GNS Construction

The Gelfand–Naimark–Segal construction turns a state on a C*-algebra into a vector expectation in a cyclic Hilbert-space representation. It constructs the Hilbert space from the state’s correlations and determines that cyclic realization uniquely up to unitary equivalence. Becoming a vector state in this representation does not imply that the original algebraic state is pure.

Required background. The Algebraic State Space defines states of a unital C*-algebra. Positive State Functionals proves the positive-functional Cauchy–Schwarz inequality. Representations defines cyclicity and unitary equivalence.

The state defines a semidefinite inner product

Let $\mathcal A$ be a unital C*-algebra and $\omega$ a state. Define

$$\langle a,b\rangle_\omega=\omega(a^*b),\qquad \mathcal N_\omega=\{a:\omega(a^*a)=0\}.$$

The form is conjugate-linear in its first argument, linear in its second, and positive semidefinite. Cauchy–Schwarz implies that every $n\in\mathcal N_\omega$ is orthogonal to every algebra element. In particular, the null set is a linear subspace, and the form descends to a positive-definite inner product on the quotient vector space $\mathcal A/\mathcal N_\omega$.

The quotient removes distinctions the state cannot detect through this quadratic form. It is not generally a quotient algebra. That distinction is essential because $\mathcal N_\omega$ is a left ideal, not necessarily a two-sided ideal.

The C*-order estimate makes left multiplication well defined

For $c,a\in\mathcal A$, the C*-order inequality $c^*c\leq\|c\|^2 1$ gives

$$0\leq a^*c^*ca\leq\|c\|^2a^*a.$$

Applying $\omega$ yields

$$\|[ca]\|_\omega^2 \leq\|c\|^2\|[a]\|_\omega^2.$$

If $a$ is null, then so is $ca$. Thus $\mathcal N_\omega$ is a left ideal, and

$$\pi_\omega(c)[a]=[ca]$$

does not depend on the representative of the class. The same estimate proves boundedness, $\|\pi_\omega(c)\|\leq\|c\|$. Therefore each $\pi_\omega(c)$ extends uniquely to the Hilbert completion $\mathcal H_\omega$ of $\mathcal A/\mathcal N_\omega$.

This is the analytic step missing from a merely formal quotient construction. Positivity alone on a general star algebra need not give bounded representing operators; the C*-norm and order estimate are doing substantive work.

Products are preserved by associativity. Moreover,

$$\langle[a],\pi_\omega(c)[b]\rangle =\omega(a^*cb) =\langle\pi_\omega(c^*)[a],[b]\rangle,$$

so $\pi_\omega(c)^*=\pi_\omega(c^*)$. Hence $\pi_\omega$ is a unital representation.

The distinguished vector recovers all expectations

Set $\Omega_\omega=[1]$. It has norm one, and

$$\pi_\omega(a)\Omega_\omega=[a],\qquad \omega(a)= \langle\Omega_\omega,\pi_\omega(a)\Omega_\omega\rangle.$$

The classes $[a]$ are dense by construction, so $\Omega_\omega$ is cyclic. This proves existence of the GNS triple $(\mathcal H_\omega,\pi_\omega,\Omega_\omega)$.

Faithfulness of the state, meaning $\omega(a^*a)=0$ only for $a=0$, implies faithfulness of this representation. The converse need not hold: a vector state on $M_n$ is not faithful for $n>1$, while its GNS representation is the faithful defining representation.

Cyclic uniqueness identifies the correct notion of equivalence

Suppose another triple $(\mathcal K,\rho,\Omega)$ realizes the same state and $\Omega$ is cyclic. Define initially

$$U[a]=\rho(a)\Omega.$$

Its inner products agree with those on the quotient:

$$\langle\rho(a)\Omega,\rho(b)\Omega\rangle =\omega(a^*b).$$

Thus $U$ is well defined and isometric. Cyclicity makes its range dense, so its continuous extension is unitary. It sends $\Omega_\omega$ to $\Omega$ and satisfies $U\pi_\omega(c)=\rho(c)U$. Uniqueness follows because its values on the dense cyclic set are fixed.

The distinguished vector matters. Equivalent representations without a specified vector can support many different states.

A mixed matrix state produces a larger cyclic space

Take $\mathcal A=M_n(\mathbb C)$ and $\omega(a)=\operatorname{Tr}(\rho a)$, with $\rho$ a density matrix of rank $r$. Map

$$[a]\longmapsto a\rho^{1/2}.$$

The Hilbert–Schmidt inner product gives exactly $\operatorname{Tr}(\rho a^*b)$, so this realizes the GNS space as matrices vanishing on the kernel of $\rho$, equivalently maps from its $r$-dimensional support to $\mathbb C^n$. Its dimension is $nr$.

The algebra acts by left multiplication; the cyclic vector is $\rho^{1/2}$. For $r>1$, right multiplication on the support supplies a nontrivial commutant, so the representation is reducible. For $r=1$, it is the defining irreducible representation.

Thus a mixed state becomes a cyclic vector in a representation with multiplicity. The pure-state theorem makes this relationship general.

Exercise: the null ideal need not be two-sided

For the vector state $\omega(a)=\langle e_1,ae_1\rangle$ on $M_2$, identify $\mathcal N_\omega$ and find $a\in\mathcal N_\omega$, $b\in M_2$ with $ab\notin\mathcal N_\omega$.

Solution

The null ideal consists of matrices with first column zero because $\omega(a^*a)=\|ae_1\|^2$. Take $a=|e_1\rangle\langle e_2|$ and $b=|e_2\rangle\langle e_1|$. Then $ab=|e_1\rangle\langle e_1|$ is not null. This is why the quotient is used as a vector space carrying left multiplication, rather than as a quotient algebra.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987, doi:10.1007/978-3-662-02520-8, Chapter 2 — GNS existence, uniqueness, and representation properties.
• Naaijkens, Pieter, Quantum Spin Systems on Infinite Lattices, Lecture Notes in Physics 933, Springer, 2017, arXiv:1311.2717 — constructive state representations and finite-dimensional examples.