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Let $G$ be a countable group with neutral element $e$. Consider the Hilbert space $$ell^2(G):=left x(t)$$ with inner product $left langle x,y right rangle=sum_tin Gx(t)overliney(t)$ for $x,yin ell^2(G)$. For each $tin G$, let $delta_tin ell^2(G)$ be defined by $delta_t(t)=1$ and $delta_t(s)=0$ if $sneq t$. The set $(delta_t)_tin G$ is an orthonormal basis for $ell^2(G)$, and $x(t)=left langle x,delta_t right rangle$ for $xin ell^2(G)$ and $tin G$.

For each $tin G$, consider the operator $U_t$ on $ell^2(G)$ given by $(U_tx)(s)=x(t^-1s)$ for $xin ell^2(G)$ and $sin G$.

Put $mathcalL(G)= U_tmid tin G''$. Consider the state $tau$ on $mathcalL(G)$ defined by $tau(T)=left langle Tdelta_e,delta_e right rangle$ for $Tin mathcalL(G)$.

Problems

1) Show that $tau(ST)=tau(TS)$ for all for all $S,Tin mathcalL(G)$.

3) Show that any isometry in $mathcalL(G)$ must be a unitary

My answers:

1) I am not sure about this part. If $S,Tin mathcalL(G)$, then we can write $S=sum_sin Galpha_sU_s$ and $T=sum_tin Gbeta_tU_t$. Can I write the multiplication
$$
ST=sum_(s,t)in Stimes Tgamma_s,tU_sU_t
$$
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 $U_sU_t=U_st$ and $U_tdelta_s=delta_ts$. Then we have
$$
tau(ST)=sum_(s,t)in Stimes Tgamma_s,ttau(U_st)=sum_(s,t)in Stimes Tgamma_s,tleft langle U_stdelta_e,delta_e right rangle=sum_(s,t)in Stimes Tgamma_s,tleft langle delta_st,delta_e right rangle=sum_(s,t)in Stimes Tgamma_s,t delta_st(e)=sum_(s,t)in Stimes Tgamma_s,t delta_ts(e)=dots=tau(TS)
$$
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 $Tin mathcalL(G)$. 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 $sum_sgamma_s,s^-1$ 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
beginalign
tau(ST)&=sum_s,t alpha_sbeta_tlangle U_stdelta_e,delta_erangle
=sum_s,t alpha_sbeta_tlangle delta_st,delta_erangle
=sum_salpha_sbeta_s^-1\
&=sum_talpha_t^-1beta_t=tau(TS).
endalign

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.