The Regular Representation

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I am taking a class this semester in which we are studying representation theory on infinite groups. We are currently focusing on compact groups. This class has so far been manageable yet incredibly abstract.

Let $G$ be a compact, metrizable group and, for $g\in G$, define $\lambda_g:L^2(G, m)\to L^2(G, m)$ by $\xi\mapsto\xi(g^{-1}\cdot)$, where $m$ is the Haar measure such that $m(G)=1$.

Claim: $\lambda_g$ is a unitary operator.

Proof: Let $\xi\in L^2(G,m)$. Then

$$\int_G|(\lambda_g(\xi))(h)|^2\,dm(h)=\int_G|\xi(g^{-1}h)|^2\,dm(h)=\int_G|\xi(h)|^2\,dm(h)<\infty.$$
Thus $\lambda_g$ is well defined.

Observe that

$$\lambda_{g^{-1}}(\lambda_g(\xi))=\lambda_{g^{-1}}(\xi(g^{-1}\cdot))=\xi(g^{-1}g\cdot)=\xi=\xi(gg^{-1}\cdot)=\lambda_g(\xi(g\cdot))=\lambda_g(\lambda_{g^{-1}}(\xi)).$$
Thus $\lambda_g$ is bijective.

Let $\eta\in L^2(G,m)$ and let $c\in\mathbb C$. Then

$$\lambda_g(c\xi+\eta)=c\xi(g^{-1}\cdot)+\eta(g^{-1}\cdot)=c\lambda_g(\xi)+\lambda_g(\eta).$$
Thus $\lambda_g$ is linear.

We showed above that $\|\lambda_g\xi\|=\|\xi\|$. Thus $\lambda_g$ is bounded.

Observe that

$$\begin{align*}\langle\lambda_g(\xi),\lambda_g(\eta)\rangle&=\int_G(\lambda_g(\xi))(h)\overline{(\lambda_g(\eta))(h)}\,dm(h)\\&=\int_G\xi(g^{-1}h)\overline{\eta(g^{-1}h)}\,dm(h)=\int_G\xi(h)\overline{\eta(h)}\,dm(h)=\langle\xi,\eta\rangle.\end{align*}$$
Thus $\lambda_g$ preserves the inner product.

Hence, $\lambda_g$ is a unitary operator. $\blacksquare$

Claim: $\lambda:G\to\mathcal U(L^2(G,m))$ defined by $g\mapsto\lambda_g$ is a continuous, unitary representation.

Proof: Let $g,h\in G$. Then

$$(\lambda(gh))(\xi)=\lambda_{gh}(\xi)=\xi(h^{-1}g^{-1}\cdot)=\lambda_g(\xi(h^{-1}\cdot))=\lambda_g(\lambda_h(\xi))=(\lambda(g)\circ\lambda(h))(\xi).$$
Thus $\lambda$ is a homomorphism.

Let $\xi,\eta\in L^2(G,m)$, define $\varphi:G\to\mathbb C$ by $g\mapsto\langle\lambda(g)\xi,\eta\rangle$, and let $\{g_n\}$ be a sequence in $G$ converging to $g$. Then

$$\begin{align*}\lim|\varphi(g_n)-\varphi(g)|&=\lim|\langle\lambda(g_n)\xi,\eta\rangle-\langle\lambda(g)\xi,\eta\rangle|\\&=\lim|\langle\lambda(g_n)\xi-\lambda(g)\xi,\eta\rangle|\\&\leq\lim\|\lambda(g_n)\xi-\lambda(g)\xi\|\|\eta\|\\&=\lim\|\xi(g_n^{-1}\cdot)-\xi(g^{-1}\cdot)\|\|\eta\|\\&=0.\end{align*}$$
Thus $\varphi$ is continuous.

Hence, $\lambda$ is a continuous, unitary representation called the regular representation. $\blacksquare$