*-Algebras

After the previous posts motivating the idea of studying probability spaces by looking at states on algebras, I will now make a start on the theory. The idea is that an abstract algebra can represent the collection of bounded, and complex-valued, random variables, with a state on this algebra taking the place of the probability measure. By allowing the algebra to be noncommutative, we also incorporate quantum probability.

I will take very small first steps in this post, considering only the basic definition of a *-algebra and positive maps. To effectively emulate classical probability theory in this context will involve additional technical requirements. However, that is not the aim here. We take a bare-bones approach, to get a feeling for the underlying constructs, and start with the definition of a *-algebra. I use ${\bar\lambda}$ to denote the complex conjugate of a complex number ${\lambda}$.

Definition 1 An algebra ${\mathcal A}$ over field ${K}$ is a ${K}$-vector space together with a binary product ${(a,b)\mapsto ab}$ satisfying

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle a(bc)=(ab)c,\smallskip\\ &\displaystyle \lambda(ab)=(\lambda a)b=a(\lambda b)\smallskip\\ &\displaystyle a(b+c)=ab+ac,\smallskip\\ &\displaystyle (a+b)c=ac+bc, \end{array}$

for all ${a,b,c\in\mathcal A}$ and ${\lambda\in K}$.

A *-algebra ${\mathcal A}$ is an algebra over ${{\mathbb C}}$ with a unary involution, ${a\mapsto a^*}$ satisfying

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle (\lambda a+\mu b)^*=\bar\lambda a^*+\bar\mu b^*,\smallskip\\ &\displaystyle (ab)^*=b^*a^*\smallskip\\ &\displaystyle a^{**}=a. \end{array}$

for all ${a,b,c\in\mathcal A}$ and ${\lambda,\mu\in{\mathbb C}}$.

An algebra is called unitial if there exists ${1\in\mathcal A}$ such that

$\displaystyle 1a=a1=a$

for all ${a\in\mathcal A}$. Then, ${1}$ is called the unit or identity of ${\mathcal A}$.

In contrast to my previous posts, I am not considering a *-algebra to contain a unit by default, and will refer to it as `unitial’ whenever the existence of a unit is required. An ${a\in\mathcal A}$ is called self-adjoint if and only if ${a^*=a}$. It can be seen that the self-adjoint elements form a real-linear subspace of the algebra, which we denote by ${\mathcal A_{\rm sa}}$. For any ${a\in\mathcal A}$, then ${a^*a}$ and ${a^*+a}$ are both self-adjoint. Furthermore, every ${a\in\mathcal A}$ can be uniquely decomposed as

$\displaystyle a=u+iv$

for ${u,v\in\mathcal A_{\rm sa}}$. Using ${a^*=u-iv}$, this is easily solved to obtain ${u=(a+a^*)/2}$ and ${v=(a-a^*)/2i}$. If the algebra is unitial, then the identity ${1}$ must be self-adjoint, as can be seen from

$\displaystyle 1^*=1^*1=(1^*1)^*=(1^*)^*=1.$

A sub-*-algebra ${\mathcal B}$ of ${\mathcal A}$ is a subset which is closed under the algebra operations. That is ${\lambda a}$, ${a+b}$, ${ab}$ and ${a^*}$ are all in ${\mathcal B}$, for any ${a,b\in\mathcal B}$ and ${\lambda\in{\mathbb C}}$. Any sub-*algebra is itself a *-algebra under these operations. An algebra is said to be commutative if the identity ${ab=ba}$ is satisfied.

Example 1 Let ${X}$ be a set and ${\mathcal A}$ be the collection of functions ${f\colon S\rightarrow{\mathbb C}}$. This is a commutative *-algebra under the pointwise operations

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle (f+g)(x)=f(x)+g(x),\smallskip\\ &\displaystyle (\lambda f)(x)=\lambda f(x),\smallskip\\ &\displaystyle (fg)(x)=f(x)g(x),\smallskip\\ &\displaystyle f^*(x)=\overline{f(x)}, \end{array}$

for ${f,g\in\mathcal A}$ and ${\lambda\in{\mathbb C}}$. The self-adjoint elements are the real-valued functions on ${X}$.

Commutative *-algebras can often be represented as sub-*-algebras of the collection of complex valued functions from some set ${X}$, although this does impose additional requirements. For example, any such algebra also satisfies ${a=0}$ whenever ${a^*a=0}$. In example 1, it can be seen that the positive elements of ${\mathcal A}$ are precisely those that can be expressed in the form ${a^*a}$.

Noncommutative *-algebras arise as collections of linear operators on an inner product space.

Example 2 If ${V}$ is a vector space over a field ${K}$, then the space of linear maps ${V\rightarrow V}$ is a ${K}$-algebra, where the algebra operations are defined by combining the linear maps in the usual way,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle (a+b)(x)=a(x)+b(x),\smallskip\\ &\displaystyle (\lambda a)(x)=\lambda a(x),\smallskip\\ &\displaystyle (ab)(x)=a(b(x)), \end{array}$

for all ${x\in V}$ and ${\lambda\in K}$. If ${V}$ is a complex vector space with inner product ${\langle\cdot,\cdot\rangle}$, let ${\mathcal A}$ be the space of linear maps ${a\colon V\rightarrow V}$ such that there exists an adjoint ${a^*\colon V\rightarrow V}$ satisfying

 $\displaystyle \langle x,ay\rangle=\langle a^*x,y\rangle$ (1)

for all ${x,y\in V}$. Then ${\mathcal A}$ is a *-algebra. In particular, if ${V}$ is a Hilbert space, then the collection ${B(V)}$ of all bounded linear maps ${V\rightarrow V}$ is a *-algebra, with involution given by the operator adjoint.

In this example, if ${a\in\mathcal A}$ satisfies ${a^*a=0}$ then

$\displaystyle \lVert ax\rVert^2=\langle ax,a x\rangle=\langle a^*ax,x\rangle=0$

so that ${a=0}$, as in example 1. Generally, we expect the property that ${a^*a=0}$ implies ${a=0}$ for the cases that we will be interested in, although I will not impose this as a condition. Any element of ${\mathcal A}$ of the form ${a^*a}$ satisfies

$\displaystyle \langle x,a^*a x\rangle=\langle ax,ax\rangle=\lVert ax\rVert^2\ge0,$

so represents a positive linear operator.

Next, we define positive linear maps on a *-agebra.

Definition 2 Let ${\mathcal A}$ be a *-algebra. Then, a linear map ${p\colon\mathcal A\rightarrow{\mathbb C}}$ is,

1. self-adjoint or real if ${p(a^*)=\overline{p(a)}}$ for all ${a\in\mathcal A}$.
2. positive if it is self-adjoint and ${p(a^*a)\ge0}$ for all ${a\in\mathcal A}$.

Example 3 Let ${(X,\mathcal E,\mu)}$ be a finite measure space, and ${\mathcal A}$ be the bounded measurable maps ${X\rightarrow{\mathbb C}}$. Then, integration w.r.t. ${\mu}$ defines a positive linear map on ${\mathcal A}$,

$\displaystyle p(f)=\int f d\mu.$

Example 4 Let ${V}$ be an inner product space, and ${\mathcal A}$ be a sub-*-algebra of the space of linear maps ${a\colon V\rightarrow V}$ as in example 2. Then, any ${\xi\in V}$ defines a positive linear map on ${\mathcal A}$,

$\displaystyle p(a)=\langle\xi,a\xi\rangle.$

Given a *-algebra ${\mathcal A}$ and a positive linear map ${p\colon\mathcal A\rightarrow{\mathbb C}}$, we can define a semi-inner product by,

 $\displaystyle \langle x,y\rangle = p(x^*y),$ (2)

for all ${x,y\in\mathcal A}$. This is only a semi-inner product, as it need not be positive definite. That is, it is possible that ${\langle x,x\rangle=0}$ for some nonzero ${x\in\mathcal A}$. The associated semi-norm is

$\displaystyle \lVert a\rVert_2=\langle a,a\rangle^{\frac12}=\sqrt{p(a^*a)}.$

I will refer to this as the ${L^2(p)}$ semi-norm on ${\mathcal A}$. If you prefer to work with a true inner product, rather than a semi-inner product, it is always possible to quotient out by the space of ${x\in\mathcal A}$ for which ${\lVert a\rVert_2=0}$. As ${\mathcal A}$ acts on itself by left-multiplication, taking ${V=\mathcal A}$ considered as a complex vector space shows that the construction of *-algebras in example 2 is quite general.

A left-ideal of a *-algebra is a subset, which is a subspace as a complex vector space, and which is closed under left-multiplication by elements of the algebra.

Lemma 3 Let ${p\colon\mathcal A\rightarrow{\mathbb C}}$ be a positive linear map on *-algebra ${\mathcal A}$. Let ${\mathcal A_0}$ denote the elements ${x\in\mathcal A}$ such that ${\lVert x\rVert_2=0}$ or, equivalently, ${p(x^*x)=0}$. This is a left-ideal of ${\mathcal A}$.

Using ${\mathcal A/\mathcal A_0}$ to denote the quotient vector space, with quotient map ${x\mapsto[x]=\mathcal A_0+x}$, then the semi-inner product uniquely defines an inner product on ${\mathcal A/\mathcal A_0}$ by

 $\displaystyle \langle[x],[y]\rangle=\langle x,y\rangle$ (3)

and ${\mathcal A}$ acts on ${\mathcal A/\mathcal A_0}$ by left-multiplication, ${a[x]=[ax]}$.

Proof: If ${x,y\in\mathcal A_0}$ and ${\lambda,\mu\in{\mathbb C}}$ then, by the triangle inequality,

$\displaystyle \lVert \lambda x+\mu y\rVert_2\le\lvert\lambda\rvert\lVert x\rVert_2+\lvert\mu\rvert\lVert y\rVert_2=0.$

So ${\lambda x+\mu y\in\mathcal A_0}$, showing that ${\mathcal A_0}$ is a vector subspace. Also, for ${a\in\mathcal A}$, Cauchy–Schwarz gives

$\displaystyle \lVert ax\rVert_2=\langle ax,ax\rangle=\langle a^*ax,x\rangle\le\lVert a^*ax\rVert_2\lVert x\rVert_2=0,$

so that ${ax}$ is in ${\mathcal A_0}$ which, therefore, is a left-ideal. Next, if ${[x]=[x^\prime]}$ then

$\displaystyle \lvert\langle x,y\rangle-\langle x^\prime,y\rangle\rvert=\lvert\langle x-x^\prime,y\rangle\rvert\le\lVert x-x^\prime\rVert_2\lVert y\rVert_2=0$

and, similarly, the value of ${\langle x,y\rangle}$ is unchanged by replacing ${y}$ with ${y^\prime}$ where ${[y]=[y^\prime]}$. So, 3 is well-defined. Furthermore, if ${\langle[x],[x]\rangle=0}$ then ${x\in\mathcal A_0}$, so that ${\mathcal A/\mathcal A_0}$ is a true inner product space. Finally, if ${[x]=0}$ then, as ${\mathcal A_0}$ is a left-ideal, ${[ax]=0}$, so we can define the action of ${\mathcal A}$ on ${\mathcal A/\mathcal A_0}$ by ${a[x]=[ax]}$. ⬜

Now, for ${a\in\mathcal A}$, consider the linear map on ${\mathcal A}$ given by left-multiplication, ${x\mapsto ax}$ (or, you can look at the action on ${\mathcal A/\mathcal A_0}$, if preferred). We denote its operator norm by ${\lVert a\rVert_\infty}$, so that,

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\lVert a\rVert_\infty&\displaystyle=\inf\left\{K\in{\mathbb R}^+\colon \lVert ax\rVert_2\le K\lVert x\rVert_2{\rm\ for\ all\ }x\in\mathcal A\right\}\smallskip\\ &\displaystyle=\sup\left\{\lVert ax\rVert_2\colon x\in\mathcal A, \lVert x\rVert_2\le1\right\}. \end{array}$ (4)

Again, this may only be a semi-norm, as it is possible that ${\lVert a\rVert_\infty=0}$ for nonzero ${a}$, and, alternatively, can be infinite. I will refer to ${\lVert\cdot\rVert_\infty}$ as the ${L^\infty(p)}$ seminorm and, sometimes, will drop the subscript and write simply ${\lVert a\rVert}$, where it is unlikely to cause confusion. I will say that ${a}$ is bounded if it acts as a bounded operator, so that ${\lVert a\rVert}$ is finite. We show that this is a C*-seminorm. Whenever I say that a *-algebra acts on a semi-inner product space, I am requiring that this is in the sense of example 2 so that, in particular, (1) holds.

Lemma 4 Let ${\mathcal A}$ be a *-algebra acting on semi-inner product space ${V}$. Then, the operator norm on ${\mathcal A}$ is an algebra semi-norm,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\lVert\lambda a\rVert = \lvert\lambda\rvert\lVert a\rVert,\smallskip\\ &\displaystyle\lVert a+b\rVert\le\lVert a\rVert+\lVert b\rVert,\smallskip\\ &\displaystyle\lVert ab\rVert\le\lVert a\rVert\lVert b\rVert, \end{array}$

for all ${a,b\in\mathcal A}$ and ${\lambda\in{\mathbb C}}$. Furthermore, the C*-inequality,

 $\displaystyle \lVert a\rVert^2\le\lVert a^*a\rVert$ (5)

holds for all ${a\in\mathcal A}$, and ${\lVert a^*\rVert=\lVert a\rVert}$.

Proof: The algebra seminorm properties are standard for any collection of operators on a normed space. For the C*-inequality, consider ${a\in\mathcal A}$ and ${x\in V}$. Cauchy–Schwarz gives

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\lVert a x\rVert^2&\displaystyle=\langle ax,ax\rangle=\langle a^*ax,x\rangle\smallskip\\ &\displaystyle\le\lVert a^*ax\rVert\lVert x\rVert\le\lVert a^*a\rVert\lVert x\rVert^2, \end{array}$

and the (5) follows. Next, cancelling ${\lVert a^*x\rVert}$ from both sides of the following

$\displaystyle \lVert a^*x\rVert^2=\langle aa^*x,x\rangle\le\lVert a\rVert\lVert a^*x\rVert\lVert x\rVert$

gives ${\lVert a^*\rVert\le\lVert a\rVert}$. Using ${a^*}$ in place of ${a}$ gives the reverse inequality. ⬜

Once it is known that a semi-norm on a *-algebra satisfies the C*-inequality, then we get various identities for free. An element ${a}$ of a *-algebra ${\mathcal A}$ is called normal if it commutes with its adjoint, ${a^*a=aa^*}$. This includes, for example, all self-adjoint elements and all unitary elements which, by definition, satisfy ${a^*a=aa^*=1}$.

Lemma 5 If ${\lVert\cdot\rVert}$ is a finite seminorm on *-algebra ${\mathcal A}$ satisfying the C*-inequality (5) then,

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\lVert a^*\rVert=\lVert a\rVert,\smallskip\\ &\displaystyle\lVert a^*a\rVert=\lVert a\rVert^2, \end{array}$ (6)

for all ${a\in\mathcal A}$. Furthermore, for normal ${a\in\mathcal A}$,

 $\displaystyle \lVert a^n\rVert=\lVert a\rVert^n$ (7)

for all ${n\ge1}$ and, more generally,

 $\displaystyle \lVert(a^*)^ra^s\rVert=\lVert a\rVert^{r+s}$ (8)

for all ${r,s\ge1}$.

Proof: The C*-inequality gives

$\displaystyle \lVert a\rVert^2\le\lVert a^*a\rVert\le\lVert a^*\rVert\lVert a\rVert.$

Cancelling ${\lVert a\rVert}$ gives ${\lVert a\rVert\le\lVert a^*\rVert}$. Replacing ${a}$ by ${a^*}$ gives the reverse inequality, so ${\lVert a\rVert=\lVert a^*\rVert}$. Then,

$\displaystyle \lVert a\rVert^2\le\lVert a^*a\rVert\le\lVert a^*\rVert\lVert a\rVert=\lVert a\rVert^2,$

giving the second equality. We now prove (7), starting with the case where ${a=a^*}$ is self-adjoint and ${n}$ is a power of 2. By what we have just shown,

$\displaystyle \lVert a^{2^{m+1}}\rVert=\lVert (a^{2^m})^2\rVert=\lVert a^{2^m}\rVert^2.$

Hence, by induction in ${m}$,

$\displaystyle \lVert a^{2^m}\rVert=\lVert a\rVert^{2^m},$

for all nonnegative integers ${m}$. This proves (7) when ${n}$ is a power of 2, and we need to extend to the case where it is any positive integer. In that case, choose ${m}$ such that ${2^m\ge n}$. Then,

$\displaystyle \lVert a\rVert^{2^m}=\lVert a^{2^m}\rVert\le\lVert a^n\rVert\lVert a\rVert^{2^m-n}.$

Assuming that ${\lVert a\rVert}$ is nonzero, cancelling ${\lVert a\rVert^{2^m-n}}$,

$\displaystyle \lVert a^n\rVert\le\lVert a\rVert^n\le\lVert a^n\rVert,$

gives the required equality. The case where ${\lVert a\rVert=0}$ is easily handled, since ${\lVert a^n\rVert\le\lVert a\rVert^n=0}$.

Now, if ${a}$ is normal then,

$\displaystyle \lVert a^n\rVert^2=\lVert (a^*)^na^n\rVert=\lVert(a^*a)^n\rVert=\lVert a^*a\rVert^n=\lVert a\rVert^{2n}.$

Finally,

$\displaystyle \lVert (a^*)^r a^s\rVert^2=\lVert (a^*)^sa^r(a^*)^ra^s\rVert=\lVert(a^*a)^{r+s}\rVert=\lVert a\rVert^{2(r+s)}$

as required. ⬜

In particular, applying this to the action of ${\mathcal A}$ on itself:

Corollary 6 If ${p\colon\mathcal A\rightarrow{\mathbb C}}$ is a positive linear map on *-algebra ${\mathcal A}$, then the ${L^\infty(p)}$ semi-norm is an algebra semi-norm satisfying the C*-identities (6), and satisfying (7,8) for bounded normal ${a\in\mathcal A}$.

Proof: Let ${\mathcal B}$ be the subalgebra of bounded elements of ${\mathcal A}$. By lemma 4, ${\lVert a^*\rVert=\lVert a\rVert}$ is finite for all ${a\in\mathcal B}$, so ${\mathcal B}$ is a *-subalgebra. As, again by lemma 4, the C*-inequality (5) holds, lemma 5 shows that the C*-identities (6) hold for bounded ${a}$ and (7,8) hold for bounded normal ${a}$. It only remains to show that (6) holds for unbounded ${a}$, but this is immediate from the C*-inequality (5) . ⬜