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 $T\in \mathcal{L}(G)$.


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 1


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.

  • $\begingroup$ 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? $\endgroup$
    – UnknownW
    Apr 1, 2019 at 12:03
  • $\begingroup$ The only way for the inner product to be 1 (and not zero) is to have $st=e$. $\endgroup$ Apr 1, 2019 at 13:51
  • $\begingroup$ 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? $\endgroup$
    – UnknownW
    Apr 1, 2019 at 20:25

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