Divergence of left invariant vector field

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Discussion Overview

The discussion revolves around the divergence of left invariant vector fields on compact Lie groups, specifically questioning why this divergence, with respect to Haar measure, is equal to zero. The scope includes theoretical aspects of Lie groups and vector fields.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions the proof of the divergence of left invariant vector fields being zero, referencing a paper that claims this result.
  • Another participant mentions that the assertion may hold for unimodular groups but suggests that it might not be universally applicable, prompting a consideration of additional hypotheses.
  • A third participant provides a relationship between divergence and the Lie derivative, suggesting that the divergence vanishes if the vector field is an element of the Lie algebra, indicating a potential method for proof.
  • A later reply expresses appreciation for the reasoning presented, indicating engagement with the technical explanation offered.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are competing views regarding the conditions under which the divergence is zero, particularly concerning the applicability to different types of Lie groups.

Contextual Notes

There are limitations regarding the assumptions made about the types of Lie groups and the implications of unimodularity, which remain unresolved in the discussion.

paweld
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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?
 
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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?
 
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.
 
Thanks fzero. I like your reasoning.
 

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