# The most outrageous (or ridiculous) conjectures in mathematics

## 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.)

### 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:

### you can’t even disprove this ridiculous idea."

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

(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).

• There 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 Jan 17 '17 at 20:25
• I 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 Jan 17 '17 at 22:19
• The answers below all look of interest, as does the question. And, to boot, this is Community Wiki. Why not keep it open? – Lucia Jan 17 '17 at 22:49
• @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 Jan 17 '17 at 23:36
• What's next - a big list with Trump tweets concerning mathematics? Does this outrageous conjecture of mine count as an example? – Franz Lemmermeyer Jan 18 '17 at 5:39

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.

• Yes, this is the spirit of the question! – Gil Kalai Jan 17 '17 at 20:18
• Dangit, beat me to it! +1. (Need to get my typing speed up!) – Noah Schweber Jan 17 '17 at 20:25
• On 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. – Joseph Van Name Jan 17 '17 at 23:07
• @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 Jan 17 '17 at 23:10
• Johnson answered [by] striking his foot with mighty force against a large stone, till he rebounded from it -- "I refute it thus." – Max Jan 19 '17 at 10:05

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).

• I'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 Jan 17 '17 at 23:14
• @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 Jan 18 '17 at 5:36
• This is nowadays known as the second Hardy-Littlewood conjecture. – Wojowu Jan 18 '17 at 13:46
• @Mark, it's not even entirely clear where or whether Hardy & Littlewood made the conjectures. See mathoverflow.net/questions/54223/whence-the-k-tuple-conjecture and mathoverflow.net/questions/30827/… and math.stackexchange.com/questions/1072194/… for some discussions. – Gerry Myerson Jan 19 '17 at 1:55
• @nigel, if 2nd Hardy-Littlewood is false, then there is a counterexample, so it's not unproveable. – Gerry Myerson Jan 24 '17 at 21:54

# $P=NP$

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

• Dear 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 Jan 17 '17 at 20:08
• @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 Jan 17 '17 at 20:09
• I 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. Jan 17 '17 at 22:51
• Fails #5. It's the negation of the almost-certainly-true $P \neq NP$ – R.. Jan 18 '17 at 3:12
• There 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 Jan 18 '17 at 16:23

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 $O(1)$.

• Let 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 Jan 17 '17 at 20:46
• Yes, 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 Jan 17 '17 at 20:53
• Of course, this "conjecture" immediately contradicts the Hardy-Littlewood prime tuplets conjecture. I am also surprised that this turns out hard to disprove. – Vesselin Dimitrov Jan 17 '17 at 20:59
• 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: mfo.de/document/1343/OWR_2013_51.pdf (see the problem session at the end). – Lucia Jan 17 '17 at 21:38
• @VesselinDimitrov: Your comment applies for the modified conjecture too. – Lucia Jan 17 '17 at 22:01

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".

• The consistency of Peano arithmetic is provable in ZF(C). Which is a stronger system, of course. – Todd Trimble Jan 18 '17 at 3:58
• Fantastic 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 Jan 18 '17 at 5:24
• In 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 Jan 18 '17 at 16:09
• @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 Jan 18 '17 at 22:07
• No, 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 Jan 18 '17 at 22:17

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).

• Whom 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 Jan 18 '17 at 13:54
• @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 Mar 13 '17 at 15:13
• @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 Mar 15 '17 at 19:46
• @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 Mar 15 '17 at 21:33
• So 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 Mar 15 '17 at 21:35

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.

• en.wikipedia.org/wiki/Vop%C4%9Bnka's_principle – Sam Hopkins Jan 18 '17 at 18:26
• This is an interesting twist! A "probably false" conjecture which now by many is probably believed to be true, or at least consistent :) – Wojowu Jan 18 '17 at 20:30
• @Wojowu - it is believed to be consistent but not believed to be true: it is known that its negation is consistent. – Adam Przeździecki Jan 18 '17 at 20:44
• @AdamPrzeździecki Negation of AC is also consistent, yet most believe it is true, so I don't get your argument. – Wojowu Jan 18 '17 at 20:55
• @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 Jan 18 '17 at 22:04

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.

• Nice answer. The first sentence seems to have one "rationally connected" too many. – potentially dense Jan 19 '17 at 14:46
• Potential density of rational points is an obstruction to unirationality . . . at least potentially :) – Jason Starr Jan 19 '17 at 18:47
• 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 supports Monica Jan 22 '17 at 9:03

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 $1$".

• Does anyone believe Siegel zeros exist? – Kimball Jan 18 '17 at 15:09
• 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 ams.org/notices/200907/rtx090700817p.pdf – Terry Tao Jan 20 '17 at 22:23
• 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: aimath.org/WWN/lrmt/articles/html/117a – Terry Tao Jan 20 '17 at 22:25
• @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 Jan 23 '17 at 9:25
• @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: projecteuclid.org/download/pdf_1/euclid.dmj/1087575083 – Terry Tao Mar 15 '17 at 0:12

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 $$|a^2-b^3|> C \sqrt{|b|}\;.$$

• It 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 Jan 21 '17 at 22:00
• So you allow C=0 ? – Wilberd van der Kallen Oct 24 '17 at 7:40
• @wilberd A sad ambiguity positive means strictly so! Otherwise we say 'non negative'. Yet I do prefer the redundant 'strictly positive'. – Jérôme JEAN-CHARLES Aug 7 at 12:45

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).

• 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 supports Monica Jan 22 '17 at 9:02
• While your definition of "conjecture" is certainly defensible, I am following the definition given in the OP. – Greg Martin Jan 24 '17 at 21:25

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!)

• This 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 Jan 21 '17 at 22:48
• Since 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 Jan 22 '17 at 6:49
• 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 that $\pi+e$ is rational? – Emil Jeřábek supports Monica Jan 22 '17 at 8:52
• @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 Jan 22 '17 at 9:36
• @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 Jan 22 '17 at 9:44

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 $p\geq 11$.

• What'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 Jan 21 '17 at 23:06
• @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 Jan 22 '17 at 3:42
• So did Tarski, or anyone else, ever conjecture that his monsters existed? At the moment this seems to be an outrageous theorem. – HJRW Jan 22 '17 at 9:01
• I 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 Jan 22 '17 at 12:43
• 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 Jan 22 '17 at 22:37

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.

• Was this conjecture made by someone before 2006? – Sam Hopkins Jan 18 '17 at 14:06
• @Sam: I heard it as a student back in the 90s – Rodrigo A. Pérez Jan 19 '17 at 3:54

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?

• See 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 Jan 21 '17 at 18:33
• Just a stupid question - what is the cup length of $SP(n)$? – Sebastian Goette Mar 14 '17 at 9:29
• The cohomology is an exterior algebra on $n$ generators, so the cup length is $n$. – Jeff Strom Mar 14 '17 at 14:31
• @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 Mar 15 '17 at 13:14

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 $\mathcal{C}$).

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

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.

$$S^6$$ 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 now 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.

• LeBrun's comment appears in the paper Orthogonal Complex Structures on $S^6$. His reasoning is that by blowing up $S^6$, one would obtain a complex three-manifold diffeomorphic to $\mathbb{CP}^3$, but not biholomorphic to it. In fact, the resulting complex structure wouldn't even be Kähler! – Michael Albanese Dec 21 '18 at 5:00

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.

• Zeeman's conjecture assert that for a 2-dimensional contractible complex $K$, $K \times I$ is collapsible. ($I$ is the unit interval.) – Gil Kalai Jan 22 '17 at 10:45
• "ACC" is overloaded – Andrews-Curtis conjecture, Ascending Chain Condition, Axiom of Countable Choice. – Gerry Myerson Jan 22 '17 at 11:52
• @GerryMyerson, changed to AC. This also agrees with Matveev's use in the given quote. – DKal Jan 23 '17 at 9:00
• If $K$ is higher dimensional (say, 3-dimensional) is the srarement of Zeeman's conjecture known to be false? – Gil Kalai Jan 23 '17 at 11:50
• Now 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 Mar 13 '17 at 12:28

Here is my favorite.

Spherical simplex is called rational, if all its dihedral angles are rational multiples of $$\pi.$$

Rational Volume Conjecture: Volume of every rational spherical simplex is a rational multiple of $$\pi^2.$$

It is true in all known examples, but is expected to be false. The conjecture first appeared in "Differential characters and geometric invariants" paper of Cheeger and Simons, p.73. (Dis)proving this conjecture would lead to interesting results in scissor congruence theory and group homology.

• What is a spherical simplex and what are its dihedral angles, if I may ask? – Fedor Petrov Jun 2 at 19:34
• @FedorPetrov I meant a geodesic simplex in the standard 3-dimensional sphere. And there one can define an angle between geodesic 2-dimensional spheres. – Daniil Rudenko Jun 2 at 20:52
• Ah, that it is 3-dimensional was not clear to me. So It has 4 vertices belonging to $\mathbb{S}^3\subset \mathbb{R}^4$? – Fedor Petrov Jun 2 at 21:13
• That’s correct. – Daniil Rudenko Jun 2 at 23:59

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.

• Is this falsifiable? What form would a falsification take? – Todd Trimble Mar 13 '17 at 11:49
• @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 Mar 13 '17 at 12:18
• No, 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 Mar 13 '17 at 12:38
• As 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 Mar 13 '17 at 12:40
• Actually, 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 Mar 13 '17 at 13:03