Lie Groups, Lie Algebras and the magic of integrating a circle by matrix powers.

Question

A wise man once told me that Lie Algebras and Lie Groups are connected in a mysterious way. I am still unsure how much of the theory I understood about it, but one particular illustrative example stuck with me:

$$\left[\begin{array}{rr} \cos(dk)&\sin(dk)\\ \sin(-dk)&\cos(dk) \end{array}\right] = \left[\begin{array}{rr} -\sin(d)&\cos(d)\\ -\cos(-d)&-\sin(d) \end{array}\right]^{k}$$

Where we recognize element-wise: $$\begin{cases}\frac{ d\cos(t)}{dt}&= -\sin(t)\\\frac{d \sin(t)}{dt} &= \cos(t)\\\frac{d\sin(-t)}{dt} &= -\cos(-t)\\\frac{d\cos(t)}{dt} &=-\sin(t)\end{cases}$$

Which I interpret as: Integrating along a curve (circle in this case) is the same as taking infinitesmal "steps" along the curves tangent. But I don't know if I may be fooling myself?

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If I'm not fooling myself, how does one go from verifying this to understanding the algebra behind and move on to understand and maybe build more complicated groups / algebras for traversing routes?

Answer 1

All you need to understand the connection between a matrix Lie group and its Lie algebra is the matrix exponential. For any closed group $G \subset \Bbb R^{n \times n}$ (i.e. for any matrix Lie group), the corresponding Lie algebra is the set of all matrices $X$ such that for all $t \in \Bbb R$, $\exp[tX] \in G$.

Let's look at your "illustrative example" in this light: $SO(2)$ can be conveniently described as $\left\{ f(\theta): \theta \in \Bbb R \right\}$, where $$ f(\theta) = \pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta} $$ and the identity occurs at $\theta = 0$. The Lie algebra is the "tangent space at the identity". Note that for any $\theta(t)$ satisfying $\theta(0) = 0$, we have $$ \frac d{dt} \left. f(\theta(t))\right|_{t = 0} = \theta'(0) \pmatrix{0 & -1\\1 & 0} $$ So, the Lie Algebra is the space consisting of the span of all multiples of $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. You can verify that this collection satisfies our other definition too: $e^{tX}$ will only be a rotation for all $t$ if $X$ is in this "tangent space".

In particular, note that $\exp\left[t\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\right] = \left(\begin{smallmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{smallmatrix}\right)$.

So, where do the infinitessimal steps come in? Recall that $$ \exp(X) = \lim_{n \to \infty} \left(I + \frac{X}{n}\right)^{n} $$ Intuitively: by taking an infinitessimal perturbation of the identity along the tangent space and applying it infinitely many times, we end up tracing out a path along the group. In fact, this path is a one-parameter subgroup.