This is a pair to

Not especially famous, long-open problems which anyone can understand

So this time I'm asking for open questions so easy to state for students of subjects such as undergraduate abstract algebra, linear algebra, real analysis, topology (etc.?) that a teacher could mention the questions almost immediately after stating very basic definitions.

If they weren't either too famous or already solved, appropriate answers might include the Koethe Nil Conjecture, the existence of odd order finite simple groups or the continuum hypothesis.


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    $\begingroup$ In other words, delete now and re-ask in a month? I agree that this is the best strategy for David, if he wants the best answers (people are kind of exhausted of this sort of question now), but it seems strange to close it against his will. $\endgroup$ – Steven Gubkin Jul 2 '12 at 19:47
  • $\begingroup$ >...some of the answers to alvarezpaiva's question would fit here (eg, normal numbers) Re normal numbers, I think not. Since one does not usually emphasize radix expansions in basic abstract algebra or real analysis and certainly one doesn't first meet such expansions at university, that question would not obviously have pedagogical value in the specified context. I mentioned Koethe's conjecture because as soon as the student has the definition of ring, ideal and left ideal a teacher can have this open problem ready to "trouble" those notions for the quickest students. $\endgroup$ – David Feldman Jul 2 '12 at 20:39
  • $\begingroup$ Thanks for the advice, but for now I'll take my chances. Moderation is good for the site, but censorship isn't. So I think it would set a bad precedent now for me to fold pre-emptively. $\endgroup$ – David Feldman Jul 2 '12 at 20:41
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    $\begingroup$ Started META: tea.mathoverflow.net/discussion/1540/… $\endgroup$ – Alexander Chervov Feb 22 '13 at 7:36
  • $\begingroup$ I deleted several of my comments dating from last July as they meanwhile completely lack context. (The 'conversation' gets sort of destroyed but this feels not so necessary, too. If your really care thee is "back-up" on meta.) $\endgroup$ – user9072 Feb 22 '13 at 23:35

10 Answers 10


The Hot spot conjecture The conjecture seems quite amazing and simple to formulate, it can be even understood by persons "from the street" seems its prediction can be tested experimentally. It is a subject of "polymath project 7". Let me quote:

The hotspots conjecture can be expressed in simple English as:

Suppose a flat piece of metal, represented by a two-dimensional bounded connected domain, is given an initial heat distribution which then flows throughout the metal. Assuming the metal is insulated (i.e. no heat escapes from the piece of metal), then given enough time, the hottest point on the metal will lie on its boundary.

In mathematical terms, we consider a two-dimensional bounded connected domain D and let u(x,t) (the heat at point x at time t) satisfy the heat equation with Neumann boundary conditions. We then conjecture that

For sufficiently large t > 0, u(x,t) achieves its maximum on the boundary of D

This conjecture has been proven for some domains and proven to be false for others. In particular it has been proven to be true for obtuse and right triangles, but the case of an acute triangle remains open. The proposal is that we prove the Hot Spots conjecture for acute triangles! Note: strictly speaking, the conjecture is only believed to hold for generic solutions to the heat equation. As such, the conjecture is then equivalent to the assertion that the generic eigenvectors of the second eigenvalue of the Laplacian attain their maximum on the boundary. A stronger version of the conjecture asserts that

For all non-equilateral acute triangles, the second Neumann eigenvalue is simple; and The second Neumann eigenfunction attains its extrema only at the boundary of the triangle.

(In fact, it appears numerically that for acute triangles, the second eigenfunction only attains its maximum on the vertices of the longest side.)


May be this problem can be mentioned when teaching determinants and in particular:

$\det(AB)= \det(A)\det(B).$

There are so-called Capelli identities which generalize this formula for specific matrices with non-commutative entries. In the paper Noncommutative determinants, Cauchy-Binet formulae, and Capelli-type identities by Sergio Caracciolo, Andrea Sportiello, Alan D. Sokal they formulate certain conjectures of the type $$\det(A)\det(B)=\det(AB+\text{correction})$$ on the page 36 (bottom), conjectures 5.1, 5.2.

I think these are quite non-trivial, but probably some smart young mathematician may solve them, given some amount of time (some months may be). I spent some amount of time thinking on them without success, and moreover let me mention that D. Zeileberger and D. Foata also failed to find a combinatorial proof of the Capelli identity of very similar type -- the one proved by Kostant-Sahi and Howe-Umeda -- see their comments in Combinatorial Proofs of Capelli's and Turnbull's Identities from Classical Invariant Theory page 9 bottom: "Although we are unable to prove the above identity combinatorially ... ". So words above are some idications of non-triviality of the conjectures.

Personally I am quite interested in a proof, probably it can give clue for further generalizations.

  • $\begingroup$ Also, a comment on the Foata-Zeilberger paper: Zeilberger's website ( math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/capelli.html ) has a list of errata that I made long ago when I was reading that paper (arguably this was an email I sent to Zeilberger, so half of the remarks are irrelevant to the reader, but the other half I believe to be quite useful). I liked the paper a lot; it was one of the first things that convinced me of the power of combinatorics to prove difficult algebraic results. $\endgroup$ – darij grinberg Jun 7 '15 at 13:07
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    $\begingroup$ Hot spots for acute triangles was indeed solved by some beginners. arxiv.org/abs/1802.01800 $\endgroup$ – Chris Judge Nov 26 '19 at 20:21

These two are probably my favourite.

We all know that cardinals are ordered by injections, i.e. $|A|\leq|B|$ if there is an injection from $A$ into $B$. But it's clear that we can also order the cardinals by surjections, $|A|\leq^\ast|B|$ if $A$ is empty, or $B$ can be mapped onto $A$.

Assuming the axiom of choice these two notions are obviously equivalent. But what happens without the axiom of choice?

Well, it is easy enough to give counterexamples that $|A|\leq^\ast|B|$ and $|A|\nleq|B|$ (e.g. Dedekind-finite sets that can be mapped onto $\omega$; $\Bbb R$ can be mapped onto $\aleph_1$ in Solovay's model).

The Partition Principle. If $|A|\leq^\ast|B|$ then $|A|\leq|B|$.

For over a century now it is open whether or not this principle implies the axiom of choice in ZF. Russell claimed to have a proof, but it was never published.

The second open problem is also related to cardinals and their order:

Assuming the axiom of choice all the cardinals are ordinals, and since there are no decreasing sequences of ordinals, the $\leq$ relation is well-founded.

But what happens without the axiom of choice? Well we know that every partial order can be realized as cardinals of some model of ZF (in fact we can assume dependent choice to hold as high as we want to). An immediate consequence is that it is consistent that the cardinals are not well-founded. Even the existence of a Dedekind-finite set implies this easily, because removing one element decreases the cardinality and so we can find a decreasing sequence of cardinals (but we can't find a decreasing chain of subsets!).

The question whether or not the assumption that there is no decreasing sequence of cardinals implies the axiom of choice in ZF is open, for about a century as well.

Unlike the previous problem where pretty much everyone feels that the answer is positive, this problem had people arguing for both sides. Some people suggested that it will imply choice, others suggested that it will not imply the axiom of choice. But no one has an answer yet.

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    $\begingroup$ I have this huge fantasy that the partition principle will not prove the axiom of choice, that would be epic in ways beyond words. I'd love for that to happen. $\endgroup$ – Asaf Karagila Feb 22 '13 at 23:46
  • $\begingroup$ In the absence of DC, one might modify your second question to ask about the status of "The ordering of cardinals is well-founded." $\endgroup$ – Andreas Blass Feb 23 '13 at 13:43
  • $\begingroup$ Andreas, I agree. But that's the historical question. An interesting question which I have never really given my full attention (for a sufficient time, anyway) is whether or not when the cardinals are not well-founded there is a decreasing sequence. The only case to verify is when there are no Dedekind-finite sets, and DC still fails. $\endgroup$ – Asaf Karagila Feb 23 '13 at 16:14

Does $\lim_{n\to\infty}R(n,n)^{1/n}$ exist? (Where $R(n,n)$ is the classical Ramsey number.)

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I think the following were already mentioned in answers to Not especially famous, long-open problems which anyone can understand

Are there circulant Hadamard matrices of degree > 4?

What is the actual value of R(5,5)?

The integer brick problem.

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    $\begingroup$ I must confess I don't see that much is gained by having this separate question in addition to the one I linked to. $\endgroup$ – Yemon Choi Jul 2 '12 at 23:21
  • $\begingroup$ Then, perhaps you could suggest that the questions should be merged ;) $\endgroup$ – user9072 Jul 2 '12 at 23:35
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    $\begingroup$ My parents have tapes of talks of mine. They play them (perhaps the first one each time) when they have trouble sleeping. $\endgroup$ – Douglas Zare Jul 3 '12 at 7:16
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    $\begingroup$ If I wasn't actually clear enough -- In my mind the difference between this question and my previous one comes down to this: what might I say to suggest to my American state university mathematics majors in their junior year what it is that professional mathematicians do, versus what might I say in the direction of accomplishing the same when speaking to parents of prospective frosh. $\endgroup$ – David Feldman Jul 5 '12 at 16:43

The Dixmier conjecture in rank $n$ asserts that any endomorphism of the $n$-th Weyl algebra (the algebra of polynomial differential operators in $n$ variables) is invertible.

See "The Jacobian Conjecture is stably equivalent to the Dixmier Conjecture" Alexei Belov-Kanel, Maxim Kontsevich

A conjecture (Kontsevich???) which says that the automorphism group of the Weyl algebra in characteristic zero is canonically isomorphic to the automorphism group of the corresponding Poisson algebra of classical polynomial symbols.

See "Automorphisms of the Weyl algebra" Alexei Belov-Kanel, Maxim Kontsevich

Kaplansky's 6th conjecture: dim(Irrep) | dim(algebra) - for semi-simple Hopf algebras

See Kaplansky's 6th conjecture: dim(Irrep) | dim(algebra) - for semi-simple Hopf algebras

Kaplansky zero-divisor conjecture Let K be a field and G a group. The so called zero-divisor conjecture for group rings asserts that the group ring K[G] is a domain if and only if G is a torsion-free group.

See What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Zero divisor conjecture and idempotent conjecture


Is there a Borel set in the plane which meets every straight line in exactly two points?

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Given $4$ involutions $\alpha_1,\alpha_2,\alpha_3,\alpha_4\in\text{Sym}(\mathbb N)$, do there exist $3$ involutions $\beta_1,\beta_2,\beta_3\in\text{Sym}(\mathbb N)$ such that $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}\subseteq\langle\beta_1,\beta_2,\beta_3\rangle$?

This is a slightly restated form of the following question from p. 231 of F. Galvin, Generating countable sets of permutations, JLMS 51, 1995:

Question 1.4. Let $E$ be an infinite set. Is every countable subset of $\operatorname{Sym}(E)$ contained in a group generated by three involutions?

  • $\begingroup$ Could you please give a reference for your question? $\endgroup$ – Alireza Abdollahi Jun 7 '15 at 9:22
  • $\begingroup$ @AlirezaAbdollahi I found a reference so I'm going to edit it into my answer. $\endgroup$ – bof Jun 7 '15 at 11:34

Given a unit square, can you find any point in the same plane, either inside or outside the square, that is a rational distance from all four corners?

Or, put another way, given a square $ABCD$ of any size, can you find a point $P$ in the same plane such that the distances $AB$, $PA$, $PB$, $PC$, and $PD$ are all integers?

Rational distance problem


  1. Guy, Richard K. Unsolved Problems in Number Theory, Vol. 1, Springer-Verlag, 2nd ed. 1991, 181-185.

  2. Barbara, Roy. "The rational distance problem", Mathematical Gazette 95, March 2011, 59-61.

  • $\begingroup$ Surely the re-wording isn't quite what you want—given a square $ABCD$ of any size, either all choices of $P$ or no choices give $AB$ integral. Was there supposed to be some scaling? $\endgroup$ – LSpice Jan 31 at 15:16

Does there exist a completely separable MAD family in ZFC? That is, an infinite family $\mathcal S$ of infinite subsets of $\mathbb N$ such that (1) the intersection of any two members of $\mathcal S$ is finite, and (2) every subset of $\mathbb N$ either contains a member of $\mathcal S$ or else is covered by finitely many members of $\mathcal S$.

It's an easy transfinite construction assuming the continuum hypothesis; that's why only a construction in ZFC counts.


Showing that if product of $ n $ Toeplitz operators is again a Toeplitz Operator.this is open and quite elementary to state. http://www.mathnet.or.kr/mathnet/thesis_file/WYLee.pdf

  • $\begingroup$ Would you be so kind to give more details ? what is n Toeplitz ? $\endgroup$ – Alexander Chervov Feb 23 '13 at 9:22
  • $\begingroup$ What are the references ? $\endgroup$ – Alexander Chervov Feb 23 '13 at 9:22
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    $\begingroup$ You reference asks for necessary and sufficient conditions that will make the product of Toeplitz operators a Toeplitz operator. $\endgroup$ – David Feldman Feb 23 '13 at 22:58
  • $\begingroup$ The link appears to be broken. $\endgroup$ – JoshuaZ Apr 24 at 17:22

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