ordering of self-adjoints

Let $𝒜$ be a $C^{*}$-algebra (http://planetmath.org/CAlgebra). Let ${𝒜}^{+}$ denote the set of positive elements of $𝒜$ and ${𝒜}_{sa}$ denote the set of self-adjoint elements of $𝒜$.

Since ${𝒜}^{+}$ is a proper convex cone (http://planetmath.org/Cone5) (see this entry (http://planetmath.org/PositiveElement3)), we can define a partial order $\le$ on the set ${𝒜}_{sa}$, by setting

$a\le b$ if and only if $b-a\in {𝒜}^{+}$, i.e. $b-a$ is positive.

Theorem - The relation $\le$ is a partial order relation on ${𝒜}_{sa}$. Moreover, $\le$ turns ${𝒜}_{sa}$ into an ordered topological vector space.

0.0.1 Properties:

• $a\le b\Rightarrow c^{*}ac\le c^{*}bc$ for every $c\in 𝒜$.

• If $a$ and $b$ are invertible and $a\le b$, then $b^{-1}\le a^{-1}$.

• If $𝒜$ has an identity element $e$, then $-\parallel a\parallel e\le a\le \parallel a\parallel e$ for every $a\in {𝒜}_{sa}$.

• $-b\le a\le b\Rightarrow \parallel a\parallel \le \parallel b\parallel$.

0.0.2 Remark:

The proof that $\le$ is partial order makes no use of the self-adjointness . In fact, $𝒜$ itself is an ordered topological vector space under the relation $\le$.

However, it turns out that this ordering relation provides its most usefulness when restricted to self-adjoint elements. For example, some of the above would not hold if we did not restrict to ${𝒜}_{sa}$.