# Are the reduced group Von Neumann algebra/ Group $C^{\ast}$ algebra functorial in the case of LCH groups

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  and  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?

• 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. Nov 3, 2013 at 7:59
• You are completely right. I will correct that. Nov 3, 2013 at 16:01
• See Paterson's book for all things Amenability related (the final question about the counit). Nov 6, 2013 at 11:53

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 $C^*_\lambda(K)$.

• 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. Nov 5, 2013 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 $H^{\perp}$. Nov 6, 2013 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. Nov 6, 2013 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.