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The traditional mathematical approach to quantum mechanics, as developed by von Neumann, is based on Hilbert spaces and unbounded self-adjoint operators. Another approach, which more closely resembles Dirac's original ideas, uses rigged Hilbert spaces and continuous self-adjoint operators.

More precisely, a rigged Hilbert space is a pair (A,μ), where A is a complex complete nuclear locally convex topological vector space and the inner product μ: Ā⊗A→C is a (continuous) morphism of such spaces that is symmetric and strictly positive (μ(x⊗y)=μ(y⊗x) and μ(x⊗x)>0 for x≠0). Here Ā⊗A denotes the unique tensor product of nuclear spaces.

Completing A with respect to μ results in a Hilbert space H together with a continuous embedding i: A→H, which has dense image, hence the name rigged Hilbert space, i.e., the Hilbert space H is “rigged” with the additional data of A. Continuous self-adjoint (with respect to μ) operators on A extend to unbounded self-adjoint operators on H. This procedure allows one to pass from the formalism of rigged Hilbert spaces to the von Neumann formalism.

Vice versa, if we assume that H is the space of L^2-sections of some vector bundle on a smooth manifold equipped with an inner product μ, then we can set A to be the space of all smooth sections of this vector bundle and let μ be the restriction of μ to A.

The advantage of the formalism of rigged Hilbert spaces lies in the fact that unbounded operators on a Hilbert space are difficult to deal with (one cannot always add or multiply them and has to be very careful about domains), whereas continuous operators on a rigged Hilbert space form an algebra.

Spectral triples, also known as unbounded Fredholm modules, were introduced by Alain Connes to study noncommutative spaces.

An odd spectral triple is a quadruple (E,H,ρ,D), where E is a C*-algebra, H is a Hilbert space, ρ: E→B(H) is a monomorphism of C*-algebras, and D is a self-adjoint unbounded operator on H with compact resolvent such that for all e∈E the commutant [D,ρ(e)] is bounded. For even spectral triples the space H is Z/2-graded, ρ is even, and D is odd. The choice of the name is motivated by the fact that every even/odd spectral triple represents a class in even/odd K-homology of E.

A typical example of an even spectral triple is given by (E,H,ρ,D), where E is the C*-algebra of continuous functions on a compact smooth Riemannian oriented spin manifold M, H is the space of L^2-sections of the spinor bundle of M, ρ acts by multiplication, and D is the (closure of) Dirac operator.

Since unbounded operators in spectral triples give rise to the same problems (absence of everywhere defined algebraic operations) one is led naturally to the following question: Can we define spectral triples using the formalism of rigged Hilbert spaces?

One might expect that rigged spectral triples should be quadruples (E,X,ρ,D), where E is an algebra belonging to a certain class of “rigged C*-algebras” (we expect all algebras of the form C^∞(M) to belong to this class; one can try certain generalizations of C*-algebras such as locally C*-algebras, perhaps with additional conditions such as nuclearity; perhaps we also need to supply an embedding of E into some C*-algebra), X is a rigged Hilbert space, ρ is a continuous representation of E on X, and D is a self-adjoint continuous endomorphism of X, possibly satisfying some additional conditions. For ordinary spectral triples the operator D is required to have compact resolvent and for all e∈E the commutator [D,ρ(e)] must be bounded. What is the right analog of this condition for rigged spectral triples?

Among desirable properties of rigged spectral triples one can name the following two:

  • One should be able to introduce a notion of homotopy of rigged spectral triples in such a way that homotopy classes of rigged spectral triples give K-homology of E.

  • Moreover, one should be able at least in some cases to complete a rigged spectral triple to an ordinary spectral triple, preserving the K-homology class.

A typical example of a rigged spectral triple should be given by (E,X,ρ,D), where E is the algebra of smooth functions on a compact smooth Riemannian oriented spin manifold M, H is the space of smooth sections of the spinor bundle of M with the natural inner product, ρ acts by multiplication, and D is the Dirac operator.

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Your definition of spectral triple is way too restrictive. One rarely takes a $C^*$-algebra, rather certain subalgebras like smooth ones. At first glance I think your proposed notion of "rigged spectral triple" is contained in the general definition of a spectral triple, see for example en.wikipedia.org/wiki/Spectral_triple. –  user5831 Nov 21 '10 at 22:59
    
@Bora: I read the Wikipedia page before posting this question, and I don't really see an answer to my question there. The Wikipedia still requires D to be an unbounded operator on the entire Hilbert space and does not explain how to express the relevant properties in terms of rigged space itself. –  Dmitri Pavlov Nov 22 '10 at 10:10
    
Dear Dmitri, maybe I am completely wrong, but as far as I understood your question, then assuming a "rigged spectral triple" $(E,X,\rho,D)$ this should be naturally extendable to a classical spectral triple $(E,H_X,D_X)$ where $H_X$ denotes the Hilbert space completion and $D_X$ the extension of $D$ to $H_X$ (we don't have to change $E$!). A priori it might happen that depending on your choice of axioms for a rigged spectral triple certain axioms of a spectral triple might not hold for the extended system, but for generic examples I would guess they hold. –  user5831 Nov 22 '10 at 12:47
    
@Bora: What you are saying is certainly true, but the question is how to state the axioms for rigged spectral triples so that after completion the usual axioms are satisfied. I certainly don't want just say that a rigged spectral triple is a tuple (E,X,ρ,D) that becomes an ordinary spectral triple after completion. –  Dmitri Pavlov Nov 22 '10 at 13:54
    
Yes, you are right. I stumbled upon your definition of spectral triple using only $C^*$-algebras and thought you want to generalise this. Sorry for the misunderstanding. –  user5831 Nov 22 '10 at 14:14

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