The Lie bracket is torsion

Posted on 20 June, 2015 by jesperg

We start with a Lie group G. An element g\in G acts on an external vector x_0 by x=gx_0. Let g be generated by a vector de at the identity e, g(t)=e^{tde}. By differentiating, with dt=1, the velocity of g is

dg=e^{tde}de=gde.

This can also be interpreted that dg is the parallel transport of de to g. The velocity of x can then be expressed as

dx=dgx_0=gdex_0=degx_0=dex,

so we see that the action of an element of the Lie algebra on a vector generates a velocity of this vector.

Now let \partial be differentiation in some independent direction and let \partial e be a corresponding vector in the Lie algebra. The \partial-derivative of de is then multiplication with \partial e:

\partial de=\partial ede.

Swapping the order of differentiation gives

d\partial e=de\partial e

and combining these gives

(d\partial-\partial d)e=de\partial e-de\partial e.

The operator d\partial-\partial d is the exterior (covariant) derivative squared D^2 and computes the torsion of the action (i.e. the connection) of the Lie group. The righthand side is of course the usual commutator in the Lie algebra. We can thus reexpress the last expression as

D^2e=[de,\partial e].

This entry was posted in Lie groups and tagged , , , , . Bookmark the permalink.

Leave a comment