# On progress in mathematics: some long-open problems and long-standing conjectures

I would like to ask a question here on Math Stack Exchange taking inspiration (and therefore combining) from two well-known threads on MathOverflow: (1) Not especially famous, long-open problems which anyone can understand; (2) The resolution of which conjecture/problem would advance Mathematics the most?

My question is:

Could you list (giving a sound motivation and reference papers) some long-standing conjectures or long-open problems in mathematics (or even in mathematical physics) that are intrinsically interesting and

(1) are not so "mainstream" (by which I mean: there is no need to mention the Millenium Problems or Riemann Hypothesis or similarly famous questions);

(2) are extremely important because their solution would imply a major progress in an area of mathematics and also in mathematics as a whole;

(3) can be stated in some appropriate, but reasonably terse, form without involving extremely abstruse concepts and terms;

(4) have been object of some (even slight) progress towards a solution in recent years.

• I would like to make this community wiki. – Dal Dec 23 '14 at 1:08
• can abc conjecture be included? – happymath Dec 23 '14 at 1:13
• @happymath we might as well (even though I guess that it is tremendously famous). – Dal Dec 23 '14 at 1:14

$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$
$\large\hspace{3in}I=4\,\pi\operatorname{arccot}$$\sqrt\phi But these techniques don't apply to every definite integral. In particular this question asks for a closed form solution to this definite integral:$$\int_{0}^{\Large\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx$$It's unknown what the closed form solution to this is, and it may be possible to nonconstructively prove it exists without actually being able to say what it is. The answer is ambiguous as to whether the particular closed form solution for this exists. Now if these sorts of definite integrals had closed form antiderivitives, one could simply apply the fundamental theorem of calculus to give closed form solutions. But the Risch algorithm tells us closed form antiderivitives to these expressions don't exist so we must resort to heuristics that apply to specific classes of definite integrals. Some of them are quite exciting, like the countour integration approach used in the first problem; but this is far short of a generic algorithm analagous to Risch for definite integrals. What's astounding to me is how little press this problem gets compared to big name problems of moderate interest and little applicable value like the Goldbach conjecture. http://mathworld.wolfram.com/DefiniteIntegral.html • What would it mean to have an algorithm for definite integrals? Any definite integral of a computable function which exists can be algorithmically approximated, so you must mean some sort of exact form. – Kevin Arlin Dec 23 '14 at 2:50 • Expressible in terms of elementary functions. Antiderivitives for many continuous functions exist, but the Risch algorithm doesn't just tell us if the antiderivitive exists but if it's expressible in terms of elementary functions. There's no similar algorithm known for definite integrals. – dezakin Dec 23 '14 at 20:37 • I guess you're talking about a definite integral with variable endpoints-what's the difference between this and finding an antiderivative, exactly? – Kevin Arlin Dec 23 '14 at 22:48 • The fundamental theorem of calculus can get you a closed form definite integral from a closed form antiderivitive, but there are closed form solutions to definite integrals for functions that don't have closed form antiderivitives. What we lack is a general algorithm for answering the question of whether these closed form solutions to definite integrals exist and what they are. We just have rules of thumb that apply to certain classes of definite integrals. The wolfram article goes into some detail. – dezakin Dec 23 '14 at 23:18 • @KevinCarlson: Not only variable endpoints. A definite integral can also contain some free parameters in its integrand. But even then it's still possible to solve numerically. – Han de Bruijn Dec 29 '14 at 15:29 Given a closed topological manifold$M$, we can ask whether$M$admits a smooth structure and whether that structure is unique. It's easy to show that it's the case in dimensions$1$and$2$, and it's nontrivial but also true in dimension$3$. It fails in dimension$4$, but there has been a lot of progress on trying to determine which$4$-manifolds are smoothable. Take$M$to be a simply-connected, closed$4$-manifold. For purely algebraic-topological reasons (although there's also a straightforward geometric proof; see, for example, Scorpan's very readable Wild World of$4$-Manfiolds),$M$is determined up to homotopy equivalence by its intersection form$i_M:H_2(M) \otimes H_2(M) \to H_4(M) = \mathbb{Z}$. (Since$\pi_1 M = 0$, the manifold$M$is clearly orientable, and we take it to be oriented.) There's another invariant, called the Kirby-Siebenmann invariant$\operatorname{ks}(M)\in H^4(M)$, that's an obstruction to smoothability. If we assume that$\operatorname{ks}(M)$vanishes, we can now ask whether which forms$i_M$correspond to smooth manifolds. The definite case is taken care of by results of Freedman and Donaldson from the '$80$s: If$i_M$is definite, then$M$is smoothable iff$i_M$is diagonalizable. (Recall that we're working over$\mathbb{Z}$throughout.) If$i_M$is indefinite and odd, then it's automatically smooth. The case of$i_M$indefinite and even, though, is currently unknown. Such an$i_M$must be of the form$i_M = H^{\oplus n} \oplus E^{\oplus 2m}$, where$H$and$Eare certain forms with \begin{align*} \dim H &= 2 & \sigma(H) &= 0;\\ \dim E &= 8 & \sigma(E) &= \pm 8 \end{align*} (The sign of the signature\sigma(E)$depends on some orientation conventions that I've been too lazy about above to define rigorously now.) If$m \leq 2n$, then it's known that$M$is not smoothable. If$m\geq 3n$, then it's known that$M$is smoothable. The general case, though, is unknown. (There are a few other pairs$(n, m)$for which the answer is known.) Since$\dim i_M = 2n + 16m$and$\sigma(i_M) = \pm 16m$, we can restate the results above as saying that$M$is smoothable if$\dim i_M \geq \frac{11}{8} |\sigma(M)|$and not smoothable if$\dim i_M \leq \frac{10}{8} |\sigma(M)|$. The$11/8$-conjecture states that the$\frac{11}{8}$is the best constant possible in the former half of the previous sentence: any$M$with$\dim M < \frac{11}{8} |\sigma(M)|$has no smooth structure. What makes this particular situation so interesting is that a lot of the tools available in higher dimensions--- the$h$-cobordism theorem, handlebody decompositions, surgery theory, etc.--- don't apply or aren't as pervasively useful in dimension$4$. On the other hand, the low dimension simplifies matters, as in the determination of$M$up to homotopy equivalence by just the form$i_M$. There are quite a bit of odd phenomena that are exclusive to dimension$4$, such as exotic$\mathbb{R}^4$s. The$\frac{11}{8}$-conjecture is an attempt to figure out what exactly is going on in these odd cases. • Just to check: when you write$\dim M$in the second-to-last paragraph, you mean$\dim i_M$, right? – tracing Dec 23 '14 at 15:02 • I do, thanks! (I suppose 'rank' is a better term, since$i_M\$ is nondegenerate by Poincare duality, but 'dimension' is common too.) – anomaly Dec 23 '14 at 15:59