Asymmetric expansion preserves hyperbolic convexity

Abstract

In an earlier paper we showed that the radial expansion of a hyperbolic convex set in the Poincaré disk about any point inside it results in a hyperbolic convex set. In this work, we generalize this result by showing that the asymmetric expansion of a hyperbolic convex set about any point inside it also results in a hyperbolic convex set.

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Correspondence to Dhruv Kohli.

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Appendix

Appendix

Proof of Lemma (3)

Since \(\theta ' = s\varDelta {\hat{\theta }}/\beta > 0\), we rewrite Eq. (6) as

$$\begin{aligned} k_g&= \frac{1}{v^3}\theta '\sqrt{EG}\left( \frac{G_r}{G}r'^2 +\frac{G_r}{2E}\theta '^2 +\frac{r'\theta ''}{\theta '} -r'' \right) . \end{aligned}$$
(54)

We denote the term outside the bracket as \(P_0\). So,

$$\begin{aligned} P_0&= \frac{1}{v^3}\theta '\sqrt{EG} = \frac{1}{v^3}\frac{s\varDelta {\hat{\theta }}}{\beta }\sinh r. \end{aligned}$$
(55)

Now, we write each term in the bracket separately and assign labels so as to group the terms which contain \({\hat{r}}'^2\), \({\hat{r}}'\varDelta {\hat{\theta }}\) and \((\varDelta {\hat{\theta }})^2\). Note that \(\beta '\) contains \(\varDelta {\hat{\theta }}\) and \(\beta ''\) contains \((\varDelta {\hat{\theta }})^2\).

$$\begin{aligned} \frac{G_r}{G}r'^2&= 2\coth ({\hat{r}}\sqrt{\beta })(\underbrace{{\hat{r}}'^2\beta }_\text {(i)} + \underbrace{\frac{{\hat{r}}^2\beta '^2}{4\beta }}_\text {(iii)}+\underbrace{{\hat{r}}{\hat{r}}'\beta '}_\text {(ii)}), \end{aligned}$$
(56)
$$\begin{aligned} \frac{G_r}{2E}\theta '^2&= \underbrace{\frac{\sinh 2{\hat{r}}\sqrt{\beta }}{2}\frac{s^2(\varDelta {\hat{\theta }})^2}{\beta ^2}}_\text {(iii)}, \end{aligned}$$
(57)
$$\begin{aligned} \frac{r'\theta ''}{\theta '}&= ({\hat{r}}'\sqrt{\beta }+\frac{{\hat{r}}\beta '}{2\sqrt{\beta }})\left( -\frac{\beta '}{\beta }\right) = -\underbrace{\frac{{\hat{r}}'\beta '}{\sqrt{\beta }}}_\text {(ii)} - \underbrace{\frac{{\hat{r}}\beta '^2}{2\beta \sqrt{\beta }}}_\text {(iii)}, \end{aligned}$$
(58)
$$\begin{aligned} r''&= \left( \underbrace{2{\hat{r}}'^2\coth {\hat{r}}}_\text {(i)} + \underbrace{(\varDelta {\hat{\theta }})^2\frac{\sinh 2{\hat{r}} }{2}}_\text {(iii)}\right) \sqrt{\beta } + \underbrace{\frac{{\hat{r}}'\beta '}{\sqrt{\beta }}}_\text {(ii)},\nonumber \\&\ \ \qquad + \underbrace{\frac{{\hat{r}}}{2}\left( \frac{\beta ''\sqrt{\beta }-\beta '^2/2\sqrt{\beta }}{\beta }\right) }_\text {(iii)}. \end{aligned}$$
(59)

We now combine the terms having the same labels. From the terms with labels (i), (ii) and (iii) we obtain \(P_1{\hat{r}}'^2\), \(P_2{\hat{r}}'\varDelta {\hat{\theta }}'\) and \(P_3(\varDelta {\hat{\theta }}')^2\) respectively. So,

$$\begin{aligned} P_1&= 2\sqrt{\beta }(\sqrt{\beta }\coth {\hat{r}}\sqrt{\beta }-\coth {\hat{r}}) \end{aligned}$$
(60)
$$\begin{aligned}&=2\sqrt{\beta }\ \frac{\psi ({\hat{r}}\sqrt{\beta })-\psi ({\hat{r}})}{{\hat{r}}}, \end{aligned}$$
(61)
$$\begin{aligned} P_2^2&= 4\left( \frac{\beta '}{\varDelta {\hat{\theta }}}\right) ^2\left( {\hat{r}}\coth {\hat{r}}\sqrt{\beta } - \frac{1}{\sqrt{\beta }}\right) ^2\end{aligned}$$
(62)
$$\begin{aligned}&= 16(\beta -s^2)(1-\beta )({\hat{r}}\coth {\hat{r}}\sqrt{\beta }-1/\sqrt{\beta })^2 \end{aligned}$$
(63)
$$\begin{aligned}&= 16(\beta -s^2)(1-\beta )\frac{[\psi ({\hat{r}}\sqrt{\beta })]^2}{\beta }, \end{aligned}$$
(64)
$$\begin{aligned} P_3&= \frac{s^2}{2\beta ^2}\sinh 2{\hat{r}}\sqrt{\beta } - \frac{\sinh 2{\hat{r}}}{2}\sqrt{\beta } - \frac{{\hat{r}}\beta ''}{2(\varDelta {\hat{\theta }})^2\sqrt{\beta }} \nonumber \\&\qquad +\frac{{\hat{r}}\beta '^2}{2(\varDelta {\hat{\theta }})^2\beta }({\hat{r}}\coth {\hat{r}}\sqrt{\beta }-\frac{1}{2\sqrt{\beta }})\end{aligned}$$
(65)
$$\begin{aligned}&= \frac{1}{2\beta \sqrt{\beta }}\left( \underbrace{\frac{s^2}{\sqrt{\beta }}\sinh 2{\hat{r}}\sqrt{\beta }}_\text {(a)} - \underbrace{\beta ^2\sinh 2{\hat{r}}}_\text {(b)} - \underbrace{2{\hat{r}}\beta (1-s^2)\cos 2{\hat{\theta }}}_\text {(d)}\right. \nonumber \\&\qquad \qquad \qquad +\left. 4{\hat{r}}(\beta -s^2)(1-\beta )({\hat{r}}\sqrt{\beta }\coth {\hat{r}}\sqrt{\beta }-\underbrace{1/2}_\text {(c)})\right) . \end{aligned}$$
(66)

We rewrite term \(\text {(d)}\) in Eq. (66) as

$$\begin{aligned} 2{\hat{r}}\beta (1-s^2)\cos 2{\hat{\theta }}&= \underbrace{2{\hat{r}}s^2}_\text {(a)} - \underbrace{2{\hat{r}}\beta ^2}_\text {(b)} + \underbrace{2{\hat{r}}(1-\beta )(\beta -s^2)}_\text {(c)} \end{aligned}$$
(67)

and substitute it back in Eq. (66) while combining the terms with the same labels to get

$$\begin{aligned} P_3&= \frac{1}{2\beta \sqrt{\beta }}\left( \frac{s^2}{\sqrt{\beta }}(\sinh 2{\hat{r}}\sqrt{\beta }-2{\hat{r}}\sqrt{\beta })- \beta ^2(\sinh 2{\hat{r}}-2{\hat{r}}) \right. \nonumber \\&\qquad \qquad \qquad + \left. 4{\hat{r}}(\beta -s^2)(1-\beta )({\hat{r}}\sqrt{\beta }\coth {\hat{r}}\sqrt{\beta }-1)\right) \end{aligned}$$
(68)
$$\begin{aligned}&= \frac{1}{2\beta \sqrt{\beta }}\left( \frac{s^2}{\sqrt{\beta }}\phi (2{\hat{r}}\sqrt{\beta }) - \beta ^2\phi (2{\hat{r}}) +4{\hat{r}}(\beta -s^2)(1-\beta )\psi ({\hat{r}}\sqrt{\beta })\right) \end{aligned}$$
(69)

\(\square \)

Proof of Lemma (4)

Using Taylor expansion about \(x=0\) we get

$$\begin{aligned} \sinh xy- y^3\sinh x -xy + xy^3&= y^3\sum _{k=1}^{\infty }\frac{x^{2k+1}(y^{2k-1}-1)}{(2k+1)!}. \end{aligned}$$
(70)

Since \(x > 0\) and \(y \in (0,1)\) each term in the above summation is negative so the overall expression is negative. \(\square \)

Lemma 7

For all \(x > 0\)

$$\begin{aligned} x^3(\coth x +x(1-\coth ^2 x)) - 6(x\coth x-1)^2 > 0. \end{aligned}$$
(71)

Proof

Since \(x > 0\) we have \(\sinh ^2 x > 0\). So we multiply by \(\sinh ^2 x\) throughout and prove the resulting inequality. Our claim is equivalent to \(I>0\), where I is defined as

$$\begin{aligned} I&= x^3(\cosh x \sinh x + x(\sinh ^2 x - \cosh ^2 x))-6(x\cosh x - \sinh x)^2 \end{aligned}$$
(72)
$$\begin{aligned}&=x^3(\sinh (2x)/2 - x) - 6(x\cosh x - \sinh x)^2 \end{aligned}$$
(73)
$$\begin{aligned}&= x^3\left( \sinh (2x)/2 - x\right) - 6\left( x^2\cosh ^2 x + \sinh ^2 x - x\sinh (2x)\right) \end{aligned}$$
(74)
$$\begin{aligned}&= x^3\left( \frac{\sinh 2x }{2} - x\right) - 6\left( x^2\frac{\cosh 2x +1}{2} + \frac{\cosh 2x -1}{2} - x\sinh 2x\right) . \end{aligned}$$
(75)

Using Taylor expansion about \(x=0\) we get

$$\begin{aligned} x^3\left( \frac{\sinh 2x }{2} - x\right)&= \frac{2x^6}{3} + \frac{2x^8}{15} + \sum _{k=3}^{\infty }\frac{2^{2k}x^{2k+4}}{(2k+1)!}, \end{aligned}$$
(76)
$$\begin{aligned} 6x^2\frac{\cosh 2x +1}{2}&= 6x^2 + 6x^4 + 2x^6 + \frac{4x^8}{15} + 6\sum _{k=3}^{\infty }\frac{2^{2k+1}x^{2k+4}}{(2k+2)!}, \end{aligned}$$
(77)
$$\begin{aligned} 6\frac{\cosh 2x -1}{2}&= 6x^2 + 2x^4 + \frac{4x^6}{15} + \frac{2x^8}{105} + 6\sum _{k=3}^{\infty }\frac{2^{2k+3}x^{2k+4}}{(2k+4)!},\end{aligned}$$
(78)
$$\begin{aligned} 6x\sinh 2x&= 12x^2 + 8x^4 + \frac{8x^6}{5} + \frac{16x^8}{105} + 6\sum _{k=3}^{\infty }\frac{2^{2k+3}x^{2k+4}}{(2k+3)!}. \end{aligned}$$
(79)

Combine the terms and observe that all terms through order \(x^8\) cancel to obtain

$$\begin{aligned} I&= \sum _{k=3}^{\infty } \frac{2^{2k+2}x^{2k+4}}{(2k+4)!}\left( (k+2)(2k+3)(k+1)\right. \nonumber \\ {}&\left. \quad - 3(2k+4)(2k+3) - 12 + 12(2k+4)\right) \end{aligned}$$
(80)
$$\begin{aligned}&= \sum _{k=3}^{\infty }\frac{2^{2k+2}x^{2k+4}}{(2k+4)!}(2k+3)(k-1)(k-2). \end{aligned}$$
(81)

Clearly, for \(k \ge 3\) all the terms in the summation are positive. So, \(I > 0\) and therefore the inequality (71) holds. \(\square \)

Proof of Lemma (5)

This is equivalent to showing that for all \(x > 0\) and \(y \in (0,1)\), \(f(x,y) > 0\) where

$$\begin{aligned} f(x,y) = \frac{\coth x - y\coth xy}{(xy\coth xy-1)(x\coth x-1)} + \frac{4(1-y^{-2})}{\sinh 2x-2x}. \end{aligned}$$
(82)

Fix \(x = a > 0\). Clearly as y tends to 1, \(f(a,y) \rightarrow 0\). So, it is sufficient to show that f(ay) is decreasing for all \(y \in (0,1)\).

$$\begin{aligned} \frac{df(a,y)}{dy}&= \frac{1}{y^3}\left( \frac{8}{\sinh 2a-2a} - \frac{a^3y^3}{a^3}\frac{(\coth ay+ay(1-\coth ^2 ay))}{(ay\coth ay-1)^2}\right) . \end{aligned}$$
(83)

Since \(ay > 0\), using Lemma (7) we obtain

$$\begin{aligned} \frac{df(a,y)}{dy}&< \frac{1}{y^3}\left( \frac{8}{\sinh 2a-2a} - \frac{6}{a^3}\right) < 0, \end{aligned}$$
(84)

where the last inequality follows by using the Taylor expansion of \(\sinh (2a)\) about 0 and noting that

$$\begin{aligned} \frac{\sinh 2a-2a}{8}&= \frac{a^3}{6} + \sum _{k=2}^{\infty }\frac{2^{2k-2}a^{2k+1}}{(2k+1)!}> \frac{a^3}{6} \ \ \text { since } a > 0. \end{aligned}$$
(85)

\(\square \)

Proof of Lemma (6)

As \(x \rightarrow 0\), \(\sin xy-y\sin x \rightarrow 0\) and its derivative with respect to x is \(y(\cos xy-\cos x) > 0\) when \(x \in (0,\pi )\) and \(y \in (0,1)\). \(\square \)

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Kohli, D., Rabin, J.M. Asymmetric expansion preserves hyperbolic convexity. J. Geom. 111, 33 (2020). https://doi.org/10.1007/s00022-020-00545-4

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Keywords

  • Hyperbolic convexity
  • Poincaré disk
  • Asymmetric expansion