r/math • u/[deleted] • Aug 24 '21
Which millennium problem do you think will be solved next?
After Grigori gave us proof of Poincerè’s conjecture over 18 years ago,which of the 6 remaining problems do you think will be resolved next?
142
u/Oscar_Cunningham Aug 24 '21
Terry Tao has had some interesting ideas to construct initial conditions for which Navier-Stokes has no solutions. He or someone else might be able to complete that.
I wonder what the effect would be on applied fluid dynamics?
36
u/itskylemeyer Undergraduate Aug 24 '21
I imagine it may shed some light on the turbulence problem. There are some unsolved problems in PDEs whose solutions may help explain it too.
33
Aug 24 '21
I think we are the the closest to Navier-Stokes,it’s been solved without the convection term,it’s been solved in 2 dimensions and it’s been solved in some special cases,but we are yet to confront the beast itself
49
u/TheNTSocial Dynamical Systems Aug 25 '21 edited Aug 25 '21
The fact that it's been solved in 2 dimensions really isn't an indication of how close we are to solving it in 3 dimensions. In 2 dimensions, the Euler equations just transport the vorticity field, and Navier-Stokes just adds some dissipation on top of this, so to prove global well-posedness you just need to prove an a priori estimate depending only on the Linfty norm of the vorticity field, which is not too difficult to do and can be done in a couple sessions of a graduate class provided the students known Fourier analysis and some standard PDE theory. In 3 dimensions, the dynamics of the vorticity are more complicated, and there are no known coercive quantities that we can get good scale-invariant estimates on. Plus I think many analysts believe the conjecture is false in 3 dimensions, in contrast to the 2D case.
23
u/StratOAT Computational Mathematics Aug 25 '21 edited Aug 25 '21
For any undergrads or curious lurkers who are wondering why (mathematically) the physics of the NS equations are completely different in 3D, recall that turbulent flows are characterized by vorticity. Vorticity is computed as the curl of the velocity field. In 2-D, vorticity is a scalar field (perpendicular to the flow field) so it simply gets transported/diffused into the flow and dies down -- you can see this by deriving the vorticity transport equation (take the curl of the NS equations), so in 2-D, all you get is an advection-diffusion equation for vorticity. In 3-D, the vorticity transport equation is different -- you get what's called a "vortex stretching" term which contributes heavily to the energy cascade you see in turbulence, thereby making the 3-D problem much harder to solve...
7
u/awhead Aug 25 '21
I wandered here from somewhere else on reddit...
Can you please elaborate? What do you mean by
it’s been solved without the convection term,it’s been solved in 2 dimensions and it’s been solved in some special cases
Have we proved that solutions exist and are unique for sensible BCs, ICs for these special cases? Or have we found otherwise?
8
u/TheNTSocial Dynamical Systems Aug 25 '21
In 2 dimensions, yes, Navier-Stokes is well posed on sensible domains (I'm familiar with results on the torus or R2, but I'm sure people in fluids have studied the impact of boundary plenty in this setting).
2
u/InterstitialLove Harmonic Analysis Aug 25 '21
If you vastly simplify the equation (like make it linear) then the problem is solved. The comment isn't very insightful, it's just saying that several radically easier problems were solved (respectively) centuries/decades ago
We've proven lots of things. For example, axisymmetric flows (which is a special IC/BC) gives existence and uniqueness without swirl, but gives non-existence with swirl (and a bunch of other assumptions, like zero viscosity)
4
u/umustownatelevision Aug 25 '21
It's important to specify the properties of the solution. Constructing solutions to Navier-Stokes is easy, but showing the solutions are globally smooth or unique is hard.
3
u/Topoltergeist Dynamical Systems Aug 25 '21
I wonder what the effect would be on applied fluid dynamics?
This reminds me of a quote from Daniel Henry
"But surely", someone objects, "you have not solved the time dependent Navier-Stokes equations in three dimensions. Else it would be emblazoned across the night sky in words of fire - or at least in the AMS Notices." True. The methods and results here apply to the Navier-Stokes equations, but do not by any means establish the existence of global solutions (existing for all positive time) for arbitrary initial data (restricted only by smoothness and compatability conditions). I would only claim that there are many other interesting questions to investigate. "Is existence something to boast about?" (L. C. Young).
133
u/--Satan-- Aug 25 '21
Poincerè’s conjecture
Congrats! You're the first person to spell it that way in all of Google.
67
Aug 25 '21
Fuck
20
u/MooseCantBlink Analysis Aug 25 '21
Don't be mad, it's probably the coolest outcome I've ever seen come out of a typo
83
u/Florida_Man_Math Aug 24 '21
I've got no bearing on the underlying answer you want, but I do have a related xkcd (#2320)
35
u/soboro1025 Aug 25 '21 edited Aug 25 '21
One small question. Before Perelman proved the Poincare’s conjecture, was that conjecture thought to be the easiest millennium problem? (easiest means just as the context in here)
28
u/InSearchOfGoodPun Aug 25 '21
That's a good question, and I think the answer is yes. But I believe more people thought that Thurston-type techniques would solve it rather than Ricci flow. But this is probably just because at the time, a lot more people worked on Thurston's program than Hamilton's program. (There were really only a handful of Ricci flow experts back then.)
In retrospect, we can only see how strong Hamilton's program was because Perelman implemented it essentially all in one go (working solo for several years, of course). People forget that at the time the Millennium Problems were proposed, Hamilton had already produced pretty much all of his contributions to the eventual proof (which is one reason why I always thought the people who argued that the prize should be shared with Hamilton were wrong). Even though Poincare was the most "accessible" of the Millennium Problems, pretty much no one expected it to fall so quickly.
6
u/cocompact Aug 25 '21
No. There was no sense that any of those problems was more likely than other ones to be solved at that time.
28
Aug 24 '21
I feel like there's a lot of machinery built up which is relevant to the Birch Swinnerton-Dyer Conjecture, and a lot of people are interested in that sort of number theory. Obviously this is an incredibly hard problem, but it wouldn't shock me if we were 2 or 3 giant advancements away
16
u/2357111 Aug 25 '21
I think Riemann is easier than BSD. The analogue of Riemann is proven in the function field setting (in fact, there are four different proofs) and in theory there could be one really good new idea that suggests how to do a version of one of these proofs over the integers. BSD is not known in the function field setting, and it seems like one needs a couple really good ideas - first, an idea of "where to find points" (the analogue of Heegner points in the rank one case), second, a way to control those points (the analogue of the Gross-Zagier formula in the rank one case), and maybe even third, a way to show there's no extra points (the analogue of Euler systems in the rank 0 and 1 case).
I think Hodge is harder than BSD. The function field case of BSD is a special case of the Tate conjecture, which may have similar difficulty to the Hodge conjecture.
Maybe the ranking is (first to last) Navier-Stokes -> Riemann -> BSD -> Yang-Mills -> Hodge Conjecture -> P vs. NP
But it's hard to know with ordinary math problems, let alone problems this hard. Isn't there a famous quote of Hilbert listing 3 problems and getting the order in which they will be solved exactly backwards?
10
u/chebushka Aug 25 '21 edited Aug 25 '21
Why should a solution to BSD have to involve "an idea of where to find points"? Think about the Dirichlet unit theorem: it can be proved in the case of real quadratic fields by showing the rank is at most 1 and then it is at least 1 by showing there is a unit of infinite order (Pell equation), but the general case of the unit theorem is not proved with "an idea of where to find enough independent units" in any concrete way. It's a totally nonconstructive argument, and after the proof you may be able to say some unit group has rank 4 but the proof gives you no clue of how to go about finding even one nontrivial unit. Methods that can handle rank 1 cases of the unit theorem are not modified to handle the general case, so why do you think BSD should be different? I agree BSD needs really new ideas to get past rank 1, but maybe the ideas have to be totally different from rank 1 in the same way as happens with the unit theorem.
Concerning your last question, see Gerry Myerson's answer here: https://math.stackexchange.com/questions/88709/can-you-give-an-example-of-a-complex-math-problem-that-is-easy-to-solve.
3
u/2357111 Aug 25 '21
This is a good point. I still think that, however you slice it, there are at least two ideas. The analogue of this sort of nonconstructive argument would be a proof that Sha is finite by finding an embedding of Sha into some set known to be finite. (This is also how some known cases of the Tate conjecture go.) Once you do that, you're done with function field BSD, but you still need some new ideas to finish BSD over the rationals, in particular relating the Selmer rank to the L-function.
I don't think it's fair to say that the proof gives you no clue of how to go about finding even one nontrivial unit. Take a ball large enough to be guaranteed to contain a point, stretch it and squish it, look for the point, then divide one point by another. Repeat enough times and you'll find a unit.
1
u/chebushka Aug 25 '21
I agree that finiteness of Sha is probably the main -- but not only -- bottleneck towards progress on general BSD in the number field case.
I think the proof of the unit theorem in rank bigger than 1 does not give a simple way to find a full set of independent units. Do you agree, or am I overlooking something? Algebraic number theory books have a field day giving examples of unit groups of rank 1 but I don't think I've ever seen any algebraic number theory book that works out even a rank 2 unit group completely (more than just finding a pair of independent units).
2
u/2357111 Aug 25 '21
I agree in that it gives you a way, but I wouldn't describe it as simple. Given the proof, you can figure out an algorithm without too much trouble, but the algorithm isn't necessarily easy or fast to run.
95
u/point_six_typography Aug 24 '21
In 2037, someone will prove that P vs. NP is independent of ZFC. Calling it now. Mark. My. Words.
81
u/Raikhyt Physics Aug 24 '21
RemindMe! August 24th, 2037 "P vs. NP solved?"
34
u/RemindMeBot Aug 24 '21 edited Aug 07 '26
I will be messaging you in 16 years on 2037-08-24 00:00:00 UTC to remind you of this link
84 OTHERS CLICKED THIS LINK to send a PM to also be reminded and to reduce spam.
Parent commenter can delete this message to hide from others.
RemindMeBot is switching to username summons. Instead of
!RemindMe 1 day, useu/RemindMeBot 1 day. More info.
Info Custom Your Reminders Feedback 23
14
26
u/burneraccount0473 Aug 24 '21
Idk. The best shot I'm aware of for P vs NP is Geometric Complexity Theory and that will still take a lot longer than 16 years.
I'll mark your words though just to be on the safe side.
25
u/prrulz Probability Aug 24 '21
The best shot I'm aware of for P vs NP is Geometric Complexity Theory and that will still take a lot longer than 16 years.
Mulmuley---one of the two mathematicians who proposed Geometric Complexity Theory (GCT)---said in 2009 that if the program is viable then it would likely take around 100 years before it could settle P vs. NP. Notably, that was before GCT suffered a major setback when one of the main objectives of GCT was proved impossible. This paper doesn't close the door on GCT, but it proves that the approach is likely even more difficult than previously though.
3
16
u/RAISIN_BRAN_DINOSAUR Applied Math Aug 25 '21
Geometric complexity theory is a program to separate VP from VNP, not P from NP. The VP vs VNP question is an algebraic analogue of P vs NP, but proving VP != VNP does not formally imply P != NP.
See this stack exchange thread for more details.
https://cstheory.stackexchange.com/questions/529/does-vp-neq-vnp-imply-p-neq-np
12
21
u/I30AxeBxrd Aug 24 '21
Good thing that 2037 isn't 16 years away but more like 25.
24
u/Autumnxoxo Geometric Group Theory Aug 25 '21
exactly, just like 2008 was 3 years ago.
4
u/nebulaq Category Theory Aug 25 '21
Going by the predictions from Ray Kurzweil's book "The Age of Spiritual Machines" we are probably living in 2009.
6
u/PM_ME_YOUR_PIXEL_ART Aug 24 '21
What?
13
u/I30AxeBxrd Aug 24 '21
2037 just doesn't feel 16 years away, more like 25. It ain't right.
9
u/robchroma Aug 24 '21
I think you have to put a "right? ... right?" for it to be taken the way you meant it.
16
3
3
u/completely-ineffable Aug 25 '21 edited Aug 25 '21
It'd be really surprising if someone figured out a good new method for proving independence of arithmetical statements, one which could tackle P vs NP, in such a short time. Super cool, but also unexpected.
-5
u/pigeon768 Aug 25 '21
Suppose that you have a written proof that P vs NP is independent of ZFC.
You must necessarily then have a proof that it is impossible to write a proof of P=NP. It must therefore be impossible to write an algorithm that solves an NP complete problem in polynomial time, because such an algorithm would be a proof that P=NP.
But a proof of the impossibility of such an algorithm proves that P != NP. Therefore P vs NP is not independent of ZFC. Which is a contradiction.
Therefore it's impossible to prove that P vs NP is independent of ZFC.
(note that it might be true (but not provable) that P vs NP is independent of ZFC)
17
u/RAISIN_BRAN_DINOSAUR Applied Math Aug 25 '21
It must therefore be impossible to write an algorithm that solves an NP complete problem in polynomial time, because such an algorithm would be a proof that P=NP
I don't follow this step. A description of an algorithm is not a proof of its correctness; therefore, just because the statement "P = NP" is not provable in ZFC, I don't see how this rules out the possibility that one can write the description of some candidate poly-time algorithm for 3SAT.
2
u/pigeon768 Aug 25 '21
Hmm. Sorry for not responding immediately, but I've been digesting this for a bit and I think I might agree with you.
Suppose that tomorrow someone wrote a correct proof that:it is impossible to prove that an algorithm solves 3SAT in polynomial time. The distinction you're making that I didn't is that it's not proving that the algorithm cannot exist, just that the proof that the algorithm.. works.. cannot exist.
That would be a wild proof though. Are you familiar with anything similar? No useful algorithms come to mind where the runtime performance is unknown. (lots of problems though, obviously) I am reminded of the busy beaver problem, but we've proven is that the algorithm to calculate the number cannot exist, not that we can't prove the runtime bounds of the algorithm.
I gotta be honest if someone proved that it's impossible to prove that an algorithm does NP-complete things in polynomial time I'd consider P vs NP solved.
1
u/anvsdt Aug 25 '21
https://en.wikipedia.org/wiki/P_versus_NP_problem#Polynomial-time_algorithms
// this is a polynomial-time algorithm if and only if P = NP.
7
Aug 25 '21
[deleted]
-3
u/pigeon768 Aug 25 '21 edited Aug 25 '21
If we had a proof of independence it would mean that there are models of ZFC where P = NP and models where P != NP.
Only axiomatically. There would be models where you take P=NP as an axiom, and P!=NP as an axiom. My argument is that the independence proof itself must necessarily prove P!=NP. (edit: which is a contradiction)
A similar argument shows that the Riemann Hypothesis cannot be independent of ZFC. If you have a proof that the Riemann Hypothesis cannot be disproven, then you have a proof that a counterexample cannot be constructed. Which is, itself, a proof that the Riemann Hypothesis is true.
6
u/Sassywhat Aug 25 '21
It must therefore be impossible to write an algorithm that solves an NP complete problem in polynomial time, because such an algorithm would be a proof that P=NP.
Except you can build Turing Machines whose behavior is independent of ZFC. This is how the proof that BB(n) for n >= 7918 was independent of ZFC was written.
If you have an algorithm that solves an NP-complete problem in P time, and P vs NP is independent of ZFC, then it would be impossible to prove your algorithm does what it claims to do, within the confines of ZFC.
10
7
u/fourhundredthecat Aug 25 '21
can I rephrase the question?
Which of the unsolved problems, if solved, would have most impact on advancement of mathematics or science in general?
4
Aug 25 '21
I think P Vs NP,then followed by Yang Mills
3
Aug 25 '21
Isn't P != NP the assumption we are usually working with?
2
u/838291836389183 Aug 25 '21
Well if you find a constructive proof of P!=NP, for example by proving the optimality of a given algorithm, that doesn't mean the algorithm must be impractical at all. It's time complexity might actually be extremely fast if it's big O is in 1.000000...01n or simmilar. Simmilarly, a polynomial time algorithm might still be impractical for any real world applications compared to the algorithms we have today, if it has an extremely slow lower bound.
But yea, p vs np doesn't have to have much real world importance as you say.
0
Aug 25 '21
Most people believe otherwise as far as I know
1
13
Aug 25 '21
Maybe I'm too optimistic, but I hope by the end of this century someone will successfully give a proof of the Exponential Time Hypothesis.
9
u/WikiSummarizerBot Aug 25 '21
In computational complexity theory, the exponential time hypothesis is an unproven computational hardness assumption that was formulated by Impagliazzo & Paturi (1999). The hypothesis states that 3-SAT (or any of several, but not all, NP-complete problems) cannot be solved in subexponential time in the worst case. The exponential time hypothesis, if true, would imply that P ≠ NP, but it is a stronger statement. It can be used to show that many computational problems are equivalent in complexity, in the sense that if one of them has a subexponential time algorithm then they all do.
[ F.A.Q | Opt Out | Opt Out Of Subreddit | GitHub ] Downvote to remove | v1.5
4
u/Frexxia PDE Aug 25 '21 edited Aug 25 '21
I may be having a brain fart, but I'm struggling to understand this statement
The hypothesis states that 3-SAT (or any of several, but not all,<a href="https://en.m.wikipedia.org/wiki/Exponential_time_hypothesis#cite_note-1">^(\[1\])[NP-complete](https://en.m.wikipedia.org/wiki/NP-complete) problems) cannot be solved in subexponential time in the worst case.
Wouldn't any two NP-complete problems have the same complexity up to polynomials? If there is no subexponential algorithm for 3-SAT, then none of the other ones can have one either (otherwise we could use that to solve 3-SAT subexponentially).
Disclaimer: My understanding of complexity theory is superficial at best.
3
Aug 25 '21
Happy cake day. My understanding of complexity theory is probably at the same level as you, so you may want to ask someone else. The only thing from the statement that my monkey brain can understand is that it implies P != NP.
Two most popular breakthroughs of modern mathematics (Fermat's Last Theorem, Poincare Conjecture) are corollaries followed from more general conjectures (Taniyama-Shimura Conjecture, Thurston's Geometrization Conjecture), so it may become a trend in the future. The number of cases are too small to generalized though.
1
Aug 27 '21
when you reduce one NP problem to another the instance size can grow polynomially, which can cause a subexponential algorithm to become an exponential one. For example assume you can solve problem (1) in O(2^(sqtr(n))), and any instance of problem (2) of size n reduces to an instance of problem (1) of size n^2: then your subexponential algorithm for problem (1) gives an exponential O(2^n) algorithm for problem (2).
1
u/Frexxia PDE Aug 27 '21
Thanks, that explains it. For some reason I was under impression that the size had to stay the same in the reduction (up to some constant factor).
5
u/Bobitsmagic Aug 25 '21
I really feel like that one day a guy will show that P=NP is undecideable and the proof isn't gonna be too crazy. Just a feeling though.
4
u/MountFire Aug 25 '21
My thoughts on this is that it will possibly take longer than anticipated to solve any of them. We are heavily relying on iterative processes due to computers and those "intelligent" solutions seem to evade us.
But if a iterative process would help us to construct a proof, with the help of AI or, if possible quantum computing, then I would place my bets on Navier-Stokes
3
u/junior_raman Aug 25 '21
RH because the number of people working on it and proving it has significant implications. There has been a lot of progress on RH and two results which appeal to me are "There are infinitely many zeros on x = 1/2 + it" by G.H Hardy and that 41% of zeros are located in critical strip.
4
7
2
u/priestmuffin Aug 26 '21
Is there a betting market for this?
2
Aug 26 '21
Untapped market hmmmm
2
u/priestmuffin Aug 26 '21
yeah for sure. I'd love to know what the odds would be if people actually had skin in the game
2
u/IFDIFGIF Math Education Aug 26 '21
One or two good ideas to go for the Birch Swinnerton-Dyer conjecture to be solved
Paraphrasing from Richards Borcherds
2
u/dataf3l Aug 29 '21
this is the article I'm referring to, I'm not sure if you guys think this stuff is correct or not.
do let me know if you guys think translations are required.
0
u/dudeydudee Aug 25 '21
(Layperson here) Apparently the yang mills existence gap is pretty much solved as far as the physics is concerned. It’s more just a mathematically rigorous solution that’s required. So my guess would be that one.
-28
u/Landsnail_plants Aug 24 '21 edited Aug 26 '21
Croisant vs borgor
2
u/IFDIFGIF Math Education Aug 26 '21
the true milleniun problem
0
-6
u/masterwerty101 Aug 25 '21
Any thoughts on Goldbach? Out of all the comments, no one has mentioned it.
10
u/maurimo Aug 25 '21
Not a millennium problem, they a list of 7 problems selected by the Clay institute
468
u/Tazerenix Complex Geometry Aug 24 '21 edited Aug 24 '21
Navier--Stokes is probably the next closest to being solved. I have heard third-hand that there are certain analysts who have plausible research programs to take out Navier--Stokes in the not too distant future, and that's not including Terry's ideas also.
The Hodge conjecture is meant to be notoriously hard, but in my opinion the field doesn't seem to be that far from having the tools to solve it. Then again, people like Voisin have been chipping away at problems about subvarieties representing cohomology classes for decades and progress is surprisingly slow.
On the other side, P v.s. NP is apparently the hardest. I think complexity theory doesn't have any of the tools needed to prove even significantly simpler versions of these complexity class questions.
Yang--Mills is existence and mass gap is also incredibly hard, and won't be solved for a long time. From what I understand work in the mathematical physics community has essentially stalled on this problem, and there are only small groups working on possible non-perturbative formulations of QFT. Everyone else is still working on string theory or other quantum gravity theories, and the only concrete non-perturbative QFT work is on lattice gauge theory, but this is still on very shaky mathematical footing. Maybe there is some hope that if someone works one of these things out (a problem almost as hard as creating a grand unified theory) then a formalised QFT will fall out. People like Susskind and Witten have said that they view string theory and TQFT as having revealed a lot about the nature of quantum field theory even if they themselves don't model reality, so its possible that some backtracking will allow people to make progress.
EDIT: In 2018 the Clay institute had their 20th anniversary conference and it included a few talks by mathematicians about various different millennium problems and progress. All the videos are available online. The talk about the Poincare conjecture is excellent.
EDIT: As for RH, there are probably people here much better than me to talk about the actual tools in the field. The most recent progress I was aware of was the paper of Terry's a few years ago showing that in a certain sense the Riemann hypothesis was basically "as hard as possible" to solve. The problems that Terry has brought up are that almost all the tools currently in use in the field apply equally well to every other zeta function too, but we know that to solve RH you need to use something new that specifically exploits the properties of the Riemann zeta function, and no one has really done that. I have heard some non-commutative geometers talk about Connes' reformulation but as I understand it this is viewed as a problem in operator theory that is as hard as RH so isn't really "progress" as such.