# Homomorphisms of *-Probability Spaces

I previously introduced the concept of a *-probability space as a pair ${(\mathcal A,p)}$ consisting of a state ${p}$ on a *-algebra ${\mathcal A}$. As we noted, this concept is rather too simplistic to properly capture a noncommutative generalisation of classical probability spaces, and I will later give conditions for ${(\mathcal A,p)}$ to be considered as a true probability space. For now, I continue the investigation of these preprobability spaces, and will look at homomorphisms in this post.

A *-homomorphism between *-algebras ${\mathcal A}$ and ${\mathcal A^\prime}$ is a map ${\varphi\colon\mathcal A\rightarrow\mathcal A^\prime}$ preserving the algebra operations,

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

for all ${a,b\in\mathcal A}$ and ${\lambda,\mu\in{\mathbb C}}$. The term `*-homomorphism’ is used to distinguish it from the concept of simple algebra homomorphisms which need not preserve the involution (the third identity above). Next, I will say that ${\varphi}$ is a homomorphism of *-probability spaces ${(\mathcal A,p)}$ and ${(\mathcal A^\prime,p^\prime)}$ if it is a *-homomorphism from ${\mathcal A}$ to ${\mathcal A^\prime}$ which preserves the state,

$\displaystyle p^\prime(\varphi(a))=p(a),$

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

Now, recall that for any *-probability space ${(\mathcal A,p)}$, we define a semi-inner product ${\langle x,y\rangle=p(x^*y)}$ on ${\mathcal A}$ and the associated ${L^2(p)}$ seminorm, ${\lVert x\rVert_2=\sqrt{p(x^*x)}}$. Homomorphisms of *-probability spaces are clearly ${L^2}$-isometries,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle\varphi(x),\varphi(y)\rangle&\displaystyle=p^\prime\left(\varphi(x)^*\varphi(y)\right)=p^\prime\left(\varphi(x^*y)\right)\smallskip\\ &\displaystyle=p(x^*y)=\langle x,y\rangle. \end{array}$

For each ${a\in\mathcal A}$, the ${L^\infty(p)}$ seminorm ${\lVert a\rVert_\infty}$ is defined as the operator norm of the left-multiplication map ${x\mapsto ax}$ on ${\mathcal A}$, considered as a vector space with the ${L^2}$ seminorm. Homomorphisms of *-probability spaces do not need to be ${L^\infty}$-isometric.

Lemma 1 If ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ is a homomorphism of *-probability spaces then, for any ${a\in\mathcal A}$,

 $\displaystyle \lVert\varphi(a)\rVert_\infty\ge\lVert a\rVert_\infty.$ (1)

Proof: By definition of the ${L^\infty(p)}$ semi-norm, there exists a sequence ${x_n\in\mathcal A}$ with ${\lVert x_n\rVert_2\le1}$ and ${\lVert ax_n\rVert_2\rightarrow\lVert a\rVert}$. Then, as ${\lVert\varphi(x_n)\rVert_2\le1}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert\varphi(a)\rVert&\displaystyle\ge\lVert\varphi(a)\varphi(x_n)\rVert_2\smallskip\\ &\displaystyle=\lVert ax_n\rVert_2\rightarrow\lVert a\rVert. \end{array}$

As a consequence of lemma 1, we can show that ${L^\infty}$-continuous homomorphisms are isometries. Recall that a linear map ${\varphi\colon V\rightarrow W}$ between seminormed spaces ${V}$ and ${W}$ is norm-continuous (or norm-bounded) iff there exists a ${K\ge0}$ such that ${\lVert\varphi(x)\rVert\le K\lVert x\rVert}$ for all ${x\in V}$.

Lemma 2 A homomorphism of *-probability spaces ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ is ${L^\infty}$-bounded if and only if it is an ${L^\infty}$-isometry. That is,

 $\displaystyle \lVert\varphi(a)\rVert_\infty=\lVert a\rVert_\infty$ (2)

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

Proof: If ${\varphi}$ is bounded then ${\Vert\varphi(a)\rVert\le K\lVert a\rVert}$ for some fixed real ${K}$. If ${\lVert a\rVert}$ is infinite, then (2) follows immediately from (1). So, suppose that ${\lVert a\rVert}$ is finite. From the C*-norm properties (corollary 6 of the post on *-algebras),

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert\varphi(a)\rVert&\displaystyle=\lVert(\varphi(a)^*\varphi(a))^n\rVert^{\frac1{2n}}=\lVert\varphi((a^*a)^n)\rVert^{\frac1{2n}}\smallskip\\ &\displaystyle\le K^{\frac1{2n}}\lVert(a^*a)^n\rVert^{\frac1{2n}}=K^{\frac1{2n}}\lVert a\rVert. \end{array}$

Letting ${n}$ go to infinity gives ${\lVert\varphi(a)\rVert\le\lVert a\rVert}$, so lemma 1 gives the result. ⬜

Most homomorphisms of interest to us will be ${L^\infty}$-continuous and, so, will be isometries. There are many cases where this is guaranteed, without having to impose continuity as an explicit requirement. For example, this is always the case for commutative algebras, and for tracial states. Recall that a state is tracial if it satisfies the identity ${p(ab)=p(ba)}$ and, in particular, all states on a commutative *-algebra are tracial.

Lemma 3 Let ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ be a homomorphism of *-probability spaces. If ${\mathcal A^\prime}$ is commutative or, more generally, if ${p^\prime}$ is tracial, then ${\varphi}$ is an ${L^\infty}$-isometry.

Proof: If ${p^\prime}$ is tracial, then applying lemma 9 of the previous post,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert\varphi(a)\rVert&\displaystyle=\lim_{n\rightarrow\infty}p^\prime(\varphi(a^*a)^n)^{\frac1{2n}}\smallskip\\ &\displaystyle=\lim_{n\rightarrow\infty}p((a^*a)^n)^{\frac1{2n}}\le\lVert a\rVert. \end{array}$

We are also guaranteed continuity in the case where the image of the homomorphism is dense.

Lemma 4 Let ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ be a homomorphism of *-probability spaces such that ${\varphi(\mathcal A)}$ is either ${L^2}$ or ${L^\infty}$ dense in ${\mathcal A^\prime}$. Then, ${\varphi}$ is an ${L^\infty}$-isometry.

Proof: First, by lemma 2 of the post on states, ${L^\infty}$ convergence is stronger than ${L^2}$ convergence so, in either case, we have that ${\varphi(\mathcal A)}$ is ${L^2}$-dense in ${\mathcal A^\prime}$. We need to show that ${\lVert\varphi(a)\rVert\le\lVert a\rVert}$, which is trivial in the case that ${\lVert a\rVert=\infty}$. So, suppose that ${\lVert a\rVert}$ is finite. For any ${x\in\varphi(\mathcal A)}$, writing ${x=\varphi(y)}$ gives

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert\varphi(a)x\rVert_2 &\displaystyle =\lVert\varphi(ay)\rVert_2=\lVert ay\rVert_2\smallskip\\ &\displaystyle \le\lVert a\rVert\lVert y\rVert_2=\lVert a\rVert\lVert x\rVert_2. \end{array}$ (3)

This needs to be extended to all ${x\in\mathcal A^\prime}$. First, by the assumption that ${\varphi(\mathcal A)}$ is ${L^2}$-dense, there exists a sequence ${x_n\in\varphi(\mathcal A)}$ converging in ${L^2}$ to ${x}$. Then, for ${y\in\mathcal A^\prime}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle y,\varphi(a)x\rangle &\displaystyle =\langle\varphi(a)^*y,x\rangle=\lim_n\langle\varphi(a)^*y,x_n\rangle\smallskip\\ &\displaystyle =\lim_n\langle y,\varphi(a)x_n\rangle\le\lim_n\lVert y\rVert_2\lVert\varphi(a)x_n\rVert_2\smallskip\\ &\displaystyle \le\lim_n\lVert y\rVert_2\lVert a\rVert\lVert x_n\rVert_2=\lVert y\rVert_2\lVert a\rVert\lVert x\rVert_2. \end{array}$

So, using ${y=\varphi(a)x}$,

$\displaystyle \lVert\varphi(a)x\rVert_2^2\le\lVert\varphi(a)x\rVert_2\lVert a\rVert\lVert x\rVert_2$

from which we obtain inequality (3), giving ${\lVert\varphi(a)\rVert\le\lVert a\rVert}$ as required. ⬜

Yet another case in which we are guaranteed continuity is where ${\mathcal A}$ is ${L^\infty}$-complete and separated. To say that ${\mathcal A}$ is separated, we mean that every ${a\in\mathcal A}$ with ${\lVert a\rVert_\infty=0}$ is zero or, equivalently, ${p}$ is nondegenerate. By taking the completion, it is always possible to extend ${\mathcal A}$ to a complete and separated *-algebra.

Lemma 5 Suppose that ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ is a homomorphism of *-probability spaces and that ${\mathcal A}$ is complete and separated under the ${L^\infty(p)}$ topology. Then, ${\varphi}$ is an ${L^\infty}$-isometry.

Proof: We first extend ${\varphi}$ to a homomorphism ${\varphi\colon{\mathbb C}\oplus\mathcal A\rightarrow{\mathbb C}\oplus\mathcal A^\prime}$,

$\displaystyle \varphi(\lambda +a)=\lambda+\varphi(a).$

Choosing any ${a\in\mathcal A}$, with finite ${L^\infty}$-norm, set ${K=\lVert a\rVert}$. Then, ${\lVert a^*a\rVert=K^2}$, which is sufficient to show that ${K^2-a^*a}$ has a square root. That is, ${K^2-a^*a=b^2}$ for some self-adjoint ${b\in{\mathbb C}\oplus\mathcal A}$. This fact can be shown using functional calculus. However, I provide a direct proof of the existence of the square root in lemma 6 below. Then,

$\displaystyle K^2-\varphi(a)^*\varphi(a)=\varphi(K^2-a^*a)=\varphi(b)^2.$

Multiplying on the right by ${x\in\mathcal A^\prime}$ and on the left by ${x^*}$,

$\displaystyle K^2x^*x-(\varphi(a)x)^*(\varphi(a)x)=(\varphi(b)x)^*(\varphi(b)x).$

So, applying ${p^\prime}$ to both sides,

$\displaystyle K^2\lVert x\rVert_2^2-\lVert\varphi(a)x\rVert_2^2\ge0,$

and ${\lVert\varphi(a)\rVert\le K=\lVert a\rVert}$ as required. ⬜

A C*-algebra is a *-algebra ${\mathcal A}$ together with a C*-norm ${\lVert\cdot\rVert}$, with respect to which ${\mathcal A}$ is complete. For example, given a bounded *-probability space ${(\mathcal A,p)}$ which is complete and separated under the ${L^\infty}$ norm, then it is a C*-algebra. More generally, if it is not bounded, then the collection of ${a\in\mathcal A}$ for which ${\lVert a\rVert_\infty}$ is finite will form a C*-algebra. The following result can then be used to take square roots, as required in the proof just given for lemma 5.

Lemma 6 Let ${\mathcal A}$ be a C*-algebra, ${a\in\mathcal A}$ be self-adjoint, and ${K\ge\lVert a\rVert}$ be real. Then, ${K+a=b^2}$ for some self-adjoint ${b\in{\mathbb C}\oplus\mathcal A}$.

Proof: By scaling, we can suppose that ${K=1}$, so that ${\lVert a\rVert\le1}$, and show that ${1+a}$ has a square root in ${{\mathbb C}\oplus\mathcal A}$. For real ${x}$, the power series expansion

$\displaystyle 1+\sum_{n=1}^\infty\binom{1/2}{n}x^n=\sqrt{1+x}$

converges absolutely when ${\lvert x\rvert < 1}$. In fact, this converges absolutely for all ${\lvert x\rvert\le1}$. Letting ${x}$ tend to -1, the right hand side goes to 0, whereas all the terms in the summation are negative. Hence,

$\displaystyle \sum_{n=1}^\infty\left\lvert\binom{1/2}{n}\right\rvert\le1.$

So, the power series expansion converges absolutely for all ${\lvert x\rvert\le1}$. Hence, we can set

$\displaystyle b=1+\sum_{n=1}^\infty\binom{1/2}{n}a^n$

which, as ${\lVert a\rVert < 1}$, the sum is absolutely convergent in ${\mathcal A}$. Since ${\mathcal A}$ is complete and separated, this has a unique limit ${b\in{\mathbb C}\oplus\mathcal A}$. Furthermore, as involution is ${L^\infty}$-continuous, any limit of self-adjoint elements must be self-adjoint. In particular, ${b}$ is self-adjoint. Next, by squaring the power series expansion,

$\displaystyle \left(1+\sum_{n=1}^\infty\binom{1/2}{n}x^n\right)^2=1+x,$

the coefficients of powers of ${x}$ on both sides must agree. So, by comparing the coefficients of powers of ${a}$ in the expansion of ${b^2}$, we obtain ${b^2=1+a}$. ⬜

I finish off this post by showing that it is possible for homomorphisms of *-probability spaces to be unbounded with respect to the ${L^\infty}$ norm.

Example 1 A homomorphism ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ of *-probability spaces and ${a\in\mathcal A}$ such that

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

This example builds on example 4 of the previous post. There, we constructed a *-probability space, which I will denote here by ${(\mathcal A^\prime,p^\prime)}$, and a self-adjoint element ${a\in\mathcal A^\prime}$ such that ${\lVert a\rVert^\prime_\infty=\infty}$, but ${p^\prime(a^n)=0}$ for all ${n\in{\mathbb N}}$. For example 1 above, let ${\mathcal A}$ be the unitial subalgebra of ${\mathcal A^\prime}$ generated by ${a}$, ${p}$ be the restriction of ${p^\prime}$ to ${\mathcal A}$, and ${\varphi\colon\mathcal A\rightarrow\mathcal A^\prime}$ be the inclusion.

In practice, we will be concerned primarily with bounded *-probability spaces — so that ${\lVert a\rVert_\infty}$ is finite for every ${a\in\mathcal A}$. Even in this case, it is possible for homomorphisms to be unbounded.

Example 2 A homomorphism ${\varphi\colon(\mathcal A,p)\rightarrow(\mathcal A^\prime,p^\prime)}$ of bounded *-probability spaces and ${a\in\mathcal A}$ such that

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

This, again, builds on an example from the previous post (example 5). There, we constructed a bounded *-probability space, denoted here by ${(\mathcal A^\prime,p^\prime)}$, and a self-adjoint element ${a\in\mathcal A^\prime}$ satisfying ${\lVert a\rVert^\prime_\infty=1}$, but ${p^\prime(a^n)=0}$ for all ${n\in{\mathbb N}}$. Example 2 follows, as above, by letting ${\mathcal A}$ be the unitial subalgebra generated by ${a}$, ${p}$ be the restriction of ${p^\prime}$ to ${\mathcal A}$, and ${\varphi\colon\mathcal A\rightarrow\mathcal A^\prime}$ be inclusion.