34.8 Order Structures of $C^{*}$-Algebras
Definition 34.8.1 (Positive).label Let $A$ be a $C^{*}$-algebra and $x \in A$, then $x$ is positive if there exists $y \in A$ such that $x = y^{*}y$.
Proposition 34.8.2.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then $x$ is positive if and only if $\sigma_{A}(x) \subset [0, \infty)$.
Proof. Using the continuous functional calculus, $x$ is positive if and only if $\Gamma_{A[x]}(x) = \text{Id}$ is positive in $C(\sigma_{A}(x); \complex)$, if and only if $\sigma_{A}(x) = \Gamma_{A[x]}(x)(\Omega(A[x])) \subset [0, \infty)$ by Proposition 33.8.2.$\square$
Proposition 34.8.3.label Let $A$ be a unital $C^{*}$-algebra and $x \in A_{sa}$, then the following are equivalent:
- (1)
$x$ is positive.
- (2)
$\sigma_{A}(x) \subset [0, \infty)$.
- (3)
There exists $\lambda \ge \norm{x}_{A}$ such that $\norm{\lambda - x}_{A} \le \lambda$.
Proof, [Lemma 11.3, Zhu93]. (1) $\Leftrightarrow$ (2): Proposition 34.8.2.
(2) $\Leftrightarrow$ (3): By assumption, $\sigma_{A}(x) \subset \real$, so Theorem 34.4.3 implies that
which is bounded above by $\lambda$ if and only if $\sigma_{A}(x) \subset [0, \infty)$.$\square$
Corollary 34.8.4.label Let $A$ be a unital $C^{*}$-algebra. For each $x, y \in A$, denote $x \ge y$ if $x - y$ is positive, then $(A, \le)$ is an ordered vector space.
Proof. By definition, the ordering is reflexive, antisymmetric, translation-invariant, and invariant under scaling by positive constants. It remains to show that $\le$ is transitive, or equivalently, the sum of two positive elements is positive.
Let $x, y \in A$ be positive, then $x + y$ is self-adjoint. Thus there exists $\lambda \ge \norm{x}_{A}$ and $\mu \ge \norm{y}_{A}$ such that $\norm{\lambda - x}_{A} \le \lambda$ and $\norm{\mu - y}_{A} \le \mu$, so $\norm{(\lambda + \mu) - (x + y)}_{A} \le \lambda + \mu$, and $x + y$ is positive by Proposition 34.8.3.$\square$
Definition 34.8.5 (Positive Square Root).label Let $A$ be a $C^{*}$-algebra and $x \in A$ be positive, then there exists a unique positive element $y \in A$ such that $y^{2} = x$. The element $y$ is the positive square root of $x$, denoted $\sqrt{x}$.
Proof. Since $x$ is positive, $\sigma_{A}(x) \subset [0, \infty)$ by Proposition 34.8.3. Therefore the square root function $f(t) = \sqrt{t}$ is defined and continuous on $\sigma_{A}(x)$. Using the continuous functional calculus, $f(x)$ is a positive element of $A$ such that $f(x)^{2} = x$.
Let $y \in A$ such that $y^{2} = x$, then by the Spectral Mapping Theorem, $y = f(y^{2}) = f(x)$, so the square root is unique.$\square$
Definition 34.8.6 (Absolute Value).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$, then $|x| = \sqrt{x^{*}x}$ is the absolute value of $x$.
Definition 34.8.7 (Positive and Negative Parts).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be self-adjoint, then there exists unique positive elements $x^{+}, x^{-} \in A$ such that
- (1)
$x = x^{+} - x^{-}$.
- (2)
$x^{+}x^{-} = x^{-}x^{+} = 0$.
The pair $(x^{+}, x^{-})$ are the positive and negative parts of $x$.
Proof. Since $x$ is self-adjoint, $\sigma_{A}(x) \subset \real$ by Proposition 34.4.6. Using the continuous functional calculus, existence is given by the functions $f^{+}(\lambda) = \lambda \vee 0$ and $f^{-}(\lambda) = \lambda \wedge 0$ and Proposition 34.8.3.
On the other hand, for each $p \in \real[z]$ with $p(0) = 0$, (2) implies that $p(x) = p(x^{+}) + p(-x^{-})$. By the Stone-Weierstrass Theorem, $f(x) = f(x^{+}) + f(-x^{-})$ for all $f \in C(\real; \real)$ with $f(0) = 0$. In particular, (1) then implies that $f^{+}(x) = f+(x^{+}) + f^{+}(-x^{-}) = f^{+}(x^{+}) = x^{+}$, and likewise $f^{-}(x) = x^{-}$. Therefore the decomposition is given uniquely by the continuous functional calculus.$\square$
Remark 34.8.1.label The condition in the sign decomposition that $x^{+}x^{-} = x^{-}x^{+} = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$.
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