The complete hyper-surfaces with zero scalar curvature in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb{R }^{n+1}$$\end{document}

Let \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M^n$$\end{document} be a complete and noncompact hyper-surface immersed in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$R^{n+1}$$\end{document}. We should show that if \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M$$\end{document} is of finite total curvature and Ricci flat, then \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M$$\end{document} turns out to be a hyperplane. Meanwhile, the hyper-surfaces with the vanishing scalar curvature is also considered in this paper. It can be shown that if the total curvature is sufficiently small, then by refined Kato’s inequality, conformal flatness and flatness are equivalent in some sense. And those results should be compared with Hartman and Nirenberg’s similar results with flat curvature assumption.

and constant mean curvature hyper-surfaces by many authors [4,5], etc., and the parametric minimal hyper-surfaces with finite total curvature by other group of mathematicians [7,18,20]. Notice that the mean curvature is just the trace of the second fundamental form. Hence, it will be equally interesting to consider other elementary symmetric functions of the second fundamental form. In particular, it is natural to ask whether hyper-surfaces with zero scalar curvature have Bernstein type property [2,6,10,17]. However, observe that the equation for the hyper-surfaces with zero mean curvature is elliptic, the analogy for surfaces with zero scalar curvature is only a degenerate elliptic equation. Thus for hyper-surfaces with the zero scalar curvature, we cannot expect the results being as nice as the ones for minimal hypersurfaces. On the other hand, we notice that there is another classical result given by Hartman and Nirenberg, which says that a complete hyper-surface with zero sectional curvature is either a hyperplane or a generalized cylinder. After we got some partial results for hypersurfaces with zero scalar curvature, the simple question we may ask is what happens if M is Ricci flat. This article will report what we have got so far along this direction.
First let us fix some standard notation. Let M n be a complete and noncompact hypersurface isometrically immersed in R n+1 . We denote the normalized mean curvature by H, the second fundamental form by B or {h i j } under a local coordinate system. For convenience, we call ( M |H | n dv) 1 n the total curvature of M. In the following, we will always assume M is orientable with a fixed orientation. Our first result is the following Bernstein type theorem: Theorem 1.1 Let M n (n > 2) be a complete and noncompact hyper-surface immersed in R n+1 with zero Ricci curvature. If the total curvature is finite, then M is a hyperplane.
Remark One should compare Theorem 1.2 with the result of Hartman and Nirenberg [9]. Notice that even M is of zero sectional curvature and M may not be a hyperplane. A typical example is the so-called generalized cylinder given by (x 1 , x 2 , . . . , x n ) ∈ R n −→ (x 1 , x 2 , . . . , x n , cosh(x n )) ∈ R n+1 . Thus our second condition is to get rid of the generalized cylinders.
Next we consider the hyper-surfaces with the vanishing scalar curvature. Observe that there are many examples of zero scalar curvature hyper-surfaces with finite total curvature which are not flat. See the examples provided by Lounie and Leite [13]. Clearly the analogy of our previous theorem with only the scalar flat assumption cannot be true. Nevertheless, we obtain the following result: (a) M is locally conformally flat; Notice that the curvature tensor can be decomposed into the Weyl tensor, Ricci tensor and scalar part. The flatness assumption means all Weyl, Ricci and scalar part vanish and Ricci flatness just simply means both Ricci and scalar part vanish. Clearly, the vanishing condition on sole one of three tensors in this decomposition is not enough to conclude the flatness. Thus it seems our above assumptions are reasonable to conclude the flatness. On the other hand, the flat hyper-surface has only one end while the circled cylinder (S 1 × R n−1 ) has two ends. In fact, with just concern of the number of ends, we have the following: then M has only one end.
The organizing of paper is as follows: in section two, we will list several useful Lemmas which will be used in the rest of paper. In section three, we will give a proof of our first result, i.e. Theorem 1.1. We should point out that what we really proved in this section is that if M is Ricci flat, then M is flat. Hence a famous result of Hartman and Nirenberg implies that M is either hyperplanes or generalized cylinders. And in section four, we prove our second result, i.e. Theorem 1.2. The main observation here is to fully use the assumption that the scalar curvature is zero. This condition implies a differential identity for mean curvature and the second fundamental form. Together with our assumption, this identity implies that M is flat hyper-surface and hence the conclusion follows as before. In the final section, we make several comments on number of ends of the hyper-surfaces under various assumptions and prove Theorem 1.3. This is motivated by similar result for either minimal hyper-surfaces or constant mean curvature hyper-surfaces. The key assumption is that the dimension is of at least three.

Several useful lemmas
First the main fact we should use for zero scalar curvature hyper-surfaces is the following Lemma: Lemma 2.1 Let M n be a hyper-surface immersed in R n+1 with scalar curvature R. Then we have Proof This is well known and can be found in any Riemannian geometry book.
Based on this identity, we have Lemma 2.2 Let M n be a hyper-surface isometrically immersed in R n+1 with constant scalar curvature R. Then the following identity holds true: Thus if R ≡ 0, then as shown in [1].
Proof Recall that for such hyper-surfaces, the curvature tensor is given by Hence differentiate the identity (2.1) to get And differentiate once again in the direction e k and sum them to get Here we have used the fact that h i j,k = h ik, j for all i, j, k. Now Ricci identity and Lemma 2.1 give Multiply h i j to both sides of above identity and sum up for i, j from 1 to n and rearrange the terms to get the identity (2.2).
To get the last inequality, we square both sides of the Eq. (2.3) and sum up with respect to k to get Notice that Cauchy-Schwartz inequality shows that Hence if H = 0, then |∇ B| 2 ≥ n 2 |∇ H | 2 . If H = 0 at some point, then h i j = 0 at that point. Hence the Eq. (2.2) shows that |∇ B| 2 = n 2 |∇ H | 2 at that point. Therefore, the desired inequality holds at all points.
Then we have

The equality holds if and only if n
Proof This is also well known. It follows from Lagrange multiplier method. For the detail, we refer readers to [15].

Lemma 2.4 Let M n be a sub-manifold immersed in R n+ p . Then for any function h
Proof This is proved by Michael and Simon [14] or Hoffman and Spruck [11], respectively.
We only need its following corollary: Proof For a function h as in Lemma 2.4, by Hölder inequality, one has Hence, this inequality and Lemma 2.4 imply ⎛ Thus the lemma is proved.
The next lemma is to get the volume control from below which will be useful in the course of proof of our main results: Lemma 2.6 Let M n (n ≥ 3) be a complete noncompact immersed hyper-surface in R n+1 . Assume that nC 1 H n < 1 where C 1 is again the constant given in Lemma 2.4. Then there exists a constant C 2 > 0 depending only upon C 1 so that

7)
for any q ∈ M, and all s ≥ 0.
Proof Take an arbitrary point p ∈ M; without loss of generality, we may assume p = 0. In the following, we let d(·, ·) be the distance function of R n+1 , and r (·, ·) the distance function of M with respect to the induced metric. We will write d(x), r (x) if the base point is 0.
Obviously d ≤ r for any two points in M. Let γ be a minimal geodesic from 0; then, By a direct computation, one can show that where η is out unit normal to the hyper-surface M and x is the position vector in R n+1 . In And also note that in any manifold, vol(∂ B(s)) = ∂ ∂r vol(B(r )) | r =s .
We thus obtain Therefore, by integrating it over the interval (0, s) and taking the exponential to get vol(B(s)) ≥ s n( Therefore, we have Hence if n = 2, we have (ν 1 + H )(ν 2 + H ) = 0, which means the sectional curvature is zero. If n ≥ 3, we conclude that (ν 1 + H ) = 0 which also implies that the sectional curvature of M vanishes. By Hartman and Nirenberg's theorem, we know that M is either R n or S 1 ×R n−1 . Then we see that the total curvature of second case is not finite. Hence we complete the proof of Theorem 1.1.

Proof of Theorem 1.2
This section is devoted to the argument for our Theorem 1.2.
If n = 2, by well-known result of Hartman and Nirenberg, any complete surfaces in R 3 with zero Gaussian curvature are either plane or cylinder. The later has infinity volume and non-zero constant mean curvature which cannot satisfy our finiteness assumption on the total integration of the square of the mean curvature. Therefore, this case follows. Thus in what follows, we assume n ≥ 3: Proof of Theorem 1.2 (d) ⇒ (a) is clear.

(a)⇒ (b):
M is locally conformally flat; then R i j,k = R ik, j since R ≡ 0. By the second Bianchi identity, one has Therefore, It is well known that the following identity holds: Combining with previous calculation, one obtains Let f be a cut-off function supported in a ball Multiplying both sides of above identity by f 2 |R i j | q with q > −1, one reaches Integrate it and use integration by parts to get which can be rewritten as Here we have used the refined Kato's inequality: This is well known since R i j is symmetric, trace free and R i j,k = R ik, j for all i, j and k. Choose q = n−4 2 . Notice that by Gauss formula, We can estimate the last term on the right-hand side as follows: where C s is Sobolev constant given in (2.5). Combine this estimate and the Eq. (4.1) to get ⎡ Therefore, if we choose where C 1 is the constant given in Lemma 2.4, then we have Thus let R → +∞ in the previous inequality; we obtain Hence, |R ji | = const. But the facts that |R ji | n 2 ∈ L 1 and M has infinite volume imply that |R ji | = 0.
Then by Theorem 1.1, h i j = 0. Therefore, Hence, the assumption (b) implies that h i j h kl,t = h i j,t h kl for all i, j, k, l and t.

Now Gauss equation implies R i j = n H h i j − h ik h k j . Thus we have
Therefore, |R i j | is a constant, since |d|R i j || ≤ |∇ R i j | = 0. Hence, the integrability of |R i j | n/2 implies that R i j = 0 which in turn implies that h i j = 0 by Theorem 1.
Hence, R i j = 0. By Theorem 1.1, M is flat.

Proof Of Theorem 1.3
In this section, we employ some methods due to Cao, Shen and Zhu to study hyper-surfaces with zero scalar curvature. We first have lower volume growth estimate as given in Lemma 2.6. Thus we can show that there exist bounded harmonic functions on such hyper-surfaces. We state it as a lemma. Proof The proof is analogy to the proof of Lemma 2 in [7]. We will provide the argument here for completeness of the paper. We first prove that for each compact set Then by the maximum principle for harmonic functions, 0 ≤ u i ≤ 1. For any j < i, we extend u j to u j : D i → R continuously such that u j = 1 or 0 on the complement D i − D j . Then u j has the same boundary condition as u i on ∂ D i . Hence by the minimality of the energy E(u i ) of u i over D i , one has the following monotonicity: Thus there exists a constant c 1 > 0 such that Therefore, we can find a harmonic function u on M such that Since Vol(D i ) → ∞, by letting i → ∞, we find that if u is a constant, then u ≡ 0 or u ≡ 1.
|∇ψ| ≤ c 2 , 0 ≤ ψ ≤ 1, and |∇ψ| vanishes outside D i 0 , then inequality (2.5) implies that As i → ∞, we find that Vol(F Proof of Theorem 1. 3 We argue by contradiction. By the construction of Lemma 5.1, we know that if M is of more than one end, then there exists a nontrivial bounded harmonic function u(x) on M which has finite total energy. For such a harmonic function u, let f (x) = |∇u|. By Bochner formula, we obtain 1 2 f 2 = |Hess u| 2 + Ric(∇u, ∇u).
Since the scalar curvature of the hyper-surface in our case is zero, by Lemma 2.1, we obtain Ric min ≥ −2(n − 1)(n − 2)H 2 .
With help of this estimate, Bochner formula takes the form, Now let ϕ be a cut-off function such that and |∇ϕ| ≤ C r with C = 2.
Multiplying ϕ 2 on both sides of the above inequality (5.5) and integrating by parts we can write it as