Semi-algebraic properties of Minkowski sums of a twisted cubic segment

We find a semi-algebraic description of the Minkowski sum A3,n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {A}_{3,n}$$\end{document} of n copies of the twisted cubic segment {(t,t2,t3)∣-1≤t≤1}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{(t,t^2,t^3)\mid -1\le t\le 1\}$$\end{document} for each integer n≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\ge 3$$\end{document}. These descriptions provide efficient membership tests for the sets A3,n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {A}_{3,n}$$\end{document}. These membership tests in turn can be used to resolve some instances of the underdetermined matrix moment problem, which was formulated by Michael Rubinstein and Peter Sarnak in order to study problems related to L-functions and their zeros.


Introduction
The zeros of L-functions are known to be able to describe various geometrical and arithmetical objects and are the subjects of several conjectures (cf. [1][2][3]). For example, the Generalized Riemann Hypothesis conjectures that all non-trivial zeros of an L-function have real part 1∕2 and the Grand Simplicity Hypothesis asserts that the imaginary parts of zeros of Dirichlet L-functions are linearly independent over ℚ (cf. [5]). L-functions can The first author was supported by the Vici grant Stabilisation in Algebra and Geometry from the Netherlands Organisation for Scientific Research (NWO). The third author was supported by DFG grant GK 1916 Kombinatorische Strukturen in der Geometrie of the German Research Foundation.
to be the Minkowski sum of n copies of C k for each integer n ≥ 1 . Then we can determine the set of tuples (t 1 , … , t n ) ∈ [−1, 1] n such that for all j ∈ {1, … , k} recursively by first computing the set of t n ∈ [−1, 1] such that In order to do the latter, we need an efficient membership test for the set A k,n for all n > k . For k ∈ {1, 2} , this is easy. In general, one way to get an efficient membership test would be to describe the sets A k,n implicitly using only equalities and inequalities involving polynomial expressions in x 1 , … , x k and unions-in other words, using semi-algebraic descriptions of the sets A k,n . In this paper, we provide exactly such descriptions in the case that k = 3.

Main results
Let n ≥ 3 be a positive integer. Our first result describes the boundary of A 3,n . We need this result in order to prove the Main Theorem. However, it also provides us with a piecewise parametrization, which is useful for rendering a visualization of A 3,n . See Fig. 1 for an example.
Before we give the semi-algebraic description of A 3,n , we first discuss the intuition behind it. As Fig. 1 for n = 5 and the interactive 3D models for n = 3, … , 20 demonstrate, the set A 3,n looks like an oyster with an upper and lower shell forming the boundary. We call these upper and lower shells B + n and B − n respectively. These two shells have identical projections to the (x, y)-plane, which we denote by B ♭ n , and both projection maps are one-to-one. This yields a first description of A 3,n : for a point (x, y, z) ∈ ℝ 3 to lie in A 3,n , it is necessary that (x, y) lies in B ♭ n . When this is the case, the point lies in A 3,n precisely when it lies below B + n and above B − n . Each of the two shells consists of n − 1 spiraling ridges. These are the sets C + k,n−k−1 and C − ,n− −1 defined below respectively. The projections of the ridges are easily visualized on B ♭ n . See Fig. 2. We can now reformulate our first description of A 3,n in the following way: let (x, y, z) ∈ ℝ 3 be a point such that (x, y) lies in B ♭ n . Then (x, y) lies in the projections of C + k,n−k−1 and C − ,n− −1 for some k, . The point (x, y, z) lies in A 3,n precisely when is lies below C + k,n−k−1 and above C − ,n− −1 . Next, we think of (x, y) as being fixed. This turns the conditions of lying below C + k,n−k−1 and above C − ,n− −1 into conditions on z. The former condition is equivalent to z being at most the biggest root of some parabola f = Az 2 + Bz + C with A > 0 and can thus be expressed as f(z) being at most 0 or z is at most the value −B∕2A where f attains its minimum. See Fig. 3. Similarly, the latter condition is equivalent to z being at least the smallest root of a parabola f = Az 2 + Bz + C with A > 0 and can be expressed as f(z) being at most 0 or z is at least the value −B∕2A where f attains its minimum. This is our description of A 3,n .
In order to state our results precisely, we define the following sets: • For all integers k, ≥ 1 and a, b ≥ 0 , we take n be the set consisting of all points (x, y) ∈ ℝ 2 such that ny ≥ x 2 and We also let ∶ ℝ 3 → ℝ 2 be the projection map sending (x, y, z) ↦ (x, y).
n be a point and z ∈ ℝ be a real number. (e) Every point on the boundary of A 3,n can be written as for some tuple (t 1 , … , t n ) ∈ [−1, 1] n . The set {t 1 , … , t n }⧵{−1, 1} has at most two elements and the tuple (t 1 , … , t n ) is unique up to permutation of its entries.
In Sect. 5, we find semi-algebraic descriptions of (in particular) C + k,a and C − ,b . To write these descriptions down, we define for all positive integers k, ≥ 1 . Note here that for all k, . We then use these descriptions together with the previous theorem to prove our main result. Take the following sets:

Main Theorem
We have A 3,n = X ∩ Y.

Structure of the paper
The first step of the proof is to show that the boundary of A 3,n is contained in the union of B + n and B − n . We start doing this by proving a result about representations of points on the boundary of A 3,n in Sect. 3. In Sect. 4, we prove the statement for n = 4 and then conclude that it holds for all n ≥ 3 . After that, in Sect. 5, we find semi-algebraic descriptions for the components that make up the boundary. And then, in Sect. 6, we study the sets B + n and B − n in more detail and prove Theorem 2.1 and the Main Theorem. We conclude the paper by discussing the problem for higher dimensions in Sect. 7.

Representations of points on the boundary of A 3,n
The goal of this and the next section is to prove that the boundary of A 3,n is contained in the union of B + n and B − n . We start with the following proposition.
1} , then every point in a small neighborhood around p is in the image of by the inverse function theorem. As this cannot happen for a point p on the boundary of A 3,n , it follows that the set {t 1 , … , t n }⧵{−1, 1} has at most two elements. ◻ From the proposition, it immediately follows that the boundary of A 3,n is contained in the union of the sets over all integers k, ≥ 1 and a, b ≥ 0 such that k + + a + b = n . So to prove that the boundary of A 3,n is contained in the union of B + n and B − n , it suffices to prove that every point that is contained in one of these sets and is not contained in B + n ∪ B − n is also not contained in the boundary of A 3,n .
Proof If the point p ′ does not lie on the boundary of A 3,n ′ , then the point p cannot lie on the boundary of and hence it can also not lie on the boundary of A 3,n . ◻ To prove that the boundary of A 3,n is contained in the union of B + n and B − n , we need to eliminate the cases where one of the following conditions holds: Using the previous lemma, we see that it suffices to eliminate the cases where and hence we first consider the boundary of A 3,n for n = 4.

The boundary of A 3,versus the boundary of A 3,n
In this section, we show that the boundary of A 3,n is contained in B + n ∪ B − n for n = 4 and conclude from this that the same statement holds for all n ≥ 3 . We need to show that no point of is contained in the boundary of A 3,4 for (k, , a, b) = (1, 1, 1, 1), (1, 2, 1, 0) , (2, 1, 0, 1), (2, 2, 0, 0). We start with the case (k, , a, b) = (2, 2, 0, 0). with the additional conditions that −1 < t 1 , t 2 , t 3 , t 4 < 1 are pairwise distinct. If this system has a solution that satisfies the additional conditions, then the point p cannot lie on the boundary of A 3,4 by Proposition 3.1. It turns out that such a solution (t 1 , t 2 , t 3 , t 4 ) can even be found when we assume that Then is a solution to the system equalities so that Here we use that |s − t| + |s + t| ≤ 2 ⋅ max(|s|, |t|) < 2 . It follows that for any point ( , ) on the circle given by that is sufficiently close to ( 1 2 (s − t), 1 2 (s − t)) also satisfies these conditions. So to conclude the proof, we simply let ( , ) be such a point with 0 ≠ ≠ ≠ 0 . ◻ Next, we take care of the case (k, , a, b) = (1, 1, 1, 1).

Lemma 4.2
Take > 0 . Then there exists an > 0 such that Assuming that d ≥ 0 and −1 ≤ t 1 , t 2 , t 3 ≤ 1 , one can check that     Proof The proof is similar to the proof of Proposition 4.4. For (u, v) in a small neighborhood of (s, t), we find points in A 3,4 above using Lemma 4.5 and we find points below using Lemma 4.6. Note here that in the latter case the role of the pair (s, t) from Lemma 4.6 is played by (v, 1). ◻ Proof The proof is similar to the proof of Proposition 4.7. For (u, v) in a small neighborhood of (s, t), we find points in A 3,4 above using Lemma 4.5 and we find points below using Lemma 4.6. Note here that in the former case the role of the pair (s, t) from Lemma 4.5 is played by (−1, u) . ◻ By combining Propositions 3.1, 4.1, 4.4, 4.7 and 4.8 , we see that the boundary of A 3,n is contained in B + n ∪ B − n for n = 4 . We now use this knowledge to prove the same for n > 4.

The semi-algebraic components of the boundary of A 3,n
Consider the sets for k, ≥ 1 . Recall that Our goal for this section is to prove the following proposition and theorem.
decomposes the polynomial f k into irreducible factors over ℚ . If k ≠ , then f k is irreducible over ℂ.

Theorem 5.2 The set
consists of all points such that f k (x, y, z) = 0 , the inequalities hold and in addition the following requirements are met: For the remainder of the section, we fix integers k, ≥ 1 and we write in order to simplify the used notation.
So we let be a point and find out when it is contained in We start by looking at the first two coordinates. So we solve the system of equations under the conditions that −1 ≤ s ≤ t ≤ 1 . Solving the system, we find that So we need to assume that k D(x, y) ≥ 0 . Adding the condition s ≤ t , we get and so the conditions −1 ≤ s and t ≤ 1 translate to As x ∈ [−(k + ), (k + )] , these conditions are equivalent to Now, also consider the third coordinate z = ks 3 + t 3 . One can check that f (x, y, z) = 0 . So if k = , then we have by Proposition 5.1 and we are done. So assume that k ≠ . Then there are a priori two possibilities for z given x and y. However, given s and t, it becomes clear that only one possibility remains. So we just need to find an inequality that selects the correct root of f (x, y, −) . One can check that So we find that when k < and when k > . This concludes the proof. ◻ 6 The sets B + n and B − n We are now ready to prove Theorem 2.1 and the Main Theorem. Let n ≥ 3 be an integer.
Recall the following notation from Sect. 2.
• We have for all integers k, ≥ 1 and a, b ≥ 0.
n consists of all points (x, y) ∈ ℝ 2 such that ny ≥ x 2 and We start by listing some properties of B + n and B − n .
(a) The map is a bijection.
Proof The proofs are similar to those of Proposition 6.1. ◻ The decomposition of B ♭ n as a union of the projections of C + 1,n−2 , … , C + n−1,0 is visualized in Fig. 2. We note that the decomposition of B ♭ n as a union of the projections of C − 1,n−2 , … , C − n−1,0 looks similar but is mirrored along the vertical axis. We can now prove Theorem 2.1.
Proof of Theorem 2. 1 We already know that (b) holds by Propositions 6.1 and 6.2 . We know that B + n , B − n ⊆ A 3,n , we know that the boundary of A 3,n is contained in B + n ∪ B − n by Theorem 4.9 and we know that the projection maps are bijections by Propositions 6.1 and 6.2 . Together these statements imply (a). Let (x, y) ∈ B ♭ n be a point. Then there exist unique numbers z + , z − ∈ ℝ such that (x, y, z + ) ∈ B + n and (x, y, z − ) ∈ B − n by Propositions 6.1 and 6.2 . Our goal is to prove that z + ≥ z − with equality if and only if (x, y) lies on the boundary of B ♭ n . Let z ∈ ℝ be a real number. From (a) and (b) it is clear that (x, y, z) ∈ A 3,n if and only if z lies between z + and z − . So z + = z − when (x, y) lies on the boundary of B ♭ n . And, to prove that z + > z − otherwise, it suffices to show that there exists a z ∈ ℝ such that z < z + and (x, y, z) ∈ A 3,n . Now, let be a point where −1 ≤ s ≤ t ≤ 1 and recall Lemmas 4.2 and 4.6 . If −1 < s < t < 1 and k = 1 , then there is a point in A 3,3 below and hence a point in A 3,n below p. If −1 < s < t ≤ 1 and k ≥ 2 , then there is a point in A 3,3 below and hence a point in A 3,n below p. Taking into account how the sets C + k,n−k−1 intersect, we find that there is a point in A 3,n below p unless s = −1 , (k, t) = (1, 1) or (s, k) = (t, n − 1) , which are exactly the cases where p projects to the boundary of B ♭ n . This proves (c) and (d). Finally, using (a), (c), and Propositions 6.1 and 6.2 , we see that the boundary of A 3,n is the disjoint union of several (but not all) sets of the form , y), where k, , a, b ≥ 0 have sum n. Given a point of the boundary, the number s is unique when k > 0 and the number t is unique when > 0 . Together with Proposition 3.1, this shows (e). ◻ Finally, we prove the Main Theorem.

Proof of the Main Theorem
Fix a point (x, y, z) ∈ ℝ 3 . The following conditions are equivalent: Take k, ∈ {1, … , n − 1} . Then, using Proposition 6.1, we see that and we similarly get using Proposition 6.2. Assume that (x, y) ∈ B ♭ n and that k, are as in (b) and (c). Using Theorem 2.1(d), we need to find conditions that express that z − ≤ z ≤ z + . We have by Theorem 5.2. So z ≤ z + when or Note here that the polynomial f k1 (x, y, −) has degree 2 in z, that its leading coefficient is positive, that z + is its highest root and that it attains its minimum at . This is visualized in Fig. 3. We also have

Higher dimensions
The Main Theorem provides a semi-algebraic description of the set A 3,n for each integer n ≥ 3 . So, a natural question to ask is: can we use the same proof strategy to find a semi-algebraic description of the sets A k,n for k > 3 ? At the moment, there still are some obstacles to doing so, which we discuss in this section. Following the same strategy as for k = 3 , we would again start by trying to find a description of the boundary of A k,n . One can check that the statement and proof of Proposition 3.1 carry over in a straightforward fashion for k > 3 , which yields a superset of the boundary. After this, one would again need to exclude points from this superset when they do not in fact lie on the boundary. In view of Theorem 2.1(d), proving that a point in A 3,n does not lie on the boundary can be done by showing that there are points in A 3,n above and below it. So an analogue of Theorem 2.1(d) for higher dimensions would be very useful. This leads to the following conjecture, which holds for k ≤ 3.

Conjecture 7.1
Let (x 1 , … , x k ) be a point in A k,n . Then the set is a closed interval.
Another approach might be through a generalization of Theorem 2.1(e). The boundary of A k,n is contained in the union of the sets over all integers 1 , … , k−1 , a, b ≥ 0 that sum to n. The uniqueness of the representation of each point on the boundary would imply that the boundary of A k,n is a disjoint union of {y ∈ ℝ|(x 1 , … , x k−1 , y) ∈ A k,n } which allows to eliminate the coordinate x 4 . So here, the problem consists of finding a semi-algebraic description of the set given x 1 , x 2 , x 3 ∈ ℝ.
If we can solve Problem 7.3, we still need to find analogues for the results in Sect. 6. These results rely on our complete understanding of the roots and extrema of parabolas. So to generalize these results, we probably need a similar level of understanding in the cases of cubics and quartics, which for now seems to be out of reach.