Commutators in the context of local Lie groups.

Question

Let $G$ be a local Lie group in the neighbourhood $V \subseteq \mathbb{C}^d$ with identity element denoted by $e \in G$. Also, let $$ t \mapsto f(t) = (f_1(t), \dots, f_d(t)) \quad \forall t \in \mathbb{C} $$ be an analytic mapping of a neighbourhood of $0 \in \mathbb{C}$ into $V$ such that $f(0) = e$. The tangent vector to the curve $f$ at $e$ is given by $$ \alpha = \left. \frac{\mathrm{d}}{\mathrm{dt}}f(t)~ \right|_{t = 0} = \left. \left( \frac{\mathrm{d}}{\mathrm{dt}}f_1(t), \dots, \frac{\mathrm{d}}{\mathrm{dt}}f_d(t) \right) \right|_{t=0} $$

Now let $g$, $h$ be analytic curves in $G$ with $g(0) = e = h(0)$, and tangent vectors $\alpha$, $\beta$ respectively. On pg. 2 of Miller's Lie Theory and Special Functions, we then have the following definition of the commutator $[\alpha, \beta]$:

Definition: The commutator of $\alpha$ and $\beta$ is the tangent vector at $e$ to the analytic curve $$ \kappa(t) = g(\tau)h(\tau)g^{-1}(\tau)h^{-1}(\tau) \quad t = \tau^{2} $$ Here by juxtaposing two elements $g(\tau) h(\tau)$ of $G$ we mean to apply the group composition rule, which is analytic in each of its $2d$-arguments.

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Question: I don't fully understand the above definition. Does this idea relate to the usual commutator in group theory, or to the Lie bracket $[X, Y] = XY - YX$? For instance, it is claimed that the commutator defined above satisfies the Jacobi identity, but I'm don't really see how one might prove this.

In general, I think I need another perspective on the content of this definition. Any help in understanding will be appreciated.

Answer 1

The way I think of it is in terms of 1-parameter subgroups. Indeed $$ [X,Y]=ad(X)Y= \partial_{s=0}Ad(\exp(sX))Y=\partial_{t=0} \partial_ {s=0}Ad(\exp(sX))\exp(tY)$$

If now you consider $h$ and $g$ to be appropriate 1-parameter subrgoups you can see that the two definitions are equivalent. However, I think for that the 1-par subgroup associated to $X$ must be in the $s$ variable and the other in the $t$ variable.

Answer 2

In fact, the Lie Algebra side is neater, in some sense. Lie brackets obey the Jacobi identity, but group elements' commutators obey the Witt-Hall identity, instead.

To find your bearings, embrace the Lie exponential; a formal representation wisecrack: $$ g(\tau)\sim e^{\tau \alpha}, \qquad h(\tau) \sim e^{\tau \beta}, $$ so $$ g(\tau)h(\tau)g^{-1}(\tau) h^{-1}(\tau)= e^{\tau \alpha}e^{\tau \beta}e^{-\tau \alpha} e^{-\tau \beta}= \\ e^{\tau ( \alpha+\beta) +\frac{\tau^2}{2}[\alpha,\beta]+O(\tau^3)} ~ e^{-\tau ( \alpha+\beta) +\frac{\tau^2}{2}[\alpha,\beta]+O(\tau^3)} = \\ e^{ \tau^2 [\alpha,\beta]+O(\tau^3)} . $$

It is these Lie brackets in the exponent that obey the Jacobi identity, provable by direct computation of associative objects (here in the Lie Algebra); and similar simple generalizations, $$ [g,[h,[k,m]]] + [h,[g,[m,k]]] + [k,[m,[g,h]]] + [m,[k,[h,g]]] = 0. $$

The group commutator versions of these are naturally much more recondite, as linked above, but you don't need to rely on them to check the Jacobi identity for the Lie algebra.