The Modular Function and L1 Group Algebras — Examples

$d\mu =a^{-2}dadb$ is a nonzero Borel measure, finite on compact sets: on a compact $K\subseteq G$ the continuous density $a^{-2}$ attains a maximum, so $\mu (K)\le {\max }_K{a}^{-2}\cdot {\lambda }_2(K)<\mathrm{\infty }$, while every nonempty open set has positive measure by [F5]. The rational rectangles contained in $G$ form a countable base, so $G$ is second countable. Under the countable choice implied by [A1], the compact-finite Borel measure $\mu$ is outer regular on Borel sets and inner regular on open sets by Locally finite Borel measures on second-countable LCH spaces are regular.

Left invariance. Fix $g_0=(a_0,b_0)\in G$. Left multiplication is $L_{g_0}(a,b)=(a_{0}a,b_0+a_{0}b)$, a $C^1$ diffeomorphism of $G$ with $\det D{L}_{g_0}=a_0^2$ at every point. For nonnegative Borel $F$, the change of variables [F4], whose countable-choice hypothesis is in force by [A1], applied to $(u,v)=L_{g_0}(a,b)$ gives ${\int }_{G}F(L_{g_0}(a,b))a^{-2}dadb={\int }_{G}F(u,v)(u\mathrm{/}{a}_0{)}^{-2}{a}_0^{-2}dudv={\int }_{G}F(u,v)u^{-2}dudv$, since the inverse map is $a=u\mathrm{/}{a}_0$, $b=(v-b_0)\mathrm{/}{a}_0$ and $\mathrm{\mid }\det D{L}_{g_0}^{-1}\mathrm{\mid }=a_0^{-2}$. Hence $\mu$ is left invariant.

Right translation scales $\mu$ by $a_0$. With the same $g_0$, right multiplication is $R_{g_0}(a,b)=(a{a}_0,b+a{b}_0)$, a $C^1$ diffeomorphism with $\det D{R}_{g_0}(a,b)=\det \left(\begin{array}{cc}{a}_0& 0\\ b_0& 1\end{array}\right)=a_0$. For nonnegative Borel $F$, the change of variables [F4], again in force by [A1], with $(u,v)=(a{a}_0,b+a{b}_0)$ gives ${\int }_{G}F(a{a}_0,b+a{b}_0)a^{-2}dadb={\int }_{G}F(u,v)(u\mathrm{/}{a}_0{)}^{-2}{a}_0^{-1}dudv=a_0{\int }_{G}F(u,v)u^{-2}dudv$, using $a=u\mathrm{/}{a}_0$, $b=v-(u\mathrm{/}{a}_0)b_0$ and $\mathrm{\mid }\det D{R}_{g_0}^{-1}\mathrm{\mid }=a_0^{-1}$. So ${\int }_{G}F(x{g}_0)d\mu (x)=a_0{\int }_{G}Fd\mu$ for every nonnegative Borel $F$, and also for every $\mu$-integrable $F$ by linearity in the real and imaginary parts.

The scalar of step 1.3 is $c(g_0)=a_0$, so ${\mathrm{\Delta }}_G(g_0)=c(g_0^{-1})$, the scalar $c(g_0^{-1})$ being the well-defined one supplied under the AC of [A1]; applying step 1.3 to $g_0^{-1}=(a_0^{-1},-b_0\mathrm{/}{a}_0)$ gives $c(g_0^{-1})=a_0^{-1}$. Hence ${\mathrm{\Delta }}_G(a_0,b_0)=a_0^{-1}$ for every $(a_0,b_0)\in G$, which is the asserted modular function.

The function $(a,b)\mapsto a^{-1}$ is not identically $1$ on $G$: at $(2,0)$ it takes the value $1\mathrm{/}2$. By [F3] the group $G$ is therefore not unimodular, while steps 1.1–1.2 show that $a^{-2}dadb$ is indeed a left Haar measure for which the computation of step 2.1 applies. ∎

The measure and the involution. By [F1], $\mu =c\cdot \text{counting}$ and $\mu$ is right invariant, so $G$ is unimodular; by [F2] the involution of $L^1(G)$ is $f^{\ast }(x)=\overline{f(x^{-1})}$.

The formula on finitely supported functions. For $f,g\in C_c(G)$ and $x\in G$, only finitely many $y$ have $f(y)\mathrm{\ne }0$, and by [F3] and [F1] $(f\ast g)(x)={\int }_{G}f(y)g(y^{-1}x)d\mu (y)=c{\sum }_{y\in G}f(y)g(y^{-1}x)$.

The series map is bilinear and bounded. For $f,g\in L^1(G)$ define $h(x):=c{\sum }_{y\in G}f(y)g(y^{-1}x)$; on this discrete group each $L^1$ class has a unique pointwise representative because every singleton has measure $c>0$. For each $x$, Cauchy--Schwarz gives ${\sum }_{y\in G}\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}<\mathrm{\infty },$ where $y\mapsto y^{-1}x$ is a bijection and $\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}$ for every summable family; this proves pointwise absolute convergence by [F1, F5]. For the $L^1$ bound, rearrange the nonnegative double sum using the finite-subsum definition of sums in [F5]: ${\sum }_{x\in G}\mathrm{\mid }h(x)\mathrm{\mid }\le c{\sum }_{x\in G}{\sum }_{y\in G}\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }=c{\sum }_{y\in G}\mathrm{\mid }f(y)\mathrm{\mid }{\sum }_{x\in G}\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }=c\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{.}$ The last equality again uses the bijection $x\mapsto y^{-1}x$. Thus $h\in L^1(G)$ and $\mathrm{\parallel }h{\mathrm{\parallel }}_1=c{\sum }_x\mathrm{\mid }h(x)\mathrm{\mid }\le c^2\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{.}$ Absolute convergence also gives bilinearity, so $(f,g)\mapsto h$ is a bounded, hence continuous, bilinear map into $L^1(G)$.

The series computes the convolution. On $C_c(G)\times C_c(G)$ the map $(f,g)\mapsto h$ of step 1.3 agrees with the $C_c$ convolution by step 1.2, and both $(f,g)\mapsto h$ and the $L^1$ convolution are continuous bilinear maps on $L^1(G)\times L^1(G)$ by steps 1.3 and [F3]. Since $C_c(G)$ is dense in $L^1(G)$ by [F4], they agree as $L^1$ classes for all $f,g\in L^1(G)$. Every singleton has positive measure $c$, so equality of classes on this discrete group is pointwise equality; hence $(f\ast g)(x)=c{\sum }_{y\in G}f(y)g(y^{-1}x)$ for all $x\in G$.

The convolution unit. Let $u:=c^{-1}{\mathbf{1}}_{\{e\}}$; then ${\int }_G\mathrm{\mid }u\mathrm{\mid }d\mu =c\cdot c^{-1}=1$, so $u\in L^1(G)$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_1=1$. By step 2.1, $(u\ast g)(x)=c{\sum }_{y\in G}u(y)g(y^{-1}x)=c\cdot c^{-1}g(x)=g(x)$ and $(g\ast u)(x)=c{\sum }_{y\in G}g(y)u(y^{-1}x)=c\cdot c^{-1}g(x)=g(x)$ for every $g\in L^1(G)$ and every $x$, the only contributing index being $y=e$, respectively $y=x$. So $u$ is a two-sided identity.

Collecting: on a discrete group with $\mu =c\cdot \text{counting}$ the group is unimodular with $f^{\ast }(x)=\overline{f(x^{-1})}$, convolution is the absolutely convergent series of step 2.1, and $c^{-1}{\mathbf{1}}_{\{e\}}$ is the norm-one convolution identity. ∎

Coefficients are square integrable. By [F8] each coefficient $f_{u,v}^{\pi }$ and $f_{w,z}^{\sigma }$ is continuous with $\mathrm{\mid }{f}_{u,v}^{\pi }\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ and $\mathrm{\mid }{f}_{w,z}^{\sigma }\mathrm{\mid }\le \mathrm{\parallel }w\mathrm{\parallel }\mathrm{\parallel }z\mathrm{\parallel }$, so both lie in $L^{\mathrm{\infty }}(G)\subseteq L^2(G)$ with $\mathrm{\parallel }{f}_{u,v}^{\pi }{\mathrm{\parallel }}_2,\mathrm{\parallel }{f}_{w,z}^{\sigma }{\mathrm{\parallel }}_2<\mathrm{\infty }$ because $\mu$ is a probability measure; the description of $L^2(G)$ used here is the one of [F3], available under the AC of [A1].

The integral formula computes convolution for $L^2$-functions. Let $f,g\in L^2(G)$ and put $h(x):={\int }_{G}f(y)g(y^{-1}x)d\mu (y)$. For each $x$ the integral converges absolutely with $\mathrm{\mid }h(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$ by Cauchy–Schwarz [F4]. The function $h$ is continuous: for $x,x_0\in G$, Cauchy–Schwarz and the measure-preserving substitution $y\mapsto w=y^{-1}{x}_0$ (measure preserving by the unimodularity and inversion invariance of [F1]) give $\mathrm{\mid }h(x)-h(x_0)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g(\cdot x_0^{-1}x)-g{\mathrm{\parallel }}_2\to 0$ as $x\to x_0$ by strong continuity of right translation [F5]. Also $\mathrm{\parallel }h{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$: the pointwise Cauchy–Schwarz bound and $\mu (G)=1$ give ${\int }_G\mathrm{\mid }h(x){\mathrm{\mid }}^{2}d\mu (x)\le {\int }_G({\int }_G\mathrm{\mid }f(y){\mathrm{\mid }}^{2}d\mu (y))({\int }_G\mathrm{\mid }g(y^{-1}x){\mathrm{\mid }}^{2}d\mu (y))d\mu (x)=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$ by left invariance of $\mu$ [F1]. Finally, on $C_c(G)\times C_c(G)$ the map $(f,g)\mapsto h$ is the $C_c$ convolution by [F2], and both $(f,g)\mapsto h$ and the $L^1$ convolution are continuous bilinear maps from $L^2(G)\times L^2(G)$ into $L^1(G)$: the first because $\mathrm{\parallel }h{\mathrm{\parallel }}_1\le \mathrm{\parallel }h{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$, the second because $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$, using $\mathrm{\parallel }t{\mathrm{\parallel }}_1\le \mathrm{\parallel }t{\mathrm{\parallel }}_2$ on the probability space $G$ and [F2], [F3]; so density of $C_c(G)$ in $L^2(G)$ [F3] forces $h=f\ast g$ for all $f,g\in L^2(G)$.

The averaged operator. Define $A:W\to V$ by $A(\xi ):={\int }_G⟨\sigma (y{)}^{-1}\xi ,z⟩\pi (y)ud\mu (y)$, the integral being computed componentwise in a basis of $V$; the integrand is continuous on the compact group, so the integral exists and $A$ is linear in $\xi$. Then $A$ intertwines $\sigma$ with $\pi$: for $g\in G$ and $\xi \in W$, substituting $y=g{y}^{\mathrm{\prime }}$ and using invariance of $\mu$ gives $A(\sigma (g)\xi )={\int }_G⟨\sigma (y{)}^{-1}\sigma (g)\xi ,z⟩\pi (y)ud\mu (y)={\int }_G⟨\sigma (g^{-1}y{)}^{-1}\xi ,z⟩\pi (y)ud\mu (y)={\int }_G⟨\sigma (y^{\mathrm{\prime }}{)}^{-1}\xi ,z⟩\pi (g{y}^{\mathrm{\prime }})ud\mu (y^{\mathrm{\prime }})=\pi (g)A(\xi )$, using $\sigma (y{)}^{-1}\sigma (g)=\sigma (y^{-1}g)$ and the substitution $y=g{y}^{\mathrm{\prime }}$.

The constant function. Since $\mu (G)=1$ and $\mathbf{1}\in L^2(G)$ by [F3], the integral formula of step 1.2 gives $(\mathbf{1}\ast \mathbf{1})(x)={\int }_G\mathbf{1}d\mu =1$ for every $x$, so $\mathbf{1}\ast \mathbf{1}=\mathbf{1}$.

The product identity. For $x\in G$, the integral formula of step 1.2 applied to the coefficients of step 1.1 gives $(f_{u,v}^{\pi }\ast f_{w,z}^{\sigma })(x)={\int }_G⟨\pi (y)u,v⟩⟨\sigma (y^{-1}x)w,z⟩d\mu (y)=⟨{\int }_G⟨\sigma (y{)}^{-1}\sigma (x)w,z⟩\pi (y)ud\mu (y),v⟩,$ where the second equality uses $\sigma (y^{-1}x)=\sigma (y{)}^{-1}\sigma (x)$ and the linearity of the inner product in its first variable; the right-hand side is $⟨A(\sigma (x)w),v⟩$ with $A$ as in step 1.3.

The trace of $A$ in the case $\sigma =\pi$. Suppose $\sigma$ is the same representation $\pi$ on $V$, of dimension $d$, so that $A$ is an endomorphism of $V$; write $R_y(\xi ):=⟨\pi (y{)}^{-1}\xi ,z⟩\pi (y)u=⟨\xi ,\pi (y)z⟩\pi (y)u$, a rank-one operator with $\mathrm{tr}{R}_y=⟨\pi (y)u,\pi (y)z⟩=⟨u,z⟩$ by unitarity of $\pi (y)$. Integrating the traces, $\mathrm{tr}A={\int }_G⟨u,z⟩d\mu (y)=⟨u,z⟩$.

The Schur step. The kernel and the image of any intertwiner are subrepresentations, by [F6]; hence if $A\mathrm{\ne }0$, irreducibility makes $A$ an isomorphism. Thus $\pi \cong ̸\sigma$ forces $A=0$. If $\pi \cong \sigma$, rescale an invertible intertwiner to a unitary one $T:W\to V$: $T^{\ast }T$ commutes with $\sigma$, hence is a positive scalar by the eigenvalue argument of [F7] and irreducibility. Then $A{T}^{-1}$ commutes with $\pi$ and is scalar by the same argument. Replacing $z$ by $Tz$ in the trace calculation of step 2.3 gives $\mathrm{tr}(A{T}^{-1})=⟨u,Tz⟩$, so $A=(⟨u,Tz⟩\mathrm{/}\dim \pi )T$.

The coefficient products. The coefficients lie in $L^2(G)\subseteq L^1(G)$ by [F3] and [F8], so step 2.2 computes their convolution. If $\pi \cong ̸\sigma$, $A=0$ by step 3.1. Otherwise step 3.1 gives $A=(⟨u,Tz⟩\mathrm{/}\dim \pi )T$, hence step 2.2 gives $(f_{u,v}^{\pi }\ast f_{w,z}^{\sigma })(x)=(⟨u,Tz⟩\mathrm{/}\dim \pi )⟨\pi (x)Tw,v⟩$. Any two unitary intertwiners differ by a scalar $c$ with $\mathrm{\mid }c\mathrm{\mid }=1$, since their quotient commutes with $\pi$ and the scalar-endomorphism argument of step 3.1 applies. Replacing $T$ by $cT$ conjugate-scales the first factor and scales the coefficient by $c$, so the product is independent of $T$.

Together with $\mathbf{1}\ast \mathbf{1}=\mathbf{1}$ from step 2.1, step 4.1 proves the coefficient formula for inequivalent and equivalent irreducible representations. For $\sigma =\pi$, taking $T=I_V$ gives $(⟨u,z⟩\mathrm{/}\dim \pi )f_{w,v}^{\pi }$. ∎

The modulus of $Jf$ and the inversion formula. For a nonnegative $f\in C_c(G)$, also $Jf\in C_c(G)$ because inversion is a homeomorphism, and $\mathrm{\mid }Jf(x)\mathrm{\mid }=f(x^{-1})$. Thus [F2] gives $\mathrm{\parallel }Jf{\mathrm{\parallel }}_1={\int }_{G}f(x^{-1})d\mu (x)={\int }_G{\mathrm{\Delta }}_G(x^{-1})f(x)d\mu (x)$, while $\mathrm{\parallel }f{\mathrm{\parallel }}_1={\int }_{G}f(x)d\mu (x)$; both integrals are finite by [F4].

A witness in the region $a>1$. Take the open rectangle $U:=(1,2)\times (0,1)$ and a compact rectangle $K:=[3\mathrm{/}2,7\mathrm{/}4]\times [1\mathrm{/}4,3\mathrm{/}4]\subseteq U$, and let $f\in C_c(G)$ be given by [F5] under the Dependent Choice derived from the assumed AC, with ${\mathbf{1}}_K\le f\le {\mathbf{1}}_U$. Then $f\ge 0$, $f\equiv ̸0$, and on $\mathrm{supp}f\subseteq U$ one has $a>1$, hence $a^{-1}>a^{-2}$ everywhere on $(1,2)$.

Comparison of norms. Since $f\ge 0$ and $\mathrm{supp}f\subseteq \{a>1\}$, step 1.1 gives $\mathrm{\parallel }Jf{\mathrm{\parallel }}_1-\mathrm{\parallel }f{\mathrm{\parallel }}_1={\int }_G({\mathrm{\Delta }}_G(x^{-1})-1)f(x)d\mu (x)={\int }_G(a-1)f(a,b)a^{-2}dadb$, where $a-1>0$ on the support of $f$ and $f>0$ on the nonempty open set where $f\ge {\mathbf{1}}_K=1$, namely on the interior of $K$. The integrand is nonnegative continuous with a strictly positive value on an open subset of $G$, and $a^{-2}dadb$ gives positive measure to every nonempty open set, so $\mathrm{\parallel }Jf{\mathrm{\parallel }}_1-\mathrm{\parallel }f{\mathrm{\parallel }}_1>0$: the norms differ.

Since $J$ changes the $L^1$ norm of the nonzero compactly supported function $f$ of step 1.2, it cannot define an isometry of $L^1(G)$; the modular involution [F3], by contrast, is isometric. The naive inversion $f\mapsto \overline{f(x^{-1})}$ is therefore not the $L^1$ involution on this nonunimodular group. ∎