Are contractive completely positive maps trace decreasing?

Question

Are contractive completely positive maps trace decreasing? More precisely, suppose that $f\colon M\to N$ is a normal cpc map between von Neumann algebras with normalised normal traces. (That is $\tau_M(1_M)=\tau_N(1_N)=1$). Do we have that $\tau_N(f(a))\leqslant \tau_M(a)$ for all $a\in M_+$?

Answer 1

For von Neumann algebras in general it is false, there is two much leeway in the choice of the traces. For instance let $M=N=\mathbb C\oplus\mathbb C$, with $f$ the identity map. Let the traces be $$ \tau_M(x,y)=\frac x4+\frac{3y}4, \ \ \tau_N=\frac{3x}4+\frac y4. $$ Then, for $a=(1,0)$, $$ \tau_M(a)=\tau_M(1,0)=\frac14<\frac34=\tau_N(1,0)=\tau_N(f(a)) $$

And with almost the same idea, we can make it fail between factors. Let $M=N=M_2(\mathbb C)$. Put $$ f\left(\begin{bmatrix}x&y\\ z&w\end{bmatrix}\right)=\frac x4+\frac{3w}4. $$ This is a state, so we can see it as a ucp map $M\to N$. And, for $a=\begin{bmatrix}0&0\\0&1\end{bmatrix}$, $$ \tau_N(f(a))=\frac34>\frac12\,\tau_N(a) $$