A Universal Triangulation for Flat Tori

Abstract

A construction due to Burago and Zalgaller (Vestnik Leningrad Univ 15:66–80, 1960; St Petersburg Math J 7(3):369–385, 1995) shows that every orientable polyhedral surface, one that is obtained by gluing Euclidean polygons, has an isometric piecewise linear embedding into Euclidean space $\mathbb {E}^3$. A flat torus, resulting from the identification of the opposite sides of a Euclidean parallelogram, is a simple example of polyhedral surface. The embeddings constructed according to Burago and Zalgaller may have a huge number of vertices, moreover distinct for every flat torus. Based on another construction of Zalgaller (J Math Sci 100(3):2228–2238, 2000. https://doi.org/10.1007/s10958-000-0007-3) and on recent works by Arnoux et al. (2021, in preparation), we exhibit a universal triangulation with 2434 triangles which can be embedded linearly on each triangle in order to realize the metric of any flat torus.

Appendix

Here, we provide the details for the proof of Lemma 3.9. From Lemma 3.7, simple computations show that the region ${{\mathcal {M}}}_{19,2}$ is bounded by the following parametrized curves:

• $\lambda _2(t) = z_2 + it$ with $t \in [0, +\infty [$ and $z_2 = \frac{2 - \sin \frac{2\pi }{19} \cot \frac{\pi }{19} + i \sin \frac{2 \pi }{19}}{19}$,

• $\beta _{2,1}(t) = \frac{2}{19} - \frac{\sin \frac{2\pi }{19}}{19 \sin \frac{\pi }{19}} e^{-it}$ with $t \in [\frac{\pi }{19}, \frac{3 \pi }{19}]$,

• $\beta _{2,2}(t) = \frac{2 + i\cot \frac{\pi }{19}}{19} - \frac{e^{i\frac{15\pi }{38}}}{19 \sin \frac{\pi }{19}} t$ with $t \in [\cos \frac{16\pi }{19}, \cos \frac{3\pi }{19}]$,

• $\rho _2(t) = w_2 + it$ with $t \in [0, +\infty [$ and $w_2 = \beta _{2,2}\left( \cos \frac{16 \pi }{19}\right)$.

While $\mathcal {M}_{19,7}$ is bounded by:

• $\lambda _7(t) = z_7 + it$ with $t \in [0, +\infty [$ and $z_7 = \frac{7 - \sin \frac{7\pi }{19} \cot \frac{\pi }{19} + i \cot \frac{\pi }{19} \left( 1 - \cos \frac{7\pi }{19}\right) }{19}$,

• $\beta _{7,1}(t) = \frac{7 + i \cot \frac{\pi }{19}}{19} - \frac{e^{i \frac{5\pi }{38}}}{19 \sin \frac{\pi }{19}} t$ with $t \in [\cos \frac{6\pi }{19}, \cos \frac{\pi }{19}]$,

• $\beta _{7,2}(t) = \frac{7}{19} - \frac{\sin \frac{7\pi }{19}}{19 \sin \frac{\pi }{19}} e^{-it}$ with $t \in [\frac{6\pi }{19}, \frac{8\pi }{19}]$,

• $\beta _{7,3}(t) = \frac{7 + i\cot \frac{\pi }{19}}{19} - \frac{e^{i \frac{5\pi }{38}}}{19 \sin \frac{\pi }{19}} t$ with $t \in [\cos \frac{11\pi }{19}, \cos \frac{8\pi }{19}]$,

• $\rho _7(t) = w_7 + it$ with $t \in [0, +\infty [$ and $w_7 = \beta _{7,3}\left( \cos \frac{11\pi }{9}\right)$.

And $\mathcal {M}_{19,13}$ is bounded by:

• $\lambda _{13}(t) = z_{13} + it$ with $t \in [0, +\infty [$ and $z_{13} = \frac{13 - \sin \frac{13\pi }{19} \cot \frac{\pi }{19} + i \cot \frac{\pi }{19} \left( 1 - \cos \frac{13\pi }{19}\right) }{19}$,

• $\beta _{13,1}(t) = \frac{13 + i \cot \frac{\pi }{19}}{19} - \frac{e^{-i\frac{7\pi }{38}}}{19 \sin \frac{\pi }{19}} t$ with $t \in [\cos \frac{5 \pi }{19}, \cos \frac{\pi }{19}]$,

• $\rho _{13}(t) = w_{13} + it$ with $t \in [0, +\infty [$ and $w_{13} = \beta _{13,1}\left( \cos \frac{5\pi }{19}\right)$.

In the sequel we denote by $\mathcal {B}(z, r)$ the closed disk of radius r centered at z and by $\mathcal {C}(z, r)$ its boundary circle. We also denote by $\Re (z)$ and $\Im (z)$, the real part and the imaginary part of z, respectively. Recall that each $g_{\delta }: z \mapsto \frac{1}{-z+\delta }$ sends the horizontal line $\{\Im (z) = h\}$ onto the circle $\mathcal {C}\left( \frac{i}{2\,h},\frac{1}{2\,h}\right)$. Furthermore, for $\delta \ge 1$, $g_{\delta }$ sends the imaginary axis onto the circle $\mathcal {C}\left( \frac{1}{2\delta }, \frac{1}{2\delta }\right)$, and the line $\{\Re (z) = \frac{1}{2}\}$ onto the circle $\mathcal {C}\left( \frac{1}{2\delta - 1}, \frac{1}{2\delta - 1}\right)$. We denote the red, blue and yellow slices of ${{\mathcal {M}}}^+_\text {short}$ on Fig. 15 by respectively ${{\mathcal {S}}}_r$, ${{\mathcal {S}}}_b$ and ${{\mathcal {S}}}_y$.

From these facts, we deduce that $g_5({{\mathcal {S}}}_r)$, the image of the red slice by $g_5$ (see Fig. 16), is bounded by four arcs of circles; one from respectively $\mathcal {C}\left( \frac{1}{9}, \frac{1}{9}\right)$, $\mathcal {C}\left( \frac{i}{66},\frac{1}{66}\right) , \mathcal {C}\left( \frac{1}{10}, \frac{1}{10}\right)$ and $\mathcal {C}\left( \frac{i}{50}, \frac{1}{50}\right)$. Similarly, the image $g_3({{\mathcal {S}}}_b)$ of the blue slice is bounded by arcs from the circles $\mathcal {C}\left( \frac{1}{5}, \frac{1}{5}\right) , \mathcal {C}\left( \frac{i}{50}, \frac{1}{50}\right) , \mathcal {C}\left( \frac{1}{6},\frac{1}{6}\right)$ and $\mathcal {C}\left( \frac{i}{24}, \frac{1}{24}\right)$. Finally, the image $g_1({{\mathcal {S}}}_y)$ of the yellow slice is bounded by arcs of the circles $\mathcal {C}(1, 1), \mathcal {C}\left( \frac{i}{24}, \frac{1}{24}\right) , \mathcal {C}\left( \frac{1}{2},\frac{1}{2}\right)$ and a segment of the line $\{ \Re (z) = \frac{1}{2} \}$. For this last slice, we note that $g_1$ sends the arc of circle $\{ e^{it} \mid \frac{\pi }{3} \le t \le \frac{\pi }{2}\}$ to the vertical line segment $\{ \frac{1}{2} + i t \mid t \in [\frac{1}{2}, \frac{\sqrt{3}}{2}] \}$.

We now proceed to prove that the three curvilinear quadrilaterals $g_5({{\mathcal {S}}}_r)$, $g_3({{\mathcal {S}}}_b)$ and $g_1({{\mathcal {S}}}_y)$ shown in Fig. 16 lie above the lower boundary of $\mathcal {M}_{19}:= \mathcal {M}_{19,2} \cup \mathcal {M}_{19, 7} \cup \mathcal {M}_{19, 13}$. (A point z lies above another point w with same real part if $\Im (z) \ge \Im (w)$.) Let us remark that for showing that such a quadrilateral lies above some boundary, it suffices to show that the bottom side and the right most side of the quadrilateral lie above this boundary as the quadrilateral is completely included in the region of the plane above these two sides.

The Red quadrilateral $g_5({{\mathcal {S}}}_r)$. Denote the vertices of this quadrilateral as in Fig. 20. From the above description, one computes $A_1 = \frac{6 + 44 i}{1479}, A_2 = \frac{5 + 33 i}{1114}, A_3 = \frac{1 + 5 i}{130}, A_4 = \frac{18 + 100 i}{2581}$. By the previous remark it is enough to show that the curvilinear sides $\overset{\frown }{A_1A_2}$ and $\overset{\frown }{A_2A_3}$ lie above the lower boundary of $\mathcal {M}_{19}$.

Fig. 20

The red quadrilateral $A_1A_2A_3A_4$

• Arc $\overset{\frown }{A_1A_2}$. The vertical line through $\beta _{2,1}(t)$ cuts $\mathcal {C}\left( \frac{i}{66}, \frac{1}{66}\right)$ (the circle containing $\overset{\frown }{A_1A_2}$) in two points. We denote by $P_1(t)$ the highest of these two points. Remark that $\overset{\frown }{A_1A_2}$ is indeed in the upper half part of $\mathcal {C}\left( \frac{i}{66}, \frac{1}{66}\right)$. We compute $P_1(t) = \frac{i + e^{i \tau }}{66}$ with $0 \le \tau \le \pi$, where $\frac{\cos \tau }{66}= \frac{2}{19} - \frac{\sin \frac{2\pi }{19}}{19 \sin \frac{\pi }{19}} \cos t$. Thus:

  $$\begin{aligned} \begin{aligned} \Im (P_1(t))&= \frac{1 + \sin \tau }{66} \ge \Im (\beta _{2,1}(t)) = \frac{\sin \frac{2\pi }{19}}{19\sin \frac{\pi }{19}} \sin t \\&\iff \frac{\sin ^2\tau }{66^2} \ge \left( -\frac{1}{66} + \frac{\sin \frac{2\pi }{19}}{19 \sin \frac{\pi }{19}}\sin t \right) ^2 \\&\iff f_1(\cos t)\ge 0, \end{aligned} \end{aligned}$$

  where

  $$\begin{aligned} f_1(x):= \frac{8x}{361}\cos \frac{\pi }{19} + \frac{2\sqrt{1 - x^2}}{627} \cos \frac{\pi }{19} - \frac{6 + 2 \cos \frac{2\pi }{19}}{361} \end{aligned}$$

  A study of $f_1$ shows that it is non negative on [0.96, 1], so that $f_1(\cos t)$ is non negative for $t \in [0, t_{red}^1:= 0.283]$. Note that $0<\pi /19<t_{red}^1<3\pi /19$. Moreover,⁷$\Re \left( z_2=\beta _{2,1}\left( \frac{\pi }{19}\right) \right) \approx 0.002 < \Re (A_1) = \frac{6}{1479}$ and $\Re (\beta _{2,1}(t_{red}^1)) \approx 0.0055 > \Re (A_2) = \frac{5}{1114}$. This shows that the arc $\overset{\frown }{A_1A_2}$ entirely lies above $\beta _{2,1}$.

• Arc $\overset{\frown }{A_2A_3}$. The vertical line through $\beta _{2,1}(t)$ cuts $\mathcal {C}\left( \frac{1}{10}, \frac{1}{10}\right)$ (the circle containing $\overset{\frown }{A_2A_3}$) in two points. Denoting by $P_2(t)$ the highest of these two points, similar computations as above lead to:

  $$\begin{aligned}{} & {} \Im (P_2(t)) \ge \Im (\beta _{2,1}(t))\\ {}{} & {} \quad \iff f_2(\cos t) := \frac{1}{1805}\frac{\sin \frac{2\pi }{19}}{\sin \frac{\pi }{19}} \cos t + \frac{18}{1805} - \frac{\sin ^2\frac{2\pi }{19}}{361 \sin ^2\frac{\pi }{19}} \ge 0. \end{aligned}$$

  Since $f_2$ is non negative on [0.55, 1], we have that $f_2(\cos t)$ is non negative for $t \in [0, 0.9]$. Note that this interval contains the interval of definition $[\frac{\pi }{19}, \frac{3 \pi }{19}]$ of $\beta _{2,1}$. Since $\Re (z_2) < \Re (A_2)$ and $\Re \left( \beta _{2,1}\left( \frac{3\pi }{19}\right) \right) \approx 0.014 > \Re (A_3) = \frac{1}{130}$, the arc $\overset{\frown }{A_2A_3}$ is included in the region above $\beta _{2,1}$. We thus conclude that $A_1A_2A_3A_4$ lies entirely above $\beta _{2,1}$.

The blue quadrilateral $g_3({{\mathcal {S}}}_b)$. Denote the vertices of this quadrilateral $B_1, B_2, B_3, B_4$ in analogy with what precedes ($B_1$ is the bottom left vertex of the quadrilateral, $B_2$ the bottom right one, $B_3$ the top right one and $B_4$ the top left one). We compute $B_1 = \frac{2 + 20 i}{505}, B_2 = \frac{3 + 25 i}{634}, B_3 = \frac{1 + 4 i}{51}, B_4 = \frac{10 + 48 i}{601}$. Again, it is enough to show that the curvilinear sides $\overset{\frown }{B_1B_2}$ and $\overset{\frown }{B_2B_3}$ lie above the lower boundary of $\mathcal {M}_{19}$.

• Arc $\overset{\frown }{B_1B_2}$. The vertical line passing by $\beta _{2,1}(t)$ cuts $\mathcal {C}\left( \frac{i}{50}, \frac{1}{50}\right)$ (the circle containing $\overset{\frown }{B_1B_2}$) in two points. Let $Q_1(t)$ denotes the highest of these two points. We have

  $$\begin{aligned}&\Im (Q_1(t)) \ge \Im (\beta _{2,1}(t)) \iff \\&f_3(\cos t) := \frac{4 \sin \frac{2\pi }{19}}{361 \sin \frac{\pi }{19}} \cos t + \frac{\sin \frac{2\pi }{19}}{475 \sin \frac{\pi }{19}} \sqrt{1 - \cos ^2 t} -\frac{4}{361} - \frac{\sin ^2\frac{2\pi }{19}}{361 \sin ^2\frac{\pi }{19}} \ge 0. \end{aligned}$$

  Furthermore, $f_3$ is non negative on [0.94, 1], so that $f_3(\cos t)$ is non negative on $[0, t_{blue}^1:= 0.3]$. Note that $0< \pi /19<0.3 < 3\pi /19$. Moreover, $\Re (z_2) \approx 0.002 < \Re (B_1) = \frac{2}{505}$ and $\Re (\beta _{2,1}(t_{blue}^1)) \approx 0.008 > \Re (B_2) = \frac{3}{634}$, which implies that $\overset{\frown }{B_1B_2}$ lies entirely above $\beta _{2,1}$.

• Arc $\overset{\frown }{B_2B_3}$. Denote by $Q_2(t)$ the highest intersection point of the vertical line passing through $\beta _{2,1}(t)$ with $\mathcal {C}\left( \frac{1}{6},\frac{1}{6}\right)$ (the circle containing $\overset{\frown }{B_2B_3}$). We have

  $$\begin{aligned}{} & {} \Im (Q_2(t)) \ge \Im (\beta _{2,1}(t)) \\ {}{} & {} \quad \iff f_4(\cos t) := - \frac{7 \sin \frac{2\pi }{19}}{1083 \sin \frac{\pi }{19}} \cos t + \frac{26}{1083} - \frac{\sin ^2\frac{2\pi }{19}}{361\sin ^2\frac{\pi }{19}} \ge 0. \end{aligned}$$

  Since $f_4$ is non negative on $[-1, 1]$ and $\Re (z_2) < \Re (B_2)$, it follows that $\overset{\frown }{B_2B_3}$ is above $\beta _{2,1}$ over the interval $[\Re (B_2), \Re \left( \beta _{2,1}\left( \frac{3\pi }{19}\right) \right) ]$. Let $\overline{\beta }_{2,2}$ be the supporting line of $\beta _{2,2}$. The point of $\overline{\beta }_{2,2}$ on the same vertical as $B_2$ is $Q_3:= \beta _{2,2}\left( \frac{1211 \sin \frac{\pi }{19}}{634 \cos \frac{15\pi }{38}} \right)$, while the point of $\beta _{2,2}$ on the same vertical line as $B_3$ is $Q_4:= \beta _{2,2}\left( t_{blue}^2:= \frac{83 \sin \frac{\pi }{19}}{51 \cos \frac{15\pi }{38}} \right)$. Observe that $t_{blue}^2 \in [\cos \frac{16\pi }{19}, \cos \frac{3\pi }{19}]$. We compute $\Im (Q_3) \approx 0.02 < \Im (B_2) = \frac{25}{634}$ and $\Im (Q_4) \approx 0.06 < \Im (B_3) = \frac{4}{51}$. By concavity of $\overset{\frown }{B_2B_3}$, we deduce that $\overset{\frown }{B_2B_3}$ lies above $\beta _{2,2}$ over $[\Re (\beta _{2,2}(\cos \frac{3\pi }{19})),\Re (B_3)]$. We conclude that $\overset{\frown }{B_2B_3}$ lies above $\beta _{2,1}\cup \beta _{2,2}$.

The yellow quadrilateral $g_1({{\mathcal {S}}}_y)$. Denote the vertices of this quadrilateral $C_1, C_2, C_3, C_4$ analogously to what precedes. One computes

$$\begin{aligned} C_1 = \frac{2 + 48 i}{577},\quad C_2 = \frac{1 + 12 i}{145}, \quad C_3 = \frac{1 + i}{2},\quad C_4 = \frac{1 + \sqrt{3} i}{2} = e^{i \frac{\pi }{3}}. \end{aligned}$$

We show that the curvilinear sides $\overset{\frown }{C_1C_2}$ and $\overset{\frown }{C_2C_3}$ lie above the lower boundary of $\mathcal {M}_{19}$.

• Arc $\overset{\frown }{C_1C_2}$. Let $R_1(t)$ be the highest intersection point of the vertical through $\beta _{2,1}(t)$ with $\mathcal {C}\left( \frac{i}{24}, \frac{1}{24}\right)$ (the circle containing $\overset{\frown }{C_1C_2}$). We have

  $$\begin{aligned}&\Im (R_1(t)) \ge \Im (\beta _{2,1}(t)) \\ {}&\quad \iff f_5(\cos t) := \frac{8 \cos \frac{\pi }{19}}{361} \cos t + \frac{\cos \frac{\pi }{19}}{114} \sqrt{1 - \cos ^2 t} - \frac{6 + 2 \cos \frac{2\pi }{19}}{361} \ge 0. \end{aligned}$$

  Since $f_5$ is non negative on [0.8, 1], it ensues that $f_5(\cos t)$ is non negative for t in [0, 0.6]. This interval contains $[\frac{\pi }{19}, \frac{3 \pi }{19}]$. As $\Re (z_2) \approx 0.002 < \Re (C_1) = \frac{2}{577}$ and $\Re \left( \beta _{2,1}\left( \frac{3\pi }{19}\right) \right) \approx 0.02 > \Re (C_2) = \frac{1}{145}$, we conclude that $\overset{\frown }{C_1C_2}$ lies entirely above $\beta _{2,1}$.

• Arc $\overset{\frown }{C_2C_3}$. First we note that if a point z with $\Re (C_3) \le \Re (z) \le \frac{1}{2}$ belongs to $\mathcal {B}\left( \frac{1}{2}, \frac{1}{2}\right)$ then it lies below the arc $\overset{\frown }{C_2C_3}$. Thus to show that $\beta _{i, j}(t)$ is below $\overset{\frown }{C_2C_3}$ it is sufficient to show that $|\beta _{i, j}(t) - \frac{1}{2}| \le \frac{1}{2}$. We have

  $$\begin{aligned} |\beta _{2,1}(t) - \frac{1}{2}| \le \frac{1}{2} \iff f_6(\cos t):= \frac{15 \sin \frac{2\pi }{19}}{361 \sin \frac{\pi }{19}} \cos t + \frac{\sin ^2\frac{2\pi }{19}}{361 \sin ^2\frac{\pi }{19}} - \frac{34}{361} \le 0. \end{aligned}$$

  Since $f_6$ is non positive on $[-1, 1]$, it follows that $f_6(\cos t)$ is always non positive. Hence $\overset{\frown }{C_2C_3}$ is above $\beta _{2,1}$ over $\left[ \Re (C_2), \Re \left( \beta _{2,1}\left( \frac{3\pi }{19}\right) \right) \right]$. Next, we show that $\overset{\frown }{C_2C_3}$ lies above $\beta _{2,2}$ in the strip $\{z\mid \Re (z)\in {[}\Re \left( \beta _{2,1}\left( \frac{3\pi }{19}\right) \right) , \Re (z_7)]\}$. Let $R_{2,7}$ be the point on $\beta _{2,2}$ with real part $\Re (z_7)$. We have $R_{2,7} = \beta _{2,2}\left( \tau _{2, 7}:= \frac{(-5 + \cos \frac{5\pi }{38} \cot \frac{\pi }{19})\sin \frac{\pi }{19}}{\sin \frac{2\pi }{19}}\right)$ and we verify that $\tau _{2, 7} \in \Big [\cos \frac{16 \pi }{19},\cos \frac{3 \pi }{19}\Big ]$ and $\Im R_{2,7} \approx 0.24 > \Im z_7 \approx 0.18$. Then, to show that $\overset{\frown }{C_2C_3}$ lies above $\beta _{2,2}$ in the above strip, it suffices by concavity of $\mathcal {B}\left( \frac{1}{2}, \frac{1}{2}\right)$ (as $\beta _{2,2}$ is a line segment) to show that $\beta _{2,2}(\cos \frac{3\pi }{19}), R_{2,7} \in \mathcal {B}\left( \frac{1}{2}, \frac{1}{2}\right)$. We indeed compute: $|\beta _{2,2}(\cos \frac{3\pi }{19}) - \frac{1}{2}|^2 \approx 0.23 < \frac{1}{4}$, $|R_{2,7} - \frac{1}{2}|^2 \approx 0.23 < \frac{1}{4}$. It remains to show that the lower boundaries of $M_{19, 7}$ and $M_{19, 13}$ lie below $\overset{\frown }{C_2C_3}$ in the strip $\{z\mid \Re (z)\in [\Re (z_7), \Re (C_3)=1/2]\}$. We have

  $$\begin{aligned}&\Big |\beta _{7,1}(t) - \frac{1}{2}\Big |^2 = |\beta _{7,3}(t) - \frac{1}{2}|^2 \le \frac{1}{4}\\ {}&\quad \iff f_7(t) := \frac{\cos ^2\frac{5\pi }{38}}{361 \sin ^2\frac{\pi }{19}} t^2 + \frac{5 \cos \frac{5\pi }{38}}{361 \sin \frac{\pi }{19}} t - \frac{84}{361} \\ {}&\qquad + \frac{\cos ^2\frac{\pi }{19} \cos \frac{\pi }{19} \sin \frac{5\pi }{38} + \sin ^2\frac{5 \pi }{38}}{361 \sin ^2\frac{\pi }{19}} \le 0. \end{aligned}$$

  As $f_7$ is non positive on the interval $[-1, 1]$, which contains the domains of $\beta _{7,1}$ and $\beta _{7,3}$, we deduce that these two curves lie entirely below $\overset{\frown }{C_2C_3}$. We then have

  $$\begin{aligned} |\beta _{7,2}(t) - \frac{1}{2}|^2 \le \frac{1}{4} \iff f_8(\cos t):= \frac{5 \cos \frac{5\pi }{38}}{361 \sin \frac{\pi }{19}}\cos t - \frac{84}{361} + \frac{\cos ^2\frac{5\pi }{38}}{361 \sin ^2\frac{\pi }{19}} \le 0. \end{aligned}$$

  Since $f_8$ is non positive on $[-1, 1]$, it follows that $f_8(\cos t)$ is non positive for all t, which shows that $\beta _{7,2}$ lies below $\overset{\frown }{C_2C_3}$. Finally, as previously noticed, since $\beta _{13,1}$ is a line segment, it suffices to show that its extremities lies below $\overset{\frown }{C_2C_3}$ by concavity. We compute: $|z_{13} - \frac{1}{2}|^2 \approx 0.244 < \frac{1}{4}$ and $|w_{13} - \frac{1}{2}| \approx 0.1 < \frac{1}{4}$. Thus $\overset{\frown }{C_2C_3}$ lies above $\beta _{13,1}$, and as$\Re (w_{13}) \approx 0.502 > \frac{1}{2} = \Re (C_3)$, we deduce that $\overset{\frown }{C_2C_3}$ is included in $M_{19}$.

This ends the proof of Lemma 3.9.