Example 41.2.8 (Arveson).label Let $\mathbb{T}= \bracsn{z \in \complex|\ |z| = 1}$, and $S = \text{span}\bracsn{1, z, \ol z}\subset C(\mathbb{T}; \complex)$, then $S$ is an operator system in $C(\mathbb{T}; \complex)$. Let
then:
- (1)
$\phi$ is a positive map.
- (2)
$\norm{\phi}_{L(C(\mathbb{T} ; \complex); M_2(\complex))}= 2$.
- (3)
$\phi$ admits no extension to a positive map on $C(\mathbb{T}; \complex)$.
As such, $M_{2}(\complex)$ is not injective with respect to operator systems and positive maps.
Proof, [Example 2.2, Po02]. (1): Let $a + bz + c \ol z \in S$, then the following are equivalent:
- (1)
$a + bz + c\ol z \ge 0$.
- (2)
$c = \ol b$ and $a \ge 2\text{Re}(b)$.
- (3)
$c = \ol b$ and $a \ge 2|b|$.
In addition, since a $2 \times 2$ matrix is positive if and only if its diagonal entries and its determinant are non-negative. Thus $\phi$ is positive.
(2): $\norm{\phi(z)}_{M_2(\complex)}= 2$.
(3): By Theorem 41.2.7, any such extension would have norm $1$.$\square$
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