The Lie Bracket

Can one reconstruct the Lie group from the Lie algebra? In general, the multiplication table of a group is determined if you know the multiplication table for its generators; why not try this with the ``infinitesimal'' generators? If you try this approach, you will find you need to know the commutators of infinitesimal elements, like $x(\epsilon)y(\epsilon)x(\epsilon)^{-1}y(\epsilon)^{-1}$.

My ``definition'' of the Lie algebra involved approximating infinitesimal generators by Taylor expansions out to the first order. In other words, I used only first order derivatives. But to the first order, the commutators are zero!

Say we approximate an ``infinitesimal'' element of the Lie group out to the second order:

$$x(\epsilon) \approx {\bf 1}+ \epsilon x'(0) + \frac{\epsilon^2}{2}x''(0)$$

If you work out the commutator, you will find expressions of the form $vw-wv$ appearing, where $v$ and $w$ belong to the Lie algebra. And one can verify that $vw-wv$ belongs in fact to the Lie algebra, as I've defined it, although $vw$ and $wv$ in general don't.

Remarkably, knowledge of these second order terms completely specifies the structure of the Lie group near the identity. That is, if the Lie algebras are isomorphic, then the Lie groups are locally isomorphic. Third and higher-order terms are not needed.