There is a quotation below:

Let $M\subset B(H)$ be a von Neumann algebra and $\{P_{i}\}_{i\in L}$ be a net of finite-rank projections which increases to the identity (in the strong operator topology). If $P_{i}$ has rank $k(i)$, then we define contractive completely positive $$\phi_{i}:M\rightarrow M_{k(i)}(\mathbb{C})\cong P_{i}B(H)P_{i}$$ by compression (i.e., $\phi_{i}(T)=P_{i}TP_{i}$) and we let $$\psi_{i}: M_{k(i)}(\mathbb{C})\rightarrow B(H)$$ be the natural inclusion maps. Since the predual of $B(H)$ is the trace class operators, a routine exercise shows that the maps $\psi_{i}\circ\phi_{i}$ converge to the identity (on all of $B(H)$, in fact) in the point-ultraweak topology.

I do know the how to do the "so call" routine exercise. Could someone give me some hints or show me more details?


Given $T_j\in B(H)$, we have by definition that $T_j\to0$ ultraweakly if $f(T_j)\to0$ for every $f$ in $B(H)_*=T(H)$. The duality here is given by $f(T)=\text{Tr}(Tf)$ for each trace-class operator $f$.

So fix $T\in B(H)$. We want to show that $P_iTP_i\to T$ ultraweakly. If $R\in B(H)$ is a finite rank operator, we can write $R=\sum_{k,j=1}^nr_{kj}E_{kj}$ for appropriate matrix units. Then $$ \text{Tr}(R(I-P_i))=\sum_{s=1}^\infty\langle R(I-P_i)e_s,e_s\rangle=\sum_{s=1}^\infty\sum_{k,j=1}^nr_{kj}\langle(I-P_i)e_s,E_{kj}e_s\rangle =\sum_{k,j=1}^nr_{kj}\langle(I-P_i)e_j,e_k\rangle\to0, $$ using that $I-P_i\to0$ in the strong operator topology (and thus in the weak operator topology), and that the sum is finite. Using that the finite-rank operators are $\|\cdot\|_1$-dense in the trace-class operators, we get that $\text{Tr}(S(I-P_i))\to0$ for all trace-class operators $S$. So $P_i\to I$ ultraweakly.

Now, writing $T-P_iTP_i=P_iT(I-P_i)+(I-P_i)TP_i+(I-P_i)T(I-P_i)$ we have, for trace-class $S$, $$ |\text{Tr}(S(T-P_iTP_i))|\leq|\text{Tr}(SP_iT(I-P_i))|+|\text{Tr}(S(I-P_i)TP_i)|+|\text{Tr}(S(I-P_i)T(I-P_i))| $$ For each term we can reason like this (I'll do the first one):

If we write $S=VS_0$ the polar decomposition, with $S_0\geq0$, then $S_0$ is trace-class. So $$ |\text{Tr}(SP_iT(I-P_i))|=|\text{Tr}(S_0^{1/2}P_iT(I-P_i)VS_0^{1/2})|\\ \leq\text{Tr}(S_0^{1/2}P_iT^*TP_iS_0^{1/2})^{1/2}\,\text{Tr}(S_0^{1/2}V^*(I-P_i)VS_0^{1/2})^{1/2}\\ \leq\|T\|\,\text{Tr}(S_0)^{1/2}\,\text{Tr}((I-P_i)VS_0V^*)^{1/2}\to0, $$ since $VS_0V^*$ is trace-class.

After working like this for the three terms above, we get $$ \lim_i\text{Tr}(S(T-P_iTP_i))=0. $$ As this occurs for any trace-class $S$, we have that $P_iTP_i\to T$ ultraweakly.

Depending on the available background, here is a much easier proof. The ultraweak topology agrees with the weak operator topology on bounded sets. As $\|P_iTP_i\|\leq\|T\|$ for all $i$, we only need to prove that $P_iTP_i\to T$ weakly. So, for $h,k\in H$, $$ |\langle P_iT(I-P_i)h,k\rangle|=|\langle T(I-P_i)h,P_ik\rangle|\leq\|T\|\,\|(I-P_i)h\|\,\|k\|\to0, $$ since $P_ih\to h$. A similar reasoning for $(I-P_i)TP_i$ and $(I-P_i)T(I-P_i)$ shows that $P_iTP_i\to T$ weakly.

| cite | improve this answer | |
  • $\begingroup$ You are welcome! $\endgroup$ – Martin Argerami Mar 24 '14 at 16:29
  • $\begingroup$ I was just reading this in Brown/Ozawa's book. I don't understand why this proof wouldn't hold for a general C*-algebra $C$ in place of $M$. In fact, couldn't we do this for $C^{**}$, then restrict the decomposition to $C$? $\endgroup$ – Merry Oct 20 '18 at 6:11
  • $\begingroup$ @Merry: what would "point-ultra weak" mean in $C $? If you mean embedding $C\subset B (H) $ and saying that the result holds then yes, it holds. $\endgroup$ – Martin Argerami Oct 20 '18 at 13:52
  • $\begingroup$ Indeed, this is what I meant - thank you very much. $\endgroup$ – Merry Oct 22 '18 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.