Not especially famous, long-open problems which higher mathematics beginners can understand - MathOverflow

Not especially famous, long-open problems which higher mathematics beginners can understand

2012-07-02T18:25:01Z

This is a pair to

http://mathoverflow.net/questions/100265/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.

Thanks

Answer by Alexander Chervov for Not especially famous, long-open problems which higher mathematics beginners can understand

2012-07-02T21:04:50Z

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.)

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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.

Answer by Yemon Choi for Not especially famous, long-open problems which higher mathematics beginners can understand

2012-07-02T23:21:03Z

I think the following were already mentioned in answers to http://mathoverflow.net/questions/100265/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.

Answer by Asaf Karagila for Not especially famous, long-open problems which higher mathematics beginners can understand

2013-02-22T23:27:31Z

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.

Answer by Koushik for Not especially famous, long-open problems which higher mathematics beginners can understand

2013-02-23T00:43:44Z

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