Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a LCH group and $\mu$ be its left Haar measure. Call $\lambda_G : G \to U(L_2(G,\mu))$ the left regular representation. We can define the reduced $C^{\ast}$ algebra and reduced Von Neumann algebra, $C_{\lambda}^{\ast} G $ and $W_{\lambda}^{\ast} G $ respectively, as the smallest $C^{\ast}$, resp. Von Neumann, subalgebras of $B(L_2(G,\mu))$ containing $\lambda_G[C_c(G)]$. It is easy to see that

$$ C_{\lambda}^{\ast} G \subset M(C_{\lambda}^{\ast} G) \subset W_{\lambda}^{\ast} G $$

and that for every $\mu$ in the space of finite Borel measures over $G$, $\lambda(\mu) \in M(C_{\lambda}^{\ast}G )$.

Question 1 Let $j : H \to G$ be a proper continuous injective homomorphism whose image $j H$ we will denote also by $H$. Let $\nu_{H} \in M_{\text{loc}}(G)$ be the locally finite measure described by its action on the compactly supported continuous functions $C_c(G)$ as:

$$ \nu_H (f) = \int_{H} f |_H (j(h)) d \mu_H (h), $$

where $\mu_H (h)$ is the Haar measure on $H$. Or equivalently $j^{\ast} \mu_H = \nu_H$. For every $\varphi \in C_c(H)$ we can define $\varphi \nu_H$ as $j^{\ast} (\varphi \mu_H)$. Since $\varphi \nu_H \in M(G)$ we have that $\lambda_G (\varphi \nu_H) \in M(C_\lambda^{\ast} G)$. Do the map ${J}$ given by:

$$ J : \lambda_H(\varphi) \mapsto \lambda_G(\varphi \nu_H) $$

extends to a normal $\ast$-homomorphism $J: W_{\lambda}^{\ast} H \to W_{\lambda}^{\ast} G$ ?. Does $J$ extends to a non degenerate $\ast$ homomorphism $J: C_{\lambda}^{\ast} H \to M( C_{\lambda}^{\ast} G )$?

Question 2 If $\alpha : G \to G$ is a continuous automorphism then, if $\varphi \in C_c(G)$, does the map $\Phi$ defined by:

$$ \Phi : \lambda_G (\varphi) \mapsto \lambda_G (\alpha_{\ast} \varphi), $$

where $\alpha_{\ast} \varphi(t) = \varphi(\alpha(t))$, extends to a normal $\ast$-homomorphism $\Phi: W_{\lambda}^{\ast} G \to W_{\lambda}^{\ast} G$?. Does $\Phi$ extends to a non degenerate $\ast$ homomorphism $\Phi: C_{\lambda}^{\ast} G \to M( C_{\lambda}^{\ast} G )$?

Question 3 If $q : G \to K$ is a continuous and open surjective group homomorphism and $q^{-1}[\{e\}] = N$ be its kernel. Then given $\phi \in C_c(G)$ we can define $P_N \phi \in C_c(K)$ as:

$$ (P_N f)(\kappa N) = \int_{N} f(\kappa \eta) d \mu_N (\eta) $$

And it is clear (as long as $\Delta_G |_N = \Delta_N$) that $P_N$ is contractive in the $L_1$ norm. Does the map $Q$ given by extension of:

$$ Q : \lambda_G(f) \mapsto \lambda_K(P_N f) $$

extends to a normal $\ast$-homomorphism $Q: W_{\lambda}^{\ast} G \to W_{\lambda}^{\ast} K$, or to a non degenerate $\ast$-homomorphism $Q: C_{\lambda}^{\ast} G \to M( C_{\lambda}^{\ast} K )$ ?.

The first two questions seem to be easy when $\Delta_G |_{j H} = \Delta_H$ and when $\alpha$ is measure preserving, respectively. While the third seems much difficult to me. Indeed, assuming $\Delta_G |_{j H} = \Delta_H$, the $G$-space $X = G/H$ has a $G$ invariant measure $\rho$ such that $(G,\mu_G) = (H \times X,\mu_H \otimes \rho)$, where the equivalence is understood as measurable spaces. That equivalence induces a unitary isometry $u: L_2(G) \to L_2(H) \otimes_2 L_2(X,\rho)$. For every $j(h) \in jH$:

$$ u \lambda_G(j(h)) = \lambda_H (h) \otimes \text{Id}_{L_2(X,\rho)} u. $$

Since $\lambda_G(j(h)) = J(\lambda_H(h))$, we have that $J$ is unitary equivalent to the tensor amplification. A very similar argument works for the second question. Just by constructing a unitary $u: L_2(G) \to L_2(G)$ given by $u(f)(t) = f(\phi(t))$ and see that $u$ intertwines $\Phi$ and the identity. Can this type of arguments be extended to the general case? Or are $\Delta_G |_{j H} = \Delta_H$ and $\alpha_{\ast}\mu_G = \mu_G$ necessary conditions for question 1 and question 2 respectively?.

I am aware that similar questions have been posted on this site, see [1] and [2] but in the context of discrete groups. In that setting the first and second questions have always a positive answer while the third has a positive answer for $C^{\ast}$ algebras if and only if $N$ is amenable and for Von Neumann algebras if and only if $N$ is finite. The converse in the $C^{\ast}$ algebra case seems to be related to the problem of characterizing amenable groups as those groups for which the co-unit $\mathcal{E} : C_\lambda^{\ast} G \to \mathbb{C}$ given by

$$ \mathcal{E}(\lambda_G(f)) = \int_G f d \mu_G $$

is bounded. Are there references for this characterization in the case of non discrete groups?

[1] The functoriality of group C* algebra structure

[2] Is the group von Neumann algebra construction functorial?

share|improve this question
1  
Nitpick: Your definition of $C^*_\lambda(G)$ is wrong-- it should be generated by $\lambda(L^1(G))$ or $\lambda(C_{00}(G))$ and not $\lambda(G)$, unless $G$ is discrete, these are different. –  Matthew Daws Nov 3 '13 at 7:59
    
You are completely right. I will correct that. –  Adrian Gonzalez-Perez Nov 3 '13 at 16:01
2  
See Paterson's book for all things Amenability related (the final question about the counit). –  Matthew Daws Nov 6 '13 at 11:53

2 Answers 2

The answer for Question 1 is "yes". I believe this to be a little subtle. Firstly, as $j:H\rightarrow G$ is proper and injective, $j(H)$ is closed in $G$, and $j:H\rightarrow j(H)$ is a homeomorphism. So wlog $H$ can be identified with a closed subgroup of $G$, with $j$ the inclusion. You sort of hint at this in the statement of your question.

Then, what is your map $J$? Well, if $\varphi\in C_c(H)$ then $f=\varphi\mu_H$ is a member of $L^1(H)$ (and such elements are dense). So $J\lambda_H(f) = \lambda_G(j^*(f))$ where $j^*$ is the pushforward $M(H)\rightarrow M(G)$. A more functional analytic way to think of this is to note that as $j$ is proper, it defines a map $\theta:C_0(G)\rightarrow C_0(H); g\mapsto g\circ j$ (as $H$ is closed, actually this map is a surjection). Then the Banach space adjoint is $j^*:M(H)=C_0(H)^* \rightarrow C_0(G)^*=M(G)$.

Now, when does $J$ extend to a normal $*$-homomorphism $W^*_\lambda(H) \rightarrow W^*_\lambda(G)$? I like to think of this in an abstract harmonic analysis framework-- the predual of $W^*_\lambda(G)$, denoted $A(G)$, can be given the structure of a commutative Banach algebra: the "Fourier Algebra" as defined by Eymard. If we regard $\lambda_G$ as a map $L^1(G)\rightarrow W^*_\lambda(G)$ then we can regard $\lambda_G^*$ as a map $A(G)\rightarrow L^1(G)^*=L^\infty(G)$, and then this actually maps into $C_0(G)$ densely (this "is" the Gelfand map of the commutative Banach algebra $A(G)$, suitably interpreted-- if $G$ is abelian, it is the Fourier transform, hence the name).

As $J$ is normal, it has a preadjoint $A(G)\rightarrow A(H)$, and if you regard these as non-closed subalgerbas of $C_0(G)$ and $C_0(H)$ respectively, we just get the map $\theta$ described above. So the question becomes equivalent to: does the map $a\mapsto a\circ j$ map $A(G)$ to $A(H)$ boundedly (hence automatically contractively).

It turns out that the answer is: "yes". It's even a quotient map-- every $A(H)$ function arises as the restriction of an $A(G)$ function (identifying $H$ as a closed subgroup of $G$). This theorem is known as "Herz restriction", and the nicest writeup I know is an MSc thesis: Cameron Zwarich's thesis, see Section 4.2.

Once we know $J$ exists, we know that $J\lambda_H(f)=\lambda_G(j^*(f))$ for all $f\in L^1(H)$, and as $j^*(f)\in M(G)$, it follows that $J\lambda_H(f)\in MC^*_\lambda(G)$, so indeed $J$ does restrict to a map $C^*_\lambda(H) \rightarrow MC^*_\lambda(G)$. The example of $\mathbb Z\subseteq\mathbb R$ shows we can't hope to get into $C^*_\lambda(G)$.

Surely the same techniques give a positive answer of Q2.

I think $\mathbb R \rightarrow \mathbb R/\mathbb Z=\mathbb T$ shows that Q3 does not have a positive answer in general; I think the Fourier transform shows that $W^*_\lambda(\mathbb R)=L^\infty(\mathbb R), W^*_\lambda(\mathbb T)=\ell^\infty(\mathbb Z)$ and the map $J$ should send $(e^{2\pi ixt})_{x\in\mathbb R}\in L^\infty(\mathbb R)$ to $(e^{2\pi int})_{n\in\mathbb Z}\in\ell^\infty(\mathbb Z)$. This doesn't exist at the $L^\infty$ level. You always have a map at the level of full $C^*$-algebras, $C^*(G)\rightarrow C^*(K)$ (surjective even, and no multiplier algebra) but amenability issues might becomes a problem trying to drop to $C^*_\lambda(K)$.

share|improve this answer
    
Thank you for pointing out the connection with the Herz restriction theorem. Indeed, my original motivation was related to the restriction of multipliers. I will look that thesis. –  Adrian Gonzalez-Perez Nov 5 '13 at 23:14
    
On Q3 the qoutient $q:\mathbb{R} \to \mathbb{R}/\mathbb{Z}$ you mention reinforces my intuition that Q3 have a positive answer iff $\text{ker}(q)$ is compact. In the abelian case that an be proved through the Poisson summation formula for LCA groups. The Fourier transform $\lambda_G: C_c(G) \to C^{\ast}_\lambda G$ intertwines the averaging operator associated to a subgroup $H \leq G$ with the restriction to its annihilator group $H^{\perp}$. –  Adrian Gonzalez-Perez Nov 6 '13 at 0:14
    
In the case above $\mathbb{Z}^{\perp} = \mathbb{Z}$. Since $\mathbb{Z}$ has measure 0 the restriction is not defined for $L_\infty(\hat{\mathbb{R}})$. Restriction can be defined at $L_{\infty}$ level when $H^{\perp}$ is open. But $H^{\perp}$ is open iff $H$ is compact. –  Adrian Gonzalez-Perez Nov 6 '13 at 0:15

The group(oid) C*-algebra construction can be made functorial with various choices of morphisms on the group(oid) side and the C*-algebra side.

See e.g. http://arxiv.org/abs/math/0511613v2, who adapt the notion of groupoid morphism to make the construction work. In particular, in the case of a group $G$ that is $\sigma$-compact, second countable, and measurewise amenable, this should answer question 1.

Another choice is made in http://www.theta.ro/jot/archive/1999-042-001/1999-042-001-005.html, who choose correspondences between both group(oid)s and their reduced C*-algebras. This can also be made to work with measured functors between group(oid)s and correspondences between C*-algebras, see http://arxiv.org/abs/math-ph/0008036.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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