A classification theorem for Helfrich surfaces

Abstract

In this paper we study the functional $\mathcal W{} _{\lambda _1,\lambda _2}$, which is the sum of the Willmore energy, $\lambda _1$-weighted surface area, and $\lambda _2$-weighted volume, for surfaces immersed in $\mathbb R ^3$. This coincides with the Helfrich functional with zero ‘spontaneous curvature’. Our main result is a complete classification of all smooth immersed critical points of the functional with $\lambda _1\ge 0$ and small $L^2$ norm of tracefree curvature, with no assumption on the growth of the curvature in $L^2$ at infinity. This not only improves the gap lemma due to Kuwert and Schätzle for Willmore surfaces immersed in $\mathbb R ^3$ but also implies the non-existence of critical points of the functional satisfying the energy condition for which the surface area and enclosed volume are positively weighted.

Appendix: Selected proofs

We collect here the proofs of several well-known formulae and results for the convenience of the reader and readability of the paper. Many of the statements contained in this appendix have appeared in a similar form in [13, 14].

Proof of (12), cf. [13, Lem. 2.3]. Let us first prove that

$$\begin{aligned} \nabla \Delta H = \Delta \nabla H - \frac{1}{4}H^2\nabla H + A*A^o*\nabla H. \end{aligned}$$

(24)

Equation (8) implies

$$\begin{aligned} \nabla _{ijk}H - \nabla _{jik}H = (A_{lj}A_{ik} - A_{il}A_{jk})g^{lm}\nabla _mH, \end{aligned}$$

so

$$\begin{aligned} \nabla _{j}\Delta H = \Delta \nabla _{j}H - (A^m_{j}H - A^m_{p}A^p_{j})\nabla _mH. \end{aligned}$$

(25)

Note that

$$\begin{aligned} A^m_j\nabla _mH&= \left( (A^o)^m_j + \frac{1}{2}\delta ^m_jH\right) \nabla _mH = \frac{1}{2}H\nabla _jH + A^o*\nabla H\\ A^m_pA^p_j\nabla _mH&= \left( (A^o)^p_j + \frac{1}{2}\delta ^p_jH\right) \left( (A^o)^m_p + \frac{1}{2}\delta ^m_pH\right) \nabla _mH\\&= \frac{1}{4}H^2\nabla _jH + A^o*A^o*\nabla H + H A^o*\nabla H, \end{aligned}$$

which when combined with (25) above gives (24). Testing (24) against $\nabla H\,\gamma ^4$ and integrating yields

$$\begin{aligned} \int _\varSigma \langle {{\nabla \Delta H}, {\nabla H}}\rangle _g \gamma ^4 \,d\mu&= \int _\varSigma \langle {{\Delta \nabla H}, {\nabla H}}\rangle _g \gamma ^4 \,d\mu \nonumber \\&- \frac{1}{4} \int _\varSigma |H|^2|\nabla H|^2\gamma ^4\,d\mu \!+\! \int _\varSigma A*A^o*\nabla H*\nabla H\, \gamma ^4\,d\mu .\qquad \end{aligned}$$

(26)

Using the divergence theorem we have

$$\begin{aligned} \int _\varSigma \langle {{\nabla \Delta H}, {\nabla H}}\rangle _g \gamma ^4 \,d\mu = -\int _\varSigma |\Delta H|^2\gamma ^4\,d\mu - 4\int _\varSigma \Delta H \langle {{\nabla H}, {\nabla \gamma }}\rangle _g\gamma ^3\,d\mu \end{aligned}$$

and

$$\begin{aligned} \int _\varSigma \langle {{\Delta \nabla H}, {\nabla H}}\rangle _g \gamma ^4 \,d\mu = -\int _\varSigma |\nabla _{(2)}H|^2\gamma ^4\,d\mu - 4\int _\varSigma \langle {{\nabla _{(2)} H}, {\nabla \gamma \nabla H}}\rangle _g\gamma ^3\,d\mu . \end{aligned}$$

Combining these identities with (26) we obtain

$$\begin{aligned}&\int _\varSigma |\nabla _{(2)}H|^2\gamma ^4\,d\mu + \frac{1}{4} \int _\varSigma |H|^2|\nabla H|^2\gamma ^4\,d\mu = \int _\varSigma |\Delta H|^2\gamma ^4\,d\mu \\&\quad + \int _\varSigma \nabla _{(2)} H*\nabla H*\nabla \gamma \ \gamma ^3\,d\mu + \int _\varSigma A*A^o*\nabla H*\nabla H\ \gamma ^4\,d\mu , \end{aligned}$$

which upon noting (6), proves (12). $\square$

Proof of (14), cf. [13, eqn. (17)]. We shall prove

$$\begin{aligned}&(1-\delta )\int _\varSigma |H|^2 |\nabla A^o|^2\gamma ^4\,d\mu + \left( \frac{1}{2}-2\delta \right) \int _\varSigma |H|^4|A^o|^2\gamma ^4\,d\mu \nonumber \\&\quad \le \left( \frac{1}{2}+3\delta \right) \int _\varSigma |H|^2|\nabla H|^2\gamma ^4\,d\mu \nonumber \\&\qquad + c\int _\varSigma \left( |A^o|^6 + |A^o|^2|\nabla A^o|^2\right) \gamma ^4\,d\mu + c_\gamma ^4c\int _{[\gamma >0]}|A^o|^2\,d\mu , \end{aligned}$$

(27)

for $\delta >0$, where $c$ is a constant depending only on $\delta$. Equation (14) follows from (27) with the choice $\delta =\frac{1}{8}$. We first begin with the following consequence of Simons’ identity:

$$\begin{aligned} \Delta A^o = S^o(\nabla _{(2)}H) + \frac{1}{2}H^2A^o - |A^o|^2A^o. \end{aligned}$$

(28)

This follows readily from (9),

$$\begin{aligned} \Delta A^o_{ij}&= \Delta A_{ij} - \frac{1}{2}g_{ij}\Delta H\\&= \nabla _{ij}H - \frac{1}{2}g_{ij}\Delta H + HA_i^lA_{lj} - |A|^2A_{ij}\\&= S^o(\nabla _{(2)}H) + HA_i^lA_{lj} - |A|^2A_{ij}\\&= S^o(\nabla _{(2)}H) + 2KA^o_{ij}, \end{aligned}$$

provided we show

$$\begin{aligned} HA_i^jA_{lj} - |A|^2A_{ij} = 2KA^o_{ij}. \end{aligned}$$

(29)

Choosing normal coordinates so that $A$ is diagonalised (at a point) with $A=\delta _{ij} k_j$ (no sum over $j$), $H=k_1+k_2$, $K=k_1k_2$, $|A|^2=k_2^2+k_1^2$ (at this point) and

$$\begin{aligned} A^o_{ij} = \left\{ \begin{array}{ll} \frac{1}{2}(k_1-k_2),&{}\quad \text{ if } \; i=j=1\\ \frac{1}{2}(k_2-k_1),&{}\quad \text{ if }\;i=j=2\\ 0,&{}\quad \text{ otherwise }, \end{array}\right. \end{aligned}$$

we have

$$\begin{aligned} (i=j=1)\quad HA_1^lA_{l1} - |A|^2A_{11}&= (k_1+k_2)k_1^2 - (k_1^2+k_2^2)k_1\\&= k_1k_2(k_1-k_2) = 2KA^o_{11}\\ (i=j=2)\quad HA_2^lA_{l2} - |A|^2A_{22}&= (k_1+k_2)k_2^2 - (k_1^2+k_2^2)k_2\\&= k_1k_2(k_2-k_1) = 2KA^o_{22}, \end{aligned}$$

and otherwise (29) holds trivially. Therefore (29) is proved. This also proves (28), since

$$\begin{aligned} K = k_1k_2 = \frac{1}{4}(k_1+k_2)^2-\frac{1}{4}(k_1-k_2)^2 = \frac{1}{4}H^2-\frac{1}{2}|A^o|^2. \end{aligned}$$

Now employing the divergence theorem and inserting (28) we compute

$$\begin{aligned} \int _\varSigma |H|^2|\nabla A^o|^2\gamma ^4\,d\mu&= - \int _\varSigma |H|^2\langle {{A^o}, {\Delta A^o}}\rangle _g\gamma ^4\,d\mu \\&- 2\int _\varSigma H\langle {{\nabla A^o}, {\nabla H\, A^o}}\rangle _g\gamma ^4\,d\mu - 4\int _\varSigma |H|^2\langle {{\nabla A^o}{\nabla \gamma \, A^o}}\rangle _g\gamma ^3\,d\mu \\&= - \int _\varSigma |H|^2\langle {{A^o}, {S^o(\nabla _{(2)}H) + \frac{1}{2}|H|^2A^o - |A^o|^2A^o}}\rangle _g\gamma ^4\,d\mu \\&- 2\int _\varSigma H\langle {{\nabla A^o}, {\nabla H\, A^o}}\rangle _g\gamma ^4\,d\mu - 4\int _\varSigma |H|^2\langle {{\nabla A^o}{\nabla \gamma \, A^o}}\rangle _g\gamma ^3\,d\mu \\&= - \int _\varSigma |H|^2\langle {{A^o}, {\nabla _{(2)}H}}\rangle _g\gamma ^4\,d\mu - \frac{1}{2}\int _\varSigma |H|^4|A^o|^2\gamma ^4\,d\mu + \int _\varSigma |H|^2|A^o|^4\gamma ^4\,d\mu \\&- 2\int _\varSigma H\langle {{\nabla A^o}, {\nabla H\, A^o}}\rangle _g\gamma ^4\,d\mu - 4\int _\varSigma |H|^2\langle {{\nabla A^o}{\nabla \gamma \, A^o}}\rangle _g\gamma ^3\,d\mu \\&= \int _\varSigma |H|^2\langle {{\nabla ^* A^o}, {\nabla H}}\rangle _g\gamma ^4\,d\mu + 2\int _\varSigma H\langle {{A^o}, {\nabla H\nabla H}}\rangle _g\gamma ^4\,d\mu \\&- \frac{1}{2}\int _\varSigma |H|^4|A^o|^2\gamma ^4\,d\mu \!+\! \int _\varSigma |H|^2|A^o|^4\gamma ^4\,d\mu \!-\! 2\int _\varSigma H\langle {{\nabla A^o}, {\nabla H\, A^o}}\rangle _g\gamma ^4\,d\mu \\&+ 4\int _\varSigma |H|^2\langle {{A^o}{\nabla \gamma \nabla H}}\rangle _g\gamma ^3\,d\mu - 4\int _\varSigma |H|^2\langle {{\nabla A^o}, {\nabla \gamma \, A^o}}\rangle _g\gamma ^3\,d\mu . \end{aligned}$$

Noting (6), we estimate the right hand side to obtain for $\delta > 0$

$$\begin{aligned}&\int _\varSigma |H|^2|\nabla A^o|^2\gamma ^4\,d\mu + \frac{1}{2}\int _\varSigma |H|^4|A^o|^2\gamma ^4\,d\mu \\&\quad = \frac{1}{2}\int _\varSigma |H|^2|\nabla H|^2\gamma ^4\,d\mu + 2\int _\varSigma H\langle {{A^o}, {\nabla H\nabla H}}\rangle _g\gamma ^4\,d\mu \\&\qquad + \int _\varSigma |H|^2|A^o|^4\gamma ^4\,d\mu - 2\int _\varSigma H\langle {{\nabla A^o}, {\nabla H\, A^o}}\rangle _g\gamma ^4\,d\mu \\&\qquad + 4\int _\varSigma |H|^2\langle {{A^o}, {\nabla \gamma \nabla H}}\rangle _g\gamma ^3\,d\mu - 4\int _\varSigma |H|^2\langle {{\nabla A^o}, {\nabla \gamma \, A^o}}\rangle _g\gamma ^3\,d\mu \\&\quad \le \left( \frac{1}{2}+3\delta \right) \int _\varSigma |H|^2|\nabla H|^2\gamma ^4\,d\mu + \frac{1}{\delta }\int _\varSigma |A^o|^2|\nabla H|^2\gamma ^4\,d\mu \\&\qquad + \delta \int _\varSigma |H|^4|A^o|^2\gamma ^4\,d\mu + \frac{1}{4\delta }\int _\varSigma \left( |A^o|^6 + 4|A^o|^2|\nabla A^o|^2\right) \gamma ^4\,d\mu \\&\qquad + \delta \int _\varSigma |H|^2|\nabla A^o|^2\gamma ^4\,d\mu + c_\gamma ^2\frac{8}{\delta }\int _\varSigma |H|^2|A^o|^2\gamma ^2\,d\mu \\&\quad \le \left( \frac{1}{2}+3\delta \right) \int _\varSigma |H|^2|\nabla H|^2\gamma ^4\,d\mu + \delta \int _\varSigma |H|^2|\nabla A^o|^2\gamma ^4\,d\mu \\&\qquad + 2\delta \int _\varSigma |H|^4|A^o|^2\gamma ^4\,d\mu + \frac{1}{4\delta }\int _\varSigma \left( |A^o|^6 + 20|A^o|^2|\nabla A^o|^2\right) \gamma ^4\,d\mu \\&\qquad + c_\gamma ^4\frac{16}{\delta ^3}\int _{[\gamma >0]}|A^o|^2\,d\mu . \end{aligned}$$

Absorbing the second and third terms from the right hand side into the left finishes the proof of (27). $\square$

Proof of (16), cf. [13, Prop. 2.4]. Let us first note that Eq. (28) allows us to estimate

$$\begin{aligned}&\int _\varSigma |\Delta A^o|^2\gamma ^4\,d\mu + c\int _\varSigma |A|^2|\nabla A|^2\gamma ^4\,d\mu + \int _\varSigma \nabla _{(2)}A^o*\nabla A^o*\nabla \gamma \ \gamma ^3\,d\mu \\&\quad \le c\int _\varSigma |\nabla _{(2)}H|^2\gamma ^4\,d\mu + \frac{1}{2}\int _\varSigma |\nabla _{(2)}A^o|^2\gamma ^4\,d\mu \\&\qquad + c\int _\varSigma \left( |A|^2|\nabla A|^2 + |A|^4|A^o|^2\right) \gamma ^4\,d\mu + c_\gamma ^2c\int _\varSigma |\nabla A^o|^2\gamma ^2\,d\mu . \end{aligned}$$

Thus (16) will be proved provided we show

$$\begin{aligned}&\int _\varSigma |\nabla _{(2)}A^o|^2\gamma ^4\,d\mu \le \int _\varSigma |\Delta A^o|^2\gamma ^4\,d\mu + c\int _\varSigma |A|^2|\nabla A|^2\gamma ^4\,d\mu \nonumber \\&\quad + \int _\varSigma \nabla _{(2)}A^o*\nabla A^o*\nabla \gamma \ \gamma ^3\,d\mu . \end{aligned}$$

(30)

This again relies upon interchange of covariant derivatives; we shall prove the following identity

$$\begin{aligned} \Delta \nabla A^o = \nabla \Delta A^o + \nabla A^o * A * A . \end{aligned}$$

(31)

Using (10), we compute

$$\begin{aligned} \nabla _{ijk}A^o_{lm}&= \nabla _{ikj}A^o_{lm} + \nabla (A^o*A*A)\nonumber \\&= \nabla _{kij}A^o_{lm} + \nabla A^o*A*A + A^o * \nabla A*A\nonumber \\&= \nabla _{kij}A^o_{lm} + \nabla A^o*A*A, \end{aligned}$$

(32)

where in the last equality we used

$$\begin{aligned} \nabla _i A_{jk} = \nabla _i A^o_{jk} + \frac{1}{2}g_{jk}\nabla _i H = \nabla _i A^o_{jk} + g_{jk}\nabla _p (A^o)_i^p. \end{aligned}$$

Tracing (32) with $g^{ij}$ gives (31), which testing against $\nabla A^o\,\gamma ^4$ and using the divergence theorem implies

$$\begin{aligned} \int _\varSigma |\nabla _{(2)}A^o|^2\gamma ^4d\mu&= - \int _\varSigma \langle {{\nabla A^o}, {\Delta \nabla A^o}}\rangle _g\gamma ^4\,d\mu - 4\int _\varSigma \langle {{\nabla \gamma \nabla A^o}, {\nabla _{(2)}A^o}}\rangle _g \gamma ^3\,d\mu \\&= - \int _\varSigma \langle {{\nabla A^o}, {\nabla \Delta A^o}}\rangle _g\gamma ^4\,d\mu - 4\int _\varSigma \langle {{\nabla \gamma \nabla A^o}, {\nabla _{(2)}A^o}}\rangle _g \gamma ^3\,d\mu \\&+ \int _\varSigma \nabla A^o * \nabla A^o * A * A\ \gamma ^4\,d\mu \\&= \int _\varSigma |\Delta A^o|^2\gamma ^4\,d\mu + 4\int _\varSigma \langle {{\nabla A^o}, {\nabla \gamma \Delta A^o}}\rangle _g \gamma ^3\,d\mu \\&- 4 \!\int _\varSigma \langle {{\nabla \gamma \nabla A^o}, {\nabla _{(2)}A^o}}\rangle _g \gamma ^3\,d\mu \!+\! \int _\varSigma \! \nabla A^o * \nabla A^o * A * A\ \gamma ^4\,d\mu \\&\le \int _\varSigma |\Delta A^o|^2\gamma ^4\,d\mu + c\int _\varSigma |A|^2|\nabla A^o|^2\gamma ^4\,d\mu \\&+ \int _\varSigma \nabla _{(2)}A^o*\nabla A^o*\nabla \gamma \ \gamma ^3\,d\mu , \end{aligned}$$

which clearly implies (30). $\square$