# 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

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