Most of Hilbert spaces appearing in real life are separable ones: even such large for the first spaces as Fock space, spaces of functions in infinite number of variables and so on. However I heard that when dealing with quantum loop gravity one encounter non-separable Hilbert space.

Could somebody explain how this space is constructed? What are natural operators, relevant for LQG, acting in such space?

I would like also to understand some conceptual reason for the apperance of nonseparable Hilbert spaces in this context: since the main idea of LQG is that the geometry is ,,quantized'' I would suspect that Hilbert spaces which occure in LQG would be finite dimensional.


1 Answer 1


The lack of separability in the kinematical Hilbert space in LQG is a mathematical artifact; it doesn't affect the physics, and it disappears when tweaking the mathematical setting.

In more detail, according to the abstract of this paper by Fairbairn & Rovelli in 2004:

In the standard construction, the kinematical Hilbert space of the diffeomorphism invariant states is non-separable. This is a consequence of the fact that the knot-space of the equivalence class of graphs under diffeomorphisms is non-countable.

They add:

However, the continuous moduli labelling these classes do not appear to affect the physics.

And then:

We investigate the possibility that these moduli could be only a consequence of a poor choice in in the fine-tuning of the mathematical setting.

To which they affirm:

We show that by simply choosing a minor extension of the functional class of the classical fields and coordinates, the moduli disappears, the knot classes become countable, and the kinematical Hilbert space becomes countable.

  • $\begingroup$ Thank you! And what was originally this set parametrizing the basis of Hilbert space? $\endgroup$
    – truebaran
    Dec 11, 2017 at 20:00
  • $\begingroup$ @truebaran: according to the linked paper, these are abstract spin networks - these are the graphs referred to in the abstract. $\endgroup$ Dec 11, 2017 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.