TITLE: How do you find the Lie algebra of a Lie group (in practice)?
QUESTION [24 upvotes]: Given a Lie group, how are you meant to find its Lie algebra?  The Lie algebra of a Lie group is the set of all the left invariant vector fields, but how would you determine them?  My group is the set of all affine maps $x \rightarrow A.x+v$ from $R^n$ to $R^n$ under function composition.

REPLY [30 votes]: Here a worked out example:
What is the Lie algebra of the group of rotations in 3-dimensional space, $SO(3)$?
Matrices $A\in SO(3)$ are defined by the property that they are invertible and that the scalar product is invariant $\langle A\vec x,A\vec y\rangle = \langle \vec x,\vec y\rangle$ for all vectors $\vec x,\vec y\in \mathbb R^3$. Expanding the latter condition into coordinates, you can see it is equivalent to $A^TA = I$.
To find the Lie algebra, take a smooth path $A(t)$ with $A(0) = I$. In first order, it can be written as
$$A(t) = I + t·H + \mathcal O(t^2) .$$
Plugging this into the condition $A^TA=I$, we get in first order
$$I = (I+t·H)^T(I+t·H) = I + t·(H^T + H) .$$
Hence, the condition on the tanget vector $H$ is that it is antisymmetric
$$ H^T = -H .$$
In other words, the Lie algebra of the rotation group $SO(3)$ consists of antisymmetric 3x3 matrices, which must have the form
$$ H = \begin{pmatrix}0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0\end{pmatrix} .$$
It is not difficult to show that exponentiating any such matrix will yield an element that preserves the scalar product. Hence, this is the whole Lie algebra.
By the way, the elements of the Lie algebra $so(3)$ are usually represented by the angular velocity $\vec\omega=(\omega_1,\omega_2,\omega_3)$ such that multiplication becomes the cross product
$$ H·\vec x = \vec\omega \times \vec x .$$

A similar method applies to your original problem. For instance, you can embed affine transformations of $\mathbb R^n$ into $GL_{n+1}(\mathbb R)$ by
$$ (x \mapsto Ax + v) \iff \begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix} \in GL_{n+1}(\mathbb R).$$
You can express this as an algebraic condition: a matrix $B\in GL_{n+1}(\mathbb R)$ represents an affine mapping of $\mathbb R^n$ if and only if
$$ (0,0,…,1)·B = (0,0,…,1) .$$
This will induce an equation for the first derivative of a path $B(t)$ and as above, you will obtain the Lie algebra as a set of matrices.