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It is well known that some problems in functional analysis and in general topology are independent from ZFC: to name a few, Kaplansky's conjecture, the existence of outer automorphisms of the Calkin algebra, the existence of a Suslin line etc. This is not too surprising, since many results in this fields require a nontrivial amount of set theory to be proved. On the other hand, differential geometry seems to be "higher up" in mathematical complexity (i.e., further away from set theoretical questions) and so it seems reasonable to me that no "natural" statement in diff. geom. is independent from ZFC.

Is that the case? That is, are there some statements which are independent from ZFC (or are conjectured to be)?

Edit: many comments (and the only, at the moment, answer) focused on computably undecidable problems. While this does indeed answer the question as formulated (because at least one instance of the problem must be independent), it is not exactly what I had in mind in that it does not offer an explicit example of an independent statement, or at least I don't see how to get one. So a more precise reformulation of my question is (but I am also very interested in more examples of computably indecidable problems): is there an explicit (in some sense of the word) statement in differential geometry that is (or is conjectured to be) independent from ZFC?

Pelota
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    Does "are these two 4-manifolds diffeomorphic" count as differential geometry? Andrei Markov Jr showed how to take two finite group presentations and build two 4-manifolds that are diffeomorphic iff the corresponding groups are isomorphic. It is possible to construct a finite presentation that presents the trivial group iff $\mathrm{Con}(\mathrm{ZFC})$ is true. By Gödel's second incompleteness theorem, it is therefore independent of ZFC if the two 4-manifolds are diffeomorphic. – Robert Furber Jan 25 '23 at 12:34
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    @RobertFurber I'd say it counts, although it feels a bit like cheating the question (in that one is basically "embedding" some group theory in diff. geom.) – Pelota Jan 25 '23 at 13:04
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    In a similar vein to Robert Furber's comment: Matt Foreman has does some work recently showing, for example, that there are diffeomorphisms of the torus such that the question of whether they one is isomorphic to its inverse (in the sense of dynamical systems) is undecidable in ZFC. He does this, roughly, by coding undecidable arithmetic statements into the problem of deciding whether such an isomorphism exists. – Will Brian Jan 25 '23 at 13:10
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    The problem of deciding whether a finite presentation presents the trivial is undecidable in the sense of the theory of Computation. That is I think different than being independent of ZFC. Unfortunately undecidable is used with two meanings in math. I don't think see why trivially is independent of ZFC. – Benjamin Steinberg Jan 25 '23 at 13:10
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    Actually I guess this answer https://math.stackexchange.com/questions/465189/does-an-undecidable-decision-problem-have-a-zfc-independent-instance says every Turing undecidable problem has a ZFC independent instance so I guess it is ok. – Benjamin Steinberg Jan 25 '23 at 13:14
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    @BenjaminSteinberg I thought someone might say that. ZFC has a recursively enumerable set of axioms, so we can make a Turing machine $T$ that enumerates all formal proofs starting from the ZFC axioms and halts if it reaches a proof of $0 = 1$. Then $\mathrm{Con}(\mathrm{ZFC})$ is equivalent to the halting problem for $T$. – Robert Furber Jan 25 '23 at 13:15
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    @BenjaminSteinberg We really use the halting problem, not just computational undecidability. There are undecidable problems that are weaker than the halting problem, and the proof wouldn't work with those. – Robert Furber Jan 25 '23 at 13:16
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    @RobertFurber, this seems to be roughly the idea of the link I gave above but I think you are arguing slightly backwards. If ZFC has a proof the group is trivial, you can enumerate the proofs and find it and if ZFC has a proof the group is nontrivial you can enumerate it and find it. So at least one finite group presentation has triviality independent of ZFC I guess assuming ZFC is consistent. – Benjamin Steinberg Jan 25 '23 at 13:19
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    @BenjaminSteinberg Every computably undecidable decision problem is saturated with logical undecidability, over any base theory, since otherwise we could solve the problem by searching for proofs. In this sense, any computably undecidable problem of differential geometry will provide an answer to the question, even if one uses much stronger theories than ZFC. – Joel David Hamkins Jan 25 '23 at 13:20
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    @JoelDavidHamkins, yes I realized that after my first comment when I Googled. – Benjamin Steinberg Jan 25 '23 at 13:21
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    I don't have an explicit example, but one place to look would be in the work of Nabutovsky and Weinberger. For example, in The fractal nature of Riem/Diff I, they show that the space of Riemannian metrics on a smooth manifold is extremely complicated. There are many uncomputable phenomena lurking inside. But possibly you would also regard this as "cheating" by embedding group theory into differential geometry. – Timothy Chow Jan 25 '23 at 13:28
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    @BenjaminSteinberg I will summarize my comments into an answer, but I'm not arguing "slightly backwards", I'm giving an argument that only assumes consistency of ZFC rather than soundness. – Robert Furber Jan 25 '23 at 19:07
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    Maybe of interest: Model and Set-Theoretic Aspects of Exotic Smoothness Structures on R – Mohammad Golshani Jan 26 '23 at 05:43
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    Not sure if this counts: https://doi.org/10.1017/bsl.2020.36 – Steve Huntsman Jan 26 '23 at 21:17
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    These undecidability problems are very explicit. There is a specific problem you can completely write down constructively. – Trebor Jan 27 '23 at 03:24

2 Answers2

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[Using the comments for context on undecidability/independence of ZFC]

A computably undecidable problem is whether or not a homology sphere has a metric of positive scalar curvature [Page 79 of Computers, Rigidity, and Moduli].

Steve Huntsman
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  • Great answer, thank you! I'll wait a few more days to see if somebody else has something to add before accepting. – Pelota Jan 26 '23 at 16:10
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An explicit construction from the undecidability problem is this. There is a computer program $C$ that can check whether a (completely formal) proof in ZFC is valid, i.e. whether each step is either an axiom or follows from previous steps by a set of logical rules. Proofs are simply finite length strings drawn from a finite alphabet, so there is a program $S$ that outputs every (valid or invalid) proof one by one, ordered by length and alphabetical order. Now use $C$ to filter out the invalid proofs. This gives you a program that outputs every valid proof. Program $H$ runs through every proof, and when it encounters a proof of $p \land \neg p$, it halts. All of these are very easy but tedious to write out, if you have some basic programming knowledge. Whether this program halts is obviously independent of ZFC.

There is a very constructive proof that constructs a finite presentation of a group, such that whether a particular element $w$ is equal to the identity is equivalent to some program halting. This is done by using a lot of generators to guard the boundaries of symbols, so that there is only one potential path of equational reasoning that the given element is the identity. This path mimics the execution of a program. This construction is explicit, and you can check all the construction details if you have enough time.

Finally, there are ways to construct a 4-manifold for any given finite presentation of a group, so that it is diffeomorphic to another 4-manifold if and only if adding the equation $w = 1$ gives an isomorphic group. This result is mentioned in the comments.

Trebor
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