Haar measure

If $G$ is a topological group, a *Haar measure* is a translation-invariant measure on the Borel sets of $G$. The archetypal example of Haar measure is the Lebesgue measure on the (additive group underlying) cartesian space $\mathbb{R}^n$.

The proper generality in which to discuss Haar measure is where the topological group $G$ is assumed to be locally compact Hausdorff, and from here on we assume this. (For topological groups, the Hausdorff assumption is rather mild; it is equivalent to the $T_0$ separation condition. See the discussion at uniform space.)

Let $C_c(G)$ denote the vector space of continuous real-valued functionals with compact support on $G$. This is a locally convex topological vector space where the locally convex structure is specified by the family of seminorms

$\rho_K(f) = \sup_{x \in K} |f(x)|,$

$K$ ranging over compact subsets of $G$. Recall that a Radon measure on $G$ may be described as a continuous linear functional

$\mu: C_c(G) \to \mathbb{R}$

which is *positive* in the sense that $\mu(f) \geq 0$ whenever $f \geq 0$. This defines a measure $\hat{\mu}$ on the $\sigma$-algebra of Borel sets in the usual sense of measure theory, where

$\hat{\mu}(B) = sup \{\mu(f): supp(f) = K \subseteq B, \rho_K(f) = 1\}$

By abuse of notation, we generally conflate $\mu$ and $\hat{\mu}$.

A **left Haar measure** on $G$ is a nonzero Radon measure $\mu$ such that

$\mu(g B) = \mu(B)$

for all $g \in G$ and all Borel sets $B$.

Any locally compact Hausdorff topological group $G$ admits a Haar mesaure that is unique up to scalar multiple. This result was first proven by Weil. A proof by be found in these online notes by Rubinstein-Salzedo.

The left and the right Haar measure may or may not coincide, groups for which they coincide are called **unimodular**. Consider the matrix subgroup

$G := \left\{ \left.\, \begin{pmatrix} y & x \\ 0 & 1 \end{pmatrix}\,\right|\, x, y \in \mathbb{R}, y \gt 0 \right\}$

The left and right invariant measures are, respectively,

$\mu_L = y^{-2} \,\mathrm{d}x \,\mathrm{d}y,\quad \mu_R = y^{-1} \,\mathrm{d}x \,\mathrm{d}y$

and so G is not unimodular.

Abelian groups are obviously unimodular; so are compact groups and discrete groups.

Revised on September 21, 2016 06:47:05
by Hadleigh Frost?
(129.67.185.227)