# How do you find the Lie algebra of a Lie group (in practice)?

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.

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Embed your group into GL_n and see if you can figure out what elements of gl_n exponentiate into it. – Qiaochu Yuan Feb 13 '11 at 13:25
@Qiaochu Yuan, you mean in GL_n+1 with each map as {A x, 0 1} where 0 is a row of n zeros? – Zeophlite Feb 13 '11 at 14:06
Sorry, I didn't intend the two values of n to be the same. – Qiaochu Yuan Feb 13 '11 at 14:16
Please consult the wonderful book on Matrix Lie Groups/Algebra's by Brian Hall – Matt Calhoun Feb 13 '11 at 16:48

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.

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Okay this makes sense but my example doesn't have the extra condition of A^t.A = I. So does that mean that all H are acceptable? That conflicts with Theo's answer. – Zeophlite Feb 13 '11 at 14:42
@Zeophlite: There's no conflict. You should consider smooth paths $\gamma: (-\varepsilon,\varepsilon) \to \operatorname{Mat}_{n \times n}(\mathbb{R})$ such that $\gamma(t) \in G$ for all $t$, $\gamma(0) = I$ and then take $\dot{\gamma}(0)$. In your example and the notation I use in my answer, you can take $\gamma(t) = \begin{pmatrix} \exp(Xt) & ty \\\ 0 & 1 \end{pmatrix}$. – t.b. Feb 13 '11 at 15:29
I added some information on your particular example, though I am reluctant to present the solution in full. – Greg Graviton Feb 13 '11 at 16:37
@HaukeStrasdat: Sure. Basically you just differentiate with respect to $t$ at the point $t=0$. In other words, you calculate $\frac{d}{dt}|_{t=0} A(t)^TA(t) = H^T+H$ where $H=\frac{d}{dt}_{t=0}A(t)$. The expression "first order" refers to "first order in the Taylor expansion". In other words, the formula $A(t) = I + tH + \mathcal O(t^2)$ is just the Taylor expansion of $A(t)$ "up to first order". Admittedly, this kind of language takes a while to get used to, but it's very intuitive once you get the hang of it. – Greg Graviton Feb 17 '12 at 19:43
@HaukeStrasdat: Yes, you set $t=0$ after differentiation. In practice, one often keeps the $t$ around before differentiation and says that the expression is "to first order in $t$", which means that one silently drops all terms $t^n$ where $n>1$. This has the same effect as differentiating at $t=0$, but this way of thinking is very useful if you are interested in the higher-order terms. – Greg Graviton Feb 22 '12 at 7:20

To expand Qiaochu's comment: Your $G$ can be seen as the (closed) subgroup of $\operatorname{GL}_{n+1}(\mathbb{R})$ given by the matrices of the form $$\begin{pmatrix} A & v \\\ 0 & 1 \end{pmatrix}$$ with the usual group multiplication and acting on the affine subspace of vectors of the form $\begin{pmatrix} x \\\ 1 \end{pmatrix}$. If you understand why $\mathfrak{gl}_{n}(\mathbb{R}) = \operatorname{Mat}_{n}(\mathbb{R})$ with the commutator bracket $[X,Y] = XY - YX$ then you should have no difficulty in proving that $\mathfrak{g} = \operatorname{Lie} G$ is given by the matrices of the form $\begin{pmatrix} X & y \\\ 0 & 0 \end{pmatrix}$ with the commutator bracket.

For me, the easiest way to determine the Lie algebra of a matrix group is to think of it as the tangent space at the identity, constructed by tangent vectors of paths through the identity.

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All my text books skip the reason or proof why gln(R)=Matn(R) – Zeophlite Feb 13 '11 at 14:34
@Zeophlite: Really? Well, $\operatorname{GL}_{n}(\mathbb{R}) = \det^{-1}(\mathbb{R} \smallsetminus \{0\})$ is an open subset of $\operatorname{Mat}_{n}(\mathbb{R})$, hence its tangent space at the identity is all of it. Probably the easiest way to identify the Lie bracket $[X,Y]$ explicitly with the commutator bracket is to use the definition for vector fields applied to the special paths $\exp(tX)$ and $\exp(tY)$, where $\exp$ is the matrix exponential. – t.b. Feb 13 '11 at 14:45
I think I've seen that argument before, but I don't understand it (A open subset of B :. $T_{e}A$ = B). What do you mean by "special" vector fields? – Zeophlite Feb 13 '11 at 14:53
@Zeophlite: I meant "special paths", sorry. An open subset $U$ of $\mathbb{R}^{n}$ is a manifold for trivial reasons. The tangent space at a point $p \in U$ is $\mathbb{R}^{n}$ because the paths $p + tx$ where $x \in \mathbb{R}^{n}$ are inside $U$ for $t$ small enough. – t.b. Feb 13 '11 at 15:04

Like Qiaochu says, thinking of this as a matrix group is a good first step. Then you need to find one-parameter smooth paths $A(t)$ through the unit element $I$ corresponding to every tangent direction, where $A(0)=I$. Taking the derivative of these smooth paths at $0$ yields a basis for the Lie algebra.

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Correct me if I'm wrong, but When you say A(t), you aren't refering to my A above (as in A.x+v) but to a map A:t->GL_n+1, right? The identity of the affine maps is A=I, v=0, which is when embedded in GL_n+1 is I_n+1. Then a path A(t) = I_n+1 + t*X where X is an element of my embedding. Assuming that's correct, now what do I do? The derivative at t=0 is X. Don't I need to find the left invariant vector fields? – Zeophlite Feb 13 '11 at 14:33
Yes, my A is different. Oops for using confusing notation. We are using here the fact that the tangent space at $I$ is in 1-1 correspondence with left-invariant vector fields, and the fact that we are embedded in GL means that the Lie bracket of vector fields will correspond to the associative bracket of matrices: $[X,Y]=XY-YX$. – Grumpy Parsnip Feb 13 '11 at 15:03