Let $A=C^\*\_r(G)$ be the reduced group $C^\*$-algebra of a finitely generated group. Consider the map $M\_2(A)\to M\_2(A)$,
$$\left[\begin{array}{ll}a&b\\c&d \end{array}\right]\mapsto \left[\begin{array}{ll}tr(a)I&b\\c&tr(d) I \end{array}\right].$$
$I$ is the identity in $A$ and $tr(a)$ is the trace of $a$.


Is this map completely positive?



Completely positive map defined by the trace
This map is not even positive, let alone completely positive.


This comes from the following lemma: A matrix $\left[\begin{array}{cc} I & a \\ a^{\ast} & b\end{array}\right]$ is positive in $M\_{2}(A)$ iff $a^{\ast}a \leqslant b$. Therefore the matrix $\left[\begin{array}{cc} I & a \\ a^{\ast} & a^{\ast}a\end{array}\right]$ is positive and its image, $\left[\begin{array}{cc} I & a \\ a^{\ast} & tr(a^{\ast}a) I\end{array}\right]$, is positive iff $a^{\ast}a \leqslant tr(a^{\ast}a) I$. This means exactly that $\|a^{\ast}a\| \leqslant tr(a^{\ast}a)$, which happens very rarely. Indeed, denote $a^{\ast}a$ by $x$, then we get $tr(x) = \|x\|$, so $tr(\|x\|1 - x)=0$. Since $x\leqslant \|x\|1$, faithfulness of the trace yields $x=\|x\|1$, which means $a^{\ast}a=\|a\|^2 1$ (by traciality also $aa^{\ast} = \|a\|^2 1$), so $\frac{a}{\|a\|}$ is unitary. For all other elements you get a counterexample.


