Unitary representation of $G$ induces representation of $L^1(G)$

Question

I am reading Davidson's '$C^*$ algebras by example'. In chapter VII regarding group $C^*$ algebras, he makes the following claim which I do not understand:

When $\pi$ is a unitary representation of a Hausdorff, locally compact group $G$, it induces a representation of $L^1(G)$ by integration: $$\tilde \pi(f)=\int f(t)\pi(t)dt$$

Here, a unitary representation is a representation of $G$ on a subgroup of unitaries in $B(H)$, where $H$ is some Hilbert space.

I do not understand how exactly this gives a representation (and in fact how it is even defined). First of all, if $\pi(t)$ is an operator, how do we integrate over it with respect to the function $f$? Doesn't this only makes sense if $\pi(t)\in \mathbb C$? And why is the LHS even an operator? What Hilbert space does it act on and what does it do?

I'd appreciate any clarification.

Thanks in advance!

Answer 1

You are used to talking about representations of groups on vector spaces, but you can also talk about representations of $\mathbb C$-algebras.

Let $\pi$ be a unitary representation of $G$ on a Hilbert space $V$. The space $L^1(G)$ is a $\mathbb C$-algebra (usually without identity), where multiplication is given by convolution. A representation of $L^1(G)$ on $V$ is an algebra homomorphism from $L^1(G)$ into the ring $\mathcal B(V)$ of bounded linear operators on $V$.

How do we define $\tilde{\pi}: L^1(G) \rightarrow \mathcal B(V)$? Given $f \in L^1(G)$ and a fixed element $v \in V$, $\tilde{\pi}(f)$ should be defined to send $v$ to some other element of $V$. What we do is consider the bounded linear map $V \rightarrow \mathbb C$ given by

$$w \mapsto \int\limits_G \langle f(g) \pi(g)v,w \rangle \space dg.$$

By the Riesz representation theorem, there exists a unique element $\tilde{\pi}(f)v$ of $V$ such that

$$\langle \tilde{\pi}(f)v,w \rangle = \int\limits_G \langle f(g)\pi(g)v,w \rangle \space dg \tag{1}$$

for all $w \in W$. This is how $\tilde{\pi}(f)$ is defined as a function from $V$ to itself, and the map $f \mapsto \tilde{\pi}(f)$ is indeed an algebra homomorphism from $L^1(G)$ to $\mathcal B(V)$, i.e. a representation. By abuse of notation, we take the integral inside the inner product to say that

$$\tilde{\pi}(f)v = \int\limits_G f(g)\pi(g)v \space dg \tag{2}$$

and to abuse notation even further, we write

$$\tilde{\pi}(f) = \int\limits_G f(g)\pi(g) \space dg.$$

**The vector-valued integral (2) is called a Gelfand-Pettis integral.

Answer 2

It is trivial to integrate functions where the codomain is in a Banach space. If you look at how you define integration, whether it's Riemann or Lebesgue, all you require for the codomain is to be a complete topological vector space; because all you do with the codomain is to take linear combinations and limits.

The fact that $\pi$ is unitary makes $\|\pi(t)\|=1$ for all $t$, so that doesn't harm taking limits in the integral. And $f\in L^1$ guarantees that the integral exists.

The integral $$ \int_G f(t)\pi(t)\,dt $$ is in $B(H)$, since $B(H)$ is closed under taking linear combinations and limits, so the integral stays within. It's also trivial to see how it acts: $$ \tilde\pi(f)\xi=\int_G f(t)\pi(t)\xi\,dt, $$ where now the integral is calculated over $G$, for the function $t\longmapsto f(t)\pi(t)\xi$ with codomain in $H$.