# What is the Lie algebra of the Euclidean group?

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.

• Can you show that every matrix of that form is in the Lie Algebra? Commented Nov 25, 2016 at 14:43
• Yes, I think I can show inclusion in that direction but I relied on the matrix exponential map. I would rather do so without the map as the text I'm using has yet to introduce it, suggesting that there is another way without relying on such machinery. Commented Nov 25, 2016 at 14:57
• There's definitely a quick way to show inclusion in that direction without the exponential map. I'm not sure how to get the reverse inclusion without a suitable theorem, though Commented Nov 25, 2016 at 15:19
• What is the theorem you're thinking of for the reverse inclusion? Commented Nov 25, 2016 at 15:36
• I don't know any of the names and I don't care to flip through my textbooks, but something to the effect of "if $M$ is an element of the Lie Algebra, then $\exp(tM)$ must be a one parameter subgroup of the Lie Group". That way, we can exclude certain elements from the Lie Algebra. Commented Nov 25, 2016 at 15:46

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.
• If $A'(0) = I$, how come, in the last line, the lower right block is $B$? Commented Nov 25, 2016 at 16:08
• @Omnomnomnom For the converse direction, how do you prove that $\gamma(t)$ is the exponential of $(b,B)$ for all $t$? I initially thought you were implying that $\gamma(t)$ is a one-parameter subgroup, but that doesn't seem to be the case. Commented Sep 25, 2019 at 22:37
• @Andrew $\gamma$ is not a one parameter subgroup, it's an arbitrary differentiable path. Commented Sep 26, 2019 at 7:20