Finding precisely what the group needs seems a mildly interesting project to me.

Same here, that’s the reason that I included the counterexample in the first place. My main motivation is to understand more about quantum information and connecting the Kochen-Specker theorem, Gleason’s theorem, Bell’s inequality and the fact that one can prove that the latter are maximally violated in AQFT. (Quantum stochastic calculus itself is of course very interesting, too, and Parthasarathy’s book seems to be one of the best sources to learn about it).

Maybe contrasting the counterexample with the proof will shed some light on the question, at the next opportunity I will start to add it.

]]>@Tim,

no, I just had the wrong idea. I suppose I was just wondering what properties of the unitary groups can be used and isolated, rather than saying ’there are more rotations’.

the rotational symmetry is enough to get the higher Fourier coefficients of the weight function f vanish

this is closer, but what does ’rotational symmetry is enough’ mean? I would like to think that we can describe analogous results for arbitrary linear groups, by considering some sort of constrained system or something, such that the symmetry group is cut down. Finding precisely what the group needs seems a mildly interesting project to me.

]]>I added a link to a PDF version of Gleason’s original paper.

]]>I wrote that page learning the material myself (still processing it), so I do have the same questions as you do :-) But I think I can comment on two of them:

…this is due to the difference between the symmetry groups of 2- and 3-dimensional (and presumably higher-dimensional) Hilbert spaces…

My impression from both the counterexample and the proof given in Parthasarathy’s book is that in three dimensions, the rotational symmetry is enough to get the higher Fourier coefficients of the weight function $f$ vanish (that can be defined in three dimensions just like in the counterexample) so that the “linearity” follows. An elaboration of the proof could make this more precise.

(The proof itself is given for three real dimensions and extended by induction to arbitrary real/complex dimensions).

Can we say this is due to the fact that (the component of the identity of) B(ℝ 2) is not 1-connected?

I don’t think so, in $\infty$ dimensions the unitary group is 1-connected, in finite dimensions it is not 1-connected, there does not seem to be a connection to Gleason’s theorem failing in two dimensions (or do I misunderstand you?).

]]>at Gleason’s theorem we have a result that holds for Hilbert spaces of dimension* $\gt 2$. There is a heuristic explanation that this is due to the difference between the symmetry groups of 2- and 3-dimensional (and presumably higher-dimensional) Hilbert spaces, and gives a counterexample using $\mathbb{R}^2$. Can we say this is due to the fact that (the component of the identity of) $B(\mathbb{R}^2)$ is not 1-connected? I’m assuming $B(\mathbb{R}^2) = O(2)$ here.

(*)Do we need to take real dimension, as the counterexample seems to indicate, or does the theorem also hold for complex dimension? What about other base fields?

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