Distinct Distances on Non-Ruled Surfaces and Between Circles

Abstract

We improve the current best bound for distinct distances on non-ruled 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 $\Omega (\min {\{m^{2/3}n^{2/3},m^2,n^2\}})$.

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

$$\begin{aligned} 96r(p+r \cos \alpha ) \cos \beta \sin ^2\alpha (-r&\,+p \cos \alpha + r \cos ^2\alpha + q \sin \alpha +r \sin ^2\alpha )\\&\cdot \sin \beta (\cos ^2\alpha \cos ^2\beta \sin ^2\alpha + \cos ^2\beta \sin ^4\alpha + \sin ^2\beta ). \end{aligned}$$

The above expression can be simplified by noting that

$$\begin{aligned} -r + p \cos \alpha + r \cos ^2\alpha + q \sin \alpha + r \sin ^2\alpha&= p \cos \alpha + q \sin \alpha \quad \text { and}\\ \cos ^2\alpha \cos ^2\beta \sin ^2\alpha + \cos ^2\beta \sin ^4\alpha + \sin ^2\beta&= \cos ^2\beta \sin ^2\alpha + \sin ^2\beta . \end{aligned}$$

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

$$\begin{aligned} \frac{\partial }{\partial x} \log |f(x)| = {\left\{ \begin{array}{ll} \displaystyle \frac{\partial }{\partial x} \log f(x) \qquad &{} \text { if } f(x)> 0, \\ \displaystyle \frac{\partial }{\partial x} \log \left( -f(x)\right) \qquad &{} \text { if } f(x)< 0. \end{array}\right. }\end{aligned}$$

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 xy-plane or the xz-plane. Since we also assumed H does not contain lines parallel to the x-axis, 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 xz-plane, 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

$$\begin{aligned} p \cos \alpha + q \sin \alpha =0 \quad \text { and } \quad -2 q \cos ^2\alpha + p \cos \alpha \sin \alpha - q \sin ^2\alpha =0. \end{aligned}$$

We rearrange the first equation as $p \cos \alpha = -q \sin \alpha$. Plugging this into the second equation gives

$$\begin{aligned} 0 = -2 q \cos ^2\alpha + p \cos \alpha \sin \alpha - q \sin ^2\alpha = -2 q \cos ^2\alpha - 2q \sin ^2\alpha = -2q. \end{aligned}$$

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 xy-plane, to the xz-plane, or to the yz-plane.

• H does not contain lines parallel to the x-axis.

• 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 xy-plane. As discussed in the introduction, if H is the xy-plane 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 xy-plane, 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 xz-plane. 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 x-axis 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 yz-plane. 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 $16q(3p^2-8r^2)$. 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 x-axis. 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 xy-plane and xz-plane, 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 $-8p(p-r)^2 \cos \gamma$.

• The coefficient of $s^3 t^5$ is $-128q(p-r)r\cos \gamma \sin \gamma$.

Since $p+r$ and $p-r$ cannot simultaneously be zero, the first two coefficients imply that $p=0$. Since $r>0$, we get that $p-r\ne 0$. 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 x-axis.

Note that $\sin \alpha \ne 0$, since otherwise H is parallel to the xy-plane. Similarly, $\sin \beta \ne 0$, since otherwise H is parallel to the xz-plane. 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

$$\begin{aligned} -32 \cos ^2\alpha \cos \beta \sin ^2\alpha \sin \beta (\cos ^2\beta \sin ^2\alpha + \sin ^2\beta ). \end{aligned}$$

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 xy-plane, 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 xy-plane. We may assume that $\cos \beta \ne 0$, since this is the case where H is parallel to the yz-plane. 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.