Almost Sure

14 January 20

The GNS Representation

Filed under: Probability Theory — George Lowther @ 12:06 AM
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As is well known, the space of bounded linear operators on any Hilbert space forms a *-algebra, and (pure) states on this algebra are defined by unit vectors. Considering a Hilbert space {\mathcal H}, the space of bounded linear operators {\mathcal H\rightarrow\mathcal H} is denoted as {B(\mathcal H)}. This forms an algebra under the usual pointwise addition and scalar multiplication operators, and involution of the algebra is given by the operator adjoint,

\displaystyle  \langle x,a^*y\rangle=\langle ax,y\rangle

for any {a\in B(\mathcal H)} and all {x,y\in\mathcal H}. A unit vector {\xi\in\mathcal H} defines a state {p\colon B(\mathcal H)\rightarrow{\mathbb C}} by {p(a)=\langle\xi,a\xi\rangle}.

The Gelfand-Naimark–Segal (GNS) representation allows us to go in the opposite direction and, starting from a state on an abstract *-algebra, realises this as a pure state on a *-subalgebra of {B(\mathcal H)} for some Hilbert space {\mathcal H}.

Consider a *-algebra {\mathcal A} and positive linear map {p\colon\mathcal A\rightarrow{\mathbb C}}. Recall that this defines a semi-inner product on the *-algebra {\mathcal A}, given by {\langle x,y\rangle=p(x^*y)}. The associated seminorm is denoted by {\lVert x\rVert_2=\sqrt{p(x^*x)}}, which we refer to as the {L^2}-seminorm. Also, every {a\in\mathcal A} defines a linear operator on {\mathcal A} by left-multiplication, {x\mapsto ax}. We use {\lVert a\rVert_\infty} to denote its operator norm, and refer to this as the {L^\infty}-seminorm. An element {a\in\mathcal A} is bounded if {\lVert a\rVert_\infty} is finite, and we say that {(\mathcal A,p)} is bounded if every {a\in\mathcal A} is bounded.

Theorem 1 Let {(\mathcal A,p)} be a bounded *-probability space. Then, there exists a triple {(\mathcal H,\pi,\xi)} where,

  • {\mathcal H} is a Hilbert space.
  • {\pi\colon\mathcal A\rightarrow B(\mathcal H)} is a *-homomorphism.
  • {\xi\in\mathcal H} satisfies {p(a)=\langle\xi,\pi(a)\xi\rangle} for all {a\in\mathcal A}.
  • {\xi} is cyclic for {\mathcal A}, so that {\{\pi(a)\xi\colon a\in\mathcal A\}} is dense in {\mathcal H}.

Furthermore, this representation is unique up to isomorphism: if {(\mathcal H^\prime,\pi^\prime,\xi^\prime)} is any other such triple, then there exists a unique invertible linear isometry of Hilbert spaces {L\colon\mathcal H\rightarrow\mathcal H^\prime} such that

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle \pi^\prime(a)=L\pi(a)L^{-1},\smallskip\\ &\displaystyle \xi^\prime=L\xi. \end{array}

The GNS representation is constructed by taking a Hilbert space completion of {\mathcal A} under the {L^2} semi-inner product. Rather than proving theorem 1 in one go, I will first show a few preliminary lemmas from which the full result will follow. Any triple {(\mathcal H,\pi,\xi)} satisfying the conclusion of theorem 1 will be called a (or, the) GNS representation of {(\mathcal A,p)}. First, assuming that the GNS representation exists, then it is called faithful if all {a\in\mathcal A} satisfy {\pi(a)=0} only when {a=0}. This occurs precisely when the the state {p} is nondegenerate and, in this case, {\pi} identifies {\mathcal A} with a *-subalgebra of {B(\mathcal H)}.

Lemma 2 Let {(\mathcal A,p)} be a bounded *-probability space with GNS representation {(\mathcal H,\pi,\xi)}. Then,

  1. the map {x\mapsto\pi(x)\xi} is an {L^2}-isometry from {\mathcal A} to {\mathcal H}.
  2. {\pi} is an {L^\infty}-isometry, so that {\lVert\pi(a)\rVert=\lVert a\rVert_\infty}.
  3. {\pi} has kernel {\{a\in\mathcal A\colon\lVert a\rVert_\infty=0\}}.
  4. the representation is faithful if and only if {p} is nondegenerate.

Proof: That {x\mapsto\pi(x)\xi} is an {L^2}-isometry follows from,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle\pi(x)\xi,\pi(y)\xi\rangle&\displaystyle=\langle\xi,\pi(x)^*\pi(y)\xi\rangle\smallskip\\ &\displaystyle=\langle\xi,\pi(x^*y)\xi\rangle\smallskip\\ &\displaystyle=p(x^*y)=\langle x,y\rangle. \end{array}

Next, as {\xi} is cyclic for {\mathcal A}, the inequality

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert\pi(a)\pi(x)\xi\rVert&\displaystyle=\lVert ax\rVert_2\le\lVert a\rVert_\infty\lVert x\rVert_2\smallskip\\ &\displaystyle=\lVert a\rVert_\infty\lVert\pi(x)\xi\rVert \end{array}

gives {\lVert\pi(a)\rVert\le\lVert a\rVert_\infty}. Similarly,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert ax\rVert_2&\displaystyle=\lVert\pi(a)\pi(x)\xi\rVert\smallskip\\ &\displaystyle\le\lVert\pi(a)\rVert\lVert\pi(x)\xi\rVert\smallskip\\ &\displaystyle=\lVert\pi(a)\rVert\lVert x\rVert_2 \end{array}

gives {\lVert a\rVert_\infty\le\lVert\pi(a)\rVert}, so {\pi} is an {L^\infty}-isometry. The second statement is immediate, as {\pi(a)=0} iff {\lVert a\rVert_\infty=\lVert\pi(a)\rVert=0}. The third statement is also immediate, as {\pi} is faithful iff its kernel is {\{0\}} and {p} is nondegenerate iff {a=0} whenever {\lVert a\rVert_\infty=0}. ⬜

Now, consider a nonnegative linear map {p\colon\mathcal A\rightarrow{\mathbb C}}. As this does not have to be a state, and elements of {\mathcal A} might not be {L^\infty}-bounded, the full GNS representation as described by theorem 1 need not exist. However, it is still possible to define the Hilbert space {\mathcal H} by taking the {L^2}-completion of {\mathcal A}. Note that, if the GNS representation does exist, then {\iota x\equiv\pi(x)\xi} satisfies the properties of the isometry defined by the following result.

Lemma 3 Let {\mathcal A} be a *-algebra and {p\colon\mathcal A\rightarrow{\mathbb C}} be a positive linear map. Then, there exists a Hilbert space {\mathcal H} and a linear isometry {\iota\colon\mathcal A\rightarrow\mathcal H} with dense image.

Proof: As previously explained, we make {\mathcal A} into a semi-inner product space by {\langle x,y\rangle=p(x^*y)}. Then, we take {\iota\colon\mathcal A\rightarrow\mathcal H} to be its completion. ⬜

If we introduce the condition that every {a\in\mathcal A} is {L^\infty}-bounded, then the *-homomorphism {\pi} can be constructed.

Lemma 4 Let {\mathcal A} be a *-algebra and {p\colon\mathcal A\rightarrow{\mathbb C}} be a positive linear map such that every {a\in\mathcal A} is {L^\infty(p)}-bounded. If {\iota\colon\mathcal A\rightarrow\mathcal H} is as in lemma 3 then there is a unique *-homomorphism {\pi\colon\mathcal A\rightarrow B(\mathcal H)} satisfying

\displaystyle  \pi(a)\iota x=\iota ax (1)

for all {a\in\mathcal A}. Furthermore, {\lVert\pi(a)\rVert=\lVert a\rVert_\infty}.

Proof: As {a\in\mathcal A} is bounded, {x\mapsto ax} is a bounded linear map on {\mathcal A} with operator norm {\lVert a\rVert_\infty}. By continuous linear extension, there is a unique {\pi(a)\in B(\mathcal H)} satisfying (1), and has operator norm {\lVert\pi(a)\rVert=\lVert a\rVert_\infty}. That {\pi} is a *-homomorphism is immediate from the definitions. For example,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle\iota y,\pi(a^*)\iota x\rangle&\displaystyle=\langle\iota y,\iota a^*x\rangle=p(y^*a^*x)\smallskip\\ &\displaystyle=\langle\iota ay,\iota x\rangle=\langle\pi(a)\iota y,\iota x\rangle \end{array}

so that {\pi(a^*)=\pi(a)^*}

Alternatively, if it is assumed that {p} is a state or, equivalently, {(\mathcal A,p)} is a *-probability space, then the distinguished element {\xi\in\mathcal H} can be constructed. To simplify matters, to handle the case where {\mathcal A} is not unitial, we use the fact that {p} uniquely extends to a state on the unitial algebra {{\mathbb C}\oplus\mathcal A} by taking {p(\lambda+a)=\lambda+p(a)} for {\lambda\in{\mathbb C}}. In fact, by lemma 10 of the post on states, {\mathcal A} is {L^2}-dense in {{\mathbb C}\oplus\mathcal A}.

Lemma 5 Let {(\mathcal A,p)} be a *-probability space and let {\iota\colon\mathcal A\rightarrow\mathcal H} be as in lemma 3. Then, there exists a unique {\xi\in\mathcal H} satisfying

\displaystyle  p(x)=\langle\xi,\iota x\rangle (2)

for all {x\in\mathcal A}. Furthermore, {\lVert\xi\rVert=1} and, if {\mathcal A} is unitial, {\xi=\iota1}. More generally, {\iota} uniquely extends to an {L^2}-isometry {{\mathbb C}\oplus\mathcal A\rightarrow\mathcal H}, in which case {\xi=\iota1}.

Proof: Uniqueness of {\xi} is immediate from (2) and the requirement that {\{\iota x\colon x\in\mathcal A\}} is dense in {\mathcal H}. When {\mathcal A} is unitial, then taking {\xi=\iota1} gives

\displaystyle  \langle\xi,\iota x\rangle=\langle\iota 1,\iota x\rangle=p(1^*x)=p(x).

In the non-unitial case, by existence and uniqueness of bounded linear extensions, {\iota} uniquely extends to an isometry {{\mathbb C}\oplus\mathcal A\rightarrow\mathcal H}. Then, as above, {\xi=\iota1} and, hence,

\displaystyle  \lVert\xi\rVert^2=\langle\xi,\iota1\rangle=p(1)=1.

In particular, if we have a GNS representation {(\mathcal H,\mathcal A,\xi)}, then {\iota x=\pi(x)\xi} satisfies the requirements of lemma 5, and we see that {\xi} is necessarily a unit vector.

Corollary 6 Let {(\mathcal A,p)} be a bounded *-probability space with GNS representation {(\mathcal H,\pi,\xi)}. Then, {\lVert\xi\rVert=1}.

The existence of the GNS representation follows from what we have shown so far.

Lemma 7 Let {(\mathcal A,p)} be a bounded *-probability space, and assume the notation of lemmas 4 and 5. Then, {(\mathcal H,\pi,\xi)} satisfies the requirements of the GNS representation of theorem 1, and {\iota x=\pi(x)\xi}.

Proof: By definition, {\mathcal H} is a Hilbert space and {\pi\colon\mathcal A\rightarrow B(\mathcal H)} is a *-homomorphism. Extend {\iota} to an {L^2}-isometry {{\mathbb C}\oplus\mathcal A\rightarrow\mathcal H}. By lemma 4, {\pi} extends to a *-homomorphism {{\mathbb C}\oplus\mathcal A\rightarrow B(\mathcal H)} satisfying (1). Then,

\displaystyle  \pi(a)\xi=\pi(a)\iota1=\iota a. (3)

Hence

\displaystyle  \langle\xi,\pi(a)\xi\rangle=\langle\iota1,\iota a\rangle=p(1^*a)=p(a),

as required. Finally, (3) shows that {\{\pi(a)\xi\colon a\in\mathcal A\}=\iota(\mathcal A)}, which is dense in {\mathcal H}. ⬜

To easily handle non-unitial algebras, we note that GNS representations of {\mathcal A} automatically extend to GNS representations of the unitial algebra {{\mathbb C}\oplus\mathcal A}.

Lemma 8 Let {(\mathcal A,p)} be a bounded *-probability space with GNS representation {(\mathcal H,\pi,\xi)}. Then, {\pi} uniquely extends to a *-homomorphism {\tilde\pi\colon{\mathbb C}\oplus\mathcal A\rightarrow B(\mathcal H)}, in which case {(\mathcal H,\tilde\pi,\xi)} is a GNS representation for {({\mathbb C}\oplus\mathcal A,p)}.

Proof: Any extension {\tilde\pi} satisfies

\displaystyle  \tilde\pi(1)\pi(a)\xi=\tilde\pi(1)\tilde\pi(a)\xi=\pi(a)\xi

so, as {\xi} is cyclic for {\mathcal A}, {\tilde\pi(1)=1}. Hence, {\tilde\pi(\lambda+a)=\lambda+\pi(a)} is the unique extension of {\pi} to a *-homomorphism from {{\mathbb C}\oplus\mathcal A}. As {\lVert\xi\rVert=1} by corollary 6,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \langle\xi,\tilde\pi(\lambda+a)\xi\rangle&\displaystyle=\lambda\langle\xi,\xi\rangle+\langle\xi,\pi(a)\xi\rangle\smallskip\\ &\displaystyle=\lambda+p(a)=p(\lambda+a) \end{array}

as required. ⬜

Next, the GNS representation is functorial. A homomorphism between *-probability spaces is a state preserving *-homomorphism of their *-algebras, and these canonically induce isometries of their GNS Hilbert spaces.

Lemma 9 Let {\varphi} be a homomorphism of bounded *-probability spaces {(\mathcal A,p)} and {(\mathcal A^\prime,p^\prime)}, which have GNS representations {(\mathcal H,\pi,\xi)} and {(\mathcal H^\prime,\pi^\prime,\xi^\prime)} respectively. Then, there exists a unique isometric linear map {L\colon\mathcal H\rightarrow\mathcal H^\prime} satisfying

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle L\pi(a)=\pi^\prime(\varphi(a))L,\smallskip\\ &\displaystyle L\xi=\xi^\prime, \end{array} (4)

for all {a\in\mathcal A}.

Proof: Let {V=\{\pi(a)\xi\colon a\in\mathcal A\}} which, by definition, is a dense subspace of {\mathcal H}. Combining equations (4),

\displaystyle  L\pi(a)\xi=\pi^\prime(\varphi(a))\xi^\prime, (5)

which uniquely determines {L} on {V} and, by continuity, this uniquely determines {L}. Conversely, we can use (5) to construct {L} on {V}. We need to show that this is an isometry and, to be well-defined, that the right-hand-side of (5) is zero whenever {\pi(a)\xi=0}. Using

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \lVert\pi^\prime(\varphi(a))\xi^\prime\rVert^2 &\displaystyle=\langle\xi^\prime,\pi^\prime(\varphi(a))^*\pi^\prime(\varphi(a))\xi^\prime\rangle\smallskip\\ &\displaystyle=\langle\xi^\prime,\pi^\prime(\varphi(a^*a))\xi^\prime\rangle\smallskip\\ &\displaystyle=p^\prime(\varphi(a^*a)) =p(a^*a)\smallskip\\ &\displaystyle=\langle\xi,\pi(a^*a)\xi\rangle\smallskip\\ &\displaystyle=\lVert\pi(a)\xi\rVert^2, \end{array}

we see that {L} is well-defined and an isometry. Hence, by continuous linear extension it uniquely extends to an isometry {L\colon\mathcal H\rightarrow\mathcal H^\prime}. Next, we show that (4) is satisfied. Using

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle L\pi(a)\pi(b)\xi&\displaystyle=L\pi(ab)\xi=\pi^\prime(\varphi(ab))\xi^\prime\smallskip\\ &\displaystyle=\pi^\prime(\varphi(a))\pi^\prime(\varphi(b))\xi^\prime\smallskip\\ &\displaystyle=\pi^\prime(\varphi(a))L\pi(b)\xi, \end{array}

we see that the first identity of (4) holds on {V} and, by continuity, holds on all of {\mathcal H}. Finally, by extending the constructions above to {{\mathbb C}\oplus\mathcal A} and {{\mathbb C}\oplus\mathcal A^\prime}, we can wlog assume that {\mathcal A} and {\mathcal A^\prime} are unitial. Then,

\displaystyle  L\xi=L\pi(1)\xi=\pi^\prime(\varphi(1))\xi^\prime=\xi^\prime

as required. ⬜

Finally, we put together the previous steps to complete the proof of theorem 1.

Proof of Theorem 1: The existence of the GNS representation was proven in lemma 7, so only uniqueness remains. Suppose that {(\mathcal H,\pi,\xi)} and {(\mathcal H^\prime,\pi^\prime,\xi^\prime)} are two GNS representations. Applying lemma 9 to the identity map {\varphi} on {\mathcal A} gives a unique isometry {L\colon\mathcal H\rightarrow\mathcal H^\prime} satisfying {L\pi(a)=\pi^\prime(a)L} and {L\xi=\xi^\prime}. Similarly, there is a unique isometry {L^\prime\colon\mathcal H^\prime\rightarrow\mathcal H} satisfying {L^\prime\pi^\prime(a)=\pi(a)L^\prime} and {L^\prime\xi^\prime=\xi}. Then, {\tilde L=L^\prime L} satisfies {\tilde L\pi(a)\xi=\pi(a)\xi} and, hence, is the identity map. Similarly, {LL^\prime} is the identity, so {L} is invertible. ⬜

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