# The GNS Representation

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. ⬜