2. Preliminaries

In this section, we will review the moving frame method for surfaces in $\mathbb{C}P^{2}$ following Chern and Wolfson (for more details, see [Reference Chern and Wolfson4]). In the paper, we will adopt the following ranges of indices:

\begin{equation*} 0\leqslant a,b,c\leqslant 2, \,1\leqslant i,j,k\leqslant 2,\, 3\leqslant \alpha,\beta \leqslant 4,\,1\leqslant A, B\leqslant 4. \end{equation*}

Let $\langle,\rangle$ be the hermitian product in $\mathbb{C}^{3}$ , that is,

\begin{equation*} \langle Z,W\rangle =\sum \limits _{l=1}^{3}Z_{l}\overline {W}_{l} \end{equation*}

for any $Z,W\in \mathbb{C}^{3}$ , where $\overline{W}$ denotes the conjugate of $W$ . Let $\mathbb{C}P^{2}$ be the complex projective plane with its canonical Fubini–Study metric of constant holomorphic sectional curvature 4. Then,

\begin{equation*} \mathbb {C}P^{2}=\{[Z_{0}]=\Pi (Z_{0})|Z_{0}=(z_{1},z_{2},z_{3})\in \mathbb {C}^{3}-\{0\},|Z_{0}|=1 \}, \end{equation*}

where $\Pi \;:\; S^{5}\rightarrow \mathbb{C}P^{2}$ is the Hopf projection. We denote $g$ by its Fubini–Study metric and $J$ by its complex structure induced by $\mathbb{C}^{3}$ on $\mathbb{C}P^{2}$ . Then,

(2.1) \begin{equation} \begin{aligned} g=\langle dZ_{0},dZ_{0}\rangle -\langle dZ_{0},Z_{0}\rangle \langle Z_{0},dZ_{0}\rangle. \end{aligned} \end{equation}

The Kähler form $\Omega$ on $\mathbb{C}P^{2}$ is defined by:

(2.2) \begin{equation} \begin{aligned} \Omega (u,v)=g(Ju,v), \hbox{ for any } u,v\in \Gamma (T(\mathbb{C}P^{2})). \end{aligned} \end{equation}

Let $\{Z_{0},Z_{1},Z_{2}\}$ be a unitary frames in $\mathbb{C}^{3}$ . Then, we have

(2.3) \begin{equation} \begin{aligned} \langle Z_{a},Z_{b}\rangle =\delta _{a\overline{b}},\ dZ_{a}= \sum \limits _{b}\psi _{a\overline{b}}Z_{b}, \end{aligned} \end{equation}

where $\psi _{a\overline{b}}=\overline{\psi _{\overline{a}b}}$ is connection 1-form and satisfies structure equation:

(2.4) \begin{equation} \begin{aligned} d\psi _{a\overline{b}}= \sum \limits _{c}\psi _{a\overline{c}}\wedge \psi _{c\overline{b}}. \end{aligned} \end{equation}

Moreover, the Fubini–Study metric (2.1) can be written as:

(2.5) \begin{equation} \begin{aligned} g=\sum \limits _{i}\psi _{0\overline{i}}\overline{\psi _{0\overline{i}}}. \end{aligned} \end{equation}

On the other hand, let $\{\varsigma _{i}\}$ be a unitary frames in $\mathbb{C}P^{2}$ with dual frames $\{\omega _{i}\}$ , the structure equation of $\mathbb{C}P^{2}$ can be written as:

(2.6) \begin{equation} \begin{aligned} d\omega _{i}=\sum \limits _{j}\omega _{i\overline{j}}\wedge \omega _{j},\, \omega _{i\overline{j}}+\omega _{\overline{j}i}=0,\,\omega _{\overline{j}i} =\overline{\omega _{j\overline{i}}}, \end{aligned} \end{equation}

with $\{\omega _{i\overline{j}}\}$ being the unitary connection forms with respect to $\{\omega _{i}\}$ . We have then

(2.7) \begin{equation} \begin{aligned} g=\sum \limits _{i}\omega _{i}\overline{\omega _{i}} =\sum \limits _{i}\omega _{i}\omega \overline{_{i}}. \end{aligned} \end{equation}

Combining with (2.5) and (2.7) and (2.4) and (2.6), we get

(2.8) \begin{equation} \begin{aligned} \omega _{i}=\psi _{0\overline{i}}, \end{aligned} \end{equation}

(2.9) \begin{equation} \begin{aligned} \omega _{i\overline{j}} =\delta _{ij}\psi _{0\overline{0}}-\psi _{j\overline{i}}. \end{aligned} \end{equation}

Let $f\;:\; M\rightarrow \mathbb{C}P^{2}$ be an isometric immersion of a compact surface $M$ in the complex projective plane $\mathbb{C}P^{2}$ . The Kähler function $C$ on $M$ is defined by:

(2.10) \begin{equation} \begin{aligned} f^{\ast }\Omega =C \textrm{d} M, \end{aligned} \end{equation}

where $\textrm{d} M$ is the area form on $M$ . The surface $f$ is holomorphic, anti-holomorphic, or Lagrangian, respectively, depending on $C\equiv 1,-1$ , or $0$ . Now, we consider $M\subset \mathbb{C}P^{2}$ first from the Riemannian geometry version and then from the complex version due to Chern and Wolfson [Reference Chern and Wolfson4].

Let us choose a (new) local orthonormal frame $\{e_{A}\}$ of $\mathbb{C}P^{2}$ with its dual $\{\theta _{A}\}$ such that restricting to $M$ , $\{e_{i}\}$ is a local orthonormal basis of $TM$ . Then, we have restricted to $M$

\begin{equation*} \theta _{\alpha }=0, \,\theta _{i\alpha }=\sum \limits _{j}h_{ij}\theta _{j}. \end{equation*}

The second fundamental form $II$ and the mean curvature vector $H$ are defined by:

\begin{equation*} II=\sum \limits _{\alpha ij}h^{\alpha }_{ij}\theta _{i}\otimes \theta _{j}\otimes e_{\alpha },\, H=\frac {1}{2}\sum \limits _{\alpha i}h^{\alpha }_{ii}e_{\alpha } =\sum \limits _{\alpha }H^{\alpha }e_{\alpha }. \end{equation*}

Let $\overline{\nabla }$ be the Riemannian connection of $\mathbb{C}P^{2}$ , and let $\nabla$ and $\nabla ^{\perp }$ be the induced connection and normal connection of $M$ , respectively. The covariant derivative and Laplacian of $H$ on the normal bundle of $M$ are, respectively, defined as:

\begin{equation*} \sum \limits _{i}H^{\alpha }_{,i}\theta _{i}=dH^{\alpha }+\sum \limits _{\beta }H^{\beta }\theta _{\beta \alpha }, \end{equation*}

\begin{equation*} \sum \limits _{j}H^{\alpha }_{,ij}\theta _{j}=dH^{\alpha }_{,i} +\sum \limits _{j}H^{\alpha }_{,j}\theta _{ji} +\sum \limits _{\beta }H^{\beta }_{,i}\theta _{\beta \alpha }, \end{equation*}

\begin{equation*} \triangle ^{\perp }H^{\alpha }=\sum \limits _{i}H^{\alpha }_{,ii}. \end{equation*}

Let $\overline{R}_{ABCD}$ be the Riemannian curvature tensor of $\mathbb{C}P^{2}$ , we have then

(2.11) \begin{equation} \begin{aligned} \overline{R}_{ABCD} &=g(e_{A},e_{C})g(e_{B},e_{D})-g(e_{A},e_{D})g(e_{B},e_{C}) +g(Je_{A},e_{C})g(Je_{B},e_{D})\\[5pt] &\quad -g(Je_{A},e_{D})g(Je_{B},e_{C})+2g(Je_{A},e_{B})g(Je_{C},e_{D}). \end{aligned} \end{equation}

Now, let us recall the complex version of the geometry of $M$ which is due to Chern and Wolfson [Reference Chern and Wolfson4]. Set $\phi =\theta _{1}+i\theta _{2}$ . It defines a complex structure on $M$ . The induced metric on $M$ is of the form:

(2.12) \begin{equation} \begin{aligned} f^{\ast }g=\sum \limits _{i}\omega _{i}\overline{\omega _{i}}=\phi \overline{\phi }. \end{aligned} \end{equation}

Then after a normalization of $\{\varsigma _i\}$ if necessary ([Reference Chern and Wolfson4, p. 66]), we can assume that there exist complex-valued smooth functions $s,t$ which satisfy $|s|^{2}+|t|^{2}=1$ such that

(2.13) \begin{equation} \begin{aligned} \omega _{1}=s\phi,\, \omega _{2}=t\overline{\phi }. \end{aligned} \end{equation}

In particular, setting $|s|=\cos \dfrac{\alpha }{2}, \,|t|=\sin \dfrac{\alpha }{2}$ , one has then $C=\cos \alpha$ with $0\leq \alpha \leq \pi$ . Now we have, along $M$ ,

(2.14) \begin{equation} \begin{aligned} \overline{s}\omega _{1}+t\overline{\omega _{2}}=\phi, \end{aligned} \end{equation}

(2.15) \begin{equation} \begin{aligned} \overline{t}\omega _{1}-s\overline{\omega _{2}}=\theta _{3}+i\theta _{4}=0. \end{aligned} \end{equation}

Taking exterior derivative of the first term of (2.15), we get

(2.16) \begin{equation} \begin{aligned} ((sd\overline{t}-\overline{t}ds)+s\overline{t}(\omega _{1\overline{1}} +\omega _{2\overline{2}}))\wedge \phi +\omega _{1\overline{2}}\wedge \overline{\phi }=0. \end{aligned} \end{equation}

Set

(2.17) \begin{equation} \begin{aligned} (sd\overline{t}-\overline{t}ds)+s\overline{t}(\omega _{1\overline{1}} +\omega _{2\overline{2}})=a\phi +b\overline{\phi }, \end{aligned} \end{equation}

(2.18) \begin{equation} \begin{aligned} \,\omega _{1\overline{2}}=b\phi +c\overline{\phi }, \end{aligned} \end{equation}

then the complex-valued second fundamental forms can be given by:

(2.19) \begin{equation} \begin{aligned} II^{\mathbb{C}}=a\phi ^{2}+2b\phi \overline{\phi }+c\overline{\phi }^{2}. \end{aligned} \end{equation}

Lemma 2.1. The coefficients $a,\,b,\,c$ of the complex-valued second fundamental forms $II^{\mathbb{C}}$ satisfy

(2.20) \begin{equation} \begin{aligned} &\textit{Re}(a+c)+\textit{Re}(2b)=h_{11}^{3}, \,\textit{Im}(c-a)=h_{12}^{3}, \,\textit{Re}(2b)-\textit{Re}(a+c)=h_{22}^{3},\\[5pt] &\textit{Im}(a+c)+\textit{Im}(2b)=h_{11}^{4},\,\textit{Re}(a-c)=h_{12}^{4},\,\textit{Im}(2b)-\textit{Im}(a+c)=h_{22}^{4}, \end{aligned} \end{equation}

where $\text{Re}$ and $\text{Im}$ denote, respectively, the real and imaginary parts.

Proof. Taking exterior derivative of the second term of (2.15), we get

\begin{equation*} \begin {aligned} &d(\theta _{3}+i\theta _{4})\\[5pt] =&-\frac {1}{2}\left(\frac {h_{11}^{3}-h_{22}^{3}+2h_{12}^{4} +i(h_{11}^{4}-h_{22}^{4}-2h_{12}^{3})}{4}\phi +\frac {h_{11}^{3}+h_{22}^{3}+i(h_{11}^{4}+h_{22}^{4})}{4}\overline {\phi }\right)\wedge \phi \\[5pt] &-\frac {1}{2}\left(\frac {h_{11}^{3}+h_{22}^{3}+i(h_{11}^{4}+h_{22}^{4})}{4}\phi +\frac {h_{11}^{3}-h_{22}^{3}-2h_{12}^{4} +i(h_{11}^{4}-h_{22}^{4}+2h_{12}^{3})}{4}\overline {\phi }\right)\wedge \overline {\phi }. \end {aligned} \end{equation*}

Then from (2.15), (2.16), (2.17), and (2.18), we have that

\begin{equation*} \begin {aligned} &a=\frac {h_{11}^{3}-h_{22}^{3}+2h_{12}^{4} +i(h_{11}^{4}-h_{22}^{4}-2h_{12}^{3})}{4},\\[5pt] &b=\frac {h_{11}^{3}+h_{22}^{3}+i(h_{11}^{4}+h_{22}^{4})}{4},\\[5pt] &c=\frac {h_{11}^{3}-h_{22}^{3}-2h_{12}^{4} +i(h_{11}^{4}-h_{22}^{4}+2h_{12}^{3})}{4}, \end {aligned} \end{equation*}

and (2.20) follows.

Remark 2.2. Note that we can re-choose the unitary coframe $\{\omega _{1},\omega _{2}\}$ such that

(2.21) \begin{equation} \begin{aligned} s=\cos \frac{\alpha }{2}, \,t=\sin \frac{\alpha }{2}. \end{aligned} \end{equation}

The Kähler angle $\alpha$ is smooth at the points with $0\lt \alpha \lt \pi$ . At the points $\alpha =0$ or $\pi$ , $\alpha$ could be only continuous. Moreover, under the assumption of (2.21), we can obtain by (2.13) that

\begin{equation*} \left \{\begin {split}\omega _1&=\cos \frac {\alpha }{2}\theta _{1}+\sin \frac {\alpha }{2}\theta _{3} +i\!\left(\!\cos \frac {\alpha }{2}\theta _{2}+\sin \frac {\alpha }{2}\theta _{4}\right),\\[5pt] \omega _2&=\sin \frac {\alpha }{2}\theta _{1}-\cos \frac {\alpha }{2}\theta _{3} +i\!\left(\!-\!\sin \frac {\alpha }{2}\theta _{2}+\cos \frac {\alpha }{2}\theta _{4}\right), \end {split}\right. \end{equation*}

and hence

\begin{equation*} \left \{\begin {split}\varsigma _1&=\frac {1}{2}\left(\!\cos \frac {\alpha }{2}e_{1}+\sin \frac {\alpha }{2}e_{3} -i\!\left(\!\cos \frac {\alpha }{2}e_{2}+\sin \frac {\alpha }{2}e_{4}\right)\right),\\[5pt] \varsigma _2&=\frac {1}{2}\left(\!\sin \frac {\alpha }{2}e_{1}-\cos \frac {\alpha }{2}e_{3} -i\!\left(\!-\!\sin \frac {\alpha }{2}e_{2}+\cos \frac {\alpha }{2}e_{4}\right)\right). \end {split}\right .\end{equation*}

So, we have

\begin{equation*}J\!\left(\!\cos \frac {\alpha }{2}e_{1}+\sin \frac {\alpha }{2}e_{3}\right) =\cos \frac {\alpha }{2}e_{2}+\sin \frac {\alpha }{2}e_{4},\ J\!\left(\!\sin \frac {\alpha }{2}e_{1}-\cos \frac {\alpha }{2}e_{3}\right) =-\sin \frac {\alpha }{2}e_{2}+\cos \frac {\alpha }{2}e_{4},\end{equation*}

from which we get

\begin{equation*} \left \{\begin {split}Je_{1}&=C e_{2}+\sqrt {1-C^{2}} e_{4},\ Je_{2}=-C e_{1}-\sqrt {1-C^{2}} e_{3},\\[5pt] Je_{3}&=-C e_{4}+\sqrt {1-C^{2}} e_{2},\ Je_{4}=C e_{3}-\sqrt {1-C^{2}} e_{1}. \end {split}\right .\end{equation*}

This is exactly we assume in (1.2) of Section 1.

3. Euler–Lagrange equation of Helfrich functional

Let $f(p,t)\;:\; M\times (\!-\!\epsilon,\epsilon )\rightarrow \mathbb{C}P^{2}$ be a variation of $M$ with $f_{0}(p)=f(p)$ . Here, we denote by $f_t(p)=f(p,t)\;:\;M\rightarrow \mathbb{C}P^{2}$ for $t\in (\!-\!\epsilon,\epsilon )$ . Let $\{x_{1},x_{2},t\}$ be a local coordinate system around the point $(p,0)$ such that $\{\textrm{d} f\!\left(\dfrac{\partial }{\partial x_{1}}\right),\textrm{d} f\!\left(\dfrac{\partial }{\partial x_{2}}\right)\}|_p$ is an orthonormal basis of $T_{p}M$ . Set $\bar V=\textrm{d} f_{t}\!\left(\dfrac{\partial }{\partial t}\right), X_{i}=\textrm{d} f_{t}\!\left(\frac{\partial }{\partial x_{i}}\right)$ . Then, we have the induced metric of $ f_{t}$ and its area form as follows:

\begin{equation*}(g_{t})_{ij}=g_{t}(X_{i},X_{j}), \textrm {d} M_{t}=\sqrt {G_{t}}\textrm {d} x_{1}\wedge \textrm {d} x_{2} \hbox { with } G_{t}=\text {det}(g_{t}).\end{equation*}

Then $(g_0)_{ij}(p)=g_{ij}(p)\delta _{ij}$ . Set $\tilde e_{i}=\textrm{d} f_{t}|_{t=0} \left(\dfrac{\partial }{\partial x_{i}}\right)$ and $V=\textrm{d} f_{t}|_{t=0}\left(\dfrac{\partial }{\partial t}\right)=V^{\top }+V^{\perp }$ with $V^{\top }\in \Gamma (TM)$ and $V^{\perp }\in \Gamma (T^{\perp }M)$ .

We first consider $\dfrac{\partial (\sqrt{G_{t}})}{\partial t}|_{t=0}$ and $\dfrac{\partial |H|^{2}}{\partial t}|_{t=0}$ . It is well known that

(3.1) \begin{equation} \begin{aligned} \frac{\partial (\sqrt{G_{t}})}{\partial t}|_{t=0} =(\text{div}V^{\top }-2\langle H,V\rangle )\sqrt{G_{t}}. \end{aligned} \end{equation}

Now we consider $\dfrac{\partial |H|^{2}}{\partial t}|_{t=0}$ . It follows the definition of the mean curvature vector that

\begin{equation*} 2H_{t}=\sum \limits _{ij}g_{t}^{ij}h(X_{i},X_{j}), \end{equation*}

where $((g_{t})^{ij})$ denotes the inverse matrix of $((g_{t})_{ij})$ . Thus, we have at the point $p$ that

(3.2) \begin{equation} \begin{aligned} \frac{\partial }{\partial t}(2H_{t})|_{t=0}=\sum \limits _{ij}\bigg (\frac{\partial (g_{t})^{ij}}{\partial t}|_{t=0} h(\tilde e_i,\tilde e_{j})+\delta _{ij}\frac{\partial h(X_{i},X_{j})}{\partial t}|_{t=0}\bigg ). \end{aligned} \end{equation}

Differentiating the formula $\sum \limits _{j}g^{ij}g_{jk}=\delta _{ik}$ and using the fact that $\left[\dfrac{\partial }{\partial t}, \dfrac{\partial }{\partial x_{i}}\right]=0$ , we get

(3.3) \begin{equation} \begin{aligned} \frac{\partial (g_{t})^{ij}}{\partial t}|_{t=0} &=-\frac{\partial (g_{t})_{ij}}{\partial t}|_{t=0} =-\overline{\nabla }_{V}g(\tilde e_i,\tilde e_{j})\\[5pt] &=-\overline{\nabla }_{V^{\top }}g(\tilde e_i,\tilde e_{j}) +(\overline{\nabla }_{\tilde e_i}V^{\bot },\tilde e_{j}) +( \tilde e_i,\overline{\nabla }_{\tilde e_{j}}V^{\bot }))\\[5pt] &=-\nabla _{V^{\top }}g(\tilde e_i,\tilde e_{j}) +2g(h(\tilde e_i,\tilde e_{j}),V^{\bot }). \end{aligned} \end{equation}

Also, we have

(3.4) \begin{equation} \begin{aligned} \delta _{ij}\frac{\partial h(X_{i},X_{j})}{\partial t}|_{t=0} &=\overline{\nabla }_{V}(\overline{\nabla }_{\tilde e_i}\tilde e_i -\nabla _{\tilde e_i}\tilde e_i)\\[5pt] &=\overline{\nabla }_{V^{\top }}\overline{\nabla }_{\tilde e_i}\tilde e_i -\overline{\nabla }_{V^{\top }}\nabla _{\tilde e_i}\tilde e_i\\[5pt] & \quad +\overline{\nabla }_{\tilde e_i}\overline{\nabla }_{\tilde e_i}V^{\bot } +\overline{R}(V^{\bot },\tilde e_i)\tilde e_i-\overline{\nabla }_{V^{\bot }}\nabla _{\tilde e_i}\tilde e_i. \end{aligned} \end{equation}

Since $\dfrac{\partial |H|^{2}}{\partial t}=g\left(H,2\dfrac{\partial H}{\partial t}\right)$ , we only need to know the normal part of (3.4). So, by a direct computation we have

(3.5) \begin{equation} \begin{aligned} \left(\sum \limits _{i}(\overline{\nabla }_{V^{\top }}\overline{\nabla }_{\tilde e_i}\tilde e_i -\overline{\nabla }_{V^{\top }}\nabla _{\tilde e_i}\tilde e_i)\right)^{\bot } =\left(\sum \limits _{i}\overline{\nabla }_{V^{\top }}h(\tilde e_i,\tilde e_i)\right)^{\bot } =2\nabla _{V^{\top }}^{\bot }H, \end{aligned} \end{equation}

(3.6) \begin{equation} \begin{aligned} &\left(\sum \limits _{i}(\overline{\nabla }_{\tilde e_i}\overline{\nabla }_{\tilde e_i}V^{\bot } +\overline{R}(V^{\bot },\tilde e_i)\tilde e_i-\overline{\nabla }_{V^{\bot }}\nabla _{\tilde e_i}\tilde e_i)\right)^{\bot }\\[5pt] &=\sum \limits _{i}(\!-\!h(\tilde e_i,A_{V^{\bot }}(\tilde e_i)) +\nabla _{\tilde e_i}^{\bot }\nabla _{\tilde e_i}^{\bot }V^{\bot } +(\overline{R}(V^{\bot },\tilde e_i)\tilde e_i)^{\bot } -\nabla _{\nabla _{\tilde e_i}\tilde e_i}^{\bot }V^{\bot })\\[5pt] &=\triangle ^{\bot }V^{\bot }+\sum \limits _{i}(\!-\!g(h(\tilde e_i,\tilde e_{j}),V^{\bot })h(\tilde e_i,\tilde e_{j}) +(\overline{R}(V^{\bot },\tilde e_i)\tilde e_i)^{\bot }). \end{aligned} \end{equation}

Here, we used the fact that $(\overline{\nabla }_{V^{\bot }}\nabla _{\tilde e_i}\tilde e_i)^{\bot } =\nabla _{\nabla _{\tilde e_i}\tilde e_i}^{\bot }V^{\bot }$ . Finally, substituting (3.3) and (3.4) into (3.2) and using (3.5), (3.6), and (2.11), we obtain at the point $p$ that

(3.7) \begin{align} \frac{\partial |H|^{2}}{\partial t}|_{t=0} &=g\!\left(H,2\frac{\partial H}{\partial t}\right) \nonumber \\[5pt] &=\sum \limits _{\alpha }\Big (H^{\alpha }\triangle ^{\bot }V^{\alpha } +V^{\alpha }(5-3C^{2})H^{\alpha } +\sum \limits _{ij\beta }V^{\alpha }H^{\beta }h_{ij}^{\alpha }h_{ij}^{\beta } \nonumber \\[5pt] & \quad -\sum \limits _{ijk}H^{\alpha }h_{ij}^{\alpha }V^{k}\nabla _{\tilde e_{k}}g_{ij}\Big ) +\sum \limits _{i}V^{i}\nabla _{\tilde e_i}|H|^{2}. \nonumber \\[5pt] &=\sum \limits _{\alpha }\left (H^{\alpha }\triangle ^{\bot }V^{\alpha } +V^{\alpha }(5-3C^{2})H^{\alpha } +\sum \limits _{ij\beta }V^{\alpha }H^{\beta }h_{ij}^{\alpha }h_{ij}^{\beta }\right) +\sum \limits _{i}V^{i}\nabla _{\tilde e_i}|H|^{2}. \end{align}

Here, we used

\begin{equation*} \sum \limits _{ijk}V^{k}H^{\alpha }h_{ij}^{\alpha }\nabla _{\tilde e_{k}}g_{ij} =\sum \limits _{ik}(\!-\!V^{k}\nabla _{\tilde e_{k}}(H^{\alpha }h_{ii}^{\alpha })-H^{\alpha }h_{ii}^{\alpha }\text {div}V^{\top } +\text {div}(H^{\alpha }h_{ii}^{\alpha }V^{\top }))=0. \end{equation*}

Next, we consider $\dfrac{\partial (C_{t})^2}{\partial t}|_{t=0}$ . First for an oriented orthonormal basis $\{e_1, e_2\}$ of $TM$ , set

\begin{equation*} X_{V}=\Omega (V,e_{2})e_{1}+\Omega (e_{1},V)e_{2}.\end{equation*}

It is direct to check that $X_{V}$ is independent of the choice $\{e_{1},e_{2}\}$ , and hence it defines a smooth vector field on $M$ with

(3.8) \begin{equation} \begin{aligned} \text{div}X_{V}=\nabla _{e_{1}}\Omega (V,e_{2})+\nabla _{e_{2}}\Omega (e_{1},V). \end{aligned} \end{equation}

The definition of the Kähler function means that

\begin{equation*} C_{t}=\frac {\Omega (X_{1},X_{2})}{\sqrt {G_{t}}}. \end{equation*}

So we get, at the point $p$ ,

(3.9) \begin{equation} \begin{aligned} \frac{\partial (C_{t})}{\partial t}|_{t=0} =\frac{\partial \Omega (X_{1},X_{2})}{\partial t}|_{t=0} -C\frac{\partial (\sqrt{G_{t}})}{\partial t}|_{t=0}. \end{aligned} \end{equation}

Since $[\tilde e_1,\tilde e_2]|_p=0$ and $[V,\tilde e_i]|_p=0$ for $1\leq i\leq 2$ , we have at the point $p$ that

(3.10) \begin{equation} \begin{aligned} \frac{\partial \Omega (X_{1},X_{2})}{\partial t}|_{t=0} &=\Omega (\overline{\nabla }_{\tilde e_{1}}V,\tilde e_{2})+\Omega (\tilde e_{1},\overline{\nabla }_{\tilde e_{2}}V)\\[5pt] &=\overline{\nabla }_{\tilde e_{1}}\Omega (V,\tilde e_{2})-\Omega (V,\overline{\nabla }_{\tilde e_{1}}\tilde e_{2}) +\overline{\nabla }_{\tilde e_{2}}\Omega (\tilde e_{1},V)-\Omega (\overline{\nabla }_{\tilde e_{2}}\tilde e_{1},V)\\[5pt] &=\nabla _{\tilde e_{1}}\Omega (V,\tilde e_{2})+\nabla _{\tilde e_{2}}\Omega (\tilde e_{1},V)=\text{div}(X_V). \end{aligned} \end{equation}

Substituting (3.1) and (3.10) into (3.9), we obtain at the point $p$ that

(3.11) \begin{equation} \begin{aligned} \frac{\partial C^{2}}{\partial t}|_{t=0} =\Big (2C\,\text{div}X_{V} -2C^{2}(\text{div}V^{\top }-2\langle H,V\rangle )\Big ). \end{aligned} \end{equation}

Noting that the right sides of (3.7) and (3.11) are independent of the coordinates and hence valid at any point of $M$ . Thus, from (3.1), (3.7), and (3.11), we get

(3.12) \begin{equation} \begin{aligned} \frac{d}{dt}|_{t=0}(\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)) &=\int _{M}\frac{\partial }{\partial t}(|H|^{2}+\lambda _{1}+\lambda _{2} C^{2})|_{t=0}\textrm{d} M +\int _{M}(|H|^{2}+\lambda _{1}+\lambda _{2} C^{2})\frac{\partial (\textrm{d} M)}{\partial t}|_{t=0}\\[5pt] &=\int _{M}\bigg (\sum \limits _{\alpha }\bigg (H^{\alpha }\triangle ^{\perp }V^{\alpha } +[(5-2\lambda _{1}-(3-2\lambda _{2})C^{2}-2|H|^{2})H^{\alpha } +\sum \limits _{\beta ij}h_{ij}^{\alpha }h_{ij}^{\beta }H^{\beta }]V^{\alpha }\bigg )\\[5pt] &\quad +\sum \limits _{i}V^{i}\nabla _{\tilde e_i}|H|^{2} +2\lambda _{2}C\,\text{div}X_{V}+(\lambda _{1}+\lambda _{2}C^{2}+|H|^{2})\text{div}V_{\top }\bigg )\textrm{d} M. \end{aligned} \end{equation}

Furthermore, it follows from the divergence theorem that

(3.13) \begin{equation} \begin{aligned} \int _{M}H^{\alpha }\triangle ^{\perp } V^{\alpha }\textrm{d} M =\int _{M}V^{\alpha }\triangle ^{\perp } H^{\alpha }\textrm{d} M. \end{aligned} \end{equation}

(3.14) \begin{equation} \begin{aligned} \int _{M}\sum \limits _{i}V^{i}\nabla _{\tilde e_i}|H|^{2}\textrm{d} M =-\int _{M}|H|^{2}\text{div}(V^{\top })\textrm{d} M. \end{aligned} \end{equation}

(3.15) \begin{equation} \begin{aligned} \int _{M}2C\,\text{div}X_{V} =-2\int _{M}\langle \nabla C, X_{V}\rangle \textrm{d} M =-2\int _{M}\Big (\nabla _{e_{1}}C \cdot \Omega (V, e_{2}) +\nabla _{e_{2}}C \cdot \Omega (e_{1}, V)\Big )\textrm{d} M. \end{aligned} \end{equation}

(3.16) \begin{equation} \begin{aligned} \int _{M}C^{2}\text{div}V^{\top }\textrm{d} M =-2\int _{M}\Big (\nabla _{e_{1}}C \cdot \Omega (V^{\top },e_{2})+ \nabla _{e_{2}}C \cdot \Omega (e_{1},V^{\top })\Big )\textrm{d} M. \end{aligned} \end{equation}

Substituting (3.13)–(3.16) into (3.12) and noting

(3.17) \begin{align} \nabla _{e_{1}}C \cdot \Omega (V^{\perp },e_{2})+\nabla _{e_{2}}C \cdot \Omega (e_{1},V^{\perp }) &=-\nabla _{e_{1}}C\cdot \langle Je_{2}, V^{\perp }\rangle +\nabla _{e_{2}}C \cdot \langle Je_{1},V^{\perp }\rangle \nonumber \\[5pt] &=\sqrt{1-C^{2}}\langle (\nabla _{e_{1}}C\,e_{3}+\nabla _{e_{2}}C\,e_{4}),V\rangle, \end{align}

we obtain that

\begin{align*} \frac {d}{dt}|_{t=0}(\mathcal {H}_{\lambda _{1},\lambda _{2}}(f)) &=\int _{M}\bigg (\sum \limits _{\alpha }V^{\alpha }[\triangle ^{\perp } H^{\alpha } +(5-2\lambda _{1}-(3-2\lambda _{2})C^{2}-2|H|^{2})H^{\alpha }\nonumber \\[5pt] &\quad +\sum \limits _{\beta ij}h_{ij}^{\alpha }h_{ij}^{\beta }H^{\beta }] -2\lambda _{2}\sqrt {1-C^{2}}(C_{,1}V^{3}+C_{,2}V^{4})\bigg )\textrm {d} M, \end{align*}

where $C_{,i}\,(1\leq i\leq 2)$ denote the first covariant derivatives of $C$ . This implies that the Euler–Lagrange equation of $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)$ is

(3.18) \begin{equation} \begin{cases} \triangle ^{\perp }H^{3}+(5-2\lambda _{1}-(3-2\lambda _{2})C^{2}-2|H|^{2})H^{3}+ \sum \limits _{\beta ij}H^{\beta }h_{ij}^{\beta }h_{ij}^{3}-2\lambda _{2}\sqrt{1-C^{2}}C_{,1}=0,\\[12pt] \triangle ^{\perp }H^{4}+(5-2\lambda _{1}-(3-2\lambda _{2})C^{2}-2|H|^{2})H^{4}+ \sum \limits _{\beta ij}H^{\beta }h_{ij}^{\beta }h_{ij}^{4}-2\lambda _{2}\sqrt{1-C^{2}}C_{,2}=0. \end{cases} \end{equation}

This gives the proof of Theorem 1.1.

Combing this we have

4. Homogeneous tori in $\mathbb{C}P^{2}$

In this section, we consider the homogeneous tori in $\mathbb{C}P^{2}$ .

4.1. The geometry of homogeneous tori

The definition of the homogeneous torus in $\mathbb{C}P^{2}$ is given by the image of the Hopf projection:

\begin{equation*} T_{r_{1},r_{2},r_{3}}=\{\Pi (Z_{0})\in \mathbb {C}P^{2}|Z_{0}=(z_{1},z_{2},z_{3}), |z_{l}|=r_{l}, l=1,2,3\}, \end{equation*}

for positive numbers $r_{1}, r_{2}, r_{3}$ that satisfy $r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=1$ . In this case, we also call $T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}$ the Clifford torus. We note that Ma, Mironov, and Zuo in [Reference Ma, Mironov and Zuo11] gave a basis of the period module for the homogeneous tori. Here, we discuss the conformal structure of the homogeneous tori for completeness.

Taking into account the definition of $T_{r_{1},r_{2},r_{3}}$ , we assume that the homogeneous coordinate of $T_{r_{1},r_{2},r_{3}}$ is

\begin{equation*} Z_{0}=(r_{1},r_{2}e^{i\varphi },r_{3}e^{i\psi }),\hbox { i.e., } dZ_{0}=(0,ir_{2}e^{i\varphi },ir_{3}e^{i\psi }), \end{equation*}

where $\varphi,\,\psi \in \mathbb{R}$ . Then we have from (2.1) that

Lemma 4.1. The induced metric of $T_{r_{1},r_{2},r_{3}}$ in $\mathbb{C}P^{2}$ is

(4.1) \begin{equation} \begin{aligned} g=r_{2}^{2}r_{3}^{2}\bigg (\frac{1-r_{2}^{2}}{r_{3}^{2}}\textrm{d} \varphi ^{2} +\frac{1-r_{3}^{2}}{r_{2}^{2}}\textrm{d}\psi ^{2}-2\textrm{d} \varphi \textrm{d}\psi \bigg ). \end{aligned} \end{equation}

Setting

(4.2) \begin{equation} \begin{cases} u=\frac{\sqrt{1-r_{2}^{2}}}{r_{3}}\varphi -\frac{r_{3}}{\sqrt{1-r_{2}^{2}}}\psi,\\[12pt] v=\frac{r_{1}}{r_{2}\sqrt{1-r_{2}^{2}}}\psi, \end{cases} \hbox{i.e. } \begin{cases} \varphi =\frac{r_{3}}{\sqrt{1-r_{2}^{2}}}u+\frac{r_{2}^{2}r_{3}}{r_{1}\sqrt{1-r_{2}^{2}}}v,\\[12pt] \psi =\frac{r_{2}\sqrt{1-r_{2}^{2}}}{r_{1}}\psi. \end{cases} \end{equation}

Then

\begin{equation*} \begin {aligned} g=r_{2}^{2}r_{3}^{2}(du^{2}+dv^{2})=r_{2}^{2}r_{3}^{2}|\textrm {d} z|^{2}, \end {aligned} \end{equation*}

and hence $z=u+iv$ gives a complex coordinate of $T_{r_{1},r_{2},r_{3}}$ . Also, from (4.2) we get a basis $(\omega _{1}, \omega _{2})$ of the period module for $T_{r_{1},r_{2},r_{3}}$ with

\begin{equation*} \omega _{1}=\frac {2\pi \sqrt {1-r_2^2}}{r_3}, \,\omega _{2}=\frac {2\pi r_1^2}{r_3\sqrt {1-r_2^2}}+i\frac {2\pi r_1}{r_2\sqrt {1-r_2^2}}. \end{equation*}

So without loss of generality, we assume $r_{3}\geqslant r_{1}\geqslant r_{2}$ , $k=\frac{r_{3}}{r_{1}}$ , then

\begin{equation*} \tau =\frac {\omega _{2}}{\omega _{1}}=\frac {1}{k^{2}+1}+i\frac {k}{r_{2}(k^{2}+1)}, \hbox { with } \{\tau |0\lt \text {Re} \,\tau \leqslant \frac {1}{2},\, |\tau |\geqslant 1\} \end{equation*}

gives the module space for $T_{r_{1},r_{2},r_{3}}$ (see the shaded part of Figure 1).

Figure 1. $\tau$ -plane.

In particular, for Clifford torus $T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}$ , $\tau =\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ .

Now, we are going to consider the second fundamental forms of $T_{r_{1},r_{2},r_{3}}$ in $\mathbb{C}P^{2}$ .

Lemma 4.2. The second fundamental forms of $T_{r_{1},r_{2},r_{3}}$ satisfy

(4.3) \begin{equation} \left ( \begin{array}{c@{\quad}c} h_{11}^{3}&h_{12}^{3}\\[5pt] h_{21}^{3}&h_{22}^{3} \end{array} \right ) = \left ( \begin{array}{c@{\quad}c} 0 & \frac{-r_{2}}{\sqrt{1-r_{2}^{2}}}\\[12pt] \frac{-r_{2}}{\sqrt{1-r_{2}^{2}}}&\frac{r_{1}^{2}-r_{3}^{2}}{r_{1}r_{3}\sqrt{1-r_{2}^{2}}} \end{array} \right ), \end{equation}

(4.4) \begin{equation} \left ( \begin{array}{c@{\quad}c} h_{11}^{4}&h_{12}^{4}\\[5pt] h_{21}^{4}&h_{22}^{4} \end{array} \right ) = \left ( \begin{array}{c@{\quad}c} \frac{2r_{2}^{2}-1}{r_{2}\sqrt{1-r_{2}^{2}}} & 0\\[12pt] 0&\frac{r_{2}}{\sqrt{1-r_{2}^{2}}} \end{array} \right ). \end{equation}

Proof. Set $\phi =r_{2}r_{3}(\textrm{d} u+i\textrm{d} v)$ , then the induced metric of $T_{r_{1},r_{2},r_{3}}$ can be written as:

\begin{equation*}g=\phi \cdot \bar {\phi }.\end{equation*}

Let $\{Z_{0},Z_{1},Z_{2}\}$ be a unitary frames in $\mathbb{C}^{3}$ . Then $\textrm{d} Z_{0}=\sum \limits _{a}\psi _{0\overline{a}}Z_{a}$ with $\psi _{a\overline{b}}=\overline{\psi _{\overline{a}b}}$ . From this, we get

(4.5) \begin{align} \psi _{0\overline{1}}Z_{1}+\psi _{0\overline{2}}Z_{2} &=\textrm{d} Z_{0}-\psi _{0\overline{0}}Z_{0} \nonumber \\[5pt] &=\textrm{d} Z_{0}-h(dZ_{0},Z_{0})Z_{0} \nonumber \\[5pt] &=\left(\frac{-r_{3}-ir_{1}r_{2}}{2\sqrt{1-r_{2}^{2}}},\frac{i\sqrt{1-r_{2}^{2}}e^{i\varphi }}{2}, \frac{(r_{1}-ir_{2}r_{3})e^{i\psi }}{2\sqrt{1-r_{2}^{2}}}\right)\phi \nonumber \\[5pt] &\quad +\left(\frac{r_{3}-ir_{1}r_{2}}{2\sqrt{1-r_{2}^{2}}},\frac{i\sqrt{1-r_{2}^{2}}e^{i\varphi }}{2}, \frac{(\!-\!r_{1}-ir_{2}r_{3})e^{i\psi }}{2\sqrt{1-r_{2}^{2}}}\right)\overline{\phi }. \end{align}

And hence we obtain from $|Z_{1}|=|Z_{2}|=1$ that

(4.6) \begin{equation} \begin{aligned} Z_{1}=\left(\frac{-r_{3}-ir_{1}r_{2}}{\sqrt{2(1-r_{2}^{2})}},\frac{i\sqrt{1-r_{2}^{2}}e^{i\varphi }}{\sqrt{2}}, \frac{(r_{1}-ir_{2}r_{3})e^{i\psi }}{\sqrt{2(1-r_{2}^{2}})}\right),\\[5pt] Z_{2}=\left(\frac{r_{3}-ir_{1}r_{2}}{\sqrt{2(1-r_{2}^{2}})},\frac{i\sqrt{1-r_{2}^{2}}e^{i\varphi }}{\sqrt{2}}, \frac{(\!-\!r_{1}-ir_{2}r_{3})e^{i\psi }}{\sqrt{2(1-r_{2}^{2})}}\right). \end{aligned} \end{equation}

Let $\{\omega _{i}\}$ be an unitary coframe in $\mathbb{C}P^{2}$ such that restricting to $M$

(4.7) \begin{equation} \begin{aligned} \omega _{1}=s\phi,\, \omega _{2}=t\overline{\phi }. \end{aligned} \end{equation}

Then from (2.8) and (4.5), we have

(4.8) \begin{equation} \begin{aligned} s=t=\frac{\sqrt{2}}{2}. \end{aligned} \end{equation}

Now, using the fact $\psi _{a\overline{b}}=\langle dZ_{a},Z_{b}\rangle$ , then from (2.9) we get

\begin{equation*} \omega _{1\overline {1}}=\psi _{0\overline {0}}-\psi _{1\overline {1}} =\frac {(3r_{2}^{2}-1)i}{2}d\varphi +\frac {(3r_{3}^{2}-1)i}{2}d\psi, \end{equation*}

\begin{equation*} \omega _{2\overline {2}}=\psi _{0\overline {0}}-\psi _{2\overline {2}} =\frac {(3r_{2}^{2}-1)i}{2}d\varphi +\frac {(3r_{3}^{2}-1)i}{2}d\psi, \end{equation*}

\begin{equation*} \omega _{1\overline {2}}=-\psi _{2\overline {1}} =-\frac {(1-r_{2}^{2})i}{2}d\varphi -\frac {2r_{1}r_{2}r_{2}r_{3}+i(r_{2}^{2}r_{3}^{2}-r_{0}^{2})}{2(1-r_{2}^{2})}d\psi, \end{equation*}

and so

\begin{equation*} \begin {aligned} s\overline {t}(\omega _{1\overline {1}} +\omega _{2\overline {2}})&=\left(\frac {2r_{3}^{2}+r_{2}^{2}-1}{4r_{1}r_{3}\sqrt {(1-r_{2}^{2})}} +\frac {i(3r_{2}^{2}-1)}{4r_{2}\sqrt {1-r_{2}^{2}}}\right)\phi \\[5pt] &\quad +\left(\frac {-(2r_{3}^{2}+r_{2}^{2}-1)}{4r_{1}r_{3}\sqrt {(1-r_{2}^{2})}} +\frac {i(3r_{2}^{2}-1)}{4r_{2}\sqrt {1-r_{2}^{2}}}\right)\overline {\phi }, \end {aligned} \end{equation*}

\begin{equation*} \begin {aligned} \omega _{1\overline {2}} &=\left(\frac {-(2r_{3}^{2}+r_{2}^{2}-1)}{4r_{1}r_{3}\sqrt {(1-r_{2}^{2})}} +\frac {i(3r_{2}^{2}-1)}{4r_{2}\sqrt {1-r_{2}^{2}}}\right)\phi \\[5pt] &\quad +\left(\frac {2r_{3}^{2}+r_{2}^{2}-1}{4r_{1}r_{3}\sqrt {(1-r_{2}^{2})}} -\frac {i(r_{2}^{2}+1)}{4r_{2}\sqrt {1-r_{2}^{2}}}\right)\overline {\phi }. \end {aligned} \end{equation*}

Thus, we obtain from (2.17) and (2.18) that the coefficients of the complex-valued second fundamental forms (see (2.19)) of $T_{r_{1},r_{2},r_{3}}$ are

(4.9) \begin{equation} \begin{cases} a=\dfrac{r_{3}^{2}-r_{1}^{2}}{4r_{1}r_{3}\sqrt{1-r_{2}^{2}}} +\dfrac{i(3r_{2}^{2}-1)}{4r_{2}\sqrt{1-r_{2}^{2}}},\\[16pt] b=\dfrac{r_{1}^{2}-r_{3}^{2}}{4r_{1}r_{3}\sqrt{1-r_{2}^{2}}} +\dfrac{i(3r_{2}^{2}-1)}{4r_{2}\sqrt{1-r_{2}^{2}}},\\[16pt] c=\dfrac{r_{3}^{2}-r_{1}^{2}}{4r_{1}r_{3}\sqrt{1-r_{2}^{2}}} -\dfrac{i(r_{2}^{2}+1)}{4r_{2}\sqrt{1-r_{2}^{2}}}. \end{cases} \end{equation}

Using Lemma 2.1, (4.3) and (4.4) follow directly.

Remark 4.3. The above proof shows that $s=t=\dfrac{\sqrt{2}}{2}$ , that is, $\cos \alpha =\dfrac{\sqrt{2}}{2}$ and the Kähler function $C=\cos \alpha =0$ , which also implies that the homogeneous torus in $\mathbb{C}P^{2}$ is Lagrangian.

Now we discuss the classification the homogeneous Helfrich tori in $\mathbb{C}P^{2}$ .

Theorem 4.4. $T_{r_{1},r_{2},r_{3}}$ is a Helfrich surface if and only if

1. 1. When $0\leqslant \lambda _{1}\leqslant \dfrac{5}{2}$ , $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}.$

2. 2. When $\lambda _{1}\gt \dfrac{5}{2}$ , $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}$ or $T_{r_{1},r_{2},r_{3}}=T_{\sqrt{\frac{1}{4\lambda _{1}-9}}, \sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}},\sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}}}$ .

Proof. By Lemma 4.2, we get the coefficients of the mean curvature of $T_{r_{1},r_{2},r_{3}}$ as follows:

(4.10) \begin{equation} \begin{aligned} H^{3}=\frac{r_{1}^{2}-r_{3}^{2}}{2r_{1}r_{3}\sqrt{1-r_{2}^{2}}},\,\, H^{4}=\frac{(3r_{2}^{2}-1)}{2r_{2}\sqrt{1-r_{2}^{2}}}. \end{aligned} \end{equation}

And hence the norm square of the mean curvature is

(4.11) \begin{equation} \begin{aligned} |H|^{2}=\frac{(1-r_{1}^{2})(1-r_{2}^{2})(1-r_{3}^{2})}{4r_{1}^{2}r_{2}^{2}r_{3}^{2}}-2. \end{aligned} \end{equation}

Thus, substituting (4.10) and (4.11) into the Euler–Lagrange equation (3.18) and using the fact $C=\cos \alpha \equiv 0$ , we get that

\begin{equation*} \begin {cases} \left \{\left (12-4\lambda _{1}+(4\lambda _{1}-8)r_{2}^{2}\right ) r_{1}^{2}r_{2}^{2}r_{3}^{2}+2r_{2}^{2}(r_{1}^{4}+r_{3}^{4}) -(1-r_{1}^{2})(1-r_{2}^{2})^{2}(1-r_{3}^{2})\right \}(r_{1}^{2}-r_{3}^{2})=0,\\[5pt] \left \{\left (10-4\lambda _{1}+(4\lambda _{1}-8)r_{2}^{2}\right ) r_{1}^{2}r_{2}^{2}r_{3}^{2}+2r_{1}^{2}r_{3}^{2} -(1-r_{1}^{2})(1-r_{2}^{2})^{2}(1-r_{3}^{2})\right \}(3r_{2}^{2}-1) +2r_{2}^{4}(r_{1}^{2}-r_{3}^{2})=0,\\[5pt] r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=1. \end {cases} \end{equation*}

By solving the equation above, we obtain

1. 1. When $0\leqslant \lambda _{1}\leqslant \dfrac{5}{2}$ , $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}.$

2. 2. When $\lambda _{1}\gt \dfrac{5}{2}$ , $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}} \,\text{or} \,T_{r_{1},r_{2},r_{3}}=T_{\sqrt{\frac{1}{4\lambda _{1}-9}}, \sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}},\sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}}}.$

4.2. The Helfrich energy of homogeneous tori

Proposition 4.5. The Helfrich energy of $T_{r_{1},r_{2},r_{3}}$ is

(4.12) \begin{equation} \begin{aligned} \mathcal{H}_{\lambda _{1},\lambda _{2}}(f) &=\int _{M}(|H|^{2}+\lambda _{1}+\lambda _{2} C^{2})\textrm{d} M\\[5pt] &=\frac{((1-r_{1}^{2})(1-r_{2}^{2})(1-r_{3}^{2}) +(4\lambda _{1}-8)r_{1}^{2}r_{2}^{2}r_{3}^{2})\pi ^{2}}{r_{1}r_{2}r_{3}}. \end{aligned} \end{equation}

Proof. From Lemma 4.1, we have $\textrm{d} T_{r_{1},r_{2},r_{3}}=r_{1}r_{2}r_{3}\textrm{d}\varphi \textrm{d}\psi$ . By using of (4.11), (4.12) follows directly.

Now we consider the lower bound of $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)$ for $T_{r_{1},r_{2},r_{3}}$ .

Theorem 4.6. The Helfrich energy of $T_{r_{1},r_{2},r_{3}}$ satisfies

1. 1. When $0\leqslant \lambda _{1}\leqslant 3$ ,

   \begin{equation*}\mathcal {H}_{\lambda _{1},\lambda _{2}}(f)\geqslant \frac {4\lambda _{1}\pi ^{2}}{3\sqrt {3}},\end{equation*}
   and the equality holds if and only if $r_{1}=r_{2}=r_{3}=\dfrac{\sqrt{3}}{3}$ .

2. 2. When $\lambda _{1}\gt 3$ ,

   \begin{equation*}\mathcal {H}_{\lambda _{1},\lambda _{2}}(f)\geqslant \frac {(4\lambda _{1}-8)\pi ^{2}}{\sqrt {4\lambda _{1}-9}},\end{equation*}
   and the equality holds if and only if $r_{1}=\sqrt{\dfrac{1}{4\lambda _{1}-9}},\,r_{2}=r_{3} =\sqrt{\dfrac{2\lambda _{1}-5}{4\lambda _{1}-9}}.$

Proof. Computing the extreme value of (4.12) under the constraint $r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=1$ yields

\begin{equation*} \begin {cases} (4\lambda _{1}-8)(r_{1}^{2}-r_{2}^{2})r_{1}^{2}r_{2}^{2}r_{3}^{2} +(1+r_{1}^{2})(1-r_{2}^{2})(1-r_{3}^{2})r_{2}^{2} -(1-r_{1}^{2})(1+r_{2}^{2})(1-r_{3}^{2})r_{1}^{2}=0,\\[5pt] (4\lambda _{1}-8)(r_{1}^{2}-r_{3}^{2})r_{1}^{2}r_{2}^{2}r_{3}^{2} +(1+r_{1}^{2})(1-r_{2}^{2})(1-r_{3}^{2})r_{3}^{2} -(1-r_{1}^{2})(1-r_{2}^{2})(1+r_{3}^{2})r_{1}^{2}=0,\\[5pt] r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=1. \end {cases} \end{equation*}

A straightforward calculation shows that

1. 1. When $0\leqslant \lambda _{1}\leqslant \dfrac{5}{2}$ ,

   \begin{equation*}T_{r_{1},r_{2},r_{3}}=T_{\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}}, \,\mathcal {H}_{\lambda _{1},\lambda _{2}}(f)=\frac {4\lambda _{1}\pi ^{2}}{3\sqrt {3}}.\end{equation*}

2. 2. When $\lambda _{1}\gt \dfrac{5}{2}$ ,

   \begin{equation*}T_{r_{1},r_{2},r_{3}}=T_{\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}}, \,\mathcal {H}_{\lambda _{1},\lambda _{2}}(f)=\frac {4\lambda _{1}\pi ^{2}}{3\sqrt {3}},\end{equation*}
   or
   \begin{equation*}T_{r_{1},r_{2},r_{3}}=T_{\sqrt {\frac {1}{4\lambda _{1}-9}}, \sqrt {\frac {2\lambda _{1}-5}{4\lambda _{1}-9}},\sqrt {\frac {2\lambda _{1}-5}{4\lambda _{1}-9}}}, \,\mathcal {H}_{\lambda _{1},\lambda _{2}}(f)=\frac {(4\lambda _{1}-8)\pi ^{2}}{\sqrt {4\lambda _{1}-9}}.\end{equation*}

For the case of $0\leqslant \lambda _{1}\leqslant \dfrac{5}{2}$ , it is obvious that $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)\geqslant \dfrac{4\lambda _{1}\pi ^{2}}{3\sqrt{3}}$ and equality holds if and only if $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}$ . In the second case, if $\dfrac{5}{2}\lt \lambda _{1}\leqslant 3$ , since $\dfrac{4\lambda _{1}\pi ^{2}}{3\sqrt{3}} \leqslant \dfrac{(4\lambda _{1}-8)\pi ^{2}}{\sqrt{4\lambda _{1}-9}}$ , we have that $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)\geqslant \frac{4\lambda _{1}\pi ^{2}}{3\sqrt{3}}$ and equality holds if and only if $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}$ . If $\lambda _{1}\gt 3$ , $\dfrac{4\lambda _{1}\pi ^{2}}{3\sqrt{3}} \gt \dfrac{(4\lambda _{1}-8)\pi ^{2}}{\sqrt{4\lambda _{1}-9}}$ , and hence $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)\geqslant \frac{(4\lambda _{1}-8)\pi ^{2}}{\sqrt{4\lambda _{1}-9}}$ , the equality holds if and only if $T_{r_{1},r_{2},r_{3}}=T_{\sqrt{\frac{1}{4\lambda _{1}-9}}, \sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}},\sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}}}$ .

Remark 4.7. For the case of $\lambda _{1}=3$ in the above proof, $T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}} =T_{\sqrt{\frac{1}{4\lambda _{1}-9}}. \sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}},\sqrt{\frac{2\lambda _{1}-5}{4\lambda _{1}-9}}}$ , This implies that $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)\geqslant \dfrac{4\lambda _{1}\pi ^{2}}{3\sqrt{3}}$ , and equality holds if and only if $T_{r_{1},r_{2},r_{3}}=T_{\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}}$ .

Remark 4.8. In [Reference Montiel and Urbano14], Montiel and Urbano introduced the conformal invariants $W^{-}(F)$ and $W^{+}(F)$ for compact surfaces in $\mathbb{C}P^{2}$ . In view of the Helfrich functional $\mathcal{H}_{\lambda _{1},\lambda _{2}}(f)$ , we have

1. 1. When $\lambda _{1}=2, \lambda _{2}=0$ , $\mathcal{H}_{2,0}(F)=W^{-}(F)=\int _{M}(|H|^{2}+2)\textrm{d} M$ ;

2. 2. When $\lambda _{1}=0, \lambda _{2}=6$ , $\mathcal{H}_{0,6}(F)=W^{+}(F)=\int _{M}(|H|^{2}+6C^{2})\textrm{d} M$ ;

3. 3. When $\lambda _{1}=1, \,\lambda _{2}=3$ , $\mathcal{H}_{1,3}(F)=W(F)=\int _{M}(|H|^{2}+1+3C^{2})\textrm{d} M$ , that is, the Willmore functional.

Thus, by using the Euler–Lagrange equation of $\mathcal{H}_{\lambda _{1},\lambda _{2}}(F)$ , we have the following corollary.

Corollary 4.9. ([Reference Hu and Li8]) The Euler–Lagrange equation of $W(F)$ is

\begin{equation*} \begin {cases} \triangle ^{\perp }H^{3}+(3+3C^{2}-2|H|^{2})H^{3}+ \sum \limits _{\beta, i,j}h_{ij}^{3}h_{ij}^{\beta }H^{\beta }-6\sqrt {1-C^{2}}C_{,1}=0,\\[12pt] \triangle ^{\perp }H^{4}+(3+3C^{2}-2|H|^{2})H^{4}+ \sum \limits _{\beta, i,j}h_{ij}^{4}h_{ij}^{\beta }H^{\beta }-6\sqrt {1-C^{2}}C_{,2}=0. \end {cases} \end{equation*}

Corollary 4.10. ([Reference Hu and Li8]) The Euler–Lagrange equation of $W^{-}(F)$ is

\begin{equation*} \triangle ^{\perp }H^{\alpha }+(1-3C^{2}-2|H|^{2})H^{\alpha }+ \sum \limits _{\beta, i,j}h_{ij}^{\alpha }h_{ij}^{\beta }H^{\beta }=0. \end{equation*}

Corollary 4.11. The Euler–Lagrange equation of $W^{+}(F)$ is

\begin{equation*} \begin {cases} \triangle ^{\perp }H^{3}+(5+9C^{2}-2|H|^{2})H^{3}+ \sum \limits _{\beta, i,j}h_{ij}^{3}h_{ij}^{\beta }H^{\beta }-12\sqrt {1-C^{2}}C_{,1}=0,\\[12pt] \triangle ^{\perp }H^{4}+(3+3C^{2}-2|H|^{2})H^{4}+ \sum \limits _{\beta, i,j}h_{ij}^{4}h_{ij}^{\beta }H^{\beta }-12\sqrt {1-C^{2}}C_{,2}=0. \end {cases} \end{equation*}

Let us consider the homogeneous tori $T_{r_{1},r_{2},r_{3}}$ in $\mathbb{C}P^{2}$ , then it follows from Theorems 4.4 and 4.6 that

Corollary 4.12. $T_{r_{1},r_{2},r_{3}}$ is a critical surface for $W(F),\,W^{-}(F)$ , or $W^{+}(F)$ if and only if it is the Cliford torus $T_{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}}$ .

Corollary 4.13. Considering the homogeneous tori $T_{r_{1},r_{2},r_{3}}$ in $\mathbb{C}P^{2}$ . Then $W^{-}(F)\geqslant \dfrac{8\pi ^{2}}{3\sqrt{3}}$ (see also [Reference Montiel and Urbano14]) and $W^{+}(F)\geqslant 0$ , and the equalities hold if and only if $T_{r_{1},r_{2},r_{3}}$ is the Cliford torus $T_{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}}$ .