Any isometry in $\mathcal{L}(G)$ must be a unitary

Let $$G$$ be a countable group with neutral element $$e$$. Consider the Hilbert space $$\ell^2(G):=\left \{ x:G\to \mathbb{C}\mid \sum_{t\in G}|x(t)|^2<\infty \right \}$$ with inner product $$\left \langle x,y \right \rangle=\sum_{t\in G}x(t)\overline{y(t)}$$ for $$x,y\in \ell^2(G)$$. For each $$t\in G$$, let $$\delta_t\in \ell^2(G)$$ be defined by $$\delta_t(t)=1$$ and $$\delta_t(s)=0$$ if $$s\neq t$$. The set $$(\delta_t)_{t\in G}$$ is an orthonormal basis for $$\ell^2(G)$$, and $$x(t)=\left \langle x,\delta_t \right \rangle$$ for $$x\in \ell^2(G)$$ and $$t\in G$$.

For each $$t\in G$$, consider the operator $$U_t$$ on $$\ell^2(G)$$ given by $$(U_tx)(s)=x(t^{-1}s)$$ for $$x\in \ell^2(G)$$ and $$s\in G$$.

Put $$\mathcal{L}(G)=\{ U_t\mid t\in G\}''$$. Consider the state $$\tau$$ on $$\mathcal{L}(G)$$ defined by $$\tau(T)=\left \langle T\delta_e,\delta_e \right \rangle$$ for $$T\in \mathcal{L}(G)$$.

Problems

1) Show that $$\tau(ST)=\tau(TS)$$ for all for all $$S,T\in \mathcal{L}(G)$$.

3) Show that any isometry in $$\mathcal{L}(G)$$ must be a unitary

1) I am not sure about this part. If $$S,T\in \mathcal{L}(G)$$, then we can write $$S=\sum_{s\in G}\alpha_sU_s$$ and $$T=\sum_{t\in G}\beta_tU_t$$. Can I write the multiplication $$ST=\sum_{(s,t)\in S\times T}\gamma_{s,t}U_sU_t$$ where $$\gamma_{s,t}$$'s are complex numbers in terms of $$\alpha$$ and $$\beta$$? If not, how would you write it mathematically? Assume that this is true. I have shown that $$U_{s}U_{t}=U_{st}$$ and $$U_t\delta_s=\delta_{ts}$$. Then we have $$\tau(ST)=\sum_{(s,t)\in S\times T}\gamma_{s,t}\tau(U_{st})=\sum_{(s,t)\in S\times T}\gamma_{s,t}\left \langle U_{st}\delta_e,\delta_e \right \rangle=\sum_{(s,t)\in S\times T}\gamma_{s,t}\left \langle \delta_{st},\delta_e \right \rangle=\sum_{(s,t)\in S\times T}\gamma_{s,t} \delta_{st}(e)=\sum_{(s,t)\in S\times T}\gamma_{s,t} \delta_{ts}(e)=\dots=\tau(TS)$$ Is this correct?

I am not sure about the last problem. Could you help me with it? The problem 2) was as follows: Show that $$\tau(T^* T)=0$$ implies $$T=0$$ for all $$T\in \mathcal{L}(G)$$. I have solved this one. Just informing it if it is relevant for the last problem.

Is some informations are missing, please let me know.

For part 1, the way you are doing looks problematic to me, because where your dots start you would have $$\sum_s\gamma_{s,s^{-1}}$$ and you need to relate them to the corresponding coefficients for $$U_{ts}$$.
Rather, first note that it is enough to work with finite sums, since $$\tau$$ is (obviously!) wot-continuous. With finite sums (and thus no need to worry about convergence) you have \begin{align} \tau(ST)&=\sum_{s,t} \alpha_s\beta_t\langle U_{st}\delta_e,\delta_e\rangle =\sum_{s,t} \alpha_s\beta_t\langle \delta_{st},\delta_e\rangle =\sum_s\alpha_s\beta_{s^{-1}}\\ &=\sum_t\alpha_{t^{-1}}\beta_t=\tau(TS). \end{align}
For part 3, part 2 (which says that $$\tau$$ is faithful) is the essential bit. If $$S$$ is an isometry, you have $$S^*S=I$$. Then $$\tau(I-SS^*)=\tau(I)-\tau(SS^*)=\tau(I)-\tau(S^*S)=1-1=0.$$ Note also that $$SS^*\leq I$$, since it is positive and its norm is 1 (because $$\|SS^*\|=\|S^*\|^2=\|S\|^2=\|S^*S\|=1$$). Any positive element is of the form $$T^*T$$ for some $$T$$. So $$I-SS^*=0$$, and $$SS^*=I$$, making $$S$$ a unitary.
• Thank you. Where does this come from $\sum_{s,t} \alpha_s\beta_t\langle \delta_{st},\delta_e\rangle =\sum_s\alpha_s\beta_{s^{-1}}$? Could you please elaborate a little more on the last paragraph? – UnknownW Apr 1 at 12:03
• The only way for the inner product to be 1 (and not zero) is to have $st=e$. – Martin Argerami Apr 1 at 13:51
• Okay, thank you. I have one last question: Is $S$ belongs to $\mathcal{L}(G)$ would it be okay to write $S=\sum_{s\in G}\alpha_sU_s$ or should it be written in another from? – UnknownW Apr 1 at 20:25