Hypersurfaces with null higher order anisotropic mean curvature

Given a positive function $F$ on $\mathbb S^n$ which satisfies a convexity condition, for $1\leq r\leq n$, we define for hypersurfaces in $\mathbb{R}^{n+1}$ the $r$-th anisotropic mean curvature function $H_{r; F}$, a generalization of the usual $r$-th mean curvature function. We call a hypersurface is anisotropic minimal if $H_F=H_{1; F}=0$, and anisotropic $r$-minimal if $H_{r+1; F}=0$. Let $W$ be the set of points which are omitted by the hyperplanes tangent to $M$. We will prove that if an oriented hypersurface $M$ is anisotropic minimal, and the set $W$ is open and non-empty, then $x(M)$ is a part of a hyperplane of $\mathbb R^{n+1}$. We also prove that if an oriented hypersurface $M$ is anisotropic $r$-minimal and its $r$-th anisotropic mean curvature $H_{r; F}$ is nonzero everywhere, and the set $W$ is open and non-empty, then $M$ has anisotropic relative nullity $n-r$.


Introduction
Let : S → R + be a smooth function which satisfies the following convexity condition: where S is the standard unit sphere in R +1 , 2 denotes the intrinsic Hessian of on S , denotes the identity on S , and >0 means that the matrix is positive definite. We consider the map : S → R +1 , → ( ) + (grad S ) ; (2) its image = (S ) is a smooth, convex hypersurface in R +1 called the Wulff shape of (see [1][2][3][4][5][6][7][8][9]). When ≡ 1, the Wulff shape is just S . Now let : → R +1 be a smooth immersion of an oriented hypersurface. Let : → S denote its Gauss map. The map ] = ∘ : → is called the anisotropic Gauss map of .
is called the -Weingarten operator, and the eigenvalues of are called anisotropic principal curvatures. Let be the elementary symmetric functions of the anisotropic principal curvatures 1 , 2 , . . . , : We set 0 = 1. The th anisotropic mean curvature ; is defined by ; = / , also see Reilly [10]. := 1; is called the anisotropic mean curvature. When ≡ 1, is just the Weingarten operator of hypersurfaces, and ; is just the th mean curvature of hypersurfaces which has been studied by many authors (see [11][12][13][14]). Thus, the th anisotropic mean curvature ; generalizes the th mean curvature of hypersurfaces in the ( +1)-dimensional Euclidean space R +1 .
We say that : → R +1 is anisotropic -minimal if +1; = 0. For ∈ , we define V( ) = dim ker( ). We call V = min ∈ V( ) the anisotropic relative nullity; it generalized the usual relative nullity. → Q +1 of a hypersurface into an ( + 1)-dimensional space form with constant sectional curvature , we denote by where for every ∈ , (Q ) is the totally geodesic hypersurface of Q +1 tangent to ( ) at ( ). So, in the case of = 0, is the set of points which are omitted by the hyperplanes tangent to ( ). We will study immersion with nonempty. In this direction, Hasanis and Koutroufiotis (see [15]) proved the following.

Theorem 1. Let
: → Q 3 be a complete minimal immersion with ≥ 0. If is nonempty, then is totally geodesic.
Later, in [16], Alencar and Frensel extended the result above assuming an extra condition. They proved the following.

Theorem 2. Let :
→ Q +1 be an oriented, minimally immersed hypersurface. If is open and nonempty, then is totally geodesic.
In [17], Alencar and Batista studied hypersurfaces with null higher order mean curvature; they proved the following. We note that, Alencar in [18] provides examples of nontotally geodesic minimal hypersurfaces in R 2 , ≥ 4, with nonempty ; in [17], Alencar and Batista provides examples of 1-minimal hypersurfaces with 1 ̸ = 0 everywhere in R 2 , ≥ 5, with nonempty but V ̸ = −1. These examples show that it is necessary to add an extra hypothesis.
In this paper, we prove the anisotropic version of Theorems 2 and 3 for an immersion : → R +1 . Explicitly, we prove the following two theorems.
We define * : then * is a Minkowski norm on R +1 . In fact, as proved in [19], * : R +1 \ {0} → R is smooth and we have the following.
, and the equality holds if and only if = 0, or = 0 or = for some > 0.

Lemma 7.
For any ∈ R +1 \ {0} and ∈ R +1 one has and the equality holds if and only if there exists ≥ 0 such that = .

A Connection on Hypersurfaces of Minkowski Space
Let : → R +1 be an oriented hypersurface in the Euclidean space R +1 and denote ] : → its anisotropic Gauss map.
Let ∇ be the standard connection on the ( + 1)-dimensional Euclidean space R +1 . For vector fields , on , we decompose ∇ as the tangent part ∇ and the anisotropic normal part II ( , )] with respect to the inner product ] . That is, where ] (∇ , ]) = 0.
We also have the Weingarten formula: where we have used (9). It is easy to verify that ∇ is a torsion free connection on and II is a symmetric second order covariant tensor field on . We call II the anisotropic second fundamental form.
Let { } =1 be a local frame of and { } =1 its dual frame.
where is any smooth vector field on . Define by = ; then grad = .
From the definition of ( ), we have (32)

Lemma 14. One has
Proof. From the definition of , we have the following calculation: We define an operator ; : ∞ ( ) → ∞ ( ) by In the sequel, we will need the following lemma. Item (a) is essentially the content of Lemma 1.1 and Equation (1.3) in [24], while item (b) is quoted as Proposition 1.5 in [25].  Another important result is as follows (see [26]).

Lemma 16.
Let : → R +1 be an oriented hypersurface, and ∈ . The result below is standard, so we omit the proof. It is easy to see that Δ is an elliptic differential operator.

Proof of Theorems 4 and 5
We fix a point ∈ as the origin of R +1 . Without loss of generality, we assume, for each ∈ , ]( ) is the anisotropic unit normal vector of ( ) at ( ) such that ⟨ ( ), ]( )⟩ ]( ) > 0 (otherwise we consider the function − instead). This gives an orientation to ; indeed, the component of the position vector perpendicular (with respect to the inner product ] ) to defines a never zero, anisotropic normal, vector field on , such that the support function = ⟨ ( ), ]( )⟩ ]( ) is positive on .
Using Lemma 15(a) we have that is semidefinite. Since ; does not vanish, we have that ; is positive or negative, because ( ) ; = Trace( ), where ( ) = ( − ) . Now we use Lemma 16 and obtain the following: Using the information above, we claim that +2; ≡ 0.
Thus we conclude that +2; ≡ 0. Now, we use Lemma 16(b) to conclude that ; = 0 for ≥ + 1 and so that V ≥ − . Since ; does not change sign we have that V = − .