# Erratum: Breaking the 3/2 barrier for unit distances in three dimensions (International Mathematics Research Notices (2019) 20 (6235-6284) DOI: 10.1093/imrn/rnx336)

Micha Sharir, Joshua Zahl*

*Corresponding author for this work

Research output: Contribution to journalComment/debate

## Abstract

We wish to correct an error in the paper “Breaking the 3/2 barrier for unit distances in three dimensions.” Lemma 3.2 of this paper contained an incorrect claim about the intersection patterns of ellipses obtained by projecting circles from three dimensions to the plane. In this erratum we state and prove a correct version of this statement. We wish to correct an error in the paper “Breaking the 3/2 barrier for unit distances in three dimensions.” Lemma 3.2 of this paper claimed that if C and C' are two circles in ℝ3 that intersect at two points p and q, then under some mild assumptions, the projections of the shorter circular arcs between p and q to the x1x2-plane will always intersect an even number of times (i.e., either zero or two times, excluding the projections of p and q). In particular, Lemma 3.2 claimed that the arcs must lift to a depth cycle in ℝ4. While this claim was wrong as stated, it is true after applying a suitable orthogonal transformation, provided the arcs are sufficiently short. In this erratum, we give a simple construction that describes the problem, and present a new statement and proof of Lemma 3.2, and a new proof of Corollary 3.2, that address this problem. The rest of the argument in the original paper [2] remains unchanged. A counterexample. Let C and C' be two circles in ℝ3, not lying in the same plane, that meet at two distinct points p and q, let (Formula presented) ⊂ C (resp., (Formula presented) ⊂ C') be the shorter (closed) arc of C (resp., C') with endpoints p and q. Let τp and τq be the unit tangents to (Formula presented) at p and q, respectively, each pointing away from (Formula presented). Define τp' and τq' analogously for (Formula presented). Let Hp (resp., Hq) be the plane spanned by τp and τp' (resp., by τq and τq' ). Let γp (resp., γq) denote the great circle on &#x1D52 that is parallel to Hp (resp., Hq). The great circles γp and γq partition &#x1D52 into four quadrants (slices), and there exists precisely one slice with the following property: for each direction v in that slice, the orthogonal projection πv in direction (Formula presented) onto some plane v is such that the projection of τp lies counterclockwise to that of τp', and the projection of τq also lies counterclockwise to that of τq'. It is then easy to check that in this case the relative interiors of the projections of (Formula presented) and (Formula presented) intersect exactly once. This contradicts the claim originally made in [2, Lemma 3.2]. See Figure 1 for an illustration. Revised claims and proofs. Lemma 3.2. There is a finite set of unit vectors V ⊂ ℝ3 so that the following holds. Let C and C' be circles in ℝ3 that intersect at the two distinct points p and q. Let (Formula presented) ⊂ C be the shorter (closed) arc of C with endpoints p and q, and similarly for (Formula presented) ⊂ C'. Suppose that the length of (Formula presented) is at most 1/100 the circumference of C, and similarly for (Formula presented). Then there is a vector v ∈ V so that the orthogonal projection πv: ℝ3 → v is injective on (Formula presented) ∪ (Formula presented). Proof. Let V be a set of unit vectors, so that every nonzero vector in ℝ3 makes an angle ≤ 1 degree with one of the vectors in V. The set V is clearly finite. Let C and C', p and q, and (Formula presented) ⊂ C, (Formula presented) ⊂ C' be as in the statement of the lemma. We claim that there exists a vector v ∈ V so that the projection πv: ℝ3 → v is injective on (Formula presented) ∪ (Formula presented). First, if C and C' are contained in a common plane S ⊂ ℝ3, select v ∈ V to be any vector not parallel to S. Since v is not parallel to S, the projection πv: S → v is injective, and thus πv is injective on (Formula presented) ∪ (Formula presented). Next, if C and C' are not coplanar, let S ⊂ ℝ3 be the unique sphere containing C and C'; such a sphere must exist since C and C' intersect at two points. Select v ∈ V to be a vector that makes angle ≤ 1 degree with the normal vector of S at p. Observe that radius(S) ≥ max(radius(C), radius(C')). Let (Formula presented) ⊂ S be the spherical cap of diameter (1/10)radius(S) centered at p. We have that (Formula presented) ⊂ (Formula presented) and (Formula presented) ⊂ (Formula presented). The projection πv: (Formula presented) → v is injective (this is a simple geometry exercise), and thus πv is injective on (Formula presented) ∪ (Formula presented). Corollary 3.2. Let C be a set of n circles in ℝ3. Suppose that at most B circles can lie in a common algebraic variety of degree ≤ 400. Then the circles in C can be cut into Oε(n4/3+ε + nB1/2+ε) pseudo-segments. Proof. First, cut each circle in C into 100 arcs of equal length; we call this the first cutting. Next, let V be the set of unit vectors from Lemma 3.2. For each v ∈ V, let Ov : ℝ3 → ℝ3 be an orthogonal transformation sending v to the vector (0, 0, 1). We will identify (0, 0, 1) with the x1x2-plane. Define Ov(C) = {Ov(C): C ∈ C}; this is a set of n circles in ℝ3, and {π(0,0,1)(C): C ∈ Ov(C)} is a collection of n ellipses in the x1x2-plane (by perturbing the vectors in V slightly if necessary, we can ensure that fp 4-18-202 of the projected circles are line segments). For each v ∈ V, we cut each of the corresponding ellipses at their two points of x2-vertical tangency. This corresponds to 2|V|n = O(n) cuts to the original set of circles. We call this the second cutting. Now, observe that if C, C' ∈ C are circles that intersect at two distinct points p, q ∈ ℝ3, then one of the following two properties holds. (A) At most one of the four arcs from C and C' connecting p and q remains uncut. (B) The (shorter) arc (Formula presented) ⊂ C between p and q has length at most 1/100 the circumference of C, and similarly for the shorter arc (Formula presented) ⊂ C'; each of the longer arcs has been cut. If (A) occurs, then the first cutting has ensured that at most one of the segments from C or C' contains both p and q, and the pseudo-segment property holds for this scenario. Suppose then that (B) occurs, and let (Formula presented) ⊂ C and (Formula presented) ⊂ C' be the shorter arcs of C and C', respectively, with endpoints p and q. By Lemma 3.2, there is a unit vector v ∈ V so that the projection πv: ℝ3 → v is injective on (Formula presented) ∪ (Equation presented).

Original language English 1795-1800 6 International Mathematics Research Notices 2023 2 https://doi.org/10.1093/imrn/rnab119 Published - 1 Jan 2023

### Funding

FundersFunder number
Israel Science Foundation

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