# States on *-Algebras

So far, we have been considering positive linear maps on a *-algebra. Taking things a step further, we want to consider positive maps which are normalized so as to correspond to expectations under a probability measure. That is, we require ${p(1)=1}$, although this is only defined for unitial algebras. I use the definitions and notation of the previous post on *-algebras.

Definition 1 A state on a unitial *-algebra ${\mathcal A}$ is a positive linear map ${p\colon\mathcal A\rightarrow{\mathbb C}}$ satisfying ${p(1)=1}$.

Examples 3 and 4 of the previous post can be extended to give states.

Example 1 Let ${(X,\mathcal E,\mu)}$ be a probability space, and ${\mathcal A}$ be the bounded measurable maps ${X\rightarrow{\mathbb C}}$. Then, integration w.r.t. ${\mu}$ defines a state on ${\mathcal A}$,

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

Example 2 Let ${V}$ be an inner product space, and ${\mathcal A}$ be a *-algebra of the space of linear maps ${a\colon V\rightarrow V}$ as in example 2 of the previous post, and including the identity map ${I}$. Then, any ${\xi\in V}$ with ${\lVert\xi\rVert=1}$ defines a state on ${\mathcal A}$,

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

As I am not considering algebras to be unitial by default, it is desirable to generalise definition 1 to include the non-unitial case. It is always possible to extend a *-algebra ${\mathcal A}$ to a unitial *-algebra, by taking the direct sum ${{\mathbb C}\oplus\mathcal A}$. This consists of pairs ${\lambda\oplus a}$ for ${\lambda\in{\mathbb C}}$ and ${a\in\mathcal A}$, which I will denote more simply as ${\lambda+a}$. The algebra operations are defined as

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

for ${a,b\in\mathcal A}$ and ${\lambda,\mu\in{\mathbb C}}$. It can be seen that this makes ${{\mathbb C}\oplus\mathcal A}$ into a unitial *-algebra, with unit ${1+0}$, and ${\mathcal A}$ can be identified with the sub-*-algebra of elements ${0+a}$ for ${a\in\mathcal A}$.

In order to define a state ${p\colon\mathcal A\rightarrow{\mathbb C}}$, we will require ${p}$ to be a positive linear map. We then ask whether ${p}$ can be extended to a state on the unitial *-algebra ${{\mathbb C}\oplus\mathcal A}$, satisfying definition 1. Any such extension is uniquely determined by the normalisation condition ${p(1+0)=1}$, so that

 $\displaystyle p(\lambda+a)=\lambda+p(a).$ (1)

For this to be a state, we just need to show that it is positive.

Lemma 2 Let ${\mathcal A}$ be a *-algebra and ${p\colon\mathcal A\rightarrow{\mathbb C}}$ be a positive linear map. Defining

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\lVert p\rVert&\displaystyle=\inf\left\{K\in{\mathbb R}^+\colon\lvert p(a)\rvert ^2\le Kp(a^*a){\rm\ for\ all\ }a\in\mathcal A\right\},\smallskip\\ &\displaystyle=\sup\left\{\lvert p(a)\rvert^2\colon a\in\mathcal A, p(a^*a)\le1\right\}, \end{array}$ (2)

then ${p}$ extends to a state on ${{\mathbb C}\oplus\mathcal A}$ if and only if ${\lVert p\rVert\le1}$.

Proof: We extend ${p}$ to ${{\mathbb C}\oplus\mathcal A}$ by (1), and just need to verify when this is positive. For ${\lambda+a\in{\mathbb C}\oplus\mathcal A}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle p\left((\lambda+a)^*(\lambda+a)\right)&\displaystyle=p\left(\bar\lambda\lambda+(\bar\lambda a+\lambda a^*+a^*a)\right)\smallskip\\ &\displaystyle=\lvert\lambda\rvert^2+2\Re[\bar\lambda p(a)]+p(a^*a). \end{array}$

By standard properties of quadratics, this is nonnegative for all ${\lambda}$ iff

$\displaystyle \lvert p(a)\rvert^2\le p(a^*a).$

By the definition of ${\Vert p\rVert}$, this holds for all ${a\in\mathcal A}$ iff ${\lVert p\rVert\le1}$. ⬜

In the case where the algebra is unitial, then the norm ${\lVert p\rVert}$ defined by (2) is given simply by ${p(1)}$.

Lemma 3 If ${p\colon\mathcal A\rightarrow{\mathbb C}}$ is a positive linear map on unitial *-algebra ${\mathcal A}$ then ${\lVert p\rVert=p(1)}$.

Proof: By Cauchy–Schwarz,

$\displaystyle \lvert p(a)\rvert^2=\lvert p(1^*a)\rvert^2\le p(1^*1)p(a^*a)=p(1)p(a^*a),$

with equality when ${a=1}$, giving ${\lVert p\rVert=p(1)}$. ⬜

This suggests the following definition of state on a *-algebra, which is both consistent with the unitial case and ensures that the state can be extended to ${{\mathbb C}\oplus\mathcal A}$.

Definition 4 A state on a *-algebra ${\mathcal A}$ is a positive linear map ${p\colon\mathcal A\rightarrow{\mathbb C}}$ with ${\lVert p\rVert=1}$.

A *-algebra ${\mathcal A}$ together with a state ${p\colon\mathcal A\rightarrow{\mathbb C}}$ reflects the concept of a classical probability space, with ${\mathcal A}$ representing the algebra of random variables and ${p}$ being the expectation. It also incorporates the extension to quantum probability, by allowing the algebra to be noncommutative. However, it is much too simplistic to serve as a definition of a probability space in keeping with the classical, commutative, theory. In the Kolmogorov axiomatisation, random variables are given by measurable functions, which are closed under various operations, such as taking limits of increasing sequences of variables. We do not simply allow any algebra of functions, such as the polynomials ${{\mathbb C}[X]}$, to represent the space of all random variables. To simplify the statements in this post, however, I adopt some terminology for a pair ${(\mathcal A,p)}$ consisting of a *-algebra ${\mathcal A}$ and a state ${p\colon\mathcal A\rightarrow{\mathbb C}}$. I use NC’ as an abbreviation for noncommutative’, although the algebras in question are not actually required to be noncommutative. We are simply incorporating the noncommutative case in addition to the commutative algebras of classical probability theory.

Definition 5 A *-probability space (or NC preprobability space)[1] is a pair ${(\mathcal A,p)}$, where ${\mathcal A}$ is a *-algebra and ${p\colon\mathcal A\rightarrow{\mathbb C}}$ is a state.

If ${\mathcal A}$ is a unitial *-algebra and ${p\colon\mathcal A\rightarrow{\mathbb C}}$ is a (nontrivial) positive linear map, then ${\lVert p\rVert=p(1)}$ will be a positive real number. Multiplying by ${\lVert p\rVert^{-1}}$ will give a state. This shows that positive linear maps defined on a unitial *-algebra can be uniquely expressed as a scalar multiple of a state. The situation for non-unitial *-algebras is a bit different. Similar to infinite classical measures, it is possible for ${\lVert p\rVert}$ to be infinite, in which case ${p}$ cannot be scaled to give a state.

Example 3 Let ${\mathcal A}$ be the commutative non-unitial *-algebra of complex polynomials ${f\in{\mathbb C}[X]}$ with zero constant term, ${f(0)=0}$.

Define ${p\colon\mathcal A\rightarrow{\mathbb C}}$ so that ${p(f)=f^\prime(0)}$ is the coefficient of ${X}$ in ${f}$. Then, ${p(f^*f)=0}$ for all ${f\in\mathcal A}$, so that ${p}$ is positive and ${\lVert p\rVert=\infty}$. Hence, ${p}$ is not a multiple of a state on ${\mathcal A}$.

The expectation, ${L^2}$ semi-norm, and ${L^\infty}$ semi-norms, as defined in the previous post, are ordered in the same way as for classical probability spaces.

Lemma 6 If ${(\mathcal A,p)}$ is a *-probability space and ${a\in\mathcal A}$ then,

$\displaystyle \lvert p(a)\rvert\le\lVert a\rVert_2\le\lVert a\rVert_\infty.$

Proof: First, by the condition that ${\lVert p\rVert=1}$,

$\displaystyle \lvert p(a)\rvert^2\le p(a^*a)=\lVert a\rVert_2^2$

giving the left hand inequality. Applying the same inequality with ${a^*a}$ in place of ${a}$,

$\displaystyle p(a^*a)\le \lVert a^*a\rVert_2\le\lVert a^*\rVert\lVert a\rVert_2=\lVert a\rVert\lVert a\rVert_2.$

Cancelling ${\lVert a\rVert_2}$ from both sides gives the right hand inequality. ⬜

In classical probability spaces, it is common to define the ${L^n}$-norm of a random variable ${X}$ by ${\lVert X\rVert_n={\mathbb E}[\lvert X\rvert^n]^{\frac1n}}$, where ${n}$ can be any real number in the range ${[1,\infty)}$. A useful and well-known feature of the ${L^n}$ norms is that they are increasing in ${n}$. Looking, now, at *-probability spaces, the random variables are replaced by elements of an algebra ${a}$. We would like to look at the equivalent norms in this setting, so would be something like ${p(\lvert a\rvert^n)^{\frac1n}}$ with the notation of this post. In a general *-algebra, the absolute value ${\lvert a\rvert}$ is not defined. However, for integer ${n}$, then ${\lvert a\rvert^{2n}}$ can be written as ${(a^*a)^n}$, which is a well-defined element of the algebra.

Lemma 7 Let ${(\mathcal A,p)}$ be a *-probability space. Then, for any ${a\in\mathcal A}$, the sequence ${p((a^*a)^n)^{\frac1{2n}}}$ is increasing in positive integer ${n}$.

In fact, as was previously noted, for a self-adjoint element ${a}$ of a *-algebra ${\mathcal A}$, and state ${p\colon\mathcal A\rightarrow{\mathbb C}}$, there exists a probability measure ${\mu}$ on the real line ${{\mathbb R}}$ such that

$\displaystyle p(f(a))=\int fd\mu$

for all polynomials ${f\in{\mathbb C}[X]}$. In particular, for any ${a\in\mathcal A}$, then ${a^*a}$ is self-adoint, so we can let ${\mu}$ be its distribution, which will be supported on ${[0,\infty)}$. Then,

$\displaystyle p((a^*a)^n)^{\frac1{2n}}=\left(\int_0^\infty\lvert x\rvert^{2n} d\mu(x)\right)^{\frac1{2n}}.$

Hence, lemma 7 can be implied from the classical case. It is good, however, to have a direct proof using *-algebras. In fact, the proof of lemma 7 is a consequence of lemma 8 below.

A sequence ${s_n}$ of nonnegative reals, over nonnegative integer ${n}$, is log-convex if it satisfies

$\displaystyle s_n\le s_i^\alpha s_j^\beta$

for all ${i < n < j}$ and reals ${\alpha+\beta=1}$ with ${\alpha i+\beta j=n}$. Equivalently, ${\alpha=(j-n)/(j-i)}$ and ${\beta=(n-i)/(j-i)}$. In particular, choosing ${i=n-1}$ and ${j=n+1}$,

 $\displaystyle s_n\le\sqrt{s_{n-1}s_{n+1}}.$ (3)

It can be seen that (3) over the range ${n\ge1}$ is equivalent to log-convexity.

Lemma 8 Let ${(\mathcal A,p)}$ be a *-probability space. For any ${a\in\mathcal A}$, define the sequence

$\displaystyle s_n=p((a^*a)^n)$

for positive integer ${n}$, and ${s_0=1}$. Then, ${s_n}$ is log-convex in ${n\ge0}$.

Proof: By passing to ${{\mathbb C}\oplus\mathcal A}$ if necessary, we may suppose that ${\mathcal A}$ is unitial. Then define a sequence ${a_n\in\mathcal A}$ by ${a_0=1}$ and

$\displaystyle a_{n+1}=\begin{cases} aa_n,&{\rm if\ }n{\rm\ is\ even},\\ a^*a_n,&{\rm if\ }n{\rm\ is\ odd}. \end{cases}$

Using ${a_{n+2}=a^*aa_n}$ for even ${n}$, we see that ${a_{n}=(a^*a)^{n/2}}$, so is self-adjoint. Hence, if m and n are even then ${a_m^*a_n=(a^*a)^{(m+n)/2}}$. Similarly, if m and n are both odd then,

$\displaystyle a_m^*a_n=a_{m-1}^*a^*a_n=a_{m-1}^*a_{n+1}=(a^*a)^{(m+n)/2}.$

Hence, by Cauchy–Schwarz, for all ${n\ge1}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle s_n &\displaystyle=p(a_{2n})=p(a_{n-1}^*a_{n+1})\smallskip\\ &\displaystyle\le\sqrt{p(a_{n-1}^*a_{n-1})p(a_{n+1}^*a_{n+1})}\smallskip\\ &\displaystyle=\sqrt{s_{n-1}s_{n+1}}. \end{array}$

This is equivalent to log-convexity of ${s_n}$, and implies that ${s_n^{1/2n}}$ is increasing in ${n\ge1}$ (exercise!). ⬜

As lemma 7 shows that ${p((a^*a)^n)^{\frac1{2n}}}$ is increasing in ${n}$, the limit as ${n}$ goes to infinity will converge to a nonnegative limit (although it could be infinite). This gives an alternative approach to defining an ${L^\infty}$ norm on the algebra. In fact, in many cases, it will give the same result as the ${L^\infty(p)}$ semi-norm defined above. Interestingly, Terry Tao uses exactly this definition of the norm in his posts on NC probability, although he does concentrate on tracial states.

A state ${p\colon\mathcal A\rightarrow{\mathbb C}}$ is called tracial if it satisfies ${p(ab)=p(ba)}$ for all ${a,b\in\mathcal A}$. In particular, if the algebra is commutative, then every state is tracial. Note, however, that the property of being tracial is much weaker than commutativity of the underlying algebra. For example, if ${p}$ is tracial then,

$\displaystyle p(abc)=p(bca)=p(cab)$

for all ${a,b,c\in\mathcal A}$. That is, we can take cyclic permutations without affecting the state. This does not apply for non-cyclic permutations so, ${p(abc)}$ will generally differ from ${p(acb)}$. Straightforward examples of tracial states are given by (multiples of) the trace of an nxn complex matrix.

Lemma 9 Let ${(\mathcal A,p)}$ be a *-probability space and ${a\in\mathcal A}$. Then,

$\displaystyle \lVert a\rVert_\infty\ge\lim_{n\rightarrow\infty}p((a^*a)^n)^{\frac1{2n}}.$

Furthermore, if ${\mathcal A}$ is commutative or, more generally, if ${p}$ is tracial, then we have equality,

$\displaystyle \lVert a\rVert_\infty=\lim_{n\rightarrow\infty}p((a^*a)^n)^{\frac1{2n}}.$

Proof: For brevity, I will use ${a_2}$ to denote ${a^*a}$ in the proof. Using corollary 6 of the previous post, we have

$\displaystyle \lVert a\rVert=\lVert a_2\rVert^{\frac12}=\lVert a_2^n\rVert^{\frac1{2n}}\ge p(a_2^n)^{\frac1{2n}},$

giving the first inequality. We now assume that the state is tracial, and show that equality is obtained. Choosing any ${x\in\mathcal A}$ with ${\lVert x\rVert_2=1}$, we show that

 $\displaystyle \lVert ax\rVert_2\le p(x^*a_2^{2^n}x)^{2^{-n-1}}$ (4)

for all nonnegative integer ${n}$. This follows from induction. The case with ${n=0}$ is immediate, from the definition of the ${L^2}$ norm. Next, suppose that (4) holds for some ${n}$. Using Cauchy–Schwarz,

$\displaystyle p(x^*a_2^{2^n}x)\le p(x^*a_2^{2^n}a_2^{2^n}x)^{\frac12}=p(x^*a_2^{2^{n+1}}x)^{\frac12}$

we see that (4) also holds for ${n+1}$. So, by induction, (4) holds for all nonnegative integer ${n}$ as claimed. Now, using the fact that ${p}$ is tracial, and applying Cauchy–Schwarz,

$\displaystyle p(x^*a_2^{2^n}x)=p(a_2^{2^n}xx^*)\le p(a_2^{2^{n+1}})^{\frac12}\lVert xx^*\rVert_2.$

Substituting this into (4) gives

$\displaystyle \lVert ax\rVert_2\le p(a_2^{2^{n+1}})^{2^{-n-2}}\lVert xx^*\rVert_2^{2^{-n}}.$

Now let ${n}$ go to infinity,

$\displaystyle \lVert ax\rVert_2\le\lim_{n\rightarrow\infty}p(a_2^{2^n})^{2^{-n-1}}=\lim_{n\rightarrow\infty}p(a_2^n)^{\frac1{2n}}.$

The previous lemma shows that for tracial states, at least, the ${L^\infty(p)}$ semi-norm coincides with ${\lim_np((a^*a)^n)^{\frac1{2n}}}$. For non-tracial states, however, this can fail, and can fail quite dramatically.

Example 4 A *-probability space ${(\mathcal A,p)}$ and ${a\in\mathcal A}$ such that, for all ${n}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle p((a^*a)^n)=0,\smallskip\\ &\displaystyle\lVert a\rVert_\infty=\infty. \end{array}$

Let ${V}$ be the vector space of measurable and square integrable functions ${\psi\colon{\mathbb R}\rightarrow{\mathbb C}}$ with compact support (up to almost-everywhere equivalence), and use the inner product

$\displaystyle \langle \phi,\psi\rangle=\int\overline{\phi(x)}\psi(x)dx.$

Let ${\xi=1_{[0,1]}\in V}$, so that ${\lVert\xi\rVert=1}$. Fixing a measurable and locally bounded ${f\colon{\mathbb R}\rightarrow{\mathbb R}}$, we define the multiplication by ${f}$‘ operator ${M_f}$, and for all ${t\in{\mathbb R}}$, define translation by ${t}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle (M_f\psi)(x)=f(x)\psi(x),\smallskip\\ &\displaystyle (U_t\psi)(x)=\psi(x-t). \end{array}$

It can be seen that these have adjoints ${M_f^*=M_f}$ and ${U_t^*=U_{-t}}$. So, ${M_f}$ and ${\{U_t\colon t\in{\mathbb R}\}}$ generate a (unitial) *-algebra ${\mathcal A}$ of linear operators on ${V}$. Define the state

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

We now choose ${f}$ to be identically zero on the interval ${[0,1]}$, so that ${p((M_f^*M_f)^n)=0}$ for all ${n}$. If, in addition, ${f}$ is not essentially bounded on ${{\mathbb R}}$, then ${\lVert M_f\rVert=\lVert f\rVert_\infty=\infty}$. We can take ${f(x)=\max(x-1,0)}$, and ${a=M_f}$ gives the required example.

It could be argued that the previous example involves unbounded elements of the algebra so, maybe, equivalence of the two norms may hold if the elements are bounded. I will say that a *-probability space ${(\mathcal A,p)}$ is bounded if every ${a\in\mathcal A}$ is ${L^\infty(p)}$ bounded. Even in this case, the equivalence of the two norms can fail in as bad a way as possible.

Example 5 A bounded *-probability space ${(\mathcal A,p)}$ and ${a\in\mathcal A}$ such that, for all ${n}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle p((a^*a)^n)=0,\smallskip\\ &\displaystyle\lVert a\rVert_\infty=1. \end{array}$

This example follows in the same way as in the previous one except that, here, we choose ${f}$ with essential supremum ${\lVert f\rVert_\infty=1}$. For example, ${f=1_{[1,\infty)}}$. Then, ${M_f}$ and ${\{U_t\colon t\in{\mathbb R}\}}$ are bounded operators, so ${\mathcal A}$ is bounded. As before, ${p((M_f^*M_f)^n)=0}$ and, now, ${\lVert M_f\rVert_\infty=1}$.

Although I am not requiring algebras to be unitial, *-probability spaces do always contain a sequence approximating a unit in the ${L^2}$ sense.

Lemma 10 Let ${(\mathcal A,p)}$ be a *-probability space. Then, there exists a sequence ${u_n\in\mathcal A}$ such that ${\lVert u_n\rVert_2=1}$ and ${p(u_n)\rightarrow1}$.

For any such sequence,

1. ${u_n}$ is ${L^2}$-Cauchy
2. ${p(au_n)\rightarrow p(a)}$ for all ${a\in\mathcal A}$.
3. ${p(u_n^*au_n)\rightarrow p(a)}$ for all bounded ${a\in\mathcal A}$.
4. ${au_n\rightarrow a}$ in ${L^2(p)}$ for all bounded ${a\in\mathcal A}$.
5. if ${\mathcal A}$ is unitial then ${u_n\rightarrow1}$, wrt ${L^2(p)}$.
6. ${u_n\rightarrow1}$ in ${{\mathbb C}\oplus\mathcal A}$, wrt ${L^2(p)}$.

Proof: By definition (2) of the norm ${\lVert p\rVert}$, there exists a sequence ${u_n\in\mathcal A}$ with ${\lVert u_n\rVert_2\le1}$ and ${\lvert p(u_n)\rvert^2\rightarrow\lVert p\rVert=1}$. By scaling, we can choose ${\lVert u_n\rVert_2=1}$. Then, we can multiply ${u_n}$ by ${\omega\in{\mathbb C}}$ with ${\lvert\omega\rvert=1}$ and ${\omega p(u_n) > 0}$, to ensure that ${p(u_n) > 0}$. So, ${p(u_n)\rightarrow1}$ as required.

Extending the state to ${{\mathbb C}\oplus\mathcal A}$, as given by lemma 2,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\lVert u_n-1\rVert_2^2&\displaystyle=p(u_n^*u_n+1-u_n-u_n^*)\smallskip\\ &\displaystyle=2-p(u_n)-\overline{p(u_n)}\rightarrow0. \end{array}$

This shows several things: that ${u_n\rightarrow1+0}$ in ${{\mathbb C}\oplus\mathcal A}$ wrt ${L^2(p)}$ and, hence, ${u_n}$ is ${L^2}$-Cauchy. If ${\mathcal A}$ is itself unitial, then it shows that ${u_n\rightarrow1}$ in ${\mathcal A}$. Now, for any ${a\in\mathcal A}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\lvert p(au_n-a)\rvert&\displaystyle=\lvert p(a(u_n-1))\rvert\smallskip\\ &\displaystyle\le\lVert a^*\rVert_2\lVert u_n-1\rVert_2\rightarrow0. \end{array}$

giving the second point.

Next, suppose that ${a\in\mathcal A}$ is bounded. Then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lvert p(u_n^*au_n)-p(u_n^*au_m)\rvert &\displaystyle= \lvert p(u_n^*a(u_n-u_m))\rvert\smallskip\\ &\displaystyle\le\lVert u_n\rVert_2\lVert a\rVert_\infty\lVert u_n-u_m\rVert_2\smallskip\\ &\displaystyle=\lVert a\rVert_\infty\lVert u_n-u_m\rVert_2 \end{array}$

Letting ${m}$ go to infinity,

$\displaystyle \lvert p(u_n^*au_n)-p(u_n^*a)\rvert\le\lVert a\rVert_\infty\lVert u_n-1\rVert_2$

Then, let ${n}$ go to infinity to obtain,

$\displaystyle \lvert \lim_np(u_n^*au_n)-p(a)\rvert=0,$

giving the third point. Finally, we compute

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert au_n-a\rVert_2^2 &\displaystyle=p(u_n^*a^*au_n)-p(u_n^*a^*a)-p(a^*au_n)+p(a^*a)\smallskip\\ &\displaystyle\rightarrow p(a^*a)-p(a^*a)-p(a^*a)+p(a^*a)=0, \end{array}$

so that ${au_n\rightarrow a}$ in ${L^2(p)}$. ⬜

Suppose that ${\mathcal A}$ is contained in a larger *-algebra ${\mathcal A^\prime}$, and that a state ${p}$ extends to state ${p^\prime}$ on ${\mathcal A^\prime}$. We consider how the ${L^2}$ and ${L^\infty}$ semi-norms of these two NC probability spaces compare. I use prime a superscript (${{}^\prime}$) to denote constructions in the larger *-algebra. It is straightforward to show that the ${L^2}$ semi-norm and inner product agree. For ${x,y\in\mathcal A}$,

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

so that ${\lVert x\rVert_2=\lVert x\rVert_2^\prime}$. On the other hand, the fact that the ${L^\infty}$ seminorm involves taking a supremum (equation 3 of the previous post) over the elements of the algebra, means that it can increase when we pass to the larger algebra.

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert a\rVert_\infty &\displaystyle=\sup\left\{\lVert ax\rVert_2\colon x\in\mathcal A,\ \lVert x\rVert_2\le1\right\}\smallskip\\ &\displaystyle \le\sup\left\{\lVert ax\rVert_2^\prime\colon x\in\mathcal A^\prime,\ \lVert x\rVert_2^\prime\le1\right\}\smallskip\\ &\displaystyle =\lVert a\rVert_\infty^\prime. \end{array}$

The extreme case is given by example 4, where ${\lVert a\rVert_\infty=0}$ in the sub-*-algebra generated by ${a}$, but ${\lVert a\rVert^\prime_\infty=\infty}$ in the larger *-algebra. When dealing with non-unitial *-algebras, a useful trick is to embed it in the larger, unitial, *-algebra ${{\mathbb C}\oplus\mathcal A}$. It is important, then, to know that the ${L^\infty}$ norm remains unchanged.

Lemma 11 Let ${(\mathcal A,p)}$ be a *-probability space. Then, for any ${a\in\mathcal A}$, its ${L^\infty(p)}$ semi-norm is the same computed in ${{\mathbb C}\oplus\mathcal A}$ as in ${\mathcal A}$.

Proof: Let us denote the ${L^\infty}$ semi-norm of ${a}$ computed in ${{\mathbb C}\oplus\mathcal A}$ by ${\theta(a)}$. As ${\mathcal A}$ is a subalgebra of ${{\mathbb C}\oplus\mathcal A}$, we have ${\lVert a\rVert_\infty\le\theta(a)}$ for all ${a\in\mathcal A}$, so only the reverse inequality needs to be shown. As it is immediate when ${\lVert a\rVert_\infty}$ is infinite, we assume that ${a\in\mathcal A}$ is bounded. Then, for ${\lambda+x\in{\mathbb C}\oplus\mathcal A}$, let ${u_n\in\mathcal A}$ be as in lemma 10.

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert a(\lambda+x)\rVert_2 &\displaystyle =\lim_n\lVert a(\lambda u_n+x)\rVert_2\smallskip\\ &\displaystyle \le\lVert a\rVert_\infty\lim_n\lVert\lambda u_n+x\rVert_2\smallskip\\ &\displaystyle =\lVert a\rVert_\infty\lVert\lambda+x\rVert_2, \end{array}$

so that ${\theta(a)\le\lVert a\rVert_\infty}$ as required. ⬜

In general, if ${a}$ is an element of a *-probability space with zero operator norm, ${\lVert a\rVert_\infty=0}$, then we can effectively treat ${a}$ as if it is zero. The collection of all such elements can be characterised in various ways.

Lemma 12 Let ${(\mathcal A,p)}$ be a *-probability space. For ${a\in\mathcal A}$, the following are equivalent.

1. ${\lVert a\rVert_\infty=0}$.
2. ${p(xay)=0}$ for all ${x,y\in\mathcal A}$.
3. ${p(x^*ax)=0}$ for all ${x\in\mathcal A}$.

If ${\mathcal A}$ is commutative or, more generally, ${p}$ is tracial, then these are also equivalent to the following,

1. ${\lVert a\rVert_2=0}$.
2. ${p(ax)=0}$ for all ${x\in\mathcal A}$.

Proof: 1 ⇒ 2: The second statement follows from the inequality,

$\displaystyle \lvert p(xay)\rvert=\langle x^*,ay\rangle\le\lVert x^*\rVert_2\lVert a\rVert_\infty\lVert y\rVert_2=0.$

2 ⇒ 3: The third statement is just a special case of the second.

3 ⇒ 2: Writing ${F(x)}$ for ${\langle x^*,ax\rangle}$, the polarization identity gives,

$\displaystyle p(xay)=\left(F(x^*+y)-F(x^*-y)-iF(x^*+iy)+iF(x^*-iy)\right)/4.$

As the third statement says that ${F=0}$, the second statement follows.

2 ⇒ 1: The first statement follows from

$\displaystyle \lVert ax\rVert_2^2=p((x^*a^*)ax)=0.$

We have shown that the first three statements are equivalent. Now, consider the case where ${p}$ is tracial.

4 ⇒ 5: If ${p}$ is tracial, the fifth statement follows from,

$\displaystyle \lvert p(ax)\rvert=\lvert p(xa)\rvert\le\lVert x^*\rVert_2\lVert a\rVert_2.$

5 ⇒ 4: If ${p}$ is tracial, the fourth statement follows from,

$\displaystyle \lVert a\rVert_2^2=p(a^*a)=p(ax)$

with ${x=a^*}$.

5 ⇒ 2: If ${p}$ is tracial then the second statement follows from,

$\displaystyle p(xay)=p(a(yx))=0.$

Note that, so far, we have made no use of the fact that ${p}$ is a state. In the case where ${\mathcal A}$ is unitial, then this is not required at all and, if ${\mathcal A}$ is non-unitial, it is only needed to show that the last two statements follow from the first three.

2 ⇒ 5: If ${p}$ is a state then the fifth statement follows from,

$\displaystyle \lvert p(ax)\rvert^2\le p((x^*a^*)ax)=0.$

It would be useful to have a term to describe a state on a *-algebra for which ${a=0}$ whenever ${\lVert a\rVert_\infty=0}$. The word faithful would seem to fit, but this is already taken to describe the property that ${a=0}$ whenever ${\lVert a\rVert_2=0}$. Except in certain cases, such as for tracial states, this is a much stronger condition than we want. So, instead, I will say that a state ${p\colon\mathcal A\rightarrow{\mathbb C}}$ for which ${a=0}$ whenever ${\lVert a\rVert_\infty=0}$ is nondegenerate or separating, and similarly say that the *-probability space ${(\mathcal A,p)}$ is nondegenerate or separated. This captures the idea that it is equivalent to the ${L^\infty(p)}$ topology being separated (i.e., it is T0, T1, or Hausdorff, which are equivalent conditions for a topological vector space).

It is always possible to turn a *-probability space into a nondegenerate one, by quotienting out by the *-ideal of elements satisfying ${\lVert a\rVert_\infty=0}$. A subset ${\mathcal I}$ of a *-algebra ${\mathcal A}$ is called a *-ideal if it is a subspace as a complex vector space, and is closed under involution and by multiplication by elements of the algebra. Then, the quotient ${\mathcal A/\mathcal I}$ is the collection of equivalence classes ${[a]=\mathcal I+a}$, for ${a\in\mathcal A}$. It is made into a *-algebra by,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle [a]+[b]=[a+b],\smallskip\\ &\displaystyle \lambda[a]=[\lambda a],\smallskip\\ &\displaystyle [a][b]=[ab],\smallskip\\ &\displaystyle [a]^*=[a^*], \end{array}$

and we have the canonical *-homomorphism ${a\mapsto[a]}$.

Lemma 13 Let ${(\mathcal A,p)}$ be a *-probability space, and ${\mathcal N}$ denote the set of ${a\in\mathcal A}$ with ${\lVert a\rVert_\infty=0}$. Then, ${\mathcal N}$ is a *-ideal, so that the quotient ${\mathcal A/\mathcal N}$ is a *-algebra, and ${p(a)=0}$ for all ${a\in\mathcal N}$.

Furthermore, ${p}$ factors uniquely through a nondegenerate state on ${\mathcal A/\mathcal N}$ given by,

$\displaystyle p^\prime([a])=p(a)$

for all ${a\in\mathcal A}$.

Proof: If ${a,b\in\mathcal N}$, ${c\in\mathcal A}$, and ${\lambda\in{\mathbb C}}$ then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle \lVert\lambda a\rVert=\lvert\lambda\rvert\lVert a\rVert=0,\smallskip\\ &\displaystyle \lVert a+b\rVert\le\lVert a\rVert+\lVert b\rVert=0,\smallskip\\ &\displaystyle \lVert a^*\rVert=\lVert a\rVert=0,\smallskip\\ &\displaystyle \lVert ac\rVert\le\lVert a\rVert\lVert c\rVert=0, \end{array}$

showing that ${\lambda a}$, ${a+b}$, ${a^*}$ and ${ac}$ are all in ${\mathcal N}$. Hence, ${\mathcal N}$ is a *-ideal. Next, if ${a\in\mathcal N}$ and ${u_n\in\mathcal A}$ is the sequence given by lemma 10 then,

$\displaystyle p(a)=\lim_{n\rightarrow\infty}p(u_n^*au_n)=0.$

We just need to show that the extension of ${p}$ to ${\mathcal A/\mathcal N}$ is nondegenerate. However, if ${\lVert [a]\rVert_\infty^\prime=0}$ then ${\lVert a\rVert_\infty=0}$, so that ${a\in\mathcal N}$ and ${[a]=0}$ as required. ⬜

1 Note: In the initial version of this post, I exclusively used the term NC preprobability space’. This was updated to use the term `*-probability space’, which is a bit cleaner and more consistent with the terminology of C*-probability and W*-probability spaces to be introduced in a later post, and mirrors the definitions of *-algebras, C*-algebras and W*-algebras.

## 4 thoughts on “States on *-Algebras”

1. Stephen Crowley says:

This is a really interesting post, thank you, I wrote a paper covering definitions of lots of the things described here a while back , http://vixra.org/abs/1203.0011 Best, –Stephen

2. In the quantum group community it is common to use notation that refers to virtual objects. Are you aware of this fashion? It is rather seductive.

1. I am not entirely sure…I think you might be referring to considering a non-commutative algebra as functions on some underlying space (or group, etc), although this is only really correct if the algebra is commutative. Is that along the correct lines?