C*-Algebras

C*-Algebras: Definitions and Examples

$\ast$-Algebra & $\ast$-Homomorphism

A $\ast$-algebra $\mathcal{A}$ is an algebra over $\mathbb{C}$ equipped with an involution $\ast :\text{A}\to \text{A}$ such that $a^{\ast \ast }=a$, $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a+b{)}^{\ast }=a^{\ast }+b^{\ast }$, $(\lambda a{)}^{\ast }=\bar{\lambda }{a}^{\ast }$ for all $a,b\in A$ and $\lambda \in \text{C}$.
A $\ast$-homomorphism between $\ast$-algebras is an algebra homomorphism also preserving the involution. A bijective $\ast$-homomorphism is called a $\ast$-isomorphism.

Banach $\ast$-Algebra & C*-Algebra

A Banach $\ast$-algebra is a Banach algebra $\mathcal{A}$ that is also a $\ast$-algebra, with the involution $\ast$ such that $\Vert a^{\ast }\Vert =\Vert a\Vert$ for all $a\in \mathcal{A}$. If $\mathcal{A}$ is unital, we also require $\Vert 1_{\text{A}}\Vert =1$.
A C*-algebra is a Banach $\ast$-algebra $\mathcal{A}$ that satisfies the C*-identity:
$\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2\text{for all }a\in \mathcal{A}.$

Remark

If a C*-algebra $\mathcal{A}$ has a unit $1$, then $\Vert 1\Vert =1$. This follows from the C*-identity: $\Vert 1^{\ast }1\Vert =\Vert 1{\Vert }^2$. Since $1^{\ast }=1$, this gives $\Vert 1\Vert =\Vert 1{\Vert }^2$, and since $1\ne 0$, we must have $\Vert 1\Vert =1$.

e.g.

1. The complex numbers $\mathbb{C}$ with the involution being conjugation is a C*-algebra.
2. For a locally compact Hausdorff space $X$, the algebra $C_0(X)$ of continuous complex-valued functions vanishing at infinity is a C*-algebra with pointwise operations and the involution $f^{\ast }(x)=\overline{f(x)}$. The norm is the supremum norm.
3. For a Hilbert space \H, the algebra $B(\stackrel{˝}{)}$ of bounded linear operators on $H$ is a unital C*-algebra, where the involution is the operator adjoint.
   

$\ast$-Subalgebra & C*-Subalgebra

Let $\mathcal{A}$ be a $\ast$-algebra. A $\ast$-subalgebra $\mathcal{B}$ of $\mathcal{A}$ is a $\ast$-subalgebra if it is closed under the involution, i.e., $b^{\ast }\in \mathcal{B}$ whenever $b\in \mathcal{B}$.
Similarly, a C*-subalgebra of a C*-algebra $\mathcal{A}$ is a norm-closed $\ast$-subalgebra of $\mathcal{A}$.

Special Elements

Let $\mathcal{A}$ be a $\ast$-algebra. An element $a\in \mathcal{A}$ is called:

• Self-adjoint (or Hermitian) if $a=a^{\ast }$
• Normal if $a^{\ast }a=a{a}^{\ast }$
• A projection if $p=p^{\ast }=p^2$

If $\mathcal{A}$ is unital, an element $u\in \mathcal{A}$ is called:

• Unitary if $u^{\ast }u=u{u}^{\ast }=1_{\text{A}}$
• An isometry if $u^{\ast }u=1_{\text{A}}$
  

Fundamental Properties

Properties of Unital C*-Algebras

Let $\mathcal{A}$ be a unital C*-algebra.

1. If $a\in \mathcal{A}$ is invertible, then $a^{\ast }$ is invertible and $(a^{\ast }{)}^{-1}=(a^{-1}{)}^{\ast }$.
2. For any $a\in \mathcal{A}$, the spectrum of its adjoint is given by $\sigma (a^{\ast })=\{\bar{\lambda }\mid \lambda \in \sigma (a)\}$.
3. Any $a\in \mathcal{A}$ can be uniquely written as $a=\text{Re}(a)+i\text{Im}(a)$, where $\text{Re}(a)=\frac{a+a^{\ast }}{2}$ and $\text{Im}(a)=\frac{a-a^{\ast }}{2i}$ are self-adjoint.
4. If $a$ is self-adjoint ($a=a^{\ast }$), then its spectrum $\sigma (a)$ is a subset of the real numbers $\mathbb{R}$, and $r(a)=\Vert a\Vert$.
5. If $u$ is unitary, then $\Vert u\Vert =1$ and its spectrum $\sigma (u)$ is a subset of the unit circle $\{z\in \mathbb{C}\mid |z|=1\}$.
   

Uniqueness of the C*-Norm

There is at most one norm on a $\ast$-algebra that makes it a C*-algebra.

Proof Suppose $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are two norms on a $\ast$-algebra $\mathcal{A}$ that both satisfy the C*-identity. For any $a\in \mathcal{A}$, the element $a^{\ast }a$ is self-adjoint. For any self-adjoint element $h$, its norm is equal to its spectral radius, $\Vert h\Vert =r(h)$. Since the spectral radius depends only on the algebraic structure, it is independent of the norm. Thus, for any $a\in \mathcal{A}$: $\Vert a{\Vert }_1^2=\Vert a^{\ast }a{\Vert }_1=r(a^{\ast }a)=\Vert a^{\ast }a{\Vert }_2=\Vert a{\Vert }_2^2$This implies $\Vert a{\Vert }_1=\Vert a{\Vert }_2$. $\square$

Unitarization

If a C*-algebra $\mathcal{A}$ does not have a unit, it’s possible to embed it into a larger, unital C*-algebra.

Unitarization

Let $\mathcal{A}$ be a non-unital Banach $\ast$-algebra. The unitarization of $\mathcal{A}$ is the algebra $\tilde{\mathcal{A}}=\mathcal{A}\times \mathbb{C}$, with operations:

• Addition: $(a,\lambda )+(b,\mu )=(a+b,\lambda +\mu )$
• Multiplication: $(a,\lambda )(b,\mu )=(ab+\mu a+\lambda b,\lambda \mu )$
• Involution: $(a,\lambda {)}^{\ast }=(a^{\ast },\bar{\lambda })$

$\tilde{\mathcal{A}}$ is a unital *-algebra with unit $(0,1)$.

The C*-Norm on the Unitization

Let $\mathcal{A}$ be a non-unital C*-algebra. The norm on its unitization $\tilde{\mathcal{A}}$ defined by
$\Vert (a,\lambda ){\Vert }_{\tilde{\mathcal{A}}}=\sup \{\Vert ax+\lambda x\Vert :x\in \mathcal{A},\Vert x\Vert \le 1\}$
makes $\tilde{\mathcal{A}}$ a C*-algebra.

Proof This norm is the operator norm of the left multiplication operator $L_a+\lambda I$ acting on $\mathcal{A}$. It can be shown that this norm is complete and satisfies the C*-identity. The key step in proving the C*-identity is showing $\Vert (a,\lambda ){\Vert }^2\le \Vert (a,\lambda {)}^{\ast }(a,\lambda )\Vert$. $\square$

Homomorphisms

$\ast$-Homomorphisms are Contractive

Any $\ast$-homomorphism $\phi :\mathcal{A}\to \mathcal{B}$ between C*-algebras is a contraction, i.e., $\Vert \phi \Vert \le 1$.

Proof For any self-adjoint element $a=a^{\ast }\in \mathcal{A}$, we have $\sigma (\phi (a))\subseteq \sigma (a)$. This implies $r(\phi (a))\le r(a)$. For a C*-algebra, the norm of a self-adjoint element equals its spectral radius. So, $\Vert \phi (a)\Vert \le \Vert a\Vert$. For a general element $a$, we have: $\Vert \phi (a){\Vert }^2=\Vert \phi (a{)}^{\ast }\phi (a)\Vert =\Vert \phi (a^{\ast }a)\Vert \le \Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$Thus, $\Vert \phi (a)\Vert \le \Vert a\Vert$. $\square$

$\ast$-Isomorphisms are Isometries

Any $\ast$-isomorphism between C*-algebras is an isometry.

Characters on Abelian C*-Algebras

Any character (non-zero multiplicative linear functional) $\tau$ on an abelian C*-algebra $\mathcal{A}$ is a $\ast$-homomorphism.

Proof WLOG, assume $\mathcal{A}$ is unital. Let $h=h^{\ast }$ be a self-adjoint element. One can show that $\tau (h)$ must be a real number. For an arbitrary $a\in \mathcal{A}$, write it as $a=\text{Re}(a)+i\text{Im}(a)$. Then: $\tau (a^{\ast })=\tau (\text{Re}(a)-i\text{Im}(a))=\tau (\text{Re}(a))-i\tau (\text{Im}(a))$Since $\tau (\text{Re}(a))$ and $\tau (\text{Im}(a))$ are real, this is equal to $\overline{\tau (\text{Re}(a))+i\tau (\text{Im}(a))}=\overline{\tau (a)}$. $\square$

Spectral Permanence

Let $\mathcal{A}$ be a unital C*-algebra and let $\mathcal{B}$ be a C*-subalgebra of $\mathcal{A}$ containing the unit of $\mathcal{A}$. Then for any element $b\in \mathcal{B}$, its spectrum in $\mathcal{B}$ is the same as its spectrum in $\mathcal{A}$.
${\sigma }_{\mathcal{B}}(b)={\sigma }_{\mathcal{A}}(b).$

Proof The inclusion ${\sigma }_{\mathcal{A}}(b)\subseteq {\sigma }_{\mathcal{B}}(b)$ is always true. To prove the other direction, one shows that if $b$ is invertible in $\mathcal{A}$, it must also be invertible in $\mathcal{B}$. This is first proven for self-adjoint elements, leveraging the fact that for such elements $h$, ${\sigma }_{\mathcal{B}}(h)\subseteq \mathbb{R}$, which implies that $\mathbb{C}\setminus {\sigma }_{\mathcal{B}}(h)$ is connected. Since the spectra must agree for self-adjoint elements, one can extend the argument to arbitrary elements $b\in \mathcal{B}$ by considering the self-adjoint element $b^{\ast }b$. $\square$