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

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?

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?

• What's the relationship between the first and second display equations exactly? What does "traversing routes" mean here? Is the question really, how does one compute the Lie algebra of $SO(2)$ in terms of the standard matrix representation? May 16, 2017 at 14:55
• Traversing routes mean ways to calculate where I end up if I follow a surface of some differentiable manifold. Treating the functions in the left hand side matrix as functions of t, differentiating them to show they are primitive functions to the right hand side matrix. Oh wait, I forgot to subtract start point of integration (angle 0), didn't I? May 16, 2017 at 15:13

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.
• Yes!! That's an important point. In this case, note that $$f'(\theta(t)) = f(\theta(t)) \left[\theta'(t) \pmatrix{0 & -1\\1&0}\right]$$ so we just end up shifting the tangent space along by the group element $f(\theta(t))$. May 16, 2017 at 15:19