Divergence of left invariant vector field

[paweld]

Let's assume that a compact Lie group and left invariant vector filed X are given.
I wonder why the divergence (with respect to Haar measure) of this field has to
be equall 0. I found such result in one paper but I don't know how to prove it.
Any suggestions?

 

[mathwonk]

searching the web i have found assertions that this holds for unimodular groups, but perhaps not in general. could there be another hypothesis you haven't mentioned?

 

[fzero]

Given a volume form $\omega$, then the divergence of a vector field $X$ is related to the Lie derivative as:

$(\mathrm{div}~X) \omega = \mathcal{L}_X \omega.$

In the case of a Lie group, there are presumably a number of ways to show that the RHS vanishes iff $X$ is an element of the Lie algebra. In particular, we can argue that it's natural from the point of view of the Lie algebra generating the isometries of the Lie group manifold.

 

[paweld]

Thanks fzero. I like your reasoning.

 

[blue_raver22]

The divergence of a left invariant vector field on a compact Lie group is indeed equal to 0, and this can be proven using the properties of Haar measure. Haar measure is a unique left invariant measure on a compact Lie group, meaning that it is preserved under left translations by elements of the group. This property allows us to simplify the calculation of the divergence of a left invariant vector field.

First, we can rewrite the divergence as the integral of the dot product of the vector field and the gradient of a function f with respect to Haar measure. This can be expressed as:

div(X) = ∫<X, ∇f> dμ

Since the group is compact, we can choose a basis for the vector space of left invariant vector fields. Let {X1, X2, ..., Xn} be the basis, and let {f1, f2, ..., fn} be the corresponding basis for the space of smooth functions on the group. Then, we can write the vector field X as a linear combination of the basis vectors:

X = a1X1 + a2X2 + ... + anXn

where ai are smooth functions on the group. Now, we can substitute this expression for X into the above integral and use the linearity of the integral to get:

∫<X, ∇f> dμ = ∫<a1X1 + a2X2 + ... + anXn, ∇f> dμ

= a1∫<X1, ∇f> dμ + a2∫<X2, ∇f> dμ + ... + an∫<Xn, ∇f> dμ

= a1∫<∇f1, ∇f> dμ + a2∫<∇f2, ∇f> dμ + ... + an∫<∇fn, ∇f> dμ

= 0

since the inner product of two gradients is always 0 (by the fundamental theorem of calculus).

Therefore, we have shown that the divergence of a left invariant vector field is equal to 0 with respect to Haar measure. This result is important in many areas of mathematics and physics, as it allows us to simplify calculations and make use of the powerful properties of Haar measure.