Theorem 4. Let \(M^5\) be a closed \(5\)-dimensional minimal hypersurface in the unit sphere \(\mathbf{S}^6\) such that \(S, f_3,f_4\) are constants. Suppose in addition \[\label{tpvcemjd} A(r)=\sum_{i,j,k\in I_r} s^{ijk}\left(\left(s_1^{ijk}\right)^2+2s_2^{ijk}-\sigma_2\right) \left(2 s_1^{ijk} s_2^{ijk} -3s_3^{ijk}+2\sigma_3\right)>0\qquad{(1)}\] for \(r=1,2,\cdots,5\) at every point where the principal curvatures consist of at least four distinct values, or exactly three distinct values with multiplicities \((2,2,1)\). Then \(M^5\) is isoparametric.
Minimal Hypersurfaces with constant scalar curvature in \(\mathbf{S}^6\)
May 19, 2026
Abstract
In this paper, we propose certain assumptions on the principal curvatures for a closed minimal hypersurface \(M^5\) in \(\mathbf{S}^6\) to be isoparametric, provided that the functions \(S, f_3,f_4\) are constants. Our result removes the nonnegative scalar curvature assumption as in Tang and Yan [1]. Finally, as a rigidity result, if \(M^5\subset \mathbf{S}^6\) has a point with exactly two distinct principal curvatures, then it must be a Clifford torus.
1 Introduction↩︎
Exploring the relationship between geometric invariants and the structure of manifolds or submanifolds has always been an important problem in global differential geometry. In 1968, Simons [2] gave an integral formula for the squared norm \(S\) of the second fundamental form. Shortly thereafter, Chern et al. [3] and Lawson [4] independently obtained the rigidity result when \(S = n\), as shown in the theorem below.
Theorem 1. \((\)[2]–[4]\()\) Let \(M^n\subset\mathbf{S}^{n+1}\) be a closed minimal immersed hypersurface, then \[\int_{M}(S-n)S\geq 0.\] In particular, for \(S\leq n\), one has either \(S\equiv0\) or \(S\equiv n\) on \(M^n\). Moreover, for \(S\equiv0\), \(M^n\) is the equatorial \(n\)-spheres in \(\mathbf{S}^{n+1}\); for \(S\equiv n\), \(M^n\) is the Clifford tori \(\mathbb{S}^k\left(\sqrt{\frac{k}{n}}\right)\times\mathbb{S}^{n-k}\left(\sqrt{\frac{n-k}{n}}\right)\), \(1\leq k\leq n-1\).
Based on the above work, Chern [5] proposed the following famous conjecture regarding compact minimal hypersurfaces in a sphere.
Chern conjecture. Let \(M^n\) be an \(n\)-dimensional compact minimal hypersurface in the unit sphere \(\mathbf{S}^{n+1}\) with constant
scalar curvature \(R\). Then the set of all possible values of the scalar curvature (equivalently, \(S\)) of \(M^n\) is a discrete set in \(\mathbb{R}\).
Chern’s conjecture can be decomposed into several problems of pinching the scalar curvature of compact minimal hypersurfaces in the unit sphere. The first difficulty is the second gap problem, which states that, under the assumptions of Chern conjecture, if \(n \leq S\leq 2n\), then either \(S = n\) or \(S = 2n\). Peng and Terng [6], [7] completely solved the second gap problem for the case \(n = 3\), and \(S = 6\) can be realized by Cartan minimal isoparametric hypersurfaces in the unit sphere \(\mathbf{S}^4\). Furthermore, for general dimension \(n\), under the assumptions of Chern conjecture, they also proved that if \(S > n\), then necessarily \(S > n + \frac{1}{12n}\), which for the first time gave a breakthrough in the second pinching problem. Subsequently, Yang and Cheng [8] advanced the second gap problem to \(\frac{n}{3}\); Suh and Yang [9] improved Peng and Terng’s result to \(\frac{3n}{7}\). For a more detailed introduction to Chern conjecture and related problems, we refer the reader to [1], [10]–[13].
Up to now, all known closed minimal hypersurfaces in spheres with constant scalar curvature are isoparametric. Based on this, Verstraelen, Montiel, Ros and Urbano [14] first proposed a stronger version of Chern conjecture, namely:
Stronger Chern conjecture. Let \(M^n\) be a closed, minimally immersed hypersurface of the unit sphere \(\mathbf{S}^{n+1}\) with constant scalar curvature. Then \(M^n\) is isoparametric.
In 1993, Chang [15] proved the above version of Chern’s conjecture for the case \(n = 3\). In fact, without requiring minimality, de Almeida and Brito [16] proved the following theorem.
Theorem 2. \((\)[16]\()\) Let \(M^3\subset\mathbf{S}^4\) be a closed hypersurface with constant mean curvature \(H\) and constant nonnegative scalar curvature \(R\). Then \(M^3\) is isoparametric.
Soon thereafter, Chang [17], Cheng and Wan [18] independently proved that under the assumptions of the above theorem, one always has \(R \geq 0\), thereby generalizing the aforementioned theorem.
In the case \(n=4\), Lusala et al. [19], [20] proved that closed minimal Willmore hypersurfaces with nonnegative constant scalar curvature in \(\mathbf{S}^5\) are isoparametric, where the Willmore assumption here is equivalent to \(f_3 = 0\). Deng et al. [21] removed the nonnegative scalar curvature assumption and generalized this result. In addition, Tang and Yang [22] proved that if the number of distinct principal curvatures is fixed, then any closed minimal hypersurface \(M^n \subset \mathbf{S}^{n+1}\) with constant \(3\)-rd mean curvature \(H_3\) and constant nonnegative scalar curvature \(R\) must be isoparametric.
In the case \(n = 6\), Scherfner et al. [23] proved that closed hypersurfaces in \(\mathbf{S}^7\) with constant nonnegative scalar curvature are isoparametric if \(H = f_3 = f_5 = 0\) and \(f_4 = \text{const.}\), which is listed as in [24].
For general dimension \(n\), based on the method of [16], Tang, Wei and Yan [13] and Tang and Yan [1] proved the following theorem.
Theorem 3. \((\)[1]\()\)Let \(M^n\) \((n>3)\) be a closed hypersurface in the unit sphere \(\mathbf{S}^{n+1}\). If \(R\geq 0\) and \(\sum_{i=1}^{n}\lambda_{i}^k\) \((k=1,\cdots,n-1)\) are constants for principal curvatures \(\lambda_{1}\leq\lambda_{2}\leq \cdots \leq \lambda_{n}\), then \(M^n\) is isoparametric.
When \(n = 4\), the result of the above theorem was also obtained in [25]. Furthermore, for the minimal case, Cheng and Li [26] proved that if the number of distinct principal curvatures is constant, the assumption \(R \geq 0\) on \(M^4\) in Theorem 3 is redundant. Based on the result of Cheng and Li [26], using the method in [16], He, Xu and Zhao [27] removed the requirement on the number of distinct principal curvatures in [26]. This also shows, in the four dimensional minimal case, that the Theorem 3 of Tang and Yan does not require the assumption of nonnegative scalar curvature. Therefore, one can naturally propose the following question: for general dimensions, can the assumption \(R \geq 0\) in Theorem 3 also be removed? This would bring the result closer to the statement of the Chern conjecture.
In this paper, for \(n = 5\) in the minimal case, we propose a new assumption on the principal curvatures (see \((\ref{Ass})\)) that can replace the requirement of nonnegative scalar curvature in Theorem 3. The method that we use is generalizing the 3-form \(\Phi\) in [26] to an \((n-1)\)-form (see \((\ref{Phi})\)) and performing a crucial simplification on the differential of this \((n-1)\)-form when \(n=5\) (see Section 3). Throughout this paper, we adopt the conventions that \(\sigma_3\geq 0\) and \(\lambda_{1}\leq \lambda_{2}\leq \lambda_{3}\leq \lambda_{4}\leq \lambda_{5}.\) For three distinct indices \(1\leq i<j<k\leq 5\), we define \[\qquad\quad s_1^{ijk}=\lambda_{i}+\lambda_j+\lambda_k,\;\;s_2^{ijk}=\lambda_i\lambda_j+\lambda_i\lambda_k+\lambda_j\lambda_k,\;\;s_3^{ijk}=\lambda_i\lambda_j\lambda_k,\] \[\begin{gather} \qquad\quad s^{ijk} =(\lambda_{i}-\lambda_{j})^2(\lambda_{i}-\lambda_{k})^2(\lambda_{j}-\lambda_{k})^2. \end{gather}\] For \(r=1,2,\cdots,5\), we set \(I_r=\{1,2,3,4,5\}\textbackslash\{r\}\).
The main result of this paper is the following theorem.
Now we give another characterization of \((\ref{Ass})\). Lemma 8 (see Section 2) implies that at each point of the hypersurface \(M^5\), we have \[A(5)= -\Bigl\langle \left(z_4q_4, z_3q_3, z_2q_2, z_1q_1\right), \left(\lambda_{4}z_4q_4, \lambda_{3}z_3q_3, \lambda_{2}z_2q_2, \lambda_{1}z_1q_1\right) \Bigr\rangle.\] Therefore, \(A(5) > 0\) is equivalent to the angle between vectors \((z_4q_4, z_3q_3, z_2q_2, z_1q_1)\) and \((\lambda_{4}z_4q_4, \lambda_{3}z_3q_3, \lambda_{2}z_2q_2, \lambda_{1}z_1q_1)\) lying between \(\frac{\pi}{2}\) and \(\pi\), where the two vectors are further constrained by \((\ref{vec1})\) and \((\ref{vec2})\), respectively. The remaining \(A(i)\), \(i\in I_5\), are similar.
There are many configurations of principal curvatures fit into assumption \(\eqref{Ass}\). However, for the four principal curvature configurations in the following corollary, the scalar curvature \(R\) may be not nonnegative.
Corollary 5. Let \(M^5\) be a closed \(5\)-dimensional minimal hypersurface in the unit sphere \(\mathbf{S}^6\) such that \(S, f_3,f_4\) are constants. At each point, suppose either there are at most three distinct principal curvatures with multiplicities not equal to \((2,2,1)\), or the configuration of principal curvatures belong to one of the following four types:
\(\lambda_{1}<\lambda_{2}<\lambda_{3}=\lambda_{4}<0<\lambda_{5}\);
\(\lambda_{1}=\lambda_{2}<\lambda_{3}<\lambda_{4}<\lambda_{5}\);
\(\lambda_{1}<\lambda_{2}=\lambda_{3}<\lambda_{4}<0<\lambda_{5}\);
\(\lambda_1<\lambda_{2}<\lambda_{3}<\lambda_{4}<0<\lambda_{5}\).
Then \(M^5\) is isoparametric.
Here we point out that in \((2)\), if \(\lambda_{3}\) is nonnegative, \(\sigma_3\) may be less than zero. In all other cases, \(\sigma_3\) is always nonnegative. This is also illustrated in the proof of the corollary. Furthermore, we obtain a global rigidity result.
Theorem 6. Let \(M^5\subset\mathbf{S}^6\) be a closed minimal hypersurface with constant scalar curvature \(R\) and constant \(4\)-th mean curvature \(H_4\). Suppose there is a point with two distinct principal curvatures of multiplicities \((m_1,m_2)\). If \((m_1,m_2)=(3,2)\), suppose in addition that the \(3\)-rd mean curvature \(H_3\) is constant. Then \(S=5\) and \(M^5\) is the Clifford torus \(\mathbb{S}^2(\sqrt\frac{2}{5})\times\mathbb{S}^3(\sqrt\frac{3}{5})\) or \(\mathbb{S}^1(\sqrt\frac{1}{5})\times\mathbb{S}^4(\sqrt\frac{4}{5})\).
The rest of this paper is organized as follows. In Section 2, we give some preliminaries and show some lemmas of this paper. In Section 3, we show our core lemma (Lemma 13). In Section 4, we prove Theorem 4, Corollary 5 and Theorem 6.
2 Preliminary↩︎
In this section, we assume that \(M^n\) is connected and oriented. Otherwise, we can discuss on each connected component of \(M^n\) or on the double covering of \(M^n\).
2.1 Basic Knowledge↩︎
Let \(f:M^n\rightarrow \mathbf{S}^{n+1}\) be an \(n\)-dimensional immersed hypersurface, and let \(\{e_1,e_2,\cdots,e_{n+1}\}\) be an oriented local orthonormal frame fields of \(\mathbf{S}^{n+1}\) such that \(e_1,e_2,\cdots,e_n\) are tangent to \(M^n\). We use \(\{\theta_i,i=1,2,\cdots,n\}\) and \(\{\omega_{ij}, 1\leq i,j\leq n\}\) to denote the dual \(1\)-form and connection \(1\)-form corresponding to \(\{e_1,e_2,\cdots,e_n\}\), respectively. Then the structure equations of \(M^n\) are given by: \[\begin{cases} d\theta_i=\sum_{j=1}^{n}\omega_{ij}\wedge\theta_j,\\ d\omega_{ij}=\sum_{k=1}^{n}\omega_{ik}\wedge\omega_{kj}-R_{ij}, \end{cases}\] where \(R_{ij}=\frac{1}{2}\sum_{k,l=1}^{n}R_{ijkl}\theta_k\wedge\theta_l\) denotes the curvature \(2\)-forms of \(M^n\).
Let \(\operatorname{II}=\sum_{i,j=1}^{n}h_{ij}\theta_i\otimes\theta_j\) denotes the second fundamental form, then the mean curvature is given by \[H=\frac{1}{n}\sum_{i=1}^nh_{ii}.\] Let \(S=|\operatorname{II}|^2=\sum_{i,j=1}^{n}h_{ij}^2\) be the square length of the second fundamental form. Then the Gauss equation implies that \[\label{gauss} R_{ijkl}=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}+h_{ik}h_{jl}-h_{il}h_{jk},\tag{1}\] \[R=n(n-1)+n^2H^2-S,\] where \(R\) is the scalar curvature of \(M\).
Define the covariant derivative \(\nabla\operatorname{II}\) of \(\operatorname{II}\) \((\)with component \(h_{ijk}\) \()\) by \[\sum_{m=1}^{n}h_{ijm}\theta_m=dh_{ij}+\sum_{m=1}^nh_{mj}\omega_{mi}+\sum_{m=1}^nh_{im}\omega_{mj}.\] Then by Codazzi equation we have \[\label{coda} h_{ijk}=h_{ikj}\;for \;i,j,k=1,2,\cdots,n,\tag{2}\] it implies immediately that \(h_{ijk}\) is symmetric, and when \(M\) is minimal, from [6] we know \[\label{DeltaS} \frac{1}{2}\Delta S=(n-S)S+\sum_{i,j,k=1}^nh_{ijk}^2.\tag{3}\]
Next we exterior differentiate the above formula and define \(h_{ijkl}\) by \[\sum_{m=1}^{n}h_{ijkm}\theta_m=dh_{ijk}+\sum_{m=1}^{n}h_{mjk}\omega_{mi}+\sum_{m=1}^{n}h_{imk}\omega_{mj}+\sum_{m=1}^{n}h_{ijm}\omega_{mk},\] and we define \(f_3\) and \(f_4\) as \[f_3=\sum_{i,j,k=1}^nh_{ij}h_{jk}h_{ki},\quad f_4=\sum_{i,j,k,l=1}^{n}h_{ij}h_{jk}h_{kl}h_{li}.\]
For an arbitrary fixed point \(x\in M^n\), we take an orthonormal frame such that \(h_{ij}=\lambda_{i}\delta_{ij}\) at \(x\), for all \(i,j=1,2,\cdots,n\). Then at this point \(x\), we have \[f_3=\sum_{i=1}^{n}\lambda_{i}^3,\quad f_4=\sum_{i=1}^{n}\lambda_{i}^4,\quad H=\frac{1}{n}\sum_{i=1}^{n}\lambda_{i},\quad S=\sum_{i=1}^{n}\lambda_{i}^2\] and we define the smooth function \[h=\sum_{i=1}^{n}\lambda_{i}^n.\]
Definition 7. [16] The combination \((U,\theta)\) is admissible if
\(U\) is an open subset of \(Y\), where \(Y\) is given in \((\ref{space})\);
\(\theta=(\theta_1,\theta_{2},\cdots,\theta_n)\) is a smooth orthonormal coframe field on \(U\);
\(\theta_{1}\wedge\theta_{2}\wedge\cdots\wedge\theta_n=vol\) on \(U\), where vol is the volume form of \(U\);
\(\operatorname{II}=\sum_{i=1}^{n}\lambda_{i}\theta_i\otimes\theta_i.\)
In this paper, we choose a proper system on \(M^n\) such that \((U,\theta)\) is admissible, at this time, the connection form \(\omega_{ij}\) on \(U\) are uniquely determined and \(h_{ij}=\lambda_{i}\delta_{ij}.\) In this admissible chart, we suppose \(\lambda_{1}\leq \lambda_{2}\leq \cdots\leq \lambda_n\). Then we define an \((n-1)\)-form \(\Phi\) as follows, which is the key point of our proof. \[\label{Phi} \Phi=\sum_{\sigma}S(\sigma)(\lambda_{i_{n-1}}+\lambda_{i_n})\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\theta_{i_{n-1}}\wedge\omega_{i_{n-1}i_{n}},\tag{4}\] where \(\sigma(1,\cdots,n)=(i_1,\cdots,i_n)\) is a permutation and \(S(\sigma)\) is the sign of \(\sigma\). By [16], we know the \((n-1)\)-form \(\Phi\) is globally well-defined on \(M^n\). In fact, every “\(\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\theta_{i_{n-1}}\wedge\omega_{i_{n-1}i_{n}}\)” is well-defined.
Let \(\sigma_r:\mathbb{R}^n\rightarrow\mathbb{R}\) be the elementary symmetric functions defined by \[\sigma_r(\lambda_{1},\cdots,\lambda_{n})=\sum_{i_1<i_2<\cdots<i_r}\lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_r}\;for\;1\leq r\leq n.\] and then define \(r\)-th mean curvature by \[H_r=\frac{1}{C^r_n}\sigma_r.\]
Now we define the region \(\Omega\) as follows.
\[\label{Om} \Omega=\left\{x\in M^n\middle|\sum_{i=1}^{n}\lambda_i^j(x)=c_j,\;\forall j=1,2,\cdots,n-1\,\text{and}\;\lambda_{1}(x)<\lambda_{2}(x)<\cdots<\lambda_{n}(x)\right\},\tag{5}\] where \(c_1,c_2,\cdots,c_{n-1}\) are constants.
The functions \(\lambda_{i}\) \((i=1,2,\cdots,n)\) are smooth on \(\Omega\). Thus we have \[\label{dlam} d\lambda_{i}=\sum_{j=1}^{n}\lambda_{ij}\theta_j,\tag{6}\] and \(\lambda_{ij}\) are smooth functions on \(\Omega\). In addition, we express connection coefficients of the connection form \(\omega_{ij}\) as \[\label{conform} \omega_{ij}=\sum_{k=1}^{n}\Gamma_{ijk}\theta_k,\tag{7}\] where \(\Gamma_{ijk}=\omega_{ij}(e_k)\) for \(i,j=1,2,\cdots,n\). From [13], we obtain \[\label{Gamma} h_{iik}=\lambda_{ik}\;and\;h_{ijk}=\left(\lambda_{i}-\lambda_{j}\right)\Gamma_{ijk}\;for\;i\neq j\tag{8}\] and furthermore \[\label{lam1} \lambda_{ij}=(-1)^{n+1}\frac{h_j}{n}\cdot\frac{1}{\prod_{k=1; k\neq i}^{5} (\lambda_k - \lambda_i)},\tag{9}\] where \(h_j\) is defined by \[\label{dh} dh=\sum_{j=1}^{n}h_j\theta_j.\tag{10}\]
In the following lemma, we provide another expression for \(A(r)\). We define \[v_k=\prod_{i,j=1;i<j;i,j\neq k}^4(\lambda_{i}-\lambda_{j}),\quad v=\prod_{i,j=1;i<j}^{4}(\lambda_{i}-\lambda_{j})\;and\ t_k=(-1)^k\prod_{i=1;i\neq k}^{4}(\lambda_{i}-\lambda_k).\]
Lemma 8. Define \(q_i\) and \(p_i\) \((i=1,2,3,4)\) by equation \((\ref{qk})\) and \((\ref{p})\), separately, we have \[A(5)=-\left(\lambda_{4}v_4^2q_4^2+\lambda_{3}v_3^2q_3^2+\lambda_{2}v_2^2q_2^2+\lambda_{1}v_1^2q_1^2\right).\] Moreover, \(A(i)\) \((i=1,2,3,4)\) is obtained by replacing \(\lambda_{i}\) in \(A(5)\) with \(\lambda_{5}\).
Proof. Since \[\label{vec1} v_4q_4-v_3q_3+v_2q_2-v_1q_1=0,\tag{11}\] from \((\ref{eql})\) and\((\ref{Al})\), we obtain \[\begin{align} A(5)=&v_4^2p_4q_4+v_3^2p_3q_3+v_2^2p_2q_2+v_1^2p_1q_1\\ =&-v_4^2q_4\left(t_4+\lambda_{4}q_4\right)-v_3^2q_3\left(-t_3+\lambda_{3}q_3\right)-v_2^2q_2\left(t_2+\lambda_{2}q_2\right)-v_1^2q_1\left(-t_1+\lambda_{1}q_1\right)\\ =&-v\left(v_4q_4-v_3q_3+v_2q_2-v_1q_1\right)-\lambda_{4}v_4^2q_4^2-\lambda_{3}v_3^2q_3^2-\lambda_{2}v_2^2q_2^2-\lambda_{1}v_1^2q_1^2\\ =&-\left(\lambda_{4}v_4^2q_4^2+\lambda_{3}v_3^2q_3^2+\lambda_{2}v_2^2q_2^2+\lambda_{1}v_1^2q_1^2\right). \end{align}\]
Besides, by direct calculation, the following equation holds. \[\label{vec2} -\lambda_{4}v_4q_4+\lambda_{3}v_3q_3-\lambda_{2}v_2q_2+\lambda_{1}v_1q_1=3v>0.\tag{12}\] ◻
2.2 Preparation for the proof of Theorem 4↩︎
Firstly, when \(n=5\), we give the characteristic polynomial \(F(x)\) and its relationship with the smooth function \(h\).
\[F(x)=\prod_{i=1}^5(x-\lambda_{i})=x^5-\sigma_1x^4+\sigma_2x^3-\sigma_3x^2+\sigma_4x-\sigma_5.\]
From Newton’s formula and the assumptions of Theorem 4, it follows that \(\sigma_i\) \((i=1,2,3,4)\) are constants, and \[\sigma_5=\frac{h}{5}+C_h,\] where \(C_h=-\frac{1}{6}Sf_3\) is a constant.
Let \[F_0(x)=x^5-\sigma_1x^4+\sigma_2x^3-\sigma_3x^2+\sigma_4x.\] Obviously, \(F_0(x)\) is a well-determined polynomial of degree \(5\), and \[F(x)=F_0(x)-\frac{h}{5}-C_h.\]
Since \(M^5\) is closed, we have that the range of \(h\) is a closed interval, denoted by \(\operatorname{Im}h=[a_0,b_0]\), \(a_0\leq b_0\).
From the definition of \(\Omega\), it follows that \(F(x)\) has \(5\) distinct real roots \(\lambda_{1}<\lambda_{2}<\lambda_{3}<\lambda_{4}<\lambda_{5}\) on \(\Omega\). By Rolle’s theorem, there exist \(\tau_i\) \((i=1,2,3,4)\) lying between these roots such that \[F_0'(\tau_i)=F'(\tau_i)=0 \;for\;i=1,2,3,4.\] Therefore, \(\tau_i\) \((i=1,2,3,4)\) are the extreme points of \(F_0(x)\). Define \[b'=\min\{F_0(\tau_1),F_0(\tau_3)\}\;and\;a'=\max\{F_0(\tau_2),F_0(\tau_4)\},\] as shown in the Figure 1.
From the fact that \(F(x)\) has \(5\) distinct roots on \(\Omega\), we have \(b'>a'\). Furthermore, as described in [1], for any \(\xi\in [a_0,b_0]\), the equation \[F_0(x)-\frac{1}{5}\xi-C_h=0\] has \(5\) real roots. Let \(b=5\left(b'-C_h\right)\) and \(a=5\left(a'-C_h\right)\). We have \[\operatorname{Im}h=[a_0,b_0]\subset[a,b].\]
For the case \(a_0>a\), \(b_0<b\), combined with the proof in this paper, Theorem 4 can be proved by an argument as in [13]. For the other cases, as described in [1], it suffices to consider that \(a_0=a\) and \(b_0=b\). In this case, we have \(\operatorname{Im}h=[a,b]\) and when \(h=a\) (or \(b\)), \(F_0(x)\) attains the maximum of all the local minimum values (or the minimum of all the local maximum values).
Next, following [16], we perform a region division on the manifold: \(M^5=X\cup Y\cup Z\), where \[\label{space} \begin{align} &X:=\{x\in M^5: h(x)=a \}=h^{-1}(a),\\ &Y:=\{x\in M^5: a<h(x)<b\},\\ &Z:=\{x\in M^5: h(x)=b\}=h^{-1}(b). \end{align}\tag{13}\] Assume \(Y\neq \emptyset\), otherwise the conclusion obviously holds. If \(\Omega\neq \emptyset\), then \(Y\subset \Omega\). Now, we introduce some notations: for \(0<\epsilon<\frac{b-a}{2}\), write \[\begin{align} &X_\epsilon:=\{x\in M^5: a<h(x)<a+\epsilon\},\\ &Y_\epsilon:=\{x\in M^5: a+\epsilon\leq h(x)\leq b-\epsilon\},\\ &Z_\epsilon:=\{x\in M^5: b-\epsilon<h(x)<b\}, \end{align}\] and then \(Y=X_\epsilon\cup Y_\epsilon\cup Z_\epsilon\).
At the end of this subsection, we will present several lemmas used in the proof of Theorem 4. The following lemma gives the differential of the \((n-1)\)-form \(\Phi\).
Lemma 9. For a minimal hypersurface \(M^n\) in the unit sphere \(\mathbf{S}^{n+1}\), we have \[\label{eqdPhi} \begin{align} d\Phi =& (-1)^n \,2\, \Bigg( \left(n-2\right)! \,f_3 + \sum_{\sigma} \frac{\left(\lambda_{i_{n-1} i_n} + \lambda_{i_ni_n}\right) \lambda_{i_{n-1} i_n}}{\lambda_{i_n} - \lambda_{i_{n-1}}} \\ &\qquad\qquad +\sum_{\sigma} \lambda_{i_n} \sum_{k=1; k \neq i_{n-1}, i_n}^{n} \frac{\lambda_{k i_n} \lambda_{i_{n-1}i_n}}{\left(\lambda_k - \lambda_{i_n}\right)\left(\lambda_{i_{n-1}} - \lambda_{i_n}\right)} \Bigg)\cdot vol. \end{align}\qquad{(2)}\]
Proof. The differential of \(\Phi\) can be calculated by parts as follows: \[\label{comdPhi} \begin{align} d\Phi=&\sum_{\sigma}S(\sigma)\,d\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\wedge\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\omega_{i_{n-1}i_{n}}\\ &+\sum_{\sigma}S(\sigma)\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)d\left(\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\omega_{i_{n-1}i_{n}}\right)\\ :=&\phi+\psi. \end{align}\tag{14}\]
Using \((\ref{dlam}-\ref{Gamma})\), we have \[\begin{align} \phi_1&:=\sum_{\sigma}S(\sigma)\,d\lambda_{i_{n-1}}\wedge\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\left(\sum_{k=1}^n\Gamma_{i_{n-1}i_nk}\,\theta_k\right)\\ &=\sum_{\sigma}S(\sigma)\left(\sum_{j=1}^{n}\lambda_{i_{n-1}j}\,\theta_j\right)\wedge\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\left(\sum_{k=1}^n\,\Gamma_{i_{n-1}i_nk}\,\theta_k\right)\\ &=\sum_{\sigma}S\left(\sigma\right)\Bigg(\lambda_{i_{n-1}i_{n-1}}\,\theta_{i_{n-1}}\wedge\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\Gamma_{i_{n-1}i_ni_n}\,\theta_{i_n}\\ &\qquad\qquad\qquad+\lambda_{i_{n-1}i_n}\,\theta_{i_n}\wedge\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\Gamma_{i_{n-1}i_ni_{n-1}}\,\theta_{i_{n-1}}\Bigg)\\ &=(-1)^n\,\sum_{\sigma}\Bigg(\frac{\lambda_{i_{n-1}i_{n-1}}\,\lambda_{i_ni_{n-1} }-\lambda_{i_{n-1} i_n}\,\lambda_{i_{n-1} i_n}}{\lambda_{i_{n-1}}-\lambda_{i_n}}\Bigg)\cdot vol. \end{align}\] Similarly, \[\begin{align} \phi_2:=&\sum_{\sigma}S(\sigma)\,d\lambda_{i_{n}}\wedge\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge\omega_{i_{n-1}i_{n}}\\ =&(-1)^n\,\sum_{\sigma}\Bigg(\frac{\lambda_{i_{n}i_{n-1}}\lambda_{i_ni_{n-1}}-\lambda_{i_ni_n}\lambda_{i_{n-1}i_n}}{\lambda_{i_{n-1}}-\lambda_{i_n}}\Bigg)\cdot vol.\end{align}\] Thus, \[\label{phi} \phi=\phi_1+\phi_2=(-1)^n\,2\,\sum_{\sigma}\left(\frac{\left(\lambda_{i_{n-1} i_n}+\lambda_{i_ni_n}\right)\lambda_{i_{n-1} i_n}}{\lambda_{i_n}-\lambda_{i_{n-1}}}\right)\cdot vol.\tag{15}\]
According to [13], we obtain \[\begin{align} \psi_1:=&\sum_{\sigma}S(\sigma)\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)d\left(\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\right)\wedge\omega_{i_{n-1}i_{n}}\\ =& (-1)^n\, \sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right) \sum_{k=1; k \neq i_{n-1}, i_n}^{n} \Bigg( -\frac{\lambda_{k i_{n-1}}\lambda_{i_n i_{n-1}}}{(\lambda_k - \lambda_{i_{n-1}})(\lambda_{i_{n-1}} - \lambda_{i_n})}\\ &\qquad\qquad\quad+ \frac{h_{k i_{n-1} i_n}^2}{(\lambda_k - \lambda_{i_{n-1}})(\lambda_{i_{n-1}} - \lambda_{i_n})} + \frac{\lambda_{k i_n}\lambda_{i_{n-1} i_n}}{(\lambda_k - \lambda_{i_n})(\lambda_{i_{n-1}} - \lambda_{i_n})} \\ &\qquad\qquad\qquad \qquad\qquad\qquad\qquad \left. - \frac{h^2_{k i_{n-1} i_n}}{(\lambda_k - \lambda_{i_n})(\lambda_{i_{n-1}} - \lambda_{i_n})} \right)\cdot vol, \end{align}\] and since \[\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^{n}\frac{h_{ki_{n-1}i_n}^2}{(\lambda_k-\lambda_{i_{n-1}})(\lambda_{i_{n-1}}-\lambda_{i_n})}=0,\] we get \[\label{psi1} \psi_1=(-1)^n\,2\,\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^{n}\frac{\lambda_{ki_n}\lambda_{i_{n-1}i_n}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{i_n})}\cdot vol.\tag{16}\]
It also follows from [13] that \[\label{psi2} \begin{align} \psi_{2}&=(-1)^n\sum_{\sigma}S(\sigma)\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\\ & \qquad\qquad\qquad \qquad\qquad\qquad \qquad\wedge \left(\sum_{k=1;k\neq i_{n-1},i_n}^{n}\omega_{i_{n-1}k}\wedge\omega_{ki_n}\right)\\ &=(-1)^n\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^{n}\left(\frac{\lambda_{i_{n-1}k}\,\lambda_{i_nk}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{k})}\right.\\ & \qquad\qquad\qquad\qquad\qquad\qquad \qquad\quad \left.-\frac{h_{ki_{n-1}i_n}^2}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{k})}\right)\cdot vol\\ &=(-1)^n\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^{n}\frac{\lambda_{i_{n-1}k}\,\lambda_{i_nk}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{k})}\cdot vol, \end{align}\tag{17}\] where the last equation follows from the following equation: \[\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^{n}\frac{h_{ki_{n-1}i_n}^2}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_k)}=0,\] whose proof can refer to [26].
Moreover, from \((\ref{gauss})\), when \(H=0\) we have \[\sum_{i,j=1;i\neq j}^{n}\left(\lambda_{i}+\lambda_j\right)R_{ijij}=-2f_3.\] Therefore, we obtain \[\label{psi3} \begin{align} \psi_{3}&=(-1)^n\,\sum_{\sigma}S(\sigma)\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\wedge R_{i_{n-1}i_n}\\ &=(-1)^n\,\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)R_{i_{n-1}i_ni_{n-1}i_n}\,\theta_{1}\wedge\theta_{2}\wedge\cdots\wedge\theta_{n-1}\wedge \theta_{n}\\ &=2\,(-1)^{n+1}\,(n-2)!\,f_3\cdot vol. \end{align}\tag{18}\]
By \((\ref{psi1}-\ref{psi3})\), we have \[\label{psi} \begin{align} \psi&=\psi_1+\psi_2-\psi_3\\ &=\sum_{\sigma}S(\sigma)\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)d\left(\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\right)\wedge\omega_{i_{n-1}i_n}\\ &\qquad\qquad\qquad\qquad+(-1)^n\,\theta_{i_1}\wedge\theta_{i_2}\wedge\cdots\wedge\theta_{i_{n-2}}\\ &\qquad\qquad\qquad\qquad\qquad\wedge\left(\sum_{k=1;k\neq i_{n-1},i_n}^{n}\omega_{i_{n-1}k}\wedge\omega_{ki_n}-R_{i_{n-1}i_n}\right)\\ &=(-1)^n\left(2\,\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^n\frac{\lambda_{ki_n}\lambda_{i_{n-1}i_n}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{i_n})}\right.\\ & \qquad\qquad-\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^{n}\frac{\lambda_{i_{n-1}k}\lambda_{i_nk}}{(\lambda_k-\lambda_{i_n})(\lambda_{k}-\lambda_{i_{n-1}})}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad+2\,(n-2)!\,f_3\Bigg)\cdot vol\\ &=(-1)^n\,2\,\sum_{\sigma}\lambda_{i_n}\sum_{k=1;k\neq i_{n-1},i_n}^n\frac{\lambda_{ki_n}\lambda_{i_{n-1}i_n}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{i_n})}\cdot vol.\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad+(-1)^n\,2\,(n-2)!\,f_3\cdot vol\\ \end{align}\tag{19}\] The last equality above holds because \[\begin{align}&2\sum_{\sigma}\lambda_{i_{n-1}}\sum_{k=1;k\neq i_{n-1},i_n}^n\frac{\lambda_{ki_n}\lambda_{i_{n-1}i_n}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{i_n})}\\ &\qquad\qquad\qquad=\sum_{\sigma}\left(\lambda_{i_{n-1}}+\lambda_{i_n}\right)\sum_{k=1;k\neq i_{n-1},i_n}^n \frac{\lambda_{i_{n-1}k}\lambda_{i_nk}}{(\lambda_k-\lambda_{i_n})(\lambda_{i_{n-1}}-\lambda_{i_n})}.\end{align}\]
Finally, combining \((\ref{comdPhi}, \ref{phi},\ref{psi})\), we complete the proof of the lemma. ◻
The following lemma gives the \(5\)-form \(dh\wedge\Phi\). Since the lemma can be obtained by direct calculation, we only provide a brief proof here. For detailed computations, we refer the reader to [27].
Lemma 10. For a hypersurface \(M^5\) in the unit sphere \(\mathbb{S}^6\), we have \[\label{u} dh\wedge\Phi=\sum_{i=1}^{5}u_ih_i^2\cdot vol,\qquad{(3)}\] where \[\label{v} \begin{align} u_i=&-\frac{6}{5}\sum_{k=1;k\neq i }^{5}\frac{\lambda_{k}+\lambda_{i}}{\left(\lambda_{k}-\lambda_{i}\right)^2\prod_{j=1;j\neq k, i}^5\left(\lambda_{j}-\lambda_{i}\right)}\;for\; i=1,2,\cdots,5. \end{align}\qquad{(4)}\]
Proof. We express \(\Phi\) as follows: \[\Phi=6\sum_{i,j=1;i<j}^{5}(\lambda_{i}+\lambda_{j})\theta_k\wedge\theta_l\wedge\theta_m\wedge\omega_{ij}\] for \(k,l,m\) are distinct and for a permutation \(\sigma'(1,2,3,4,5)=(k,l,m,i,j)\), \(S(\sigma')=1\) holds.
According to \((\ref{conform}, \ref{dh})\), through direct calculation, we obtain \[dh\wedge\Phi=6\sum_{i,j=1;i<j}^{5}(\lambda_{i}+\lambda_{j})\left(-h_1\Gamma_{ijj}+h_2\Gamma_{iji}\right)\cdot vol.\] Replacing \(\Gamma_{ijj}\) and \(\Gamma_{iji}\) by \((\ref{coda}, \ref{Gamma}, \ref{lam1})\), we obtain the conclusion. ◻
Using Lemma 10, we now give a lemma on the boundedness of \(u_i\).
Lemma 11. There exists a constant \(C>0\) depending only on \(c_1, c_2, c_3,c_4\), satisfied \[\begin{align} u_i \leq C\;on\;X_\epsilon\;and\;u_i \geq -C\;on \;Z_\epsilon. \end{align}\]
Proof. Define \[h^{-1}(b)=(\beta_1,\beta_2,\beta_3,\beta_4,\beta_5) \;and\; h^{-1}(a)=(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5).\] Then Figure 1 shows that the multiplicities of \(\alpha_{i}\) or \(\beta_i\) \((i=1,2,\cdots,5)\) is at most \(2\) as \(h\rightarrow a\) or \(h\rightarrow b\). However, at a point where there are three distinct principal curvatures with multiplicities \((2, 2, 1)\), we have \(A(r) = 0\) for any \(r = 1, 2,\cdots, 5\), which contradicts \((\ref{Ass})\). Moreover, if \(\lambda_{1}<\lambda_{2}<\lambda_{3}<\lambda_{4}=\lambda_{5}\), then we have \[A(3)=-2\lambda_{5}\prod_{i,j=1,2,5;i<j}{\left(\lambda_{i}-\lambda_{j}\right)}^2pol_{A(3)}^2<0\] which also contradicts assumption \((\ref{Ass})\). Here \(pol_{A(3)}\) is a homogeneous polynomial of degree two in \(\lambda_{i}\) \((i=1,2,5)\). Consequently, such points cannot exist on the hypersurface \(M^5\). Hence, only the following cases are possible.
By Figure 1 , when \(h\rightarrow a\), we have \(\alpha_{1}<\alpha_2=\alpha_{3}<\alpha_{4}<\alpha_{5}\). From \((\ref{v})\), we can write \(u_1\) as \[\begin{align} u_1=&-\frac{6}{5}\Bigg( \frac{pol_{v_1}}{{\left(\lambda_{1}-\lambda_{2}\right)}^2{\left(\lambda_{1}-\lambda_{3}\right)}^2\left(\lambda_{2}-\lambda_{4}\right)\left(\lambda_{2}-\lambda_{5}\right)\left(\lambda_{3}-\lambda_{4}\right)\left(\lambda_{3}-\lambda_{5}\right)}\\ &+\frac{\lambda_{1}+\lambda_{4}}{(\lambda_{1}-\lambda_{4})^2\prod_{i=2,3,5}(\lambda_{i}-\lambda_{4})}+\frac{\lambda_{1}+\lambda_{5}}{(\lambda_{1}-\lambda_{5})^2\prod_{i=2,3,4}(\lambda_{i}-\lambda_{5})}\Bigg) \end{align}\] where \(pol_{v_1}\) is a polynomial of \(\lambda_{i}\) \((i=1,2,\cdots,5)\), thus \(u_1\) is bounded. Similarly, \(u_4\) and \(u_5\) are bounded. For \(u_2,u_3\), from \((\ref{v})\) we know when \(h\rightarrow a\), they tend to \(-\infty\) if \(\alpha_{2}=\alpha_{3}<0\), which can be obtained by the assumption \(A(5)>0\). More precisely, it follows from \[A(5)=-2\alpha_{3}\prod_{i,j=1,3,4;i<j}{\left(\alpha_{i}-\alpha_{j}\right)}^2pol_{A(5)}^2>0\] that \(\alpha_{2}=\alpha_{3}<0\). Here \(pol_{A(5)}\) is a homogeneous polynomial of degree two in \(\alpha_{i}\) \((i=1,3,4)\). Thus, \(u_2\) and \(u_3\) have upper bounds.
By Figure 1, When \(h\rightarrow b\), we have \[\beta_{1}=\beta_2<\beta_{3}<\beta_{4}<\beta_{5}\;or\; \beta_{1}<\beta_2<\beta_{3}=\beta_{4}<\beta_{5}.\] Then, similar to the discussion for \(h\rightarrow a\), we have that \(u_3,u_4,u_5\) \((or\;u_1,u_2,u_5)\) are bounded. Moreover, from \(A(5)>0\), we obtain \(\beta_{1}=\beta_2<0\) \((or\;\beta_{3}=\beta_4<0)\), and therefore, \(u_1,u_2\rightarrow +\infty\) \((or\;u_3,u_4\rightarrow +\infty)\). ◻
The following Lemma 12 was first proved by de Almeida and Brito [16] for the case \(n=3\), and then Tang and Yan [1] pointed out that it also holds for arbitrary dimension \(n\).
Lemma 12. Suppose \(u:M^5\rightarrow\mathbb{R}\) is smooth and \(m=min_{M^5}u\). If \(D_\epsilon=u^{-1}([m,m+\epsilon])\), then \[\lim_{\epsilon\to 0}\int_{D_\epsilon}|\Delta u|\cdot vol=0.\] In particular, \[\lim_{\epsilon\to 0}\int_{M^5-Y_\epsilon}|\Delta h|\cdot vol=0,\;if\;X\cup Z\neq \emptyset.\]
3 Simplification of \(d\Phi\)↩︎
In this section, we give a more explicit expression for \(d\Phi\) when \(n=5\), which will be used in the proof of Theorem 4.
Lemma 13. For a minimal hypersurface \(M^5\) in the unit sphere \(\mathbf{S}^{6}\), we have \[\label{key32form} \begin{align} d\Phi =-12\left( 3\,\sigma_3 +\frac{1}{25\prod_{i,j=1;i<j}^5(\lambda_{i}-\lambda_{j})^2}\sum_{r=1}^5A(r)h_r^2\right)\cdot vol. \end{align}\qquad{(5)}\]
Proof. Let \[\label{letL} \begin{align} d\Phi =& (-1)^n \,2 \,\Big( \left(n-2\right)!\, f_3 + \mathcal{L}\Big)\cdot vol\\ =&(-1)^n \,2\, \left( \left(n-2\right)!\, f_3 + \frac{(n-2)!}{n^2}\sum_{r=1}^nL(r)h_r^2\right)\cdot vol. \end{align}\tag{20}\] Then it follows from \((\ref{eqdPhi})\) and \((\ref{lam1})\) that \[\label{L} \begin{align} \mathcal{L}=&(n-2)!\sum_{\substack{p,q,r=1 \\ p,q,r \text{ are distinct}}}^{n}\lambda_{r}\frac{\lambda_{qr}\lambda_{pr}}{(\lambda_{q}-\lambda_{r})(\lambda_{p}-\lambda_{r})}+(n-2)!\sum_{\substack{p,r=1 \\ p\neq r}}^n\frac{(\lambda_{pr}+\lambda_{rr})\lambda_{pr}}{(\lambda_{r}-\lambda_{p})}\\ =&\frac{(n-2)!}{n^2}\sum_{r=1}^n\left(\sum\limits_{p=1; p\neq r}^{n} \frac{1}{(\lambda_r - \lambda_p) \left( \prod_{k=1; k\neq p}^{n} (\lambda_k - \lambda_p)\right)^2} \right.\\ &\qquad\qquad+ \sum\limits_{p=1; p\neq r}^{n}\frac{1}{(\lambda_r - \lambda_p) \prod_{k=1;k\neq p}^{n} (\lambda_k - \lambda_p) \prod_{l=1;l\neq r}^{n} (\lambda_l - \lambda_r)}\\ &\left.+\sum\limits_{\substack{p,q=1\\ p\neq q;p,q\neq r}}^{n} \frac{\lambda_r}{(\lambda_r-\lambda_p)(\lambda_r-\lambda_q)\prod_{k=1;k\neq p}^{n} (\lambda_k - \lambda_p) \prod_{l=1;l\neq q}^{n} (\lambda_l - \lambda_q)} \right)h_r^2. \end{align}\tag{21}\]
Next, we only consider the case \(n=5\). Firstly, we introduce some notations. Let \[\xi_k=\sum_{i,j=1;i<j;i,j\neq k }^{4}\lambda_{i}\lambda_{j}+\lambda_k\lambda_{5}-(\lambda_k+\lambda_{5})\sum_{i=1;i\neq k}^{4}\lambda_{i}+\lambda_{k}^2+\lambda_{5}^2\] and \[\begin{align} a_k=&-\prod_{i,j=1;i<j;i,j\neq k}^4(\lambda_{i}-\lambda_{j})^2\prod_{i=1;i\neq k}^4(\lambda_{i}-\lambda_{5})\xi_k \end{align}\] for \(k=1,2,3,4.\)
Write \[\begin{align} a^{(kl)}=(-1)^{k+l+1}\prod_{\substack{i,j=1;i<j\\(i,j)\neq (k,l),(k,5),(l,5)}}^5(\lambda_{i}-\lambda_{j})(\lambda_{m}-\lambda_{n})(\lambda_{m}-\lambda_{5})(\lambda_{n}-\lambda_{5}) \end{align}\] for \(k<l\), \(m<n\) and \(\{k,l,m,n\}=\{1,2,3,4\}\).
Then we define \[\mathcal{A}=-2\lambda_{5}\sum_{k,l=1;k<l}^{4}a^{(kl)}+\sum_{k=1}^{4}a_k,\] and after a very complicated series of calculations, we obtain \[\label{eqLl} L(5)=\frac{\mathcal{A}}{ \prod_{i,j=1;i<j}^5(\lambda_{i}-\lambda_{j})^2}.\tag{22}\]
Let \(\lambda=\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}\). Since \(H=0\), substituting \(\lambda_{5}\) with \(-\lambda\) in \(\mathcal{A}\), we obtain \[\begin{align} \mathcal{A} = &2\lambda\sum_{k,l=1;k<l}^4(-1)^{k+l+1}b^{(kl)} -\sum_{k=1}^{4}q_k\prod_{i,j=1;i<j;i,j\neq k}^4(\lambda_{i}-\lambda_{j})^2\prod_{i=1;i\neq k}^{4}(\lambda_{i}+\lambda)\\ \end{align}\]
where \[\begin{align} b^{(kl)}=\prod_{\substack{i,j=1;i<j;(i,j)\neq (k,l)}}^4(\lambda_{i}-\lambda_{j})(\lambda_{m}-\lambda_{n})(\lambda_{m}+\lambda)^2(\lambda_{n}+\lambda)^2 \end{align}\] for \(k<l,m<n\), \(\{k,l,m,n\}=\{1,2,3,4\}\),
and \[\label{qk} \begin{align} q_k &= \lambda_{k}^2 +2\sum_{i=1;i\neq k}^{4}\lambda_{i}^2+\sum_{i=1;i\neq k}^{4}\lambda_{i}\lambda_{k}+5\sum_{i,j=1;i<j;i,j\neq k}^{4}\lambda_{i}\lambda_{j} \end{align}\tag{23}\] for \(k=1,2,3,4.\)
Notice that \[(\lambda+\lambda_{i})(\lambda+\lambda_{j})=q_4-(\lambda_{m}-\lambda_{n})(2\lambda_{i}+2\lambda_{j}+\lambda_{k})\] for \(\{i,j,k\}=\{1,2,3\}\) and \(\{i,j,m,n\}=\{1,2,3,4\}\).
Substituting these three equalities into \(b^{(kl)}\) \((k<l,k,l=1,2,3,4)\), we obtain \[\label{lnew1} \begin{align} &b^{(34)}-b^{(24)}+b^{(14)}= \prod_{i,j=1;i<j}^3(\lambda_i- \lambda_j)^2\left(q_4^2 - (3\lambda-2\lambda_4)\prod_{i=1}^{3}(\lambda_i - \lambda_4) \right). \end{align}\tag{24}\]
More precisely, this is because, when viewing \(q_4\) as a variable, the coefficient of the square term of \(q_4\) in \(b^{(34)}-b^{(24)}+b^{(14)}\) is \[\begin{align} T_{squ} &=\sum_{k=1}^3(-1) ^{k+1}\prod_{\substack{i,j=1;i<j;(i,j)\neq (k,4)}}^4(\lambda_{i}-\lambda_{j})(\lambda_{m}-\lambda_{n})=\prod_{i,j=1;i<j}^3(\lambda_{i}-\lambda_{j})^2, \end{align}\] where \(m<n\), \(\{k,m,n\}=\{1,2,3\}\). Moreover, the coefficients of the mixed term of \(q_4\) and the constant term of \(q_4\) in \(b^{(34)}-b^{(24)}+b^{(14)}\) are respectively
\[\begin{align} T_{mix} =&-2\prod_{i,j=1;i<j}^{4} (\lambda_i - \lambda_j) \left(\sum_{i,j=1;i<j}^{3}(-1)^{i+j+1}(\lambda_{i}-\lambda_{j})\right.\\ &\qquad\qquad\qquad\cdot(2\lambda_{i}+2\lambda_{j}+\lambda_{k})\Bigg)= 0\;for\;\{i,j,k\} =\{1,2,3\} \end{align}\] and \[\begin{align} T_{con} =&\prod_{i,j=1;i<j}^{4} (\lambda_i - \lambda_j) \left(\sum_{k,l=1;k<l}^{3}(-1)^{k+l+1}(\lambda_k - \lambda_l)(\lambda_m- \lambda_n)(2\lambda_k + 2\lambda_l+ \lambda_m)^2\right)\\ = &-\prod_{i,j=1;i<j}^3(\lambda_i- \lambda_j)^2 \prod_{i=1}^3(\lambda_i- \lambda_4) (3\lambda-2\lambda_{4}), \end{align}\] for \(m<n\), \(\{k,l,m,n\}=\{1,2,3,4\}\).
Similarly, we have \[\label{lnew2} \begin{align} &b^{(34)}+b^{(23)}-b^{(13)}= \prod_{i<j;i,j=1,2,4}(\lambda_i- \lambda_j)^2\left(q_3^2 -\prod_{i=1,2,4}(\lambda_i - \lambda_3) (3\lambda-2\lambda_3)\right),\\ &b^{(23)}+b^{(12)}-b^{(24)}= \prod_{i<j;i,j=1,3,4}(\lambda_i- \lambda_j)^2\left(q_2^2+\prod_{i=1,3,4}(\lambda_i - \lambda_2) (3\lambda-2\lambda_2)\right),\\ &b^{(12)}-b^{(13)}+b^{(23)}= \prod_{i<j;i,j=2,3,4}(\lambda_i- \lambda_j)^2\left(q_1^2 -\prod_{i=2,3,4}(\lambda_i - \lambda_1) (3\lambda-2\lambda_1)\right). \end{align}\tag{25}\]
By adding the four equations in \((\ref{lnew1})\) and \((\ref{lnew2})\), and observing that the sum of the terms independent of \(q_k^2\) \((k=1,2,3,4)\) is zero, we have \[\label{l} \mathcal{A}=\sum_{k=1}^{4}\left(\prod_{i,j=1;i<j;i,j\neq k}^4(\lambda_i - \lambda_j)^2\right)\left(\lambda q_k^2-\prod_{i=1;i\neq k}^4(\lambda+\lambda_{i})q_k\right).\tag{26}\]
At last, by some direct calculations, we get
\[\label{eql} \begin{align} \mathcal{A}=\sum_{k=1}^{4}\prod_{\substack{i,j=1;i<j;i,j\neq k}}^4\left(\lambda_{i}-\lambda_{j}\right)^2p_kq_k, \end{align}\tag{27}\] and in which \[\label{p} p_k=-2\lambda\lambda_{k}\left(\lambda-\lambda_{k}\right)-\prod_{i=1;i\neq k}^{4}\lambda_{i}\;for\;k=1,2,3,4\tag{28}\] .
Now we show that \(\mathcal{A}=A(5)\). Firstly, from \(H=0\), we can obtain \[\label{eqp4} p_l=2s_1^{ijk}s_2^{ijk}-3s_3^{ijk}+2\sigma_3,\tag{29}\] \[\label{eqq4new} q_l = 2\left(s^{ijk}_1\right)^2+s^{ijk}_2- \lambda_{l}\lambda_{5}.\tag{30}\] for \(\{i,j,k,l\}=\{1,2,3,4\}\).
Since for \(H=0\), we have \[\label{eqC1} \sigma_2=\lambda_{l}\lambda_{5}-\left(s^{ijk}_1\right)^2+s^{ijk}_2.\tag{31}\] for \(\{i,j,k,l\}=\{1,2,3,4\}\). Hence, from \((\ref{eqq4new})\) and \((\ref{eqC1})\), we get \[\label{eqq4} q_l = \left(s^{ijk}_1\right)^2+2s^{ijk}_2-\sigma_2.\tag{32}\]
Based on the combination of \(( \ref{eql},\ref{eqp4},\ref{eqq4})\), we obtain \[\label{Al} A(5)=\mathcal{A}.\tag{33}\]
Therefore, according to \((\ref{eqLl})\), we have \[L(5)=\frac{A(5)}{ \prod_{i,j=1;i<j}^5(\lambda_{i}-\lambda_{j})^2}.\]
Besides, it follows from \((\ref{L})\) that \(L(r)\) \((r\in I_5)\) is obtained from \(L(5)\) by replacing \(\lambda_{r}\) with \(\lambda_{5}\), and \(A(r)\) \((r\in I_5)\) we given in \((\ref{Ass})\) is also can be obtained from \(A(5)\) in the same way. Hence, we have \[\label{eqAr} L(r)=\frac{A(r)}{ \prod_{i,j=1;i<j}^5(\lambda_{i}-\lambda_{j})^2}.\tag{34}\] for \(r=1,2,\cdots,5\).
Consequently, combining \((\ref{letL}, \ref{eqAr})\), we obtain \[\begin{align} d\Phi =& -12\left( f_3 +\frac{1}{25}\sum_{r=1}^5L(r)h_r^2\right)\cdot vol\\ =& -12 \left( 3\,\sigma_3 +\frac{1}{25\prod_{i,j=1;i<j}^5(\lambda_{i}-\lambda_{j})^2}\sum_{r=1}^5A(r)h_r^2\right)\cdot vol. \end{align}\] ◻
4 proof of Theorems↩︎
4.1 Proof of Theorem 4↩︎
Proof. When \(\Omega\) is empty, it follows from [13] that all principal curvatures of \(M^5\) are constants, and the theorem holds. Therefore, in the following we only discuss the case that \(\Omega\neq \emptyset\).
Under our convention that \(\sigma_3\geq 0\) and \((\ref{Ass}), (\ref{key32form})\), we have \[\label{int1} \int_{Y}-d\Phi \geq 0.\tag{35}\]
As in [16], for any smooth function \(\eta:(a,b)\rightarrow \mathbb{R}\) with compact support, we apply Stokes’ theorem to \[d\left(\left(\eta\circ h\right)\Phi\right)=\left(\eta\circ h\right)d\Phi+\left(\eta'\circ h\right)dh\wedge\Phi\] and obtain
\[\label{int2} \int_{Y}\left(\eta\circ h\right)d\Phi+\int_{Y}\left(\eta'\circ h\right)dh\wedge\Phi =0.\tag{36}\]
Given a small \(\epsilon>0\), we choose a smooth function \(\eta_{\epsilon}:\mathbb{R}\rightarrow\mathbb{R}\) such that
\(0\leq\eta_{\epsilon}\leq 1\);
\(\eta_{\epsilon}(t)=0\) for \(a\leq t \leq a+\frac{\epsilon}{n}\) or \(b-\frac{\epsilon}{n}\leq t \leq b\);
\(\eta_{\epsilon}(t)=1\) for \(a+\epsilon\leq t\leq b-\epsilon\);
\(\eta_{\epsilon}'(t)\geq 0\) on \((-\infty,\frac{a+b}{2})\), and \(\eta_{\epsilon}'(t)\leq 0\) on \((\frac{a+b}{2},+\infty)\).
Then it follows from \(( \ref{u},\ref{int1},\ref{int2})\) that \[\label{int} \begin{align} 0&\leq\int_{Y}-\left(\eta_{\epsilon}\circ h \right)d\Phi =\int_Y\left(\eta_{\epsilon}'\circ h\right )dh\wedge \Phi\\ &=\int_Y\left(\eta_{\epsilon}'\circ h\right)\sum_{i=1}^nu_ih_i^2 \cdot vol\leq\int_{Y}C\left|\eta_{\epsilon}'\circ h\right|\left|dh\right|^2\cdot vol, \end{align}\tag{37}\] where the last inequality follows from Lemma 11.
Moreover, for any smooth function \(\gamma:\mathbb{R}\rightarrow\mathbb{R}\), applying Stokes’ theorem to \[d^*\left(\left(\gamma\circ h\right)dh\right)=\left(\gamma'\circ h\right)\left|dh\right|^2\cdot vol+\left(\gamma\circ h\right)\Delta h \cdot vol\] yields \[\label{int3} \int_{M^5}\left(\gamma'\circ h\right)\left|dh\right|^2\cdot vol+\int_{M^5}\left(\gamma\circ h\right)\Delta h \cdot vol=0.\tag{38}\]
Next, we construct a new smooth function \(\gamma_\epsilon\) from \(\eta_{\epsilon}\), defined as follows. \[\gamma_\epsilon = \begin{cases} \eta_\epsilon - 1 & \text{on } \left( -\infty, \frac{a+b}{2} \right], \\ 1 - \eta_\epsilon & \text{on } \left[ \frac{a+b}{2}, +\infty \right). \end{cases}\] It is obvious that \(\gamma_\epsilon'=\left|\eta_\epsilon'\right|\). Then it follows from \((\ref{int3})\) that \[\int_Y\left|\eta'_\epsilon\circ h\right| \left|dh\right|^2\cdot vol=-\int_{M^5}\left(\gamma_\epsilon\circ h\right)\Delta h \cdot vol\leq \int_{M^5}\left|\gamma_\epsilon\circ h\right|\left|\Delta h\right|\cdot vol.\]
From the construction of \(\gamma_\epsilon\), we have \(|\gamma_\epsilon|\leq 1\) and \(\gamma_\epsilon\circ h=0\) on \(Y_\epsilon\). Thus by Lemma 12, we obtain \[\begin{align} \lim_{\epsilon\to0}\int_{M^5}|\gamma_\epsilon\circ h||\Delta h| \cdot vol=\lim_{\epsilon\to0}\int_{M^5-Y_\epsilon}|\gamma_\epsilon\circ h||\Delta h| \cdot vol\leq\lim_{\epsilon\to0}\int_{M^5-Y_\epsilon}|\Delta h| \cdot vol=0. \end{align}\]
Furthermore, \[\lim_{\epsilon\to0}\int_{Y}\left|\eta_\epsilon'\circ h\right|\left|dh\right|^2\cdot vol=0.\] Together with \((\ref{int})\), we obtain \[\lim_{\epsilon\to0}\int_{Y}-\left(\eta_\epsilon\circ h\right)d\Phi=0.\]
Finally, according to the convention \(\sigma_3\geq 0\), \((\ref{key32form})\) and assumption \((\ref{Ass})\), we have \[0\leq \int_{Y_{\epsilon'}}\frac{12}{25\prod_{i,j=1;i<j}^5(\lambda_{i}-\lambda_{j})^2}\sum_{r=1}^5A(r)h_r^2\cdot vol\leq \int_{Y}-\left(\eta_\epsilon\circ h\right)d\Phi\] for all \(0<\epsilon\leq\epsilon'<\frac{b-a}{2}\). Thus we must have \(h_r=0\) on \(Y\) for any \(r=1,2,\cdots, 5\). Thus, \(h\) is constant on \(Y\), and therefore constant on \(M^5\).
We remark that if \((\lambda_1,\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5})=(-6,-5,1,3,7)\), we have \[A(i) > 0\,(i\in I_5)\;and\;A(5) < 0,\] so not all \(\lambda_i\) \((i=1,2,\cdots,5)\) satisfy \((\ref{Ass})\). ◻
4.2 proof of Corollary 5↩︎
Proof. We need only show that for the four principal curvature configurations \((1)\), \((2)\), \((3)\) and \((4)\), the inequality \(A(r)>0\) holds for every \(r=1,2,\cdots,5\).
Case (1): \(\lambda_{1}<\lambda_{2}<\lambda_{3}=\lambda_{4}<0<\lambda_{5}\). In this case, substituting \(\lambda_{4}\) for \(\lambda_{3}\) in \(A(5)\), we get \[\begin{align} A(5)=-2\lambda_{4}\prod_{i,j=1,2,4;i<j}{\left(\lambda_{i}-\lambda_{j}\right)}^2pol_1^2, \end{align}\] where \[pol_1=2{\lambda_{1}}^2+5\lambda_{1}\lambda_{2}+6\lambda_{1}\lambda_{4}+2{\lambda_{2}}^2+6\lambda_{2}\lambda_{4}+4{\lambda_{4}}^2>0.\]
Besides, \(A(1)\) and \(A(2)\) are the results of replacing either \(\lambda_{1}\) or \(\lambda_{2}\) in \(A(5)\) with \(\lambda_{5}\), respectively. Thus it is clear that \(A(1), A(2)\) and \(A(5) >0\).
Next, replacing \(\lambda_{5}\) with \(-\lambda_{1}-\lambda_{2}-2\lambda_{4}\), we obtain \[\begin{align} A(3)=-{\left(\lambda_{1}-\lambda_{2}\right)}^2pol_1^2\cdot pol_2\left(\lambda_1,\lambda_{2},\lambda_{4}\right), \end{align}\] where \[\begin{align} pol_2\left(\lambda_1,\lambda_{2},\lambda_{4}\right)=&{\lambda_{1}}^4\lambda_{2}+{\lambda_{1}}^4\lambda_{4}+2{\lambda_{1}}^3{\lambda_{2}}^2+6{\lambda_{1}}^3\lambda_{2}\lambda_{4}+4{\lambda_{1}}^3{\lambda_{4}}^2+2{\lambda_{1}}^2{\lambda_{2}}^3\\ &+9{\lambda_{1}}^2{\lambda_{2}}^2\lambda_{4}-5{\lambda_{1}}^2\lambda_{2}{\lambda_{4}}^2-4{\lambda_{1}}^2{\lambda_{4}}^3+\lambda_{1}{\lambda_{2}}^4+6\lambda_{1}{\lambda_{2}}^3\lambda_{4}\\ &-5\lambda_{1}{\lambda_{2}}^2{\lambda_{4}}^2-34\lambda_{1}\lambda_{2}{\lambda_{4}}^3-16\lambda_{1}{\lambda_{4}}^4+{\lambda_{2}}^4\lambda_{4}+4{\lambda_{2}}^3{\lambda_{4}}^2\\ &-4{\lambda_{2}}^2{\lambda_{4}}^3-16\lambda_{2}{\lambda_{4}}^4+47{\lambda_{4}}^5, \end{align}\] and \(A(4)\) is obtained by replacing \(\lambda_{4}\) in \(A(3)\) with \(\lambda_{3}\).
Now we claim that \(pol_2(\lambda_1,\lambda_{2},\lambda_{4})<0\), and thus \(A(3)>0\). Note that \(pol_2\) is a polynomial of degree \(5\), so it suffices to prove that \(f(\lambda') = -pol_2(\lambda_1,\lambda_{2},\lambda_{4}) > 0\) for \(\lambda'=(\lambda_1',\lambda_{2}',\lambda_{4}')\) with \(\lambda_1' > \lambda_2' > \lambda_4' > 0\), where we set \[\lambda_{1}'=-\lambda_{1},\lambda_{2}'=-\lambda_{2}\;and\;\lambda_{4}'=-\lambda_{4}.\]
Let \[x_1=\frac{\lambda_{1}'}{\lambda_{4}'}\;and\;x_2=\frac{\lambda_{2}'}{\lambda_{4}'}.\] Then \(x_1 > x_2 > 1\), and \(f(\lambda') =( \lambda_4')^5 g\left(x_1,x_2\right)\), where \[\begin{align} g\left(x_1,x_2\right)=& {x_{1}}^4x_{2}+{x_{1}}^4+2{x_{1}}^3{x_{2}}^2+6{x_{1}}^3x_{2}+4{x_{1}}^3+2{x_{1}}^2{x_{2}}^3\\ &+9{x_{1}}^2{x_{2}}^2-5{x_{1}}^2x_{2}-4{x_{1}}^2+x_{1}{x_{2}}^4+6x_{1}{x_{2}}^3-5x_{1}{x_{2}}^2\\ &-34x_{1}x_{2}-16x_{1}+{x_{2}}^4+4{x_{2}}^3-4{x_{2}}^2-16x_{2}+47. \end{align}\]
Therefore, it only remains to show that \(g(x_1,x_2) > 0\) for all \(x_1 > x_2 > 1\). By direct calculation, \[\begin{align} &\frac{\partial g}{\partial x_1}=4{x_{1}}^3x_{2}+4{x_{1}}^3+6{x_{1}}^2{x_{2}}^2+18{x_{1}}^2x_{2}+12{x_{1}}^2\\ &\qquad\quad+4x_{1}{x_{2}}^3+18x_{1}{x_{2}}^2-10x_{1}x_{2}-8x_{1}+{x_{2}}^4\\ &\qquad\quad+6{x_{2}}^3-5{x_{2}}^2-34x_{2}-16,\\ &\frac{\partial^2g}{\partial x_1^2}=12{x_{1}}^2x_{2}+12{x_{1}}^2+12x_{1}{x_{2}}^2+36x_{1}x_{2}+24x_{1}\\ &\qquad\quad+4{x_{2}}^3+18{x_{2}}^2-10x_{2}-8. \end{align}\]
Since \(x_1>x_2>1\), it is obvious that \(\frac{\partial^2g}{\partial x_1^2}>0\). Thus \(\frac{\partial g}{\partial x_1}\) is strictly increasing in \(x_1\), and hence \[\begin{align} \frac{\partial g}{\partial x_1}&\left(x_1,x_2\right)>\frac{\partial g}{\partial x_1}\left(x_2,x_2\right)\\ &=\left(x_2-1\right)\left(15{x_{2}}^3+61{x_{2}}^2+58x_{2}+16\right)>0. \end{align}\]
Thus, it follows that \[\begin{align} g&(x_1,x_2)>g(x_2,x_2)\\ &\qquad={\left(x_{2}-1\right)}^2\left(6{x_{2}}^3+35{x_{2}}^2+62x_{2}+47\right)>0. \end{align}\]
Therefore, the claim holds. Completely analogously, we can prove that \(A(4)>0\). Finally, we point out that in this case, \(\sigma_3\geq 0\) is obvious.
Case (2): \(\lambda_{1}=\lambda_{2}<\lambda_{3}<\lambda_{4}<\lambda_{5}\). In this case, we first prove that \(\sigma_3\geq 0\). Substituting \(\lambda_{5}=-\lambda\) and \(\lambda_{1}=\lambda_{2}\) into \(\sigma_3\), we get \[\sigma_3=-\left(2\lambda_{2}+\lambda_{3}\right){\lambda_{4}}^2-\left(4{\lambda_{2}}^2+4\lambda_{2}\lambda_{3}+{\lambda_{3}}^2\right)\lambda_{4}-2\lambda_{2}{\lambda_{3}}^2-4{\lambda_{2}}^2\lambda_{3}-2{\lambda_{2}}^3,\]
In the situation \(\lambda_{3}<\lambda_{4}<0<\lambda_{5}\), the conclusion holds trivially.
In the situation \(\lambda_{3}<0<\lambda_{4}<\lambda_{5}\), the conclusion follows from \(-\left(2\lambda_{2}+\lambda_{3}\right)>0\) and \(\Delta=-8{\lambda_{2}}^3\lambda_{3}-8{\lambda_{2}}^2{\lambda_{3}}^2+{\lambda_{3}}^4<0.\)
Then we prove that \(A(r)>0\) for any \(r=1,2,\cdots,5\). Similar to the Case (1), substituting \(\lambda_{1}\) with \(\lambda_{2}\), we can directly obtain \(A(3),A(4), A(5)>0\) and for \(A(1)\) \((or\;A(2))\), replacing \(\lambda_{2}\) \((or\;\lambda_{1})\) with \(-\frac{1}{2}(\lambda_{3}+\lambda_{4}+\lambda_{5})\), we get \[32A(1)=32A(2)=\prod_{i,j=3,4,5;i<j}{({\lambda}_{i}-{\lambda}_{j})}^{2}pol_3(\lambda_3,\lambda_{4},\lambda_{5}),\] where \[\begin{align} pol_3(\lambda_3,\lambda_{4},\lambda_{5})=&47{\lambda}_{3}^{5}+267{\lambda}_{3}^{4}{\lambda}_{4}+267{\lambda}_{3}^{4}{\lambda}_{5}+582{\lambda}_{3}^{3}{\lambda}_{4}^{2}+1060{\lambda}_{3}^{3}{\lambda}_{4}{\lambda}_{5}+582{\lambda}_{3}^{3}{\lambda}_{5}^{2}\\ &+582{\lambda}_{3}^{2}{\lambda}_{4}^{3}+1570{\lambda}_{3}^{2}{\lambda}_{4}^{2}{\lambda}_{5}+1570{\lambda}_{3}^{2}{\lambda}_{4}{\lambda}_{5}^{2}+582{\lambda}_{3}^{2}{\lambda}_{5}^{3}+267{\lambda}_{3}{\lambda}_{4}^{4}\\ &+1060{\lambda}_{3}{\lambda}_{4}^{3}{\lambda}_{5}+1570{\lambda}_{3}{\lambda}_{4}^{2}{\lambda}_{5}^{2}+1060{\lambda}_{3}{\lambda}_{4}{\lambda}_{5}^{3}+267{\lambda}_{3}{\lambda}_{5}^{4}+47{\lambda}_{4}^{5}\\ &+267{\lambda}_{4}^{4}{\lambda}_{5}+582{\lambda}_{4}^{3}{\lambda}_{5}^{2}+582{\lambda}_{4}^{2}{\lambda}_{5}^{3}+267{\lambda}_{4}{\lambda}_{5}^{4}+47{\lambda}_{5}^{5}. \end{align}\]
Thus it is clear that \(A(1)=A(2)>0\) when \(0<\lambda_{3}<\lambda_{4}<\lambda_{5}\). Now we claim that \(pol_3(\lambda_3,\lambda_{4},\lambda_{5})>0\) when \(\lambda_{3}<0<\lambda_{4}<\lambda_{5}\) or \(\lambda_{3}<\lambda_{4}<0<\lambda_{5}\) and then \(A(1)=A(2)>0\). Firstly, we notice that \[0>\lambda_{3}>-\frac{1}{3}(\lambda_{4}+\lambda_{5}), \lambda_{4}>-\frac{1}{4}\lambda_{5}\;and\;\lambda_{4}+\lambda_{5}>0.\] The third partial derivative of \(pol_3(\lambda_3,\lambda_{4},\lambda_{5})\) with respect to \(\lambda_{3}\) is \[\begin{align} \frac{\partial^3 pol_3}{\partial \lambda_{3}}=&2820{\lambda_{3}}^2+6408\lambda_{3}\left(\lambda_{4}+\lambda_{5}\right)+3492{\lambda_{4}}^2+6360\lambda_{4}\lambda_{5}+3492{\lambda_{5}}^2\\ >&2820{\lambda_{3}}^2-2160\left(\lambda_{4}+\lambda_{5}\right)^2+3492{\lambda_{4}}^2+6360\lambda_{4}\lambda_{5}+3492{\lambda_{5}}^2\\ =&2820{\lambda_{3}}^2+1020\left(\lambda_{4}+\lambda_{5}\right)^2+312{\lambda_{4}}^2+312{\lambda_{5}}^2 >0, \end{align}\] thus \[\begin{align} \frac{\partial^2 pol_3}{\partial \lambda_{3}}&>\frac{\partial^2 pol_3}{\partial \lambda_{3}}\left(-\frac{1}{3}(\lambda_{4}+\lambda_{5}),\lambda_{4},\lambda_{5}\right)\\ &=\frac{8672\lambda_{4}^3}{27}+\frac{7376{\lambda_{4}}^2\lambda_{5}}{9}+\frac{7376\lambda_{4}{\lambda_{5}}^2}{9}+\frac{8672{\lambda_{5}}^3}{27}\\ &=\left(\lambda_{4}+\lambda_{5}\right)\left(\frac{8672{x_{4}}^2}{27}+\frac{13456x_{4}x_{5}}{27}+\frac{8672{x_{5}}^2}{27}\right)>0. \end{align}\]
From this we obtain \[\begin{align} \frac{\partial pol_3}{\partial \lambda_{3}}&> \frac{\partial pol_3}{\partial \lambda_{3}}\left(-\frac{1}{3}(\lambda_{4}+\lambda_{5}),\lambda_{4},\lambda_{5}\right)\\ &=\frac{2944{\lambda_{4}}^4}{81}+\frac{17824{\lambda_{4}}^3\lambda_{5}}{81}+\frac{9488{\lambda_{4}}^2{\lambda_{5}}^2}{27}+\frac{17824\lambda_{4}{\lambda_{5}}^3}{81}+\frac{2944{\lambda_{5}}^4}{81}\\ &=\frac{16}{81}\lambda_{5}^4\left(\frac{\lambda_{4}}{\lambda_{5}}+1\right)\left(4\frac{\lambda_{4}}{\lambda_{5}}+1\right)\left(46\left(\frac{\lambda_{4}}{\lambda_{5}}\right)^2+ 83\frac{\lambda_{4}}{\lambda_{5}} + 46\right)>0, \end{align}\] therefore \[\begin{align} pol_3\left(\lambda_3,\lambda_{4},\lambda_{5}\right)&>pol_3\left(-\frac{1}{3}(\lambda_{4}+\lambda_{5}),\lambda_{4},\lambda_{5}\right)\\ &=\frac{64\left(\lambda_{4}+\lambda_{5}\right){\left(4{\lambda_{4}}^2+17\lambda_{4}\lambda_{5}+4{\lambda_{5}}^2\right)}^2}{243}>0. \end{align}\] The claim holds.
Case (3): \(\lambda_{1}<\lambda_{2}=\lambda_{3}<\lambda_{4}<0<\lambda_{5}\). Likewise, as in Case (1), substituting \(\lambda_{2}\) with \(\lambda_{3}\), we can directly obtain \(A(1),A(4),A(5)>0\), and in \(A(2)\) \((or\;A(3))\), replacing \(\lambda_{5}\) by \(-\lambda_{1}-2\lambda_{3}-\lambda_{4}\) \((or\;\lambda_{1}-2\lambda_{2}-\lambda_{4})\), we obtain \[\begin{align} A(2)=- (\lambda_1 - \lambda_4)^2 pol^2\cdot pol_4\left(\lambda_1,\lambda_{3},\lambda_{4}\right), \end{align}\] where \[\begin{align} &pol=2\lambda_1^2 + 6\lambda_1\lambda_3 + 5\lambda_1\lambda_4 + 4\lambda_3^2 + 6\lambda_3\lambda_4 + 2\lambda_4^2,\\ &pol_4\left(\lambda_1,\lambda_{3},\lambda_{4}\right)=\lambda_1^4\lambda_3 + \lambda_1^4\lambda_4 + 4\lambda_1^3\lambda_3^2 + 6\lambda_1^3\lambda_3\lambda_4+ 2\lambda_1^3\lambda_4^2- 4\lambda_1^2\lambda_3^3 \\ &\qquad\qquad\qquad\qquad- 5\lambda_1^2\lambda_3^2\lambda_4 + 9\lambda_1^2\lambda_3\lambda_4^2+ 2\lambda_1^2\lambda_4^3- 16\lambda_1\lambda_3^4 - 34\lambda_1\lambda_3^3\lambda_4 \\ &\qquad\qquad\qquad\qquad- 5\lambda_1\lambda_3^2\lambda_4^2 + 6\lambda_1\lambda_3\lambda_4^3 + \lambda_1\lambda_4^4+ 47\lambda_3^5 - 16\lambda_3^4\lambda_4\\ &\qquad\qquad\qquad\qquad - 4\lambda_3^3\lambda_4^2 + 4\lambda_3^2\lambda_4^3 + \lambda_3\lambda_4^4, \end{align}\] and \(A(3)\) is obtained by replacing \(\lambda_{3}\) in \(A(2)\) with \(\lambda_{2}\).
Since \(pol_4\left(\lambda_1,\lambda_{3},\lambda_{4}\right)\) is homogeneous, we can set \(d=-\lambda_{1},\lambda_3=-1,c=-\lambda_{4}\), so that \(d>1>c>0\) and \(pol_4\left(\lambda_1,\lambda_{3},\lambda_{4}\right)=-f(c,d)\), where \[\begin{align} f(c,d)=&c^4d+c^4+2c^3d^2+6c^3d+4c^3+2c^2d^3+9c^2d^2\\ &-5c^2d-4c^2+cd^4+6cd^3-5cd^2-34cd\\ &-16c+d^4+4d^3-4d^2-16d+47. \end{align}\]
Let \(a=d-1>0,\, b=1-c\in (0,1)\). Then we get \[\begin{align} &g(a,b)=f(c,d)=\left(2-b\right)a^4+\left(2b^2-14b+20\right)a^3\\ &\qquad+\left(-2b^3+21b^2-55b+50\right)a^2+\left(b^4-14b^3+55b^2-50b\right)a\\ &\qquad\quad+2b^4-20b^3+50b^2. \end{align}\]
Since \[2-b>0,\,2b^2-14b+20>0,\, -2b^3+21b^2-55b+50>0\] and\[\begin{align} \Delta=&\left(b^4-14b^3+55b^2-50b\right)^2-4\left(-2b^3+21b^2-55b+50\right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\left(2b^4-20b^3+50b^2\right)\\ =&-b^2\left(b - 5\right)^2\left(- b^4 + 2b^3 + 67b^2 - 260b + 300\right)<0, \end{align}\] we have \(g(a,b)>0\), and therefore \(A(2)>0\). Analogously, \(A(3)>0\).
Case (4): \(\lambda_{1}<\lambda_{2}<\lambda_{3}<\lambda_{4}<0<\lambda_{5}\). In this case, it is clear that \(A(5) > 0\) by Lemma 8. For the other \(A(r)\) \((r=1,2,3,4)\), we only prove \(A(4) > 0\), and the others can be obtained similarly.
Replacing \(\lambda_5\) by \(-\lambda\), Lemma 8 yields
\[\label{A4} \begin{align} A(4) =& \lambda\prod_{i,j=1;i<j}^3\left(\lambda_{i}-\lambda_{j}\right)^{2}q_4^2-\sum_{k=1}^{3}\lambda_{k}\rho_k^2 \left(\lambda_{m}-\lambda_{n}\right)^{2}\left(\lambda_m+\lambda\right)^{2}\left(\lambda_{n}+\lambda\right)^{2} \end{align}\tag{39}\] for \(\{m,n,k \}= \{1,2,3\}\), where \[\rho_k=\sum_{i=1;i\neq k}^{3}\left(\lambda_{i}^2+\lambda_{i}\lambda_{4}\right)+\sum_{i,j=1,i<j}^{3}\lambda_{i}\lambda_{j}-2\left(\lambda_k^2+\lambda_{4}^2\right)-3\lambda_{k}\lambda_{4}\] for \(k=1,2,3\).
Write \[\begin{align} pol_5=&-\lambda_{2}(\lambda+\lambda_{3})^{2}(\lambda+\lambda_{1})^{2}\rho_2^2+ (\lambda_{1}+\lambda_{2})(\lambda_{1}-\lambda_{2})^{2}(\lambda_{2}-\lambda_{3})^{2}q_4^2,\\ pol_6=&- \lambda_{3}(\lambda+\lambda_{2})^{2}(\lambda+\lambda_{1})^{2}\rho_3^2+4\lambda_{3}(\lambda_{1}-\lambda_{3})^{2}(\lambda_{2}-\lambda_{3})^{2}q^2_4.\; \end{align}\] We now prove separately that \(pol_5 > 0\) and \(pol_6 > 0\). Let \[d=-\lambda_{1},\quad c=-\lambda_{2}, \quad b=-\lambda_{3},\quad a=-\lambda_{4}.\] Then \(d>c>b>a>0\) and we have \[\begin{align} pol_5=&c{\left(a+b+c+2d\right)}^2{\left(a+2b+c+d\right)}^2\\ &\qquad\cdot{\left(-2a^2+ab-3ac+ad+b^2+bc+bd-2c^2+cd+d^2\right)}^2\\ &-\left(c+d\right){\left(b-c\right)}^2{\left(c-d\right)}^2\\ &\qquad\cdot{\left(a^2+ab+ac+ad+2b^2+5bc+5bd+2c^2+5cd+2d^2\right)}^2, \end{align}\] \[\begin{align} pol_6=&b\left (a+b+c+2d\right)^2 \left(a+b+2c+d\right)^2\left(pol^{(1)}_{6}\right)^2 - 4b \left(b-c\right)^2 \left(b-d\right)^2\left(pol^{(2)}_{6}\right)^2.\\ \end{align}\] where \[pol^{(1)}_{6}=-2a^2-3ab+ac+ad-2b^2+bc+bd+c^2+cd+d^2>0,\] \[pol^{(2)}_{6}=a^2+ab+ac+ad+2b^2+5bc+5bd+2c^2+5cd+2d^2>0.\]
Therefore, for \(pol_6>0\), it suffices to show that \[\begin{align} pol_{6}^{fac} =&\left(a+b+c+2d\right) \left(a+b+2c+d\right)pol^{(1)}_{6}\\ &- 2\left(b-c\right) \left(b-d\right)pol^{(2)}_{6}>0. \end{align}\]
Taking the second derivative of both sides of \(pol_{6}^{fac}\) with respect to \(d\) yields \[\begin{align} &\frac{\partial^2 pol_{6}^{fac} }{\partial d}=12 b^2 + 34 b c + 54 b d + 8 a b - 2 c^2 \\ &\qquad\qquad+ 18 c d + 22 a c + 24 d^2 + 30 a d >0. \end{align}\] Thus \[\begin{align} \frac{\partial pol_{6}^{fac} }{\partial d}>&\left.\frac{\partial pol_{6}^{fac} }{\partial d}\right|_{d=c}=-5 a^3 - 10 a^2 b - 5 a^2 c\\ &- 12 a b^2 + 9 a b c + 48 a c^2 - 11 b^3 \\ &\quad+ 15 b^2 c + 78 b c^2 + 18 c^3 >0. \end{align}\]
Consequently, we have \[\begin{align} pol_{6}^{fac} >&\left. pol_{6}^{fac} \right|_{d=c}=(-2a^4-7a^3b+9ac^3)\\ &+(-20a^2bc+9abc^2+11bc^3)+(-6b^4+bc^3+4b^2c^2+bc^3)\\ &+(-10a^3c+10ac^3)+(-12a^2b^2+ac^3+11b^2c^2)+(-9ab^3+9bc^3)\\ &+(-5a^2c^2+4bc^3+ac^3)+(-24ab^2c+13bc^3+11ac^3)\\ &+(-22b^3c+9c^4+13bc^3) >0. \end{align}\]
For \(pol_5>0\), by direct calculation, we obtain \[\frac{\partial^7 pol_5}{\partial d}=20160\left(5ac+9bc+8cd-b^2+4c^2\right)>0,\] and thus \[\begin{align} \frac{\partial^6 pol_5}{\partial d}>&\left.\frac{\partial^6 pol_5}{\partial d}\right|_{d=c}=2880\left( 6.25a^2 c - a b^2 + 32.5a b c\right. \\ &\left.+48.5 a c^2 - 5b^3 + 20.25b^2 c + 96.5 b c^2 + 58.25 c^3\right)>0. \end{align}\]
By analogy, \[\begin{align} \frac{\partial^5 pol_5}{\partial d}>&\left.\frac{\partial^5 pol_5}{\partial d}\right|_{d=c}> 600\left(117c^4-4a^3c-a^2b^2-2.8ab^3-6.6b^4\right)>0,\\ \frac{\partial^4 pol_5}{\partial d}>&\left.\frac{\partial^4 pol_5}{\partial d}\right|_{d=c}>-1392a^4c-48a^3b^2-2784a^3bc\\ &-7200a^3c^2-288a^2b^3-1752a^2c^3-336ab^4-480b^5+19224c^5>0,\\ \frac{\partial^3 pol_5}{\partial d}>&\left.\frac{\partial^3 pol_5}{\partial d}\right|_{d=c}>-120 a^5 c - 6 a^4 b^2 - 1236 a^4 b c - 2058 a^4 c^2 - 12 a^3 b^3 \\ &- 1452 a^3 b^2 c - 7128 a^3 bc^2 - 7308 a^3 c^3 - 30 a^2 b^4 - 3360 a^2 b c^3 \\ &- 7008 a^2 c^4 - 24 a b^5 - 24 b^6 + 38142 b^2 c^4 + 24012 b c^5 + 3402 c^6>0,\\ \frac{\partial^2 pol_5}{\partial d}>&\left.\frac{\partial^2 pol_5}{\partial d}\right|_{d=c}>2c\Big[- 22 a^5 b - 289 a^4 b^2 - 792 a^4 b c - 444 a^4 c^2- 242 a^3 b^3 \\ &+\left(- 1752 a^3 c^3 - 3392 a^2 b c^3+ 5444 b^3 c^3\right) +\left(- 468 a c^5+ 2196 b c^5 \right)\\ &- 1718 a^3 b^2 c+\left(- 2023 a^2 c^4 + 5342 b^2 c^4 \right)- 3238 a^3 b c^2 - 1065 a^2 b^2 c^2\\ &+ 2777 b^4 c^2 + 162 c^6\Big]>0,\\ \frac{\partial pol_5}{\partial d}>&\left.\frac{\partial pol_5}{\partial d}\right|_{d=c}=c (a-b)^2 (2a+b+2c)^2 (2a+4b+4c) (a+b+3c)^2\\ & - 2c (a-b) (a+2b+2c)^2 (2a+b+2c) (a+b+3c)^3\\ & + c (a-b)^2 (a+2b+2c)^2 (2a+b+2c)^2 (4a+4b+12c)>0, \end{align}\] and hence we have \[pol_5>\left.pol_5\right|_{d=c}= c{\left(a-b\right)}^2{\left(a+2b+2c\right)}^2{\left(2a+b+2c\right)}^2{\left(a+b+3c\right)}^2>0.\]
In summary, it follows from \((\ref{A4})\) that \[A(4)> \left(\lambda_{1}-\lambda_{3}\right)^2pol_5+\left(\lambda_{1}-\lambda_{2}\right)^2pol_6>0.\] ◻
4.3 Proof of Theorem 6↩︎
Proof. We first point out that, from Newton’s formula, it follows that the constancy of \(H_3\) is equivalent to that of \(f_3\), and when the scalar curvature \(R\) is constant, \(H_4\) is constant if and only if \(f_4\) is constant. By our convention \(\sigma_3\geq 0\), there are no points whose principal curvatures are of the two types: \(\lambda_{1}<\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}\) and \(\lambda_{1}=\lambda_{2}<\lambda_{3}=\lambda_{4}=\lambda_{5}\). Thus we only have the following two cases.
Case (1): One is a triple real root, and the other is a double real root. In this case, we assume \[\lambda_{1}=\lambda_{2}=\lambda_{3}=\zeta<\lambda_{4}=\lambda_{5}=-\frac{3}{2}\zeta.\]
After taking the covariant derivative on both sides of conditions \(H=0\) and \(S=constant\), we have \[\begin{cases} h_{11k} +h_{22k}+h_{33k}+h_{44k}+h_{55k}=0,\\ \zeta\left(h_{11k} +h_{22k}+h_{33k}\right)-\frac{3}{2}\zeta\left(h_{44k}+h_{55k}\right)=0, \end{cases}\] Thus \[h_{11k} +h_{22k}+h_{33k}=0\;and\;h_{44k}+h_{55k}=0\; for\;k=1,2,\cdots,5.\]
Since \(H=0\) and \(S\) is constant, we get \((H)_{mm}=0\) and \((S)_{mm}=0\), so we have \[\label{H2} \sum_{i=1}^{5}h_{iimm}=0\;for\;m=1,2,\cdots,5.\tag{40}\] \[\label{S2} \sum_{i=1}^{5}h_{ii}h_{iimm}+\sum_{i,j=1}^{5}h_{ijm}^2=0\;for\;m=1,2,\cdots,5.\tag{41}\]
For convenience, we write \[Y_1=h_{114}^2+h_{115}^2+h_{124}^2+h_{125}^2+h_{134}^2+h_{135}^2.\] Taking \(m=1\) in \((\ref{H2})\) and \((\ref{S2})\), we get \[\label{S} \begin{align} \sum_{i=1}^{5}h_{ii}h_{ii11}+\sum_{i,j=1}^{5}h_{ij1}^2=&\zeta\left(h_{1111} +h_{2211}+h_{3311}\right)-\frac{3}{2}\zeta\left(h_{4411}+h_{5511}\right)\\ &+h_{111}^2+h_{221}^2+h_{331}^2+h_{441}^2+h_{551}^2\\ &+2\left(h_{112}^2+h_{113}^2+h_{123}^2+h_{145}^2+Y_1\right)\\ =&-\frac{5}{2}\zeta\left(h_{4411}+h_{5511}\right)+h_{111}^2+h_{221}^2+h_{331}^2+h_{441}^2\\ &+h_{551}^2+2\left(h_{112}^2+h_{113}^2+h_{123}^2+h_{145}^2+Y_1\right)\\ =&0. \end{align}\tag{42}\]
Next, since \(f_4=\sum_{i,j,k,l=1}^{5}h_{ij}h_{jk}h_{kl}h_{li}\) is constant, we obtain \((f_4)_{mm}=0\), i.e., \[\label{f42} \sum_{i=1}^{5}\lambda_{i}^3h_{iimm}+3\sum_{i,j=1}^{5}\lambda_{i}^2h_{ijm}^2=0\;for\;m=1,2,\cdots,5.\tag{43}\] Taking \(m=1\) in \((\ref{H2})\) and \((\ref{f42})\), we get \[\label{f4} \begin{align} \sum_{i=1}^{5}\lambda_{i}^3h_{ii11}+&\sum_{i,j=1}^{5}3\lambda_{i}^2h_{ij1}^2=\zeta^3(h_{1111} +h_{2211}+h_{3311})-\frac{27}{8}\zeta^3(h_{4411}+h_{5511})\\ &+3\zeta^2\sum_{i}(h_{11i}^2+h_{12i}^2+h_{13i}^2+\frac{9}{4}h_{14i}^2+\frac{9}{4}h_{15i}^2)\\ =&-\frac{35}{8}\zeta^3\left(h_{4411}+h_{5511}\right)+3\zeta^2\Big[h_{111}^2+h_{221}^2+h_{331}^2+2\big(h_{112}^2+h_{113}^2\\ &+h_{123}^2\big)+\frac{9}{4}\left(h_{441}^2+h_{551}^2\right)+\frac{9}{2}h_{145}^2+\frac{13}{4}Y_1\Big]\\ =&0. \end{align}\tag{44}\]
Then subtracting \(\frac{7}{4}\zeta^2(\ref{S})\) from \((\ref{f4})\), we get \[\frac{5}{4}\left(h_{111}^2+h_{221}^2+h_{331}^2\right)+5\left(h_{441}^2+h_{551}^2\right)+\frac{5}{2}\left(h_{112}^2+h_{113}^2+h_{123}^2\right)+10h_{145}^2+\frac{25}{4}Y_1=0.\] Thus \[\begin{cases} h_{111}=h_{221}=h_{331}=h_{441}=h_{551}=0,\\ h_{112}=h_{113}=h_{123}=h_{145}=0,\\ h_{114}=h_{115}=h_{124}=h_{125}=h_{134}=h_{135}=0. \end{cases}\]
By the same argument, for \(m=5\), we obtain \[h_{552}=h_{553}=h_{554}=h_{555}=h_{225}=h_{335}=h_{445}=0\] and \[h_{235}=h_{245}=h_{345}=0.\]
For \(m=2\), we obtain \[h_{222}=h_{332}=h_{442}=h_{223}=h_{224}=h_{225}=h_{234}=0.\]
For \(m=3\), we obtain \[h_{333}=h_{443}=h_{334}=0.\]
For \(m=4\), we obtain \[h_{444}=0.\]
Case (2): One is a simple real root, and the other is a quadruple real root. In this case, we assume \[\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=\mu<\lambda_{5}=-4\mu.\]
Similar to Case (1), taking the covariant derivatives of \(H=0\) and \(S=constant\) yields
\[\begin{cases} h_{11k} +h_{22k}+h_{33k}+h_{44k}+h_{55k}=0,\\ \mu\left(h_{11k} +h_{22k}+h_{33k}+h_{44k}\right)-4\mu h_{55k}=0, \end{cases}\] thus \[h_{11k} +h_{22k}+h_{33k}+h_{44k}=0\;and\;h_{55k}=0\;for\;k=1,2,\cdots,5.\]
Since \(f_3=\sum_{i,j,k=1}^{5}h_{ij}h_{jk}h_{ki}\) is constant, we have \((f_3)_{mm}=0\), and we get \[\label{case2f3} \sum_{i=1}^{5}\lambda_{i}^2h_{iimm}+2\sum_{i,j=1}^{5}\lambda_{i}h_{ijm}^2=0\;for\;m=1,2,\cdots,5.\tag{45}\]
For convenience, we write \[Y_2=h_{112}^2+h_{113}^2+h_{114}^2+h_{123}^2+h_{124}^2+h_{134}^2.\]
For \(m=1\) in \((\ref{H2})\) and \((\ref{case2f3})\), we get \[\label{f3} \begin{align} \sum_{i=1}^{5}\lambda_{i}^2h_{ii11}+&\sum_{i,j=1}^{5}2\lambda_{i}h_{ij1}^2=\mu^2\left(h_{1111} +h_{2211}+h_{3311}+h_{4411}\right)\\ &+16\mu^2h_{5511}+2\mu\sum_{i}\left(h_{11i}^2+h_{12i}^2+h_{13i}^2+h_{14i}^2-4h_{15i}^2\right)\\ =&15\mu^2h_{5511}+2\mu\Big[\left(h_{111}^2+h_{221}^2+h_{331}^2+h_{441}^2\right)\\ &+2Y_2-3\left(h_{115}^2+h_{125}^2+h_{135}^2+h_{145}^2\right)\Big]\\ &=0. \end{align}\tag{46}\]
Then, taking \(m=1\) in \((\ref{H2})\) and \((\ref{f42})\), we get \[\label{case2f4} \begin{align} \sum_{i=1}^{5}\lambda_{i}^3h_{ii11}+&\sum_{i,j=1}^{5}3\lambda_{i}^2h_{ij1}^2=\mu^3\left(h_{1111} +h_{2211}+h_{3311}+h_{4411}\right)\\ &-64\mu^3h_{5511}+3\mu^2\sum_{i}\left(h_{11i}^2+h_{12i}^2+h_{13i}^2+h_{14i}^2+16h_{15i}^2\right)\\ =&-65\mu^3h_{5511}+3\mu^2\Big[\left(h_{111}^2+h_{221}^2+h_{331}^2+h_{441}^2\right)\\ &+2Y_2+17\left(h_{115}^2+h_{125}^2+h_{135}^2+h_{145}^2\right)\Big]\\ =&0. \end{align}\tag{47}\]
Next, adding \(\frac{13}{3}\mu(\ref{f3})\) and \((\ref{case2f4})\), we obtain
\[\frac{35}{3}\left(h_{111}^2+h_{221}^2+h_{331}^2+h_{441}^2\right) +\frac{70}{3}Y_2 +25\left(h_{115}^2+h_{125}^2+h_{135}^2+h_{145}^2\right)=0.\]
Finally, in complete analogy with Case (1), we can obtain that all \(h_{ijk}=0\) for \(i,j,k=1,2,\cdots,5.\)
In summary, for the above two cases, we have proved that \(h_{ijk}=0\) for all \(i,j,k=1,2,\cdots,5\). Furthermore, from \((\ref{DeltaS})\), we have \(S(S-5)=0\). Noting that \(S>0\), it follows that \(S\equiv5\) and \(M^5\) is either the Clifford torus \(\mathbb{S}^2(\sqrt\frac{2}{5})\times\mathbb{S}^3(\sqrt\frac{3}{5})\) or \(\mathbb{S}^1(\sqrt\frac{1}{5})\times\mathbb{S}^4(\sqrt\frac{4}{5})\). ◻
Acknowledgments. The author would like to thank Professor Tang Zizhou and Professor Ge Jianquan for helpful discussion.
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