Let $ x: M^{2}\rightarrow S^{2+p} $ be a $ 2- $dimensional submanifold in the $ (2+p)- $dimensional unit sphere $ S^{2+p} $ without umbilic points. Let $ \{e_{i}\} $ be a local orthonormal basis for the first fundamental form $ I = d x \cdot d x $ with dual basis $ \{\theta_{i}\} $. Let $ I I = \sum\limits_{i, j, \alpha} h_{i j}^{\alpha} \theta_{i} \theta_{j} e_{\alpha} $ be the Moebius second fundamental form and $ H = \sum\limits_{\alpha} H^{\alpha} e_{\alpha} $ be the mean curvature vector of $ x $, where $ \{e_{\alpha}\} $ is a local orthonormal basis for the normal bundle of $ x $. Define positive function $ \rho^{2} = 2(\sum\limits_{\alpha, i, j}\left(h_{i j}^{\alpha}\right)^{2}-2\|H\|^{2}) $, the g$ = \rho^{2}I $ is Moebius metric and is a Moebius invariant, the normalized scalar curvature of g will be denoted by $ R $ and is called the normalized Moebius scalar curvature. Three basic Moebius invariants of $ x $, Moebius form $ \Phi = \sum\limits_{i, \alpha} C_{i}^{\alpha} \theta_{i} e_{\alpha} $, Blaschke tensor $ A = \rho^{2} \sum\limits_{i, j} A_{i j} \theta_{i} \theta_{j} $ and the Moebius second fundamental form $ B = \rho^{2}\sum\limits_{i, j}B_{ij}\theta_{i}\theta_{j} $, are defined by$ ([1]) $

$ \begin{align} C_{i}^{\alpha} = -\rho^{-2}(H_{, i}^{\alpha}+\sum\limits_{j}(h_{i j}^{\alpha}-H^{\alpha} \delta_{i j}) e_{j}(\log \rho)), \end{align} $

(1.1)

$ \begin{align} A_{i j} = &-\rho^{-2}(H e s s_{i j}(\log \rho)-e_{i}(\log \rho) e_{j}(\log \rho)-\sum\limits_{\alpha} H^{\alpha} h_{i j}^{\alpha}) \\ &-\frac{1}{2} \rho^{-2}\left(\|\nabla \log \rho\|^{2}-1+\|H\|^{2}\right) \delta_{i j}, \end{align} $

(1.2)

$ \begin{align} B_{i j}^{\alpha} = \rho^{-1}\left(h_{i j}^{\alpha}-H^{\alpha} \delta_{i j}\right), \end{align} $

(1.3)

where $ Hess_{ij} $ and $ \nabla $ are the Hessian-matrix and the gradient with respect to the induced metric $ I = d x \cdot d x $. Let $ \nabla^{\bot} $ be normal connection, and the $ H^{\alpha}_{, i} $ is defined by $ \nabla^{\bot}H = H^{\alpha}_{, i}\theta_{i}e_{\alpha} $. Moreover, we introduce the trace-free Blaschke tensor

$ \begin{align} \widetilde{A} = A-\frac{1}{2}\operatorname{tr}A\cdot \rm{g}, \quad\widetilde{A}_{i j} = A_{i j}-\frac{1}{2} \sum\limits_{k} A_{k k} \delta_{i j}, \quad\|\widetilde{A}\|^{2} = \sum\limits_{i, j}\left(\widetilde{A}_{i j}\right)^{2}, \end{align} $

(1.4)

$ \|\widetilde{A}\| = 0 $ if and only if $ A $ is isotropic tensor.

Hu and Li studied the dimension of submanifold is $ m\geq3 $, the Moebius form $ \Phi = 0 $, and the constant scalar curvature. In this paper, we proved the dimension of submanifold is $ m = 2 $, the Moebius form $ \Phi = 0 $, and $ \operatorname{Det}A = c(const)>0 $, we get the following theorem.

Theorem 1.1 Let $ x: M^{2}\rightarrow S^{2+p} $ be a $ 2- $dimensional compact submanifold in the $ (2+p)- $dimensional unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $ and $ 0<\operatorname{Det}A = c(const)<\frac{1}{64} $, then

$ \begin{align} \int_{M}\left(\operatorname{tr} A-\frac{1}{4}\right)\|\widetilde{A}\|^{2}dM \leq 0. \end{align} $

(1.5)

In particular, if $ \operatorname{tr}A\geq\frac{1}{4} $, then either

$ \begin{align} \widetilde{A}\equiv0, \end{align} $

(1.6)

and $ x(M^{2}) $ is Moebius equivalent to a minimal submanifold with constant curvature in $ S^{2+p} $; or

$ \begin{align} \operatorname{tr}A = \frac{1}{4}, \end{align} $

(1.7)

and $ x(M^{2}) $ is Moebius equivalent to $ S^{1}(r)\times S^{1}(\sqrt{\frac{1}{1+c^{2}}-r^{2}}) $ in $ S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $, where $ r^{2} = \frac{2-\sqrt{1-64c}}{4(1+c^{2})} $.

In this section, we give the Moebius invariants and review its structural equations for surfaces in $ S^{2+p} $, for details we refer to[2].

Let $ R_{1}^{4+p} $ be the Lorentzian space with inner product

$ \begin{align} \langle x, y\rangle = -x_{0} y_{0}+x_{1} y_{1}+\cdots+x_{3+p} y_{3+p}, \end{align} $

(2.1)

where $ x = \left(x_{0}, x_{1}, \cdots, x_{3+p}\right) $, and $ y = \left(y_{0}, y_{1}, \cdots, y_{3+p}\right) $. Let $ x:M^{2}\rightarrow S^{2+p} $ be an umbilic-free surface immersed in $ S^{2+p} $. We define the Moebius position vector $ Y: M^{2} \rightarrow R_{1}^{4+p} $ of $ x $ by

$ \begin{align} Y = \rho(1, x), \quad \rho^{2} = 2(\sum\limits_{\alpha, i, j}\left(h_{i j}^{\alpha}\right)^{2}-2\|H\|^{2})>0. \end{align} $

(2.2)

Then we have the following.

Theorem 2.1(see[2]) Two submanifolds $ x $, $ \widehat{x} $ : $ M \rightarrow S^{n} $ are Moebius equivalent if and only if there exists T in the Lorentz group $ O(n+1, 1) $ in $ R^{4+p}_{1} $ such that $ Y = \widehat{Y} T $.

From Theorem 2.1, we know that the 2-form

$ \begin{align} \rm{g} = \langle d Y, d Y\rangle = \rho^{2} d x\cdot d x, \end{align} $

(2.3)

is a Moebius invariant(see[1]). Let $ \Delta $ be the Laplace operator with respect to g. Then we have $ \langle\Delta Y, \Delta Y\rangle = 1+4 K, $ where $ K $ is the sectional curvature of g ([1]). By defining

$ \begin{align} N = -\frac{1}{2} \Delta Y-\frac{1}{8}(1+4 K) Y, \end{align} $

(2.4)

then we have([1])

$ \begin{align} \langle\Delta Y, Y\rangle = -2, \quad\langle\Delta Y, d Y\rangle = 0, \end{align} $

(2.5)

$ \begin{align} \langle Y, Y\rangle = 0, \quad\langle N, Y\rangle = 1, \quad\langle N, N\rangle = 0. \end{align} $

(2.6)

Let $ \left\{E_{1}, E_{2}\right\} $ be a local orthonormal basis for $ (M^{2}, \rm{g}) $ with dual basis $ \left\{\omega_{1}, \omega_{2}\right\} $, write $ Y_{i} = E_{i}(Y) $, then

$ \begin{align} \langle Y_{i}, Y\rangle = \langle Y_{i}, N\rangle = 0, \quad\langle Y_{i}, Y_{j}\rangle = \delta_{i j}, \quad 1 \leq i, j \leq 2. \end{align} $

(2.7)

Let $ V $ be the orthogonal complement of $ \operatorname{span}\left\{Y, N, Y_{1}, Y_{2}\right\} $ in $ R_{1}^{4+p} $. Then we have the orthogonal decomposition

$ \begin{align} R_{1}^{4+p} = \operatorname{span}\{Y, N\} \oplus \operatorname{span}\left\{Y_{1}, Y_{2}\right\} \oplus V. \end{align} $

(2.8)

Let $ \{E_{\alpha}\} $ be an orthonormal basis of $ V $, where

$ \begin{align} E_{\alpha} = \left(H^{\alpha}, H^{\alpha} x+e_{\alpha}\right), \quad 3 \leq \alpha \leq 2+p. \end{align} $

(2.9)

Then $ \left\{Y, N, Y_{1}, Y_{2}, E_{3}, \cdots, E_{2+p}\right\} $ forms a moving frame in $ R_{1}^{4+p} $ along $ M^{2} $. The structure equations are given by

$ \begin{align} d Y = Y_{1} \omega_{1}+Y_{2} \omega_{2}, \end{align} $

(2.10)

$ \begin{align} d N = (A_{11} \omega_{1} +A_{12} \omega_{2} )Y_{1}+(A_{21} \omega_{1} +A_{22} \omega_{2}) Y_{2}+\sum\limits_{\alpha} (C_{1}^{\alpha} \omega_{1}+C_{2}^{\alpha} \omega_{2}) E_{\alpha}, \end{align} $

(2.11)

$ \begin{align} d Y_{1} = - (A_{1 1} \omega_{1} + A_{1 2} \omega_{2}) Y-\omega_{1} N+ \omega_{1 1} Y_{1}+\omega_{1 2} Y_{2}+\sum\limits_{ \alpha}( B_{1 1}^{\alpha} \omega_{1} +B_{1 2}^{\alpha} \omega_{2}) E_{\alpha}, \end{align} $

(2.12)

$ \begin{align} d Y_{2} = - (A_{2 1} \omega_{1} + A_{2 2} \omega_{2}) Y-\omega_{2} N+ \omega_{21} Y_{1}+\omega_{2 2} Y_{2}+\sum\limits_{ \alpha}( B_{2 1}^{\alpha} \omega_{1} +B_{2 2}^{\alpha} \omega_{2}) E_{\alpha}, \end{align} $

(2.13)

$ \begin{align} d E_{\alpha} = -\sum\limits_{i} C_{i}^{\alpha} \omega_{i} Y-\sum\limits_{j} (B_{1j}^{\alpha} \omega_{j} Y_{1}+B_{2j}^{\alpha} \omega_{j} Y_{2})+\sum\limits_{\beta} \omega_{\alpha \beta} E_{\beta}, \end{align} $

(2.14)

where the coefficients $ \omega_{ij} $ belong to the connection form of the Moebius metric g, and we have the symmetries $ A_{ij} = A_{ji}, \, B_{ij} = B_{ji} $. It is clear that

$ \begin{align} A = A_{11} \omega_{1} \otimes \omega_{1}+A_{22} \omega_{2} \otimes \omega_{2}+2A_{12} \omega_{1} \otimes \omega_{2}, \end{align} $

(2.15)

$ \begin{align} B = (B_{11}^{\alpha} \omega_{1} \otimes \omega_{1}+B_{22}^{\alpha} \omega_{2} \otimes \omega_{2}+2B_{12}^{\alpha} \omega_{1} \otimes \omega_{2})E_{\alpha}, \end{align} $

(2.16)

$ \begin{align} \Phi = (C_{1}^{\alpha} \omega_{1}+C_{2}^{\alpha} \omega_{2})E_{\alpha}, \end{align} $

(2.17)

are Moebius invariants, and

$ \begin{align} B^{\alpha}_{11}+B^{\alpha}_{22} = 0, \quad \sum\limits_{\alpha}[(B^{\alpha}_{11})^{2}+2(B^{\alpha}_{12})^{2}+(B^{\alpha}_{22})^{2}] = \frac{1}{2}. \end{align} $

(2.18)

Define the covariant derivatives of $ A $, $ B $ and $ \Phi $ by ([1])

$ \begin{align} A_{ij, 1}\omega_{1}+ A_{ij, 2}\omega_{2} = dA_{ij}+A_{i1}\omega_{1j}+A_{i2}\omega_{2j}+A_{1j}\omega_{1i}+A_{2j}\omega_{2i}, \end{align} $

(2.19)

$ \begin{align} B_{ij, 1}^{\alpha} \omega_{1}+B_{i j, 2}^{\alpha} \omega_{2} = d B_{i j}^{\alpha}+B_{i1}^{\alpha} \omega_{1j}+B_{i2}^{\alpha} \omega_{2j}+B_{1j}^{\alpha} \omega_{1 i}+B_{2j}^{\alpha} \omega_{2i}+\sum\limits_{\beta} B_{i j}^{\beta} \omega_{\beta \alpha}, \end{align} $

(2.20)

$ \begin{align} C_{i, 1}^{\alpha} \omega_{1}+C_{i, 2}^{\alpha} \omega_{2} = d C_{i}^{\alpha}+C_{1}^{\alpha} \omega_{1i}+C_{2}^{\alpha} \omega_{2i}+\sum\limits_{\beta} C_{i}^{\beta} \omega_{\beta \alpha}. \end{align} $

(2.21)

The integrability conditions for the structure equations $ (2.10)-(2.14) $ are given by ([1])

$ \begin{align} A_{i 1, 2}-A_{i 2, 1} = \sum\limits_{\alpha}\left(B_{i 2}^{\alpha} C_{1}^{\alpha}-B_{i 1}^{\alpha} C_{2}^{\alpha}\right), \end{align} $

(2.22)

$ \begin{align} C_{1, 2}^{\alpha}-C_{2, 1}^{\alpha} = \sum\limits_{k}\left(B_{1 k}^{\alpha} A_{k 2}-B_{k 2}^{\alpha} A_{k 1}\right), \end{align} $

(2.23)

$ \begin{align} B_{i 1, 2}^{\alpha}-B_{i 2, 1}^{\alpha} = \delta_{i 1} C_{2}^{\alpha}-\delta_{i 2} C_{1}^{\alpha}, \end{align} $

(2.24)

$ \begin{align} R_{1212} = -\frac{1}{4}+\operatorname{tr}A, \end{align} $

(2.25)

$ \begin{align} R_{\alpha \beta 1 2} = -2\left(B_{22}^{\alpha} B_{12}^{\beta}-B_{12}^{\alpha} B_{22}^{\beta}\right), \end{align} $

(2.26)

$ \begin{align} \sum\limits_{i} B_{i j, i}^{\alpha} = -C_{j}^{\alpha}, \end{align} $

(2.27)

where $ R_{1212} $ and $ R_{\alpha\beta 12} $ denote the sectional curvature of g and the normal curvature of the normal connection. Set $ K = R_{1212} $. The second covariant derivative of $ A_{ij} $ and $ B_{ij}^{\alpha} $ are defined by ([1])

$ \begin{align} A_{i j, k1} \omega_{1}+A_{i j, k2} \omega_{2} = &d A_{i j, k}+A_{1j, k} \omega_{1i}+A_{2j, k} \omega_{2i}+ A_{i1, k} \omega_{1j}+A_{i2, k} \omega_{2j}\\ &+A_{i j, 1} \omega_{1 k}+A_{i j, 2} \omega_{2k}, \end{align} $

(2.28)

$ \begin{align} B_{i j, k1}^{\alpha} \omega_{1}+B_{i j, k2}^{\alpha} \omega_{2} = &d B_{i j, k}^{\alpha}+B_{1j, k}^{\alpha} \omega_{1i}+B_{2j, k}^{\alpha} \omega_{2i}+ B_{i1, k}^{\alpha} \omega_{1j}+B_{i2, k}^{\alpha} \omega_{2j}\\ &+ B_{i j, 1}^{\alpha} \omega_{1 k}+B_{i j, 2}^{\alpha} \omega_{2 k}+\sum\limits_{\beta} B_{i j, k}^{\beta} \omega_{\beta \alpha}. \end{align} $

(2.29)

Let $ x: M^{2}\rightarrow S^{2+p} $ be a submanifold in $ S^{2+p} $ without umbilic points, the Moebius metric is $ \rm{g} = \rho^{2}dx\cdot dx $, and so the canonical lift of $ x $ is given by $ Y = \rho(k, x)([1]) $. Then along with $ M^{2} $, we can choose a moving frame $ \left\{Y, N, Y_{1}, Y_{2}, E_{3}, \cdots, E_{2+p}\right\} $ in $ R_{1}^{4+p} $, and we replace $ E_{\alpha} $ in $ (2.9) $ by $ E_{\alpha} = \left(H^{\alpha} k, e_{\alpha}+H^{\alpha} x\right) $. For the Moebius invariants $ A $, $ B $ and $ \Phi $ appearing in the structure equation $ (2.11)-(2.14) $, by calculation, we can get the expression $ (1.1) $ for $ \Phi $, $ (1.3) $ for $ B $ and $ (1.2) $ should be changed to([3])

$ \begin{align} A_{i j} = &-\rho^{-2}\left(H e s s_{i j}(\log \rho)-e_{i}(\log \rho) e_{j}(\log \rho)-\sum\limits_{\alpha} H^{\alpha} h_{i j}^{\alpha}\right) \\ &-\frac{1}{2} \rho^{-2}\left(\|\nabla \log \rho\|^{2}-\frac{1}{k^{2}}+\|H\|^{2}\right) \delta_{i j}. \end{align} $

(3.1)

Lemma 3.1 For any positive constants $ k>a>0 $, the torus $ x_{a, k}: M_{a, k} = S^{1}(a) \times S^{1}(b) \rightarrow S^{3}(k) $, $ a^{2}+b^{2} = k^{2} $, choose unit frame field $ \{e_{1}\} $ and $ \{e_{2}\} $ in $ S^{1}(a) $ and $ S^{1}(b) $ respectively, the Moebius invariants components of the torus are as follows:

$ \begin{align} \Phi\equiv0, \end{align} $

(3.2)

$ \begin{align} K = 0, \end{align} $

(3.3)

$ \begin{align} A_{11} = \frac{3}{8}-\frac{a^{2}}{2k^{2}}, \, \, A_{22} = -\frac{1}{8}+\frac{a^{2}}{2k^{2}}, \, \, A_{12} = 0, \end{align} $

(3.4)

$ \begin{align} B_{11} = -\frac{1}{2}, \, \, B_{22} = \frac{1}{2}, \, \, B_{12} = 0. \end{align} $

(3.5)

Proof We write $ R^{4} = R^{2}\times R^{2} $ and let $ x_{1}:S^{1}\rightarrow R^{2} $ and $ x_{2}:S^{1}\rightarrow R^{2} $ be standard embeddings of the unit sphere, then $ x = a(x_{1}, 0)+b(0, x_{2}) $. The unit normal vector of $ M_{a, k} = S^{1}(a) \times S^{1}(b) $ in $ S^{3}(k) $ is given by $ e_{3} = \frac{b}{k}\left(x_{1}, 0\right)-\frac{a}{k}\left(0, x_{2}\right) $; the second fundamental form of $ M_{a, k} $ is given by $ I I = -d x d e_{3} = \frac{ab}{k}\left(-d x_{1} \cdot d x_{1}+d x_{2}\cdot d x_{2}\right) $; the Euclidean induced metric of $ M^{2} $ is given by $ I = a^{2} d x_{1} \cdot d x_{1}+b^{2} d x_{2}\cdot d x_{2} $. Choose an orthonormal frame $ \{e_{1}, e_{2}\} $ on $ TM^{2} $ with dual frame $ \{\theta_{1}, \theta_{2}\} $ such that $ d(ax_{1}) = \theta_{1}e_{1} $ and $ d(bx_{2}) = \theta_{2}e_{2}, $ then we have

$ \begin{align} I = \theta_{1}^{2}+\theta_{2}^{2}, \quad I I = -\frac{b}{ka} \theta_{1}^{2}+\frac{a}{k b} \theta_{2}^{2}, \end{align} $

(3.6)

$ \begin{align} h_{i j} = \lambda_{i} \delta_{i j}, \quad \lambda_{1} = -\frac{b}{k a}, \quad \lambda_{2} = \frac{a}{k b}. \end{align} $

(3.7)

From $ (3.7) $ we see that

$ \begin{align} H: = \frac{1}{2} (\lambda_{1}+\lambda_{2}) = \frac{a^{2}-b^{2}}{2 ab k}, \end{align} $

(3.8)

$ \begin{align} S: = \lambda_{1}^{2}+\lambda_{2}^{2} = \frac{a^{4}+b^{4}}{a^{2} b^{2} k^{2}}, \end{align} $

(3.9)

$ \begin{align} \rho^{2}: = 2\left(S-2 H^{2}\right) = \frac{k^{2}}{a^{2} b^{2}}. \end{align} $

(3.10)

The Moebius metric g is given by

$ \begin{align} \rm{g} = \rho^{2} d x\cdot d x = \frac{k^{2}}{a^{2} b^{2}}\left(\theta_{1}^{2}+\theta_{2}^{2}\right) = \omega_{1}^{2}+\omega_{2}^{2}, \end{align} $

(3.11)

and

$ \begin{align} \omega_{i} = \frac{k}{ab} \theta_{i}, \quad E_{i} = \frac{ab}{k} e_{i}, \quad Y_{i} = \left(0, e_{i}\right), \quad E_{3} = \left(H k, e_{3}+H x\right), \quad 1 \leq i \leq 2. \end{align} $

(3.12)

$ \begin{align} B_{i j} = b_{i} \delta_{i j}, \quad b_{1} = -\frac{1}{2}, \quad b_{2} = \frac{1}{2}. \end{align} $

(3.13)

From $ (3.1), (3.8), (3.9) $ and $ (3.10) $, we have

$ \begin{align} A_{i j} = -\frac{1}{2} \rho^{-2}\left(H^{2} \delta_{i j}-\frac{1}{k^{2}}-2 H h_{i j}\right) = a_{i} \delta_{i j}, \end{align} $

(3.14)

where

$ \begin{align} a_{1} = \frac{3}{8}-\frac{a^{2}}{2 k^{2}}, \quad a_{2} = -\frac{1}{8}+\frac{a^{2}}{2 k^{2}} . \end{align} $

(3.15)

Then we have

$ \begin{align} \operatorname{tr}A = a_{1}+a_{2} = \frac{1}{4}, \end{align} $

(3.16)

from $ \operatorname{tr}A = \frac{1}{4}(1+4K) $, thus

$ \begin{align} K = 0. \end{align} $

(3.17)

So the conclusion holds.

Now, we suppose that $ \Phi = 0, $ $ K = 0 $ and $ \operatorname{tr}A = \frac{1}{4}>0 $, then we have

$ \begin{align} A_{i1, 2} = A_{i2, 1}, \quad B^{\alpha}_{i1, 2} = B_{i2, 1}^{\alpha}, \end{align} $

(3.18)

$ \begin{align} B^{\alpha}_{i1}A_{1j}+B^{\alpha}_{i2}A_{2j} = B_{1j}^{\alpha}A_{1i}+B_{2j}^{\alpha}A_{2i}. \end{align} $

(3.19)

From $ (1.4) $, we see $ \tilde{A}_{i j} = A_{i j}-\frac{1}{2} \operatorname{t r} A \delta_{i j}, $ and define

$ \begin{align} \|A\|^{2} = A_{11}^{2}+2A_{12}^{2}+A_{22}^{2}, \quad\|\widetilde{A}\|^{2} = \tilde{A}_{11}^{2}+2\tilde{A}_{12}^{2}+\tilde{A}_{22}^{2}, \end{align} $

(3.20)

then we have

$ \begin{align} \|A\|^{2} = \|\widetilde{A}\|^{2}+\frac{1}{2}(\operatorname{t r} A)^{2}, \end{align} $

(3.21)

$ \begin{align} \operatorname{t r} A^{3} = a_{1}^{3}+a_{2}^{3}. \end{align} $

(3.22)

Choose a basis $ \{E_{i}\} $ such that $ (A_{ij}) $ is diagonalized, i.e.,

$ \begin{align} A_{ij} = a_{i}\delta_{ij}. \end{align} $

(3.23)

Then we have

$ \begin{align} \tilde{A}_{i j} = \tilde{a}_{i} \delta_{i j}, \quad \tilde{a}_{1} = a_{1}-\frac{1}{2} ( a_{1}+a_{2}), \quad\tilde{a}_{2} = a_{2}-\frac{1}{2} ( a_{1}+a_{2}), \end{align} $

(3.24)

$ \begin{align} \|\tilde{A}\|^{2} = \tilde{a}_{1}^{2}+\tilde{a}_{2}^{2}, \quad\|A\|^{2} = a_{1}^{2}+a_{2}^{2}, \quad \tilde{a}_{1}+\tilde{a}_{2} = 0, \end{align} $

(3.25)

$ \begin{align} \|A\|^{2} = \|\widetilde{A}\|^{2}+\frac{1}{2}\left( a_{1}+a_{2}\right)^{2}, \end{align} $

(3.26)

$ \begin{align} a_{1}^{3}+a_{2}^{3} = \tilde{a}_{1}^{3}+\tilde{a}_{2}^{3}+\frac{3}{2}\left( a_{1}+a_{2}\right)(\tilde{a}_{1}^{2}+\tilde{a}_{2}^{2})+\frac{1}{4}\left( a_{1}+a_{2}\right)^{3}, \end{align} $

(3.27)

$ \begin{align} \tilde{a}_{i} = a_{i}-\frac{1}{8}(1+4 K), \quad \operatorname{t r} A = a_{1}+a_{2} = \frac{1}{4}(1+4 K). \end{align} $

(3.28)

Proposition 3.2 Let $ x: M^{2}\rightarrow S^{2+p} $ be a compact submanifold in the unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $, $ \|\nabla A\|^{2} = \sum\limits_{i, j, k}A^{2}_{ij, k} $, we have the following equation

$ \begin{align} \frac{1}{2} \Delta\|A\|^{2} = \|\nabla A\|^{2}-\sum\limits_{\alpha}\left[\operatorname{tr}\left(A B^{\alpha}\right)\right]^{2}+\operatorname{tr}A(\|A\|^{2}-2\operatorname{Det}A)+\sum\limits_{i, j} A_{i j}(\operatorname{tr} A)_{i j}. \end{align} $

(3.29)

Proof

$ \begin{align} A_{i j, 11}+ A_{i j, 22} = & A_{i j, 11}- A_{i 1, j 1}+ A_{i 1, j 1}- A_{i 1, 1 j}+ A_{1i, 1j}-A_{11, i j}+A_{i j, 22}\\ &- A_{i 2, j 2}+ A_{i 2, j 2}- A_{i 2, 2 j}+ A_{2i, 2j}-A_{22, i j}+(\operatorname{t r} A)_{i j}, \end{align} $

(3.30)

$ \begin{align} \sum\limits_{i, j} (A_{i j} A_{i j, 11}+A_{i j} A_{i j, 22}) = &-2 \sum\limits_{i, j} A_{i j} C_{i}^{\alpha} C_{j}^{\alpha}-2 \sum\limits_{\alpha}\left\|A B^{\alpha}-B^{\alpha} A\right\|^{2}-\sum\limits_{\alpha}\left[\operatorname{tr}\left(A B^{\alpha}\right)\right]^{2} \\ &-\sum\limits_{i, j, } A_{i j} B_{i j}^{\alpha} (C_{1, 1}^{\alpha}+C_{2, 2}^{\alpha})+2 \operatorname{tr} A^{3}-\operatorname{tr} A\|A\|^{2}\\ &-2 \sum\limits_{i, j} A_{i j} (C_{1}^{\alpha} B_{i j, 1}^{\alpha}+C_{2}^{\alpha} B_{i j, 2}^{\alpha} )+\sum\limits_{\alpha}(\operatorname{t r} A)[(C_{1}^{\alpha})^{2}+(C_{2}^{\alpha})^{2}]\\ &+\sum\limits_{i, j} A_{i j}(\operatorname{t r} A)_{i j}, \end{align} $

(3.31)

then

$ \begin{align} \frac{1}{2} \Delta\|A\|^{2} = & \|\nabla A\|^{2}+\sum\limits_{i, j} A_{i j} \Delta A_{ij} \\ = & \|\nabla A\|^{2}-2 \sum\limits_{\alpha} \sum\limits_{i, j}\left(A_{i j}-\frac{1}{2} \operatorname{tr} A \delta_{i j} C_{i}^{\alpha} C_{j}^{\alpha}\right)-2 \sum\limits_{\alpha}\left\|A B^{\alpha}-B^{\alpha} A\right\|^{2} \\ &-\sum\limits_{\alpha}\left[\operatorname{tr}\left(A B^{\alpha}\right)\right]^{2}-\sum\limits_{\alpha} \sum\limits_{i, j, k} A_{i j} B_{i j}^{\alpha}(C_{1, 1}^{\alpha}+C_{2, 2}^{\alpha})+2 \operatorname{tr} A^{3}-\operatorname{tr} A\|A\|^{2} \\ &-2 \sum\limits_{\alpha} \sum\limits_{i, j, k} A_{i j}(C_{1}^{\alpha} B_{i j, 1}^{\alpha}+C_{2}^{\alpha} B_{i j, 2}^{\alpha} )+\sum\limits_{i, j} A_{i j}(\operatorname{t r} A)_{i j}. \end{align} $

(3.32)

From $ \Phi = 0 $, we obtain

$ \begin{align} \frac{1}{2} \Delta\|A\|^{2} = \|\nabla A\|^{2}-\sum\limits_{\alpha}\left[\operatorname{tr}\left(A B^{\alpha}\right)\right]^{2}+\operatorname{tr}A(\|A\|^{2}-2\operatorname{Det}A)+\sum\limits_{i, j} A_{i j}(\operatorname{tr} A)_{i j}. \end{align} $

(3.33)

Lemma 3.3 Let $ x: M^{2}\rightarrow S^{2+p} $ be a compact submanifold in the unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $, we have

$ \begin{align} 0\geq\int_{M}\|\nabla A\|^{2}-\|\nabla \operatorname{tr}A\|^{2}+2(\operatorname{tr}A-\frac{1}{4})\|\widetilde{A}\|^{2}, \end{align} $

(3.34)

where equality holds if and only if $ B^{\alpha} = \lambda_{\alpha}\widetilde{A} (\widetilde{A}\neq0) $.

Proof Let $ L:C^{\infty}(M)\rightarrow C^{\infty}(M) $ be $ L $ operator([5]), and $ L(f) $ is defined by $ L(f): = (A_{ij}-\operatorname{tr}A\delta_{ij})f_{, ij} $, from$ (3.18) $ we have

$ \begin{align} \sum\limits_{j} A_{i j, j}-(\operatorname{t r} A)_{j} = \sum\limits_{j}A_{i j, j}-\sum\limits_{i}A_{i i, j} = 0, \end{align} $

(3.35)

thus, $ L^{2} $ self-adjointness, i.e., $ (\rm{g}, L(f)) = (f, L(\rm{g})) $. In particular, $ (L(f), 1) = 0, $ i.e., $ \int_{M} L(f) = 0 $. Then

$ \begin{align} L(\operatorname{tr} A) & = \left(A_{i j}-\operatorname{tr} A \delta_{i j}\right)(\operatorname{tr} A)_{i j} = A_{i j}(\operatorname{tr} A)_{i j}-\operatorname{tr} A \Delta \operatorname{tr} A \\ & = A_{i j}(\operatorname{tr} A)_{i j}-\frac{1}{2} \Delta(\operatorname{tr} A)^{2}+\|\nabla \operatorname{tr} A\|^{2}, \end{align} $

(3.36)

then we have

$ \begin{align} \frac{1}{2} \Delta\left[\|A\|^{2}-(\operatorname{tr} A)^{2}\right] = & \|\nabla A\|^{2}-\|\nabla \operatorname{tr} A\|^{2}-\sum\limits_{\alpha}(\operatorname{tr}(A B ^{\alpha}))^{2} +\\ 2\operatorname{tr} A^{3}-\operatorname{tr} A\|A\|^{2}+L(\operatorname{tr} A). \end{align} $

(3.37)

Integrating both sides of the above equation, according to the properties of $ \Delta $ and L, we get

$ \begin{align} 0 = \int_{M}[\|\nabla A\|^{2}-\|\nabla \operatorname{tr}A\|^{2}-\sum\limits_{\alpha}(\operatorname{tr}(AB^{\alpha}))^{2}+\operatorname{tr}A(\|A\|^{2}-2\operatorname{Det}A)]dM. \end{align} $

(3.38)

From Cauchy-Schwarz Inequality, we see

$ \begin{align} -\left(\operatorname{t r}\left(A B^{\alpha}\right)\right)^{2} = -\left(\operatorname{tr}\left(\widetilde{A} B^{\alpha}\right)\right)^{2}\geq-\frac{1}{2}\|\widetilde{A}\|^{2}, \end{align} $

(3.39)

where the equality holds in $ (3.39) $ if and only if $ B^{\alpha} = \lambda_{\alpha}\widetilde{A} (\widetilde{A}\neq0) $. It is clear that $ \|A\|^{2}-2\operatorname{Det}A = 2\|\widetilde{A}\|^{2} $. Thus, we have

$ 0\geq\int_{M}\|\nabla A\|^{2}-\|\nabla \operatorname{tr}A\|^{2}+2(\operatorname{tr}A-\frac{1}{4})\|\widetilde{A}\|^{2}. $

Lemma 3.4 Let $ x: M^{2}\rightarrow S^{2+p} $ be a compact submanifold in the unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $, $ \operatorname{Det}A = c>0 $, we have

$ \begin{align} \|\nabla A\|^{2}-\|\nabla \operatorname{tr}A\|^{2}\geq0, \end{align} $

(3.40)

where equality holds if and only if $ \nabla A = 0 $.

Proof

$ \begin{align} \|A\|^{2}-(\operatorname{tr}A)^{2} = -2\operatorname{Det}A = -2c, \end{align} $

(3.41)

i.e.,

$ \begin{align} A^{2}_{11}+2A^{2}_{12}+A^{2}_{22} = (A_{11}+A_{22})^{2}-2c. \end{align} $

(3.42)

Take the derivative of the left hand side of the equation, we have

$ \begin{align} \sum\limits_{i, j}A_{ij, k}^{2} = 2\sum\limits_{i, j}A_{ij, k}A_{ij}. \end{align} $

(3.43)

Take the derivative of the right hand side of the equation, we have

$ \begin{align} (\sum\limits_{i}A_{ii})^{2}_{k} = 2(\operatorname{tr}A)(\operatorname{tr}A)_{k}, \end{align} $

(3.44)

thus

$ \begin{align} \sum\limits_{i, j, k}A_{ij, k}A_{ij} = \operatorname{tr}A(\operatorname{tr}A)_{k}. \end{align} $

(3.45)

Square both ends of the above equation, we have

$ \begin{align} (\sum\limits_{i, j, k}A_{ij, k}A_{ij})^{2} = \|\nabla \operatorname{tr}A\|^{2}(\operatorname{tr}A)^{2}. \end{align} $

(3.46)

On the left hand side, we use the Cauchy-Schwarz inequality to get

$ \begin{align} (\sum\limits_{i, j, k}A_{ij, k}A_{ij})^{2}\leq\|\nabla A\|^{2}\|A\|^{2}, \end{align} $

(3.47)

taking $ (3.47) $ in $ (3.46) $, we have

$ \begin{align} \|\nabla A\|^{2}\|A\|^{2}\geq\|\nabla \operatorname{tr}A\|^{2}(\operatorname{tr}A)^{2}, \end{align} $

(3.48)

$ \begin{align} \|\nabla A\|^{2}\|A\|^{2}\geq\|\nabla \operatorname{tr}A\|^{2}(\|A\|^{2}+2c), \end{align} $

(3.49)

$ \begin{align} (\|\nabla A\|^{2}-\|\nabla \operatorname{tr}A\|^{2})\|A\|^{2}\geq2c\|\nabla \operatorname{tr}A\|^{2}\geq0. \end{align} $

(3.50)

Then we have

$ \begin{align} \|\nabla A\|^{2}-\|\nabla \operatorname{tr}A\|^{2}\geq0, \end{align} $

(3.51)

where the equality holds if and only if $ \nabla A = 0 $.

From the above lemma, we can get the following theorem:

Lemma 3.5 Let $ x: M^{2}\rightarrow S^{2+p} $ be a compact submanifold in the unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $, $ \operatorname{Det}A = c(const)>0 $, we have the following inequality

$ \int_{M}\left(\operatorname{tr} A-\frac{1}{4}\right)\|\widetilde{A}\|^{2}dM \leq 0. $

In particular, when $ \operatorname{tr}A\geq\frac{1}{4} $, then either $ \widetilde{A} = 0 $ or $ \operatorname{tr}A = \frac{1}{4} $, and we have $ B^{\alpha} = \lambda_{\alpha}\widetilde{A}, \, \, \nabla A = 0. $

Theorem 4.1 Let $ x: M^{2}\rightarrow S^{2+p} $ be a $ 2- $dimensional compact submanifold in the $ (2+p)- $dimensional unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $, and $ 0<\operatorname{Det}A = c(const)<\frac{1}{64} $, then

$ \begin{align} \int_{M}\left(\operatorname{tr} A-\frac{1}{4}\right)\|\widetilde{A}\|^{2}dM \leq 0. \end{align} $

(4.1)

In particular, if $ \operatorname{tr}A\geq\frac{1}{4} $, then either

$ \begin{align} \widetilde{A}\equiv0, \end{align} $

(4.2)

and $ x(M^{2}) $ is Moebius equivalent to a minimal submanifold with constant curvature in $ S^{2+p} $; or

$ \begin{align} \operatorname{tr}A = \frac{1}{4}, \end{align} $

(4.3)

and $ x(M^{2}) $ is Moebius equivalent to $ S^{1}(r)\times S^{1}(\sqrt{\frac{1}{1+c^{2}}-r^{2}}) $ in $ S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $, where $ r^{2} = \frac{4+\sqrt{1-64c}}{8} $.

Lemma 4.2 Let $ x: M^{2}\rightarrow S^{2+p} $ be a $ 2- $dimensional compact submanifold in the $ (2+p)- $dimensional unit sphere $ S^{2+p} $ with vanishing Moebius form $ \Phi $, if $ \|\widetilde{A}\|\neq0, \operatorname{tr}A = \frac{1}{4} $ and $ 0<\operatorname{Det}A = c(const)<\frac{1}{64} $, then the $ x(M^{2}) $ is Moebius equivalent to $ S^{1}(r)\times S^{1}(\sqrt{\frac{1}{1+c^{2}}-r^{2}}) $ in $ S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $, where $ r^{2} = \frac{2-\sqrt{1-64c}}{4(1+c^{2})} $.

Proof From $ \operatorname{tr}A = \frac{1}{4}, \, \, \operatorname{Det}A = c $, we have

$ \begin{align} a_{1}+a_{2} = \frac{1}{4}, \quad a_{1}a_{2} = c. \end{align} $

(4.4)

Which implies

$ \begin{align} a_{1} = \frac{1+\sqrt{1-64c}}{8}, \quad a_{2} = \frac{1-\sqrt{1-64c}}{8}, \end{align} $

(4.5)

then we get

$ \begin{align} \widetilde{a}_{1} = \frac{\sqrt{1-64c}}{8}, \quad \widetilde{a}_{2} = -\frac{\sqrt{1-64c}}{8}. \end{align} $

(4.6)

From Lemma 3.5, we have

$ \begin{align} B^{\alpha} = \lambda_{\alpha}\widetilde{A}, \end{align} $

(4.7)

i.e,

$ \begin{align} (B^{\alpha}_{11})^{2}+2(B^{\alpha}_{12})^{2}+(B^{\alpha}_{22})^{2} = \sum\limits_{\alpha}\lambda_{\alpha}^{2}[(\widetilde{A}_{11})^{2}+(\widetilde{A}_{22})^{2}] = \frac{1-64c}{32}\sum\limits_{\alpha}\lambda_{\alpha}^{2}, \end{align} $

(4.8)

by use of $ (2.18) $, we see

$ \begin{align} (\frac{1}{32}-2c)\sum\limits_{\alpha}\lambda_{\alpha}^{2} = \frac{1}{2}. \end{align} $

(4.9)

We claim that we can choose the normal frame field $ \{E_{\alpha}\} $, such that

$ \begin{align} B_{ij}^{3} = \lambda \widetilde{A}_{ij}, \quad B_{ij}^{\alpha} = 0, \quad \lambda = \frac{4}{\sqrt{1-64c}}, \quad \forall i, j;\alpha\geq4. \end{align} $

(4.10)

In fact, we can choose a new orthonormal frame $ \{\bar{e}_{\alpha}\} $ in the normal bundle $ N(M^{2}) $ such that $ \bar{e}_{3} = \frac{\sum\limits_{\alpha}\lambda_{\alpha}e_{\alpha}}{\sqrt{\sum\limits_{\alpha}\lambda_{\alpha}^{2}}} $, and then define a new orthonormal frame $ \{\bar{E}_{\alpha}\} $ in the Moebius normal bundle by $ \bar{E}_{\alpha} = \left(\bar{H}_{\alpha}, \bar{H}^{\alpha} x+\bar{e}_{\alpha}\right) $, where $ \sum\limits_{\alpha} H^{\alpha} e_{\alpha} = \sum\limits_{\alpha} \bar{H}^{\alpha} \bar{e}_{\alpha} = H $ is the mean curvature vector of $ M^{2} $, then $ \bar{E}_{3} = \frac{\sum\limits_{\alpha} \lambda_{\alpha} E_{\alpha}}{\sqrt{\sum\limits_{\beta} \lambda_{\beta}^{2}}} $ and with respect to $ {\bar{E}_{\alpha}} $. If $ \{e_{\alpha}\} $, $ \{\bar{e}_{\alpha}\} $ are two orthonormal frames in the normal bundle with $ e_{\alpha} = \sum\limits \sigma_{\alpha \beta} \bar{e}_{\beta} $, where $ (\sigma_{\alpha\beta}) $ is an orthogonal matrix, then we have $ E_{\alpha} = \sum\limits_{\beta} \sigma_{\alpha \beta} \bar{E}_{\beta} $. From $ B_{i j}^{\alpha} = \rho^{-1}\left(h_{i j}^{\alpha}-H^{\alpha} \delta_{i j}\right) $, when $ \alpha\geq4 $, we have

$ \begin{align} h_{i j}^{\alpha} = H^{\alpha} \delta_{i j}, \end{align} $

(4.11)

that means that $ \operatorname{span}\left\{e_{4}, e_{5}, \cdots, e_{2+p}\right\} $ is totally umbilical in the normal bundle $ N(M^{2}) $. From $ (4.10) $ we have

$ \begin{align} \omega_{3 \alpha} = \theta_{3 \alpha} \equiv 0, \quad \forall \alpha \end{align} $

(4.12)

where $ \theta_{\alpha\beta} $ is the Euclidean normal connection of $ N(M^{2})([1]) $.

From $ (2.18) $, we have $ B^{3}_{11}+B^{3}_{22} = 0, \frac{1}{2} = (B^{3}_{11})^{2}+(B^{3}_{22})^{2}, $ which implies

$ \begin{align} B^{3}_{11} = -\frac{1}{2}, \quad B^{3}_{22} = \frac{1}{2}. \end{align} $

(4.13)

From $ (2.19) $ and $ \nabla A = 0 $, we have

$ \begin{align} A_{11}\omega_{12}+A_{22}\omega_{21} = 0, \end{align} $

(4.14)

$ \begin{align} (a_{1}-a_{2})\omega_{12} = 0, \end{align} $

(4.15)

thus, we have

$ \begin{align} \omega_{12} = 0, \end{align} $

(4.16)

$ \begin{align} d\omega_{1} = 0, \quad d\omega_{2} = 0. \end{align} $

(4.17)

Hence, by [1, Theorem 1] and $ (4.11), (4.12) $, we can conclude that $ x(M^{2}) $ is located in some sphere $ S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $ which are totally umbilical submanifold of $ S^{n} $.

Since a Riemannian universal coverage is locally equidistant and not general, we can assume that $ M^{2} $ is simply connected. From above, $ TM^{2} $ has the following decomposition $ TM^{2} = V_{1}\oplus V_{2}, $ where $ V_{1} $ and $ V_{2} $ are the 1-dimensional eigenspace of the Blaschke tensor $ A $ with eigenvalues $ a_{1} $ and $ a_{2} $.

Form (4.17), we can get that the eigenspaces $ V_{1} $ and $ V_{2} $ are both integrable. We can write $ M = M_{1}\times M_{2} $, where $ M_{1} $ and $ M_{2} $ are both $ 1- $dimensional submanifold. We can define $ \rm{g}_{1} = \omega_{1}^{2} $ and $ \rm{g}_{2} = \omega_{2}^{2} $, then we have

$ \begin{align} (M^{2}, \rm{g}) = (M_{1}, \rm{g}_{1})\times(M_{2}, \rm{g}_{2}). \end{align} $

(4.18)

From $ (2.12) $ and $ (4.16) $, we have

$ \begin{align} dY_{1} = (-a_{1}Y-N+b_{1}E_{3})\omega_{1}, \end{align} $

(4.19)

$ \begin{align} dY_{2} = (-a_{2}Y-N+b_{2}E_{3})\omega_{2}, \end{align} $

(4.20)

$ \begin{align} \langle dY_{1}, dY_{1}\rangle = (2a_{1}+b_{1}^{2})(\omega_{1})^{2} = \frac{2+\sqrt{1-64c}}{4}(\omega_{1})^{2}, \end{align} $

(4.21)

$ \begin{align} \langle dY_{2}, dY_{2}\rangle = (2a_{2}+b_{2}^{2})(\omega_{2})^{2} = \frac{2-\sqrt{1-64c}}{4}(\omega_{2})^{2}. \end{align} $

(4.22)

Since $ M^{2} $ is compact submanifold, $ M_{1} $ and $ M_{2} $ are also compact submanifold.

Then we consider $ \bar{x}:S^{1}(r_{1})\times S^{1}(r_{2})\rightarrow S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $ being a torus, where $ r_{1}^{2} = \frac{2-\sqrt{1-64c}}{4(1+c^{2})}, \, \, r_{2}^{2} = \frac{2+\sqrt{1-64c}}{4(1+c^{2})} $. Let $ \widetilde{x}:M^{2}\rightarrow S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $, from $ (3.13), (3.15), (4.5), (4.13) $, $ \widetilde{x} $ and $ \bar{x} $ have the same Moebius shape. From the proof of Lemma 3.1 we see that the Moebius metric $ \bar{\rm{g}} $ of $ \bar{x} $ is given by $ \bar{\rm{g}} = \bar{\rho}^{2}d\bar{x}\cdot d\bar{x} = \bar{\rm{g}}_{1}+\bar{\rm{g}}_{2} $, where for $ \bar{x} = r_{1}\overline{(x_{1}, 0)}+r_{2}\overline{(0, x_{2})}, $

$ \begin{align} \bar{\rm{g}}_{1} = \frac{4}{2+\sqrt{1-64c}}d\bar{x}_{1}\cdot d\bar{x}_{1}, \quad \bar{\rm{g}}_{2} = \frac{4}{2-\sqrt{1-64c}}d\bar{x}_{2}\cdot d\bar{x}_{2}. \end{align} $

(4.23)

From $ (4.21)-(4.23) $, we have a $ 1- $dimensional manifold with the same curvature as $ (S^{1}(r_{1}), \bar{\rm{g}}_{1}) $ and $ (M_{1}, \rm{g}_{1}) $, $ (s^{1}(r_{2}), \bar{\rm{g}}_{2}) $ and $ (M_{2}, \rm{g}_{2}) $ are also $ 1- $dimensional manifolds with the same curvature. Thus there exist isometries $ \psi_{1}:(M_{1}, \rm{g}_{1})\rightarrow (S^{1}(r_{1}), \bar{\rm{g}}_{1}) $ and $ \psi_{2}:(M_{2}, \rm{g}_{2})\rightarrow (S^{1}(r_{2}), \bar{\rm{g}}_{2}) $, let $ \psi = (\psi_{1}, \psi_{2}) $, then $ \psi:M^{2}\rightarrow S^{1}(r_{1})\times S^{1}(r_{2}) $ holds the Moebius metric and the Moebius shape operator. $ (M^{2}, \rm{g}) $ and $ (S^{1}(r_{1})\times S^{1}(r_{2}), \bar{\rm{g}}) $ have the same Moebius metric, Moebius second fundamental form, Moebius shape operator, Blaschke tensor, so the $ x(M^{2}) $ is Moebius equivalent to $ S^{1}(r)\times S^{1}(\sqrt{\frac{1}{1+c^{2}}-r^{2}}) $ in $ S^{3}(\frac{1}{\sqrt{1+c^{2}}}) $, where $ r^{2} = \frac{2-\sqrt{1-64c}}{4(1+c^{2})} $.

This completes the proof of the main theorem.