Is the Haar measure on a real Lie group absolutely continuous with respect to a Lebesgue measure?
I just got introduced to the Haar measure and Lie groups. I know measure theory, but my background in group theory and differential geometry is limited. I know that the statement is true for the Haar measure on the translation group and the orthogonal group. In the end, I am interested in the Lebesgue-Radon-Nikodym decomposition of such a Haar measure.
Some observations:
A Lie group is always locally compact.
In every coordinate system the (a) Haar measure will be of form $f(x_1, \ldots, x_n) dx_1\ldots dx_n$, where $f$ is a smooth positive function. So we can say that it is equivalent to the Lebesgue measure (although the Lebesgue measure depends on the coordinate system, while the Haar measure does not).
We can get an explicit form for the Haar measure in a coordinate system around say $e$. It is helpful to keep in mind the Haar measure for the group $(\mathbb{R}^{\times}, \cdot)$ which is $\frac{d x}{|x|}$. Now, for the coordinate system around $e$, let $f(0,0, \ldots, 0)=1$, then $f(x)=f(x_1, \ldots, x_n)=\frac{1}{|J(L_{x})|}$, where $L_x$ is the left multiplication in the Lie group by the element corresponding to $x$, and the Jacobian is considered in the given coordinate system at the point $(0,0,\ldots, 0)$. All this is to make the measure invariant under all the left multiplications.