# Diffeomorphisms that let the Haar measure invariant and null divergent

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, $$$$\label{integraljacobiano} \int_0^1\operatorname{div}(X)(T_s(x)) ds=0,$$$$ for every $$x\in G$$. But I cannot go any further.

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