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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?

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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.

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  • $\begingroup$ 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

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