Positive Elements and Ideals in C*-Algebras

Positive Elements

Positive Element

An element $a$ in a unital C*-algebra $\mathcal{A}$ is said to be positive if it is self-adjoint and its spectrum $\sigma (a)$ consists of non-negative real numbers, i.e., $\sigma (a)\subseteq [0,\infty )$. The set of all positive elements in $\text{A}$ is denoted by ${\text{A}}^{+}$.

e.g. Let $X$ be a locally compact Hausdorff space. Then $f\in C_0(X)$ is positive if and only if either $f(x)\ge 0$ for all $x\in X$ (i.e. $\sigma (f)\subset {\mathbb{R}}_{+}$) or $f=\overline{g}g$ for some $g\in C_0(X)$.

Our main goal in this section is to show that the set of positive elements is precisely the set of elements of the form $x^{\ast }x$ for some $x\in \text{A}$.

Proposition

For any element $a$ in a C*-algebra $\text{A}$, $a$ is positive if and only if $a=b^{\ast }b$ for some $b\in \mathcal{A}$.

Proof We prove this in two steps.

1. Step 1: We show that if $-x^{\ast }x$ is a positive element, then $x$ must be zero.
   If $-x^{\ast }x\in A^{+}$, its spectrum $\sigma (-x^{\ast }x)$ is contained in $[0,\infty )$. We also know that for any $x$, the spectrum of $x^{\ast }x$ is contained in $[0,\infty )$. Since $\sigma (-x^{\ast }x)=\{-\lambda \mid \lambda \in \sigma (x^{\ast }x)\}$, the only way for both conditions to hold is if $\sigma (x^{\ast }x)=\{0\}$. Since $x^{\ast }x$ is self-adjoint, its norm is equal to its spectral radius. Therefore, $\Vert x^{\ast }x\Vert =r(x^{\ast }x)=0$, which implies $x^{\ast }x=0$. By the C*-identity, $\Vert x{\Vert }^2=\Vert x^{\ast }x\Vert =0$, so $x=0$.
2. Step 2: Let $b=x^{\ast }x$. As $b$ is self-adjoint, we can decompose it into its positive and negative parts: $b=b^{+}-b^{-}$, where $b^{+},b^{-}\in A^{+}$ and $b^{+}{b}^{-}=0$. Our goal is to show that $b^{-}=0$.
   Consider the element $c=x{b}^{-}$. We compute $-c^{\ast }c$: $-c^{\ast }c=-(x{b}^{-}{)}^{\ast }(x{b}^{-})=-b^{-}{x}^{\ast }x{b}^{-}=-b^{-}b{b}^{-}$Substituting $b=b^{+}-b^{-}$, we get: $-c^{\ast }c=-b^{-}(b^{+}-b^{-})b^{-}=-b^{-}{b}^{+}{b}^{-}+(b^{-}{)}^3$
   Since $b^{+}{b}^{-}=0$, we have $b^{-}{b}^{+}=(b^{+}{b}^{-}{)}^{\ast }=0$, so the first term vanishes. Thus, $-c^{\ast }c=(b^{-}{)}^3$. Since $b^{-}$ is positive, $(b^{-}{)}^3$ is also positive. We now have that $-c^{\ast }c$ is a positive element. By Step 1, this implies $c=0$. If $c=x{b}^{-}=0$, then $c^{\ast }c=(x{b}^{-}{)}^{\ast }(x{b}^{-})=b^{-}{x}^{\ast }x{b}^{-}=b^{-}b{b}^{-}=(b^{-}{)}^3=0$. Since $b^{-}$ is positive, $(b^{-}{)}^3=0$ implies $b^{-}=0$.
   Therefore, $b=b^{+}$, which means $x^{\ast }x$ is a positive element.

$\square$

Unique Positive Square Root

An element $a\in \text{A}$ is positive if and only if there exists a unique positive element $b\in {\text{A}}^{+}$ such that $a=b^2$. This element $b$ is called the positive square root of $a$, denoted by $a^{1/2}$.

Proof

• Existence: If $a\in {\text{A}}^{+}$, then $\sigma (a)\subseteq [0,\infty )$. The square root function $f(t)=\sqrt{t}$ is continuous on $\sigma (a)$. By the continuous functional calculus, we can define $b=f(a)=a^{1/2}$.
  • Since $f$ is a real-valued function, $b$ is self-adjoint.
  • The spectrum of $b$ is $\sigma (b)=f(\sigma (a))=\{\sqrt{\lambda }\mid \lambda \in \sigma (a)\}\subseteq [0,\infty )$, so $b\in A^{+}$.
  • By the properties of functional calculus, $b^2=(f\cdot f)(a)$. Since $(f\cdot f)(t)=(\sqrt{t}{)}^2=t$, we have $b^2=a$.
• Uniqueness: Suppose there is another element $c\in A^{+}$ such that $c^2=a$. Since $c$ is a polynomial in itself ($c^2=a$), it commutes with $a$. The element $b=a^{1/2}$ is a limit of polynomials in $a$, so $c$ also commutes with $b$. Now consider the commutative C*-algebra $B=C^{\ast }(b,c)$ generated by $b$ and $c$. By the Gelfand representation theorem, there is a $\ast$-isomorphism $\hat{\cdot }:B\to C(\Omega )$ for some compact Hausdorff space $\Omega$. We have ${\hat{b}}^2=\hat{a}$ and ${\hat{c}}^2=\hat{a}$. Since $b,c\in A^{+}$, their Gelfand transforms $\hat{b}$ and $\hat{c}$ are non-negative real-valued functions. From ${\hat{b}}^2={\hat{c}}^2$, we can conclude that $\hat{b}=\hat{c}$. Since the Gelfand map is an isomorphism, this implies $b=c$.

$\square$

Absolute Value and Positive/Negative Parts

For any self-adjoint element $a\in \text{A}$, we define:

• The absolute value of $a$ as $|a|:=(a^2{)}^{1/2}$.
• The positive part of $a$ as $a^{+}:=\frac{1}{2}(a+|a|)$.
• The negative part of $a$ as $a^{-}:=\frac{1}{2}(|a|-a)$.

These elements are positive and satisfy $a=a^{+}-a^{-}$ and $a^{+}{a}^{-}=0$.

The Partial Order on Self-Adjoint Elements

Lemma

If $a,b\in {\text{A}}^{+}$, then $a+b\in {\text{A}}^{+}$.

Proof Since $a$ and $b$ are positive, they are self-adjoint, so $a+b$ is self-adjoint. The proof that $\sigma (a+b)\subseteq [0,\infty )$ uses the fact that for $x\in A_{sa}$, $x\in A^{+}$ if and only if $\Vert \lambda \cdot 1-x\Vert \le \lambda$ for all $\lambda \ge \Vert x\Vert$. Let $s=\Vert a\Vert$ and $t=\Vert b\Vert$. Then we have $\Vert s\cdot 1-a\Vert \le s$ and $\Vert t\cdot 1-b\Vert \le t$.
By the triangle inequality: $\Vert (s+t)\cdot 1-(a+b)\Vert =\Vert (s\cdot 1-a)+(t\cdot 1-b)\Vert \le \Vert s\cdot 1-a\Vert +\Vert t\cdot 1-b\Vert \le s+t$Since $s+t\ge \Vert a+b\Vert$, this shows that $a+b$ is positive. $\square$

The Partial Order on ${\text{A}}_{\text{sa}}$

Let ${\text{A}}_{\text{sa}}$ be the set of self-adjoint elements in $\text{A}$. For $a,b\in {\text{A}}_{\text{sa}}$, we write $a\le b$ if $b-a\in {\text{A}}^{+}$. This defines a partial order on ${\text{A}}_{\text{sa}}$.

Ideals and Quotient Algebras

Ideals and Quotients of C*-Algebras

Let $I$ be a closed two-sided ideal in a C*-algebra $\text{A}$. Then:

1. $I$ is a self-adjoint ideal (i.e., a $\ast$-ideal).
2. The quotient algebra $\text{A}/I$ is a C*-algebra with the quotient norm.
   

To prove this, we first need a result about approximate identities.

Approximate Identity

For any $a\in A$ and $ϵ>0$, there exists an element $e\in I\cap A^{+}$ with $\Vert e\Vert \le 1$ such that $\Vert a-ae\Vert <ϵ$ and $\Vert a-ea\Vert <ϵ$. Such a family of elements is called an approximate identity for $A$. If $I$ is an ideal, it contains an approximate identity for itself.

Proof of the Theorem

1. $I$ is a $\ast$-ideal: We need to show that if $a\in I$, then $a^{\ast }\in I$.
   Since $a\in I$ and $I$ is a right ideal, $a^{\ast }a\in I$. Because $I$ is an ideal, it contains an approximate identity $\{e_n\}$. These $e_n$ can be constructed via functional calculus on elements like $a^{\ast }a$, and since $I$ is closed, these elements $e_n$ lie in $I$.
   We know that ${\lim }_{n\to \infty }\Vert a-a{e}_n\Vert =0$. Then, using the property that the involution is an isometry: $\Vert a^{\ast }-e_n{a}^{\ast }\Vert =\Vert (a-a{e}_n{)}^{\ast }\Vert =\Vert a-a{e}_n\Vert$So, ${\lim }_{n\to \infty }\Vert a^{\ast }-e_n{a}^{\ast }\Vert =0$. Each element $e_n{a}^{\ast }$ is in $I$ because $e_n\in I$ and $a^{\ast }\in A$. Since $a^{\ast }$ is the limit of a sequence of elements in $I$ and $I$ is a closed subspace, we must have $a^{\ast }\in I$. Thus, $I$ is a $\ast$-ideal.
2. $A/I$ is a C*-algebra: Since $I$ is a closed ideal in a Banach algebra $A$, the quotient $A/I$ is a Banach algebra with the quotient norm $\Vert x+I\Vert ={\inf }_{b\in I}\Vert x+b\Vert$. Since $I$ is a $\ast$-ideal, $A/I$ is a Banach $\ast$-algebra. We only need to verify the C*-identity: $\Vert a^{\ast }a+I\Vert =\Vert a+I{\Vert }^2\text{for all }a\in A$First, for any $b\in I$: $\Vert a^{\ast }a+I\Vert ={\inf }_{c\in I}\Vert a^{\ast }a+c\Vert \le \Vert (a+b{)}^{\ast }(a+b)\Vert =\Vert a+b{\Vert }^2$Taking the infimum over all $b\in I$ gives $\Vert a^{\ast }a+I\Vert \le \Vert a+I{\Vert }^2$.
   For the reverse inequality, let $\{e_n\}$ be an approximate identity for $I$. Then $\Vert a+I\Vert ={\lim }_{n\to \infty }\Vert a(1-e_n)\Vert$. We compute the square: $\Vert a(1-e_n){\Vert }^2=\Vert (a(1-e_n){)}^{\ast }a(1-e_n)\Vert =\Vert (1-e_n)a^{\ast }a(1-e_n)\Vert$Since $e_n\in I$, the element $c_n=e_n{a}^{\ast }a(1-e_n)+a^{\ast }a{e}_n$ is in $I$. Thus, $(1-e_n)a^{\ast }a(1-e_n)=a^{\ast }a-c_n$.Therefore, for each $n$: $\Vert (1-e_n)a^{\ast }a(1-e_n)\Vert =\Vert a^{\ast }a-c_n\Vert \ge {\inf }_{c\in I}\Vert a^{\ast }a-c\Vert =\Vert a^{\ast }a+I\Vert$So, $\Vert a(1-e_n){\Vert }^2\ge \Vert a^{\ast }a+I\Vert$ for all $n$. Taking the limit as $n\to \infty$, we get: $\Vert a+I{\Vert }^2={\lim }_{n\to \infty }\Vert a(1-e_n){\Vert }^2\ge \Vert a^{\ast }a+I\Vert$Combining the two inequalities, we obtain the C*-identity.

$\square$