Representations of Observable Algebras

A representation realizes an abstract observable algebra as bounded operators on a Hilbert space while preserving products and adjoints. Faithfulness, cyclicity, and irreducibility describe different aspects of that realization. Keeping them separate prevents an abstract Hilbert-space isomorphism or a convenient cyclic vector from being mistaken for physical equivalence of representations.

Required background. C*-Algebras supplies the abstract norm and spectrum. The Algebraic State Space supplies expectation functionals independently of their realization.

A star homomorphism preserves more than multiplication

A representation is a complex-linear star homomorphism

$$\pi:\mathcal A\longrightarrow\mathcal B(\mathcal H).$$

It is nondegenerate when the span of $\pi(\mathcal A)\mathcal H$ is dense in $\mathcal H$. For a unital algebra, nondegeneracy is equivalent to $\pi(1)=I$, and that convention is used below. A degenerate representation may include a subspace on which every algebra element acts as zero.

Every C*-representation is contractive. For a unital representation, an inverse in $\mathcal A$ maps to an inverse of the represented operator, so

$$\sigma(\pi(a))\subseteq\sigma(a).$$

For a positive element, spectral radius equals norm; applying this to $a^*a$ gives $\|\pi(a)\|\leq\|a\|$. In particular, no separate boundedness hypothesis is needed for a star homomorphism between C*-algebras.

Its kernel is a closed two-sided star ideal. A faithful representation has zero kernel and is isometric. This is a C*-theorem: the induced injective map from $\mathcal A/\ker\pi$ preserves the C*-norm. Faithful unital representations also preserve spectra by spectral permanence.

A representation can discard possible observable values

Let $\mathcal A=C([0,1])$ and $a(x)=x$. Evaluation at $x_0$ is a one-dimensional representation,

$$\pi_{x_0}(f)=f(x_0).$$

It sends the abstract observable with spectrum $[0,1]$ to the scalar $x_0$, whose spectrum is $\{x_0\}$. Its kernel consists of functions vanishing at that point. This is a valid irreducible representation, but it is not faithful.

Multiplication on $L^2([0,1],dx)$ is faithful because the reference measure has full support. It retains the spectrum $[0,1]$ but is reducible: subspaces supported in measurable subsets are invariant.

These examples show that faithfulness and irreducibility are independent. A representation may preserve all algebra elements without acting irreducibly, or act irreducibly while discarding a large ideal. On a simple algebra, every nonzero representation is faithful, but that additional hypothesis must be stated.

Cyclicity concerns one orbit, not every invariant subspace

A vector $\Omega$ is cyclic if

$$\overline{\pi(\mathcal A)\Omega}=\mathcal H.$$

The closure already includes linear combinations because $\mathcal A$ is a linear space. The constant vector $1$ is cyclic for continuous multiplication on $L^2([0,1])$: continuous functions are dense in that Hilbert space. Yet this representation is reducible.

A vector state specifies $\omega(a)=\langle\Omega,\pi(a)\Omega\rangle$. Restricting to the cyclic subspace generated by $\Omega$ preserves every expectation of that state. The GNS theorem shows that this restricted cyclic realization is unique up to a unitary that also carries the distinguished vector.

The distinction matters for mixed states. A state can become a vector expectation in a larger cyclic representation without becoming pure as a functional on the original algebra. Its purity is governed by irreducibility of that GNS representation, not by the existence of a representing vector.

Equivalence must intertwine the labeled algebra

Representations $\pi_1,\pi_2$ are unitarily equivalent when a unitary $U$ satisfies

$$U\pi_1(a)U^*=\pi_2(a) \qquad(a\in\mathcal A).$$

An arbitrary unitary between their Hilbert spaces does not suffice. The same algebra element must correspond under the comparison.

One useful invariant is the commutant: unitary equivalence conjugates $\pi_1(\mathcal A)'$ onto $\pi_2(\mathcal A)'$. Another is the kernel. Different point evaluations above have different kernels and cannot be equivalent, despite both using the Hilbert space $\mathbb C$.

A weaker comparison asks whether two representations support the same normal state functionals on $\mathcal A$. This leads to quasi-equivalence and is appropriate when multiplicity is irrelevant. Superselection in Algebraic Language distinguishes it from unitary equivalence and disjointness.

Normality itself is relative to the represented von Neumann algebra. A state realized as a vector in its own GNS representation need not be normal relative to a different representation’s closure. Choosing a vector realization does not erase that distinction.

Exercise: equivalent spectra do not establish equivalence

Let $\pi_1(f)=\operatorname{diag}(f(0),f(1))$ and $\pi_2(f)=\operatorname{diag}(f(0),f(0),f(1))$ on $\mathbb C^2$ and $\mathbb C^3$. Do they have the same kernel and represented spectra for every $f$? Are they unitarily equivalent?

Solution

Both kernels consist of functions vanishing at $0$ and $1$, and both represented spectra are $\{f(0),f(1)\}$. They are not unitarily equivalent: their Hilbert dimensions differ, and the multiplicity of evaluation at zero differs. Spectral sets and kernels omit representation multiplicity. They do support the same state functionals $p f(0)+(1-p)f(1)$, illustrating why a weaker equivalence is useful.

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 — representations, kernels, cyclicity, and equivalence.
• Naaijkens, Pieter, Quantum Spin Systems on Infinite Lattices, Lecture Notes in Physics 933, Springer, 2017, arXiv:1311.2717 — algebraic quantum states and their Hilbert-space realizations.