Haar measure on a Lie group

Remarks

Since $L_h^* \omega = \omega$ for all $h \in G$, the measure $\mu_\omega$ is invariant under left translations, so it is a left Haar measure of $G$; changing $\omega_e$ multiplies it by a positive constant.

Example: on $G = GL_n(\R)$, open in $M_n(\R) = \R^{n^2}$, the measure $\abs{\det X}^{-n} \, dX$, with $dX$ Lebesgue measure, is a left Haar measure. Left multiplication $X \mapsto AX$ is linear with determinant $(\det A)^n$, which cancels the change of $\abs{\det X}^{-n}$.

Example and non-example: on $G = \set{(a, b) : a > 0, b \in \R}$ with $(a, b)(a', b') = (aa', ab' + b)$, the measure $a^{-2}\, da\, db$ is a left Haar measure. The measure $a^{-1}\, da\, db$ is not left-invariant: left translation by $(c, 0)$ maps $(a, b)$ to $(ca, cb)$, with Jacobian $c^2$, and multiplies $a^{-1}\, da\, db$ by $c$. It is a right Haar measure instead.

A right Haar measure arises in the same way from right-invariant forms. The group is unimodular, meaning that left and right Haar measures coincide, if and only if $\abs{\det \operatorname{Ad}(g)} = 1$ for all $g \in G$, where $\operatorname{Ad}$ is the adjoint representation.