Abstract
We improve the current best bound for distinct distances on nonruled algebraic surfaces in \({\mathbb {R}}^3\). In particular, we show that n points on such a surface span \(\Omega (n^{32/39{\varepsilon }})\) distinct distances, for any \({\varepsilon }>0\). Our proof adapts the proof of Székely for the planar case, which is based on the crossing lemma. As part of our proof for distinct distances on surfaces, we also obtain new results for distinct distances between circles in \({\mathbb {R}}^3\). Consider two point sets of respective sizes m and n, such that each set lies on a distinct circle in \({\mathbb {R}}^3\). We characterize the cases when the number of distinct distances between the two sets can be \(O(m+n)\). This includes a new configuration with a small number of distances. In any other case, we prove that the number of distinct distances is \(\Omega (\min {\{m^{2/3}n^{2/3},m^2,n^2\}})\).
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Notes
We use \(O_{v_1, v_2, \dots , v_k}\) to represent the usual bigO notation where the constant of proportionality depends on the variables \(v_1, \dots , v_k\). We define \(\Omega _{v_1, v_2, \dots , v_k}\) and \(\Theta _{v_1, v_2, \dots , v_k}\) symmetrically.
Whenever we refer to open sets, we mean open according to the Euclidean topology (rather than the Zariski toplogy).
Wolfram Research, Inc., Mathematica, version 12.0, Champaign, IL (2019).
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Acknowledgements
We are grateful to Frank de Zeeuw for many useful discussions, including help with Theorem 5.2 and with the circle constructions. We thank Toby Aldape, Jingyi (Rose) Liu, MinhQuan Vo, and the anonymous referees for helping to improve this paper. We also thank Sara Fish for inspirational support. The first author would like to thank everyone involved in the 2019 CUNY REU for motivating her with their passion and engaging with her in many helpful discussions.
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This research project was done as part of the 2019 CUNY Combinatorics REU, supported by NSF awards DMS1802059 and DMS1851420.
This research was completed while the first author was at Caltech. Supported by Caltech’s Summer Undergraduate Research Fellowships (SURF) Program and the Olga TausskyTodd Award.
Supported by NSF award DMS1802059.
Appendices
Appendix A: The Mathematica Code
In this appendix, we describe the Mathematica program that was used in the proof of Theorem 1.4(b). Listing 1 contains the code that is used for the general case of the proof. Lines 1–7 define the parametrizations described in (6) and (8). Lines 8–9 define the function \(\rho (s,t)\). Line 10 is the derivative test of Lemma 5.3. Finally, line 11 shows the coefficient of a specific term.
The code of Listing 1 leads to expressions somewhat more involved than the ones stated in Sect. 5. For example, the coefficient of \(s^5t\) produced by Mathematica is
The above expression can be simplified by noting that
Looking at Lemma 5.3, it may seem as if we forgot to include the absolute value in Listing 1. For some \(f\in {\mathbb {R}}[x]\), consider
In either case the derivative is \(f'(x)/f(x)\). We may thus ignore the absolute value in Lemma 5.3.
The special cases at the end of Sect. 5 require minor changes in the code from Listing 1. For example, the case of \(\cos \beta =0\) is obtained by changing lines 2–3 to \(v_2=(0,1,0)\). The case of \(C_2\) being centered at the origin is obtained by removing “\(p+\)” for line 5 and “\(q+\)” from line 7.
Appendix B: Analysis of the Derivative Test
In this appendix, we complete the proof of Theorem 1.4(b) by showing that g(s, t) as defined in (10) cannot vanish everywhere, and also address the special cases. To do so, we note that g(s, t) is a rational function, and it suffices to show that there exist monomials in the numerator whose coefficients cannot simultaneously vanish. We consider several such monomials:

The coefficient of \(s^5t\) is
$$\begin{aligned} 96r(p + r \cos \alpha ) \cos \beta \sin ^2\alpha (p \cos \alpha + q \sin \alpha ) \sin \beta (\cos ^2\beta \sin ^2\alpha + \sin ^2\beta ).\end{aligned}$$ 
The coefficient of \(s^5t^9\) is
$$\begin{aligned} 96r(p  r \cos \alpha ) \cos \beta \sin ^2\alpha (p \cos \alpha + q \sin \alpha ) \sin \beta (\cos ^2\beta \sin ^2\alpha + \sin ^2\beta ). \end{aligned}$$ 
The coefficient of \(s^3t^9\) is
$$\begin{aligned}{} & {} 64r(p  r \cos \alpha ) \cos \beta \sin \alpha \sin \beta (\cos ^2\beta \sin ^2\alpha + \sin ^2\beta ) \\{} & {} \cdot \,(2 q \cos ^2\alpha + p \cos \alpha \sin \alpha  q \sin ^2\alpha ). \end{aligned}$$ 
The coefficient of \(s^3t\) is
$$\begin{aligned}{} & {} 64r(p + r \cos \alpha ) \cos \beta \sin \alpha \sin \beta (\cos ^2\beta \sin ^2\alpha + \sin ^2\beta )\\{} & {} \cdot \,(2 q \cos ^2\alpha + p \cos \alpha \sin \alpha  q \sin ^2\alpha ). \end{aligned}$$
By definition, \(r\ne 0\). Recall that H (the plane containing \(C_2\)) is not parallel to the xyplane or the xzplane. Since we also assumed H does not contain lines parallel to the xaxis, we have that \(\sin \alpha \ne 0\) and \(\sin \beta \ne 0\). Indeed, this is easy to see when recalling that \(V'_1\) and \(V'_2\) from (7) span the directions of H. This implies that \(\cos ^2\beta \sin ^2\alpha +\sin ^2\beta \ne 0\). Since we assume that H is not parallel to the xzplane, we have that \(\sin \beta \ne 0\). We now also assume that the center of \(C_2\) is not the origin, that \(\cos \alpha \ne 0\), and that \(\cos \beta \ne 0\). These special cases are addressed after completing the general cases.
Since \(\cos \alpha \ne 0\), it is not possible for \(p  r \cos \alpha \) and \(p + r \cos \alpha \) to be zero simultaneously. Thus, the only way for all of the four above coefficients to equal zero is to have both
We rearrange the first equation as \(p \cos \alpha = q \sin \alpha \). Plugging this into the second equation gives
That is, \(q=0\). We then have \(p \cos \alpha = q \sin \alpha =0\), or \(p=0\). This contradicts the assumption that \(C_2\) is not centered at the origin.
By the above, it is impossible for the four above coefficients to be zero simultaneously. This implies that g(s, t) is not identically zero in any open neighborhood \(N\subset {\mathbb {R}}^2\). By Lemma 5.3, we get that \(\rho (s,t)\) cannot be rewritten as \(\varphi _1(\varphi _2(s)+\varphi _3(t))\) for every \((x,y)\in N\). Then, Lemma 5.1 implies the assertion of the theorem.
The Special Cases. In the above proof, we assume:

The plane H is not parallel to the xyplane, to the xzplane, or to the yzplane.

H does not contain lines parallel to the xaxis.

The center of \(C_2\) is not the origin.

\(\cos \beta \ne 0\).

\(\cos \alpha \ne 0\).
We now address each of these special cases, in the above order.
We first consider the case where H is parallel to the xyplane. As discussed in the introduction, if H is the xyplane and \(C_1\) and \(C_2\) are not concentric, then \(D({\mathcal {P}}_1,{\mathcal {P}}_2) = \Omega (\min {\{m^{2/3}n^{2/3},m^2,n^2\}})\). When H is parallel to the xyplane, we can combine the above with Lemma 2.7, to obtain the following. If \(C_1\) and \(C_2\) are not aligned then \(D({\mathcal {P}}_1,{\mathcal {P}}_2) =\Omega (\min {\{m^{2/3}n^{2/3},m^2,n^2\}})\).
Next consider the case where H is parallel to the xzplane. In this case, we can rewrite \(V_1 = (1,0,0)\) and \(V_2 = (0,0,1)\), which significantly simplifies (8). The Mathematica program then implies that the numerator of g(s, t) is \(4 q(1 + t^2) (1 + s^2)\). This expression is not identically zero unless \(q=0\). If \(q=0\) then the center of \(C_2\) is on the xaxis and H contains the origin. That is, if \(q=0\) then \(C_1\) and \(C_2\) are perpendicular. We conclude that either \(D({\mathcal {P}}_1,{\mathcal {P}}_2) =\Omega (\min {\{m^{2/3}n^{2/3},m^2,n^2\}})\) or the circles are perpendicular.
We move to the case where H is parallel to the yzplane. We rewrite \(V_1 = (0,1,0)\) and \(V_2 = (0,0,1)\). In this case, the coefficient of \(s^2\) in the numerator of g(s, t) is 4pr. The coefficient of \(s^3t^4\) is \(16q(3p^28r^2)\). For both of these coefficients to be zero, we must have \(p=q=0\), implying that the two circles are perpendicular. Once again, either \(D({\mathcal {P}}_1,{\mathcal {P}}_2) =\Omega (\min {\{m^{2/3}n^{2/3},m^2,n^2\}})\) or the circles are perpendicular.
We next assume that H contains lines parallel to the xaxis. In this case, we can write \(V_1 = (1,0,0)\) and \(V_2=(0,\cos \gamma ,\sin \gamma )\). Since H is not parallel to the xyplane and xzplane, we have that \(\sin \gamma \ne 0\) and \(\cos \gamma \ne 0\). We consider the following coefficients of g(s, t):

The coefficient of s (with no t) is \(8p(p+r)^2 \cos \gamma \).

The coefficient of \(st^6\) is \(8p(pr)^2 \cos \gamma \).

The coefficient of \(s^3 t^5\) is \(128q(pr)r\cos \gamma \sin \gamma \).
Since \(p+r\) and \(pr\) cannot simultaneously be zero, the first two coefficients imply that \(p=0\). Since \(r>0\), we get that \(pr\ne 0\). Then, the third coefficient implies that \(q=0\). That is, we are in the special case where \(V_1=(1,0,0)\), \(V_2=(0,\cos \gamma ,\sin \gamma )\), and \(p=q=0\). In this case, the coefficient of \(s^6t^2\) is \(4\cos \gamma \sin ^2\gamma \). This coefficient is never zero, which completes the case of lines parallel to the xaxis.
Note that \(\sin \alpha \ne 0\), since otherwise H is parallel to the xyplane. Similarly, \(\sin \beta \ne 0\), since otherwise H is parallel to the xzplane. We next consider the case where \(C_2\) is centered at the origin. That is, we set \(p=q=0\). In this case the coefficient of \(t^3\) (with no s factor) in the numerator of g(s, t) is
For this expression to be zero, we must have either \(\cos \alpha =0\) or \(\cos \beta =0\). If \(\cos \alpha =0\), then H is perpendicular to the xyplane, so the two circles are perpendicular. If \(\cos \beta =0\), then \(V_2=(0,1,0)\). Running the program once again gives that the coefficient of \(s^6 t^6\) is \(8 \cos \alpha \sin ^2\alpha \). For this coefficient to vanish, we again require \(\cos \alpha =0\), so the circles are again perpendicular.
Consider the case where \(\cos \beta =0\). In this case \(\sin \beta =1\), so \(V_2 = (0,1,0)\). As before, we run the Mathematica program to find coefficients in the numerator of g(s, t). The coefficient of \(s^5t^3\) is \(128p r \cos \alpha \). For this coefficient to vanish, we require either \(p=0\) or \(\cos \alpha =0\). Consider the case where \(\cos \alpha =0\). In this case, the coefficient of \(t^{10}\) (and no factor of s) is \(4pr\sin ^2\alpha \). Since \(\sin \alpha =1\), we get that \(p=0\). That is, in either case we have that \(p=0\).
We continue the case where \(\cos \beta =0\) and \(p=0\). We may assume that \(q\ne 0\), since we already handled the case where the center of \(C_2\) is the origin. The coefficient of \(t^9\) is \(16r^2q \cos \alpha \sin \alpha \). This coefficient vanishes if and only if \(\cos \alpha =0\). In this case \(V_1=(0,0,1)\) and running the program again leads to the numerator being \(8 q(1 + t^2)s\). Since \(q\ne 0\), this numerator does not vanish identically.
Finally, we assume that \(\cos \alpha =0\). In this case, \(V_1=(0,0,1)\), so H is perpendicular to the xyplane. We may assume that \(\cos \beta \ne 0\), since this is the case where H is parallel to the yzplane. As usual, we consider coefficients of terms in the numerator of g(s, t). The coefficient of \(s^3t^3\) is \(384p q r \cos \beta \sin \beta \). For this coefficient to vanish, we require either \(p=0\) or \(q=0\). When \(p=0\), the numerator becomes \(4 q(1 \!+\! t^2) (\cos \beta + s^2 \cos \beta + 2 s \sin \alpha )\). When \(q=0\), the numerator becomes \(4 p r(1 \!+\! t^2) (1 + s^2) \sin \beta \). In either case, the numerators is zero if and only if \(p=q=0\), which is a case we already handled.
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Mathialagan, S., Sheffer, A. Distinct Distances on NonRuled Surfaces and Between Circles. Discrete Comput Geom 69, 422–452 (2023). https://doi.org/10.1007/s0045402200449x
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DOI: https://doi.org/10.1007/s0045402200449x
Keywords
 Distinct distances
 Incidences
 Crossing lemma
 Combinatorial geometry
Mathematics Subject Classification
 52C10