On stable hypersurfaces with vanishing scalar curvature

Abstract

We will prove that there are no stable complete hypersurfaces of $\mathbb {R}^4$ with zero scalar curvature, polynomial volume growth and such that $\frac{(-K)}{H^3}\ge c>0$ everywhere, for some constant $c>0$, where K denotes the Gauss-Kronecker curvature and $H$ denotes the mean curvature of the immersion. Our second result is the Bernstein type one there is no entire graphs of $\mathbb {R}^4$ with zero scalar curvature such that $\frac{(-K)}{H^3}\ge c>0$ everywhere. At last, it will be proved that, if there exists a stable hypersurface with zero scalar curvature and $\frac{(-K)}{H^3}\ge c>0$ everywhere, that is, with volume growth larger than polynomial growth of order four, then its tubular neighborhood is not embedded for suitable radius.

Appendix

Let us prove the following fact established in the Introduction:

Let $x:M^3\rightarrow \mathbb {R}^4$ be an isometric immersion with zero scalar curvature. If $H$ and $K$ denotes the mean curvature and Gauss-Kronecker curvature, respectively, then

$$\begin{aligned} 0\le \dfrac{-K}{H^3}\le \dfrac{4}{27}\ \text{ everywhere } \text{ on } \ M. \end{aligned}$$

In fact, let $(\lambda _1,\lambda _2,\lambda _3)=t\omega$ where $\omega \in \mathbb {S}^{2}$. Using (1.1), we can see that $R,H$ and $K$ are homogeneous polynomials. It implies $H(t\omega )=tH(\omega ), \ R(t\omega )=t^2R(\omega ),\ K(t\omega )=t^3K(\omega )$ and hence

$$\begin{aligned} \dfrac{K}{H^3}(t\omega )=\dfrac{K}{H^3}(\omega ). \end{aligned}$$

Then the behavior of $\frac{K}{H^3}$ depends only of its values on the sphere $\mathbb {S}^{2}.$ Since $N:=\{(\lambda _1,\lambda _2,\lambda _3)\in \mathbb {R}^3; R=\lambda _1\lambda _2+\lambda _1\lambda _3+\lambda _2\lambda _3=0\}$ is closed and $\mathbb {S}^{2}$ is compact, we obtain that $N_\omega =N\cap \mathbb {S}^{2}$ is compact, see Fig. 1. Then, $\frac{K}{H^3}:N_\omega \rightarrow \mathbb {R}$ is a continuous function with compact domain. The claim then follow from the Weierstrass maxima and minima theorem. Upperbound $\frac{4}{27}$ can be found using Lagrange multipliers method.

Fig. 1

Representation of the domain $N_\omega$ of $\frac{K}{H^3}$ over $\mathbb {S}^2,$ considering this function as an algebraic function of the eigenvalues. This domain is the intersection of one of the plane $\lambda _1+\lambda _2+\lambda _3=1$ with $\mathbb {S}^2.$ The hypothesis cuts off only three small neighbourhoods around the coordinate axis