{ 1 & 0\\ t\mathbf b & A(t) } \implies \gamma'(0) = \pmatrix{ 0&0\\\mathbf b & B } $$ thus, we have both inclusions and the sets are equal.