Induced Unitary Representations of Locally Compact Groups — Examples

When $H=\{e\}$, the rho covariance imposes no restriction, and $\rho =1$ is valid. The averaging map $T_H$ is identity and the Weil formula becomes ${\int }_{G}f(x)dx={\int }_{G}f(x)d{\mu }_{\rho }(x)$. Radon uniqueness [F2] identifies ${\mu }_{\rho }$ with left Haar measure.

Covariance is empty for the trivial subgroup, so the dense model is $C_c(G)$ and its completion is $L^2(G)$. The density cocycle is $D_g(x)=1$, hence the induced action is $F(x)\mapsto F(g^{-1}x)$, exactly [F3]. The construction theorem [F1] supplies strong continuity and unitarity. ∎

Since $\mathrm{\Gamma }$ is discrete, [F1] gives ${\mathrm{\Delta }}_{\mathrm{\Gamma }}=1$. The function $\rho (x)={\mathrm{\Delta }}_G(x{)}^{-1}$ obeys $\rho (x\gamma )={\mathrm{\Delta }}_{\mathrm{\Gamma }}(\gamma ){\mathrm{\Delta }}_G(\gamma {)}^{-1}\rho (x)$, so it is a rho-function by [F5]. Its cocycle is constant: $D_g(x\mathrm{\Gamma })=\rho (g^{-1}x)\mathrm{/}\rho (x)={\mathrm{\Delta }}_G(g)$.

Let ${\mu }_{\rho }$ be its quotient measure. Since $G\mathrm{/}\mathrm{\Gamma }$ is compact, $0<{\mu }_{\rho }(G\mathrm{/}\mathrm{\Gamma })<\mathrm{\infty }$: finiteness is Radon compact-finiteness and positivity follows from full support. The pushforward $g_{\ast }{\mu }_{\rho }$ has the same total mass as ${\mu }_{\rho }$, while [F2] gives $g_{\ast }{\mu }_{\rho }={\mathrm{\Delta }}_G(g){\mu }_{\rho }$. Therefore ${\mathrm{\Delta }}_G(g)=1$ for every $g$, and $G$ is unimodular.

Now ${\mathrm{\Delta }}_G{\mathrm{\mid }}_{\mathrm{\Gamma }}={\mathrm{\Delta }}_{\mathrm{\Gamma }}=1$, so [F3] gives a nonzero invariant Radon measure on $G\mathrm{/}\mathrm{\Gamma }$; compactness makes it finite. For $\rho =1$ its cocycle is identically one. Scalar covariance says $F(x\gamma )=F(x)$, so sections are exactly functions on $G\mathrm{/}\mathrm{\Gamma }$, and the action is $F(q)\mapsto F(g^{-1}q)$ with the quotient $L^2$ norm. This is the quasi-regular representation, as claimed by [F4]. ∎

Counting measure is left and right invariant on a finite group, so both modular functions are $1$ and $\rho =1$ satisfies the rho covariance. The finite quotient formula is the partition of a finite sum into cosets: ${\sum }_{x\in G}a(x)={\sum }_{xH\in G\mathrm{/}H}{\sum }_{h\in H}a(xh)$. Thus the quotient measure is counting measure, and the general covariance and left action reduce to [F1].

If $x$ is replaced by $x{h}_0$, then $F(x{h}_0)=\sigma (h_0{)}^{-1}F(x)$, so unitarity makes $\mathrm{\parallel }F(x{h}_0)\mathrm{\parallel }=\mathrm{\parallel }F(x)\mathrm{\parallel }$. The stated norm and inner product are therefore independent of coset representatives.

Left multiplication permutes the finite set $G\mathrm{/}H$, and [F1] shows it preserves the covariance law; reindexing the finite sum proves $\mathrm{\parallel }\mathrm{\Pi }(g)F\mathrm{\parallel }=\mathrm{\parallel }F\mathrm{\parallel }$. The covariant subspace is closed in the complete finite product $V^G$, so completion adds no vectors. This is exactly the algebraic induced module with the stated invariant inner product. ∎

The subgroup $H=\{(0,a):a>0\}$ is closed because it is the zero set of the continuous first-coordinate map, and it is a subgroup by the multiplication law. The map $(b,a)\mapsto b$ is continuous and constant on right $H$-cosets; its fibers are exactly those cosets, since $(b,a)(0,t)=(b,at)$. It therefore descends to a continuous bijection $\bar{q}:G\mathrm{/}H\to \mathbb{R}$. The inverse $b\mapsto (b,1)H$ is continuous as a composition of the continuous section $b\mapsto (b,1)$ with the quotient map, so $\bar{q}$ is a homeomorphism. Under this identification the left action of $(b_0,a_0)$ is $T(x)=b_0+a_{0}x$. Let $m$ be a left Haar Radon measure on $(\mathbb{R},+)$, supplied by [F1]. For each such $T$, the measure ${\nu }_T(E)=m(T^{-1}E)$ on Borel sets is Radon because $T$ is a homeomorphism. It is translation invariant: $T^{-1}(E+u)=T^{-1}(E)+u\mathrm{/}{a}_0$, so left invariance of $m$ applies. Thus [F2] gives ${\nu }_T=c_{T}m$ for some $c_T>0$. Hence every group translate scales $m$ by a positive finite constant, so its null sets are preserved and its Radon measure class is quasi-invariant.

On the two-dimensional group manifold use the measure $d{\mu }_G(b,a)=a^{-2}dbda$, defined by the positive continuous density in [F3] relative to two-dimensional Lebesgue measure. By [A1, A2], AC supplies countable choice, so [F8] makes the base measure Radon; the density is bounded above on each compact set, so the weighted measure is locally finite and inherits regularity from the base measure. Left multiplication by $(b_0,a_0)$ sends $(b,a)$ to $(b_0+a_{0}b,a_{0}a)$ with Jacobian $a_0^2$; substituting these coordinates gives $a^{-2}dbda=(a_{0}a{)}^{-2}d(b_0+a_{0}b)d(a_{0}a)$, so ${\mu }_G$ is left invariant. Therefore it is a left Haar measure by [F7]. For $h=(0,t)\in H$, right multiplication sends $(b,a)$ to $(b,at)$; with $A=at$, $a^{-2}dbda=t{A}^{-2}dbdA$, and hence ${\int }_{G}f(xh)d{\mu }_G(x)=t{\int }_{G}f(x)d{\mu }_G(x)$ for $f\in C_c(G)$. Applying [F4] to $h^{-1}=(0,t^{-1})$ yields ${\mathrm{\Delta }}_G(0,t)=t^{-1}$. On the other hand [F5] gives ${\mathrm{\Delta }}_H(0,t)=1$, since $H$ is abelian. Taking any $t\mathrm{\ne }1$, the modular functions disagree on $H$; [F6] rules out every nonzero invariant Radon measure on $G\mathrm{/}H$. ∎