Haar measure

A left Haar measure on a locally compact group that is a Hausdorff space is a nonzero regular Borel measure, finite on compact sets, satisfying $\mu (gA)=\mu (A)$ for every Borel set $A$. Such a measure exists and is unique up to a positive scalar. Right Haar measure instead satisfies $\mu (Ag)=\mu (A)$; the two need not coincide. On a compact group, normalization gives total mass one and invariance on both sides. See Tornier, Haar Measures, Definition 2.1 and Theorem 2.2.

For a finite group, normalized Haar measure is uniform counting; for the circle group, it is normalized angle integration on the circle. These examples turn symmetry into a rule for averaging.

Generate a real $n$-by-$n$ matrix $G$ with independent entries having the standard normal distribution. Take its QR decomposition with positive diagonal in the upper triangular matrix factor, $G=QR$. The orthogonal matrix $Q$ has normalized Haar measure on the full orthogonal group $O(n)$. If a numerical factorization returns $Q_0{R}_0$, set $S=\mathrm{diag}(\mathrm{sgn}{R}_{0,ii})$ and return $Q_{0}S$. The signs are nonzero almost surely: successively drawn Gaussian columns have probability zero of lying in the proper linear subspace spanned by earlier columns.

For correctness, every fixed orthogonal matrix $U$ preserves the standard Gaussian column density and column independence, so $UG$ and $G$ have the same probability distribution. The positive-diagonal QR decomposition is unique: two decompositions would give an orthogonal matrix that is also an upper triangular matrix with positive diagonal, hence the identity. Therefore $Q(UG)=UQ(G)$ and $Q(G)$ has a left-invariant probability law. Uniqueness of normalized Haar measure on the compact group $O(n)$ identifies that law. Each determinant sign has probability $1/2$ by reflection invariance.

On a $d$-dimensional Lie group, take the coframe dual to a basis of left-invariant vector fields. Its top wedge is nonvanishing and invariant under left translations, so its absolute density defines a left Haar measure. If the frame has coordinate matrix $A(x)$, the density is $|\det A(x){|}^{-1}d{x}^1\cdots d{x}^d$. This makes explicit computation of Haar measure a determinant calculation.

For nonzero nonnegative $u,h\in C_c(G)$, let $(f:u)$ be the infimum of ${\sum }_j{a}_j$ over finite dominations $f\le {\sum }_j{a}_{j}u(x_j^{-1}\cdot )$ with $a_j>0$. Composition of covers gives $(f:u)\le (f:h)(h:u)$, so $I_u(f)$ is uniformly bounded while $I_u(h)=1$. It is invariant under left translation, homogeneous, monotone and subadditive. As the support of $u$ shrinks to the identity, its additivity defect tends to zero: use the continuous proportions $f_i/(f_1+f_2+\delta p)$, with a cutoff $p$ positive on both supports, whose oscillations are small on every translate of that small support. In a locally compact sigma-compact metrizable group, diagonal convergence on a countable dense family, followed by uniform approximation on fixed compact supports, gives a positive invariant linear functional on all $C_c(G)$. The Riesz representation on compactly supported continuous functions turns it into Haar measure. Local compactness is essential for the compactly supported cutoffs and small-support test functions.

For a compact group $G$, positive normalized linear functionals on $C(G)$ form a nonempty weak-star topology compact convex set. For finitely many translations, apply the Schauder-Tychonoff fixed-point theorem to the average of their induced affine dual maps. A fixed functional is stationary under this average. For each real continuous $f$, its orbit function $h(g)=\varphi (l_{g}f)$ is continuous on the compact subgroup generated by those translations and satisfies a finite averaging identity. At a maximum, every summand must attain that same maximum; iteration and density of the positive-word semigroup force $h$ to be constant. Thus the functional is fixed by each of the finitely many translations. Compactness and the finite intersection property now give a functional invariant under every translation. The Riesz-Markov-Kakutani representation theorem turns it into a normalized Haar measure, without assuming commutativity of the group.

A Haar integral integrates a function against Haar measure. On a compact group, normalizing the measure to total mass one makes it an averaging operation. Translation invariance then implies that the average of a nontrivial continuous unitary character is zero: translating the integral multiplies it by a character value different from one.

For a class function on SU(2) and normalized Haar measure, the displayed formula reduces a group integral to its conjugacy-class angle. The weight follows by identifying SU(2) with the unit sphere in four real dimensions: writing a unit quaternion as $\cos \theta +\mathbf{n}\sin \theta$ gives spherical volume proportional to ${\sin }^2\theta d\theta d\mathbf{n}$. Since ${\int }_0^{\pi }{\sin }^2\theta d\theta =\pi /2$, normalization gives $2/\pi$.

The hemisphere graph $u_0=\pm \sqrt{1-|\mathbf{u}{|}^2}$ of SU(2) as the three-sphere has induced metric $g_{ij}={\delta }_{ij}+u_i{u}_j/u_0^2$, whose determinant is $1/u_0^2$. Its volume density is therefore $d^{3}u/|u_0|$. Fixed left or right multiplication preserves the Euclidean norm on the four Pauli coefficients, so the round volume is Haar measure. Both hemispheres have volume ${\pi }^2$, giving total volume $2{\pi }^2$ and normalized density $d^{3}u/(2{\pi }^2|u_0|)$ on each hemisphere. The apparent equatorial divergence is an integrable coordinate singularity.

For an open identity neighbourhood $V$ in a compact topological group, a finite left covering net is a finite set $F$ such that the left translates $fV$, $f\in F$, cover the group. Compactness gives such finite sets; minimum cardinality gives a minimal one. Left translation preserves minimality because $G=FV$ implies $G=gFV$.

Two minimal left covering nets for the same identity neighbourhood $V$ can be matched bijectively so that paired translates intersect. Join $f\in F$ to $f^{\prime }\in F^{\prime }$ when $fV\cap f^{\prime }V\ne \varnothing$. For $S\subseteq F$, its neighbours $N(S)$ cover every point covered by $S$, so $(F\setminus S)\cup N(S)$ still covers the group. Minimality forces $|N(S)|\ge |S|$. Hall's marriage theorem supplies the matching, and an intersection $fv=f^{\prime }{v}^{\prime }$ gives $f^{-1}{f}^{\prime }=v{v}^{\prime -1}\in V{V}^{-1}$. Averaging a continuous function over matched nets therefore gives values differing by at most its uniform oscillation on right displacements in $V{V}^{-1}$. This is the finite combinatorial step in constructing Haar measure on a compact group.