Locally Strongly Convex Affine Hypersurfaces with Semi-parallel Cubic Form

Abstract

In this paper, we investigate the locally strongly convex affine hypersurfaces with semi-parallel cubic form relative to the Levi-Civita connection of affine metric. We obtain two results on such hypersurfaces which admit at most one affine principal curvature of multiplicity one: (1) classify these being not affine hyperspheres; (2) classify these affine hyperspheres with constant scalar curvature. For the latter, by proving the parallelism of their cubic forms we translate the classification into that of affine hypersurfaces with parallel cubic form, which has been completed by Hu-Li-Vrancken (J Differ Geom 87:239–307, 2011).

1 Introduction

The classical equiaffine differential geometry is mainly concerned with geometric properties of hypersurfaces in affine space, that are invariant under unimodular affine transformations. Let $\mathbb {R}^{n+1}$ be the $(n+1)$-dimensional real unimodular affine space. On a non-degenerate hypersurface immersion of $\mathbb {R}^{n+1}$, it is well known how to induce an affine connection $\nabla$, an affine shape operator S whose eigenvalues are called affine principal curvatures, and a symmetric bilinear form h, called the affine metric. From a local point of view, there are two natural tensors, namely the difference tensor K which is defined as the difference between $\nabla$ and the Levi-Civita connection ${\hat{\nabla }}$ of h, and the cubic form $C:=\nabla h$. The classical Pick-Berwald theorem states that the cubic form or difference tensor vanishes, if and only if the hypersurface is a non-degenerate hyperquadric. In that sense, the cubic form or difference tensor plays the role as the second fundamental form for submanifolds of real space forms.

In the same style as the Pick-Berwald theorem, geometric conditions on the cubic form and difference tensor have been used to classify natural classes of affine hypersurfaces by many geometers in the past decades, see e.g. [5,6,7, 13, 17,18,19,20, 32, 35]. Among them, one of the most interesting developments may be the classification of locally strongly convex affine hypersurfaces with parallel cubic form relative to ${\hat{\nabla }}$. In this subject, F. Dillen, L. Vrancken, et al. obtain the classifications for lower dimensions in [11, 15, 21, 27], and finally Z. Hu, H. Li and L. Vrancken complete the classification for all dimensions as follows:

Theorem 1.1

(cf. [23]) Let M be an n-dimensional ($n\ge 2$) locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{\nabla }} C=0$. Then, M is either a hyperquadric (i.e., $C=0$) or a hyperbolic affine hypersphere with $C\not =0$; in the latter case either

• (i) M is obtained as the Calabi product of a lower dimensional hyperbolic affine hypersphere with parallel cubic form and a point, or

• (ii) M is obtained as the Calabi product of two lower dimensional hyperbolic affine hyperspheres with parallel cubic form, or

• (iii) $n=\tfrac{1}{2}m(m+1)-1,\ m\ge 3$, (M, h) is isometric to $\textbf{SL}(m,\mathbb {R})/\textbf{SO}(m)$, and M is affinely equivalent to the standard embedding $\textbf{SL}(m,\mathbb {R})/\textbf{SO}(m)\hookrightarrow \mathbb {R}^{n+1}$, or

• (iv) $n=m^2-1,\ m\ge 3$, (M, h) is isometric to $\textbf{SL}(m,\mathbb {C})/\textbf{SU}(m)$, and M is affinely equivalent to the standard embedding $\textbf{SL}(m,\mathbb {C})/\textbf{SU}(m)\hookrightarrow \mathbb {R}^{n+1}$, or

• (v) $n=2m^2-m-1,\ m\ge 3$, (M, h) is isometric to $\textbf{SU}^*\big (2m\big )/\textbf{Sp}(m)$, and M is affinely equivalent to the standard embedding $\textbf{SU}^*\big (2m\big )/\textbf{Sp}(m)\hookrightarrow \mathbb {R}^{n+1}$, or

• (vi) $n=26$, (M, h) is isometric to $\textbf{E}_{6(-26)}/\textbf{F}_4$, and M is affinely equivalent to the standard embedding $\textbf{E}_{6(-26)}/\textbf{F}_4\hookrightarrow \mathbb {R}^{27}$.

As that did in [23, 24], we say that an affine hypersurface is of semi-parallel (resp. parallel) cubic form relative to the Levi-Civita connection of affine metric if ${\hat{R}}\cdot C=0$ (resp. ${\hat{\nabla }} C=0$), where ${\hat{R}}$ is the curvature tensor of affine metric, and the tensor ${\hat{R}}\cdot C$ is defined by

$$\begin{aligned} \begin{array}{lll} {\hat{R}}(X,Y)\cdot C={\hat{\nabla }}_X{\hat{\nabla }}_{YC}-{\hat{\nabla }_Y{\hat{\nabla }}_XC}- {\hat{\nabla }}_{[X,Y]}C \end{array} \end{aligned}$$

(1.1)

for tangent vector fields X, Y. Obviously, the parallelism of cubic form implies its semi-parallelism, the converse is not true, we refer to Remark 3.1 for the counter-examples.

In this paper, we investigate locally strongly convex affine hypersurfaces with semi-parallel cubic form relative to the Levi-Civita connection of affine metric. First, we prove that if all the affine principal curvatures of the hypersurface have multiplicity more than one, then the hypersurface is an affine hypersphere. If further assume that its affine metric is of constant scalar curvature, by proving the parallelism of the cubic form we translate the classification into that of Theorem 1.1. More precisely, let H, $\triangle$ and ${\hat{Ric}}$ be the affine mean curvature, Laplacian operator and Ricci curvature of affine metric h, respectively, we can state the first main result as follows.

Theorem 1.2

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 2$. Assume that $M^n$ does not admit any affine principal curvature of multiplicity one. Then $M^n$ is either a hyperquadric (i.e., $C=0$) or a hyperbolic affine hypersphere with non-positive scalar curvature $\kappa$ and $C\not =0$. Moreover, there hold

$$\begin{aligned}&2\triangle \kappa =\Vert \hat{\nabla }C\Vert _h^2, \end{aligned}$$

(1.2)

$$\begin{aligned}&(n+1)\kappa H=\Vert {\hat{R}}\Vert _h^2+\Vert {\hat{Ric}}\Vert _h^2, \end{aligned}$$

(1.3)

where $\Vert \cdot \Vert _h$ denotes the tensorial norm with respect to h. If additionally assume that $\kappa$ is constant for $n\ge 3$, then $\hat{\nabla }C=0$, and $M^n$ is affinely equivalent to one of the examples in Theorem 1.1.

Remark 1.1

For a locally strongly convex affine hypersurface $M^n$, Theorems 1.1 and 1.2 imply that:

• (1) If $n\ge 3$, it is an affine hypersphere with ${\hat{R}}\cdot C=0$ and constant scalar curvature if and only if ${\hat{\nabla }} C=0$.

• (2) If $n=2$, it is an affine sphere with ${\hat{R}}\cdot C=0$ if and only if ${\hat{\nabla }} C=0$.

We conjecture that any locally strongly convex affine hypersphere with ${\hat{R}}\cdot C=0$ must satisfy ${\hat{\nabla }} C=0$.

Second, if the hypersurface admits exactly one affine principal curvature of multiplicity one, then the number of its distinct affine principal curvatures is either three or two (i.e., the hypersurface is quasi-umbilical), which are further classified, respectively. These results are given precisely in the following theorems.

Theorem 1.3

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 3$. If it admits exactly one affine principal curvature of multiplicity one, then the number of its distinct affine principal curvatures is either two or three.

Theorem 1.4

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 5$. Assume that there are exactly three distinct affine principal curvatures $\mu _1, \mu _2, \mu _3$ of multiplicity $(1, n_2, n_3)$ with $n_2\ge 2$ and $n_3\ge 2$, respectively. Then $(M^n,h)$ is locally isometric to the warped product $\mathbb {R_+} \times M_2\times _{t}M_3$, and each $\mu _i$ is a function which depends only on t such that $\mu _2\mu _3\ne 0$,

$$\begin{aligned} t^2(\mu _1-\mu _2)(\mu _2-\mu _3)^2=2(\mu _1-\mu _3)(\mu _2+\mu _3). \end{aligned}$$

Moreover, $M^n$ is affinely equivalent to

$$\begin{aligned} F(t,p_2,p_3)=(\gamma _2(t)\phi _2(p_2),\gamma _3(t)\phi _3(p_3)), \end{aligned}$$

where $\gamma _2,\ \gamma _3$ are nonzero functions satisfying

$$\begin{aligned} \gamma '_2=\tfrac{1}{2}(\mu _3-\mu _2)t\gamma _2,\ \gamma '_3=\tfrac{2\mu _3}{(\mu _3-\mu _2)t}\gamma _3, \end{aligned}$$

and $\phi _i:M_i\rightarrow \mathbb {R}^{n_i+1}$ is a locally strongly convex proper affine hypersphere with the affine mean curvature $H_i$ for $i=2,3$, which are nonzero constant defined by

$$\begin{aligned} H_2=\mu _2-\tfrac{1}{4}(\mu _2-\mu _3)^2t^2,\ H_3=\mu _3t^2+\tfrac{2}{\mu _2-\mu _3}, \end{aligned}$$

$\phi _2$ is an ellipsoid if $H_2>0$ and is of semi-parallel cubic form otherwise, whereas $\phi _3$ is an ellipsoid if $H_3\ge 1$.

Theorem 1.5

Let $M^n$ be a locally strongly convex quasi-umbilical affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 3$. Then $(M^n,h)$ is locally isometric to the warped product $\mathbb {R_+} \times _{f}M_2$, and $M^n$ is affinely equivalent to one of the immersions explicitly described in Theorem 6.1, where the warped function $f(t)=1$ or t.

Remark 1.2

The examples in Theorems 1.4 and 1.5 are the generalized Calabi compositions of affine hyperspheres in some special forms. The construction method of such examples initially originates from E. Calabi [8], and now has been extended and characterized by F. Dillen, L. Vrancken, Z. Hu, H. Li, et al. in [1, 2, 12, 22].

This paper is organized as follows. In Sect. 2, we briefly review the local theory of equiaffine hypersurfaces, some results and concepts of warped product manifolds. In Sect. 3, we begin with the Tsinghua principle to study the properties of the hypersurfaces involving the affine principal curvatures, the difference tensor and the eigenvalue distributions of affine shape operator, and present the proof of Theorem 1.2. Based on these properties, in Sect. 4 we obtain Theorem 1.3 by showing the number of the affine principal curvatures being three or two. In either case, we prove the warped product structure, discuss all the possibilities of the immersion and complete the proofs of Theorems 1.4 and 1.5 in last two sections, respectively.

2 Preliminaries

In this section, we briefly recall the local theory of equiaffine hypersurfaces. For more details, we refer to the monographs [26, 29].

Let $\mathbb {R}^{n+1}$ denote the standard $(n+1)$-dimensional real unimodular affine space that is endowed with its usual flat connection D and a parallel volume form $\omega$, given by the determinant. Let $F: M^{n} \rightarrow \mathbb {R}^{n+1}$ be an oriented non-degenerate hypersurface immersion. On such a hypersurface, up to a sign there exists a unique transversal vector field $\xi$, called the affine normal. A non-degenerate hypersurface equipped with the affine normal is called an (equi)affine hypersurface, or a Blaschke hypersurface. Denote by X, Y, Z, W the tangent vector fields on $M^{n}$ from now on. By the affine normal we have

$$\begin{aligned}&D_XF_*Y=F_*\nabla _XY+h(X,Y)\xi , \quad \quad \quad \quad \quad&(\text {Gauss formula}) \end{aligned}$$

(2.1)

$$\begin{aligned}&D_X\xi =-F_*SX, \quad \quad \quad \quad \quad&(\text {Weingarten formula}) \end{aligned}$$

(2.2)

which induce on $M^n$ the affine connection $\nabla$, a symmetric bilinear form h, called the affine metric, the affine shape operator S whose eigenvalues are called affine principal curvatures, and the cubic form $C:=\nabla h$. An affine hypersurface is called locally strongly convex if h is definite, we always choose $\xi$, up to a sign, such that h is positive definite. We call a locally strongly convex affine hypersurface quasi-umbilical if it admits exactly two distinct affine principal curvatures, one of which is simple.

Let $\hat{\nabla }$ be the Levi-Civita connection of the affine metric h. The difference tensor K is defined by

$$\begin{aligned} K(X,Y):=\nabla _XY-\hat{\nabla }_XY. \end{aligned}$$

(2.3)

We also write $K_XY$ and $K_X=\nabla _X-\hat{\nabla }_X$. Since both $\nabla$ and $\hat{\nabla }$ have zero torsion, K is symmetric in X and Y. It is related to the totally symmetric cubic form C by

$$\begin{aligned} C(X, Y, Z)=-2h(K(X, Y), Z), \end{aligned}$$

(2.4)

which implies that the operator $K_X$ is symmetric relative to h. Moreover, K satisfies the apolarity condition, namely, $\textrm{tr}\, K_X=0$ for all X.

The curvature tensor ${\hat{R}}$ of affine metric h, S and K are related by the following Gauss and Codazzi equations:

$$\begin{aligned} \begin{array}{lll} {\hat{R}}(X,Y)Z&{}=\tfrac{1}{2}[h(Y,Z)SX-h(X,Z)SY+h(SY,Z)X-h(SX,Z)Y]\\[2mm] &{}\quad -[K_X,K_Y]Z, \end{array} \end{aligned}$$

(2.5)

$$\begin{aligned} \begin{array}{lll} &{}(\hat{\nabla }_XK)(Y,Z)-(\hat{\nabla }_YK)(X,Z)\\[2mm] &{}\qquad \qquad =\frac{1}{2}[h(Y,Z)SX-h(X,Z)SY-h(SY,Z)X+h(SX,Z)Y], \end{array} \end{aligned}$$

(2.6)

$$\begin{aligned} (\hat{\nabla }_XS)Y-(\hat{\nabla }_YS)X=K(SX,Y)-K(SY,X), \end{aligned}$$

(2.7)

where, by definitions, $[K_X,K_Y]Z=K_XK_YZ-K_YK_XZ$, and

$$\begin{aligned} \begin{array}{lll} &{}{\hat{R}}(X,Y)Z=\hat{\nabla }_X\hat{\nabla }_YZ -\hat{\nabla }_{Y}\hat{\nabla }_XZ-\hat{\nabla }_{[X,Y]}Z,\\[2mm] &{}(\hat{\nabla }_XK)(Y,Z)=\hat{\nabla }_X(K(Y,Z)) -K(\hat{\nabla }_XY,Z)-K(Y,\hat{\nabla }_XZ),\\[2mm] &{}(\hat{\nabla }_XS)Y=\hat{\nabla }_X(SY)-S\hat{\nabla }_XY. \end{array} \end{aligned}$$

Contracting Gauss equation (2.5) we obtain

$$\begin{aligned} \chi =H+J, \end{aligned}$$

(2.8)

where $J=\tfrac{1}{n(n-1)}h(K,K)$, $H=\tfrac{1}{n}\textrm{tr}\,S$, $\chi =\tfrac{\kappa }{n(n-1)}$ and $\kappa$ are the Pick invariant, affine mean curvature, normalized scalar curvature and scalar curvature of h, respectively. Recall the second covariant differentiation of K, defined by

$$\begin{aligned} \begin{array}{lll} {\hat{\nabla }}^2_{X,Y}K={\hat{\nabla }_X{\hat{\nabla }}_YK}-{\hat{\nabla }}_ {{\hat{\nabla }}_XY}K, \end{array} \end{aligned}$$

and the following Ricci identity:

$$\begin{aligned} \begin{array}{lll} &{}({\hat{\nabla }}^2_{X,Y}K)(Z,W)-({\hat{\nabla }}^2_{Y,X}K)(Z,W)=({\hat{R}}(X, Y)\cdot K)(Z, W)\\ &{}\quad ={\hat{R}}(X, Y)K(Z, W)-K({\hat{R}}(X,Y)Z, W)-K(Z, {\hat{R}}(X,Y)W). \end{array} \end{aligned}$$

The affine hypersurface $M^n$ is called an affine hypersphere if $S=H\, id$. Then it follows from (2.7) that H is constant if $n\ge 2$. $M^n$ is said to be a proper (resp. improper) affine hypersphere if H is nonzero (resp. zero). Moreover, a locally strongly convex affine hypersphere is called parabolic, elliptic or hyperbolic according to $H=0$, $H>0$ or $H<0$, respectively. For affine hyperspheres, the Gauss and Codazzi equations reduce to

$$\begin{aligned} \begin{aligned} {\hat{R}}(X,Y)Z=H[h(Y,Z)X-h(X,Z)Y]-[K_{X},K_{Y}]Z, \end{aligned} \end{aligned}$$

(2.9)

$$\begin{aligned} \begin{aligned} (\hat{\nabla }_XK)(Y,Z)=(\hat{\nabla }_YK)(X,Z). \end{aligned} \end{aligned}$$

(2.10)

We collect the following two results for later use.

Theorem 2.1

(cf. Theorem 6.2 of [24]) A locally strongly convex affine surface $M^2$ in $\mathbb {R}^{3}$ satisfies ${\hat{R}}\cdot C=0$ if and only if either $M^2$ is locally a quadric or $(M^2,h)$ is flat.

Theorem 2.2

(cf. Theorem 1 of [1]) Let $M^{m+1}$, $m\ge 2$, be a locally strongly convex affine hypersurface of the affine space $\mathbb {R}^{m+2}$ such that its tangent bundle is an orthogonal sum, with respect to the affine metric h, of two distributions: a one-dimensional distribution $\mathcal {D}_1$ spanned by a unit vector field T and an m-dimensional distribution $\mathcal {D}_2$, such that

$$\begin{aligned}&K(T,T)=\lambda _1 T,\quad K(T, X)=\lambda _2 X, \\&ST=\mu _1 T,\quad SX=\mu _2 X, \quad \forall \ X\in \mathcal {D}_2. \end{aligned}$$

Then either $M^{m+1}$ is an affine hypersphere such that $K_T=0$ or is affinely congruent to one of the following immersions:

• (1) $f(t,x_1,\dots ,x_m)=(\gamma _1(t),\gamma _2(t)g_2(x_1,\dots ,x_m))$, for $\gamma _1,\gamma _2$ such that

  $$\begin{aligned} \epsilon \gamma _1'\gamma _2(\gamma _1'\gamma _2''- \gamma _1''\gamma _2')<0; \end{aligned}$$

• (2) $f(t,x_1,\dots ,x_m)=\gamma _1(t)C(x_1,\dots ,x_m)+\gamma _2(t)e_{m+1}$, for $\gamma _1, \gamma _2$ such that

  $$\begin{aligned} \textrm{sgn}\Big (\gamma _1'\gamma _2''-\gamma _1''\gamma _2'\Big ) =\textrm{sgn}(\gamma _1'\gamma _1)\ne 0; \end{aligned}$$

• (3) $f(t,x_1,\dots ,x_m)=C(x_1,\dots ,x_m)+\gamma _2(t)e_{m+1}+\gamma _1(t)e_{m+2}$, for $\gamma _1, \gamma _2$ such that

  $$\begin{aligned} \textrm{sgn}(\gamma _1'\gamma _2''-\gamma _1''\gamma _2')=\textrm{sgn} (\gamma _1')\ne 0. \end{aligned}$$

Here $g_2:{\mathbb {R}}^m\rightarrow {\mathbb {R}}^{m+1}$ is a proper affine hypersphere centered at the origin with affine mean curvature $\epsilon$, and $C:{\mathbb {R}}^{m} \rightarrow {\mathbb {R}}^{m+2}$ is an improper affine hypersphere, given by $C(x_1,\dots ,x_m)=(x_1,\dots ,x_m, p(x_1,\dots ,x_m),1)$, with the affine normal $e_{m+1}$.

Finally, we review some notions of warped product manifolds and subbundles. For Riemannian manifolds $(B,g_B)$, $(M_1,g_1)$, ..., $(M_k,g_k)$ and positive functions $f_1,\ldots , f_k:B\rightarrow \mathbb {R}$, the manifold $M:=B\times M_1\times \cdots \times M_k$ equipped with the metric $h=g_B\oplus f_1^2g_1\oplus \cdots \oplus f_k^2g_k$ is a warped product manifold with warped functions $f_i$, denoted by $B\times _{f_1}M_1\times \cdots \times _{f_k}M_k$. Let $\hat{\nabla }$ be the Levi-Civita connection of a Riemannian manifold (M, h). A subbundle $E\subset TM$ is called auto-parallel if $\hat{\nabla }_XY\in E$ holds for all $X,\ Y\in E$. Whereas a subbundle E is called totally umbilical if there exists a vector field $V\in E^\perp$ such that $h(\hat{\nabla }_XY,Z)=h(X,Y)h(V,Z)$ for all $X,\ Y\in E$ and $Z\in E^\perp$, here we call V the mean curvature vector of E. If, moreover, $h(\hat{\nabla }_XV,Z)=0$ holds, we say that E is spherical. We conclude this section by the decomposition theorem of Riemannian manifolds.

Theorem 2.3

(cf. Theorem 4 of [28]) Let M be a Riemannian manifold, and let $TM=\bigoplus _{i=0}^kE_i$ be an orthogonal decomposition into nontrivial vector subbundles such that $E_i$ is spherical and $E_{i}^\bot$ is autoparallel for $i=1, \ldots , k$. Then, for every point $p\in M$ there is an isometry $\psi$ of a warped product $M_0\times _{f_1}M_1\times \cdots \times _{f_k}M_k$ onto a neighbourhood of p in M such that $\psi (\{p_0\}\times \cdots \times \{p_{i-1}\}\times M_i\times \{p_{i+1}\}\times \cdots \times \{p_k\})$ is an integral manifold of $E_i$ for $i=0, \ldots , k$ and all $p_0\in M_0, \ldots , p_k\in M_k$.

3 Properties of Affine Hypersurfaces with ${\hat{R}}\cdot C=0$

From this section on, when we say that an affine hypersurface has semi-parallel cubic form, it always means that ${\hat{R}}\cdot C=0$, equivalently ${\hat{R}}\cdot K=0$. Then, by the Ricci identity of K we have

$$\begin{aligned} {\hat{R}}(X, Y)K(Z, W)=K({\hat{R}}(X,Y)Z, W)+K(Z, {\hat{R}}(X,Y)W). \end{aligned}$$

(3.1)

In fact, by (2.4) and the Ricci identities of C and K, the equivalence above follows from the following formula:

$$\begin{aligned} \begin{array}{lll} &{}({\hat{R}}\cdot C)(U,V,X,Y,Z)=({\hat{R}}(U,V)\cdot C)(X,Y,Z)\\ &{}\quad =-C(X,Y,{\hat{R}}(U,V)Z)-C(X,{\hat{R}}(U,V)Y,Z)-C({\hat{R}}(U,V)X,Y,Z)\\ &{}\quad =2[h(K_XY,{\hat{R}}(U,V)Z)+h(K_X{\hat{R}}(U,V)Y,Z)+h(K_Y{\hat{R}}(U,V)X,Z)]\\ &{}\quad =-2h({\hat{R}}(U,V)K_XY-K_X{\hat{R}}(U,V)Y-K_Y{\hat{R}}(U,V)X,Z)\\ &{}\quad =-2h(({\hat{R}}(U,V)\cdot K)(X,Y),Z). \end{array} \end{aligned}$$

(3.2)

Remark 3.1

Besides examples in Theorem 1.1, we see from (3.1) that all flat affine hypersurfaces satisfy ${\hat{R}}\cdot C=0$. Therefore, to see the examples whose cubic forms are semi-parallel but not parallel, we refer to Remark 6.2 in [24] for such flat surfaces, and Theorem 4.1 in [3] for the flat and quasi-umbilical affine hypersurfaces.

In what follows, if no other stated, we always assume that $M^n$ is a locally strongly convex affine hypersurface with semi-parallel cubic form. First, by using the Codazzi equations for both the shape operator and the difference tensor, we obtain some linear equations involving the components of the difference tensor and affine principal curvatures as follows.

Lemma 3.1

Let $M^n$ $(n\ge 2)$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$. Denote by $\{e_1,\ldots , e_n\}$ the orthonormal frame of $M^n$, where $e_i$ are the eigenvector fields of the shape operator S with corresponding eigenvalues $\mu _i$, $i=1,\ldots , n$. Then, for any $i, j, k, \ell$, there holds

$$\begin{aligned} \begin{aligned}&(\mu _k-\mu _i)[\delta _{j\ell }K(e_k,e_i)+h(K(e_k,e_i),e_\ell )e_j]\\&\quad +(\mu _i-\mu _j)[\delta _{k\ell }K(e_i,e_j)+h(K(e_i,e_j),e_\ell )e_k]\\&\quad +(\mu _j-\mu _k)[\delta _{i\ell }K(e_j,e_k)+h(K(e_j,e_k),e_\ell )e_i]=0. \end{aligned} \end{aligned}$$

(3.3)

Proof

By the second covariant differentiation of K and (3.2) it holds that

$$\begin{aligned} \begin{aligned} ({\hat{\nabla }}^2_{W,X}K)(Y,Z)-({\hat{\nabla }}^2_{X,W}K)(Y,Z)=({\hat{R}}(W,X)\cdot K)(Y,Z)=0. \end{aligned} \end{aligned}$$

(3.4)

On the other hand, direct calculations show that

$$\begin{aligned} \begin{aligned}&\sigma _{W,X,Y}\{({\hat{\nabla }}^2_{W,X}K)(Y,Z)-({\hat{\nabla }}^2_{X,W}K)(Y,Z)\}\\&\quad =\sigma _{W,X,Y}\{({\hat{\nabla }}^2_{W,X}K)(Y,Z)-({\hat{\nabla }}^2_{W,Y}K)(X,Z)\}, \end{aligned} \end{aligned}$$

(3.5)

where $\sigma _{W,X,Y}$ denotes the cyclic summation over W, X, Y. Moreover, by the second covariant differentiation of K we have (see also (3.3) in [3])

$$\begin{aligned} \begin{aligned}&({\hat{\nabla }}^2_{W,X}K)(Y,Z)-({\hat{\nabla }}^2_{W,Y}K)(X,Z)\\&\quad =(\hat{\nabla }_W\hat{\nabla }K)(X,Y,Z)-(\hat{\nabla }_W\hat{\nabla }K)(Y,X,Z)\\&\quad =\tfrac{1}{2}\{h(Y,Z)(\hat{\nabla }_WS)X-h(X,Z)(\hat{\nabla }_WS)Y\\&\qquad -h((\hat{\nabla }_WS)Y,Z)X+h((\hat{\nabla }_WS)X,Z)Y\}, \end{aligned} \end{aligned}$$

(3.6)

where the last equality follows from the covariant differentiation of (2.6) along W. Together with (2.7), by (3.4) and (3.5) we see that

$$\begin{aligned} \begin{aligned} 0&=2\sigma _{W,X,Y}\{({\hat{\nabla }}^2_{W,X}K)(Y,Z)-({\hat{\nabla }}^2_{W,Y}K)(X,Z)\}\\&=h(Y,Z)(K(SW,X)\!-\!K(SX,W))\!+\!h(W,Z)(K(SX,Y)\!-\!K(SY,X))\\&\quad +h(X,Z)(K(SY,W)\!-\!K(SW,Y))\!+\!h(K(SY,W)\!-\!K(SW,Y),Z)X\\&\quad +h(K(SW, X)-K(SX,W),Z)Y+h(K(SX,Y)-K(SY,X),Z)W. \end{aligned} \end{aligned}$$

(3.7)

Finally, by taking $X=e_i, Y=e_j$, $W=e_k, Z=e_\ell$ in (3.7) we have (3.3). $\square$

Remark 3.2

The technique used in Lemma 3.1, is based on the Tsinghua principle due to H. Li, L. Vrancken and X. Wang [3]. For some tensor, it allows one to take the cyclic permutation of the covariant derivative of its Codazzi equation, use the Ricci identity in an indirect way and express the tensor in a conveniently chosen frame, see [4, 9, 10, 14, 25] for its applications in various purposes.

By the notations of Lemma 3.1, we always denote by $\mathfrak {D}(\mu _i)$ the eigenvalue distribution of S corresponding to the eigenvalue $\mu _i$ and by $n_i$ its dimension. Note that the conclusion of Lemma 3.1 is the same as Lemma 3.1 of [3], although the assumptions are different. Therefore, following the proof of Lemma 3.2 in [3], by (3.3) we obtain the same results as below.

Lemma 3.2

The difference tensor K satisfies:

• (i) If $\mu _i\ne \mu _j$ and $n_i,n_j\ge 2$, then $K(e_i,e_j)=0$.

• (ii) If $n_j=1$ and $n_i\ge 2$, then there exist functions $\lambda ^j_i:=h(K_{e_i}e_i,e_j)$ depending on the choice of $\mu _i,\mu _j$ such that $K(e_j, e_i)=\lambda ^j_ie_i$.

• (iii) If there are at least two different eigenvalues $\mu _i\ne \mu _k$ such that $n_i,n_k\ge 2$ and $n_j=1$, then there exists a differentiable function ${\bar{\lambda }}_j$ such that it holds that $(\mu _j-\mu _i)\lambda ^j_i=(\mu _j-\mu _k) \lambda ^j_k={\bar{\lambda }}_j$.

Furthermore, by Codazzi equation (2.7) we have

$$\begin{aligned} \begin{aligned}&e_i(\mu _j)e_j-e_j(\mu _i)e_i+\mu _j\hat{\nabla }_{e_i}e_j-\mu _i\hat{\nabla }_{e_j}e_i\\&\quad =S\hat{\nabla }_{e_i}e_j-S\hat{\nabla }_{e_j}e_i+(\mu _i-\mu _j)K(e_i, e_j). \end{aligned} \end{aligned}$$

(3.8)

By multiplying this with the eigenvector $e_k$, we get the following lemma.

Lemma 3.3

It holds that

• (i) $e_i(\mu _j)=(\mu _j-\mu _i)h(\hat{\nabla }_{e_j}e_j-K_{e_j}e_j,e_i)$ for $k=j\ne i$;

• (ii) $e_i(\mu _j)\delta _{jk}-e_j(\mu _i)\delta _{ik}+(\mu _j-\mu _k)h(\hat{\nabla }_{e_i}e_j,e_k)$ $=(\mu _i-\mu _k)h(\hat{\nabla }_{e_j}e_i,e_k) +(\mu _i-\mu _j)h(K_{e_i}e_j,e_k)$ for any i, j, k.

By taking $e_i, e_j\in \mathfrak {D}(\mu _i)$ and $e_k\in \mathfrak {D}(\mu _i)^\bot$ in Lemma 3.3 (ii), we see that each eigenvalue distribution $\mathfrak {D}(\mu _i)$ forms an integrable subbundle. Similarly, taking $e_i, e_j\in \mathfrak {D}(\mu _i)$ in Lemma 3.3 (i) for $i\ne j$, we get that each eigenvalue of multiplicities more than one is constant on its integral submanifolds.

Next, for more information we also denote by ${\tilde{\mu }}_1,\ldots , {\tilde{\mu }}_r$ the eigenvalue functions of affine shape operator S with the multiplicity one, and by $u_1,\ldots , u_r$ the corresponding unit eigenvector fields. Let $\mu _1,\ldots , \mu _s$ be the eigenvalues of higher multiplicities, and $v^i_1,\ldots , v^i_{n_i}$ be the orthonormal eigenvector fields of $\mu _i$, which span the distribution $\mathfrak {D}(\mu _i)$ for $i=1,\ldots , s$. Note from Lemma 3.2 that $K(v^i_j,v^i_k)\in \bigoplus _{j=1}^r$ Span$(u_j) \bigoplus \mathfrak {D}(\mu _i)$. Define tensors $L^i:\mathfrak {D}(\mu _i)\times \mathfrak {D}(\mu _i)\rightarrow \mathfrak {D}(\mu _i)$ given by

$$\begin{aligned} L^i(X,X')=K(X,X')-\sum _{j=1}^r\lambda ^j_ih(X,X')u_j,\ i=1,\ldots , s, \end{aligned}$$

(3.9)

which are the projection of K onto the distribution $\mathfrak {D}(\mu _i)$, then we are ready to prove the next two results.

Lemma 3.4

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$. If the multiplicity of affine principal curvature $\mu _i$ is more than one, then

• (i) The eigenvalue distribution $\mathfrak {D}(\mu _i)$ is integrable, on which $\mu _i$ is constant.

• (ii) $L^i$ is totally symmetric and satisfies the apolarity condition.

• (iii) For any $X,X',X''\in \mathfrak {D}(\mu _i)$,$W\in \mathfrak {D}(\mu _i)^\bot$, there hold ${\hat{R}}(X, X')\cdot L^i=0$, and

  $$\begin{aligned} \begin{aligned}&{\hat{R}}(X, X')X''=(\mu _i-\sum ^r_{j=1}(\lambda ^j_i)^2)[h(X',X'')X-h(X,X'')X']\\&\quad \ -[L^i_{X}, L^i_{X'}]X'',\\&{\hat{R}}(X, X')W=0. \end{aligned} \end{aligned}$$

  (3.10)

• (iv) Assume that $\mathfrak {D}(\mu _i)$ is spherical. Denote by ${\hat{R}}^{\bot }$ the curvature tensor of the connection $\hat{\nabla }^{\bot }$ on the integral manifold of $\mathfrak {D}(\mu _i)$ induced from $(M^n,h)$, and by $\rho _iT$ the mean curvature vector with the unit vector $T\in \mathfrak {D}(\mu _i)^{\bot }$. Then

  $$\begin{aligned} \begin{aligned}&({\hat{R}}^{\bot }(X,X')\cdot L^i)(X'',{\tilde{X}})\\&\quad =\rho _i^2\{h(X,X'')L^i(X',{\tilde{X}}) -h(X',X'')L^i(X,{\tilde{X}})\\&\qquad +h(X,{\tilde{X}})L^i(X',X'')-h(X',{\tilde{X}})L^i(X,X'')\\&\qquad +h(X',L^i(X'',{\tilde{X}}))X-h(X,L^i(X'',{\tilde{X}}))X'\}, \end{aligned} \end{aligned}$$

  (3.11)

  where ${\hat{R}}^{\bot }(X,X')X''={\hat{R}}(X,X')X'' +\rho _i^2(h(X',X'')X-h(X,X'')X')$. In particular, if $\mathfrak {D}(\mu _i)$ is auto-parallel, i.e., $\rho _i=0$, then ${\hat{R}}^{\bot }\cdot L^i=0$.

Proof

The previous analysis after Lemma 3.3 gives the proof of the part (i). Note that $h(L^i(v^i_j,v^i_k), v^i_\ell )=h(K(v^i_j,v^i_k), v^i_\ell )$ is totally symmetric. The apolarity condition yields $\sum _{j=1}^rK(u_j,u_j)+\sum ^s_{\ell =1}\sum ^{n_\ell }_{p=1}K(v^\ell _p, v^\ell _p)=0$. Then, for arbitrary $v^i_q\in \mathfrak {D}(\mu _i)$, by (3.9) and Lemma 3.2 there holds

$$\begin{aligned} \begin{aligned}&\sum ^{n_i}_{p=1}h(L^i(v^i_p, v^i_p),v^i_q)=\sum ^{n_i}_{p=1}h(K(v^i_p, v^i_p),v^i_q)\\&\quad =-\sum _{\ell \ne i}\sum ^{n_\ell }_{p=1}h(K(v^{\ell }_p, v^{\ell }_p),v^i_q) -\sum _{j=1}^rh(K(u_j,u_j),v^i_q)=0. \end{aligned} \end{aligned}$$

(3.12)

Therefore, $\sum ^{n_i}_{p=1}L^i(v^i_p, v^i_p)=0$, i.e., the tensor $L^i$ satisfies the apolarity condition. We have proved part (ii).

Denote also by $L^i_XX'=L^i(X,X')$. By the total symmetry of $L^i$, Lemma 3.2 and (3.9) we have

$$\begin{aligned} \begin{aligned} ~[L^i_{X},L^i_{X'}]X''&=L^i(X,L^i(X',X''))-L^i(X',L^i(X,X''))\\&=K(X,L^i(X',X''))-K(X',L^i(X,X''))\\&=K(X,K(X',X''))-K(X',K(X,X''))\\&\quad -\sum _j\lambda ^j_i[h(X',X'')K(u_j,X)-h(X,X'')K(u_j,X')]\\&=[K_X,K_{X'}]X''-\sum _j(\lambda ^j_i)^2[h(X',X'')X-h(X,X'')X']. \end{aligned} \end{aligned}$$

Together with Gauss equation (2.5), by Lemma 3.2 we further have (3.10). Combining this with (3.1) we deduce that

$$\begin{aligned} \begin{aligned} {\hat{R}}(X, X')L^i(X'',{\tilde{X}})&={\hat{R}}(X, X')K(X'',{\tilde{X}}) -\sum _j\lambda ^j_ih(X'',{\tilde{X}}){\hat{R}}(X, X')u_j\\&=K({\hat{R}}(X, X')X'',{\tilde{X}})+K(X'',{\hat{R}}(X, X'){\tilde{X}})\\&=L^i({\hat{R}}(X, X')X'',{\tilde{X}})+L^i(X'',{\hat{R}}(X, X'){\tilde{X}})\\&\quad +\sum _{j}\lambda ^j_i[h({\hat{R}}(X, X')X'',{\tilde{X}})+h(X'',{\hat{R}}(X, X'){\tilde{X}})]u_j\\&=L^i({\hat{R}}(X, X')X'',{\tilde{X}})+L^i(X'',{\hat{R}}(X, X'){\tilde{X}}). \end{aligned} \end{aligned}$$

This shows the part (iii).

For part (iv), as $\mathfrak {D}(\mu _i)$ is spherical, we have

$$\begin{aligned} \hat{\nabla }_XX'=\hat{\nabla }^{\bot }_XX'+\rho _ih(X,X')T, \ \hat{\nabla }_XT=-\rho _iX, \ X(\rho _i)=0. \end{aligned}$$

It is well known that the projection $\hat{\nabla }^{\bot }_{X'}X$ of $\hat{\nabla }_{X'}X$ on $\mathfrak {D}(\mu _i)$ defines a connection that turns out to be the Levi-Civita connection of the induced metric on the integral manifold of $\mathfrak {D}(\mu _i)$ from $(M^n,h)$. Then, by the definition of curvature tensor and ${\hat{R}}(X, X')\cdot L^i=0$, direct computations give (3.11). $\square$

Proposition 3.1

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$. If the multiplicity of affine principal curvature $\mu _i$ is more than one, then, by the notations as above, it holds that either $L^i=0$, or $L^i\ne 0$ and

$$\begin{aligned} \mu _i-\sum ^r_{j=1}(\lambda ^j_i)^2<0. \end{aligned}$$

(3.13)

In particular, if $M^n$ is an affine hypersphere, then $M^n$ is either a hyperquadric (i.e., $C=0$), or a hyperbolic affine hypersphere with $C\ne 0$.

Proof

Set $\lambda =\mu _i-\sum ^r_{j=1}(\lambda ^j_i)^2$. It follows from Lemma 3.4 that $L^i$ is totally symmetric and satisfies the apolarity condition. Moreover,

$$\begin{aligned} \begin{aligned}&{\hat{R}}(X, X')L^i(X'',{\tilde{X}})=L^i({\hat{R}}(X, X')X'',{\tilde{X}})+L^i(X'',{\hat{R}}(X, X'){\tilde{X}}),\\&{\hat{R}}(X, X')X''=\lambda [h(X',X'')X-h(X,X'')X']-[L^i_{X},L^i_{X'}]X'' \end{aligned} \end{aligned}$$

(3.14)

for any vector fields $X,X',X''\in \mathfrak {D}(\mu _i)$.

Assume that $L^i\ne 0$. Fix a point $p\in M^n$, we now choose an orthonormal basis of $\mathfrak {D}(\mu _i)(p)$ with respect to the affine metric h in the following way. Let $U_p\mathfrak {D}(\mu _i):=\{u\in \mathfrak {D}(\mu _i)(p)|h(u,u)=1\}$. Since h is positive definite, $U_p\mathfrak {D}(\mu _i)$ is compact. We define a function $f(u)=h(L^i_uu,u)$ on $U_p\mathfrak {D}(\mu _i)$. Let $e_1$ be an element of $U_p\mathfrak {D}(\mu _i)$ at which f attains an absolute maximum. Since $L^i\ne 0$, we have $f(e_1)>0$.

Let $u\in U_p\mathfrak {D}(\mu _i)$ such that $h(e_1,u)=0$, and define another function $g(t)=f(e_1\cos t+u\sin t)$. Then we have $g'(0)=3h(L^i_{e_1}e_1,u)$, $g''(0)=6h(L^i_{e_1}u,u)-3f(e_1)$. Since g attains an absolute maximum at $t=0$, we have $g'(0)=0$, thus $h(L^i_{e_1}e_1,u)=0$. Then $e_1$ is an eigenvector of $L^i_{e_1}$ with eigenvalue $\nu _1=f(e_1)>0$. Let $e_2, \ldots , e_{n_i}$ be orthonormal vectors of $\mathfrak {D}(\mu _i)(p)$, orthogonal to $e_1$, which are the remaining eigenvectors of $L^i_{e_1}$ corresponding to the eigenvalues $\nu _2,\ldots , \nu _{n_i}$, respectively.

Since $e_1$ is an absolute maximum point of f, we know that $g''(0)\le 0$. This implies that for every $j\ge 2$, we have $\nu _1-2\nu _j\ge 0$. From the apolarity condition of $L^i_{e_1}$ we have

$$\begin{aligned} \nu _1+\nu _2+\cdots +\nu _{n_i}=0. \end{aligned}$$

(3.15)

By applying (3.14) we have

$$\begin{aligned} \begin{aligned}&\nu _1{\hat{R}}(e_1, e_j)e_1={\hat{R}}(e_1, e_j)L^i(e_1,e_1)=2L^i({\hat{R}}(e_1, e_j)e_1,e_1),\\&{\hat{R}}(e_1, e_j)e_1=-\lambda e_j-[L^i_{e_1},L^i_{e_j}]e_1 =(-\lambda -\nu _j^2+\nu _1\nu _j)e_j, \end{aligned} \end{aligned}$$

which imply that

$$\begin{aligned} (\nu _1-2\nu _j)(-\lambda -\nu _j^2+\nu _1\nu _j)=0. \end{aligned}$$

(3.16)

If $\nu _1=2\nu _j$ for all $j\ge 2$, then (3.15) implies that $\nu _1=0$, this is a contradiction to $\nu _1=f(e_1)>0$. Hence, there exists an integer $k\in \{1,\ldots , n_i-1\}$ such that, after rearranging the ordering,

$$\begin{aligned} \nu _2=\cdots =\nu _k=\tfrac{1}{2}\nu _1,\ \nu _{k+1}<\tfrac{1}{2}\nu _1, \ldots , \nu _{n_i}<\tfrac{1}{2}\nu _1. \end{aligned}$$

(3.17)

Moreover, if $j>k$, we see from (3.16) that

$$\begin{aligned} -\lambda -\nu _j^2+\nu _1\nu _j=0. \end{aligned}$$

(3.18)

Subtracting this for $j,\ell >k$, we have

$$\begin{aligned} (\nu _j-\nu _{\ell })(\nu _1-\nu _j-\nu _{\ell })=0. \end{aligned}$$

Note from (3.17) that $\nu _1-\nu _j-\nu _{\ell }>0$. Thus $\nu _{k+1}=\cdots =\nu _{n_i}:=\nu _0$. Then, it follows from (3.15) and (3.18) that

$$\begin{aligned} \begin{aligned} \nu _0=-\tfrac{k+1}{2(n_i-k)}\nu _1,\quad -\lambda =\tfrac{(k+1)(2n_i-k+1)}{4(n_i-k)^2}\nu _1^2>0. \end{aligned} \end{aligned}$$

Hence, (3.13) follows. The first part has been proved.

If $M^n$ is an affine hypersphere, from Lemma 3.4 we see that $\mathfrak {D}(\mu _i)(p)$ is the tangent space $T_pM^n$ with $n_i=n$, the projection tensor $L^i$ is nothing but K, $\mu _i=H$, and $r=0$ (i.e., $\sum ^r_{j=1}(\lambda ^j_i)^2=0$). Following the same process as above, we see that either $K=0$ or $K\ne 0$ and $H<0$. The conclusion follows. $\square$

Remark 3.3

For the affine hypersphere, Proposition 3.1 extends the result of Proposition 2.1 in [15] from $\hat{\nabla }C$ to ${\hat{R}}\cdot C=0$. The technique, which is employed to construct a typical orthonormal basis on $\mathfrak {D}(\mu _i)(p)$, was introduced by Ejiri [16] and has been extended and widely applied for various purposes, see e.g. [5, 7, 9,10,11, 15, 21, 23, 30, 31, 33, 34].

Finally, we conclude this section by proving Theorem 1.2.

Completion of Theorem 1.2’s Proof

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 2$, whose affine principal curvatures are all of the multiplicity at least two. Then, for $n=2, 3$, $M^n$ is an affine hypersphere.

For $n\ge 4$, we assume that there exist at least two different affine principal curvatures, namely $\mu _i$ and $\mu _j$. Then, $M^n$ is not an affine hypersphere, and thus $K\ne 0$. By Gauss equation (2.5) and Lemma 3.2 (i) we see that, for any unit vector fields $X, X'\in \mathfrak {D}(\mu _i)$, $Y\in \mathfrak {D}(\mu _j)$,

$$\begin{aligned} \begin{aligned}&K(X,Y)=0,\ K(X,X')\in \mathfrak {D}(\mu _i),\\&{\hat{R}}(X,Y)Y=\tfrac{1}{2}(\mu _i+\mu _j)X,\ {\hat{R}}(X,Y)X=-\tfrac{1}{2}(\mu _i+\mu _j)Y. \end{aligned} \end{aligned}$$

By (3.1) we further obtain that

$$\begin{aligned} \begin{aligned} 0&={\hat{R}}(X,Y)K(Y,X)=K({\hat{R}}(X,Y)Y,X)+K(Y,{\hat{R}}(X,Y)X)\\&=\tfrac{1}{2}(\mu _i+\mu _j)(K_XX-K_YY). \end{aligned} \end{aligned}$$

(3.19)

As $K\ne 0$, there must exist a unit vector $X_0\in \mathfrak {D}(\mu _{i_0})$ for some $\mu _{i_0}$ such that $h(K_{X_0}X_0,X_0)\ne 0$. Let $X=X_0$ in (3.19), by multiplying this with $X_0$, we have

$$\begin{aligned} \mu _{i_0}+\mu _j=0 \end{aligned}$$

(3.20)

for any $\mu _j \ne \mu _{i_0}$. This means that there are exactly two different affine principal curvatures, namely $\mu _1$ and $\mu _2$, and $\mu _1=-\mu _2$.

For any unit vector fields $X, X'\in \mathfrak {D}(\mu _1)$ and $Y,Y'\in \mathfrak {D}(\mu _2)$, by Lemma 3.3 (i) we have $X(\mu _1)=-X(\mu _2)=0$. Similarly, $Y(\mu _2)=-Y(\mu _1)=0$. Hence, $\mu _1$ and $\mu _2$ are constant. Then, taking $e_i=X, e_j=Y, e_k=X'$ and $e_i=Y, e_j=X, e_k=Y'$ respectively in Lemma 3.3 (ii), by Lemma 3.2 (i) we have

$$\begin{aligned} h(\hat{\nabla }_{X}Y,X')=0,\quad h(\hat{\nabla }_{Y}X,Y')=0, \end{aligned}$$

(3.21)

which imply that $\hat{\nabla }_{Y}X, \hat{\nabla }_{X}X'\in \mathfrak {D}(\mu _1)$ and $\hat{\nabla }_{X}Y, \hat{\nabla }_{Y}Y'\in \mathfrak {D}(\mu _2)$. Together with the Codazzi equation (2.6) we see that

$$\begin{aligned} 0=h((\hat{\nabla }_XK)(Y,X)-(\hat{\nabla }_YK)(X,X),Y) =\tfrac{1}{2}(\mu _1-\mu _2)=\mu _1, \end{aligned}$$

(3.22)

which implies that $\mu _2=-\mu _1=0$. This is a contradiction to $\mu _1\ne \mu _2$. Therefore, $M^n$ has a single affine principal curvature, i.e., it is an affine hypersphere.

In summary, $M^n$ is an affine hypersphere for $n\ge 2$. Then, Proposition 3.1 implies that $M^n$ is either a hyperquadric or a hyperbolic affine hypersphere with $C\ne 0$.

For such affine hyperspheres, denote by $\{e_1, \ldots , e_{n}\}$ an orthonormal frame relative to h, set $A_{ijk}=h(K_{e_i}e_j,e_k)$, we see from (2.10) that the components of first covariant differentiation $A_{ijk,\ell }$ are totally symmetric. It follows from ${\hat{R}}\cdot K=0$ and the apolarity condition that the components of second covariant differentiation $A_{ijk,\ell s}$ are symmetric and trace-free in any two indices. Then, by $n(n-1)J=h(K,K)=\sum (A_{ijk})^2$ we have

$$\begin{aligned} \begin{aligned} \tfrac{1}{2}n(n-1)\triangle J=\sum (A_{ijk,\ell })^2+\sum A_{ijk}A_{ijk,\ell \ell }=\sum (A_{ijk,\ell })^2, \end{aligned} \end{aligned}$$

(3.23)

where, by (2.4) there holds (cf. (7) of [6])

$$\begin{aligned} \begin{aligned} A_{ijk,\ell }=h((\hat{\nabla }_{e_\ell }K)(e_i,e_j),e_k) =-\tfrac{1}{2}(\hat{\nabla }_{e_\ell }C)(e_i,e_j,e_k). \end{aligned} \end{aligned}$$

Together with $n(n-1)J=\kappa -n(n-1)H$, we have (1.2). Recall the formula (3.32) in [26] for the Laplacian of J on affine hyperspheres:

$$\begin{aligned} \tfrac{1}{2}n(n-1)\triangle J=\sum (A_{ijk,\ell })^2+ \sum ({\hat{R}}_{ij})^2+\sum ({\hat{R}}_{ijk\ell })^2-(n+1)\kappa H. \end{aligned}$$

(3.24)

Combining with (3.23) we obtain (1.3), which implies that $\kappa H\ge 0$, and thus $\kappa \le 0$ on the hyperbolic affine hypersphere.

Furthermore, for $n=2$, it follows from Theorem 2.1 that $M^2$ is affinely equivalent to either a quadric or a flat affine sphere, and thus $\kappa$ is constant. Together with the assumption that $\kappa$ is constant for $n\ge 3$, we see from (1.2) that $\hat{\nabla }C=0$ for $n\ge 2$. Then, $M^n$ is affinely equivalent to one of the examples in Theorem 1.1. $\square$

4 Proof of Theorem 1.3

Let $M^n$ be a locally strongly convex affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 3$. Assume that $\mu _1,\ldots ,\mu _m$ are the m distinct affine principal curvatures of multiplicity $(1,n_2,\ldots ,n_m)$ with $m\ge 2$ and $n_i\ge 2$, respectively. Then, $M^n$ is not an affine hypersphere, and thus $K\ne 0$.

Assume that $m\ge 3$, and thus $n\ge 5$. For our purpose, it is sufficient to prove $m=3$. In Lemmas 3.2-3.4, as $r=1$ we always omit the upper index $j=1$ of $\lambda ^j_i$ for simplicity. Denote by T the unit eigenvector field of the affine principal curvature $\mu _1$, by Lemma 3.2 we have

$$\begin{aligned} \begin{aligned}&ST=\mu _1T,\ SX=\mu _iX,\ SY=\mu _jY,\\&K_{T}T=\lambda _1T,\ K_{T}X=\lambda _iX,\ K_{T}Y=\lambda _jY,\\&K(X, Y)=0,\ \forall \ X\in \mathfrak {D}(\mu _i), \ \forall \ Y\in \mathfrak {D}(\mu _j), \end{aligned} \end{aligned}$$

(4.1)

where

$$\begin{aligned} (\mu _1-\mu _i)\lambda _i=(\mu _1-\mu _j)\lambda _j={\bar{\lambda }}_1,\ \ i\ne j\ge 2. \end{aligned}$$

(4.2)

In the following, if no other stated, we always assume the unit vector fields

$$\begin{aligned} X, X'\in \mathfrak {D}(\mu _i),\ Y, Y'\in \mathfrak {D}(\mu _j),\ i\ne j\ge 2. \end{aligned}$$

From the Gauss equation (2.5), by (4.1) we have

$$\begin{aligned} \begin{aligned}&{\hat{R}}(T,X)X=(\lambda ^2_i-\lambda _1\lambda _i+\tfrac{1}{2}(\mu _1+\mu _i))T,\\&{\hat{R}}(T,X)T=-(\lambda ^2_i-\lambda _1\lambda _i+\tfrac{1}{2}(\mu _1+\mu _i))X,\\&{\hat{R}}(Y,X)Y=(\lambda _i\lambda _j-\tfrac{1}{2}(\mu _i+\mu _j))X,\\&{\hat{R}}(Y,X)X=(\tfrac{1}{2}(\mu _i+\mu _j)-\lambda _i\lambda _j)Y,\\&{\hat{R}}(T,X)Y=0. \end{aligned} \end{aligned}$$

(4.3)

Notice that $K_YY-\lambda _jT\in \mathfrak {D}(\mu _j)$, by (3.1) and (4.3) we can compute

$$\begin{aligned} \begin{aligned}&{\hat{R}}(T,X)K(X,T)=K({\hat{R}}(T,X)X,T)+K(X,{\hat{R}}(T,X)T),\\&{\hat{R}}(Y,X)K(X,Y)=K({\hat{R}}(Y,X)X,Y)+K(X,{\hat{R}}(Y,X)Y),\\&{\hat{R}}(T,X)K(Y,Y)=2K({\hat{R}}(T,X)Y,Y) \end{aligned} \end{aligned}$$

(4.4)

to obtain respectively that

$$\begin{aligned} \begin{aligned}&(\lambda ^2_i-\lambda _1\lambda _i+\tfrac{1}{2}(\mu _1+\mu _i))(K_XX+(\lambda _i-\lambda _1)T)=0,\\&(\lambda _i\lambda _j-\tfrac{1}{2}(\mu _i+\mu _j))(K_XX-K_YY)=0,\\&(\lambda ^2_i-\lambda _1\lambda _i+\tfrac{1}{2}(\mu _1+\mu _i))\lambda _j X=0. \end{aligned} \end{aligned}$$

(4.5)

Remark 4.1

${\bar{\lambda }}_1\ne 0$, and it follows from (4.2) that $\lambda _2,\ldots , \lambda _m$ are all nonzero and distinct. Otherwise, if ${\bar{\lambda }}_1=0$, by (4.2) and the apolarity condition we see that $\lambda _i=0$ for all $i\ge 1$, thus $K_{T}=0$. As $K\ne 0$, there must exist a unit vector $X_0\in \mathfrak {D}(\mu _{i_0})$ for some eigenvalue $\mu _{i_0}\ne \mu _1$ such that $h(K_{X_0}X_0,X_0)\ne 0$. Taking the inner product with $X_0$ of the first two equations in (4.5) for $X=X_0$, we have

$$\begin{aligned} \mu _1+\mu _{i_0}=\mu _j+\mu _{i_0}=0, \end{aligned}$$

thus $\mu _j=\mu _1=-\mu _{i_0}$, i.e., $m=2$. This is a contradiction to $m\ge 3$.

By multiplying the last two equations in (4.5) respectively with T and X, by Remark 4.1 we have

$$\begin{aligned} \begin{aligned}&\mu _i+\mu _j=2\lambda _i\lambda _j\ne 0,\ i\ne j>1, \\&\mu _1+\mu _i=2\lambda _i(\lambda _1-\lambda _i),\ i>1. \end{aligned} \end{aligned}$$

(4.6)

By subtracting these equations, we further obtain that

$$\begin{aligned} \begin{aligned}&\mu _1-\mu _j=2\lambda _i(\lambda _1-\lambda _i-\lambda _j)\ne 0,\\&\mu _i-\mu _j=2(\lambda _i-\lambda _j)(\lambda _1-\lambda _i-\lambda _j)\ne 0. \end{aligned} \end{aligned}$$

(4.7)

Remark 4.2

$m\le 4$. If there exist three different affine principal curvatures $\mu _i$, $\mu _j$, $\mu _k$ of multiplicity more than one, then from (4.2) and (4.6) we have

$$\begin{aligned} \begin{aligned} \frac{\lambda _j}{\lambda _k}=\frac{\mu _1-\mu _k}{\mu _1-\mu _j} =\frac{\mu _i+\mu _j}{\mu _i+\mu _k}, \end{aligned} \end{aligned}$$

(4.8)

which further implies that

$$\begin{aligned} \mu _1=\mu _i+\mu _j+\mu _k. \end{aligned}$$

(4.9)

Then $\mu _k$ is uniquely determined by (4.9) for fixed $\mu _i$ and $\mu _j$. Therefore, $m\le 4$.

By taking special vector fields in Lemma 3.3, we can obtain that

$$\begin{aligned} \begin{aligned}&X(\mu _1)=(\mu _1-\mu _i)h(\hat{\nabla }_{T}T, X),\ X(\mu _i)=0,\\&Y(\mu _i)h(X,X')=(\mu _j-\mu _i)h(\hat{\nabla }_{X}Y, X'),\\&(\mu _i-\mu _j)h(\hat{\nabla }_{T}X,Y)=(\mu _1-\mu _j)h(\hat{\nabla }_{X}T,Y),\\&(\mu _j-\mu _i)h(\hat{\nabla }_{T}Y,X)=(\mu _1-\mu _i)h(\hat{\nabla }_{Y}T,X),\\&T(\mu _i)h(X,X')=(\mu _1-\mu _i)h(\hat{\nabla }_{X}T+\lambda _iX, X'). \end{aligned} \end{aligned}$$

(4.10)

By the Codazzi equation (2.6) and (4.1), taking the inner product of

$$\begin{aligned} (\hat{\nabla }_XK)(T,T)=(\hat{\nabla }_{T}K)(X,T)+\tfrac{1}{2}(\mu _i-\mu _1)X \end{aligned}$$

(4.11)

with T, $X'$ and Y, respectively, we see that

$$\begin{aligned} \begin{aligned}&X(\lambda _1)=(\lambda _1-2\lambda _i)h(\hat{\nabla }_{T}T,X),\\&(\lambda _1-2\lambda _i)h(\hat{\nabla }_{X}T,X') =h((T(\lambda _i)+\tfrac{1}{2}(\mu _i-\mu _1))X-K_X\hat{\nabla }_{T}T,X'),\\&(\lambda _i-\lambda _j)h(\hat{\nabla }_{T}X,Y) =(\lambda _1-2\lambda _j)h(\hat{\nabla }_{X}T,Y). \end{aligned} \end{aligned}$$

(4.12)

By changing the role of X, Y in the last equation of (4.12), we also get

$$\begin{aligned} (\lambda _j-\lambda _i)h(\hat{\nabla }_{T}Y,X) =(\lambda _1-2\lambda _i)h(\hat{\nabla }_{Y}T,X). \end{aligned}$$

(4.13)

Then we are ready to prove the following results.

Lemma 4.1

It holds that

$$\begin{aligned} \begin{aligned}&h(\hat{\nabla }_{T}X,Y)=h(\hat{\nabla }_{X}T,Y) =h(\hat{\nabla }_{Y}T,X)=0,\\&\hat{\nabla }_{X}T=-\rho _iX, \ \rho _i:=\tfrac{T(\mu _i)}{\mu _i-\mu _1}+\lambda _i, \end{aligned} \end{aligned}$$

(4.14)

where $X\in \mathfrak {D}(\mu _i)$, $Y\in \mathfrak {D}(\mu _j)$, $i\ne j>1$.

Proof

From (4.10), (4.12) and (4.13) we have

$$\begin{aligned} \begin{aligned}&(\mu _i-\mu _j)h(\hat{\nabla }_{T}X,Y)=(\mu _1-\mu _j)h(\hat{\nabla }_{X}T,Y) =(\mu _1-\mu _i)h(\hat{\nabla }_{Y}T,X),\\&(\lambda _i-\lambda _j)h(\hat{\nabla }_{T}X,Y) =(\lambda _1-2\lambda _j)h(\hat{\nabla }_{X}T,Y) =(\lambda _1-2\lambda _i)h(\hat{\nabla }_{Y}T,X). \end{aligned} \end{aligned}$$

(4.15)

Assume on the contrary that $h(\hat{\nabla }_{T}X,Y)\ne 0$, then (4.15) imply that $h(\hat{\nabla }_{X}T,Y)$ and $h(\hat{\nabla }_{Y}T,X)$ are nonzero, too. And we further see from (4.15) that

$$\begin{aligned} \begin{aligned}&(\mu _1-\mu _j)(\lambda _i-\lambda _j)=(\mu _i-\mu _j)(\lambda _1-2\lambda _j),\\&(\mu _1-\mu _i)(\lambda _i-\lambda _j)=(\mu _i-\mu _j)(\lambda _1-2\lambda _i). \end{aligned} \end{aligned}$$

By subtracting these equations, we get $(\mu _i-\mu _j)(\lambda _i-\lambda _j)=0$. This is a contraction to $\lambda _i\ne \lambda _j$. Therefore, $h(\hat{\nabla }_{T}X,Y)=0$. Then, the conclusions follow from the first line equation in (4.15), and the last equation of (4.10). $\square$

By the Codazzi equation (2.6) and (4.1), taking the inner product of

$$\begin{aligned} (\hat{\nabla }_{T}K)(X,Y)-(\hat{\nabla }_XK)(T,Y)=0 \end{aligned}$$

(4.16)

with $X'$ and Y, respectively, by Lemma 4.1 we have

$$\begin{aligned} \begin{aligned}&(\lambda _i-\lambda _j)h(\hat{\nabla }_{X}Y,X') =h(K_XX',\hat{\nabla }_{T}Y)=-\lambda _ih(X,X')h(\hat{\nabla }_{T}T,Y),\\&X(\lambda _j)=h(K_YY,\hat{\nabla }_{X}T-\hat{\nabla }_{T}X) =\lambda _jh(\hat{\nabla }_{T}T,X). \end{aligned} \end{aligned}$$

(4.17)

Lemma 4.2

It holds that

$$\begin{aligned} \begin{aligned}&\hat{\nabla }_{T}T=0, \ h(\hat{\nabla }_{X}Y,X')=0,\\&T(\lambda _i)=(2\lambda _i-\lambda _1)\rho _i +\tfrac{1}{2}(\mu _1-\mu _i),\\&X(\lambda _1)=X(\lambda _i)=X(\mu _1)=X(\mu _i)=0,\\&\rho _i\lambda _j-\rho _j\lambda _i+\tfrac{1}{2}(\mu _i-\mu _j)=0, \end{aligned} \end{aligned}$$

(4.18)

where $X, X'\in \mathfrak {D}(\mu _i)$, $Y\in \mathfrak {D}(\mu _j)$, $i\ne j>1$.

Proof

Since $h(\hat{\nabla }_{T}T,T)=0$, there exist unit vector fields $V^i_0\in \mathfrak {D}(\mu _i)$ such that

$$\begin{aligned} \hat{\nabla }_{T}T=a_2V^2_0+\cdots +a_mV^m_0 \end{aligned}$$

for some differential functions $a_i$. Then, we see from the first equation of (4.12) and (4.17) that

$$\begin{aligned} V^i_0(\lambda _1)=a_i(\lambda _1-2\lambda _i),\ V^i_0(\lambda _j)=a_i\lambda _j,\ j\ne i>1. \end{aligned}$$

(4.19)

Recall from the apolarity condition that $\lambda _1+n_2\lambda _2+\cdots +n_m\lambda _m=0$, then

$$\begin{aligned} V^i_0(\lambda _i)=(1+2/n_i)a_i\lambda _i. \end{aligned}$$

(4.20)

Let $\{V^i_0,\ldots ,V^i_{n_i-1}\}$ be an orthonormal frame of $\mathfrak {D}(\mu _i)$ for $i>1$. Taking $X=V^i_j, X'=L^i(V^i_0,V^i_j)$ in the last equation of (4.10), by Lemma 3.4 we obtain

$$\begin{aligned} \begin{aligned}&(\mu _1-\mu _i)h(\hat{\nabla }_{V^i_j}T, L^i(V^i_0,V^i_j)) =(T(\mu _i)+(\mu _i-\mu _1)\lambda _i)h(V^i_j,L^i(V^i_0,V^i_j)),\\&(\mu _1-\mu _i)\sum _{j=0}^{n_i-1}h(\hat{\nabla }_{V^i_j}T, L^i(V^i_0,V^i_j))\\&\quad =(T(\mu _i)+(\mu _i-\mu _1)\lambda _i)\sum _{j=0}^{n_i-1}h(V^i_0,L^i(V^i_j,V^i_j))=0. \end{aligned} \end{aligned}$$

(4.21)

Considering the Codazzi equation (2.6) of the following form

$$\begin{aligned} h((\hat{\nabla }_{V^i_j}K)(V^i_0,T),V^i_j)=h((\hat{\nabla }_{V^i_0}K)(V^i_j,T),V^i_j), \ j\ne 0 \end{aligned}$$

(4.22)

we get

$$\begin{aligned} V^i_0(\lambda _i)=h(K_{V^i_j}V^i_j,\hat{\nabla }_{V^i_0}T) -h(K_{V^i_0}V^i_j,\hat{\nabla }_{V^i_j}T),\ j=1,\ldots , n_i-1, \end{aligned}$$

which together with (4.21) further shows that

$$\begin{aligned} \begin{aligned}&(n_i-1)V^i_0(\lambda _i)=\sum _{j=1}^{n_i-1}h(K_{V^i_j}V^i_j,\hat{\nabla }_{V^i_0}T) -\sum _{j=1}^{n_i-1}h(K_{V^i_0}V^i_j,\hat{\nabla }_{V^i_j}T)\\&\quad =-h(K_{V^i_0}V^i_0,\hat{\nabla }_{V^i_0}T) -\sum _{j=1}^{n_i-1}h(K_{V^i_0}V^i_j,\hat{\nabla }_{V^i_j}T)\\&\quad =-\sum _{j=0}^{n_i-1}h(\hat{\nabla }_{V^i_j}T, L^i(V^i_0,V^i_j))=0. \end{aligned} \end{aligned}$$

(4.23)

It follows from (4.20) that $a_i\lambda _i=0$, thus $a_i=0$. Then $\hat{\nabla }_{T}T=0$. Together with (4.17), (4.10) and (4.12), by (4.14) we have (4.18) except the last equation.

Finally, we consider the Codazzi equation (2.6) of the following form

$$\begin{aligned} h((\hat{\nabla }_{Y}K)(X,X),Y)=h((\hat{\nabla }_{X}K)(Y,X),Y)+\tfrac{1}{2}(\mu _j-\mu _i). \end{aligned}$$

(4.24)

Since $h(\hat{\nabla }_{X}X, Y)=0$, and similarly $h(\hat{\nabla }_{Y}Y, X)=0$, then by (4.1) direct computations from (4.24) show the last equation of (4.18). $\square$

Lemma 4.3

If the number m of distinct affine principal curvatures is at least three, then $m=3$.

Proof

By Remark 4.2 it is sufficient to prove $m\ne 4$. On the contrary, assume $m=4$, let $\mu _2,\mu _3,\mu _4$ be the three different affine principal curvatures of multiplicity more than one. For any $X\in \mathfrak {D}(\mu _2)$, $Y\in \mathfrak {D}(\mu _3)$, $Z\in \mathfrak {D}(\mu _4)$, by (4.1) we consider the Codazzi equation (2.6) of the following form

$$\begin{aligned} h((\hat{\nabla }_{Y}K)(X,Z),T)=h((\hat{\nabla }_{X}K)(Y,Z),T) \end{aligned}$$

to obtain that

$$\begin{aligned} (\lambda _3-\lambda _4)h(\hat{\nabla }_{X}Y,Z) =(\lambda _2-\lambda _4)h(\hat{\nabla }_YX,Z). \end{aligned}$$

(4.25)

It follows from Lemma 3.3 (ii) that

$$\begin{aligned} (\mu _3-\mu _4)h(\hat{\nabla }_{X}Y,Z) =(\mu _2-\mu _4)h(\hat{\nabla }_YX,Z) =(\mu _2-\mu _3)h(\hat{\nabla }_ZX,Y). \end{aligned}$$

(4.26)

First, we claim that

$$\begin{aligned} h(\hat{\nabla }_{X}Y,Z)=h(\hat{\nabla }_YX,Z)=h(\hat{\nabla }_ZX,Y)=0. \end{aligned}$$

(4.27)

On the contrary, assume that $h(\hat{\nabla }_{X}Y,Z)\ne 0$, then the linear homogeneous system of equations (4.25) and (4.26) has nonzero solutions, thus its determinant vanishes:

$$\begin{aligned} (\mu _2-\mu _4)(\lambda _3-\lambda _4) =(\mu _3-\mu _4)(\lambda _2-\lambda _4). \end{aligned}$$

(4.28)

By the first equation of (4.6) we have

$$\begin{aligned} \mu _2-\mu _4=2\lambda _3(\lambda _2-\lambda _4),\ \mu _3-\mu _4=2\lambda _2(\lambda _3-\lambda _4), \end{aligned}$$

which together with (4.28) imply that $\lambda _2=\lambda _3$, a contradiction to Remark 4.1. Therefore, $h(\hat{\nabla }_{X}Y,Z)=0$. Together with (4.26) the claim (4.27) follows.

Next, we consider the Gauss equations for unit vector fields $X\in \mathfrak {D}(\mu _i)$, $Y\in \mathfrak {D}(\mu _j)$. From (4.27), Lemmas 4.1 and 4.2 we see that

$$\begin{aligned} \hat{\nabla }_{Y}Y-\rho _jT\in \mathfrak {D}(\mu _j),\ \hat{\nabla }_{X}Y\in \mathfrak {D}(\mu _j),\ \hat{\nabla }_{Y}X, \hat{\nabla }_{T}X\in \mathfrak {D}(\mu _i), \end{aligned}$$

which imply that $h(\hat{\nabla }_X\hat{\nabla }_YY,X) =-h(\hat{\nabla }_YY,\hat{\nabla }_XX)=-\rho _i\rho _j$. Then, by straightforward computation we obtain

$$\begin{aligned} \begin{aligned} h({\hat{R}}(X,Y)Y,X)&=h(\hat{\nabla }_X\hat{\nabla }_YY-\hat{\nabla }_Y\hat{\nabla }_XY,X) -h(\hat{\nabla }_{\hat{\nabla }_XY}Y-\hat{\nabla }_{\hat{\nabla }_YX}Y,X)\\&=-\rho _i\rho _j+h(\hat{\nabla }_XY,\hat{\nabla }_YX)-0 -h(Y,\hat{\nabla }_{\hat{\nabla }_YX}X)\\&=-\rho _i\rho _j. \end{aligned} \end{aligned}$$

(4.29)

On the other hand, it follows from (4.3) and (4.6) that $h({\hat{R}}(X,Y)Y,X)=0$, thus $\rho _i\rho _j=0$, which means that $\rho _2\rho _3=\rho _3\rho _4=\rho _2\rho _4=0$. Then at least two of $\rho _2, \rho _3, \rho _4$ are zero locally. Without loss of generality, we assume that $\rho _2=\rho _3=0$. From the last equation of (4.18) for $i=2, j=3$ we have $\mu _2=\mu _3$, a contradiction to $\mu _2\ne \mu _3$. $\square$

By Lemma 4.3 we finish the proof of Theorem 1.3.

5 Proof of Theorem 1.4

In this section, we continue the analysis of Sect. 4 for $m=3$ to complete the proof of Theorem 1.4. Let $F:M^n\rightarrow \mathbb {R}^{n+1}$ be a locally strongly convex affine hypersurface with ${\hat{R}}\cdot C=0$ and $n\ge 5$. Assume that there are exactly three distinct affine principal curvatures $\mu _1, \mu _2, \mu _3$ of multiplicity $(1, n_2, n_3)$ with $n_2\ge 2$ and $n_3\ge 2$, respectively.

First, we will prove the warped product structure of $(M^n, h)$. By the apolarity condition we have

$$\begin{aligned} \lambda _1+n_2\lambda _2+n_3\lambda _3=0, \end{aligned}$$

(5.1)

which together with the equations of the third line in (4.18) gives that

$$\begin{aligned} X(\lambda _i)=Y(\lambda _i)=X(\mu _i)=Y(\mu _i)=0,\ i=1, 2, 3 \end{aligned}$$

(5.2)

for any $X\in \mathfrak {D}(\mu _2)$, $Y\in \mathfrak {D}(\mu _3)$. Then, we can show the following lemma.

Lemma 5.1

There hold that

$$\begin{aligned} \begin{aligned}&X(\rho _i)=Y(\rho _i)=0,\ \forall \ X\in \mathfrak {D}(\mu _2),\ \forall \ Y\in \mathfrak {D}(\mu _3), \\&T(\lambda _i)=(2\lambda _i-\lambda _1)\rho _i+\tfrac{1}{2}(\mu _1-\mu _i),\\&T(\mu _i)=(\mu _i-\mu _1)(\rho _i-\lambda _i),\ \ i=2, 3,\\&\rho _2\lambda _3-\rho _3\lambda _2+\tfrac{1}{2}(\mu _2-\mu _3)=0,\\&\rho _2\rho _3=0,\ T(\rho _2)=\rho _2^2,\ T(\rho _3)=\rho _3^2. \end{aligned} \end{aligned}$$

(5.3)

Proof

Let $\{X_1,\ldots ,X_{n_2}\}$ (resp. $\{Y_1,\ldots ,Y_{n_3}\}$) be an orthonormal frame of $\mathfrak {D}(\mu _2)$ (resp. $\mathfrak {D}(\mu _3)$). Lemmas 4.1 and 4.2 imply that

$$\begin{aligned} \begin{aligned}&\hat{\nabla }_XY=\sum b_j Y_j,\ \ \hat{\nabla }_Y X=\sum a_iX_i,\\&[X, X']=\hat{\nabla }_XX'-\hat{\nabla }_{X'}X\in \mathfrak {D}(\mu _2), \end{aligned} \end{aligned}$$

(5.4)

which simplify the Gauss equation as

$$\begin{aligned}&0={\hat{R}}(X, X')T=\hat{\nabla }_X(-\rho _2 X') -\hat{\nabla }_{X'}(-\rho _2X)+\rho _2[X, X']\\&\quad =-X(\rho _2)X'+X'(\rho _2)X, \end{aligned}$$

so we have $X(\rho _2)=0$. Similarly, we get $Y(\rho _3)=0$.

Analogously, using (5.4), from

$$\begin{aligned} 0&={\hat{R}}(X, Y)T=-\hat{\nabla }_X(\rho _3 Y)+\hat{\nabla }_Y (\rho _2 X)-\sum _j b_j\hat{\nabla }_{Y_j}T+\sum _i a_i\hat{\nabla }_{X_i}T\\&=-X(\rho _3)Y+Y(\rho _2)X, \end{aligned}$$

we get $X(\rho _3)=Y(\rho _2)=0$. Together with Lemmas 4.1 and 4.2 we have proved (5.3) except the equations of last line in (5.3).

By the same computations as that did in (4.29), we have $\rho _2\rho _3=0$. Analogously, it follows from (4.3) and (4.6) that ${\hat{R}}(X, T)T=0$. On the other hand, by Lemmas 4.1 and 4.2 we also have

$$\begin{aligned} {\hat{R}}(X, T)T&=\hat{\nabla }_T(\rho _2 X)-\hat{\nabla }_{\hat{\nabla }_XT}T+\hat{\nabla }_{\hat{\nabla }_TX}T\\&=T(\rho _2)X+\rho _2\hat{\nabla }_TX-\rho _2^2X-\rho _2\hat{\nabla }_TX\\&=(T(\rho _2)-\rho _2^2)X. \end{aligned}$$

Thus, $T(\rho _2)=\rho _2^2$. Similarly, we get $T(\rho _3)=\rho _3^2$. $\square$

Now, it follows from Lemma 4.1 that

$$\begin{aligned} \hat{\nabla }_XT=-\rho _2 X,\quad \hat{\nabla }_YT=-\rho _3 Y. \end{aligned}$$

Together with previous lemmas in Sect. 4 we see that $\mathfrak {D}(\mu _i)$ ($i=1,2,3$) are integrable, and both $\mathfrak {D}(\mu _1)\oplus \mathfrak {D}(\mu _3)$ and $\mathfrak {D}(\mu _1)\oplus \mathfrak {D}(\mu _2)$ are auto-parallel. Moreover, one can show that $\mathfrak {D}(\mu _2)$ (resp. $\mathfrak {D}(\mu _3)$) is spherical with the mean curvature vector $\rho _2 T$ (resp. $\rho _3 T$). Therefore, by Theorem 2.3 we conclude that $M^n$ is locally a warped product $\mathbb {R} \times _{f_2}M_2\times _{f_3}M_3$, where $\mathbb {R}$, $M_2$ and $M_3$ are, respectively, integral manifolds of the distributions $\mathfrak {D}(\mu _1), \mathfrak {D}(\mu _2)$ and $\mathfrak {D}(\mu _3)$. The warping functions $f_2$ and $f_3$ are determined by

$$\begin{aligned} \rho _i=-T(\ln f_i),\ i=2,\ 3. \end{aligned}$$

By the warped product structure, we always take the local coordinates $\{t,x_i,y_j\}$ on $M^n$ such that $\tfrac{\partial }{\partial t}=T$, span$\{\tfrac{\partial }{\partial x_1},\ldots ,\tfrac{\partial }{\partial x_{n_2}}\}=\mathfrak {D}(\mu _2)$ and span$\{\tfrac{\partial }{\partial y_1},\ldots ,\tfrac{\partial }{\partial y_{n_3}}\}=\mathfrak {D}(\mu _3)$, and also let $X, X'\in \mathfrak {D}(\mu _2)$ and $Y,Y'\in \mathfrak {D}(\mu _3)$ for convention. Then we see from (5.2) and (5.3) that all the functions $\mu _i, \lambda _i, \rho _j$ and $f_j$ depend only on t. Denote by $\partial _t()=(\cdot )'$, they are related by (4.2), (4.6) and (5.3):

$$\begin{aligned} \begin{aligned}&(\mu _1-\mu _2)\lambda _2=(\mu _1-\mu _3)\lambda _3\ne 0,\ \mu _2+\mu _3=2\lambda _2\lambda _3\ne 0,\\&\mu _1+\mu _i=2\lambda _i(\lambda _1-\lambda _i),\ \rho _i'=\rho _i^2,\ i=2, 3,\\&\rho _2\rho _3=0,\ \rho _2\lambda _3-\rho _3\lambda _2+\tfrac{1}{2}(\mu _2-\mu _3)=0. \end{aligned} \end{aligned}$$

(5.5)

By the equations of last line in (5.5), without loss of generality, from now on we assume $\rho _2=0$ locally, thus $\rho _3\ne 0$. Then, we can solve from the equations above for the warping function $f_3$ and $\rho _3$ to get that, up to a translation and a direction of the parametric t,

$$\begin{aligned} f_2=1,\ f_3=t,\ \rho _3=-\tfrac{1}{t}, \end{aligned}$$

(5.6)

where locally we take $t>0$. Together with (5.5) we further see that

$$\begin{aligned} \begin{aligned} \lambda _2=\tfrac{1}{2}(\mu _2-\mu _3)\rho _3^{-1},\ \lambda _3 =\tfrac{\mu _2+\mu _3}{\mu _2-\mu _3}\rho _3,\ \lambda _1-\lambda _2=\tfrac{\mu _1+\mu _2}{\mu _2-\mu _3}\rho _3. \end{aligned} \end{aligned}$$

(5.7)

Second, we will show some properties for the functions as above in next two lemmas. Recall from Remark 4.1 that $\lambda _2, \lambda _3$ are nonzero and distinct, we can prove the similar results for $\mu _2$ and $\mu _3$ as follows.

Lemma 5.2

Locally, both $\mu _2$ and $\mu _3$ are nonzero and distinct.

Proof

As $\mu _1, \mu _2, \mu _3$ are distinct, by $\rho _2=0$ we see from the equations of third line in (5.3) that $T(\mu _2)\ne 0$, thus $\mu _2$ cannot vanish identically, locally let $\mu _2\ne 0$.

Assume that $\mu _3=0$. Then $\mu _2\mu _1\ne 0$. From Lemma 5.1 and (5.5) we see that

$$\begin{aligned} \begin{aligned}&\lambda _3=\rho _3,\ (\mu _1-\mu _2)\lambda _2=\mu _1\lambda _3,\\&2\lambda _3\lambda _2=\mu _2,\ \lambda _1-\lambda _2=\tfrac{\mu _1+\mu _2}{2\lambda _2}, \end{aligned} \end{aligned}$$

(5.8)

which together with (5.1) imply that

$$\begin{aligned} \begin{aligned}&\lambda _2=\tfrac{1}{2}\mu _2\rho _3^{-1},\qquad \qquad \ \lambda _1=\tfrac{1}{2}\mu _2\rho _3^{-1}+(1+\mu _1/\mu _2)\rho _3,\\&\mu _1-\mu _2=2\rho _3^2\mu _1/\mu _2,\quad \ \lambda _1=-\tfrac{1}{2}n_2\mu _2\rho _3^{-1}-n_3\rho _3. \end{aligned} \end{aligned}$$

(5.9)

Therefore, it holds that

$$\begin{aligned} \begin{aligned}&\tfrac{1}{2}(n_2+1)\mu _2\rho _3^{-1}+(n_3+1)\rho _3+\tfrac{\mu _1}{\mu _2}\rho _3=0,\\&\tfrac{1}{2}(\mu _2-\mu _1)\rho _3^{-1}+\tfrac{\mu _1}{\mu _2}\rho _3=0. \end{aligned} \end{aligned}$$

By subtracting these equations we get

$$\begin{aligned} \mu _1 +n_2\mu _2=-2(n_3+1)\rho _3^2\ne 0, \end{aligned}$$

(5.10)

which together with the third equation of (5.9) shows that

$$\begin{aligned} \frac{\mu _1 +n_2\mu _2}{\mu _1-\mu _2}=-\frac{n_3+1}{\mu _1/\mu _2}. \end{aligned}$$

(5.11)

Then we see that $\kappa _0:=\mu _1/\mu _2$ is the solution of the quadric equation

$$\begin{aligned} \kappa _0^2+n\kappa _0-n_3-1=0. \end{aligned}$$

It follows from this and (5.10) that $\kappa _0$ is a constant, $\kappa _0\notin \{0,1,2,-n_2\}$, and

$$\begin{aligned} \mu _2=-\frac{2(n_3+1)}{n_2+\kappa _0}\rho _3^2,\ \mu _1=-\frac{2(n_3+1)\kappa _0}{n_2+\kappa _0}\rho _3^2. \end{aligned}$$

(5.12)

By taking the derivative on both sides of $\mu _2=2\lambda _2\rho _3$ in (5.8), we see from (5.3) and (5.9) that

$$\begin{aligned} \mu _2'=2\rho _3^2\lambda _2+\rho _3(\mu _1-\mu _2)=\mu _1\rho _3. \end{aligned}$$

On the other hand, by (5.12) we have $\mu _2'=2\rho _3\mu _2=\tfrac{2}{\kappa _0}\mu _1\rho _3$. Combining this with the equation above, as $\kappa _0\ne 2$, we get $\mu _1\rho _3=0$, a contradiction to $\mu _1\rho _3\ne 0$. Therefore, $\mu _3\ne 0$. $\square$

Furthermore, we see from (5.5) that

$$\begin{aligned} \rho _3\lambda _2=\tfrac{1}{2}(\mu _2-\mu _3)=\mu _2-\lambda _2\lambda _3 =\lambda _2\lambda _3-\mu _3, \end{aligned}$$

(5.13)

which implies that

$$\begin{aligned} \mu _2=\lambda _2(\lambda _3+\rho _3),\ \ \mu _3=\lambda _2(\lambda _3-\rho _3). \end{aligned}$$

(5.14)

Then, it holds that

$$\begin{aligned} \mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2 =\lambda _2(\rho _3-\lambda _3)(\rho _3+\lambda _3-\lambda _2) =\mu _3(\lambda _2-\rho _3-\lambda _3). \end{aligned}$$

(5.15)

Now, we are ready to prove the following lemma.

Lemma 5.3

Set $H_2=\mu _2-\lambda _2^2$, $H_3=1+(\mu _3-\lambda _3^2)/\rho _3^2$. Then $H_2$ and $H_3$ are nonzero constant. Moreover,

$$\begin{aligned} \begin{aligned}&4\mu _3H_2+(\mu _2-\mu _3)^2H_3=0,\\&\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2=(\rho _3-\lambda _3)H_2\ne 0. \end{aligned} \end{aligned}$$

(5.16)

Proof

By (5.3) and the equations of second line in (5.5) we can check that

$$\begin{aligned} (\mu _2-\lambda _2^2)'=0,\ (\tfrac{\mu _3-\lambda _3^2}{\rho _3^2})'=0. \end{aligned}$$

Therefore, $H_2$ and $H_3$ are all constant. By Lemma 5.2 and (5.14) we have

$$\begin{aligned} \rho _3-\lambda _3\ne 0,\ H_2=\lambda _2(\rho _3+\lambda _3-\lambda _2),\ H_3=(\rho _3-\lambda _3)(\rho _3+\lambda _3-\lambda _2)\rho _3^{-2}. \end{aligned}$$

(5.17)

Assume that $H_2=0$. Then we see from (5.17) and $\lambda _2\ne 0$ that

$$\begin{aligned} \begin{aligned}&\rho _3=\lambda _2-\lambda _3\ne 0,\ \mu _2=\lambda _2^2,\\&\mu _3-\lambda _3^2+\rho _3^2=\rho _3^2H_3=0. \end{aligned} \end{aligned}$$

(5.18)

Taking the derivative on both sides of $\rho _3=\lambda _2-\lambda _3$, by (5.3) and (5.14) we have

$$\begin{aligned} \rho _3^2=(\lambda _1-\lambda _2-2\lambda _3)\rho _3, \end{aligned}$$

which together with (5.18) implies that

$$\begin{aligned} \lambda _1=2\lambda _2+\lambda _3. \end{aligned}$$

(5.19)

Combining with (5.1) and (5.18) we obtain

$$\begin{aligned} \lambda _3=-\tfrac{n_2+2}{n_3+1}\lambda _2,\ (\lambda _3-\lambda _2)^2=(\tfrac{n+2}{n_3+1})^2\lambda _2^2=\rho _3^2, \end{aligned}$$

which further show that

$$\begin{aligned} \mu _2=\lambda _2^2=(\tfrac{n_3+1}{n+2})^2\rho _3^2,\ \mu _3=\lambda _3^2-\rho _3^2=-\tfrac{(n_3+1)(n+n_2+4)}{(n+2)^2}\rho _3^2. \end{aligned}$$

By combining this with (5.13) we have

$$\begin{aligned} \lambda _2=\tfrac{1}{2}(\mu _2-\mu _3)\rho _3^{-1}=\tfrac{n_3+1}{n+2}\rho _3,\ \lambda _3=-\tfrac{n_2+2}{n+2}\rho _3, \end{aligned}$$

(5.20)

which imply that

$$\begin{aligned} \lambda _2'=\tfrac{n_3+1}{n+2}\rho _3^2. \end{aligned}$$

(5.21)

On the other hand, by (5.19) and (5.20) we see from the second equation of (5.3) and the first equation of (4.7) that

$$\begin{aligned} \lambda _2'=\tfrac{1}{2}(\mu _1-\mu _2)=\lambda _3\lambda _2 =-\tfrac{(n_2+2)(n_3+1)}{(n+2)^2}\rho _3^2. \end{aligned}$$

Together with (5.21) we have $\rho _3=0$. This contradiction shows that $H_2\ne 0$.

Now, by $H_2\ne 0$, (5.13)-(5.15) and (5.17) we obtain (5.16), which together with Lemma 5.2 implies $H_3\ne 0$. $\square$

Finally, based on previous lemmas, we can prove Theorem 1.4.

Completion of Theorem 1.4’s Proof

Define a vector field by

$$\begin{aligned}&g_3=M(\lambda _2\xi +\mu _2 T), \end{aligned}$$

(5.22)

where M(t) is a nonzero solution of the equation $M'+M(\lambda _1-\lambda _2)=0$. Then direct computations give that

$$\begin{aligned} \begin{aligned}&D_T g_3=(M'+M(\lambda _1-\lambda _2))(\lambda _2\xi +\mu _2T)=0,\\&D_X g_3=M(-\mu _2\lambda _2X+\mu _2(\hat{\nabla }_X T+K(X, T)))=0,\\&D_Y g_3={g_3}_*Y=-M(\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2)Y,\\&D_{Y'}D_Yg_3=-M(\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2)\\&\qquad \qquad \quad \cdot [\hat{\nabla }^{\bot }_{Y'}Y +L^3(Y, Y')+h(Y,Y')(\xi +(\rho _3+\lambda _3)T)], \end{aligned} \end{aligned}$$

(5.23)

where $\hat{\nabla }^{\bot }_{Y'}Y=\hat{\nabla }_{Y'}Y-\rho _3h(Y,Y')T$ is the projection of $\hat{\nabla }_{Y'}Y$ on $\mathfrak {D}(\mu _3)$, and $L^3$ is the projection tensor of K on $\mathfrak {D}(\mu _3)$ defined by (3.9).

Similarly, define another vector field

$$\begin{aligned} g_2=N((\lambda _3-\rho _3)\xi +\mu _3 T), \end{aligned}$$

(5.24)

where N(t) is a nonzero solution of $N'+N(\rho _3+\lambda _1-\lambda _3)=0$. It holds that

$$\begin{aligned} \begin{aligned}&D_Tg_2=(N'+N(\rho _3+\lambda _1-\lambda _3))((\lambda _3-\rho _3)\xi +\mu _3T)=0,\\&D_Y g_2=N(-\mu _3(\lambda _3-\rho _3)Y+\mu _3(\hat{\nabla }_Y T+K(Y,T)))=0,\\&D_X g_2={g_2}_*X=N(\mu _3\lambda _2+\mu _2(\rho _3-\lambda _3))X,\\&D_{X'}D_Xg_2=N(\mu _3\lambda _2+\mu _2(\rho _3-\lambda _3))\\&\qquad \qquad \quad \cdot [\hat{\nabla }^{\bot }_{X'}X +L^{2}(X, X') +h(X, X')(\xi +\lambda _2T)], \end{aligned} \end{aligned}$$

(5.25)

where $\hat{\nabla }^{\bot }_{X'}X$ is the projection of $\hat{\nabla }_{X'}X$ on $\mathfrak {D}(\mu _2)$, and $L^2$ is the projection tensor of K on $\mathfrak {D}(\mu _2)$ defined by (3.9).

From Lemma 5.3 we have $\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2\ne 0$. Then, by (5.23) (resp. (5.25)) we see that $g_3$ (resp. $g_2$) is an immersion from the integral manifold $M_3$ (resp. $M_2$) of $\mathfrak {D}(\mu _3)$ (resp. $\mathfrak {D}(\mu _2)$) into the affine space. Moreover, by (5.14), (5.22) and (5.23) there holds

$$\begin{aligned} D_{Y'}D_Y g_3={g_3}_*(\hat{\nabla }^{\bot }_{Y'}Y +L^3(Y, Y')) -h(Y,Y')(\mu _3-\lambda _3^2+\rho _3^2)g_3\in \mathfrak {D}(\mu _3)+\textrm{span}(g_3), \end{aligned}$$

where $\mu _3-\lambda _3^2+\rho _3^2=H_3\rho _3^2\ne 0$. It follows from Lemma 3.4 that $L^{3}$ satisfies apolarity condition, and ${\hat{R}}^{\bot }(Y,Y')\cdot L^3\ne 0$ in general, thus $g_3$ is a proper affine hypersphere with affine metric $\rho _3^{2}h=f_3^{-2}h$ (cf. (5.6)), affine mean curvature $H_3$, and difference tensor $L^3$. It follows from Proposition 3.1 that $g_3$ is an ellipsoid if $\mu _3-\lambda _3^2=\rho _3^2(H_3-1)\ge 0$, i.e., $H_3\ge 1$.

Similarly, we have

$$\begin{aligned} D_{X'}D_X g_2={g_2}_*(\hat{\nabla }^{\bot }_{X'}X +L^2(X, X')) -h(X,X')(\mu _2-\lambda _2^2)g_2\in \mathfrak {D}(\mu _2)+\textrm{span}(g_2), \end{aligned}$$

where $\mu _2-\lambda _2^2 =H_2\ne 0$. Then, we see from $\rho _2=0$ and Lemma 3.4 that $L^{2}$ satisfies apolarity condition and ${\hat{R}}^{\bot }\cdot L^2=0$. Therefore, $g_2$ is a proper affine hypersphere with affine metric h, affine mean curvature $H_2$, and difference tensor $L^2$. Hence, $g_2$ has semi-parallel cubic form. It follows from Proposition 3.1 that $g_2$ is an ellipsoid if $H_2>0$.

Let $\beta _1(t)$ and $\beta _2(t)$ be functions such that

$$\begin{aligned} \beta _1'=-\beta _2,\quad \beta _2'=1+\beta _1\mu _1-\beta _2\lambda _1. \end{aligned}$$

Denote by $\delta _1=1+\mu _2\beta _1-\lambda _2\beta _2$ and $\delta _2=1+\mu _3\beta _1+(\rho _3-\lambda _3)\beta _2$. It follows from Lemma 5.3 that $\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2\ne 0$. Then, by choosing the initial conditions for $\beta _1$ and $\beta _2$ appropriately we can let $\delta _1(0)=\delta _2(0)=0$. Moreover, from (4.6) and (5.3) we see that

$$\begin{aligned} \delta _1'=-\lambda _2\delta _1,\ \delta _2'=(\rho _3-\lambda _3)\delta _2. \end{aligned}$$

Therefore, by the initial conditions we have $\delta _1=\delta _2=0$ identically.

Now, straight computations from above show that

$$\begin{aligned}&D_X(\beta _1 \xi +\beta _2 T)=X,\ \ \forall \ X\in \mathfrak {D}(\mu _2),\\&D_Y(\beta _1 \xi +\beta _2 T)=Y,\ \ \forall \ Y\in \mathfrak {D}(\mu _3),\\&D_T(\beta _1 \xi +\beta _2 T)=T. \end{aligned}$$

Then, up to a translation constant, we can write $F:M^{n}\rightarrow \mathbb {R}^{n+1}$ as

$$\begin{aligned} F=\beta _1\xi +\beta _2 T. \end{aligned}$$

From (5.22) and (5.24), as $\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2\ne 0$, by (5.15) and (5.17) we can uniquely express $\xi$ and T to obtain

$$\begin{aligned} F(t,x,y)=\gamma _2(t)g_2(x)+\gamma _3(t)g_3(y), \end{aligned}$$

(5.26)

where $x=(x_1,\ldots ,x_{n_2})$, $y=(y_1,\ldots ,y_{n_3})$,

$$\begin{aligned}&\gamma _2(t)=\tfrac{N^{-1}}{\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2} =\tfrac{1}{H_3N\lambda _2\rho _3^2},\\&\gamma _3(t)=\tfrac{-M^{-1}}{\mu _2(\rho _3-\lambda _3)+\mu _3\lambda _2} =-\tfrac{1}{H_3M\lambda _2\rho _3^2}. \end{aligned}$$

By (5.3), (5.5) and (5.14), direct computations show that

$$\begin{aligned} \gamma '_2(t)=\lambda _2\gamma _2(t),\ \gamma '_3(t)=(\lambda _3-\rho _3)\gamma _3(t), \end{aligned}$$

where, it follows from (5.6) and (5.7) that

$$\begin{aligned} \lambda _2=\tfrac{1}{2}(\mu _3-\mu _2)t,\ \lambda _3-\rho _3=\tfrac{2\mu _3}{(\mu _3-\mu _2)t}. \end{aligned}$$

Furthermore, we put $\rho _3=-1/t$ and (5.7) into the first equation of (5.5) to get

$$\begin{aligned} t^2(\mu _1-\mu _2)(\mu _2-\mu _3)^2=2(\mu _1-\mu _3)(\mu _2+\mu _3). \end{aligned}$$

By (5.7) and Lemma 5.3 we can rewrite the nonzero constants $H_2$ and $H_3$ by

$$\begin{aligned} H_2=\mu _2-\tfrac{1}{4}(\mu _2-\mu _3)^2t^2,\ H_3=\mu _3t^2+\tfrac{2}{\mu _2-\mu _3}. \end{aligned}$$

Summing above, we have completed the proof of Theorem 1.4. $\square$

6 Proof of Theorem 1.5

Let $F: M^n\rightarrow \mathbb {R}^{n+1}$ be a locally strongly convex quasi-umbilical affine hypersurface with ${\hat{R}}\cdot C=0$ and $n\ge 3$. Denote by $\mu _1,\mu _2$ the two distinct affine principal curvatures of multiplicity $(1, n-1)$, respectively. Then, $M^n$ is not an affine hypersphere. Let T be the unit eigenvector field of the affine principal curvature $\mu _1$. As before, omit the upper index $j=1$ for $\lambda ^j_i$ in Lemma 3.2, we have

$$\begin{aligned} \begin{aligned}&ST=\mu _1T,\quad \quad SX=\mu _2X,\\&K_{T}T=\lambda _1T,\quad K_{T}X=\lambda _2X,\ \forall \ X\in \mathfrak {D}(\mu _2), \end{aligned} \end{aligned}$$

(6.1)

where by apolarity condition it holds that

$$\begin{aligned} \lambda _1+(n-1)\lambda _2=0. \end{aligned}$$

(6.2)

Remark 6.1

$\lambda _1, \lambda _2$ are distinct and nonzero, $M^n$ is affinely equivalent to one of the three classes of immersions in Theorem 2.2 by taking $m=n-1$. In fact, it follows from (6.1) that $M^n$ satisfies the conditions of Theorem 2.2. In the proof of Theorem 2.2 in [1], it was shown in Lemma 3 that if $\lambda _2=0$, then $K_T=0$ and $M^n$ is an affine hypersphere. In our situation, by $M^n$ being not an affine hypersphere we can exclude this possibility in Theorem 2.2, and obtain the conclusions.

Next, for more information we will show the warped product structure and discuss all the possibilities of the immersion. By (6.1) we see from (2.5) that

$$\begin{aligned} {\hat{R}}(X,T)T=(\lambda ^2_2-\lambda _1\lambda _2+\tfrac{1}{2}(\mu _1+\mu _2))X \end{aligned}$$

for any unit vector field $X\in \mathfrak {D}(\mu _2)$. As $\lambda _2\ne 0$, it follows from (6.2) that $\lambda _1-2\lambda _2\ne 0$. Then, by (3.1) we can compute

$$\begin{aligned} \begin{aligned} h({\hat{R}}(X,T)K(T,T),X)=2h(K({\hat{R}}(X,T)T,T),X) \end{aligned} \end{aligned}$$

to obtain that

$$\begin{aligned} \begin{aligned} \lambda ^2_2-\lambda _1\lambda _2+\tfrac{1}{2}(\mu _1+\mu _2)=0, \end{aligned} \end{aligned}$$

(6.3)

which together with (6.2) implies that

$$\begin{aligned} \mu _1+\mu _2=-2n\lambda ^2_2<0. \end{aligned}$$

(6.4)

In the proof of Theorem 2.2 in [1], together with (6.3) it was shown that

$$\begin{aligned} \begin{aligned}&\hat{\nabla }_{T}T=0,\ \hat{\nabla }_{X}T=-\alpha X,\ T(\alpha )=\alpha ^2,\\&X(\alpha )=X(\mu _1)=X(\mu _2)=X(\lambda _2)=0,\ \forall \ X\in \mathfrak {D}(\mu _2),\\&T(\lambda _2)=(n+1)\lambda _2\alpha +\tfrac{1}{2}(\mu _1-\mu _2),\\&T(\mu _2)=(\mu _2-\mu _1)(\alpha -\lambda _2). \end{aligned} \end{aligned}$$

(6.5)

Therefore, $\mathfrak {D}(\mu _1)$ is auto-parallel and the distribution $\mathfrak {D}(\mu _2)$ is spherical with the mean curvature vector $\alpha T$. It follows from Theorem 2.3 that $M^n$ is locally a warped product $\mathbb {R} \times _{f}M_2$, where $\mathbb {R}$ and $M_2$ are, respectively, integral manifolds of the distributions $\mathfrak {D}(\mu _1)$ and $\mathfrak {D}(\mu _2)$. The warping function f is determined by $\alpha =-T(\ln f)$. As before, we take the local coordinate $\{t,x_1,\ldots ,x_{n-1}\}$ on $M^n$ such that $\tfrac{\partial }{\partial t}=T$, span$\{\tfrac{\partial }{\partial x_1},\ldots ,\tfrac{\partial }{\partial x_{n-1}}\}=\mathfrak {D}(\mu _2)$. Hence all functions $\mu _i, \lambda _i, \alpha$ and f depend only on t.

Denote by $\partial _t()=(\cdot )'$, we have $\alpha =-f'/f$. By solving from the equations above for f and $\alpha$, we get that, up to a translation and a direction of the parametric t,

$$\begin{aligned} f=1,\ \alpha =0;\ \textrm{or} \ f=t,\ \alpha =-\tfrac{1}{t}, \end{aligned}$$

(6.6)

where locally we take $t>0$. From (6.4)-(6.6) we can check that $(\mu _2-\lambda _2^2)'=0$ if $f=1$, and $((\mu _2-\lambda _2^2)/\alpha ^2)'=0$ if $f=t$. Therefore, by (6.4) and (6.6) we have a constant:

$$\begin{aligned} H_0= \left\{ \begin{array}{lll} \mu _2-\lambda _2^2=\tfrac{\mu _1+(2n+1)\mu _2}{2n},\ {} &{}\textrm{if} \ f=1,\\ 1+(\mu _2-\lambda _2^2)/\alpha ^2=1+\tfrac{\mu _1+(2n+1)\mu _2}{2n}t^2, &{}\textrm{if} \ f=t. \end{array} \right. \end{aligned}$$

(6.7)

Finally, based on the proof of Theorem 2.2 and (6.6) we will follow the computations in [3] for three kinds of immersion in Theorem 2.2 to prove the following theorem, which give the explicit expressions of the immersions in Theorem 1.5.

Theorem 6.1

Let $M^n$ be a locally strongly convex quasi-umbilical affine hypersurface in $\mathbb {R}^{n+1}$ with ${\hat{R}}\cdot C=0$ and $n\ge 3$. Denote by $\mu _1, \mu _2$ the two distinct affine principal curvatures of multiplicity $(1, n-1)$, respectively. Then, $(M^n,h)$ is locally isometric to the warped product $\mathbb {R_+} \times _{f}M_2$, where $f(t)=1$ or t. Moreover, $H_0$ defined by (6.7) is a constant, and $M^n$ is affinely equivalent to one of the following hypersurfaces:

• (1) The immersion $(\gamma _1(t), \gamma _2(t)g_2(x_1,\ldots ,x_{n-1}))$ if $H_0\mu _2\ne 0$ and $f(t)=1$, where $\gamma _1'\gamma _2^{n}=1$, $\gamma _2$ is explicitly given in (6.10), $g_2$ is a hyperbolic affine hypersphere with semi-parallel cubic form if $H_0<0$, or an ellipsoid if $H_0>0$.

• (2) The immersion $(\gamma _1(t), \gamma _2(t)g_2(x_1,\ldots ,x_{n-1}))$ if $H_0\mu _2\ne 0$ and $f(t)=t$, where $\gamma _1'\gamma _2^{n}=t^{n+1}$, $\gamma _2$ is a positive solution to the differential equation

  $$\begin{aligned} \gamma _2=k(t)^{1/(n+1)},\ t^2k''(t)-(n+1)tk'(t)+(n+1)H_0k(t)=0, \end{aligned}$$

  $g_2$ is a locally strongly convex proper affine hypersphere with affine mean curvature $H_0$, and it is an ellipsoid if $H_0\ge 1$.

• (3) The immersion $(\gamma _1(t)x,\tfrac{1}{2}\gamma _1(t)\sum _{i=1}^{n-1}x_i^2 +\gamma _2(t),\gamma _1(t))$ if $\mu _2\ne 0$, $H_0=0$ and $f(t)=1$, where

  $$\begin{aligned} \gamma _1=((n+1)t)^{\tfrac{1}{n+1}}, \gamma _2=\tfrac{t((n+1)t)^{(n+2)/(n+1)}}{4n+6}, x=(x_1,\ldots ,x_{n-1}). \end{aligned}$$

• (4) The immersion $(\gamma _1(t)x, \gamma _1(t)g(x)+\gamma _2(t),\gamma _1(t))$ if $\mu _2\ne 0$, $H_0=0$ and $f(t)=t$, where

  $$\begin{aligned} \qquad \gamma _1=(\tfrac{n+1}{n+2} t^{n+2}+c_1)^{\tfrac{1}{n+1}}, \gamma _2'=\tfrac{n+1}{n+2}\gamma _1'\ln t-\tfrac{\gamma _1}{(n+2)t}, x=(x_1,\ldots ,x_{n-1}), \end{aligned}$$

  $c_1$ is a constant, and g(x) is a convex function whose graph immersion is a parabolic affine hypersphere.

• (5) The immersion $(x_1,\ldots ,x_{n-1}, g(x_1,\ldots ,x_{n-1})-\tfrac{1}{n+2}\ln t,\tfrac{1}{n+2}t^{n+2})$ if $\mu _2=0$, where the warped function $f(t)=t$ and g is a convex function whose graph immersion is a parabolic affine hypersphere.

Proof

We continue the analysis as above. First, we remark that $\mu _2=0$ if and only if $\alpha =\lambda _2$. In fact, it follows from the last equation of (6.5) that $\alpha =\lambda _2$ if $\mu _2=0$. If $\alpha =\lambda _2$, by taking its derivative on both sides, we see from (6.4) and (6.5) that $\mu _2=0$. Therefore, $\mu _2\ne 0$ if and only if $\alpha \ne \lambda _2$. Then, by $H_0$ defined by (6.7) we divide our discussions into three cases:

Case I: $H_0\mu _2\ne 0$;   Case II: $\mu _2\ne 0$, $H_0=0$;   Case III: $\mu _2=0$.

Case I. In this case, by (6.6) and (6.7) we have

$$\begin{aligned} \mu _2^2+(\alpha -\lambda _2)^2\ne 0,\ \mu _2-\lambda _2^2+\alpha ^2\ne 0. \end{aligned}$$

(6.8)

Then it was shown in [1] that $M^n$ is locally given by

$$\begin{aligned} F(t, x_1,\ldots ,x_{n-1})=(\gamma _1(t), \gamma _2(t)g_2(x_1,\ldots ,x_{n-1})), \end{aligned}$$

where $g_2$ is the proper affine hypersphere. The same proof in [1] implies that the projection tensor $L^2$ of the difference tensor on $\mathfrak {D}(\mu _2)$ (cf. (3.9)) is the difference tensor of $g_2$, and $g_2$ has the affine metric $f^{-2}h$, affine mean curvature $(\mu _2-\lambda _2^2+\alpha ^2)f^2=H_0\ne 0$ (cf. (6.6)-(6.8)) and the affine normal $-H_0 g_2$. Then, by the computations of this immersion on page 292-294 in [3] we take $\lambda =-H_0$ in (4.3) of [3], and deduce that

$$\begin{aligned} \begin{aligned}&\gamma _1'\gamma _2^{n}=f^{n+1},\ \gamma _2=k(t)^{1/(n+1)},\\&f^2k''(t)-(n+1)ff'k'(t)+(n+1)H_0k(t)=0, \end{aligned} \end{aligned}$$

(6.9)

where k(t) and $\gamma _2$ are positive functions.

If $f=1$, then $\alpha =0$. By Lemma 3.4 (iv) the integral manifold $M_2$ of $\mathfrak {D}(\mu _2)$ is totally geodesic and ${\hat{R}}^{\bot }\cdot L^2=0$, i.e., $g_2$ has semi-parallel cubic form. Proposition 3.1 and (6.7) further imply that $g_2$ is an ellipsoid if $H_0=\mu _2-\lambda _2^2>0$. Moreover, (6.9) reduces to $\gamma _1'\gamma _2^{n}=1$ and $\gamma _2=k(t)^{1/(n+1)}$, where $k''(t)+(n+1)H_0k(t)=0$. Solving this equation we obtain that

$$\begin{aligned} \gamma _2= \left\{ \begin{array}{lll} (c_1e^{\sqrt{-(n+1)H_0}t}+c_2e^{-\sqrt{-(n+1)H_0}t})^{\tfrac{1}{n+1}}, &{}\textrm{if}\ H_0<0,\\ (c_1\cos (\sqrt{(n+1)H_0}t)+c_2\sin (\sqrt{(n+1)H_0}t))^{\tfrac{1}{n+1}}, &{}\textrm{if}\ H_0>0, \end{array} \right. \end{aligned}$$

(6.10)

where the constants $c_1, c_2$ are chosen such that $\gamma _2>0$. This is the immersion (1).

If $f=t$, then $\alpha =-1/t$. We see from (6.7) and Proposition 3.1 that $g_2$ is an ellipsoid if $\mu _2-\lambda _2^2=\alpha ^2(H_0-1)\ge 0$, i.e., $H_0\ge 1$. Then, (6.9) reduces to $\gamma _1'\gamma _2^{n}=t^{n+1}$, $\gamma _2=k(t)^{1/(n+1)}$ and

$$\begin{aligned} \begin{aligned} t^2k''(t)-(n+1)tk'(t)+(n+1)H_0k(t)=0. \end{aligned} \end{aligned}$$

(6.11)

In particular, if k(t) is a power function of t, we deduce that

$$\begin{aligned} \gamma _2(t)= \left\{ \begin{array}{lll} c_1t^{\tfrac{n+2}{2(n+1)}}, &{}\textrm{if} \ H_0=\tfrac{(n+2)^2}{4(n+1)},\\ (c_2t^{\tau _1}+c_3t^{\tau _2})^{\tfrac{1}{n+1}}, &{}\textrm{if} \ H_0<\tfrac{(n+2)^2}{4(n+1)},\\ 0, &{}\textrm{if} \ H_0>\tfrac{(n+2)^2}{4(n+1)}, \end{array} \right. \end{aligned}$$

(6.12)

where $c_1$ is a positive constant, $c_2, c_3$ are chosen such that $\gamma _2>0$, and $\tau _1, \tau _2$ are the solutions of the quadric equation $\tau ^2-(n+2)\tau +(n+1)H_0=0$. This is the immersion (2).

Case II. In this case, by (6.6) and (6.7) we have

$$\begin{aligned} \mu _2^2+(\alpha -\lambda _2)^2\ne 0,\ \mu _2-\lambda _2^2+\alpha ^2= 0. \end{aligned}$$

It was shown in [1] that $M^n$ is locally given by

$$\begin{aligned} \begin{aligned} F(t,x)=(\gamma _1(t)x, \gamma _1(t)g(x)+\gamma _2(t),\gamma _1(t)), \end{aligned} \end{aligned}$$

where $x=(x_1,\ldots ,x_{n-1})$, and g(x) is a convex function whose graph immersion is a parabolic affine hypersphere. As before, the same proof in [1] implies that the projection tensor $L^2$ of the difference tensor on $\mathfrak {D}(\mu _2)$ (cf. (3.9)) is the difference tensor of g. It follows from the computations of such hypersurfaces on page 294 of [3] that $\gamma _1,\gamma _2$ satisfy

$$\begin{aligned} (\gamma _1'\gamma _2''-\gamma _1''\gamma _2')f^2=\gamma _1\gamma _1',\ f=|\gamma _1^n\gamma _1'|^{1/(n+1)}. \end{aligned}$$

(6.13)

If $f=1$, then $\alpha =0$, and thus $\mu _2-\lambda _2^2=0$, it follows from Proposition 3.1 that $L^{2}=0$, which together with (3.10) of Lemma 3.4 (iii) implies that $M^n$ is a flat and quasi-umbilical affine hypersurface. We see from Theorem 4.1 of [3] that this is the immersion (3).

If $f=t$, by (6.13) we have $(\gamma _1^{n+1})'=\varepsilon (n+1)t^{n+1}$, $\varepsilon \in \{-1,1\}$, which gives that $\gamma _1^{n+1}=\tfrac{n+1}{n+2}\varepsilon t^{n+2}+c_1$. By applying an affine reflection we may assume $\gamma _1>0$, then put $\varepsilon =1$ and $\gamma _1=(\tfrac{n+1}{n+2} t^{n+2}+c_1)^{1/(n+1)}$. By (6.13) we get $(\gamma _2'/\gamma _1')'=\tfrac{n+1}{n+2}t^{-1}+ c_1t^{-n-3}$, which yields that

$$\begin{aligned} \gamma _2'/\gamma _1'=\tfrac{n+1}{n+2}\ln t -\tfrac{ c_1}{(n+2)t^{n+2}}+c_2. \end{aligned}$$

Then, since $\gamma _1^{n}\gamma _1'= t^{n+1}$ and $c_1=\gamma _1^{n+1}-\tfrac{n+1}{n+2} t^{n+2}$, we have

$$\begin{aligned} \gamma _2'=\tfrac{n+1}{n+2}\gamma _1'\ln t -\tfrac{\gamma _1}{(n+2)t}+(\tfrac{n+1}{(n+2)^2}+c_2)\gamma _1' \end{aligned}$$

and

$$\begin{aligned} \gamma _2=\int \big (\tfrac{n+1}{n+2}\gamma _1'\ln t -\tfrac{\gamma _1}{(n+2)t}\big )dt +(\tfrac{n+1}{(n+2)^2}+c_2)\gamma _1+c_3. \end{aligned}$$

Here, by applying equiaffine transformations we may put $c_2=-(n+1)/(n+2)^2$ and $c_3=0$. We have the immersion (4).

Case III. By $\mu _2=0$ we have $\lambda _2=\alpha$. It follows from $\lambda _2\ne 0$ and (6.4)-(6.6) that

$$\begin{aligned} \lambda _2=\alpha =-\tfrac{1}{t},\ \mu _2=0,\ f=t,\ H_0=0,\ \mu _1=-2n\alpha ^2. \end{aligned}$$

(6.14)

Moreover, it was shown in [1] that $M^n$ is locally given by

$$\begin{aligned} \begin{aligned} F(t,x)=(x, g(x)+\gamma _1(t),\gamma _2(t)), \end{aligned} \end{aligned}$$

where $x=(x_1,\ldots ,x_{n-1})$, g(x) is a convex function whose graph immersion is a parabolic affine hypersphere. It follows from the computations of such hypersurfaces on page 295 of [3] that $\gamma _1,\gamma _2$ satisfy

$$\begin{aligned} \begin{aligned} \gamma _2'^3=(\gamma _1''\gamma _2'-\gamma _1'\gamma _2'')f^{2(n+2)},\ f=\mid \gamma _2'\mid ^{1/(n+1)}. \end{aligned} \end{aligned}$$

(6.15)

Then, as $f=t$ in (6.14), we deduce that

$$\begin{aligned} \gamma _2'=\epsilon t^{n+1},\ t^2\gamma _1''-(n+1)t\gamma _1'-1=0, \end{aligned}$$

(6.16)

where $\epsilon \in \{-1,1\}$. Then, we can directly solve these equations to obtain

$$\begin{aligned} \gamma _2(t)=\tfrac{\epsilon }{n+2}t^{n+2}+c_1,\ \gamma _1(t)=-\tfrac{\ln t}{n+2}+c_2t^{n+2}+c_3. \end{aligned}$$

By applying a translation and a reflection in $\mathbb {R}^{n+1}$ we may assume that $c_1=c_3=0$, and $\gamma _2>0$, i.e., $\epsilon =1$. Also, by possibly applying an equiaffine transformation we may put $c_2=0$. Hence, we obtain the immersion (5). $\square$