# Interesting (and not sufficiently answered) questions on math.SE

In light of the generally positive response to this meta-thread, I am creating this to help advertise (and hopefully fully resolve) "interesting" questions on math.SE. I'm not really certain what the guidelines for this should be, but the following are likely uncontroversial:

1. Each answer should link directly to a single question on math.SE which has not been sufficiently resolved (and which is answerer believes would be of interest to members of the MO community).
2. Each answer should include a short summary of the question asked. There's no need to repost the math.SE question verbatim, but it should be fairly clear what the question is without following the link.
3. Feel free to downvote/vote to delete suggestions that aren't "up to snuff". Of course, commentary about why such questions aren't really of interest to the MO community would also be valuable.
4. Abusing this thread is an easy way to ensure it is closed/deleted. This thread will stay open only as long as the MO community feels it serves a useful purpose. In particular, don't post links to math.SE question that you know are open problems, and, at the other end of the spectrum, don't post links to math.SE questions that are essentially homework exercises.

(More active MO members should feel free to massage and expand these guidelines to better fit this community's expectations.)

• I think we should add that questions that the user knows are known [difficult] open problems should be off limits here. If someone asks about some well known open problem, and obviously receives no answers, it is not too productive to post it here. Of course, if a user doesn't know that a question is an open problem (e.g. they're not from the relevant field), then there's no issue here. – Asaf Karagila Nov 8 '14 at 21:17
• @Asaf: Maybe I was hoping for "good behaviour". But you're right that it's probably best to state some obvious things. – user642796 Nov 8 '14 at 21:29
• "The first rule of this thread is the first rule of this thread". :-) – Asaf Karagila Nov 8 '14 at 21:37
• I haven't thought these interesting suggestions through myself, but I see no problem in placing the thread under protection (as requested). – Todd Trimble Nov 8 '14 at 22:18
• In the case of a good question, I'd encourage people to ask the questions here at MO, rather than merely to post links via this thread. – Joel David Hamkins Nov 9 '14 at 16:11

Okay, here's one that I think is interesting and has remained unanswered for over a month. (It might also test out this system.)

Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?

If it isn't clear from the title, this question (asked by Willie Wong) asks whether the inverse of any connected bijection $\mathbb{R}^n \to \mathbb{R}^n$ is itself connected (where a mapping $f : X \to Y$ between topological spaces — not necessarily continuous — is connected if $f[A]$ is connected for every connected $A \subseteq X$).

Can someone figure out what is going on in this question about representation varieties of associative algebras? It looks like the OP may have found an error in a published article of Smalo, but I can't prove that its an error.

This question asks, given a homogenous degree $n$ polynomial map $f: \mathbb{Z}^n \to \mathbb{Z}$, how can we test if it is a norm map from a number field? I know roughly what the answer must be: Over $\mathbb{Q}^{alg}$, $f$ should factor as $n$ distinct hyperplanes and $f$ should be irreducible over $\mathbb{Q}$. That should make $f: \mathbb{Q}^n \to \mathbb{Q}$ into a norm up to scaling. But I can't quite figure out how to switch from $\mathbb{Q}$ to $\mathbb{Z}$, and the "up to scaling" is annoying.

I realize that "I know roughly what the answer should look like, but can't/don't-want-to work out the details" is not a good way to sell a problem. But it is a natural question and its been bugging me that is unanswered for so long. My hope is that, by advertising it here, someone will know a paper which already does all of this.

Here is a question which seems interesting,

What does it take to divide by $2$?

In a nutshell, we know that in $\sf ZF$, assuming classical logic, we can prove that if $A\times\{0,1\}$ and $B\times\{0,1\}$ are equipotent, then $|A|=|B|$.

How far can this bend before it breaks? Can we do the same in intuitionistic set theories?

(Minor bump: the question currently carries a 500 points bounty.)

• At a first bash, the proof feels like something that would work in a boolean topos but not a more general one (cf discussion here: golem.ph.utexas.edu/category/2014/12/…, where CSB is connected with Boolean algebras) – David Roberts Jan 27 '15 at 8:16
• @DavidRoberts: Thank you for your comment! Unfortunately, the link does't seem to work here. – Hanno Becker Feb 25 '15 at 7:25
• @HannoBecker sorry, some invisible characters ruined that url. Here's the correct one: golem.ph.utexas.edu/category/2014/12/… – David Roberts Feb 25 '15 at 8:56
• In Conway and Doyle's famous paper on division by 3 math.dartmouth.edu/~doyle/docs/three/three.pdf they say Bernstein claimed to have solved your problem constructively, but nobody ever understood his solution. Then in 1922 Sierpinski solved it. They give a reference. – John Baez May 17 '15 at 4:22
• @John: Yes, and I think Hanno mentions that. But Sierpinski treated "constructive" as meaning "without the axiom of choice". We have evolved other meanings for "constructive" since then, and that is the what Hanno is asking about. How far does this claim stretch. – Asaf Karagila May 17 '15 at 5:32

Is there an injective cubic polynomial $\mathbb Z^2\rightarrow\mathbb Z$?

This question asks exactly what its title suggests: is there a polynomial of degree $3$ in $\mathbb Z[x,y]$ whose induced map $\mathbb Z^2\rightarrow\mathbb Z$ is injective? The question is trivial for polynomials of degree $4$ (where the answer is yes) or polynomials of degree $2$ or less (where the answer is no).

I recognize that this may not be wholly in the spirit of the thread, since I have suspicions that the question is open given the existence of two unanswered questions on MO asking about polynomial bijections (one on $\mathbb Q^2$ and one on $\mathbb Z^2$) and one about surjections. However, I suspect that my question is considerably easier, as it asks only about injectivity and only about cubics (where, from my amateur knowledge, it seems like more tools are available than for general polynomials). In any case, I note that there has been some amount of research regarding the linked questions, and think that it's not unlikely that an expert in the field could have some references or insights to contribute to the question.

Does the open mapping theorem imply the Baire category theorem?

The open mapping theorem and the Baire category theorem are both theorems of ZFC, but are independent of ZF. Working in ZF, one can (apparently) prove that BCT implies OMT. Does the converse hold in ZF?

This is currently the highest-voted question with the functional-analysis tag.

• Totally. And it has a completely irrelevant answer that somehow garnered upvotes. – Asaf Karagila Dec 25 '14 at 5:10

Weil Pairing on Linear Algebraic Groups

There is of course a Weil pairing on abelian varieties (it pairs an element of the torsion with something in the torsion of the dual) which are projective algebraic groups but does a similar construction exist for linear algebraic groups?

Would someone who knows cyclotomic fields, or algebraic number theory more generally, please take a look at $A_4$ extension of $\mathbb{Q}$ ramified at one prime? The question asserts that an $A_4$ extension of $\mathbb{Q}$ ramified at only one finite prime $p$ must be totally real.

I have made some partial progress, but I feel like someone who really knows cyclotomic fields and how to use number theory software should be able to kill this.

• This now has an accepted answer. – Stopple Aug 14 at 20:48

While it hasn't received a ton of attention over at Math.SE, I'm particularly fond of my question about whether congruent rational triangles have a mutual rational dissection, since it seems so tantalizingly within reach — these triangles are constrained just enough that it feels like a proof 'should' be out there. (Even a proof of impossibility in the quadrilateral case, along the lines of the example I offered, seems like it would be a major victory.) The problem is unabashedly 'recreational' mathematics, but it's the sort that seems to touch very close to some quite serious math indeed (e.g., the parametrizations by rational points on elliptic curves in http://nyjm.albany.edu/j/1998/4-1p.pdf). I'd love to get more eyes on it if anyone has some thoughts.

Is there a direct proof that a compact unit ball implies automatic continuity?

We know that if a Banach space is infinitely dimensional, then there are discontinuous linear functionals on it. Therefore having only continuous functionals imply finite dimension (at least in the presence of choice).

At the same time, we know that a compact unit ball must imply finite dimension, and therefore every linear functional is continuous.

Is there a direct proof from the compactness of the unit ball, that every linear functional is continuous?

This is a question that I had posted on MathSE. Other than the first comment I got no other response. I tried the suggestions in the comment but could go no further than a certain point. I updated the question accordingly and offered a bounty but still did not receive an answer.I hope this thread brings attention to the post. Despite it not having many upvotes I think it is interesting and should be up here, but if anyone can give me a good reason otherwise I shall take it down. Here it is -

Does Nagata theorem hold in a field that is not algebraically closed

Nagata's theorem states that if $R$ is a finitely generated $k$ algebra and $G$ is a geometrically reductive group acting rationally on $R$ then $R^G$ is finitely generated.

The question is if it holds when $k=\mathbb R$.

Thank you.

Solving Special Function Equations Using Lie Symmetries

This question is asking for a unified, intuitive, & visual, exposition of the classical special function equations (e.g. Hypergeometric) taking ideas from Lie theory, along the lines of the example of Bessel's equation as provided in the question. Lots of interest, no answers, awesome idea! Thoughts in the comments welcome.

Given two fixed boolean algebras $\mathfrak{A}$, $\mathfrak{B}$ I define a poset, consisting of pairs $(\alpha;\beta)$ of functions $\alpha:\mathfrak{A}\rightarrow\mathfrak{B}$, $\beta:\mathfrak{B}\rightarrow\mathfrak{A}$ such that (for every $X\in\mathfrak{A}$, $Y\in\mathfrak{B}$) $$Y\sqcap^{\mathfrak{B}}\alpha(X)\ne\bot^{\mathfrak{B}} \Leftrightarrow X\sqcap^{\mathfrak{A}}\beta(Y)\ne\bot^{\mathfrak{A}}.$$ Is this poset also a boolean lattice?

If this conjecture does not hold in general, does it hold for: a. atomic boolean lattices? b. atomistic boolean lattices? c. complete boolean lattices?

For the special case when $\mathfrak{A}$, $\mathfrak{B}$ are complete atomic boolean lattices, the conjecture follows from this answer.

## protected by Todd Trimble♦Nov 8 '14 at 22:19

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).