1. Introduction
In this blog post I focus on various aspects of Lie groups theory that I find interesting.
2. A geometric interpretation of the commutator bracket
Suppose that we have
\begin{align} \label{eq:commutator} [X, Y] = Z. \end{align}What use can we make of the above? Well, for one it is certainly useful in any equation that involves mixed terms of \(X\) and \(Y\), since we can replace \(XY\) with \(YX + Z\) if we want. In order to connect this formula to geometry we need some result for the exponentials.
Let's look at a geometric example. In the spatial rotation group \(SO(3,\mathbb{R})\) we have a Lie algebra \(\mathfrak{so}(3,\mathbb{R})\) with a basis \(\{J_x, J_y, J_z\}\) where \(\exp(\theta J_x)\) corresponds to rotation about the \(x\)-axis of angle \(\theta\), and so on for the rest. We furthermore have:
\begin{align} \label{eq:so3-commutator} [J_x, J_y] = J_z. \end{align}Obviously such an equation asks for a geometric interpretation, but if someone fidgets around with a 3D shape (like a notebook) and its rotations, it is not obvious what fact \eqref{eq:so3-commutator} corresponds to for the three axis rotations.
The connection to geometry comes from the Baker-Campbell-Hausdorff formula, telling us that:
\begin{align} e^{\theta J_x}e^{\psi J_y} & = e^{\theta J_x + \psi J_y + \frac{\theta\psi}{2}[J_x, J_y] + O((\theta + \psi)\theta\psi)} \\ & = e^{\theta J_x + \psi J_y} + \frac{\theta\psi}{2}[J_x, J_y] + O((\theta + \psi)\theta\psi). \end{align}Now we can compute,
\begin{align} \label{eq:commutator-result} [e^{\theta J_x}, e^{\psi J_y}] & = \theta\psi J_z + O((\theta + \psi)\theta\psi). \end{align}Formula \eqref{eq:commutator-result} is inheretedly geometric: it's telling us that when we rotate by a \(\theta\) angle on the \(x\)-axis and a \(\psi\) angle on the \(y\)-axis, what the difference would've been if we instead did the operations in opposite order.
2.1. Other uses of the Baker-Campbell-Hausdorff formula
If one assumes \([A, [A,B]] = [B, [B, A]] = 0\), then all higher order terms in the Baker-Campbell-Hausdorff formula vanish.
For unbounded operators \(A, B\), the product \(AB\) may not be defined, e.g. if \(\dom(A)\cap\ran(B) = \emptyset\). Due to this, we may not have the luxury of speaking of a Lie group \(G\) with a Lie algebra \(\mathfrak{g}\). However, for the position and momentum operators \(X_j, P_k\) for \(1 \leq j, k \leq n\) of \(L^2(\mathbb{R}^n)\) of quantum mechanics, we have that the Weyl relations hold, which is the analogue of the above; the proof may be found in chapter 14.2 of [1].
3. Matrix groups
If \(G\) is a matrix group then \(SG := \{ g \in G : \det g = 1\}\).