The most outrageous (or ridiculous) conjectures in mathematics

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84

The purpose of this question is to collect the most outrageous (or ridiculous) conjectures in mathematics.

An outrageous conjecture is qualified ONLY if:

1) It is most likely false

(Being hopeless is NOT enough.)

2) It is not known to be false

3) It was published or made publicly before 2006.

4) It is Important:

(It is based on some appealing heuristic or idea; refuting it will be important etc.)

5) IT IS NOT just the negation of a famous commonly believed conjecture.

As always with big list problems please make one conjecture per answer. (I am not sure this is really a big list question, since I am not aware of many such outrageous conjectures. I am aware of one wonderful example that I hope to post as an answer in a couple of weeks.)

Very important examples where the conjecture was believed as false when it was made but this is no longer the consensus may also qualify!

Shmuel Weinberger described various types of mathematical conjectures. And the type of conjectures the question proposes to collect is of the kind:

On other times, I have conjectured to lay down the gauntlet: “See,

you can’t even disprove this ridiculous idea."

Summary of answers (updated March, 13, 2017):

  1. Berkeley Cardinals exist

  2. There are at least as many primes between $2$ to $n+1$ as there are between $k$ to $n+k-1$

  3. P=NP

  4. A super exact (too good to be true) estimate for the number of twin primes below $n$.

  5. Peano Arithmetic is inconsistent.

  6. The set of prime differences has intermediate Turing degree.

  7. Vopěnka's principle.

  8. Siegel zeros exist.

  9. All rationally connected varieties are unirational.

  10. Hall's original conjecture (number theory).

  11. Siegel's disk exists.

  12. The telescope conjecture in homotopy theory.

  13. Tarski's monster do not exist (settled by Olshanski)

  14. All zeros of the Riemann zeta functions have rational imaginary part.

  15. The Lusternik-Schnirelmann category of $Sp(n)$ equals $2n-1$.

  16. The finitistic dimension conjecture for finite dimensional algebras.

  17. The implicit graph conjecture (graph theory, theory of computing)

  18. $e+\pi$ is rational.

  19. Zeeman's collapsing conjecture.

(From comments, incomplete list) 20. The Jacobian conjecture; 21. The Berman–Hartmanis conjecture 21. The conjecture that all groups are Sofic; 22 The Casas-Alvero conjecture 23. An implausible embedding into $L$ (set theory). 24. There is a gap of at most $\log n$ between threshold and expectation threshold. 25. NEXP-complete problems are solvable by logarithmic depth, polynomial-size circuits consisting entirely of mod 6 gates. 26. Fermat had a marvelous proof for Fermat's last theorem. (History of mathematics).

Gil Kalai

Posted 2017-01-17T19:55:08.770

Reputation: 13 365

2Does the Jacobian conjecture qualify? – Steve Huntsman – 2017-01-17T20:03:55.810

Hmm, I think it qualifies as a comment but not as an answer. For an actual answer I would like "most likely false" to represent a large consensus and not a personal view of the answerer. But once I asked the question my view about what qualifies is just one view in the crowd... – Gil Kalai – 2017-01-17T20:12:03.793

2The Berman–Hartmanis conjecture. – T. Amdeberhan – 2017-01-17T20:20:12.753

"In this paper we try to convince the leader that there is no good reason to believe that the Jacobian Conjecture holds. Although there are several arguments in favor of this conjecture, we show that these arguments haven't got the power to justify the statement that the Jacobian Conjecture holds in general." -van den Essen (1997): http://www.seminariomatematico.unito.it/rendiconti/cartaceo/55-4/283.pdf

– Steve Huntsman – 2017-01-17T20:20:38.477

Right! I prefer examples where it is still now commonly believed that the conjecture is false and where the proposer proposes the conjecture genuinely suggesting that it is true. But these two requirements may be too harsh. – Gil Kalai – 2017-01-17T20:22:40.550

5There is a fine line between an outrageous conjecture and a bold conjecture. But still I see the spirit of your interesting question. – Joseph O'Rourke – 2017-01-17T20:25:08.323

1It seems to me there is a conflict between "you can't even disprove this ridiculous idea" and "the proposer proposes the conjecture genuinely suggesting that it is true", so it's not clear to me what exactly you're after. – Gerry Myerson – 2017-01-17T21:59:02.647

1

Incidentally, my question at http://mathoverflow.net/q/101821/1946 was asked in the spirit of this question (but it is too recent to qualify for your 2006 requirement).

– Joel David Hamkins – 2017-01-17T22:01:05.327

9I don't think anyone has disproved the ridiculous ideas that there are only finitely many Mersenne composites, or that all the decimal digits of $\pi$ from some point on are sixes and sevens, or that the partial quotients for continued fractions of real algebraic irrationals are always bounded, but I don't think anyone has proposed any of these ideas genuinely suggesting they are true. – Gerry Myerson – 2017-01-17T22:19:13.167

I don't understand what you mean by point 5. By definition, the negation of any commonly-believed-false statement is a commonly-believed-true statement, isn't it? – Federico Poloni – 2017-01-17T22:32:11.130

19The answers below all look of interest, as does the question. And, to boot, this is Community Wiki. Why not keep it open? – Lucia – 2017-01-17T22:49:23.773

1It depends on the utility of this question. As a short term diversion to appeal to some of the forum community it serves quite well. As part of a database of questions and answers for future reference by the interested-in-mathematics consumer, I think it is too based in opinion and belongs on a blog. If the intent were to set some challenge questions to spur research, then I think the question should be reworded. As it stands now, it isn't much better than an opinion poll. Gerhard "MathOverflow Isn't Question And Opinion" Paseman, 2017.01.17. – Gerhard Paseman – 2017-01-17T23:02:19.833

5@GerhardPaseman I think it is not too uncommon for good mathematicians, working in or near an area, to nevertheless not know about various conjectures. Especially if there has been recent development in tangentially-related areas, this type of list very well might lead to some of these conjectures being refuted. I support keeping the question open. – Theo Johnson-Freyd – 2017-01-17T23:36:34.360

1And besides, I will learn things from reading it! – Theo Johnson-Freyd – 2017-01-17T23:36:51.103

@Theo, in which case, let's rewrite the question to fit both a good intent of the asker and the good intent of MathOverflow. As it is currently written, I am not sure either is achieved. Gerhard "Being Ridiculous Can Serve Research" Paseman, 2017.01.17. – Gerhard Paseman – 2017-01-18T00:30:51.203

1

I feel like a number of famous, "elementary" conjectures, while often believed to be true, have no particular (philosophical) reason to be true, and thus from a cynical perspective might be described as likely false. Examples are the Collatz conjecture (https://en.wikipedia.org/wiki/Collatz_conjecture) and the union-closed sets conjecture (https://en.wikipedia.org/wiki/Union-closed_sets_conjecture)

– Sam Hopkins – 2017-01-18T04:56:04.153

5What's next - a big list with Trump tweets concerning mathematics? Does this outrageous conjecture of mine count as an example? – Franz Lemmermeyer – 2017-01-18T05:39:21.590

@Sam, I don't know about "philosophical", but I think there are good mathematical reasons for Collatz to be true. – Gerry Myerson – 2017-01-18T05:39:27.290

1I agree with Theo on that. E. g. Socrates was, I think, The asker number one. The way I understand it, the question is about conjectures of Socratic quality. If the things Socrates asked would be only pedagogic challenges and not the things he really burningly wanted to know, he would not be Socrates. And recall what happened to him. By the way there is a Socratic badge here on MO. A golden one ;) – მამუკა ჯიბლაძე – 2017-01-18T05:40:36.183

2This one doesn't count because it status is settled, but from what I understand of the history of mirror symmetry, when the physicists first proposed it, Yau for one initially thought that it was too outrageous to be true. Along similar lines, I believe that Tao initially thought that the phenomenon of compressed sensing couldn't possibly be true, and that the Candes-Romberg-Tao paper was born out of Tao's attempts to find a disproof. – Timothy Chow – 2017-01-18T21:59:46.077

1@TimothyChow The Tao example is a really good one (and with ample documentation), even if it doesn't meet all the conditions of the OP. – Todd Trimble – 2017-01-19T02:36:27.440

1The Kahn-Kalai conjecture (the general one for Boolean functions) almost qualifies here, doesn't it – user36212 – 2017-01-19T23:58:28.360

1Not an expert on this subject, but "all groups are sofic" might potentially qualify here. – Terry Tao – 2017-01-21T22:59:49.977

I'm far from being an expert on this but perhaps Casas-Alvero conjecture is relevant. https://en.wikipedia.org/wiki/Casas-Alvero_conjecture . Although the most outrageous aspect of it is probably the date it was first conjectured (2001 !!!).

– Saal Hardali – 2017-01-21T23:40:48.920

Dear user36212, yes we both regards the conjectures (both for Boolean functions and for graphs) as fairly outrageous and would be very interested in counterexamples. Dear Timothy and Todd, these are good examples and I am certainly not too fussed about meeting all my conditions. – Gil Kalai – 2017-01-22T10:42:01.263

1@GilKalai : It's sort of too late now, but I wonder if another (and possibly better) way to phrase your question would be, what conjectures were regarded as impossibly bold or optimistic at the time they were first made? This would allow for both conjectures that have been settled and conjectures that are still open. It would also allow for bold conjectures that we've gotten used to but that were considered outrageous at first. And it would eliminate statements that nobody has ever believed. – Timothy Chow – 2017-01-22T21:26:23.793

This question reached the sidebar and as such has received a large influx of people who usually don't visit this site. Many conjectures in here are described in ways only mathematicians can understand them. I was wondering if the people who posted an answer already could clarify it in a way that people who aren't well versed in math could understand the conjecture? – Nzall – 2017-01-24T10:19:56.990

Dear @Nzall , what does it mean to reach the sidebar? – Gil Kalai – 2017-01-24T11:15:17.890

@GilKalai It means that the question has had enough activity on this site to be viewed as a "hot network question", which in turn means that it's eligible to appear in the list of questions which appear in the sidebar, beneath the meta highlights and the linked questions. This list is effectively a "best of Stack Exchange", and a lot of people tend to check out the questions on that bar, especially if the title is interesting or (like this question) clickbaity. It means that a lot of users from other exchanges on the network will look at the question and potentially give their input. – Nzall – 2017-01-24T11:26:56.900

@GilKalai The biggest consequence is that the question gets a lot of new readers, votes, comments and potentially answers, many of which haven't had much more than high school or maybe first grade college education on the topic. Making the topic a bit more understandable for those users may introduce them to aspects of mathematics they didn't know exist and may encourage them to look further on the site. – Nzall – 2017-01-24T11:31:34.520

@Nzall, certainly it could be a good idea to add elementary explanations for at least some of the answers. It is not so easy but worth the effort. – Gil Kalai – 2017-01-24T13:37:30.823

Dear Terry and Saal, indeed "all groups are sofic" is a famous conjecture which might be suitable. The Casas-Alvero conjecture strikes me as a good example as well but I dont know anything about it. – Gil Kalai – 2017-01-24T16:24:48.457

Why 2006? It seems rather arbitrary. Perhaps instead "is at least 10 years old" which will allow for more recent "outrageous" answers as time goes on. – Damien – 2017-01-25T03:36:53.823

Hi Damien, I agree with that. I also agree with Timothy's comment regarding unsettled conjectures. I am a little worried that adding settled conjectures would have made the question too board (but those can be mentioned in comments and also in other questions.) – Gil Kalai – 2017-01-25T10:09:29.913

2[Disclaimer: I haven't really done any serious maths since my degree many moons ago, and yes, I arrived here via the sidebar.] I'm surprised noone mentioned Fermat's famous "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain" quote. I guess it's not technically a conjecture, but there's an implicit conjecture in there that there exists a marvellous proof which would have been attainable before 1621, and this fits all the criteria of the question, as well as being presumably in keeping with the spirit of the question. – Adam Spiers – 2017-01-26T16:09:34.423

Adam, this was indeed a conjecture but it does not fit the question because: a) It was settled (point 2), b) it was not believed to be false (point 1). But I agree that beside the formalities Fermat's conjecture and Fermat's narrow margin claim were quite outrageous! – Gil Kalai – 2017-01-26T17:28:27.913

3@GilKalai No, you misunderstood. I was proposing Fermat's quote about his Last Theorem as an answer to the question, not the Theorem itself. The implicit conjecture that there exists a marvellous proof which would have been attainable before 1621 is not settled (point 2), and IIUC is widely believed to be false. – Adam Spiers – 2017-01-29T16:06:05.537

1Hmm, I see. This is indeed an outrageous conjecture about mathematics and its history :) – Gil Kalai – 2017-01-29T18:55:58.093

@FranzLemmermeyer I recently made a conjecture that if time goes to infinity trump will make a misspelling on twitter at some point and type abelian instead of a billion. The legend goes that this already happend and that this caused all multiplications at quantum level to be non-commutative, making the universe ridiculously hard to understand. On the other hand there are people thinking that trump tweeting about math is fake news. – M.D. – 2018-07-29T23:47:58.883

Answers

104

W. Hugh Woodin, at a 1992 seminar in Berkeley at which I was present, proposed a new and ridiculously strong large cardinal concept, now called the Berkeley cardinals, and challenged the seminar audience to refute their existence.

He ridiculed the cardinals as overly strong, stronger than Reinhardt cardinals, and proposed them in a "Refute this!" manner that seems to be in exactly the spirit of your question.

Meanwhile, no-one has yet succeeded in refuting the Berkeley cardinals.

Joel David Hamkins

Posted 2017-01-17T19:55:08.770

Reputation: 162 854

6Yes, this is the spirit of the question! – Gil Kalai – 2017-01-17T20:18:14.463

2Dangit, beat me to it! +1. (Need to get my typing speed up!) – Noah Schweber – 2017-01-17T20:25:29.843

1On the other hand, the algebras of elementary embeddings have barely been investigated, and the algebras of elementary embeddings seem like a good spot to obtain an inconsistency (hopefully at a level inconsistent with AC). We know that the algebras generated by a single elementary embedding look like, but barely anything is known about algebras generated by multiple elementary embeddings. I would wait until we better understand the algebras of elementary embeddings before declaring this problem as difficult. – 35093731895230467514051 – 2017-01-17T23:07:34.457

6@JosephVanName Well, it has been an open question for 25 years, and I know of some very smart people who have worked on it. So I think it qualifies as "difficult". But meanwhile, there is a current resurgence of interest in these very strong ZF large cardinals, and so a resolution may be close. Please go for it! – Joel David Hamkins – 2017-01-17T23:10:44.770

1@Joel David Hamkins. I will put the consistency of large cardinals beyond AC on my eventual to-do-list. – 35093731895230467514051 – 2017-01-18T00:34:06.203

2Johnson answered [by] striking his foot with mighty force against a large stone, till he rebounded from it -- "I refute it thus." – Max – 2017-01-19T10:05:58.147

The Berkeley cardinal may or may not exist, but the Stanford Cardinal has won prizes. https://en.wikipedia.org/wiki/Stanford_Cardinal

– Gerry Myerson – 2018-06-16T00:28:46.103

97

A long-standing conjecture in Number Theory is that for each positive integer $n$ there is no stretch of $n$ consecutive integers containing more primes than the stretch from 2 to $n+1$. Just looking at a table of primes and seeing how they thin out is enough to make the conjecture plausible.

But Hensley and Richards (Primes in intervals, Acta Arith 25 (1973/74) 375-391, MR0396440) proved that this conjecture is incompatible with an equally long-standing conjecture, the prime $k$-tuples conjecture.

The current consensus, I believe, is that prime $k$-tuples is true, while the first conjecture is false (but not proved to be false).

Gerry Myerson

Posted 2017-01-17T19:55:08.770

Reputation: 30 072

1Does this conjecture have a name? – Brevan Ellefsen – 2017-01-17T23:04:12.507

5I've seen it referred to as "the Hardy-Littlewood convexity conjecture". See Section 1.2.4 of the Crandall and Pomerance book, Prime Numbers: A Computational Perspective. – Gerry Myerson – 2017-01-17T23:14:42.290

What makes the consensus favor prime $k$-tuples over the Hardy-Littlewood convexity conjecture? Intuitively the former seems much less plausible than the latter. – orlp – 2017-01-18T04:08:04.693

4@orlp, there are heuristic arguments that not only do prime $k$-tuples exist but there are asymptotic formulas for how many of them there are up to any given bound, and these asymptotic formulas are always in close agreement with the numerical evidence. I don't think there are any such arguments for convexity, so it's gotta go. – Gerry Myerson – 2017-01-18T05:36:33.913

12

This is nowadays known as the second Hardy-Littlewood conjecture.

– Wojowu – 2017-01-18T13:46:54.567

Is it the case that Hardy-Littlewood really considered both the k-tuples conjecture and the convexity conjecture to be plausible? Makes me wonder what mutually implausible conjectures are also held to be believed nowadays (2017). – Mark S – 2017-01-19T00:53:09.197

1http://www.opertech.com/primes/k-tuples.html shows some numerical details. There might be a counter example to the conjecture before 10^1197 - trying to find it with current methods is quite hopeless. @MarkS Quite possible they found both conjectures plausible at the same time; a k-tuple contradicting it is quite hard to find. – gnasher729 – 2017-01-21T23:03:25.757

1And I think this conjecture being called "the second Hardy-Littlewood conjecture" - which means that if the first Hardy-Littlewood conjecture is true, the second one is false :-) – gnasher729 – 2017-01-21T23:05:06.603

Godel proved there must be unproveable true statements. One mark of such might be that A and B are both plausible, neither leads to anything outright ridiculous, but that it is proven that A implies not-B, and vice versa. – nigel222 – 2017-01-24T11:57:45.763

2@nigel, if 2nd Hardy-Littlewood is false, then there is a counterexample, so it's not unproveable. – Gerry Myerson – 2017-01-24T21:54:34.847

78

$P=NP$

Let me tick the list:

  1. Most likely false, because, as Scott Aaronson said "If $P = NP$, then the world would be a profoundly different place than we usually assume it to be."

  2. Yes, it's The Open Problem in computational complexity theory

  3. Yes, it's old

  4. It's important, again quoting Scott: "[because if it were true], there would be no special value in "creative leaps," no fundamental gap between solving a problem and recognizing the solution once it's found. Everyone who could appreciate a symphony would be Mozart; everyone who could follow a step-by-step argument would be Gauss..."

  5. It's an equality rather than a negation

Carlo Beenakker

Posted 2017-01-17T19:55:08.770

Reputation: 71 729

11Dear Carlo, I wanted to avoid "just the negation of a famous commonly believed conjecture" . (So while an equality it violates my condition 5 as I saw it :) ) But I still +1 it. – Gil Kalai – 2017-01-17T20:08:25.777

4Was it ever believed to be true I wonder? I get the impression that nowadays you are hard pressed to find anyone who believes it. Some might hence argue that it is actually the negation of a well-known conjecture, namely that P isn't NP. (posted before I saw Gil's comment) – Kevin Buzzard – 2017-01-17T20:08:27.390

12

@KevinBuzzard -- "In a 2002 poll of 100 researchers, 9 believed the answer to be yes" (from Wikipedia, with this link to the poll)

– Carlo Beenakker – 2017-01-17T20:09:21.613

2@KevinBuzzard: Dick Lipton's "P = NP" blog is broadly devoted (or at least, a frequent point of discussion) to speculating that this is true. – Daniel R. Collins – 2017-01-17T22:35:24.237

28I am downvoting in disagreement with Aaronson's hyperbolic comments. If every NP algorithm is in P but with much larger time bounds, then the consequences in 4 don't follow. And the world often is profoundly different than people assume: I don't want to declare assumptions likely simply because they're strongly held. – Matt F. – 2017-01-17T22:51:42.493

2

@MattF. -- in all fairness to Scott Aaronson, I took just one of his 10 reasons to believe $P!=NP$, and he calls this particular reason the "philosophical argument"; there are 9 more.

– Carlo Beenakker – 2017-01-17T22:55:05.697

4Without taking a stand on P vs. NP, I support the hyperbolic comments. They seem to be right on, in the metaphorical sense in which they are offered. The consequences in 4 are meant for those who consider the truly large numbers. If one is small-minded, one might miss the essence of the natural numbers. – Joel David Hamkins – 2017-01-17T23:24:46.017

Indeed, if the answer indicated ideas from Aaronson's reasons 3,6,7 instead I would probably upvote it. – Matt F. – 2017-01-17T23:41:47.297

23Fails #5. It's the negation of the almost-certainly-true $P \neq NP$ – R.. – 2017-01-18T03:12:42.747

@CarloBeenakker Thank God science and mathematics aren't Democratic. – bd1251252 – 2017-01-18T06:33:17.357

17There are much better examples from complexity theory if we allow this kind of thing. For example: "NEXP-complete problems are solvable by logarithmic depth, polynomial-size circuits consisting entirely of mod 6 gates." This is a far more dramatic illustration of our inability to prove lower bounds. – Timothy Chow – 2017-01-18T16:23:36.127

1>

  • and 4. are completely unsupported by any evidence. They are just hyperbole .
  • < – Lembik – 2017-01-19T22:33:51.973

    1

    @TimothyChow: I think this possibility was ruled out recently by Ryan Williams' breakthrough result. But there are many other, only slightly less ridiculous, possible complexity class equalities to choose from.

    – Ashley Montanaro – 2017-01-20T07:44:37.737

    4@AshleyMontanaro : No, I picked my example to be just outside what Williams was able to prove. ACC circuits have constant depth, not logarithmic depth. – Timothy Chow – 2017-01-20T18:38:00.953

    Again, quoting Scott: "It is important for it to be false due to the passionate appreciation of my self-proclamation of superiority, and also, others' proclamations I have made too! Jesus, If this is true, I'll have to accept that I and my beloved crushes are memberes of the plebe like anyone else. Scary! :'(" #FirstOrderProblems – Billy Rubina – 2017-04-23T17:55:42.280

    55

    I've heard that Roger Heath-Brown has presented the following "conjecture" at several conferences, most likely to illustrate our poor understanding of the topic more than because he actually believes it to be true.

    Let $\pi_2(x)$ denote the number of twin primes less than $x$. Then

    $\pi_2(x) = c \int_{0}^{x}\frac{dt}{\log^2 t} + O(1)$

    where $c$ is the twin prime constant.

    In other words, the twin prime asymptotic holds with error term is $O(1)$.

    Mark Lewko

    Posted 2017-01-17T19:55:08.770

    Reputation: 7 522

    5Let me try to get this right. One expects the twin prime asymptotic to hold but not with such a good error term, correct ? I think this is exactly in the spirit of the question, and a great example. Is that really so hard to disprove? – Joël – 2017-01-17T20:46:02.617

    5Yes, that's the spirit. I certainly believe the twin prime asymptotic, but it's much less clear what the size of the error term should be. Probably the truth is either $x^{1/2}$ or something like $x/\log^{k}(x)$. This "conjecture" would tell you that the twin primes are in very nearly exactly where the asymptotic tells they would be, which is quite absurd the more you think about it. – Mark Lewko – 2017-01-17T20:53:18.187

    7Of course, this "conjecture" immediately contradicts the Hardy-Littlewood prime tuplets conjecture. I am also surprised that this turns out hard to disprove. – Vesselin Dimitrov – 2017-01-17T20:59:40.060

    Vasselin, why does it contradict the Hardy-Littlewood conjecture? What version of the conjecture are you thinking of? – Joël – 2017-01-17T21:16:13.593

    7

    Indeed Heath-Brown conjectured a version of this with exactly the sense of "Refute this if you can!" However note that the asymptotic $Cx/(\log x)^2$ should be replaced by the more natural $C \int_2^x dt/(\log t)^2$, which is consistent with Hardy-Littlewood. You can find Heath-Brown's conjecture from Oberwolfach reports: https://www.mfo.de/document/1343/OWR_2013_51.pdf (see the problem session at the end).

    – Lucia – 2017-01-17T21:38:27.603

    1@Joël: Concerning the first version of the statement, with the $Cx/(\log{x})^2$ (now corrected after Lucia's comment), this is incompatible already with the most basic (qualitative) form of Hardy-Littlewood, stating that for every admissible $k$-tuple $(a_1,\ldots,a_k)$, there are infinitely many $n$ such that $n+a_i$ are simultaneously prime. For the latter in particular implies $\pi_2(x+M) - \pi_2(x) > N$ for every $N$ and arbitrarily large $x$, as soon as $M \gg_N 1$. This does not hold for the function $x/(\log{x})^2$. – Vesselin Dimitrov – 2017-01-17T21:57:03.477

    4@VesselinDimitrov: Your comment applies for the modified conjecture too. – Lucia – 2017-01-17T22:01:17.340

    2@Lucia: Ah, indeed: it doesn't make any difference. More generally, the prime tuples conjecture will preclude a formula $\pi_2(x) = f(x) + O(1)$ with a $C^1$ function $f$ of having $f'(x) \to 0$. – Vesselin Dimitrov – 2017-01-17T22:47:42.577

    @VesselinDimitrov: Your comment suggests this answer fails (or at least has trouble related to) condition 5. – R.. – 2017-01-18T03:14:44.083

    45

    I propose Edward Nelson's "conjecture" that Peano's arithmetic is inconsistent.

    First, to be honest, I am not aware that he stated it as "conjecture", using that word, but this is something he said he believed to be true, and that he wasn't able to prove (except for a little while but a mistake was discovered by Terry Tao and others independently) though he tried a lot. So a conjecture it is, in the sense this word currently has.

    It is also certainly "outrageous", in the usual sense of the word -- to many mathematicians, according to my own experience, the simple mention of it provokes disbelief, sarcasm, sometimes outright hostility.

    But let's check that it is also "outrageous" in the sense of this question. 1) It is most certainly false, or at least this is what most mathematicians, including myself, think. 2) But it is certainly not known to be false -- not now, not ever. 3) Nelson made his program public much before 2006. 4) it is obviously extremely important. 5) The negation, that is the assertion that "Peano's arithmetic is consistent" was once a conjecture by Hilbert, but since Gödel it cannot be called a conjecture anymore, since we know it cannot be proven (in a system etc.)

    Let me add that it also satisfies something Gil Kalai added in comment, namely "I prefer examples where [...] the proposer proposes the conjecture genuinely suggesting that it is true".

    Joël

    Posted 2017-01-17T19:55:08.770

    Reputation: 18 625

    13The consistency of Peano arithmetic is provable in ZF(C). Which is a stronger system, of course. – Todd Trimble – 2017-01-18T03:58:42.703

    Yes, that what I meant by "(in as system etc.)", that is "(in a system that does not contain Peano's itself)". I acknowledge it was not really clear. – Joël – 2017-01-18T04:19:06.560

    6Fantastic example! Let me just add that it possibly reveals either one more feature that might be included in the list of requirements, or the very basic explanation of why exactly such conjectures are found outrageous. This is the property of baring cruel limitations to reconciling our abilities and disabilities. I mean if you think of it there is absolutely no way to ever know whether PA is consistent or not. A finite derivation of contradiction from the induction principle indeed feels impossible but at the same time it feels so precisely because it seems to require something non-finite... – მამუკა ჯიბლაძე – 2017-01-18T05:24:43.643

    13In a similar vein, Jack Silver long conjectured that ZFC was inconsistent, and tried over a period of several decades to prove it, although unsuccessfully. His attempt to refute the existence of measurable cardinals led to the theory of Silver indiscernibles, now a fundamental part of the subject. – Joel David Hamkins – 2017-01-18T16:09:53.493

    3@JoelDavidHamkins: Dear Joel, I was not aware of the story behind the Silver indiscernibles. Do you know of some article of Silver where he explains his thoughts on the possible inconsistency of ZFC? – Burak – 2017-01-18T22:07:22.203

    3No, unfortunately, I don't know of any article of his where he states this definitively. Lacking an actual proof of inconsistency, I have understood that he was naturally reticent to state the view explicitly. And so my information is merely second-hand, and may be wrong, so please take it with a grain of salt, although I heard the statements from people at Berkeley who were in a position to know his true plan. I do know for a fact from personal experience that in his set theory lectures he described the various philosophical justifications of large cardinals as, "full of hot air." – Joel David Hamkins – 2017-01-18T22:17:38.157

    This being true would explain something preposterous I managed to prove outright, but I can't imagine what we would do if this were true. – Joshua – 2017-01-18T22:35:21.710

    I attended a conference by Edward Nelson and I agree it's a good example. – Nemo – 2017-01-20T10:53:03.760

    @Nemo. I wish I had. Though his conjecture can be called "outrageous" I would not call it "ridiculous", nor his work on the subject. In fact, while my opinion on this issue is that Peano is consistent, it is not strongly held, more something like when Socrates says in substance "I don't known anything about the gods, so why not believe in those of my city?". Actually, reading some texts of Nelson I remember having been almost convicted by them, though now I can't remember my line of thoughts then. – Joël – 2017-01-20T14:00:27.183

    1@Joël, sure, but you should have seen the other professors' faces at the conference. :) – Nemo – 2017-01-22T08:11:57.793

    What about Gentzen's proof? – Akiva Weinberger – 2017-01-22T17:38:55.110

    Gentzen's proof cannot give us the certainty that Peano is consistent since it is based on the unproved consistency of "primitive recursive arithmetic with the additional principle of quantifier-free transfinite induction up to the ordinal $\epsilon_0$", on which I know much less than on Peano. It is nice, a little bit like Lagarias's theorem that RH is equivalent to $\sigma(n)<H_n+e^{H_n} \log H_n$ is nice, but not a proof of RH. Moreover, we have hope to get a proof of RH sometimes (anyway now!) but no hope to ever get more than results of Gentzen's type concerning the consistency of Peano. – Joël – 2017-01-22T17:46:24.993

    37

    From this Math Overflow question, Joel David Hamkins wrote:

    I once heard Harvey Friedman suggest that the set of prime-differences, that is, the set of all natural numbers $n$ for which there are primes $p,q$ with $p-q=n$, as a possible candidate for all we knew for an intermediate Turing degree — a noncomputable set between $0$ and $0'$ — that was natural, not specifically constructed to have that feature.

    I've also heard others (albeit more recently than 2006) conjecture that Hilbert's 10th problem for rationals is an intermediate degree.

    Really, any conjecture that there is a natural intermediate degree is outrageous (although not exactly formal enough to refute).

    Jason Rute

    Posted 2017-01-17T19:55:08.770

    Reputation: 3 191

    5Whom did you hear suggest that Hilbert's 10th problem for rationals might be an intermediate degree? The idea came to my mind a couple of times, especially after hearing a talk by Poonen on why it seems so difficult to apply the standard approach of reducing to the Entscheidungsproblem. So I'd be curious to know what anyone else thinks on the matter. – Gro-Tsen – 2017-01-18T13:54:31.807

    Dear Jason, can you explain briefly what "intermediate degree" means? – Gil Kalai – 2017-02-07T10:58:03.427

    2

    @GilKalai He means "intermediate" in the sense of the Turing degrees. There are lots of natural examples of unsolvable problems about natural numbers (= undecidable sets) - whether a given Diophantine equation has a solution, whether a given Turing machine halts on every input, etc. - but all known examples are equivalent to some iterate of the Halting Problem. Indeed, to a certain extent we can show that there is no "easy" way to always find a Turing degree between $0^{\alpha}$ and $0^{\alpha+1}$.

    – Noah Schweber – 2017-03-13T15:13:00.713

    So call a "weakly intermediate degree" one which is not of the form $0^\alpha$ for any $\alpha$ (there's actually some dangerous subtlety here, but ignore that for now), and call an "intermediate degree" one which is strictly between $0$ and $0'$. There is currently no known example of a natural weakly intermediate degree, let alone an intermediate degree, there are theorems that finding such a thing would be hard, and there are conjectures that it would be really hard. So where can we look for examples if we're optimistic? Well the feature the hardness results/conjectures use is relativity: – Noah Schweber – 2017-03-13T15:17:09.153

    we show that it's hard to give a method for producing things between $0^\alpha$ and $0^{\alpha+1}$. So one good place to look is at problems which don't obviously relativize to higher degrees. E.g. algebraic questions arguably have this character: there is no canonical version of the rationals corresponding to a given noncomputable Turing degree, so there's no reason to believe that Hilbert's 10th for $\mathbb{Q}$ "relativizes" in any good sense, so the mentioned obstacles don't apply. Another potential type of examples is usual Hilbert's 10th, but restricted to ??? Diophantine equations. – Noah Schweber – 2017-03-13T15:20:02.453

    (Sorry for the lengthy explanation, hopefully this sort of answered your question and gave some motivation.) – Noah Schweber – 2017-03-13T15:20:22.047

    1@NoahSchweber: doesn't the solution to Post's problem give an example of a (non-natural, of course) intermediate degree between $\bf{0}$ and $\bf{0'}$? – cody – 2017-03-15T19:46:15.713

    3

    @cody No, it does not, at least not in the sense you mean! First of all, there are several different solutions to Post's problem: Friedberg-Muchnik, a low simple set, Sacks density applied to $0<0'$, Kucera's priority-free argument, etc. So right off the bat there's a ton of different constructions. But even focusing on a single one - say, the low simple set construction - things don't get better: the construction begins with a fixed enumeration of the Turing machines, and if you change that enumeration the resulting degree changes as well. (cont'd)

    – Noah Schweber – 2017-03-15T21:33:41.193

    2So really what such a construction gets you is a class of intermediate degrees, but no individual intermediate degree. (Incidentally these constructions relativize, as described above, to produce degrees between $0^\alpha$ and $0^{\alpha+1}$ for any (reasonable) $\alpha$.) – Noah Schweber – 2017-03-15T21:35:10.657

    30

    Vopěnka's Principle

    It fits here perfectly except that it has never been called a conjecture. Vopěnka himself was convinced it was wrong! But I will better just post a section from page 279 of Adámek and Rosický ``Locally presentable and accessible categories.'':

    The story of Vopěnka's principle (as related to the authors by Petr Vopěnka) is that of a practical joke which misfired: In the 1960's P. Vopěnka was repelled by the multitude of large cardinals which emerged in set theory. When he constructed, in collaboration with Z. Hedrlín and A. Pultr, a rigid graph on every set (see Lemma 2.64), he came to the conclusion that, with some more effort, a large rigid class of graphs must surely be also constructible. He then decided to tease set-theorists: he introduced a new principle (known today as Vopěnka's principle), and proved some consequences concerning large cardinals. He hoped that some set-theorists would continue this line of research (which they did) until somebody showed that the principle is nonsense. However, the latter never materialized - after a number of unsuccessful attempts at constructing a large rigid class of graphs, Vopěnka's principle received its name from Vopěnka's disciples. One of them, T. J. Jech, made Vopěnka's principle widely known. Later the consistency of this principle was derived from the existence of huge cardinals: see [Powell 1972]; our account (in the Appendix) is taken from [Jech 1978]. Thus, today this principle has a firm position in the theory of large cardinals. Petr Vopěnka himself never published anything related to that principle.

    Adam Przeździecki

    Posted 2017-01-17T19:55:08.770

    Reputation: 2 551

    4https://en.wikipedia.org/wiki/Vop%C4%9Bnka's_principle – Sam Hopkins – 2017-01-18T18:26:02.447

    3This is an interesting twist! A "probably false" conjecture which now by many is probably believed to be true, or at least consistent :) – Wojowu – 2017-01-18T20:30:34.207

    @Wojowu - it is believed to be consistent but not believed to be true: it is known that its negation is consistent. – Adam Przeździecki – 2017-01-18T20:44:31.247

    1@AdamPrzeździecki Negation of AC is also consistent, yet most believe it is true, so I don't get your argument. – Wojowu – 2017-01-18T20:55:15.990

    @Wojowu - Well, so what do you mean by "true" then? It seems that for most people "true" is synonymous to "provable in ZFC". – Adam Przeździecki – 2017-01-18T21:53:03.400

    2@AdamPrzeździecki I was refering to the somewhat platonistic meaning of "truth", regarding one's personal view of "the" universe of sets, just like one might believe CH is false even though it's not contradicting ZFC. – Wojowu – 2017-01-18T22:04:20.540

    @Wojowu I realize this has nothing to do with mathematics per se, but still - somehow I find "platonistic meaning" and "personal view" sort of contradicting each other, no? I mean, "platonistic" should be related to something as objective as possible, independent of any opinions - while "personal" presupposes openly declaring entirely subjective stance... – მამუკა ჯიბლაძე – 2017-01-19T06:12:11.353

    Nobody mentioned here that Vopenka's principle has interesting consequences in homotopy theory. A good starting point is the nLab: https://ncatlab.org/nlab/show/Vop%C4%9Bnka%27s+principle

    – Philippe Gaucher – 2017-03-13T09:07:41.000

    28

    Existence of Siegel zeros.

    1) If we are to believe (like most mathematicians do) in the generalized Riemann hypothesis, this is completely false. I wouldn't necessarily call this ridiculous or outrageous, but within the evidence we have it is rather unlikely to hold.

    2) Nonexistence of Siegel zeros is a problem wide, wide open, nowhere near close to being resolved.

    3) According to the Wikipedia article, this type of zeros was considered back in 1930s, and earlier by Landau, but I don't know if they have explicitly stated the conjecture. GRH was posed back in 1884 though.

    4) They are immensely useful in many applications, since if they exist, primes in certain arithmetic progressions "conspire" to have certain non-uniform distribution. I'm no expert, but here some uses are listed (see also this blog post by Terry Tao).

    5) It implies the negation of GRH, but the negation of a statement itself is quite an awkward statement, saying "yeah, zeros might exist, but not too close to $1$".

    Wojowu

    Posted 2017-01-17T19:55:08.770

    Reputation: 4 984

    2Does anyone believe Siegel zeros exist? – Kimball – 2017-01-18T15:09:05.293

    1@Kimball I don't know the answer to that question. – Wojowu – 2017-01-18T15:10:32.193

    6

    I don't think anyone seriously believes that they actually exist, but they nevertheless seem to have a "ghostly" quasi-existence, in that the Siegel zero hypothesis is remarkably self-consistent and leads to very interesting conclusions about the primes etc. (which are typically rather different from what the orthodox conjectures tell us), see e.g. Friedlander's Notices article at http://www.ams.org/notices/200907/rtx090700817p.pdf

    – Terry Tao – 2017-01-20T22:23:56.183

    5

    Closely related, by the way, to the Siegel zero hypothesis is the Alternative hypothesis that asserts (in blatant contradiction to the GUE hypothesis) that the zeroes of L-functions are approximately arranged in arithmetic progressions: http://www.aimath.org/WWN/lrmt/articles/html/117a/

    – Terry Tao – 2017-01-20T22:25:59.197

    2To conjecture means to publicly claim that something is most certainly true, even though no rigorous proof is known at the moment. This is quite different from just not being able to refute it with known techniques. Did anyone actually conjecture this to be true? – Emil Jeřábek – 2017-01-22T09:04:04.717

    2@Emil Jerabek: I think some people coming from Selberg zetafunctions believe that GRH holds true with the exception of possible real zeroes. I don't know whether anybody believes in a Siegel zero in the strong sense, i.e. a sequence of moduli $q_i$ and roots $\sigma_i$, where $(1-\sigma_i)\log q_i\rightarrow 0$. – Jan-Christoph Schlage-Puchta – 2017-01-23T09:25:06.583

    2

    @Jan-ChristophSchlage-Puchta Amusingly, there is a result of Sarnak and Zaharescu that says that it is inconsistent to simultaneously believe both of the assertions you mention: https://projecteuclid.org/download/pdf_1/euclid.dmj/1087575083

    – Terry Tao – 2017-03-15T00:12:34.597

    28

    The "conjecture" in algebraic geometry that all rationally connected varieties are unirational comes to mind. It's usually thrown around as a way of saying, "See, we know so little about what varieties can be unirational that we can't prove a single rationally connected variety isn't." Unirationality implies rational connectedness, but I think almost everyone believes the converse should be false.

    Some background: Algebraic geometers have been interested for a long time in proving that certain varieties are or are not rational (very roughly, figuring out which systems of polynomial equations can have their solutions parametrized.) Clemens and Griffiths showed in 1972 that a cubic hypersurface in $\mathbb{P}^4$ is irrational. Since then, there's been a lot of progress in rationality obstructions e.g., Artin-Mumford's obstruction via torsion in $H^3$, Iskovskikh-Manin on quartic threefolds, Kollar's work on rationality of hypersurfaces, and most recently, Voisin's new decomposition of the diagonal invariants which have led to major breakthroughs.

    On the other hand, unirationality has proved a far harder notion to control, and to my mind the biggest open question in this area is to find any obstruction to unirationality.

    dhy

    Posted 2017-01-17T19:55:08.770

    Reputation: 2 855

    Nice answer. The first sentence seems to have one "rationally connected" too many. – potentially dense – 2017-01-19T14:46:01.760

    4Potential density of rational points is an obstruction to unirationality . . . at least potentially :) – Jason Starr – 2017-01-19T18:47:34.197

    1To conjecture means to publicly claim that something is most certainly true, even though no rigorous proof is known at the moment. This is quite different from just not being able to refute it with known techniques. Did anyone actually conjecture this to be true? – Emil Jeřábek – 2017-01-22T09:03:32.657

    24

    I don't know about "ridiculous", but there is Hall's Conjecture in its original form:

    There is a positive constant $C$ such that for any two integers $a$ and $b$ with $a^2 \neq b^3$, one has $$|a^2-b^3|> C \sqrt{|b|}\;.$$

    T. Amdeberhan

    Posted 2017-01-17T19:55:08.770

    Reputation: 16 712

    20It should be "There is a constant $C$ such that for any two integers $a, b$, $a^2 \neq b^3$ implies $|a^2 - b^3| > C\sqrt{|b|}$." Pretty interesting conjecture. – Todd Trimble – 2017-01-21T22:00:36.330

    1So you allow C=0 ? – Wilberd van der Kallen – 2017-10-24T07:40:18.057

    21

    I was giving a talk several years ago about the conjectured linear independence (over $\Bbb Q$) of the ordinates of the zeros of the Riemann zeta function, and Lior Silberman crystallized our current lack of knowledge into a "Refute this!" statement:

    If $\zeta(x+iy)=0$, then $y\in\Bbb Q$.

    (In other words, even though the imaginary parts of the nontrivial zeros of $\zeta(s)$ are believed to be transcendental, algebraically independent, and generally unrelated to any other constants we've ever seen ... we currently can't even prove that a single one of those imaginary parts is irrational!)

    This "conjecture" can be extended to Dirichlet $L$-functions, and perhaps even further (though one needs to be careful that we don't allow a class of $L$-functions that includes arbitrary vertical shifts of some of its members).

    Greg Martin

    Posted 2017-01-17T19:55:08.770

    Reputation: 7 908

    To conjecture means to publicly claim that something is most certainly true, even though no rigorous proof is known at the moment. This is quite different from just not being able to refute it with known techniques. Did anyone actually conjecture this to be true? – Emil Jeřábek – 2017-01-22T09:02:41.897

    2While your definition of "conjecture" is certainly defensible, I am following the definition given in the OP. – Greg Martin – 2017-01-24T21:25:14.600

    19

    The telescope conjecture in homotopy theory, which says that smashing localizations are finite. This was conjectured by Ravenel in 1977. See e.g. https://math.mit.edu/conferences/talbot/2013/17-Barthel-telescope.pdf for more.


    Let me elaborate a little more. Let $E$ be any spectrum (in the homotopy-theoretic sense). Say that a spectrum $W$ is $E$-acyclic if the smash product $E\wedge W$ is zero. A spectrum $V$ is $E$-local if $[W,V]=0$ for any $E$-acyclic $W$. Bousfield proved that for any spectra $E$ and $X$, there exists an $E$-local spectrum $L_E X$ and an $E$-equivalence $f:X\to L_E X$ (i.e. $E_\ast f$ is an isomorphism). This is called the Bousfield localization of $X$.

    Alternatively, we could restrict ourselves to finite $E$-acyclic spectra. Namely, say that a spectrum $V$ is finitely $E$-local if $[W,\Sigma^{-n}V]=0$ for any finite $E$-acyclic spectrum $W$. Say that a spectrum $R$ is finitely $E$-acyclic if for any finitely $E$-local spectrum $W$, $[R,W]=0$. One can show that if $E$ and $X$ are spectra, there exists a finitely $E$-local spectrum $L^f_E X$ with a finite $E$-equivalence $X\to L^f_E X$ that is initial among maps from $X$ to finitely $E$-local spectra. The map $X\to L_E X$ factors as $X\to L^f_E X\to L_E X$ since any $E$-local spectrum is finitely $E$-local.

    A brief interlude on chromatic homotopy theory, which can be skipped by the experts: let $f$ be a formal group law over a ring $R$. This is classified by a map $MU_\ast\simeq L\to R$. One can form a functor on spaces via $E_\ast(X):=MU_\ast(X)\otimes_{MU_\ast}R$ where $R$ is made into a $MU_\ast$-module via $f$. Note that $\pi_\ast E=R$. Say that $f$ is Landweber exact if this is a homology theory (the only reason it isn't a cohomology theory in general is that it doesn't turn cofiber sequences into lexseqs). The Landweber exact functor theorem gives a condition on $f$ which makes it Landweber exact (some sequence of elements must be regular).

    An important nonexample is the $n$th Morava K-theory with $\pi_\ast K(n)\simeq \mathbf{F}_p[v_n^{\pm 1}]$ where $|v_n|=2(p^n-1)$, whose formal group law is of height exactly $n$. An important example of a Landweber exact spectrum is the Johnson-Wilson spectrum $E(n)$ (the integer $n$ is called the height) with $\pi_\ast E(n)=W(k)[[v_1,\cdots,v_{n-1}]][\beta^{\pm 1}]$ where $W(k)[[v_1,\cdots,v_{n-1}]]$ is the Lubin-Tate ring characterizing the universal deformation of a formal group law of height $n$ over a perfect field $k$ of characteristic $p>0$. The localization functors $L^f_{E(n)}$ and $L_{E(n)}$ are typically denoted $L^f_n$ and $L_n$.

    The telescope conjecture says: if $E$ is the Johnson-Wilson spectrum, then $L^f_{E(n)} = L_{E(n)}$. This implies a statement about the convergence of an Adams-Novikov spectral sequence.

    Let me make some comments on this conjecture. Both $L^f_n$ and $L_n$ are smashing, in the sense that $L^f_n X\simeq X\wedge L^f_n S^0$ and $L_n X\simeq X\wedge L_n S^0$. The telescope conjecture thus reduces to showing that $L^f_n S^0\simeq L_n S^0$. It turns out, for example, that for any $n,m\geq 0$, $L^f_n L_m X\simeq L_n L_m X$. This means that if the telescope conjecture is true at height $n$, it is true at height $n-1$. To see this, note that $L^f_n S^0\simeq L_n S^0$ implies that: $$L^f_{n-1}S^0\simeq L^f_{n-1} L^f_n S^0\simeq L^f_{n-1}L_n S^0\simeq L_{n-1}L_n S^0,$$ as desired. A nice exposition is http://www.home.uni-osnabrueck.de/mfrankland/Frankland_TelescopeConj_20110407.pdf.

    skd

    Posted 2017-01-17T19:55:08.770

    Reputation: 1 746

    10Hah! I was gonna write this. You should explain it more though! Nobody knows anything about homotopy theory outside of homotopy theory and this is why! – Jonathan Beardsley – 2017-01-18T21:30:51.893

    1@JonBeardsley details added! – skd – 2017-01-19T02:15:32.280

    1Awesome!! Now I wish someone would prove or disprove it. – Jonathan Beardsley – 2017-01-19T02:21:12.417

    1It should be added that Telescope conjecture is true for chromatic height $n=1$ but experts believe that the conjecture is false for chromatic height $n\geq 2$ for all primes! – Prasit – 2017-01-20T21:41:07.137

    13That the explanation begins 'Let $E$ be any spectrum...' also demonstrates why no one knows anything about homotopy theory outside of homotopy theory. – HJRW – 2017-01-23T09:15:44.803

    @HJRW May I say that I find this very sad? A spectrum is really nothing more esoteric than a Sobolev space or an arithmetic scheme, and they tend to pop up everywhere you want to deal with (co)homology. – Denis Nardin – 2018-02-12T12:21:42.330

    3@DenisNardin, I think we all wish that the the basic definitions of our own fields were more widely known! The bottom line is that spectra aren't standard objects outside homotopy theory (even if you think they should be), and an explanation was requested for non-homotopy-theorists. If you do want spectra to be more widely understood, rather than complaining that they aren't, a better strategy might be to be aware of your audience when communicating in a public forum. – HJRW – 2018-02-13T11:16:22.223

    19

    I'm surprised no one has mentioned it, but the first one that comes to my mind is this.

    $e+\pi$ is rational.

    I think most mathematicians would agree that it is ridiculous. It would follow from Schanuel's conjecture that it is false, but as far as I know, the conjecture is wide open, and when it comes to (ir)rationality of $e+\pi$, more or less all that is known is the (elementary) fact that either $e+\pi$ or $e\cdot\pi$ is transcendental (naturally, we expect both of them to be transcendental, so it doesn't really get us any closer to a proof).

    I'm not sure when it was made publicly, but it is very natural and unlikely to not have been considered before (I heard about it as an undergrad around 2010). I think it is quite important in that it is an obvious test case for Schanuel's conjecture, and in that it would certainly be quite shocking if it was true.

    (Caveat: I am not a specialist, so if someone more competent can contradict me, please do!)

    tomasz

    Posted 2017-01-17T19:55:08.770

    Reputation: 312

    13This was proposed, and subsequently deleted, by user Mehrdad two days ago, when it was objected that, contrary to the stipulations of the question, it's just the negation of a famous conjecture. – Gerry Myerson – 2017-01-21T22:48:07.010

    1Since two participants proposed it, perhaps we better keep it...better fitted to a hypothetical related question: "The most embarrassing mathematical statements sure to be true but not proven..." :) – Gil Kalai – 2017-01-22T06:49:33.773

    1To conjecture means to publicly claim that something is most certainly true, even though no rigorous proof is known at the moment. This is quite different from just not being able to refute it with known techniques. Did anyone actually conjecture that $\pi+e$ is rational? – Emil Jeřábek – 2017-01-22T08:52:05.680

    2@EmilJeřábek: I understand and I had some doubts, but if the quote "On other times, I have conjectured to lay down the gauntlet: >>See, you can’t even disprove this ridiculous idea.<<" doesn't apply to this, I don't know what it could possibly apply to. Is the mere technicality of not being stated as a conjecture so important? Quite a few statements listed here were apparently believed to be intuitively "obviously" false even by the people "conjecturing" them. – tomasz – 2017-01-22T09:36:22.987

    @GerryMyerson: I don't think this is a negation of a famous conjecture, if by famous conjecture you mean Schanuel's conjecture. It is a strong form of negation of the conjecture, but the same is true about all the answers related to Riemann Hypothesis here (as far a I can tell). – tomasz – 2017-01-22T09:44:54.250

    17

    For a prime $p$, an infinite group $G$ is a Tarski monster if each of its proper subgroups has order $p$.

    If I am correctly informed, then the Tarski monster was defined to demonstrate our poor understanding of infinite groups, because such monsters obviously don't exist, it should be easy to prove that they don't exist, but we cannot prove it.

    Then Olshanskii proved that Tarski monsters do exist for all large primes, and by now many people believe that "large" means something like $p\geq 11$.

    Jan-Christoph Schlage-Puchta

    Posted 2017-01-17T19:55:08.770

    Reputation: 7 628

    2What's the outrageous conjecture here? If it's the existence of Tarski monsters, this is now known to be true, and therefore doesn't satisfy item 1. – HJRW – 2017-01-21T23:06:01.887

    1@HJRW I've also had trouble keeping track for some of the answers, whether they satisfy the conditions of the OP. I'm not sure the OP is absolutely crystal clear here. But he does write, "Very important examples where the conjecture was believed as false when it was made but this is no longer the consensus may also qualify!" So this answer may be kosher according to that. – Todd Trimble – 2017-01-22T03:42:45.807

    So did Tarski, or anyone else, ever conjecture that his monsters existed? At the moment this seems to be an outrageous theorem. – HJRW – 2017-01-22T09:01:33.417

    2I do not have a reference which settles what Tarski believed, but the general context would be the existence of finitely generated infinite torsion groups. I guess that around 1910 it was generally believed that such groups do not exist, so the existence of monsters was outrageous. – Jan-Christoph Schlage-Puchta – 2017-01-22T12:43:13.040

    I guess the counterpoint between Burnside's problem and Tarski monsters is in the spirit of the question (though I read the question as asking about problems that are still open). It would be nice to know what Tarski thought. – HJRW – 2017-01-22T22:37:15.380

    16

    If the holomorphic map $f:\mathbb{C}\to\mathbb{C}$ has a fixed point $p$, and the derivative $\lambda := f'(p)$ equals $e^{2\pi i \theta}$ (with irrational $\theta$), one can ask if $f$ is conjugate to $z\mapsto\lambda\cdot z$ in a neighborhood of $p$. If it exists, the largest domain of conjugacy is called a 'Siegel disk'. Two properties to keep in mind are:

    • A Siegel disk cannot contain a critical point in its interior (boundary is ok).
    • The boundary of a Siegel disk belongs to the Julia set of $f$.

    Quadratic maps can have Siegel disks, but not for just any $\theta$; the number theoretical properties of this 'rotation number' are relevant. However, if $\theta$ is Diophantine, the boundary of the disk is well behaved (Jordan curve, quasi-circle...)

    ... But the boundary of a Siegel disk can also be wild; for instance, it can be non-locally connected. The outrageous conjecture that has been floating around is that:

    • There is a quadratic polynomial with a Siegel disk whose boundary equals the Julia set.

    Since a quadratic Julia set is symmetric with respect to the critical point, a quadratic Siegel disk would have a symmetric preimage whose boundary also equals the Julia set, but the unbounded component of the complement (Fatou set) also has boundary equal to the Julia set, so our conjectured Siegel disk would form part of a 'lakes of Wada' configuration.

    Rodrigo A. Pérez

    Posted 2017-01-17T19:55:08.770

    Reputation: 2 014

    1Was this conjecture made by someone before 2006? – Sam Hopkins – 2017-01-18T14:06:21.247

    2@Sam: I heard it as a student back in the 90s – Rodrigo A. Pérez – 2017-01-19T03:54:27.047

    14

    The Lusternik-Schnirelmann category of the Lie groups $Sp(n)$. Since $Sp(1) = S^3$, $\mathrm{cat}(Sp(1)) = 1$. In the 1960s, P. Schweitzer proved that $\mathrm{cat}(Sp(2)) = 3$. Based on this, a folklore conjecture emerged that in general $\mathrm{cat}(Sp(n)) = 2n-1$. In 2001, it was proved that $\mathrm{cat}(Sp(3)) = 5$, so maybe it's true?

    Jeff Strom

    Posted 2017-01-17T19:55:08.770

    Reputation: 7 190

    2

    See https://en.wikipedia.org/wiki/Lusternik%E2%80%93Schnirelmann_category for the definition. I suppose Jeff uses the convention of deleting one from the Wikipedia definition (also mentioned there) so that cat (sphere)=1.

    – Gil Kalai – 2017-01-21T18:33:49.877

    Just a stupid question - what is the cup length of $SP(n)$? – Sebastian Goette – 2017-03-14T09:29:53.787

    The cohomology is an exterior algebra on $n$ generators, so the cup length is $n$. – Jeff Strom – 2017-03-14T14:31:20.923

    1@JeffStrom, I would suggest evaluating the systolic category of $Sp(n)$ first. This is often easier to compute, and the two categories are equal in many cases. – Mikhail Katz – 2017-03-15T13:14:57.083

    9

    The finitistic dimension conjecture for finite dimensional algebras states that the supremum of all projective dimensions of modules having finite projective dimension is finite. It it just proven for some very special classes of algebras and in general there seems to be no reason why this should be true. References: https://arxiv.org/pdf/1407.2383v1.pdf http://www.math.uni-bonn.de/people/schroer/fd-problems-files/FD-FinitisticDimConj.pdf

    Mare

    Posted 2017-01-17T19:55:08.770

    Reputation: 3 452

    9

    Let us say a graph class $\mathcal{C}$ is small if it has at most $n^{O(n)}$ graphs on $n$ vertices. The implicit graph conjecture states that every small, hereditary graph class has an adjacency labeling scheme (ALS) with a label decoder that can be computed in polynomial time (a formal definition of ALS is given at the end of the answer).

    Initially, this was posed as question by Kannan, Naor and Rudich in their paper Implicit Representation of Graphs which appeared at STOC '88. It was restated as conjecture by Spinrad in the book Efficient Graph Representations (2003).

    It is important because it would imply the existence of space-efficient representations for all small, hereditary graph classes where querying an edge requires only polylogarithmic time with respect to the number of vertices of the graph.

    As far as I know there is no consensus about whether this conjecture should be true or not. However, from my perspective it would be an immense surprise if it holds for the following reason. The concept of adjacency labeling schemes can be defined with respect to arbitrary complexity classes. For a complexity class $\text{C}$ (more formally, a set of languages over the binary alphabet) we can define the class of graph classes $\text{GC}$ as the set of all graph classes that have an ALS with a label decoder that can be computed in $\text{C}$. It can be shown that $\text{G1EXP} \subsetneq \text{G2EXP} \subsetneq \text{G3EXP} \dots \subsetneq \text{GR} \subsetneq \text{GALL}$ where $\text{kEXP}$ is the set of languages that can be computed in time $\exp^k(\text{poly}(n))$, $\text{R}$ is the set of all decidable languages and $\text{ALL}$ is the set of all languages. I find it hard to believe that every small, hereditary graph class falls down through all these classes and just happens to sit in $\text{GP}$ (the choice just seems too arbitrary and weak). In fact, there are natural graph classes such as disk graph or line segment graphs for which it is not even known whether they are in $\text{GALL}$. Additionally, a graph class $\mathcal{C}$ is in $\text{GALL}$ iff $\mathcal{C}$ has a polynomial universal graph, i.e. a family of graphs $(G_n)_{n \in \mathbb{N}}$ such that $|V(G_n)|$ is polynomially bounded and $G_n$ contains all graphs from $\mathcal{C}$ on $n$ vertices as induced subgraph. It already seems doubtful to me that every small, hereditary graph class has such polynomial universal graphs.

    An ALS is a tuple $S=(F,c)$ where $F \subseteq \{0,1\}^* \times \{0,1\}^*$ is called label decoder and $c \in \mathbb{N}$ is the label length. A graph $G$ with $n$ vertices is represented by $S$ if there exists a labeling $\ell \colon V(G) \rightarrow \{0,1\}^{c \lceil \log n \rceil}$ such that for all $u,v \in V(G)$ it holds that $(u,v) \in E(G) \Leftrightarrow (\ell(u),\ell(v)) \in F$. A graph class $\mathcal{C}$ has an ALS $S$ if every graph in $\mathcal{C}$ can be represented by $S$ (but not necessarily every graph represented by $S$ must be in $\mathcal{C}$).

    mch

    Posted 2017-01-17T19:55:08.770

    Reputation: 1

    9

    $S^6$ has a complex structure.

    I don´t know if this apply, but this has a nice story. It has been "published" to to be true and know Atiyah has a short paper on arxiv claiming to be false, other important mathematicians has also work on this problem. According to LeBrun this would be a minor disaster.

    Gerardo Arizmendi

    Posted 2017-01-17T19:55:08.770

    Reputation: 597

    8

    Seen this, I believe really striking, example yesterday here on MO: although the question Complex vector bundles that are not holomorphic is from 2009, a recent post by algori suggests that this is still open.

    And the really ridiculous conjecture is that all topological vector bundles on $\mathbb C\mathbf P^n$ are algebraic.

    The question is described as an open problem in Okonek-Schneider-Spindler (1980) but I believe must have been asked much earlier.

    მამუკა ჯიბლაძე

    Posted 2017-01-17T19:55:08.770

    Reputation: 7 821

    7

    It seems to me that Zeeman's collapsing conjecture satisfies the criteria given. The Zeeman conjecture implies both the Poincaré conjecture (proved in 2003) and the Andrews-Curtis conjecture.

    The following is a quote from Matveev's book, where it is proved that ZC restricted to special polyhedra is equivalent to the union of PC and AC.

    Theorem 1.3.58 may cast a doubt on the widespread belief that ZC is false. If a counterexample indeed exists, then either it has a “bad” local structure (is not a special polyhedron) or it is a counterexample to either AC or PC.

    DKal

    Posted 2017-01-17T19:55:08.770

    Reputation: 121

    1Zeeman's conjecture assert that for a 2-dimensional contractible complex $K$, $K \times I$ is collapsible. ($I$ is the unit interval.) – Gil Kalai – 2017-01-22T10:45:38.873

    2"ACC" is overloaded – Andrews-Curtis conjecture, Ascending Chain Condition, Axiom of Countable Choice. – Gerry Myerson – 2017-01-22T11:52:29.247

    @GerryMyerson, changed to AC. This also agrees with Matveev's use in the given quote. – DKal – 2017-01-23T09:00:11.803

    1If $K$ is higher dimensional (say, 3-dimensional) is the srarement of Zeeman's conjecture known to be false? – Gil Kalai – 2017-01-23T11:50:06.970

    @GilKalai, yes, for dimension $\geq 3$ it seems to have been shown in the 1977 paper "Whitehead torsion, group extensions, and Zeeman's conjecture in high dimensions" by M. Cohen. (here's the link to the paper: http://www.sciencedirect.com/science/article/pii/0040938377900313)

    – DKal – 2017-01-23T17:21:25.523

    It may be also worth to mention that Adiprasito and Benedetti showed that for any contractible complex $C$ there is an $n\geq 0$ such that $C\times I^n$ is collapsible. See https://arxiv.org/pdf/1202.6606v3.pdf.

    – DKal – 2017-01-23T20:08:02.243

    1Now it states that Zermelo with Choice is false, and that a counterexample would be to either the Axiom of Choice, or something related to Power Set (and Choice, I guess)... :-P – Asaf Karagila – 2017-03-13T12:28:38.687

    3

    This one is due to Errett Bishop: "all meaningful mathematics is reducible to finite calculations with strings of $0$s and $1$s" (imho Bishop formulated this not as a conjecture but as an article of faith but that doesn't necessarily affect the truth or falsity thereof).

    A reference for Bishop's claim is his article "Crisis in contemporary mathematics" (the link is to the mathscinet review of the article) which discusses the constructivist opposition to a principle called LPO ("limited principle of omniscience") related to the law of excluded middle. The LPO is discussed starting on page 511 of the article.

    Mikhail Katz

    Posted 2017-01-17T19:55:08.770

    Reputation: 7 997

    6Is this falsifiable? What form would a falsification take? – Todd Trimble – 2017-03-13T11:49:45.543

    2@ToddTrimble, apparently Bishop thought it was (incidentally Kalai did not impose a specific falsifiability clause) otherwise he wouldn't have presented it as a meaningful assertion. I would conjecture that a proof of Fermat's last theorem that relies essentially on large cardinal hypotheses would constitute a falsification of Bishop's claim (of course MacLarty has argued that it doesn't essentially depend on such). Also if NSA is used to prove the Riemann hypothesis, many will probably interpret this as a refutation of Bishop. – Mikhail Katz – 2017-03-13T12:18:52.907

    Is the question whether "Fermat had a marvelous proof for Fermat's last theorem" falsifiable? – Mikhail Katz – 2017-03-13T12:32:26.723

    3No, a proof that adopts large cardinal hypotheses as axioms is still enacted in a script that in principle could be converted to 0's and 1's (and this is not unrealistic when one considers computer-based formalizations of even hard theorems). Actually, if you say Bishop formulated this as an article of faith, it's less clear to me that he thought it was falsifiable (and I for one don't think it is). – Todd Trimble – 2017-03-13T12:38:14.583

    1As for Fermat, it's an interesting question, but the likelihood of his having a proof could in principle be estimated if, one day, we get a better grip on the inherent complexity of any such proof. I don't think "likelihood" enters for your claim. – Todd Trimble – 2017-03-13T12:40:37.747

    1The computer-based formalisations you mention would presumably include symbols for such cardinals but Bishop would likely consider both the cardinals in question and symbols that stand for them as meaningless. Therefore I still think that if one could show that a proof of FLT depends on large cardinals this would come close to a refutation of Bishop's claim. – Mikhail Katz – 2017-03-13T12:43:46.493

    4Actually, Mikhail, I agree with your last point: that the (somewhat tautologous) finitary nature of proofs was not what Bishop had in mind when he speaks of what is "meaningful" in mathematics. (At the moment of writing, I was being rushed out of the house by my daughter who needed to get to school on time.) I'm happy at this point to let Gil decide if this answer meets his criteria -- I thought he wanted mathematical conjectures, not philosophical claims. – Todd Trimble – 2017-03-13T13:03:14.610

    1Dear Misha, this is quite interesting. When was this conjecture made and are there some relevant references/links? – Gil Kalai – 2017-03-13T17:49:34.787

    1Thanks, Gil. This is in Bishop's famous "Crisis in contemporary mathematics" text and other texts. I will add a reference to my answer. – Mikhail Katz – 2017-03-13T17:51:17.630

    1

    @MikhailKatz The link you edited into your answer is to commentary on Bishop's talk; I believe this is the article in question by Bishop.

    – Neal – 2017-03-15T14:40:24.953

    Thanks, @Neal. The link in my answer is actually to the mathscinet review of the article. This mathscinet entry also provides a link to the article itself. Incidentally, I seem to recall other articles by Bishop where the idea of mathematics as being calculations on strings of integers is discussed more explicitly. Have you seen anything related? – Mikhail Katz – 2017-03-15T14:44:07.983