On Monday, June 8, 2020 at 9:39:30 AM UTC-5, Bruno Marchal wrote:
>
>
> On 7 Jun 2020, at 13:25, Lawrence Crowell <[email protected]
> <javascript:>> wrote:
>
> On Saturday, June 6, 2020 at 5:25:23 PM UTC-5, Philip Thrift wrote:
>>
>>
>> As for Hossenfelder's fav quantum mechanics semantics, she has stated
>> many times on her blog, it's superdeterminism.
>>
>>
>> https://arxiv.org/abs/1912.06462
>>
>>
>> "A superdeterministic theory is one which violates the assumption of
>> Statistical Independence (that distributions of hidden variables are
>> independent of measurement settings). Intuition suggests that Statistical
>> Independence is an essential ingredient of any theory of science (never
>> mind physics), and for this reason Superdeterminism is typically discarded
>> swiftly in any discussion of quantum foundations. The purpose of this paper
>> is to explain why the existing objections to Superdeterminism are based on
>> experience with classical physics and linear systems, but that this
>> experience misleads us. Superdeterminism is a promising approach not only
>> to solve the measurement problem, but also to understand the apparent
>> nonlocality of quantum physics. Most importantly, we will discuss how it
>> may be possible to test this hypothesis in an (almost) model independent
>> way."
>>
>> @philipthrift
>>
>
> Superdeterminism is just a form of hidden variable theory. This invariant
> set theory of Palmer and Hossenfelder as a means of connecting nonlinearity
> with QM is interesting. The approach with Cantor sets connects with
> incomputability.
>
>
>
> The Mandelbrot Set (its complement) has been shown indecidable, but in a
> vary peculiar theory of computability, which has not so many relation with
> Turing. It is “computability in a ring”.
>
> In the Turing theory, it is an open problem if he complement of the
> rational-complex Mandelbrot set is undecidable. That is a conjecture in my
> long thesis. Penrose has come up with a similar (less precise) hypothesis.
>
>
>
The p-adic ring is what determines the trajectory of a point. where in the
case of a Cantor set the divisor of the quotient ring is the map from one
point to another. The Cantor set then has a set of orbits given by a set of
p-adic rings. The result of Matiyaesivich is there is no global method for
solving these or the Diophantine equations they correspond to. This is the
approach that I take. With the Mandelbrot set the "black bit" has periodic
orbits or maps which correspond to periods of Julius sets. The points
outside are chaotic and are in a sense "beyond chaos" and are not
computable.
>
> I prefer a more standard definition of incomputability than what P&H
> appeal to. This works invariant set theory does imply a violation of
> statistical independence, but it does so as a hidden variable.
>
> The complement of a fractal set is undecidable.
>
>
> The complement of some fractal set have been shown undecidable in a theory
> of computability on a ring. This has been shown by Blum, Smale and Shub, if
> I remember well.
>
>
>
The incomputability if with the fractal set itself. The incomputability
occurs because with a finite cut off you have uncertainty whether points or
regions are in or outside the Mandelbrot set. In this somewhat different
meaning the Mandelbrot set is considered incomputable by Blum, Smale and
Shub,
>
> A fractal set is recursively enumerable, which means we can compute it in
> a finite automata up to some point, and “in principle” a Turing machine
> that runs eternally could compute the whole thing.
>
>
> Yes, but it uses only the potential infinite. We get all element in the
> enumeration after a finite time (except that here we use computability on a
> ring, which is not so easy to compare with Turing computability).
>
>
>
>
>
> The complement of this is not computable. The complement of a recursive
> set is recursive, but the complement of a recursively enumerable set is not
> recursively enumerable and is incomputable.
>
>
> You mean “ … is not necessarily recursively enumerable”. Of course a
> complement of a recursively enumerable set can be recursively enumerable.
> That is always the case with recursive set.
>
>
Yes, if the RE set is recursive.
>
>
> The invariant set in this superdeterminism is a form of Cantor set or
> related to a fractal. The results of Matiyasevich showed that p-adic sets
> have no global solution method, where p-adic sets are equivalent to
> Diophantine equations.
>
>
>
> I would be interested in a precise statement of this, and some link to a
> proof. What has a p-adic set? Set of what?
>
> Cantor sets are related to self-reference in many ways. For example
> through the topological semantic of G and S4Grz, but also through the
> “fuzzification” of Gödel or Löb theorem, like in a paper by Grim.
>
>
>
A p-adic set is a quotient ring with the Z_p for p a prime. The Chinese
remainder theorem guarantees that all quotient rings are equivalent to the
product of quotient rings with primes that are the prime decomposition of
the quotient ring. In other words for the quotient group ℤ_n =
ℤ_{p1}×ℤ_{p2}× … ×ℤ_{p} for n = {p1}×{p2}× … ×{p} the prime factorization.
There is a lot there and quotient rings define an elementary aspect of
cohomology that leads to p-adic topology and with complex rings algebraic
geometry.
>
>
>
> This means that dynamical maps from one point to another on the Cantor set
> are not given by the same quotient group and in general there is no single
> decidable system for such maps. In effect this means it is not observable.
>
>
> What is the relation between observable and decidable? If you study my
> papers, this is the most difficult thing to do. It is possible, and
> necessary, though, by the fact intensional variant of G and G*, which makes
> the logic of the observable/predictibvle obeying a quite different logic
> than G (indeed, a quantum logic).
>
>
>
That is a part of the issue, and as I have worked things, hidden variables
are unobservable and incomputable. The superdeterminism 'tHooft advanced
and that others have taken up is really a form of hidden variable, and is
not computable.
>
>
> So, while superdeterminism violates statistical independence this is all a
> nonlocal hidden variable and thus unobservable. In ways this is where I
> depart from Hossenfelder and Palmer, where Palmer uses a different concept
> of incomputability, based on the idea of Smale et al on the need to compute
> a fractal an infinite amount.
>
>
>
> I thought you did this.
>
>
I worked this in a way similar to Hossenfelder and Palmer, but with out the
appeal to Blum, Smale and Shub.
>
>
> I appeal to the complement of a fractal, a fractal being a recursively
> enumerable set
>
>
> Set of what?
>
>
>
A fractal is a region of space that has a boundary with a Hausdorff
dimension that is not integral. So it is a set of points or orbits under
maps.
>
> and computable in a standard sense, but where the complement is not
> computable.
>
>
> The complement of a creative set (the set-definition of a universal
> machine, due to Emil Post, and done before Church and Turing) is always non
> recursively enumerable.
> The complement of any set of all theorems of an axiomatic rich enough to
> prove the axiom of RA is automatically non recursively enumerable.
>
> Thanks to the work of Myhill, we know that a theory is Turing complete
> (Turing universal) iff and only the complement in N of the set of (Gödel
> number of) its theorem is constructively not recursively enumerable.
> A set S of numbers is constructively not recursively enumerable, or called
> also productive, means that for any W_i subset of S, you can find some x in
> S, but not in W_i.. That x serves as counter-example of the recursive
> enumerability of S. You can extend the extension of S in the constructive
> transfinite, by reiterating this trasnfinitely (on the recursive ordinals,
> or beyond).
>
> A Recursively enumerable set with a productive complement is called
> creative, by Emil Post, and is the set theoretical definition of Turing
> universality, by a result of Myhill.
>
>
>
This I am not that familiar with. I tend to prefer to stay as much as
possible within more standard mathematics instead of set theory. Set theory
I will appeal to somewhat, but I prefer to stay more within algebra and
geometry.
>
> The fractal emerges from QM in a singular perturbation series and the
> complement comes with the dual of a convex set with measure L^p is L^q with
> 1/p + 1/q = 1.
>
>
>
> I fail to see what are the elements of the sets you are talking. The
> standard notion of computability concerns set of natural numbers (or of
> things encodable into finite numbers, like strings (the computer science
> one!), words, formula, etc.
>
> Bruno
>
>
The set of maps for any point is something computable or not. The Cantor
set is not because there is not a single algorithm for solving all orbits
that hop from one point to the other. This is because there is no global
solution method for all p-adic quotient rings. The elements are really
maps, maps that take a point here to there and then to elsewhere in an
iterative manner.
LC
>
>
>
> LC
>
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