Kaiyi user manual (EN)

N UMERICAL S TUDY OF K AKEYA M AXIMAL I NEQUALITIES N UMERICAL S TUDY OF K AKEYA M AXIMAL I NEQUALITIES Terry Harris , Kaiyi Huang , Tianle Chen , Daniel Alexander David , Vivek Saravanan , Marissa Stolt Terry Harris , Kaiyi Huang , Tianle Chen , Daniel Alexander David , Vivek Saravanan , Marissa Stolt University of Wisconsin - Madison University of Wisconsin - Madison Problem Introduction Greedy Algorithm The SL2 Example The Kakeya conjecture is a fundamental problem in geometric measure theory and harmonic analysis. It concerns the minimal possible size of sets that contain a unit line segment in every direction in R n .

One way to approach this problem is through the study of Kakeya maximal function inequalities. The aim of our project is to find some numerical evidence for the Kakeya maximal function conjecture in 3 dimensions. This conjecture roughly states that a collec- tion of tubes in 3-space of equal radii and separated directions is mostly disjoint. We aimed to find a way to search the parameter space for counterexamples to test whether 3 1. We choose some n (for example: n = 10 ).Let N = n 2 and ϵ = 0 .

1 , using which we calculate all v ℓ,j = v ℓ, 1 , v ℓ, 2 , q Katz, Wu, and Zahl suggested [3] that, if T is a collection of SL 2 tubes, of length 1 and radius δ , with δ -separated directions, then the inequality 1 v 2 ℓ, 1 v 2 ℓ, 2 , v ℓ, 1 = i 2 n , v ℓ, 2 = j 2 n and, every a ℓ of the form a ℓ = ( x ℓ , y ℓ , 0) , 0 x ℓ , y ℓ 1 , x ℓ , y ℓ n 0 , 1 n , 2 n , . . . , n 1 n , 1 o p Z X | T | δ 2 , T T χ T 2. Starting with a 0 = (0 , 0 , 0) , and v 0 = (0 , 0 , 1) we iterate through every a ℓ and compute the left hand side of equation (2) to find the a ℓ that maximizes the equation. We achieve this by iterating through points { 0 , 1 [0 , 1] 3 might be true with p = 2 .

There are now a few known counterexamples which show that the p above must be at most 3 / 2 , one of which is the following “hair- brush” example, due to Zahl [5]. Given a single SL 2 -tube T 0 , the hairbrush example consists of δ 2 many SL 2 δ -tubes intersecting T 0 . The tube T 0 is roughly a union of δ 1 many δ -balls, and each such δ -ball has roughly δ 1 many SL 2 -tubes passing through it. If T 0 is partitioned into segments of length δ 1 / 2 , then for each segment, there is a rectangular prism of side lengths δ 1 / 2 Ч δ 1 / 2 Ч δ containing this segment, and each point in this prism has roughly δ 1 many SL 2 tubes passing through it.

This example took some time to be found by hand, but because it happens to have a reasonably simple description, it was found fairly quickly by the RL pro- gram we used. In fact, the tube T 0 does not need to be SL 2 for the example to work. Therefore, the example found by the program showed that making T 0 an SL 2 tube was a redundant assumption. n , 1 } in the unit cube and computing the number of tubes that pass through that point. The greedy algorithm with reasonable parameters always produced the ‘ball example’. The ball example shows the sharpness of the power, p = 3 n , 2 n , . . . , n 1 2 is the correct exponent on the given bound.

We must test the bound by assigning unique starting positions to each tube care- fully because we have also shown that a random sampling approach fails be- cause it cannot achieve the sharpness of the inequality we seek. The sharpness —or worst-case scenario— usually occurs when many tubes overlap significantly within a small region, such as when all of them pass through a tiny ball. Due to the law of large numbers, randomization distributes the tubes more uniformly than we would like, and because the inequality holds with high probability, we don’t see the extreme clustering of tubes required to maximize the left-hand side of the inequality.

To methodically place the tubes we first implemented a greedy algorithm in our code and have since explored various programming approaches. 2 in equation(2). This can be seen when we integrate the indicator functions of the tubes from the family T of 1 Ч δ Ч δ -tubes with δ -separated directions over the δ ball to some power p > 1 , | T δ 2 , p p δ 3 ( δ 2 ) p Z T T χ T X Z T T χ T X . δ ball R 3 If the Kakeya maxima inequality R R 3 P T T χ T p δ 2 | T | were true (for all small δ ), this would give δ 3 2 p δ 2 | T 1 and thus p 3 / 2 , hence sharp at p = 3 / 2 .

Inequality Formula We want to test if the inequality 3 / 2 Z 1 Z 1 Z 1 N X j =1 χ T i ( x, y, z ) dx dy dz C ϵ δ ϵ Nδ 2 , (1) 0 0 0 is true, for all ϵ > 0 , for any δ > 0 , for any N 1 , where { T 1 , . . . , T N } is any finite set of δ -tubes in R 3 with δ -separated directions. The inequality quantifies how the overlaps of thin tubes in the unit cube are bounded when these tubes point in δ -separated directions. Specifically, the inequality states that the total measure of the tubes, calculated by integrating the sum of their characteristic functions raised to the 3 2 power, cannot exceed a constant times Nδ 2 up to a small loss δ ϵ .

If the tubes were completely disjoint and entirely within the unit cube, the left-hand side would be approxi- mately πδ 2 N , which matches the right-hand side up to a constant factor. Thus, the inequality suggests that the tubes are "approximately" disjoint in terms of their contribution to the integral, implying that sets containing lines in every direction must occupy a substantial volume. Fig. 1: Ball Example Fig. 2: SL 2 Hairbrush Example Reinforcement Learning Approach Next Steps To improve on the greedy algorithm, we decided to adapt a reinforcement learning approach based on [2, 4], which was originally developed to find counterexamples to conjectures in graph theory.

The key idea is to optimize over discrete structures using a reward function to guide a search process, which we were able to adapt for our use case. One way to find other examples other than the Ball example would be to consider the “Minkowski dimension” version of the conjecture rather than the Kakeya max- imal inequality. We would change the reward function to minimize the volume of the tubes, rather than maximizing the L p norm of the sum of their indicator func- tions. The conjecture is that the volume of a maximal direction separated family of tubes is 1 (where allows factors like C ϵ δ ϵ ).

In this setting, the ball exam- ple is not optimal, as it has volume 1 , but (due to the existence of area zero kakeya sets), we know that there exist collections of δ -tubes with δ -separated directions whose area can be made arbitrarily small (provided δ is sufficiently small). Converted Problem for Computer Testing 1. The search starts with a batch of candidate tube placements, each represented as a vector of ( x, y ) -plane intercepts. It is equivalent to test (1) with δ = 1 /n where n is any positive integer. Under this assumption, an equivalent version of (1) is, for any vectors a 1 , a 2 , . . . , a N in R 3 , and any δ -separated unit vectors v 1 , . . . , v N , 2.

A sequence of probability distributions over possible actions (i.e. moving a tube) guides the selection of components for these vectors. n X n X n X i a ℓ i k =1 n 3 ℓ : n , j n , k n , j n , k a ℓ , v ℓ v ℓ 1 3 / 2 3. After evaluating the reward (LHS of eq. 2) for each solution, the top-performing solu- tions are used to update the probability distributions via the cross-entropy method. n n n i =1 j =1 References C ϵ Nn ϵ 2 . (2) 4. New candidate solutions are generated using the updated distributions, iterating until the distributions converge, representing an optimal tube placement.

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