Modular function

Remarks

The number $\Delta(g)$ does not depend on the choice of $\mu$, since any other left Haar measure is $c\mu$ with $c > 0$. The map $\Delta$ is a continuous homomorphism into the multiplicative group $(0, \infty)$. In integral form, $\int_G f(xg^{-1}) \, d\mu(x) = \Delta(g) \int_G f(x) \, d\mu(x)$ for compactly supported continuous $f$.

Example: let $G$ be the group of pairs $(a, b)$ with $a > 0$ and $b \in \R$, with product $(a, b)(c, d) = (ac, ad + b)$, the group of maps $x \mapsto ax + b$. The measure $d\mu = da \, db / a^2$ is a left Haar measure. Right multiplication by $(c, d)$ sends $(a, b)$ to $(ac, ad + b)$, a map with Jacobian determinant $c$. So $\mu(A(c, d)) = \int_A c \, (ac)^{-2} a^2 \, d\mu = c^{-1} \mu(A)$, and $\Delta(c, d) = 1/c$.

For this group the measure $da \, db / a$ is a right Haar measure but not a left one, so the left and right Haar measures differ.

Some authors use the reciprocal convention, defining the modular function by $\mu(Ag) = \Delta(g)^{-1} \mu(A)$ or through right Haar measure, so formulas must be checked against the source.