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Wikipedia:Reference desk/Archives/Mathematics/2009 November 5

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November 5

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Provable or Disprovable or Independent?

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The above discussion about Godel's incompleteness theorem prompted a question I've had for a while: Is it true that any statement can either be proved from a set of axioms, disproved, or shown to be independent of those axioms?

Here's my reasoning: I've heard people suppose that, in addition to those possibilities, it might be undecidable whether or not something is undecidable, or undecidable whether or not it's undecidable that it's undecidable, and so on ad infinitum. However, if something is (undecidable)^n, then clearly there exists no proof of that statement, and there also exists no proof of its negation. But that means it is simply undecidable: it is independent of the axioms. It seems, then, that there is no possibility for a statement's undecidability to be undecidable-- given any problem, there are three options: we can prove it, disprove it, or give a proof of its independence. Is this valid? Do I need any conditions on a theory for this argument to apply? 140.114.81.68 (talk) 09:20, 5 November 2009 (UTC)—Preceding unsigned comment added by 140.114.81.68 (talk) 09:19, 5 November 2009 (UTC)[reply]

If you do things this way, you have to keep careful track of where you're proving things. For example, the Goedel sentence GPA for Peano arithmetic (PA), cannot be proved or refuted in PA. Also, PA cannot prove that GPA is independent of PA.
However the slightly stronger theory PA+"PA is consistent" can prove that GPA is independent of PA, and also that GPA is outright true.
The thing to keep in mind is that "independent" is not a truth value. It's a statement about whether something is provable or refutable from a specific formal theory, and the theory always needs to be specified. Truth sempliciter on the other hand is a much more robust notion; the Goedel sentence of a consistent theory, though not provable in that theory, is always true, and you don't have to specify the theory that makes it true: It's just true, period. Remembering this is a good anchor when learning the Goedel stuff; it helps keep you from floating off into confusion. --Trovatore (talk) 09:35, 5 November 2009 (UTC)[reply]
Also within any given consistent system of sufficient power there will be statements that we can not prove, nor prove the negation, nor give a proof that neither of the previous proofs exists. Taemyr (talk) 10:11, 5 November 2009 (UTC)[reply]
If something is (undecidable)^n in T, as you say, then it is in particular undecidable in T (because if something is decidable in T, it is also provable decidable). However, the conclusion does not follow: there is no reason why we should be able to prove in T that the statement is (undecidable)^n (indeed, we never can do that, because undecidability of anything implies consistency of T). — Emil J. 11:52, 5 November 2009 (UTC)[reply]

In a certain sense, it is true that a particular sentence φ will be either provable from PA, disprovable from PA, or independent of PA. This is just tautologous. But it does not mean that "we can prove it, disprove it, or give a proof of its independence". There are many statements of interest where we can do none of those three things at present. Also, it will usually be true that the proof of independence is carried out in a metatheory rather than the theory at hand, as others have pointed out above. — Carl (CBM · talk) 13:46, 5 November 2009 (UTC)[reply]

Relation of "contiguity space" and "proximity space"

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Are the usual meanings of "contiguity space" and "proximity space" essentially the same? At Talk:Contiguity space we have a suggestion that they are the same, mainly on the basis of a google result that included "proximity space or contiguity space". Essentially I am asking if the proposed redirect and renaming would conflict with established usage, or with what should be on Wikipedia. An article on proximity space already exists. If "contiguity space" is a separate idea we could still move out the stuff related to probability measures, but that would leave a very poor article. Melcombe (talk) 11:47, 5 November 2009 (UTC)[reply]

Lie Brackets

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I put a question about Lie brackets here. Basically, I don't understand how they are useful and so why someone would have come up with them. —Ben FrantzDale (talk) 12:16, 5 November 2009 (UTC)[reply]

The Lie bracket of vector fields has a nice geometric interpretation. Let M be a smooth Riemannian manifold and consider the C∞(M,R)-module of smooth vector fields over M. Let X and Y be smooth vector fields on M. We can consider the new vector field [X,Y] given by [X,Y] = XY − YX. If we flow along X a distance of ε and then along Y a distance of ε we reach a point p ∈ M, then if we flow first along Y a distance of ε and then along X a distance of ε we reach a point q ∈ M. The difference between p and q is given by ε2[X,Y]. More formally: "The Lie bracket [X,Y] of two vector fields X and Y measures the O(ε2) gap in an incomplete quadrilateral of O(ε) arrows made alternatively from εX and εY."[1] So it's measuring how movement is commutative. In our real world experience if we move two paces east and then two paces north we end up at the same point as if we had taken first two paces north and then two paces east. This wouldn't be the case for more exotic manifolds and more exotic vector fields. ~~ Dr Dec (Talk) ~~ 15:22, 5 November 2009 (UTC)[reply]
  1. ^ Penrose, R (2005), The Road to Reality: A Complete guide to the Laws of the Universe, Vintage Books, ISBN 0-099-44068-7
Thanks. What I don't understand is how that is useful, beyond characterizing the local non-flatness of the manifold. When would I use it to solve a problem? Would it just be to answer questions about the manifold itself? Perhaps I'm just over my head and need to fill in more understanding to ask meaningful questions :-) —Ben FrantzDale (talk) 19:34, 6 November 2009 (UTC)[reply]
Well, you've answered your own question there. The torsion and curvature tensors both use the Lie bracket in their definitions. They're both very fundamental and interesting objects in differential geometry. Asking where the Lie bracket might be useful or when you might use it to solve a problem is kind of like, to a much lesser extent, asking how addition is useful or how you might use addition to solve a problem: they're both mathematical constructions with their own uses. I know that addition is far more useful and commonplace than the Lie bracket, but I hope you get my point. ~~ Dr Dec (Talk) ~~ 19:54, 6 November 2009 (UTC)[reply]
In addition, Lie brackets are a good way of calculating the Lie derivative of one vector field with respect to another one. Lie derivative can be intuitively understood as the speed of change of one vector field in the flow of the other one; this definition (as one of many) is formalised by the last formulae of this section. — Pt (T) 19:21, 9 November 2009 (UTC)[reply]

Transporting tensors on Lie groups

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I have a question about parallel transport of tensors on Lie groups. Thanks. —Ben FrantzDale (talk) 18:02, 5 November 2009 (UTC)[reply]

The exponential map gives a mapping from a Lie algebra to its Lie group. For the connexion between the two you might like to read this section of this article. Moreover, given your last question about Lie brackets, you might like to read this section of this article. Also, you might like to read this article. ~~ Dr Dec (Talk) ~~ 18:36, 5 November 2009 (UTC)[reply]
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