Worked Problems on Haar Measure and Invariance

A Practical Strategy for Haar-Measure Problems

In explicit problems, the main tasks are to identify a measure invariant under the required translations, check its normalization, and keep track of whether invariance is on the left or the right. The last distinction matters on noncommutative groups: a left Haar measure need not be right invariant, and translating an integral on the opposite side can introduce a factor.

Throughout, a Haar measure means a nonzero, locally finite Radon measure on a locally compact group that is invariant under the specified translations. Haar measures are unique up to multiplication by a positive constant. We use the existence and uniqueness theorem for Haar measure established earlier. Thus, after verifying invariance and the Radon and local-finiteness conditions, a normalization condition determines the measure uniquely. On a compact group, the normalization convention is total mass one.

Finite Groups: Equal Mass at Every Point

Proposition: Let \(G\) be a finite group with \(N\ge1\) elements, equipped with the discrete topology and its full sigma-algebra \(\mathcal{P}(G)\). The left-invariant probability measure on \(G\) assigns mass \(1/N\) to each point. It is also right invariant.

Proof. Let \(\mu\) be a left-invariant probability measure. For any \(x,y\in G\), left translation by \(yx^{-1}\) sends \(x\) to \(y\). Invariance therefore gives \(\mu(\{x\})=\mu(\{y\})\). Since the \(N\) singletons partition \(G\), their equal masses sum to \(\mu(G)=1\), so each mass is \(1/N\). Conversely, this uniform measure is left invariant because left multiplication permutes the \(N\) points; it is right invariant because right multiplication also permutes them. The conclusion includes the trivial group \(N=1\), where the unique point has mass one. \(\square\)

Worked Example: A Haar Average on \(S_3\)

Give \(S_3\) its discrete topology and let \(\mu\) be its normalized counting measure. Define \(f:S_3\to\mathbb{R}\) by \(f(e)=2\), by \(f(\tau)=1\) for each of the three transpositions \(\tau\), and by \(f(\rho)=0\) for each of the two 3-cycles \(\rho\). The group has \(1+3+2=6\) elements, so each point has mass \(1/6\). Hence

\int_{S_3} f\,d\mu =\frac{1}{6}\left(2+1+1+1+0+0\right) =\frac{5}{6}.

Check: the sum of the six listed values is \(5\), and the total mass is \(6(1/6)=1\); their average is therefore \(5/6\). For any fixed \(g\in S_3\), the map \(x\mapsto gx\) permutes all six points, so \(\int f(gx)\,d\mu(x)\) has the same six summands and also equals \(5/6\).

Euclidean Space: Fixing the Constant

On the additive group \(\mathbb{R}^n\), left and right translation are both ordinary translation. Lebesgue measure \(m\) is translation invariant. Any Haar measure on this group is \(c\,m\) for a constant \(c>0\), by uniqueness. To fix \(c\), prescribe the measure of a set with known positive finite Lebesgue measure.

Worked Example: Normalizing Haar Measure on \(\mathbb{R}^2\)

Find the translation-invariant Radon measure \(\mu\) on the additive group \(\mathbb{R}^2\) for which the rectangle \(Q=[0,2]\times[0,3]\) has measure \(12\). Since \(\mu=c\,m\), and \(m(Q)=2\cdot3=6\), the condition \(\mu(Q)=12\) gives \(6c=12\), hence \(c=2\). Thus \(\mu=2m\). For \(R=[0,1]\times[0,4]\), we obtain

\mu(R)=2m(R)=2(1\cdot4)=8.

Check: \(R\) has Lebesgue area \(4\), so multiplying by the determined constant \(2\) gives \(8\). Translating either rectangle does not change its \(\mu\)-measure, because Lebesgue measure is translation invariant.

The same calculation works in \(\mathbb{R}^n\): if \(\mu=c\,m\) and a box \(Q=\prod_{j=1}^n[0,L_j]\), with every \(L_j>0\), is required to have measure \(M>0\), then \(m(Q)=\prod_{j=1}^nL_j\), and therefore \(c=M/\prod_{j=1}^nL_j\). The positivity and finiteness conditions ensure that this normalization determines a finite positive constant.

Tori: Integrate on One Fundamental Cube

The torus \(\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n\) is a compact additive group. Its normalized Haar probability is obtained by integrating periodic representatives over \([0,1)^n\) with Lebesgue measure. The half-open cube contains exactly one representative of each equivalence class; its boundary choices do not affect integrals, since the boundary has Lebesgue measure zero. Translation invariance follows from periodicity and translation invariance of Lebesgue measure, or equivalently from the Haar measure on the compact quotient.

Worked Example: A Haar Integral on the Circle

On \(\mathbb{T}=\mathbb{R}/\mathbb{Z}\), let \(\mu\) be normalized Haar probability and let \(f([x])=3+2\cos(2\pi x)\). This is well-defined because \(f([x+1])=f([x])\). Its integral is computed on \([0,1)\):

\int_{\mathbb{T}} f\,d\mu =\int_0^1\left(3+2\cos(2\pi x)\right)\,dx =3+\left[\frac{\sin(2\pi x)}{\pi}\right]_0^1 =3.

Check: \(\sin(2\pi)=\sin(0)=0\), so the cosine term contributes zero. Also, \(f\) is nonnegative because \(-1\le\cos(2\pi x)\le1\), giving \(1\le f([x])\le5\), consistent with an average of \(3\).

The Affine Group: Left and Right Haar Measures

Consider the locally compact group \(G=\{(b,a):b\in\mathbb{R},\ a>0\}\), with multiplication

$$ (b,a)(b',a')=(b+ab',aa'). $$

This is the group of maps \(x\mapsto ax+b\), where \(a>0\). The identity is \((0,1)\), and the inverse of \((b,a)\) is \((-b/a,1/a)\). We use the ordinary two-dimensional Lebesgue measure \(db\,da\) on the coordinate domain \(\mathbb{R}\times(0,\infty)\). The densities below are positive and continuous, so they define locally finite Radon measures; they are nonzero.

Result: On \(G\), a left Haar measure and a right Haar measure are, respectively, \(d\mu_L(b,a)=db\,da/a^2\) and \(d\mu_R(b,a)=db\,da/a\). With the convention \(\mu_L(Eg)=\Delta(g)\mu_L(E)\), the modular function is \(\Delta(b_0,a_0)=1/a_0\).

Proof. Fix \(g_0=(b_0,a_0)\in G\). Left translation sends \((b,a)\) to \((b_0+a_0b,a_0a)\). Its coordinate Jacobian determinant is \(a_0^2\). At the image point \((\widetilde b,\widetilde a)\), the density \(d\widetilde b\,d\widetilde a/\widetilde a^2\) is therefore \(a_0^2\,db\,da/(a_0a)^2=db\,da/a^2\). The change-of-variables formula proves left invariance of \(\mu_L\).

Right translation by \(g_0\) sends \((b,a)\) to \((b+ab_0,aa_0)\), with Jacobian determinant \(a_0\). The image density for \(\mu_R\) is \(d\widetilde b\,d\widetilde a/\widetilde a=a_0\,db\,da/(aa_0)=db\,da/a\), proving right invariance of \(\mu_R\). For the modular function, right translation of \(\mu_L\) gives image density \(a_0\,db\,da/(a_0a)^2=(1/a_0)db\,da/a^2\). Thus \(\mu_L(Eg_0)=(1/a_0)\mu_L(E)\) for measurable \(E\), as claimed. In particular, \(\mu_L\) is not right invariant unless \(a_0=1\). \(\square\)

For integrals, the same substitutions give the precise translation rules. For nonnegative measurable \(f:G\to[0,\infty]\), or for integrable real- or complex-valued \(f\),

\int_G f(gx)\,d\mu_L(x)=\int_G f(x)\,d\mu_L(x), \qquad \int_G f(xg)\,d\mu_L(x)=\Delta(g)^{-1}\int_G f(x)\,d\mu_L(x).

The second identity follows because \(y=xg\) has \(d\mu_L(y)=\Delta(g)d\mu_L(x)\), so \(d\mu_L(x)=\Delta(g)^{-1}d\mu_L(y)\). For \(\mu_R\), the corresponding right-translation integral is unchanged. The identities hold as equalities of extended nonnegative integrals when \(f\ge0\); for signed or complex \(f\), integrability is required.

Worked Example: Comparing Affine-Group Integrals

Let \(E=[0,1]\times[1,2]\subset G\) and \(f=\chi_E\). Since \(E\) is measurable and compact, both Haar measures give it finite mass. Direct integration gives

\int_G f\,d\mu_L =\int_1^2\int_0^1\frac{db\,da}{a^2} =\int_1^2\frac{da}{a^2} =\left[-\frac1a\right]_1^2 =\frac12, $$ $$ \int_G f\,d\mu_R =\int_1^2\int_0^1\frac{db\,da}{a} =\int_1^2\frac{da}{a} =\ln 2.

Now take \(g=(0,2)\). The condition \(xg\in E\), for \(x=(b,a)\), is \(b\in[0,1]\) and \(2a\in[1,2]\), or \(a\in[1/2,1]\). Therefore

\int_G f(xg)\,d\mu_L(x) =\int_{1/2}^1\int_0^1\frac{db\,da}{a^2} =\left[-\frac1a\right]_{1/2}^1 =1.

Check: the modular convention gives \(\Delta(g)=1/2\), so the right-translation integral must be \(\Delta(g)^{-1}\int f\,d\mu_L=2(1/2)=1\), agreeing with the direct computation.

Common Pitfall: Reversing the Modular Factor

A frequent error is to use the left Haar density \(db\,da/a^2\) and claim it is invariant under right translation. For \(g=(0,2)\), right translation changes the density by the factor \(1/2\): the image of a measurable set \(E\) has \(\mu_L(Eg)=\tfrac12\mu_L(E)\), not \(\mu_L(E)\). For integrals of the form \(\int f(xg)\,d\mu_L(x)\), the factor is the reciprocal, \(2\), because the substitution expresses the original measure \(d\mu_L(x)\) in terms of the image measure \(d\mu_L(xg)\). The set-measure factor and the integral factor are thus reciprocals under the stated convention.

When solving a Haar-measure problem, first specify left or right invariance, then verify the corresponding change of variables, and only afterward impose a normalization. On compact groups, total mass one is a natural normalization; on noncompact groups such as \(\mathbb{R}^n\) and the affine group, Haar measure has infinite total mass, so one instead normalizes a set of positive finite measure when a scale is prescribed.

Check Your Understanding

Use the stated invariance side and modular-function convention in each calculation.

On a finite group with \(12\) elements, what mass does normalized Haar measure assign to each singleton?
On \(\mathbb{R}^3\), require \([0,1]\times[0,2]\times[0,4]\) to have measure \(16\). Find the constant \(c\) such that \(\mu=c\,m\).
Compute the normalized Haar integral of \(2+\sin(2\pi x)\) on \(\mathbb{T}\).
For \(g=(b_0,a_0)\) in the affine group, state \(\Delta(g)\) under the convention \(\mu_L(Eg)=\Delta(g)\mu_L(E)\).
Using \(\mu_L=db\,da/a^2\), what factor relates \(\int f(xg)\,d\mu_L(x)\) to \(\int f\,d\mu_L\) when \(g=(b_0,3)\)?