Show that there exists $\xi_i \in l^2(\Gamma)$ with $\|\xi_i\|=1$ such that $\|\lambda_s \xi_i-\xi_i\| \to 0$

Question

Let $Γ$ be a discrete abelian group. We denote the group ring of $Γ$ by $\mathbb{C}(Γ)$, which is the set of all formal sums of the form $∑_{s∈Γ}a_ss$, where only finitely many of the scalar coefficients $a_s∈\mathbb{C}$ are nonzero.The reduced $C^∗$−algebra of $\Gamma$, denoted by $C^∗_{λ}(Γ)$, is the completion of $\mathbb{C}(Γ)$ with respect to the norm $||x||_r=||λ(x)||_{B(l^2(Γ))}$ where $λ:Γ→B(l^2(Γ))$ is defined by $λ(s)(δ_t)=δ_{st},∀s,t∈Γ$.

Suppose that $C_{\lambda}^*(\Gamma)$ has a character (i.e one dimensional representation). Show that there exists $\xi_i \in l^2(\Gamma)$ with $\|\xi_i\|=1$ such that $\|\lambda_s \xi_i-\xi_i\| \to 0$

Let $\phi: C_{\lambda}^*(\Gamma) \to \mathbb{C}$ be the one dimensional representation. Then $1=\overline{\phi}(ss^{-1})=\overline{\phi}(s)\overline{\phi}(s^{-1}) \implies 1=|\overline{\phi}(s)||\overline{\phi}(s^{-1})|$. Since $1=\phi(\lambda_s \circ \lambda_{s}^*)=\phi(\lambda_s)\phi(\lambda_{s}^*)=\phi(\lambda_s)\overline{\phi(\lambda_s)}=|\phi(\lambda_s)|^2$ gives us that $|\phi(\lambda_s)|=1$ which is equivalent to saying that $|\overline{\phi}(s)|=1 $ for all $s \in \Gamma$.

Since every state on $C_{\lambda}^*(\Gamma)$ can be approximated by vector states coming from $l^2(\Gamma)$, there exists $\xi_i \in l^2(\Gamma)$ such that $|\phi(x)-\langle x\xi_i,\xi_i\rangle| \to 0$ for $x \in C_{\lambda}^*(\Gamma)$ as $i \to \infty$. In particular, $|\phi(\lambda_s)-\langle \lambda_s\xi_i,\xi_i\rangle| \to 0$ for all $s \in \Gamma$. This along with the above observation gives us that $|\langle \lambda_s\xi_i,\xi_i\rangle| \to 1$ for all $s \in \Gamma$.

Now $$\|\lambda_s(\xi_i)-\xi_i\|^2=\langle\lambda_s(\xi_i)-\xi_i,\lambda_s(\xi_i)-\xi_i \rangle=2-\langle\lambda_s(\xi_i),\xi_i\rangle-\langle\xi_i,\lambda_s(\xi_i)\rangle$$

To prove the assertion, I need to show that $\langle \lambda_s\xi_i,\xi_i\rangle \to 1$. I am unable to conclude this from whatever I have shown till now. How do I show this?

Thanks for the help!!

Answer 1

Note that what you want to show is that $1\in\sigma(\lambda_s)$. I don't really see how to make your argument work.

An abelian group is amenable. One of the equivalent conditions for amenability (called Dixmier Property in the Wikipedia article) is precisely what you are looking for.

Reconstructing the actual sequence of vectors from the above argument looks hard: basically what one needs is a Følner sequence.

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Here is a bit more information. If you try to write things explicitly for a small group, say $\mathbb Z_2$, it is quickly clear that $\lambda(s)$ are permutations. These indeed have $1$ as an eigenvalue, and the obvious eigenvectors are those with all entries equal.

Trying to extend this idea ("all entries equal") to a infinite group doesn't work directly. For each $F\subset \Gamma$ finite, define $$ \xi_F=\frac1{\sqrt{|F|}}\sum_{t\in F} \delta_t. $$ Then $$ \|\xi_F\|^2=\frac1{|F|}\sum_{s,t\in F}\langle \delta_t,\delta_s\rangle=1. $$ Let $s\in \Gamma$. Then
$$ \|\lambda_s\xi_F-\xi_F\|^2=\frac{|sF\Delta F|}{|F|}, $$ and that's where we need the Følner sequence.

Answer 2

I claim that the character from $C_{\lambda}^*(\Gamma)$ can be assumed to take $\lambda_s$ to constant $1$ for all $s \in \Gamma$. If I am able to show this, then this would give me $|1-\langle \lambda_s \xi_i,\xi_i \rangle| \to 0$ as $i \to \infty$ for all $s \in \Gamma$. This coupled with the calculation at the end will give me the result. All that remains to be shown is the claim.

Claim: The product of two positive definite functions on $\Gamma$ coming from two different states on $C_{\lambda}^*(\Gamma)$ is a positive definite function. Moreover the positive function obtained in this way also comes from a state on $C_{\lambda}^*(\Gamma)$.

Let $\phi$ and $\psi$ be two positive definite functions on from $\Gamma$ to $\mathbb{C}$, which come from two different states $\omega_{\phi}$ and $\omega_{\psi}$ on $C_{\lambda}^*(\Gamma)$. Define $\xi:\Gamma \to \mathbb{C}$ by $\xi(s)=\phi(s)\psi(s)$ Let $F=\{s_1,s_2,\ldots,s_k\} \subset \Gamma$. Then $$\left[{\xi}(s^{-1}t)\right]_{s,t \in F}= \left[\phi(s^{-1}t)\psi(s^{-1}t)\right]_{s,t \in F}$$

This product is known as Hadamard Product. The Hadamard product of two positive definite matrices is positive definite. Let $A$ and $B$ be two positive definite matrices. Let the Hadamard product be denoted by $A \circ B$. Then $$x^* (A \circ B)x=\text{Tr}{(D_x^*AD_xB^*)}=\text{Tr}{\left(D_x^*A^{\frac{1}{2}}A^{\frac{1}{2}}D_x(B^{\frac{1}{2}})^*(B^{\frac{1}{2}})^*\right)}$$ $$=\text{Tr}{\left((B^{\frac{1}{2}})^*D_x^*A^{\frac{1}{2}}A^{\frac{1}{2}}D_x(B^{\frac{1}{2}})^*\right)}=\text{Tr}{\left((B^{\frac{1}{2}})^*D_x^*A^{\frac{1}{2}}A^{\frac{1}{2}}D_xB^{\frac{1}{2}}\right)}\ge0$$ This shows that ${\xi}$ is a positive definite function on $\Gamma$. $\overline{\omega_{\phi}}: \mathbb{C}(\Gamma) \to \mathbb{C}$ defined by $\overline{\omega_{\phi}}\left(\sum_{s}a_s s\right)=\sum_{s}a_s\phi(s)$ extends to a state on $C^*(\Gamma)$.

Since I want to define a state on $C_{\lambda}^*(\Gamma)$, I want to have a map which looks like this: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} C_{\lambda}^*(\Gamma) & & \ra{} & C_{\lambda}^*(\Gamma)\otimes 1 && \ra{} & C_{\lambda}^*(\Gamma) \otimes C^*(\Gamma) && \ra{} & \mathbb{C} \otimes C^*(\Gamma) && \ra{} && \mathbb{C} \\ \end{array} $$ which sends $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \sum_s a_s s & & \ra{} & \sum_s a_s s\otimes 1 && \ra{} & \sum_s a_s(\lambda_s \otimes s) && \ra{} &\sum_s a_s(\omega_{\psi}(\lambda_s) \otimes s) && \ra{} && \overline{\omega_{\phi}}\left(\sum_s\omega_{\psi}(\lambda_s)a_s s\right) \\ \end{array} $$

It can be checked that the map defined this way gives rise to a state.

Now Let $\Phi$ be a character on $C_{\lambda}^*(\Gamma)$. Then the map $f: \Gamma \to \mathbb{C}$ defined by $f(s)=\Phi(\lambda_s)$ is a positive definite function on $\Gamma$. Similarly the map $g: \Gamma \to \mathbb{C}$ defined by $g(s)=\Phi(\lambda_{s^{-1}})$ is a positive definite function on $\Gamma$. Hence, by the above claim, the map $h: \Gamma \to \mathbb{C}$ defined by $h(s)=f(s)g(s)$ is a positive definite function and comes from a state $\Lambda$ on $C_{\lambda}^*(\Gamma)$. Moreover $\Lambda(\lambda_s)=h(s)=1$ for all $s \in \Gamma$.