Asymmetric expansion preserves hyperbolic convexity


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.

This is a preview of subscription content, log in to check access.

Fig. 1


  1. 1.

    Alexander, S.: Local and global convexity in complete Riemannian manifolds. Pac. J. Math. 76(2), 283–289 (1976)

    MathSciNet  Article  Google Scholar 

  2. 2.

    Banchoff, T., Lovett, S.: Differential Geometry of Curves and Surfaces, 2nd edn. CRC Press, Boca Raton (2016)

    Google Scholar 

  3. 3.

    Bishop, R.L.: Infinitesimal convexity implies local convexity. Indiana Univ. Math. J. 24(2), 169–172 (1974)

    MathSciNet  Article  Google Scholar 

  4. 4.

    Kohli, D., Rabin, J.M.: Radial expansion preserves hyperbolic convexity and radial contraction preserves spherical convexity. J. Geom. 110, 40 (2019)

    MathSciNet  Article  Google Scholar 

  5. 5.

    Klen, R., Visuri, M., Vuorinen, M.: On Jordan type inequalities for hyperbolic functions. J. Inequal. Appl. 2010, 362548 (2010)

    MathSciNet  Article  Google Scholar 

  6. 6.

    Oprea, J.: Differential Geometry and Its Applications, 2nd edn. MAA, Hyderabad (2007)

    Google Scholar 

  7. 7.

    Stoker, J.J.: Differential Geometry. Wiley-Interscience, New York (1969)

    Google Scholar 

  8. 8.

    Ungar, A.A.: Möbius Transformation and Einstein Velocity Addition in the Hyperbolic Geometry of Bolyai and Lobachevsky. Nonlinear Analysis, pp. 721–770. Springer, New York (2012)

    Google Scholar 

Download references

Author information



Corresponding author

Correspondence to Dhruv Kohli.

Ethics declarations

Conflicts of interest

On behalf of all authors, the corresponding author states that there is no conflict of interest.

Additional information

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.



Proof of Lemma (3)

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

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

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

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

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 \((\varDelta {\hat{\theta }})^2\).

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

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,

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

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

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

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

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

\(\square \)

Proof of Lemma (4)

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

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

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 \(x > 0\)

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


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

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

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

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

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

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

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

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

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 \(y \in (0,1)\).

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

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

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

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

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

\(\square \)

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 \(y \in (0,1)\). \(\square \)

Rights and permissions

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Kohli, D., Rabin, J.M. Asymmetric expansion preserves hyperbolic convexity. J. Geom. 111, 33 (2020).

Download citation


  • Hyperbolic convexity
  • Poincaré disk
  • Asymmetric expansion