Left and right regular unitary representations of an LCH group

Assume AC. Let $G$ be an LCH group with a fixed left Haar measure $\mu$, let ${\mathrm{\Delta }}_G$ be the modular function of $G$ (Modular function of a locally compact group), and let $L^2(G)$ be the complex space $L^2(G,\mu ;\mathbb{C})$ with norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ and inner product $⟨\xi ,\eta ⟩={\int }_G\xi \bar{\eta }d\mu$ (Complex Haar L^p spaces and compactly supported functions, The complex $L^2$ pairing on equivalence classes). For $g\in G$ and $\xi \in L^2(G)$ define classes by $\lambda (g)\xi (x):=\xi (g^{-1}x),\rho (g)\xi (x):={\mathrm{\Delta }}_G(g{)}^{1\mathrm{/}2}\xi (xg)(x\in G)\mathrm{.}$ The assignment $\lambda$ is the left regular representation of $G$ and $\rho$ is the (modularly corrected) right regular representation.

Well-definedness and basic properties. Both formulas are representative independent: for a Borel set $E$ left invariance gives $\mu (g^{-1}E)=\mu (E)$ and the right-translation scaling ${\int }_{G}F(xg)d\mu (x)=c(g){\int }_{G}Fd\mu (x)$ with $c(g)={\mathrm{\Delta }}_G(g^{-1})>0$ (Left Haar integral and left Haar measure, Right translation scales left Haar measure) gives $\mu (Eg)=0$ exactly when $\mu (E)=0$; hence $\xi ={\xi }^{\mathrm{\prime }}$ a.e. implies $\lambda (g)\xi =\lambda (g){\xi }^{\mathrm{\prime }}$ a.e. and $\rho (g)\xi =\rho (g){\xi }^{\mathrm{\prime }}$ a.e. Both formulas are complex-linear in $\xi$ pointwise, and both preserve the norm, $\mathrm{\parallel }\lambda (g)\xi {\mathrm{\parallel }}_2=\mathrm{\parallel }\xi {\mathrm{\parallel }}_2,\mathrm{\parallel }\rho (g)\xi {\mathrm{\parallel }}_2^2={\int }_G{\mathrm{\Delta }}_G(g)\mathrm{\mid }\xi (xg){\mathrm{\mid }}^{2}d\mu (x)={\mathrm{\Delta }}_G(g){\mathrm{\Delta }}_G(g^{-1})\mathrm{\parallel }\xi {\mathrm{\parallel }}_2^2=\mathrm{\parallel }\xi {\mathrm{\parallel }}_2^2,$ using ${\mathrm{\Delta }}_G(g){\mathrm{\Delta }}_G(g^{-1})=1$ (The modular function is a continuous homomorphism); so each $\lambda (g)$ and $\rho (g)$ is a linear isometry of $L^2(G)$. The group laws hold: $\lambda (e)=\rho (e)=\mathrm{i}\mathrm{d}$, $\lambda (g)\lambda (h)=\lambda (gh)$ and $\rho (g)\rho (h)=\rho (gh)$ because ${\mathrm{\Delta }}_G$ is multiplicative and ${\mathrm{\Delta }}_G>0$, and consequently $\lambda (g{)}^{-1}=\lambda (g^{-1})$ and $\rho (g{)}^{-1}=\rho (g^{-1})$. Each $\lambda (g)$ and $\rho (g)$ is therefore an invertible linear isometry of the Hilbert space $L^2(G)$ (Hilbert space), and $g\mapsto \lambda (g)$, $g\mapsto \rho (g)$ are group homomorphisms from $G$ into the group of such operators.

Unitary representations. In the terminology of Continuous and unitary representations, a representation of $G$ on a complex Hilbert space $H$ is a group homomorphism $\pi :G\to U(H)$ into the group $U(H)$ of unitary operators on $H$ satisfying strong continuity: $g\mapsto \pi (g)\xi$ is continuous at $e$ for every $\xi \in H$ (equivalently at every point of $G$). On an infinite-dimensional $H$, unitary operator means an invertible linear isometry, so that the invertible isometries $\lambda (g),\rho (g)$ above are its unitary operators; this is the usage of Left and right regular representations on L2(G), and the finite-dimensional normal form recorded in Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces is not used here. The strong continuity of both homomorphisms follows from Strong continuity of left and modular right translations on L1 and L2: its $p=2$ left translate is exactly $\lambda (g)$, and its modular right translate is exactly $\rho (g)$. Its AC hypothesis is assumed here. Thus both assignments are unitary representations in this terminology.