On the Canham Problem: Bending Energy Minimizers for any Genus and Isoperimetric Ratio

Abstract

Building on work of Mondino–Scharrer, we show that among closed, smoothly embedded surfaces in ${\mathbb {R}}^3$ of genus g and given isoperimetric ratio v, there exists one with minimum bending energy ${\mathcal {W}}$. We do this by gluing $g+1$ small catenoidal bridges to the bigraph of a singular solution for the linearized Willmore equation $\Delta (\Delta +2)\varphi =0$ on the $(g+1)$-punctured sphere ${\mathbb {S}}^2$ to construct a comparison surface of genus g with arbitrarily small isoperimetric ratio $v\in (0, 1)$ and ${\mathcal {W}}< 8\pi$.

Appendix A. Mean Curvature on a Catenoidal Bridge

In this appendix, we estimate the mean curvature of a small catenoidal bridge in ${\mathbb {S}}^3$. This was done in [5, Example 2.15], but we summarize the argument in order to keep the exposition self-contained.

Define a parametrization $E :(0, \pi ) \times (-\pi , \pi ) \times (-\frac{\pi }{2},\frac{\pi }{2} ) \rightarrow {\mathbb {S}}^3$ by

We take $({r}, \theta , {z})$ as local coordinates for ${\mathbb {S}}^3$. The pullback metric is

$$\begin{aligned} E^* g = \cos ^2 {z}\left( \textrm{d} {r}^2 + \sin ^2 {r}\textrm{d} \theta ^2\right) + \textrm{d}{z}^2, \end{aligned}$$

and the only nonvanishing Christoffel symbols are

$$\begin{aligned}{} & {} \Gamma _{{r}{z}}^{{r}} = \Gamma _{{z}{r}}^{{r}}= \Gamma _{\theta {z}}^{\theta } = \Gamma _{{z}\theta }^{\theta } = - \tan {z}, \quad \Gamma _{\theta \theta }^{r}= - \sin {r}\cos {r}, \nonumber \\{} & {} \quad \Gamma _{{r}\theta }^{\theta } = \Gamma _{\theta {r}}^{\theta } = \cot {r}, \quad \Gamma _{{r}{r}}^{{z}} = \cos {z}\sin {z}, \quad \Gamma _{\theta \theta }^{{z}} = \sin ^2 {r}\sin {z}\cos {z}. \end{aligned}$$

(A.1)

Define a map $X: [-\underline{{{{s}}}}, \underline{{{{s}}}}] \times (-\pi , \pi ) \rightarrow {\mathbb {R}}^3$ by

$$\begin{aligned} X({{{s}}}, \vartheta )&= ({r}({{{s}}}, \vartheta ), \theta ({{{s}}}, \vartheta ), {z}({{{s}}}, \vartheta )) \nonumber \\&= (\tau \cosh {{{s}}}, \vartheta , \tau {{{s}}}), \end{aligned}$$

(A.2)

where $\underline{{{{s}}}}$ is defined by the equation $\tau \cosh \underline{{{{s}}}}= 9 \tau ^\alpha$.

Calculation shows that the pullback metric in $({{{s}}}, \vartheta )$ coordinates is

$$\begin{aligned} X^* E^* g = {r}^2( 1- \tanh ^2 {{{s}}}\sin ^2 {z}) \textrm{d}{{{s}}}^2 + \cos ^2 {z}\sin ^2 {r}\, \textrm{d}\vartheta ^2, \end{aligned}$$

(A.3)

where ${r}= {r}({{{s}}}, \vartheta )$ and ${z}= {z}({{{s}}}, \vartheta )$, and that

$$\begin{aligned} \nu =( \tanh {{{s}}}\, \partial _{{z}} - \sec ^2{z}{\,\hbox {sech}}\,{{{s}}}\, \partial _{{r}})/\sqrt{1+\tan ^2 {z}{\,\hbox {sech}}\,^2 {{{s}}}} \end{aligned}$$

is a unit normal field along the image of X. We compute the second fundamental form A of X using the formula $A = \left( X^{k}_{, \alpha \beta } + \Gamma _{lm}^{k} X^l_{, \alpha } X^{m}_{, \beta }\right) g_{kn}\nu ^n dx^\alpha dx^\beta$, where X is as in A.2, we have renamed the cylinder coordinates $(x^1, x^2): = ({{{s}}}, \vartheta )$, and Greek indices take the values 1 and 2 while Latin indices take the values 1, 2, 3, corresponding to the coordinates ${r}, \theta , {z}$. Using the preceding and the Christoffel symbols in (A.1), we find

$$\begin{aligned} A&= \frac{ \left[ \tau ^2 \tanh {{{s}}}\left( \tan {z}+ \frac{1}{2}\sinh ^2 {{{s}}}\sin 2{z}\right) - \tau \right] \textrm{d}{{{s}}}^2 +\frac{1}{2}(\sin 2{r}{\,\hbox {sech}}\,{{{s}}}+ \sin ^2 {r}\sin 2{z}\tanh {{{s}}}) \, \textrm{d}\vartheta ^2 }{ \sqrt{ 1+ \tan ^2 {z}{\,\hbox {sech}}\,^2 {{{s}}}}}\\&= (1+ O({z}^2)) \left( \tau ( - \textrm{d}{{{s}}}^2 + \textrm{d}\vartheta ^2) + O({r}^2 {z})\textrm{d}{{{s}}}^2 + O({r}^3+{r}^2 {z}) \textrm{d}\vartheta ^2\right) , \end{aligned}$$

where in the second equality we have estimated using that $\sqrt{ 1+ \tan ^2 {z}{\,\hbox {sech}}\,^2 {{{s}}}} = 1+ O({z}^2)$ and that $\frac{1}{2}\sin 2{r}{\,\hbox {sech}}\,{{{s}}}= \tau + O({r}^3)$. Finally, using that ${r}^2 g^{{{{s}}}{{{s}}}} = 1+ O({z}^2)$ and ${r}^2 g^{\vartheta \vartheta } = 1+O({z}^2 +{r}^2)$, we estimate

$$\begin{aligned} {r}^2 H = O\left( \tau {z}^2 +{r}^2 |{z}| + \tau {r}^2 \right) . \end{aligned}$$

(A.4)