# 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

My answers:

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.

## 1 Answer

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