Let $(G,A,\alpha)$ be a $C^*$-dynamical system where $G$ is a countable discrete group. When defining the reduced crossed product, one can proceed as follows:

Let $\pi$ be a faithful representation of $A$ on a Hilbert space $H$. Then we get a representation $\tilde{\pi}$ of $A$ on $\ell^2(G,H)$ such that $(\tilde{\pi}(a)\xi)(t)=\pi(\alpha_{t^{-1}}(a))\xi(t)$ for all $a\in A$, $t\in G$, and $\xi\in\ell^2(G,H)$. Let $\lambda$ be the representation of $G$ on $\ell^2(G,H)$ such that $\lambda_s\xi(t)=\xi(s^{-1}t)$ for $s,t\in G$ and $\xi\in\ell^2(G,H)$. Then $(\tilde{\pi},\lambda)$ is a covariant representation, called a regular covariant representation, and its integrated form $\tilde{\pi}\rtimes\lambda$ is a faithful representation of $AG$ on $\ell^2(G,H)$. Define the reduced norm on $AG$ by $||f||_r=||(\tilde{\pi}\rtimes\lambda)(f)||$. The completion is the reduced crossed product. It can be shown that $||f||_r=\sup ||\sigma(f)||$ for $f\in AG$, where the supremum is taken over all regular covariant representations $\sigma$ obtained from (not necessarily faithful) representations of $A$. Therefore, one could have used either one of these as the definition of $||f||_r$.

My question has to do with the $L^p$ analog of the reduced $C^*$-crossed product, which has been studied by Chris Phillips. In this case, $A$ is isometrically isomorphic to a norm-closed subalgebra of $B(L^p(X,\mu))$ for some measure space $(X,\mu)$ and $p\in[1,\infty)$. In Phillips's setup, he uses the supremum definition (following earlier work by Dirksen-de Jeu-Wortel) but the supremum is now taken over all regular covariant representations obtained from nondegenerate $\sigma$-finite contractive representations of $A$, and this gives a seminorm, so one mods out by the kernel of this seminorm and takes completion in the induced norm to get the reduced $L^p$ crossed product. I have also seen a talk by Kasparov where he uses the first definition of the reduced norm. I suppose he is thinking of $A$ as already coming with a faithful representation given by the isometric embedding. So my question is: Are the two norms the same? Or what can we say about the resulting algebras?


What happens in C*-algebras is that faithful (injective) homomorphisms are automatically isometric. The integrated form of an injective representation is again injective, and hence isometric, so one can use any such representation to define the norm on the reduced crossed product. In the context of $L^p$-operator algebras, injective does not imply isometric. If one starts with a faithful (or even isometric) representation of A on an $L^p$-space, its integrated form is injective (this needs a proof, but it's in Chris' paper), but not in general isometric. So one must really take the supremum over all regular covariant representations. I don't have an explicit example in mind, but I'm essentially certain that there are examples where these two norms could be different.

  • 1
    $\begingroup$ If $A=C(X)$ is represented on $L^p(X,\mu)$ as multiplication operators, will the integrated form be isometric? $\endgroup$ – cyc Feb 5 '16 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.