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Let $G$ be a compact Lie group with Haar measure $\mu$. Let $X\in\mathfrak{X} (G)$ be such that, if $T(x)=\exp_x(X_x)$, $$T_*\mu=\mu,$$ then $\operatorname{div}(X)=0$?

This is true when $G=S^1$, because the diffeomorfisms that let the Lebesgue measure on $S^1$ invariant are the rigid rotations and the rigid reflections, thus, such an $X$ should be constant, hence without divergence.

I'm trying this line of reasoning: if we let $T_t(x)=\exp_x(tX_x)$, and consider de Jacobian $J_t(x)$ given by $$(T_t)_*\mu=J_t(x)\mu$$ Be derivating at $t=0$ we get $$\dot{J_0}(x)=-\operatorname{div}(X)(x).$$ Since $T_0(x)=x$, we have that, by integrating, $$J_t(x)=e^{-\int_0^t\operatorname{div}(X)(T_s(x)) ds}.$$ Since $(T_1)_*\mu=\mu$, $J_1\equiv 1$ and thus, \begin{equation}\label{integraljacobiano} \int_0^1\operatorname{div}(X)(T_s(x)) ds=0, \end{equation} for every $x\in G$. But I cannot go any further.

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  • $\begingroup$ What's the meaning of ${\exp}_x$? The Lie exponential map is defined as a map $\exp : \mathfrak{g} \to G$. $\endgroup$ Commented Jun 4 at 2:14
  • $\begingroup$ @RamiroLafuente Riemannian exponential $\endgroup$
    – Gomes93
    Commented Jun 7 at 17:57
  • $\begingroup$ But then you need to specify a Riemannian metric. $\endgroup$ Commented Jun 7 at 21:15

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