On product affine hyperspheres in $\mathbb{R}^{n+1}$

TL;DR

This paper classifies locally strongly convex affine hyperspheres in Euclidean space that are locally isometric to products of constant curvature manifolds, including special cases with parallel Ricci tensor in dimensions 3 and 4.

Contribution

It provides a complete classification of certain affine hyperspheres with specific geometric structures, extending understanding of their global geometry.

Findings

Classification of affine hyperspheres as Riemannian products of constant curvature manifolds.

Explicit descriptions of hyperspheres with parallel Ricci tensor in dimensions 3 and 4.

Abstract

In this paper, we study locally strongly convex affine hyperspheres in the unimodular affine space $\mathbb{R}^{n+1}$ which, as Riemannian manifolds, are locally isometric to the Riemannian product of two Riemannian manifolds both possessing constant sectional curvatures. As the main result, a complete classification of such affine hyperspheres is established. Moreover, as direct consequences, affine hyperspheres of dimensions 3 and 4 with parallel Ricci tensor are also classified.

Equations

$$(x_1\cdots x_{n_1}{)}^2(x_{n+1}^2-x_{n_1+1}^2-\cdots -x_n^2{)}^{n_2+1}=1,$$

$$x_1^2(x_{n+1}^2-x_2^2-\cdots -x_n^2{)}^n=1,$$

$$D_X{x}_{\ast }(Y)=x_{\ast }({\nabla }_{X}Y)+h(X,Y)\xi ,$$

$$D_X\xi =-x_{\ast }(SX),$$

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

$$({\hat{\nabla }}_{X}K)(Y,Z)=({\hat{\nabla }}_{Y}K)(X,Z).$$

$$({\hat{\nabla }}^{2}K)(W,Z,X,Y):=$$

$$-(\hat{\nabla }K)(Z,{\hat{\nabla }}_{W}X,Y)-(\hat{\nabla }K)(Z,X,{\hat{\nabla }}_{W}Y).$$

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

$$\mathrm{trace}{K}_X=0,\forall X\in TM.$$

$$({\hat{\nabla }}^{2}K)(W,X,Y,Z)-({\hat{\nabla }}^{2}K)(W,Y,X,Z)=0.$$

$$\begin{array}{rl}0=& \hat{R}(W,X)K(Y,Z)-K(\hat{R}(W,X)Z,Y)+\hat{R}(X,Y)K(W,Z)\\ & -K(\hat{R}(X,Y)Z,W)+\hat{R}(Y,W)K(X,Z)-K(\hat{R}(Y,W)Z,X).\end{array}$$

$$\displaystyle\begin{split}\hat{R}(X,Y)Z=&c_{1}\big{[}h(Y_{1},Z_{1})X_{1}-h(X_{1},Z_{1})Y_{1}\big{]}\\ &+c_{2}\big{[}h(Y_{2},Z_{2})X_{2}-h(X_{2},Z_{2})Y_{2}\big{]},\end{split}$$

$$\hat{R}(X,Y)Z=H\big{[}h(Y,Z)X-h(X,Z)Y\big{]}\ \ {\rm at}\ p.$$

$$K_{X_i}{Y}_{\alpha }=\mu (X_i)Y_{\alpha },1\le i\le n_1,1\le \alpha \le n_2,$$

$$\mu (X_1{)}^2+\cdots +\mu (X_{n_1}{)}^2=-\frac{n_1}{n_2+1}H.$$

$$0=c_2\sum _{m=1}^{n_2}({\delta }_{\beta m}{Y}_{\alpha }-{\delta }_{\alpha m}{Y}_{\beta })h(K_{X_i}{Y}_{\beta },Y_m)-c_2{K}_{X_i}{Y}_{\alpha }.$$

$$h(K_{X_i}{Y}_{\alpha },Y_{\beta })=0,1\le i\le n_1,1\le \alpha \ne \beta \le n_2.$$

$$h(K_{X_i}{Y}_{\alpha },Y_{\alpha })=h(K_{X_i}{Y}_{\beta },Y_{\beta }),1\le i\le n_1,1\le \alpha ,\beta \le n_2.$$

$$0=c_2\sum _{m=1}^{n_2}({\delta }_{m\alpha }{Y}_{\beta }-{\delta }_{\beta m}{Y}_{\alpha })h(K_{X_i}{X}_j,Y_m).$$

$$h(K_{X_i}{X}_j,Y_{\alpha })=0,1\le i,j\le n_1,1\le \alpha \le n_2.$$

$$0=$$

$$=$$

$$0$$

$$=n_{1}H+\sum _{i=1}^{n_1}\mu (X_i{)}^2-\sum _{k=1}^{n_1}\sum _{i=1}^{n_1}h(K_{X_k}{X}_i,X_i)\mu (X_k).$$

$$0=\sum _{i=1}^{n_1}h(K_{X_k}{X}_i,X_i)+\sum _{j=1}^{n_2}h(K_{X_k}{Y}_j,Y_j)=\sum _{i=1}^{n_1}h(K_{X_k}{X}_i,X_i)+n_2\mu (X_k).$$

$$\mu (X_1{)}^2+\cdots +\mu (X_{n_1}{)}^2=-\frac{n_1}{n_2+1}H.$$

$$f_1(u)=h(K_{u}u,u),u\in U_p{M}_1^{n_1}\cap E_p.$$

$$g^{\prime }(0)=3h(K_{e_1}{e}_1,u),g^{\prime \prime }(0)=6h(K_{e_1}u,u)-3f_1(e_1).$$

$$h(K_{e_1}{e}_1,u)=0,h(K_{e_1}{e}_1,e_1)\ge 2h(K_{e_1}u,u),h(u,u)=1,u\perp e_1.$$

$$K_{Y_{\alpha }}{Y}_{\beta }={\delta }_{\alpha \beta }({\mu }_1{X}_1+\cdots +{\mu }_{n_1}{X}_{n_1}),1\le \alpha ,\beta \le n_2,$$

$$K_{Y_{\alpha }}{Y}_{\beta }={\delta }_{\alpha \beta }({\mu }_1{X}_1+\cdots +{\mu }_{n_1}{X}_{n_1})+\sum _{\gamma =1}^{n_2}{\theta }_{\alpha \beta }^{\gamma }{Y}_l,1\le \alpha ,\beta \le n_2.$$

Full text

2010 Mathematics Subject Classification.  Primary 53A15; Secondary 53B25, 53C25.

The first two authors were supported by NSF of China, Grant Number 11771404. The third author is a postdoctoral fellow of FWO-Flanders, Belgium.

1. Introduction

In this paper, we study locally strongly convex affine hypersurfaces in the unimodular affine space $\mathbb{R}^{n+1}$. It is well known that on a nondegenerate affine hypersurface $M^{n}$ in $\mathbb{R}^{n+1}$ there exists a canonical transversal vector field $\xi$ which is called the affine normal vector field. If all the affine normal lines of $M^{n}$ pass through a fixed point (resp. if all the affine normals are parallel), $M^{n}$ is called a proper (resp. improper) affine hypersphere. The second fundamental form $h$ associated with the affine normal vector field is called the (Blaschke) affine metric. As we consider only locally strongly convex affine hypersurfaces, the affine metric $h$ is assumed to be positive definite, and in such situation, the proper affine hyperspheres are divided into two classes, i.e., the elliptic affine hyperspheres and the hyperbolic ones.

The affine hyperspheres form a very important class of affine hypersurfaces. From the global point of view that the affine metric $h$ is complete, the improper (also called parabolic) affine hypersphere has to be the elliptic paraboloid, whereas the elliptic affine hypersphere has to be the ellipsoid. However, the class of locally strongly convex hyperbolic affine hyperspheres is very large and have been widely studied, see amongst others the works of [3, 6, 8, 13, 14, 15, 23] and also the recent monograph [17], or the survey paper [19]. Indeed, even assuming global conditions, the class of hyperbolic affine hyperspheres is surprisingly large, and one is still far from having a complete geometric understanding of them for all dimensions.

On the other hand, affine hyperspheres with constant sectional curvature are classified in [16] and [26] (see also [24, 25] for the general non-degenerate case), whereas in [12] it was further shown that all locally strongly convex Einstein affine hyperspheres in $\mathbb{R}^{5}$ are of constant sectional curvature. Contrary to the result of [12], the cases for locally strongly convex Einstein affine hyperspheres in $\mathbb{R}^{n+1}$ with $n\geq 5$ are different, and there exist Einstein affine hyperspheres which are not of constant sectional curvatures; actually, such examples occur for the standard embeddings of the noncompact symmetric spaces $\mathrm{E}_{6(-26)}/\mathrm{F}_{4}$, and $\mathrm{SL}(m,\mathbb{R})/\mathrm{SO}(m)$, $\mathrm{SL}(m,\mathbb{C})/\mathrm{SU}(m)$, $\mathrm{SU}^{*}(2m)/\mathrm{Sp}(m)$ for each $m\geq 3$ (cf. [2, 11] and [4, 5]). However, at present the complete classification of locally strongly convex Einstein affine hyperspheres in $\mathbb{R}^{n+1}$ is still an interesting and open problem.

In order to get further knowledge of the affine hyperspheres, the above mentioned facts motivate us to consider the following natural and interesting problem:

Classify all locally strongly convex affine hyperspheres which are locally isometric to the product $M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$, such that $n_{1}+n_{2}=n$ and $M_{i}^{n_{i}}(c_{i})$ is an $n_{i}$-dimensional Riemannian manifold with constant sectional curvature $c_{i}$ for $i=1,2$.

To consider this problem, we are sufficient to assume that $n\geq 3$. As the results of this paper, we have solved the above problem. More precisely, we have proved the following theorems.

Theorem 1.1**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be a locally strongly convex affine hypersphere. If $(M^{n},h)$ is locally isometric to the Riemannian product $M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$ for $n_{1}\geq 2$ and $n_{2}\geq 2$, such that $n_{1}+n_{2}=n$ and $M_{i}^{n_{i}}(c_{i})$ is an $n_{i}$-dimensional Riemannian manifold with constant sectional curvature $c_{i}$ for $i=1,2$. Then we have $c_{1}c_{2}=0$, and one of the following cases occurs:

1. (i)

$c_{1}=c_{2}=0$* and $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is locally affinely equivalent to either the paraboloid $x_{n+1}=\tfrac{1}{2}[(x_{1})^{2}+\cdots+(x_{n})^{2}]$ or $Q(1,n):\ x_{1}x_{2}\cdots x_{n+1}=1$;* 2. (ii)

$c_{1}c_{2}=0$* and $c_{1}^{2}+c_{2}^{2}\neq 0$, assuming that $c_{1}=0$ and $c_{2}\neq 0$, then $c_{2}<0$, $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is locally affinely equivalent to the Calabi composition*

[TABLE]

where $(x_{1},\ldots,x_{n+1})$ are the standard coordinates of $\mathbb{R}^{n+1}$.

Theorem 1.2**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}\ (n\geq 3)$ be a locally strongly convex affine hypersphere. If $(M^{n},h)$ is locally isometric to a Riemannian product $I\times\tilde{M}^{n-1}(c)$, with $I\subset\mathbb{R}$ and $\tilde{M}^{n-1}(c)$ an $(n-1)$-dimensional Riemannian manifold with constant sectional curvature $c\not=0$. Then we have $c<0$, and $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is locally affinely equivalent to the Calabi composition

[TABLE]

where $(x_{1},\ldots,x_{n+1})$ are the standard coordinates of $\mathbb{R}^{n+1}$.

As direct consequences of these theorems, we further have the following results.

Corollary 1.1**.**

Let $x:M^{3}\rightarrow\mathbb{R}^{4}$ be a locally strongly convex affine hypersphere with parallel Ricci tensor. Then either $M^{3}$ is an open part of a locally strongly convex hyperquadric, or $x:M^{3}\rightarrow\mathbb{R}^{4}$ is locally affinely equivalent to one of the following two hypersurfaces:

1. (i)

$x_{1}x_{2}x_{3}x_{4}=1$, 2. (ii)

$x_{1}^{2}(x_{4}^{2}-x_{2}^{2}-x_{3}^{2})^{3}=1$,

where $(x_{1},x_{2},x_{3},x_{4})$ are the standard coordinates of $\mathbb{R}^{4}$.

Corollary 1.2**.**

Let $x:M^{4}\rightarrow\mathbb{R}^{5}$ be a locally strongly convex affine hypersphere with parallel Ricci tensor. Then either $M^{4}$ is an open part of a locally strongly convex hyperquadric, or $x:M^{4}\rightarrow\mathbb{R}^{5}$ is locally affinely equivalent to one of the following hypersurfaces:

1. (i)

$x_{1}x_{2}x_{3}x_{4}x_{5}=1$, 2. (ii)

$(x_{1}x_{2})^{2}(x_{5}^{2}-x_{3}^{2}-x_{4}^{2})^{3}=1$, 3. (ii)

$x_{1}^{2}(x_{5}^{2}-x_{2}^{2}-x_{3}^{2}-x_{4}^{2})^{4}=1$,

where $(x_{1},x_{2},x_{3},x_{4},x_{5})$ are the standard coordinates of $\mathbb{R}^{5}$.

Remark 1.1*.*

The above corollaries and the main results of [7] and [9] imply that for locally strongly convex affine hyperspheres in both $\mathbb{R}^{4}$ and $\mathbb{R}^{5}$, the parallelism of the intrinsic invariant Ricci tensor and that of the extrinsic invariant cubic form are actually equivalent.

The paper is arranged as follows: In section 2, we fix notations and briefly recall the local theory of equiaffine hypersurfaces. In section 3, the most technical parts of this paper are given and we prove the crucial lemmas which imply the existence of canonical local frame so that the difference tensor can be sufficiently determined. Finally, in section 4 we complete the proof of the preceding theorems and corollaries.

2. Preliminaries

In this section, we briefly recall the local theory of equiaffine hypersurfaces. For more details, we refer to the monographs [17, 21].

Let $\mathbb{R}^{n+1}$ be the standard $(n+1)$-dimensional real unimodular affine space that is equipped with its usual flat connection $D$ and a parallel volume form given by the determinant. Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be a locally strongly convex hypersurface with affine normal $\xi$. Then, for any vector fields $X$ and $Y$ on $M^{n}$, we have

[TABLE]

where $\nabla,S$ and $h$ are the induced affine connection, the affine shape operator and the affine metric, respectively. It is well known that $M^{n}$ is an affine hypersphere if and only if $S=H\,{\rm id}$ with $H$ being a constant; moreover, $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is a proper (resp. improper) affine hypersphere if and only if $H\neq 0$ (resp. $H=0$).

Let $\hat{\nabla}$ denote the Levi-Civita connection of the affine metric $h$. The difference tensor $K$ is defined by $K(X,Y):=K_{X}Y:=\nabla_{X}Y-\hat{\nabla}_{X}Y$; it is symmetric as both connections are torsion free. Moreover, $h(K(X,Y),Z)$ is a totally symmetric cubic form. For affine hyperspheres with affine shape operator $S=H\,{\rm id}$, the Riemannian curvature tensor $\hat{R}$ of the affine metric and the difference tensor $K$ satisfy the following fundamental equations of Gauss and Codazzi:

[TABLE]

[TABLE]

As usual, we denote $(\hat{\nabla}K)(Z,X,Y):=(\hat{\nabla}_{Z}K)(X,Y)$, and define the second covariant differentiation $\hat{\nabla}^{2}K$ of $K$ by

[TABLE]

Then we have the following Ricci identity:

[TABLE]

Moreover, for unimodular affine hypersurfaces of $\mathbb{R}^{n+1}$, $K$ satisfies the so-called apolarity condition

[TABLE]

In the following, we will prove an additional relation that is very useful in our computations. To do so, we will make use of the technique introduced in [1], as the Tsinghua Principle. First, take the covariant derivative of (2.4) with respect to $W$, and use (2.4) and (2.5), to obtain straightforwardly that

[TABLE]

Then we sum over cyclic permutations of the first three vector fields in the above equation and use the Ricci identity (2.6). It follows that

[TABLE]

Additionally, if $(M^{n},h)=M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$ and applying Corollary 58 on page 89 in [22], we know that

[TABLE]

where, for $p\in M^{n}$ and $i=1,2$, $X_{i},Y_{i},Z_{i}$ are the $T_{p}M_{i}^{n_{i}}$-component of $X,Y,Z\in T_{p}M^{n}$, respectively.

3. Lemmas on the Calculations of the

Difference Tensor

In this section, we consider the $n$-dimensional locally strongly convex affine hypersphere $x:M^{n}\rightarrow\mathbb{R}^{n+1}$, such that $(M^{n},h)$ is locally isometric to a Riemannian product $M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$ for $n_{1}\geq 2$ and $n_{2}\geq 2$, $n_{1}+n_{2}=n$. Here, for $i=1,2$, $M_{i}^{n_{i}}(c_{i})$ denotes an $n_{i}$-dimensional Riemannian manifold with constant sectional curvature $c_{i}$. We first assume that $c_{1}^{2}+c_{2}^{2}\neq 0$ in this section.

Now, we would emphasize that when we dealing with the product manifold $M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, one should be aware that throughout the paper we will work with tangent vectors on $M^{n}$ denoted by $X$ and $Y$. In general, the $X$ notation (as well as $X_{i}$, $1\leq i\leq n_{1}$) will denote a tangent vector at $p=(p_{1},p_{2})\in M^{n}$, with zero component on $M_{2}^{n_{2}}$. Notice that, a priori, it means that $X$ depends on $p_{2}$ as well, not only on $p_{1}$. A corresponding meaning is given to $Y$ (or $Y_{j}$, $1\leq j\leq n_{2}$), having zero components on $M_{1}^{n_{1}}$ and depending a priory on both $p_{1}$ and $p_{2}$. One should have in mind this meaning when reading $X\in T_{p}M_{1}^{n_{1}}$, respectively, $Y\in T_{p}M_{2}^{n_{2}}$. Nonetheless, a complete understanding will be acquired with the proofs of Lemmas 4.1 and 4.2.

We begin with the following result.

Lemma 3.1**.**

If $c_{1}^{2}+c_{2}^{2}\neq 0$, then the difference tensor $K$ vanishes nowhere.

Proof.

Suppose on the contrary that the difference tensor $K$ vanishes at the point $p=(p_{1},p_{2})\in M^{n}=M_{1}^{n_{1}}\times M_{2}^{n_{2}}$. Then, from (2.3) we know that

[TABLE]

Thus $(M^{n},h)$ has constant sectional curvature $H$ at $p$.

Now, taking unit vectors $X\in T_{p}M_{1}^{n_{1}}$ and $Y=Z\in T_{p}M_{2}^{n_{2}}$ in both (2.10) and (3.1), we get $H=0$.

Next, taking unit vectors $X,Y=Z\in T_{p}M_{1}^{n_{1}}$ with $X\perp Y$ in both (2.10) and (3.1), we get $c_{1}=0$. Similarly, taking unit vectors $X,Y=Z\in T_{p}M_{2}^{n_{2}}$ with $X\perp Y$ in both (2.10) and (3.1), we get $c_{2}=0$.

Hence, $c_{1}=c_{2}=0$. This is a contradiction to that $c_{1}^{2}+c_{2}^{2}\neq 0$. ∎

Notice that if $c_{1}c_{2}=0$, then without loss of generality we can assume that $c_{1}=0$ and $c_{2}\neq 0$. Thus, in sequel we are sufficient to consider the following two cases:

Case $\mathfrak{C}_{1}$: $c_{1}=0$ and $c_{2}\neq 0$;    Case $\mathfrak{C}_{2}$: $c_{1}\neq 0$ and $c_{2}\neq 0$.

In the remaining of this section, we consider only Case $\mathfrak{C}_{1}$. In order to decide the difference tensor, first of all we have the following lemma.

Lemma 3.2**.**

For $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, let $\{X_{i}\}_{1\leq i\leq n_{1}}$ and $\{Y_{j}\}_{1\leq j\leq n_{2}}$ be orthonormal bases of $T_{p}M_{1}^{n_{1}}$ and $T_{p}M_{2}^{n_{2}}$, respectively. Then, in Case $\mathfrak{C}_{1}$, we have

[TABLE]

where $\mu(X_{i})=:\mu_{i}$ depends only on $X_{i}$ for $i=1,\ldots,n_{1}$. Moreover, it holds that

[TABLE]

Proof.

Let $\{X_{1},\ldots,X_{n_{1}}\}$ (resp. $\{Y_{1},\ldots,Y_{n_{2}}\}$) be an orthonormal basis of $T_{p}M_{1}^{n_{1}}$ (resp. $T_{p}M_{2}^{n_{2}}$). Taking $X=X_{i}$, $Y=Y_{\alpha}$ and $Z=W=Y_{\beta}\ (\alpha\neq\beta)$ in (2.9), then using (2.10) we obtain

[TABLE]

Taking the component of (3.4) on $Y_{\beta}$, we have that

[TABLE]

Taking the component of (3.4) on $Y_{\alpha}$, we have

[TABLE]

Similarly, taking $X=Y_{\alpha}$, $Y=X_{i}$, $Z=X_{j}$ and $W=Y_{\beta}$ in (2.9), then using (2.10) we obtain

[TABLE]

Let $\alpha\neq\beta$, then (3.7) implies that

[TABLE]

Combining (3.5), (3.6) and (3.8), the assertion (3.2) immediately follows.

Next, we compute the sectional curvature $K(\pi(X_{i},Y_{j}))$ of the plane $\pi$ spanned by $X_{i}$ and $Y_{j}$, for some fixed $i\in\{1,\ldots,n_{1}\}$ and $j\in\{1,\ldots,n_{2}\}$. For that purpose, using (2.10) on the one hand, and (2.3) on the other hand, together with applying (3.2), we obtain

[TABLE]

Then, taking summation over $i=1,\ldots,n_{1}$, and using (3.2), we get

[TABLE]

On the other hand, the apolarity condition implies that, for each $k=1,\ldots,n_{1}$,

[TABLE]

Therefore, from (3.9) and (3.10), we obtain

[TABLE]

This completes the proof of Lemma 3.2. ∎

Now, before going to show the next lemma, we will describe the construction of a typical orthonormal basis, which was introduced by N. Ejiri and has been widely applied, and proved to be very useful for various situations, see e.g. [10] and [18, 20]. The idea is to construct a basis from a self-adjoint operator at a point; then one extends the basis to local orthonormal vector fields. In this paper, we have the general principle as below:

For an arbitrary $p\in M^{n}=M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, let $U_{p}M_{1}^{n_{1}}=\{u\in T_{p}M_{1}^{n_{1}}\mid h(u,u)=1\}$ and $E_{p}\subset T_{p_{1}}M_{1}^{n_{1}}\times\{0\}$ a vector subspace. Since $M^{n}$ is locally strongly convex, $U_{p}M_{1}^{n_{1}}\cap E_{p}$ is compact. We define on this set the function

[TABLE]

Then there is an element $e_{1}\in U_{p}M_{1}^{n_{1}}\cap E_{p}$ at which the function $f_{1}(u)$ attains the absolute maximum. Let $u\in U_{p}M_{1}^{n_{1}}\cap E_{p}$ such that $h(u,e_{1})=0$, and define a function $g$ by g(t):=f_{1}\big{(}\cos t\,e_{1}+\sin t\,u\big{)}. Then we have

[TABLE]

Since $g$ attains an absolute maximum at $t=0$, we have $g^{\prime}(0)=0,g^{\prime\prime}(0)\leq 0$, i.e.,

[TABLE]

Analogously, we can define a function $f_{2}$ on $U_{p}M_{2}^{n_{2}}\cap\tilde{E}_{p}$, where $U_{p}M_{2}^{n_{2}}=\{u\in T_{p}M_{2}^{n_{2}}\mid h(u,u)=1\}$ and $\tilde{E}_{p}\subset\{0\}\times T_{p_{2}}M_{2}^{n_{2}}$ a vector subspace. We can choose $e_{1}\in U_{p}M_{2}^{n_{2}}\cap\tilde{E}_{p}$ such that (3.13) holds for $u\in U_{p}M_{2}^{n_{2}}\cap\tilde{E}_{p}$ with $u\perp e_{1}$.

In the following, we will apply the above principle of choosing the unit vector $e_{1}$ many times.

Now, as a supplement to Lemma 3.2, we can prove the following lemma.

Lemma 3.3**.**

Given $p=(p_{1},p_{2})\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$. Let $\{X_{i}\}_{1\leq i\leq n_{1}}$ and $\{Y_{j}\}_{1\leq j\leq n_{2}}$ be the orthonormal bases of $T_{p}M_{1}^{n_{1}}$ and $T_{p}M_{2}^{n_{2}}$, respectively. Then, in Case $\mathfrak{C}_{1}$, we have

[TABLE]

Moreover, we have $c_{2}=\frac{n+1}{n_{2}+1}H<0$.

Proof.

Let $\{X_{i}\}_{1\leq i\leq n_{1}}$ and $\{Y_{j}\}_{1\leq j\leq n_{2}}$ be orthonormal bases of $T_{p}M_{1}^{n_{1}}$ and $T_{p}M_{2}^{n_{2}}$, respectively. Then, according to Lemma 3.2, there are constants $\{\theta_{\alpha\beta}^{\gamma}\}$ such that

[TABLE]

We will show that $\theta_{\alpha\beta}^{\gamma}=0$ for $1\leq\alpha,\beta,\gamma\leq n_{2}$, or equivalently,

[TABLE]

We will prove (3.15) by contradiction.

Suppose on the contrary that (3.15) does not hold. Then, following the preceding stated procedure, we can choose a unit vector in $U_{p}M_{2}^{n_{2}}$, denoted by $\bar{Y}_{1}$, such that $\theta_{1}:=h(K_{\bar{Y}_{1}}\bar{Y}_{1},\bar{Y}_{1})>0$ is the maximum of the function $f_{2}$ defined on $U_{p_{2}}M_{2}^{n_{2}}$.

Define an operator $\mathcal{A}:T_{p}M_{2}^{n_{2}}\to T_{p}M_{2}^{n_{2}}$ by

[TABLE]

Then, it is easy to show that $\mathcal{A}$ is self-adjoint and satisfies $\mathcal{A}(\bar{Y}_{1})=\theta_{1}\bar{Y}_{1}$. We can choose orthonormal vectors in $U_{p}M_{2}^{n_{2}}$ orthogonal to $\bar{Y}_{1}$, denoted by $\bar{Y}_{2}$, $\ldots,\bar{Y}_{n_{2}}$, which are the remaining eigenvectors of the operator $\mathcal{A}$, with associated eigenvalues $\theta_{2},\ldots,\theta_{n_{2}}$, respectively. Thus, by Lemma 3.2, we get the conclusion that

[TABLE]

In order to solve $\{\theta_{i}\}$ in (3.16), taking $X=Z=\bar{Y}_{1}$ and $Y=\bar{Y}_{i}$, $2\leq i\leq n_{2}$, in (2.3), using (2.10), (3.16) and Lemma 3.2, we can obtain

[TABLE]

From (3.17) and the statement of (3.13), we obtain that

[TABLE]

Using (3.2), (3.16), (3.18) and ${\rm trace}\,K_{Y_{1}}=0$, we get

[TABLE]

Then, we have

[TABLE]

It follows that $c_{2}>\tfrac{n+1}{n_{2}+1}H$ and

[TABLE]

Next, we intend to extend $\bar{Y}_{1}\in U_{p}M_{2}^{n_{2}}$, that satisfying (3.16), to be a local unit vector field around $p\in M^{n}$. For that purpose, we first make the following Claim.

Claim 1. For every $p=(p_{1},p_{2})\in M^{n}$, the set

[TABLE]

consists of finite numbers, which are independent of the point $p\in M^{n}$.

To verify the claim, we notice that, for any fixed $p\in M^{n}$, the above discussion implies that we have $\theta_{1}\in\Omega_{p}$ with $V=\bar{Y}_{1}$. Thus, the set $\Omega_{p}$ is non-empty.

Next, assume an arbitrary $\lambda\in\Omega_{p}$ associated with $V\in U_{p}M_{2}^{n_{2}}$ such that

[TABLE]

Then we put $\tilde{Y}_{1}=V$, $\tilde{\theta}_{1}=\lambda$ and define an operator $\mathcal{B}:T_{p}M_{2}^{n_{2}}\to T_{p}M_{2}^{n_{2}}$ by

[TABLE]

It is easily seen that $\mathcal{B}$ is self-adjoint and $\mathcal{B}(\tilde{Y}_{1})=\tilde{\theta}_{1}\tilde{Y}_{1}$. Then, we may complete $\tilde{Y}_{1}$ to get an orthonormal basis $\{\tilde{Y}_{i}\}_{1\leq i\leq n_{2}}$ of $T_{p}M_{2}^{n_{2}}$ by letting $\tilde{Y}_{2},\ldots,\tilde{Y}_{n_{2}}$ to be the eigenvectors of $\mathcal{B}$, with eigenvalues $\tilde{\theta}_{2},\ldots,\tilde{\theta}_{n_{2}}$, respectively.

Similar to the proof of (3.17), we have the existence of an integer $n_{2,1}$ with $0\leq n_{2,1}\leq n_{2}-1$ such that, if necessary, after renumbering the basis, it holds

[TABLE]

Then, by ${\rm trace}\,K_{\tilde{Y}_{1}}=0$, we find that

[TABLE]

This implies that $\tilde{\theta}_{1}=\lambda$ is independent of the point $p$ and takes value of only finite possibilities. The assertion of Claim 1 immediately follows.

To extend $\bar{Y}_{1}$ differentiably to a unit vector field on a neighbourhood $U\subset M^{n}$ around $p$, which is still denoted by $\bar{Y}_{1}$, such that, at every point $q\in U$, $f_{2}$ attains an absolute maximum at $\bar{Y}_{1}(q)$, we first take differentiable $h$-orthonormal vector fields $\{E_{1},\ldots,E_{n_{2}}\}$ defined on a neighbourhood $U^{\prime}$ of $p$ and satisfying $E_{i}(q)\in T_{q}M_{2}^{n_{2}},q\in U^{\prime},1\leq i\leq n_{2}$, such that $E_{i}(p)=\bar{Y}_{i}$ for $1\leq i\leq n_{2}$. Then, we define a function $\gamma$ by

[TABLE]

where

[TABLE]

are regarded as functions on $\mathbb{R}^{n_{2}}\times U^{\prime}$: $b_{k}=b_{k}(a_{1},\ldots,a_{n_{2}},q)$.

Using (3.16) and the fact that $f_{2}$ attains an absolute maximum at $E_{1}(p)$, we then obtain that

[TABLE]

Notice that, by assumption, (3.18) and (3.19), we have $\theta_{1}>0$ and $2\theta_{k}-\theta_{1}\neq 0$ for $2\leq k\leq n_{2}$. Then, the implicit function theorem shows that there exist differentiable functions $\{a_{i}(q)\}_{1\leq i\leq n_{2}}$ defined on a neighbourhood $U^{\prime\prime}\subset U^{\prime}$ of $p$, such that

[TABLE]

Define the local vector field $V$ on $U^{\prime\prime}$ by

[TABLE]

Then, for local basis of $TM_{1}^{n_{1}}$ around $U^{\prime\prime}$, still denoted by $\{X_{i}\}_{1\leq i\leq n_{1}}$, from (3.23), (3.24) and Lemma 3.2, we have $K_{X_{i}}Y=\mu_{i}Y$ for any $Y\in TM_{2}^{n_{2}}$, and that

[TABLE]

Let us define $\|V\|=\sqrt{h(V,V)}$. Since $\|V\|(p)=1$, there exists a neighbourhood $U\subset U^{\prime\prime}$ of $p$ such that $V\not=0$ on $U$. Then, $W=\tfrac{V}{\|V\|}$ is a unit vector field on $U$ that satisfies

[TABLE]

Denote $\tilde{\theta}_{1}=\theta_{1}/\sqrt{h(V,V)}$. Then, the proof of Claim 1 implies that, as a function on $U$, $\tilde{\theta}_{1}$ takes values of finite number, which satisfy (3.22) for some $0\leq n_{2,1}\leq n_{2}-1$. This further implies from the fact $h(V,V)(p)=1$ and the continuity of the function $\theta_{1}/\sqrt{h(V,V)}$ that $h(V,V)\equiv 1$ on $U$.

Let $\bar{Y}_{1}=W$ and take orthonormal vector fields $\bar{Y}_{2},\ldots,\bar{Y}_{n_{2}}$ orthogonal to $\bar{Y}_{1}$ so that $\{\bar{Y}_{1},\ldots,\bar{Y}_{n_{1}}\}$ forms a local orthonormal basis of $TM_{2}^{n_{2}}$ on $U$. Then, according to (3.16), (3.18) and (3.20), we have a constant $\theta_{2}=\cdots=\theta_{n_{2}}$ such that the difference tensor satisfies

[TABLE]

Now, we can apply the Codazzi equation (2.4) to the basis $\{\bar{Y}_{i}\}_{1\leq i\leq n_{2}}$.

By the property $h(\hat{\nabla}_{\bar{Y}_{i}}\bar{Y}_{j},X_{k})=0$ of product manifold and (3.25), we have the following calculations:

[TABLE]

[TABLE]

Then, using $h((\hat{\nabla}_{\bar{Y}_{i}}K)(\bar{Y}_{1},\bar{Y}_{1}),\bar{Y}_{1})=h((\hat{\nabla}_{\bar{Y}_{1}}K)(\bar{Y}_{i},\bar{Y}_{1}),\bar{Y}_{1})$ for $i\geq 2$ we get $\hat{\nabla}_{\bar{Y}_{1}}\bar{Y}_{1}=0$. This and (3.27) give that $(\hat{\nabla}_{\bar{Y}_{1}}K)(\bar{Y}_{i},\bar{Y}_{1})=0$ for $1\leq i\leq n_{2}$. Thus, using (2.4) and (3.26), we can finally get

[TABLE]

It follows that $c_{2}=h(\hat{R}(\bar{Y}_{2},\bar{Y}_{1})\bar{Y}_{1},\bar{Y}_{2})=0$ and as desired we get a contradiction. Therefore, (3.15) does hold.

Finally, taking $X=\bar{Y}_{2}$ and $Y=Z=\bar{Y}_{1}$ in (2.3), with using (2.10), (3.2) and (3.14), we easily get the relation $c_{2}=\frac{n+1}{n_{2}+1}H$. This together with (3.3) further implies that $H<0$.

We have completed the proof of Lemma 3.3. ∎

For the difference tensor, besides the conclusions as stated in Lemmas 3.2 and 3.3, we shall construct in the following Lemma 3.4 a typical local orthonormal frame on $M^{n}$ so that more information of the difference tensor can be derived for Case $\mathfrak{C}_{1}$. However, the proof of Lemma 3.4 becomes more complicated when we compare it with that of Lemma 3.3.

Lemma 3.4**.**

In Case $\mathfrak{C}_{1}$, given $p\in M^{n}$, there exist local orthonormal vector fields $\{X_{i}\}_{1\leq i\leq n_{1}}$ defined on a neighbourhood $U$ of $p$, and satisfying $X_{i}(q)\in T_{q}M_{1}^{n_{1}}$ for $q\in U$ and $1\leq i\leq n_{1}$, such that the difference tensor $K$ takes the following form:

[TABLE]

where $\lambda_{i,i}$ and $\mu_{i}\ (1\leq i\leq n_{1})$ are constants, and they satisfy the relations

[TABLE]

Proof.

We give the proof by induction on the subscript $i$ of $K_{X_{i}}$. According to the general principle of induction method, this consists of two steps as below.

The first step of induction.

In this step, we should verify the assertion for $i=1$. To do so, we have to show that, around any given $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, there exist orthonormal vector fields $\{X_{i}\}_{1\leq i\leq n_{1}}$ defined on a neighbourhood $U$ of $p$ and satisfying $X_{i}(q)\in T_{q}M_{1}^{n_{1}}$ for $q\in U$ and $1\leq i\leq n_{1}$, and real numbers $\lambda_{1,1}>0$ and $\mu_{1}$, so that we have

[TABLE]

The proof of the above assertion will be divided into four claims as below.

Claim I-(1). Given $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, there exists an orthonormal basis $\{X_{i}\}_{1\leq i\leq n_{1}}$ of $T_{p}M_{1}^{n_{1}}$, real numbers $\lambda_{1,1}>0$, $\lambda_{1,2}=\cdots=\lambda_{1,n_{1}}$ and $\mu_{1}$, such that $\lambda_{1,1}$ is the maximum of $f_{1}$ defined on $U_{p}M_{1}^{n_{1}}$, and the following relations hold:

[TABLE]

Proof of Claim I-(1).

First, if for an orthonormal vectors $\{X_{i}\}_{1\leq i\leq n_{1}}$ and for any $i,j,k=1,\ldots,n_{1}$, it holds $h(K_{X_{i}}X_{j},X_{k})=0$. Then in (2.3) taking $X=X_{1}$ and $Y=Z=X_{2}$, using (2.10) and (3.2), we obtain $H=0$. This is a contradiction to Lemma 3.3.

Next, let $p\in M^{n}=M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$. We choose $X_{1}\in U_{p}M_{1}^{n_{1}}$ such that $\lambda_{1,1}=h(K_{X_{1}}X_{1},X_{1})$ is the maximum of $f_{1}(u)$ on $U_{p}M_{1}^{n_{1}}$ and it must be the case $\lambda_{1,1}>0$. Then, according to (3.2) and the statement of (3.13), we know that $X_{1}$ is an eigenvector of $K_{X_{1}}$ and we can choose orthonormal vectors $X_{2},\ldots,X_{n_{1}}\in T_{p}M_{1}^{n_{1}}$ orthogonal to $X_{1}$ such that $K_{X_{1}}X_{i}=\lambda_{1,i}X_{i}$ for $1\leq i\leq n_{1}$, and $K_{X_{1}}Y=\mu_{1}Y$ for any $Y\in T_{p}M_{2}^{n_{2}}$.

Taking in (2.3) $X=Z=X_{1}$ and $Y=X_{k}$, and using (2.10), we can obtain

[TABLE]

Similar to the proof of (3.13), we have $\lambda_{1,1}\geq 2\lambda_{1,k}$ for $2\leq k\leq n_{1}$. Thus, solving (3.32) we obtain $\lambda_{1,2}=\cdots=\lambda_{1,n_{1}}$ with

[TABLE]

Furthermore, taking in (2.3) $X=Z=X_{1}$ and $Y\in T_{p}M_{2}^{n_{2}}$ be a unit vector, using (2.10) and (3.2), we get

[TABLE]

Hence we have

[TABLE]

Finally, by (3.2), (3.33), (3.35) and ${\rm trace}\,K_{X_{1}}=0$, we obtain

[TABLE]

and therefore, we have

[TABLE]

From (3.33), (3.35) and (3.37), we have completed the proof of Claim I-(1). ∎

Claim I-(2). The real numbers described in Claim I-(1) satisfy the relations:

[TABLE]

Proof of Claim I-(2).

From (3.33), (3.35) and ${\rm trace}\,K_{X_{1}}=0$, the assertions are equivalent to that $\varepsilon_{1}=-1$. Suppose on the contrary that $\varepsilon_{1}=1$. Then we have

[TABLE]

and (3.36) implies that

[TABLE]

Put $V_{1}=\{u\in T_{p}M_{1}^{n_{1}}|\,u\perp X_{1}\}$. Then, by arguments as in the beginning of the proof for Claim I-(1) shows that the function $f_{1}\not=0$ restricting on $V_{1}\cap U_{p}M_{1}^{n_{1}}$. We rechoose a unit vector $X_{2}\in V_{1}$ such that $\lambda_{2,2}=h(K_{X_{2}}X_{2},X_{2})>0$ is the maximum of $f_{1}(u)$ restricted on $\{u\in U_{p}M_{1}^{n_{1}}|\,u\perp X_{1}\}$.

Then, according to Lemma 3.2, we can define a linear mapping $\mathcal{A}:\ V_{1}\to V_{1}$ by $\mathcal{A}(X):=K_{X_{2}}X-h(K_{X_{2}}X,X_{1})X_{1}$. It is easily seen that $\mathcal{A}$ is self-adjoint and $X_{2}$ is one of its eigenvector. We can choose orthonormal vectors $X_{3},\ldots,X_{n_{1}}\in T_{p}M_{1}^{n_{1}}$ orthogonal to $X_{2}$, which are the remaining eigenvectors of the operator $\mathcal{A}$, associated to the eigenvalues $\lambda_{2,3},\ldots,\lambda_{2,n_{1}}$, respectively. Therefore, we have

[TABLE]

Now, we can make use of (3.40) to derive the expected contradiction.

Taking in (2.3) $X=Z=X_{2}$ and $Y=X_{k}$, using (2.10) and (3.40), we can obtain

[TABLE]

Similar to the proof of (3.13), we have $\lambda_{2,2}\geq 2\lambda_{2,k}$ for $3\leq k\leq n_{1}$. Then, solving (3.13), we get $\lambda_{2,3}=\cdots=\lambda_{2,n_{1}}$ with

[TABLE]

Similarly, taking in (2.3) $X=Z=X_{2}$ and $Y\in T_{p}M_{2}^{n_{2}}$ a unit vector, using (2.10), (3.2) and (3.40), we get

[TABLE]

By using (3.38), we can reduce (3.43) to be

[TABLE]

It follows that

[TABLE]

Then, by ${\rm trace}\,K_{X_{2}}=0$, and using (3.2), (3.40), (3.42) and (3.45), we have

[TABLE]

which implies that

[TABLE]

Note that $\varepsilon_{1}=1$, from (3.37) we have

[TABLE]

Noticing that $n_{2}\geq 2$ and, by (3.39), $n_{1}\geq n_{2}+2$, we have

[TABLE]

This, together with $H<0$, implies that $\lambda_{2,2}>\lambda_{1,1}$. This is a contradiction.

Hence, we have $\varepsilon_{1}=-1$ and $\lambda_{1,2}=\cdots=\lambda_{1,n_{1}}=\mu_{1}$.

Then, by ${\rm trace}\,K_{X_{1}}=0$ we get the second assertion. ∎

Claim I-(3). For every point $p=(p_{1},p_{2})\in M^{n}$, the set

[TABLE]

consists of finite numbers, which are independent of $p\in M^{n}$.

Proof of Claim I-(3).

Claim I-(1) implies that $\Omega_{p}$ is non-empty. Assume that there exists a unit vector $V\in T_{p}M_{1}^{n_{1}}$ such that $K_{V}V=\lambda V$. Let $X_{1}:=V$ and $\lambda_{1,1}=\lambda$. Then, according to Lemma 3.2, we may complete $X_{1}$ to obtain an orthonormal basis $\{X_{i}\}_{1\leq i\leq n_{1}}$ of $T_{p}M_{1}^{n_{1}}$ such that, for each $2\leq k\leq n_{1}$, $X_{k}$ is the eigenvector of $K_{X_{1}}$ with eigenvalue $\lambda_{1,k}$.

Then we have (3.32), from which we have an integer $n_{1,1}$, $0\leq n_{1,1}\leq n_{1}-1$, such that, if necessary after renumbering the basis, we have

[TABLE]

Similarly, we have (3.35). Then, by ${\rm trace}\,K_{X_{1}}=0$, we have

[TABLE]

If $2n_{1,1}-n_{1}+1+\varepsilon_{1}n_{2}=0$, then $\lambda_{1,1}=0$.

If $2n_{1,1}-n_{1}+1+\varepsilon_{1}n_{2}<0$, then we have

[TABLE]

It follows that $\lambda_{1,1}$ has finite possibilities, and Claim I-(3) is verified. ∎

Claim I-(4). The unit vector $X_{1}\in U_{p}M_{1}^{n_{1}}$ given in Claim-I-(1) can be extended differentiably to a unit vector field, still denoted by $X_{1}$, in a neighbourhood $U\subset M^{n}$ of $p$, such that, for each $q\in U$, the function $f_{1}$ defined on $U_{q}M_{1}^{n_{1}}$ attains its absolute maximum at $X_{1}(q)$.

Proof of Claim I-(4).

Let $\{E_{1},\ldots,E_{n_{1}}\}$ be differentiable orthonormal vector fields defined on a neighbourhood $U^{\prime}$ of $p$ and satisfying $E_{i}(q)\in T_{q}M_{1}^{n_{1}},q\in U^{\prime},1\leq i\leq n_{1}$, such that $E_{i}(p)=X_{i}$ for $1\leq i\leq n_{1}$. Then, from the fact $K_{X_{1}}X_{1}=\lambda_{1,1}X_{1}$ at $p$, we define a function $\gamma$ by

[TABLE]

where

[TABLE]

are regarded as functions on $\mathbb{R}^{n_{1}}\times U^{\prime}$: $b_{k}=b_{k}(a_{1},\ldots,a_{n_{1}},q)$. Here, according to (3.37) and the proof of Claim I-(2), the maximum of $f_{1}$ defined on $U_{q}M_{1}^{n_{1}}$ is independent of $q\in U^{\prime}$, and it is equal to $\lambda_{1,1}=(n-1)\sqrt{-H/n}$.

Using (3.31) and the fact that $f_{1}$ attains the absolute maximum $\lambda_{1,1}$ at $E_{1}(p)$, we obtain that

[TABLE]

From the proof of Claim-I-(1) we have $\lambda_{1,1}>0$ and $\lambda_{1,1}>2\lambda_{1,k}$ for $2\leq k\leq n_{1}$. Then, the implicit function theorem shows that there exist differentiable functions $\{a_{i}(q)\}_{1\leq i\leq n_{1}}$, defined on a neighbourhood $U^{\prime\prime}\subset U^{\prime}$ of $p$, such that

[TABLE]

Define the local vector field $V$ on $U^{\prime\prime}$ by

[TABLE]

Then, from (3.52), (3.53) and (3.2), we get

[TABLE]

Let us define $\|V\|=\sqrt{h(V,V)}$. Since $\|V\|(p)=1$, there exists a neighbourhood $U\subset U^{\prime\prime}$ of $p$, such that $V\not=0$ on $U$, and it holds that

[TABLE]

From Claim I-(3), we know that $\tfrac{\lambda_{1,1}}{\sqrt{h(V,V)}}$ takes values of finite number. On the other hand, $\tfrac{\lambda_{1,1}}{\sqrt{h(V,V)}}$ is continuous and $h(V,V)(p)=1$. Thus $h(V,V)\equiv 1$. It follows from (3.54) that, for any point $q\in U$, the function $f_{1}$ attains its absolute maximum in $V(q)$.

Define $X_{1}:=V$ on $U$. Then we have completed the proof of Claim I-(4). ∎

Finally, having determined the unit vector field $X_{1}$ as in Claim I-(4), we can further choose orthonormal vectors $X_{2},\ldots,X_{n_{1}}$ orthogonal to $X_{1}$, defined on $U$ and satisfying $X_{i}(q)\in T_{q}M_{1}^{n_{1}},q\in U,2\leq i\leq n_{1}$. Then, it is easily seen that, combining with Lemma 3.2, Claim I-(1), Claim I-(2) and their proofs, $\{X_{1},\ldots,X_{n_{1}}\}$ turns into the desired local orthonormal vector fields so that we have completed the proof for the first step of induction.

The second step of induction

In this step, we first assume the assertion of Lemma 3.4 for all $1\leq i\leq k$, where $k\in\{1,2,\ldots,n_{1}-2\}$ is a fixed integer. Thus, we have:

Around any given $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, there exist local orthonormal vector fields $\{X_{i}\}_{1\leq i\leq n_{1}}$ defined on a neighborhood $U$ of $p$ and satisfying $X_{i}(q)\in T_{q}M_{1}^{n_{1}},q\in U,1\leq i\leq n_{1}$, such that the difference tensor $K$ takes the form:

[TABLE]

where, $\mu_{i}$ and $\lambda_{i,i}$ for $1\leq i\leq k$ are real numbers, and they satisfy the relations:

[TABLE]

Moreover, at any $q\in U$, the number $\lambda_{i,i}$ is the maximum of the function $f_{1}$ defined on

[TABLE]

for each $1\leq i\leq k$.

Then, as purpose of the second step, we should verify the assertion of Lemma 3.4 for $i=k+1$. To do so, we are sufficient to show that:

There exist an orthonormal frame $\{\tilde{X}_{i}\}_{1\leq i\leq n_{1}}$ on $TM_{1}^{n_{1}}$ around $p$, given by

[TABLE]

such that $T=(T_{l}^{t})_{k+1\leq l,t\leq n_{1}}$ is an orthogonal matrix, and the difference tensor $K$ takes the following form:

[TABLE]

where, $\mu_{i}$ and $\lambda_{i,i}$, for $1\leq i\leq k+1$, are real numbers, and they satisfy the relations

[TABLE]

Moreover, at any $q$ around $p$, the number $\lambda_{i,i}$ is the maximum of the function $f_{1}$ defined on

[TABLE]

for each $1\leq i\leq k+1$.

In order to prove the above conclusions, similar to the proof in the first step, we also divide it into the verification of the following four claims.

Claim II-(1). For any $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, there exist an orthonormal basis $\{\bar{X}_{i}\}_{1\leq i\leq n_{1}}$ of $T_{p}M_{1}^{n_{1}}$ and, real numbers $\lambda_{k+1,k+1}>0$, $\lambda_{k+1,k+2}=\cdots=\lambda_{k+1,n_{1}}$ and $\mu_{k+1}$, such that the following relations hold:

[TABLE]

Proof of Claim II-(1).

By the assumption of induction, we have local orthonormal vector fields $\{X_{i}\}_{1\leq i\leq n_{1}}$ defined on a neighborhood $U$ of $p$ and satisfying $X_{i}(q)\in T_{q}M_{1}^{n_{1}}$ for $q\in U$ and $1\leq i\leq n_{1}$, such that (3.55) and (3.56) hold. We first take $\bar{X}_{1}=X_{1}(p),\ldots,\bar{X}_{k}=X_{k}(p)$ and put

[TABLE]

Then, similar argument as in the proof of Claim I-(1) shows that when restricting on $V_{k}\cap U_{p}M_{1}^{n_{1}}$ the function $f_{1}\not=0$. Thus, we can choose a unit vector $\bar{X}_{k+1}\in V_{k}$ such that $\lambda_{k+1,k+1}=h(K_{\bar{X}_{k+1}}\bar{X}_{k+1},\bar{X}_{k+1})$ is the maximum of $f_{1}$ on $V_{k}\cap U_{p}M_{1}^{n_{1}}$ with $\lambda_{k+1,k+1}>0$.

Define a linear transformation $\mathfrak{A}:V_{k}\to V_{k}$ by

[TABLE]

It is easily seen that $\mathfrak{A}$ is self-adjoint and $\mathfrak{A}(\bar{X}_{k+1})=\lambda_{k+1,k+1}\bar{X}_{k+1}$. We can choose orthonormal vectors $\bar{X}_{k+2},\ldots,\bar{X}_{n_{1}}\in V_{k}$ orthogonal to $\bar{X}_{k+1}$, which are the remaining eigenvectors of $\mathfrak{A}$ with associated eigenvalues $\lambda_{k+1,k+2},\ldots,\lambda_{k+1,n_{1}}$, respectively. Then, by the assumption (3.55) of induction, we can show that

[TABLE]

Taking $X=Z=\bar{X}_{k+1}$ and $Y=\bar{X}_{i}$ in (2.3) for $k+2\leq i\leq n_{1}$, using (2.10) and (3.60), we can obtain

[TABLE]

Similar to the proof of (3.13), we have $\lambda_{k+1,k+1}\geq 2\lambda_{k+1,i}$ for $k+2\leq i\leq n_{1}$. Then, solving (3.61), we get $\lambda_{k+1,k+2}=\cdots=\lambda_{k+1,n_{1}}$ with

[TABLE]

Similarly, taking in (2.3) $X=Z=X_{k+1}$ and $Y\in T_{p}M_{2}^{n_{2}}$ a unit vector, then using (2.10) and (3.2), we get

[TABLE]

Hence, we have

[TABLE]

On the other hand, by applying ${\rm trace}\,K_{\bar{X}_{k+1}}=0$, we get $n_{1}-n_{2}\varepsilon_{k+1}-k-1>0$ and that

[TABLE]

From (3.62), (3.64), (3.65) and the assumption that $\mu_{1},\ldots,\mu_{k}$ are real numbers, we see that, as claimed, $\lambda_{k+1,k+2}=\cdots=\lambda_{k+1,n_{1}}$ and $\mu_{k+1}$ are also constants.

Moreover, by (3.60) and the assumption (3.55) of induction, we get the assertion that (3.59) holds. ∎

Claim II-(2). The real numbers described in Claim II-(1) satisfy the relations:

[TABLE]

Proof of Claim II-(2).

From (3.62) and (3.64), the first assertion is equivalent to showing that $\varepsilon_{k+1}=-1$. Suppose on the contrary that $\varepsilon_{k+1}=1$. Then we have

[TABLE]

Now from ${\rm trace}\,K_{\bar{X}_{k+1}}=0$ and $\lambda_{k+1,k+1}>0$ we obtain

[TABLE]

and that

[TABLE]

Put $V_{k+1}=\{u\in T_{p}M_{1}^{n_{1}}\mid u\perp\bar{X}_{1},\dots,u\perp\bar{X}_{k+1}\}$. Then (3.67) shows that $\dim V_{k+1}=n_{1}-k-1\geq n_{2}+1\geq 3$. Again, similar argument as in the proof of Claim I-(1) shows that, restricting on $V_{k+1}\cap U_{p}M_{1}^{n_{1}}$, the function $f_{1}\not=0$.

Now, by a totally similar argument as in the proof of Claim II-(1), we can choose a new orthonormal basis $\{\tilde{X}_{i}\}_{1\leq i\leq n_{1}}$ of $T_{p}M_{1}^{n_{1}}$ with $\tilde{X}_{j}=\bar{X}_{j}$ for $1\leq j\leq k+1$, such that $f_{1}$, restricting on $V_{k+1}\cap U_{p}M_{1}^{n_{1}}$, attains its maximum $\lambda_{k+2,k+2}>0$ at $\tilde{X}_{k+2}$ so that $\lambda_{k+2,k+2}=h(K_{\tilde{X}_{k+2}}\tilde{X}_{k+2},\tilde{X}_{k+2})$.

Similar as before, we define a self-adjoint operator $\mathfrak{B}:\ V_{k+1}\to V_{k+1}$ by

[TABLE]

Then $\mathfrak{B}(\tilde{X}_{k+2})=\lambda_{k+2,k+2}\tilde{X}_{k+2}$. As before we can choose orthonormal vectors $\tilde{X}_{k+3},\ldots,\tilde{X}_{n_{1}}\in V_{k+1}$, orthogonal to $\tilde{X}_{k+2}$, which are the remaining eigenvectors of $\mathfrak{B}:\ V_{k+1}\to V_{k+1}$, with associated eigenvalues $\lambda_{k+2,k+3},\ldots,\lambda_{k+2,n_{1}}$, respectively.

In this way, by using (3.59), we can show that

[TABLE]

Taking in (2.3) that $X=Z=\tilde{X}_{k+2}$ and $Y=\tilde{X}_{i}$ for $k+3\leq i\leq n_{1}$, and using (2.10), we can obtain

[TABLE]

Notice that $\lambda_{k+2,k+2}\geq 2\lambda_{k+2,i}$ for $k+3\leq i\leq n_{1}$. Then, solving (3.70), we get

[TABLE]

for $k+3\leq i\leq n_{1}$. Thus, $\lambda_{k+2,k+3}=\cdots=\lambda_{k+2,n_{1}}$.

On the other hand, taking in (2.3) $X=Z=\tilde{X}_{k+2}$ and $Y\in T_{p}M_{2}^{n_{2}}$ a unit vector, then using (2.10) and (3.2), we get

[TABLE]

From (3.66) and (3.72), we get

[TABLE]

and, equivalently,

[TABLE]

Then, from ${\rm trace}\,K_{\tilde{X}_{k+2}}=0$ and $\lambda_{k+2,k+2}>0$, we get $n_{1}-k-2>0$ and that

[TABLE]

From (3.67) and that $n_{2}\geq 2$, we have the following calculations

[TABLE]

Then, by (3.68) and (3.75), we get $\lambda_{k+2,k+2}>\lambda_{k+1,k+1}$, which is a contradiction.

Hence, as claimed we have $\varepsilon_{k+1}=-1$ and $\lambda_{k+1,k+2}=\cdots=\lambda_{k+1,n_{1}}=\mu_{k+1}$.

Finally, by ${\rm trace}\,K_{\bar{X}_{k+1}}=0$ and (3.59), we get the second assertion that

[TABLE]

This completes the proof of Claim II-(2). ∎

Claim II-(3). Under the assumptions of induction, the set

[TABLE]

consists of finite numbers, which are independent of $p\in M^{n}$.

Proof of Claim II-(3).

We first notice that, for any fixed $p\in M^{n}$, Claim II-(1) shows that $\lambda_{k+1,k+1}\in\Omega_{p,k}$ with $V=\bar{X}_{k+1}$. Thus, the set $\Omega_{p,k}$ is non-empty.

Next, with the local orthonormal vector fields $\{X_{i}\}_{1\leq i\leq n_{1}}$ around $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, given by the assumption of induction, we assume an arbitrary $\lambda\in\Omega_{p,k}$ associated with $V\in U_{p}M_{1}^{n_{1}}\setminus{\rm span}\{X_{i}(p)\}_{i=1}^{k}$ such that

[TABLE]

Then, at $p$, we put $\tilde{X}_{k+1}:=V$, $\tilde{X}_{i}=X_{i}(p)$ for $1\leq i\leq k$ and $\tilde{\lambda}_{k+1,k+1}=\lambda$.

Put $W_{k}=\{u\in T_{p}M_{1}^{n_{1}}\mid u\perp\tilde{X}_{1},\ldots,u\perp\tilde{X}_{k}\}$ and define $\mathfrak{F}:W_{k}\to W_{k}$ by

[TABLE]

Then $\mathfrak{F}$ is a self-adjoint linear transformation and that $\mathfrak{F}(\tilde{X}_{k+1})=\tilde{\lambda}_{k+1,k+1}\tilde{X}_{k+1}$. Thus, we can choose an orthonormal basis $\{\tilde{X}_{i}\}_{k+1\leq i\leq n_{1}}$ of $W_{k}$, such that

[TABLE]

Then, just like having did with equation (3.61), we have an integer $n_{1,k+1}$ with $0\leq n_{1,k+1}\leq n_{1}-(k+1)$ such that, if necessary after renumbering the basis of $W_{k}$, it holds

[TABLE]

Similar as deriving (3.64), now we also have

[TABLE]

.

Then, computing ${\rm trace}\,K_{\tilde{X}_{k+1}}=0$, gives that

[TABLE]

From (3.79) we have proved the assertion that $\lambda=\tilde{\lambda}_{k+1,k+1}$ takes values of only finite possibilities and they are independent of the point $p$. ∎

Claim II-(4). Under the assumptions of induction, the unit vector $\bar{X}_{k+1}\in T_{p}M_{1}^{n_{1}}$, determined by Claim II-(1), can be extended differentiably to a local unit vector field in a neighbourhood $U$ of $p$, denoted by $\tilde{X}_{k+1}$, such that for each $q\in U$ the function $f_{1}$, defined on

[TABLE]

attains its absolute maximum at $\tilde{X}_{k+1}(q)$.

Proof of Claim II-(4).

First of all, according to (3.65) and the proof of Claim II-(2), we notice that for any $q$ around $p$, the maximum of $f_{1}$ defined on $\mathcal{U}_{k}(q)$ is independent of $q$, and it equals to $\lambda_{k+1,k+1}=(n-k-1)\sqrt{(\sum_{l=1}^{k}\mu_{l}^{2}-H)/(n-k)}$.

Now, we choose arbitrary differentiable orthonormal vector fields $\{E_{k+1},\ldots,E_{n_{1}}\}$, defined on a neighbourhood $U^{\prime}$ of $p$ such that, for $k+1\leq i\leq n_{1}$ and $q\in U^{\prime}$, we have $E_{i}(p)=\bar{X}_{i}$ and $E_{i}(q)\in\mathcal{U}_{k}(q)$.

Next, we define a function $\gamma$ by

[TABLE]

where

[TABLE]

are regarded as functions on $\mathbb{R}^{n_{1}-k}\times U^{\prime}:\ b_{l}=b_{l}(a_{k+1},\ldots,a_{n_{1}},q)$.

Using Claim II-(1), the fact that $f_{1}$ attains its absolute maximum $\lambda_{k+1,k+1}$ at $E_{k+1}(p)$, and that

[TABLE]

where $\lambda_{k+1,i}$ is given by (3.62), we then obtain that

[TABLE]

Given that $\lambda_{k+1,k+1}>0$ and $\lambda_{k+1,k+1}-2\lambda_{k+1,l}>0$ for $k+2\leq l\leq n_{1}$, the implicit function theorem shows that in a neighbourhood $U^{\prime\prime}\subset U^{\prime}$ of $p$ there exist differentiable functions $\{a_{k+1},\ldots,a_{n_{1}}\}$ satisfying

[TABLE]

Define a local vector field $V$ on $U^{\prime\prime}$ by

[TABLE]

Then $V(p)=\bar{X}_{k+1}$, there exists a neighbourhood $U\subset U^{\prime\prime}$ of $p$, such that $V\not=0$ on $U$. Using (3.80), (3.81) and (3.2), we easily see that

[TABLE]

or, equivalently,

[TABLE]

Now, according to Claim II-(3), the function $\tfrac{\lambda_{k+1,k+1}}{\sqrt{h(V,V)}}$ takes values of only finite possibilities. On the other hand, $\tfrac{\lambda_{k+1,k+1}}{\sqrt{h(V,V)}}$ is continuous and $h(V,V)(p)=1$. Thus $h(V,V)|_{U}\equiv 1$. Let $\tilde{X}_{k+1}:=V$. Then, (3.82) with $h(V,V)=1$ implies that

[TABLE]

and for any $q\in U$, $f_{1}$ attains its absolute maximum $\lambda_{k+1,k+1}$ at $\tilde{X}_{k+1}(q)$. ∎

Let $\tilde{X}_{1}=X_{1},\ldots,\tilde{X}_{k}=X_{k}$ and choose vector fields $\tilde{X}_{k+2},\ldots,\tilde{X}_{n_{1}}$ such that, with $\tilde{X}_{k+1}$ obtained as in Claim II-(4), $\{\tilde{X}_{1},\tilde{X}_{2},\ldots,\tilde{X}_{n_{1}}\}$ is a local orthonormal frame of $TM_{1}^{n_{1}}$ defined on a neighborhood $U$ of $p$ and satisfies $X_{i}(q)\in T_{q}M_{1}^{n_{1}}$ for $q\in U$ and $1\leq i\leq n_{1}$. Then, with respect to $\{\tilde{X}_{i}\}_{1\leq i\leq n_{1}}$ and combining with Lemma 3.2, we immediately fulfil the second step of induction.

In this way, the method of induction allows us to obtain the desired orthonormal vector fields $\{X_{1},\ldots,X_{n_{1}-1}\}$ defined on a neighborhood $U$ of $p$ and satisfying $X_{i}(q)\in T_{q}M_{1}^{n_{1}}$ for $q\in U$ and $1\leq i\leq n_{1}-1$. Finally, we choose a unit vector field $X_{n_{1}}$ that is orthogonal to $\{X_{1},\ldots,X_{n_{1}-1}\}$ and that satisfies $X_{n_{1}}(q)\in T_{q}M_{1}^{n_{1}}$, such that $\lambda_{n_{1},n_{1}}\geq 0$ (if necessary we change $X_{n_{1}}$ by $-X_{n_{1}}$). Then, it is easy to see that $\{X_{1},\ldots,X_{n_{1}}\}$ are the desired orthonormal vector fields. Accordingly, we have completed the proof of Lemma 3.4. ∎

4. Proofs of the Theorems and Corollaries

First of all, continuing with the study of Case $\mathfrak{C}_{1}$ in last section, we show that the local orthonormal vector fields $\{X_{i}\}_{1\leq i\leq n_{1}}$, as determined in Lemma 3.4, consist of parallel vector fields such that $\hat{\nabla}X_{i}=0$.

Lemma 4.1**.**

The local orthonormal vector fields $\{X_{1},\ldots,X_{n_{1}}\}$, as described by Lemma 3.4, consist of parallel vector fields, i.e.,

[TABLE]

Proof.

We shall give the proof by induction on $i$. First of all, we prove $\hat{\nabla}X_{1}=0$.

In fact, for $j\geq 2$, applying (3.29), we have the following calculations

[TABLE]

[TABLE]

Now, the Codazzi equation $(\hat{\nabla}_{X_{j}}K)(X_{1},X_{1})=(\hat{\nabla}_{X_{1}}K)(X_{j},X_{1})$ gives that

[TABLE]

Then, taking the component of (4.3) in direction of $X_{1}$ for each $j\geq 2$ and using the fact that $h(\hat{\nabla}_{X_{1}}X_{1},Y)=0$ for $Y(q)\in T_{q}M_{2}^{n_{2}}$, and (3.29) again, we get $\hat{\nabla}_{X_{1}}X_{1}=0$. Substituting $\hat{\nabla}_{X_{1}}X_{1}=0$ into (4.3), and then taking its component in direction of $X_{j}$, we get $h(\hat{\nabla}_{X_{j}}X_{1},X_{k})=0$ for $2\leq j,k\leq n_{1}$. This, together with the fact that $h(\hat{\nabla}_{X_{j}}X_{1},Y)=0$ for $Y(q)\in T_{q}M_{2}^{n_{2}}$, implies that

[TABLE]

Take a unit vector field $Y$ with $Y(q)\in T_{q}M_{2}^{n_{2}}$. By a direct calculation of $h((\hat{\nabla}_{Y}K)(X_{1},X_{i}),X_{1})=h((\hat{\nabla}_{X_{1}}K)(Y,X_{i}),X_{1})$, we obtain $h(\hat{\nabla}_{Y}X_{1},X_{i})=0$ for $2\leq i\leq n_{1}$. This, together with $h(\hat{\nabla}_{Y}X_{1},Y^{\prime})=0$ for $Y^{\prime}(q)\in T_{q}M_{2}^{n_{2}}$, implies that

[TABLE]

Combining (4.4) and (4.5), we have proved the assertion $\hat{\nabla}X_{1}=0$.

Next, by induction we show that if for any fixed $2\leq i\leq n_{1}-1$ satisfying

[TABLE]

then it holds $\hat{\nabla}X_{i}=0$.

To state a proof of the above second step, we consider five cases below:

(i) By (4.6) and that $h(X_{k},X_{l})=\delta_{kl}$, we get

[TABLE]

(ii) For $j\leq i-1$, by using (3.29), (4.6) and (4.7), we can show that

[TABLE]

[TABLE]

Then, by $(\hat{\nabla}_{X_{j}}K)(X_{i},X_{i})=(\hat{\nabla}_{X_{i}}K)(X_{j},X_{i})$, for $k\geq i+1$ we obtain

[TABLE]

It follows that

[TABLE]

(iii) Similar to the above case (ii), for $j\geq i+1$, we have

[TABLE]

[TABLE]

Then, taking the $X_{i}$-components of $(\hat{\nabla}_{X_{j}}K)(X_{i},X_{i})=(\hat{\nabla}_{X_{i}}K)(X_{j},X_{i})$, with using (3.29) and (4.6), we obtain

[TABLE]

Hence, we obtain

[TABLE]

(iv) By using $(\hat{\nabla}_{X_{j}}K)(X_{i},X_{i})=(\hat{\nabla}_{X_{i}}K)(X_{j},X_{i})$ and taking its $X_{k}$-components for $j,k\geq i+1$, then applying (4.13) we obtain

[TABLE]

[TABLE]

(v) If $Y$ is a unit vector field with $Y(q)\in T_{q}M_{2}^{n_{2}}$, by a direct calculation of $h((\hat{\nabla}_{Y}K)(X_{i},X_{k}),X_{i})=h((\hat{\nabla}_{X_{i}}K)(Y,X_{k}),X_{i})$ for $i+1\leq k$, we obtain

[TABLE]

For $Y,Y^{\prime}$ with $Y(q),Y^{\prime}(q)\in T_{q}M_{2}^{n_{2}}$, we have $h(\hat{\nabla}_{X_{j}}X_{i},Y)=h(\hat{\nabla}_{Y}X_{i},Y^{\prime})=0$. Hence, combining (4.7), (4.10) and (4.13)–(4.15), we finally get

[TABLE]

Therefore, by induction we have proved that

[TABLE]

Finally, for vector fields $X,Y$ with $X(q)\in T_{q}M_{1}^{n_{1}},\,Y(q)\in T_{q}M_{2}^{n_{2}}$ and $k\leq n_{1}-1$, from (4.17) it is easily seen the following

[TABLE]

so that it holds also $\hat{\nabla}X_{n_{1}}=0$.

We have completed the proof of Lemma 4.1. ∎

Moreover, we have the following further conclusion.

Lemma 4.2**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be an $n$-dimensional locally strongly convex affine hypersphere such that Case $\mathfrak{C}_{1}$ in section 3 occurs, then the difference tensor is parallel, i.e., $\hat{\nabla}K=0$.

Proof.

Let $\{X_{1},\ldots,X_{n_{1}}\}$ be the local orthonormal vector fields as described by Lemma 3.4. Then Lemma 4.1 shows that

[TABLE]

On the other hand, as $(M^{n},h)=M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$, by Proposition 56 in p.89 of [22], we can choose local orthonormal vector fields $\{Y_{1},\ldots,Y_{n_{2}}\}$ with $Y_{i}(q)\in T_{q}M_{2}^{n_{2}}$, such that

[TABLE]

Then, using (4.18), (4.19) and properties of the difference tensor established by Lemmas 3.2, 3.3 and 3.4, direct calculations immediately give the assertion that $\hat{\nabla}K=0$. ∎

Theorem 4.1**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be an $n$-dimensional locally strongly convex affine hypersphere such that Case $\mathfrak{C}_{1}$ in section 3 occurs. Then $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is locally affinely equivalent to the Calabi composition

[TABLE]

where $(x_{1},\ldots,x_{n+1})$ are the standard coordinates of $\mathbb{R}^{n+1}$.

Proof.

By Lemma 3.4 and Lemma 4.2, we can apply Theorem 4.1 of [10] with $X_{1}$ being regarded as $e_{1}$ there. Then $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is a Calabi product of a point $G_{1}$ with a hyperbolic affinesphere $G_{2}^{\prime}:\tilde{M}_{1}^{n-1}\to\mathbb{R}^{n}$ with parallel cubic form and affine mean curvature $H_{2}$, so that we have the decomposition $M^{n}=I_{1}\times\tilde{M}_{1}^{n-1}$, $I\subset\mathbb{R}$, and the parametrization

[TABLE]

Moreover, the affine metric of $G_{2}^{\prime}:\tilde{M}_{1}^{n-1}\to\mathbb{R}^{n}$ is $(\mu^{2}_{1}-H)h|_{T\tilde{M}_{1}^{n-1}}$ (cf. [10]).

Notice that $T\tilde{M}_{1}^{n-1}={\rm span}\{X_{2},\ldots,X_{n_{1}},Y_{1},\ldots,Y_{n_{2}}\}$. Let us denote by $K^{1}$ the difference tensor of $G_{2}^{\prime}:\tilde{M}_{1}^{n-1}\to\mathbb{R}^{n}$, then from the proof of Theorem 4.1 of [10] and Lemmas 3.3 and 3.4, we can derive that $K^{1}$ has the expressions as follows:

[TABLE]

Notice also that, up to scaling a constant multiple, $\{X_{2},\ldots,X_{n_{1}},Y_{1},\ldots,Y_{n_{2}}\}$ are the orthomormal basis of the affine metric of $G_{2}^{\prime}:\tilde{M}_{1}^{n-1}\to\mathbb{R}^{n}$. Applying Theorem 4.1 in [10] once again by regarding $X_{2}$ as $e_{1}$ there, then $G_{2}^{\prime}:\tilde{M}_{1}^{n-1}\to\mathbb{R}^{n}$ is a Calabi product of a point $G_{2}$ with a hyperbolic affinesphere $G_{3}^{\prime}:\tilde{M}_{2}^{n-2}\to\mathbb{R}^{n-1}$ with parallel cubic form, so that we have the decomposition $\tilde{M}_{1}^{n-1}=I_{2}\times\tilde{M}_{2}^{n-2}$, $I_{2}\subset\mathbb{R}$, and the further parametrization

[TABLE]

Continuing in this way $n_{1}$ times, we finally see that $M^{n}=M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, with $M_{1}^{n_{1}}\cong I_{1}\times I_{2}\times\cdots\times I_{n_{1}}$, and $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ has a parametrization

[TABLE]

where, $(s_{1},\ldots,s_{n_{1}})\in M_{1}^{n_{1}}$, $\{G_{i}\}_{1\leq i\leq n_{1}}$ are constant vectors and $G_{n_{1}+1}^{\prime}:M_{2}^{n_{2}}\to\mathbb{R}^{n_{2}+1}$ is a hyperbolic affine hypersphere with parallel cubic form.

Furthermore, from the above procedure of induction, it can be easily seen that $G_{n_{1}+1}^{\prime}:M_{2}^{n_{2}}\to\mathbb{R}^{n_{2}+1}$ has vanishing difference tensor. This implies that $G_{n_{1}+1}^{\prime}:M_{2}^{n_{2}}\to\mathbb{R}^{n_{2}+1}$ is a hyperboloid. Therefore, up to an affine transformation, there exist constant vectors $G_{n_{1}+1},\ldots,G_{n+1}$ such that

[TABLE]

where $y_{1}^{2}+\cdots+y_{n_{2}}^{2}-y_{n_{2}+1}^{2}=-1$.

Combining (4.22) and (4.23), we finally see that, up to an affine transformation, $x:M^{n}=M_{1}^{n_{1}}\times M_{2}^{n_{2}}\to\mathbb{R}^{n+1}$ can be written as

[TABLE]

Hence, $x:M^{n}\to\mathbb{R}^{n+1}$ is affinely equivalent to the affine hypersphere (4.20). ∎

Next, we consider Case $\mathfrak{C}_{2}$ as stated in section 3 such that $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ is an $n$-dimensional locally strongly convex affine hypersphere with $(M^{n},h)=M_{1}^{n_{1}}(c_{1})\times M_{2}^{n_{2}}(c_{2})$, $n_{1}\geq 2,\ n_{2}\geq 2$ and $c_{1}c_{2}\neq 0$. Then, similar to that in Lemma 3.2 for the proof of (3.8), we can obtain the following result.

Lemma 4.3**.**

For $p\in M_{1}^{n_{1}}\times M_{2}^{n_{2}}$, let $\{X_{i}\}_{1\leq i\leq n_{1}}$ and $\{Y_{j}\}_{1\leq j\leq n_{2}}$ be orthonormal bases of $T_{p}M_{1}^{n_{1}}$ and $T_{p}M_{2}^{n_{2}}$, respectively. Then, in Case $\mathfrak{C}_{2}$, the difference tensor satisfies

[TABLE]

Moreover, we have the following further conclusion.

Theorem 4.2**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be a locally strongly convex product affine hypersphere, then Case $\mathfrak{C}_{2}$ in section 3 does not occur.

Proof.

If otherwise, we assume that Case $\mathfrak{C}_{2}$ does occur. Then, as by Lemma 3.1 the difference tensor $K$ vanishes nowhere, we may assume that for an arbitrary fixed $p\in M^{n}=M_{1}^{n_{1}}\times M_{2}^{n_{2}}$ there exists $X\in T_{p}M_{1}^{n_{1}}$ such that $K_{X}\neq 0$. Now, similar to the proof for the first step of induction in the proof of Lemma 3.4, we can show that around $p\in M^{n}$ there exist local orthonormal vector fields $\{X_{1},\ldots,X_{n_{1}}\}$ with $X_{i}(q)\in T_{q}M_{1}^{n_{1}},1\leq i\leq n_{1}$, such that the difference tensor takes the form

[TABLE]

where, $\lambda_{1}$ and $\lambda_{2}$ are real numbers with $\lambda_{1}>0$ and $\lambda_{1}+(n-1)\lambda_{2}=0$. Then, similar to the proof of (3.28), we can show that $\hat{\nabla}_{X_{i}}X_{1}=0$ for $1\leq i\leq n_{1}$. It follows that $\hat{R}(X_{1},X_{2})X_{1}=0$, which is a contradiction to that $c_{1}c_{2}\neq 0$. ∎

The Completion of Theorem 1.1’s Proof.

If $c_{1}=c_{2}=0$, it follows from (2.10) that $(M^{n},h)$ is flat. Then, according to the result of [26], we get the assertion (i) of Theorem 1.1.

If $c_{1}^{2}+c_{2}^{2}\neq 0$, we have two cases: Case $\mathfrak{C}_{1}$ and Case $\mathfrak{C}_{2}$, as preceding described.

If Case $\mathfrak{C}_{1}$ occurs, then by Theorem 4.1, we obtain the hypersphere as stated in (ii) of Theorem 1.1. Moreover, according to Theorem 4.2, Case $\mathfrak{C}_{2}$ does not occur.

We have completed the proof of Theorem 1.1. ∎

Next, we come to give the proof of Theorem 1.2. First of all, similar to the proof of Lemma 3.1, we can obtain the following result.

Lemma 4.4**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}\ (n\geq 3)$ be a locally strongly convex affine hypersphere such that $(M^{n},h)$ is locally isometric to a Riemannian product $\mathbb{R}\times M_{2}^{n-1}(c_{2})$, where $M_{2}^{n-1}(c_{2})$ is an $(n-1)$-dimensional Riemannian manifold with constant sectional curvature $c_{2}\neq 0$. Then the difference tensor $K$ of $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ vanishes nowhere.

Next, similar to the proofs of (3.5) and (3.6), we have

Lemma 4.5**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be a locally strongly convex affine hypersphere as described in Lemma 4.4. For $p\in M^{n}$, assume that $\{Y_{\alpha}\}_{1\leq\alpha\leq n-1}$ is an orthonormal basis of $T_{p}M_{2}^{n-1}$ and $X\in T_{p}\mathbb{R}$ is a unit vector, then we have

[TABLE]

where $\mu(X)=:\mu$ depends only on $X$.

Now, we will prove a lemma which plays the same important role as Lemma 3.4.

Lemma 4.6**.**

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ be a locally strongly convex affine hypersphere as described in Lemma 4.4 with $S=H\,{\rm id}$. Then, around any point $p\in M^{n}$, there exists a local orthonormal frame $\{X_{1},Y_{1},\ldots,Y_{n-1}\}$ on $M^{n}$ with $X_{1}(q)\in T_{q}\mathbb{R}$ and $Y_{\alpha}(q)\in T_{q}M_{2}^{n-1},\ 1\leq\alpha\leq n-1$, such that the difference tensor of $x:M^{n}\rightarrow\mathbb{R}^{n+1}$ takes the following form

[TABLE]

Moreover, we have $c_{2}=(n+1)H/n<0$ and $\hat{\nabla}K=0$.

Proof.

Around any point $p\in M^{n}=I\times M_{2}^{n-1}$, we take local unit vector fields $X$ and $Y$, with $X(q)\in T_{q}\mathbb{R}$ and $Y(q)\in T_{q}M_{2}^{n-1}$. Then, similar to the proof of (3.8), and applying Lemma 4.5, we obtain

[TABLE]

Moreover, by using (2.3) and the fact $\hat{R}(Y,X)X=0$, we have

[TABLE]

On the other hand, by ${\rm trace}\,K_{X}=0$, we get $\lambda+(n-1)\mu=0$. This together with (4.29) implies that, if necessary replacing $X$ by $-X$,

[TABLE]

Similar to the proof of Lemma 3.3, we can also show that

[TABLE]

Since $(M^{n},h)=\mathbb{R}\times M_{2}^{n-1}(c_{2})$, by Proposition 56 in p.89 of [22], we can take an orthonormal frame $\{X_{1},Y_{1},\ldots,Y_{n-1}\}$ on $M^{n}$ with $X_{1}=X$, such that

[TABLE]

Then, w.r.t $\{X_{1},Y_{\alpha}\}$, (4.27) immediately follows from the preceding conclusions.

Next, using (4.27), we can apply (2.3) and (2.10), with $X=Y_{2}$ and $Y=Z=Y_{1}$, to obtain that $c_{2}=(n+1)H/n<0$.

Finally, similar to the proof of $\hat{\nabla}X_{1}=0$ in Lemma 4.1, by (2.4) and (4.27), we can show that $\hat{\nabla}X_{1}=0$. From this, together with (4.27) and (4.31), we can show by direct calculations that $\hat{\nabla}K=0$. ∎

The Completion of Theorem 1.2’s Proof.

Under the assumptions of Theorem 1.2, we can apply Lemma 4.6, then as a direct consequence of Theorem 4.1 in [10] we easily get the assertion.∎

Proof of Corollaries.

Let $x:M^{n}\rightarrow\mathbb{R}^{n+1}$, with $n=3$ (resp. $n=4$), be a locally strongly convex affine hypersphere whose Ricci tensor is parallel with respect to the Levi-Civita of the affine metric. Then, by the classical de Rham-Wu’s decomposition theorem [27], $(M^{n},h)$ is locally isometric to a Riemannian product of Einstein manifolds.

If $n=3$, then either $(M^{3},h)$ is Einstein and thus $M^{3}$ is of constant sectional curvature, or $(M^{3},h)$ is locally isometric to a Riemannian product $\mathbb{R}\times\tilde{M}^{2}$, where $\tilde{M}^{2}$ is a Riemannian manifold with constant sectional curvature. For both of these cases, according to [26] and Theorem 1.2, we obtain Corollary 1.1.

If $n=4$, then either $(M^{4},h)$ is Einstein, or $(M^{4},h)$ is locally isometric to a Riemannian product $\mathbb{R}\times\tilde{M}^{3}$, or $(M^{4},h)$ is locally isometric to a Riemannian product $M_{1}^{2}\times M_{2}^{2}$, where $\tilde{M}^{3}$, $M_{1}^{2}$ and $M_{2}^{2}$ are Riemannian manifolds with constant sectional curvature. Then, for each of these three cases, applying the results of [12], Theorem 1.1 and Theorem 1.2, we obtain Corollary 1.2. ∎