A Kakeya set is a subset of \R^n containing a segment of unit length in every direction. Kakeya conjecture states that any such set must have the full dimension. For simplicity, we will be talking about the Minkowski (box-counting) dimension.

The conjecture is still widely open for any d>2 (proved for d=2 by Davies). The best known lower bound being 5/2+eps in dimension 3, 3 in dimension 4 by the result of Wolff, and (2-sqrt{2})*(n-4)+3 in higher dimensions by this result of Katz and Tao. A stronger conjecture that any Kakey set has positive Lebesgue measure is wrong, a counterexample was first given by Besicovitch.

Here is a short proof sketch of the 4n/7 lower bound, which is asympototically only slightly worse than the bound of Katz and Tao.

1)Let $S \subset \R^n$ be a Kakeya set. Discretize the problem by considering a lattice (\eps\Z)^n of mesh \eps and taking A \subet (\eps\Z)^n to be the set of all lattice points which are \eps close (in the L^\infty metric) to at least one point in the set. Since, one can take roughly \eps^{1-n} ~unit vectors in the lattice which are 5\eps separated from each other, the resulting set has the property that A-A contains at least of order (1/\eps)^{n-1} elements all close to unit length. The goal is to lower-bound the cardinality of A.

2)Notice that a trivial bound |A-A|<|A|^2 (there are at most |A|^2 pairs of points) already implies a bound of (n-1)/2 on the Minkowski dimension of S. To improve upon it, notice that the original set S must also contain a point close to the mid-points of the segments. So if \Gamma=(V, E) is the graph having A as the set of vertices and edges for different segments of almost unit length (and so the number of edges |E| is at least of order \eps^{1-n}), then we have that it's enough to show that A+_Gamma A (i.e. we are allowed to add elements connected by an edge) is large. Morover, choosing slightly coarser grid and then choosing points not at the ends of the segment but closer to the middle we arrive at the following problem:

Let \Gamma be a graph as before and assume that A-_\Gamma A is large. Then the set
X(A):=\cup{|p|, |q| < 10, p+q\not= 0}\{pA +_\Gamma qA\} is also large.

The idea is that if (x, y) is a segment then px+qy up to a scaling factor of p+q gives a point spliting this segment into a given ratio. Possibly the point is outside the segment, that's why one would want to choose points closer to the middle.

The connection with additive combinatorics outlined above must have been first noticed by Bourgain.

3)Now, if the number of vertices is almost as small as |E|^{1/2} (which we want to rule out), then the graph is dense and must contain a lot of cycles of length 4, so let's try to upper bound this number of cycles. For simplicity, if the graph G (embedded in \R^n) has two edges which are equal as vectors we delete one of them, so that each of order (\eps)^{1-n} almost unit vectors now corresponds exactly to one edge of \Gamma.

Let x, y, z, t form a cycle in \Gamma. Observe that we can reconstruct all four points knowing only
(x-2y, 2y-3z, 3z-t). Indeed, summing all three values we get x-t, which allows us to reconstruct x and t, and then we can reconstruct y and z easily.

This means that the number of 4-cycles in the graph is at most |X(A)|^3. Application of Cauchy–Schwarz inequality twice shows that any graph with n vertices and m>>n edges has at least m^4/n^4 cycles of length 4. Putting this together we see that |X(A)|^3>|E|^4/|A|^4. Trivially, X(A) also contains A, so we deduce that |X(A)|^7>|E|^4. Since the number of edges is of order \eps^{1-n}, this implies that X(A) has size at least \eps^{-4(1-n)/7} which gives a lower bound of 4(n-1)/7 on the Minkowski dimension of the original set S.

A simple application of the power product trick then turns the bound 4(n-1)/7 into 4n/7.

The above argument was explained to me by Zeev Dvir. Quite a bit longer argument is given in his survey Incidence Theorems and Their Applications, see Section 4.1. there.

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One can further "project" the problem from the lattice to \Z arriving at the following conjecture, known as the arithmetic Kakeya conjecture (see this paper by Green and Rusza for several equivalent formulations), which would imply the original Kakeya conjecture:

Conjecture: There exists a sequence of constants \eps_k tending to zero such that the following holds. Let N be a large positive integer and let A\subset \Z be a set of integers containing a k-term arithmetic progression with difference d for each d=1, 2, 3, \dots, N. Then |A|>N^{1-\eps_k} provided N is large enough.

In fact, as observed by Bourgain, to deduce the original Kakeya conjecture it would be enough to show the above with k=N^\eta for arbitrary small \eta, showing a bound of N^{1-o(1)}.

Conjecture: Given n points A_1,...A_n on the plane, it is possible to choose n new points B_1,...,B_n distinct from them, such that any disk containing at least two points among A_j's also contains at least one of the B_j's.

This is a [possibly incorrect] strengthening of Conjecture 1 from https://arxiv.org/pdf/1907.01617.pdf

Also mentioned in that paper: Every Delaunay triangulation contains a perfect matching. This is a weak version of the problem I mentioned recently: Prove that for every 2n points on the plane there exists a set of n disjoint circles each of them containing exactly two of the given points.
Indeed, the Delaunay trianglulation contains an edge PQ if and only if there exists a disk containing P and Q but no other points from the set.

Here is a problem communicated to me by Daniel Zhu:

Let A_1…A_n be finite sets and let P be their product. A sub-box of P is defined to be a product \prod B_j, in which each of B_j’s is a proper subset of the corresponding set A_j. Prove that P cannot be split into fewer than 2^n sub-boxes.

Solution: For each element of each of A_j’s attach a random variable in the field with two elements and for an element in P attach the product of n corresponding random variables (one from each coordinate). Now, take this variables uniformly at random conditioned on each set having sum of variables equal to 1. Then the sum in each sub-box is 1 with probability 2^{-n} (as any proper subset of A_j has sum 0 with probability 1/2) and since the sum in the whole box P is 1 we must have at least 2^n boxes by the union bound.

Apparently, this was first proven in this article by Noga Alon et.al. According to the article, the question was asked by Petar Markovic

Educational aspects of mathematics:

Today I read on wiki that Euler, in 1755, managed to compute 20 digits of \pi within one hour by using some Machin-like formula. Seems a bit of a cheating as I'm sure he had known many digits of pi by heart, but anyway.

So today in the calculus class that I'm teaching here at Princeton, I chose a similar quest adopting the difficulty to the class: We computed \sqrt{26} with a relative error of less than 10^{-7}! Also took us an hour almost. Apparently, students seemed way more excited about this way of learning what linearization / differential is

Today I came across Murmurations of Elliptic Curves (some fascinating plots inside!) and Peter Sarnak's take on this phenomenon, in the format of a 24-page-long handwritten letter!

Does there exist a fixed natural number k such that any finite set of points on the plane has a line through at least two of its points where the number of points on either side of this line differ by at most k?

This question was first raised by Kupitz (and apparently later by many others) and remains open to this day. The best example known is a configuration requiring k to be at least 2. In the opposite direction, Pinchasi proved that k=O(\log\log n) works, where n is the cardinality of the set of points. This improved on the previous bounds of n/3 by Kupitz, O(\sqrt n) Alon, and O(\log n) by Perles.

The proof of Pinchasi doesn't use much about straight lines and, in fact, extends to any generalised configuration, which is collection of points together with a pseudoline arrangement such that each pseudoline contains at least two points and there is a pseudoline through any pair of points. Recall that a pseudoline arrangement is a collection of two-way unbounded simple curves any two of which meet in at most one point.

Very recently, David Conlon and Jeck Lim proved that for generalised configurations the result of Pinchasi is essentially optimal: There exists c>0 such that for large enough n there is a generalised configurations where the number of points on either side of each pseudoline differ by at least c* \log \log n. They also conjecture that similar bound may be true for the original question of Kupitz.

Apparently, taking functions f(x)=x+sin(2x)/2 and g(x):=f(x)*e^{sin(x)} one has

f'(x)=2*cos(x)^2 and g'(x)=e^{sin(x)}*cos(x)*[2cos(x)+x+sin(2x)/2]

So naïvely applying L'Hôpital's rule and cancelling cos(x) one would get that

\lim_{x --> \infty} f'(x)/g'(x) = 0 and so \lim_{x --> \infty} f(x)/g(x) = 0,

which is clearly nonsense. This "counterexample" is due to Otto Stolz. Here is a more general construction.

Here is a story of one particular PhD General Examination at Princeton. I am kinda glad I didn't have to go through it! Never have I ever knew enough of maths to pass this sort of examination.

Mathematically interesting from the story: the question "Given, positive integers N, A, B, does there exist a divisor of N which is in [A, B]" is NP-complete. I haven't managed to google the reduction to the subset sum and the naïve approach seems to require being able to deterministically produce large primes which is hard. So if anyone knows the answer and/or is more proficient in googling, let me know.

Also, apparently Peter Sarnak thinks that factoring is in P. Not sure how wide-spread this belief is.

Marden's theorem states that if complex numbers a, b, c forming a triangle on the complex plane are roots of some cubic polynomial P(x), then roots p, q of its derivate P'(x) are exactly the foci of the unique ellipse inscribed in the triangle and touching its sides at the midpoints of edges.

Here is a short proof which came out of a discussion with Daniel Zhu (though I'd thought everyone knew roughly this proof). Without loss of generality we assume that P is monic.

1)Note that (p-a)(q-a)=P'(a)/3 and (c-a)(b-a)=P'(a) which implies that p and q are isogonally conjugate in the angle BAC. Similarly, for the other two angles of the triangle. This means that P and Q are isogonally conjugate in ABC and hence there is an ellipse E with foci P, Q inscribed in ABC (P, Q are inside the triangle by the Gauss-Lucas theorem).

2)Note that (a+b+c)/3=(p+q)/2 is the unique root of P''(x), so the center of this ellipse is the barycenter of the triangle. Hence, concidering an affine transformation which maps ellipse E to a circle, we obtain a new triangle in which the incribed circle has center coinsiding with the barycenter, so this triangle will be an equilateral triangle. Inscribed circle of the equilateral triangle touches its sides at midpoints so the same is true for the original ellipse.

Apparently, Terrence Tao believes (see his lecture on youtube and also this paper of his) that Navier-Stokes equations can blow up in finite time. Do many people believe that?

His argument is roughly: Water is "Turing complete" so it should be possible to somehow push all the energy into smaller and smaller part of space making the solution blow up (as the equation is supercritical, whatever that means).

Anyone more knowledgeable about these things is very welcome to explain to me how solid this heuristics is!

My coauthor Zhi-Wei Sun is known for publishing a plethora of conjectures every few years, many of them are mathematically meaningless (i.e. not shedding any light on anything, sometimes clearly way too hard for the current methods, etc).

Most of his conjectures involve prime numbers (typical example: any integer n>1 can be written as a+b such that a+2^b prime), infinite sums involving binomial coefficients, polynomials and exponents, or sums modulo a power of a prime number. There is some not yet well understood correspondence between infinite sums having explicit values and their p-adic analogues where trancated sum up to p-1 can be evaluated modulo some power of p.

Here is an example of a paper of Sun with ~100 new conjectures of the second and third kind. Some of the conjectures of Sun are proved by very elementary methods. Some of them are proved by literally asking Mathematica "Prove this identity" and waiting for an hour, some are proved using non-trivial ideas. But usually all the methods are rather ad hoc and don't provide much insight.

I recently came across a paper by Kam Cheong Au which explains what methods exist for proving these sort of conjectures and then explains a new connection between some of them. I found most of the paper hard to read but the introduction is very nice. Au also proves dozens of Conjectures of Sun along the way

One of the conjectures Au proves

Back to more reasonable mathematics.

One of the beautiful and fascinating things about planar percolation is the existence of rational critical exponents, i.e. powers in the polynomial decay of quantities at criticality. For instance, consider this statement:

Theorem: On the hexagonal grid, color each cell black or white with probability 1/2. Then the probability that the origin is connected to a cell at distance N from it decays as N^{-5/48+o(1)}.

This result was first proved here as a consequence of conformal invariance (e.g. Cardy's formula) proved by Smirnov (here is a modern exposition of his proof). Some other exponents, for example, the full-plane two-arm exponent (corresponding to the probability of having both a white and a black path of cells to distance N), are easier to compute, see this paper. All computations of arm exponents followed the same strategy: use Cardy's formula to show SLE_6 convergence of interfaces and then compute the corresponding "continuous" version of exponents for the SLE_6 process. At this point it usually doesn't matter if \kappa=6 or not.

The question then boils down to some PDE computation which an observable of SLE_\kappa satisfies. One prominent exponent which was not computed in this way is the so-called backbone exponent: corresponding to the probability of having two disjoint arms to distance N. The problem isn't that the stochastic calculus is harder for this case but rather that the PDE question couldn't be answered. Here is the PDE question (see page 16 here):

Let T be the triangle given by {x>0, y>0, x+y<2*\pi}. Consider an operator

D:=3(\partial_x^2-2\partial_x\partial_y+\partial_y^2)+\cot{(x/2)}\partial_x-\cot{(y/2)}\partial_y

Then the backbone exponent is the unique \lambda > 0 such that -\lambda is the eigenvalue corresponding to a non-negative function G(x, y) on T with boundary conditions

G(x, 2*\pi-x) = 0, G(0, y) = 1, \partial_y G(x, 0) = 0.

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The authors, smart as they are, couldn't solve this PDE explicitly and so no formula for the backbone exponent was known. It had been assumed to be 6/17 (though I struggle to find any explicit conjectures about it from mathematicians, only from some physicists from 80's and 90's) but later simulations showed this is likely not the case. Very recently the following result was proved using Liouville quantum gravity (LQG) techniques:

Theorem: Backbone exponent is equal to (x^2-1)/12, where x \in (2, 3) is the unique root of
\sqrt{3}*x/4+\sin(2\pi x / 3) = 0.

Given how complicated the proof is, a natural question to ask is: Could this implicitly given value suggest something about the solution of the PDE, leading to a much simpler derivation of the exponent?

Here is the PDE question from the paper, in case I copied it with typos

Some news from the world of additive combinatorics: polynomial Freiman–Ruzsa conjecture in \F_2^n was recently proved by Timothy Gowers, Ben Green, Freddie Manners, and Terrence Tao. Moreover, the constants are very explicit; they prove that

If a set A\subset \F_2^n has doubling constant at most K (i.e. |A+A| <= K|A|) then A can be covered by at most 2*K^12 cosets of some subgroup H\subset \F_2^n of size at most |A|.

See also a paper A Counterexample to a Strong Variant of the Polynomial Freiman-Ruzsa Conjecture in Euclidean Space for a discussion of the analogous conjecture in \R^n.

Given an n-dimensional cube, any k-dimensional central section which is close to being a sphere (I.e sandwiched between a sphere and its dilate by a factor of 1.1 say) must have k<5*log(n)

Proof sketch: Let T: L^2(\R^k) --> L^\infty (\R^n) be the embedding together with a dilation mapping the boundary of the section to almost the unit sphere of \R^k. Then |x|_{L^2(\R^k)} and |Tx|_{L^\infty(\R^n)} are within 1.1 factor from each other for any vector x in the section. Consider T*, a mapping from the dual spaces, i.e. T*: L^1 (\R^n) --> L^2(\R^k). Then e_j\in L^1 (\R^n) is mapped to some vector y_j\in L^2(\R^k) which is almost of norm 1. We claim that any point y in the unit sphere of L^2(\R^k) is close to one of \pm y_j which ensures that n (half of the number of points \pm y_j) is exponentially large in terms of k.

Indeed, any point x on the boundary of the section has one coordinate (in \R^n) equal to \pm 1, which means that |<x, e_j>|=1. Since T is almost an isometry, Tx has large L^2 scalar product with \pm y_j, which implies that Tx is close to \pm y_j. Since Tx can approximate, within a factor of 1.1, any point on the unit sphere of L^2(\R^k), this completes the proof.

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This shows that cubes are asymptotically the worst for Dvoretzky's theorem: Any centrally symmetric convex body in \R^n has a section of dimention at least C(\eps)*log(n) which is 1+\eps close to a sphere (This quantitative version was only later shown by Milman, who also showed that a random section works).

In some sense a cube is a fundamental counterexmaple: If you're OK with the section being close to either a sphere or a cube (i.e. to an L^2 or an L^\infty unit sphere) then N.Alon and Milman show that a much large dimensional section (with dimension k=e^{C(\eps)*\sqrt{\log{n}}}) is possible.

Let {x} denote the fractional part of x. Then for any positive real numbers x_1,\dots, x_n one has
\sum_i\sum_j {x_i/x_j} < 9n^2/14 and the constant 9/14 is asymptotically sharp.

The inequality is taken from here, an elementary proof is given by Fedya, and a different approach which leads to a weaker bound is outlined by another Fedya.