Stable Bernstein in Dimension Two

The proof presented here is based on the method of Fischer–Colbrie and Schoen [FCS80]. The argument has three main ingredients:

1. Properties of the Schrödinger operator \(-\Delta+q\), or equivalently of the equation \((\Delta-q)g=0\), on complete Riemannian manifolds.

2. Properties of the operator \(\Delta-a K\) on conformal metrics on the disc, where \(K\) is the Gauss curvature function and \(a\) is a constant.

3. Classification of the topology of stable minimal surfaces in 3-manifolds with non-negative scalar curvature.

Properties of differential operators on complete Riemannian manifolds

Let \((M, g)\) be a complete \(n\)-dimensional Riemannian manifold, and let \(q\) be a smooth function on \(M\). For any bounded domain \(D \subset M\) with smooth boundary, we denote by \(\lambda_{1}(D)\lt{}\lambda_{2}(D) \leq \lambda_{3}(D) \leq \cdots\) the Dirichlet eigenvalues of the Schrödinger operator \(-\Delta+q\) on \(D\), where \(\Delta\) is the Laplace–Beltrami operator with respect to the metric \(g\). Thus \(\lambda_1(D)\) is the bottom of the quadratic form associated with the equation \((\Delta-q)g=0\). The standard variational characterization of the first eigenvalue is given by

where \(|\nabla f|_{g}\) denotes the norm of the gradient of \(f\) with respect to the metric \(g\), and \(\mathrm{dvol}_{g}\) is the volume form induced by \(g\). The following is a fundamental property:

We now state the main result of this section.

Proof. (i) \(\Rightarrow\) (ii). This is a consequence of Lemma 5.1.2 since, for any bounded domain \(D \subset M\) and any point \(x_{0} \in M\) we can choose \(R\) large enough so that the ball \(B_{R}(x_{0})=\{x \in M: \mathrm{dist}(x, x_{0})\lt{}R\}\) satisfies \(B_{R}(x_{0}) \setminus \bar{D} \neq \emptyset\) and we have \(\lambda_{1}(B_{R}(x_{0})) \geq 0\) by hypothesis.

\(ii\) \(\Rightarrow\) (iii). To prove the existence of a positive solution \(g\) of \(\Delta g-q g=0\) we fix a point \(x_{0} \in M\). For each \(R\gt{}0\) we consider the problem

Since \(\lambda_{1}(B_{R}(x_{0}))\gt{}0\), the Fredholm alternative thus implies the existence of the above problem.

We now prove that \(u\gt{}0\) on \(B_{R}(x_{0})\). It follows from the strong maximum principle that if \(u \geq 0\) on \(B_{R}(x_{0})\), then \(u\gt{}0\) on \(B_{R}(x_{0})\). Suppose now that \(\Omega=\{x \in B_{R}(x_{0}): u(x)\lt{}0\} \neq \emptyset\). Hence \(\Omega \subset B_{R}(x_{0})\) is a bounded domain and thus, by Lemma 5.1.2, \(\lambda_{1}(\Omega)\gt{}0\). Since \(\Delta u-q u=0\) on \(\Omega\) and \(u=0\) on \(\partial \Omega\), we would have \(u \equiv 0\) in \(\Omega\) contradicting the unique continuation property. We have shown that \(u\gt{}0\) on \(B_{R}(x_{0})\).

We now set \(g_{R}(x)=u(x_{0})^{-1} u(x)\) for \(x \in M\). We have seen that \(g_{R}\) satisfies

From the Harnack inequality, it follows that on any ball \(B_{\sigma}(x_{0})\), there is a constant \(C\) depending only on \(\sigma\) and \(M\) (independent of \(R\)) such that, for \(R\gt{}2\sigma\)

It now follows from standard elliptic theory that all derivatives of \(g_{R}\) are bounded uniformly (independent of \(R\)) on compact subsets of \(M\). We may therefore choose a sequence \(R_{i} \to \infty\) so that \(g_{R_{i}}\) converges along with its derivatives on any compact subset of \(M\), and by taking a diagonal sequence we can arrange that \(g_{R_{i}}\) along with its derivatives, converges uniformly on compact subsets of \(M\) to a function \(g\) satisfying \(\Delta g-q g=0\) and \(g(x_{0})=1\). Since \(g\) is not identically zero and \(g \geq 0\) the strict maximum principle implies that \(g\gt{}0\). This finishes the proof that (ii) \(\Rightarrow\) (iii).

\(iii\) \(\Rightarrow\) (i). If \(g\gt{}0\) satisfies \(\Delta g-q g=0\) on \(M\) we define a new function \(w=\log g\). We now calculate

Let \(f\) be any function with compact support on \(M\). Multiplying (5.1.2) by \(f^{2}\) and integrating by parts, we obtain

Applying the Schwarz inequality and the arithmetic-geometric mean inequality we have

Putting this into the above equation and canceling the terms \(\int_{M} f^{2}|\nabla w|^{2}\) we obtain

If \(D\) is any bounded domain and \(f\) is any function with support in \(D\) we have shown that

It now follows from (5.1.1) that \(\lambda_{1}(D) \geq 0\). This finishes the proof of Theorem 5.1.3. ◻

The last part of the proof actually yields

The Operator \(\Delta-aK\) on Surfaces

Let \(M\) be the unit disc in the complex plane endowed with the metric \(\mathrm{d}s^{2}=\mu(z)|d z|^{2}\). We assume \(\mathrm{d}s^{2}\) is a complete metric. Let \(K\) denote the Gaussian curvature of \(M\) and \(\Delta\) the metric Laplacian, i.e., \(\Delta f=\mu^{-1}(f_{xx}+f_{yy})\), where \(z=x+i y\). The well-known formula for \(K\) is \(K=-\frac{1}{2} \Delta \log \mu\). We shall prove the following theorem.

This is the key analytic input for determining the conformal type of stable minimal surfaces in \(\mathbb{R}^3\). Fischer–Colbrie and Schoen [FCS80] prove the case \(a=1\), while the method of do Carmo and Peng [dCP79] gives the stated range \(a\gt{}\frac{1}{2}\). For the Poincaré metric on the disc the critical value is \(\frac{1}{4}\). Under the additional assumption \(K\leq 0\), Kawai [Kaw88] proves the corresponding nonexistence result for \(a\gt{}\frac{1}{4}\).

Proof. We define a function \(h\) by \(h=\mu^{-1/2}\). We see from the definition of \(K\) that \(\Delta \log h=K\), i.e.,

In particular, \(h\) satisfies

Let \(D \subset M\) be a bounded domain, and let \(\zeta\) be a smooth function on \(M\) with compact support in \(D\). We now calculate

So we have

Now define a smooth function \(\zeta(r)\) for \(r \in \mathbb{R}\) which satisfies

If \(r\) measures the metric distance to \(0\), and \(R\) is any positive number, then \(\zeta(r)\) defines a Lipschitz function on \(M\) with support in \(B_{R}(0)\). A standard approximation argument justifies this choice of \(\zeta\) in (5.1.5). Then

Putting this into (5.1.5) we have

Since \(\mu(z)|d z|^{2}\) is a complete metric on the disc, \(\mu\) cannot be a constant function. Therefore, \(|\nabla h|^{2}\) is not identically zero on \(M\). Thus, by choosing \(R\) sufficiently large in (5.1.7), we conclude that \(\lambda_{1}(B_{R}(0))\lt{}0\). By Theorem 5.1.3 this implies that there is no positive solution of \(\Delta g-aKg=0\) on \(M\). This completes the proof of Theorem 5.1.5. ◻

The next result is an extension of Theorem 5.1.5 with an additional non-negative potential. It follows directly from Theorem 5.1.3, Theorem 5.1.5, and formula (5.1.1).

Complete Stable Minimal Surfaces in 3-Manifolds

The stability of \(M\) is given by the following inequality:

where \(f\) is any function having compact support on \(M\) and \(\mathrm{Ric}(e_{3})\) is the Ricci curvature of \(N\) in the direction of \(e_{3}\). We now do the rearrangement described in Schoen–Yau [SY79a]. The Gauss curvature equation says that

where \(K\) is the intrinsic Gaussian curvature of \(M\) and \(K_{ij}\) is the sectional curvature of \(N\) for the section determined by \(e_{i}\) and \(e_{j}\).

Using minimality and symmetry of \(h_{ij}\) we have

Inequality (5.1.8) may then be written in the form

where \(S\) is the scalar curvature of \(N\) given by \(S=2(K_{12}+K_{13}+K_{23})\).

Set

According to (5.1.1), this inequality is equivalent to

for every bounded domain \(D \subset M\), i.e. to Theorem 5.1.3 with \(q=-P\). The associated equation is given by the stability operator

We now classify the stable minimal surfaces in three-manifolds of non-negative scalar curvature.

Before giving the proof of Theorem 5.1.7 we state the following corollary for the case when \(N\) is \(\mathbb{R}^{3}\). This implies the classical Bernstein theorem [Ber17] for complete minimal graphs in \(\mathbb{R}^{3}\).

Proof. In this case the stability operator (5.1.10) becomes \(\Delta-2K\) and by Theorem 5.1.7 we know that \(M\) is conformally either \(\mathbb{C}\) or \(\Lambda\). By Theorem 5.1.3 there is a positive function \(g\) on \(M\) satisfying \(\Delta g-2K g=0\). If \(M\) is conformal to \(\Lambda\) we may lift \(g\) to the universal covering \(\mathbb{C}\) of \(\Lambda\). In either case we have a metric on \(\mathbb{C}\) with \(K \leq 0\) and a positive \(g\) satisfying \(\Delta g-2K g=0\). Thus \(\Delta g \leq 0\), and \(g\) is a positive superharmonic function on \(\mathbb{C}\) which must be constant. Therefore \(K\) is identically zero and hence \(\sum_{i,j} h_{ij}^{2}=-2K\) is identically zero. Consequently \(M\) is a plane. ◻

Observe that each of the four possibilities of Theorem 5.1.7 does occur. For example, \(S^{2} \times \mathbb{R}\) has positive scalar curvature and has a stable \(S^{2}\), \(T^{2} \times \mathbb{R}\) is flat and has a stable \(T^{2}\). We can choose a metric on \(\mathbb{C}\) of positive Gaussian curvature and by crossing with \(\mathbb{R}\) construct a metric of positive scalar curvature on \(\mathbb{R}^{3}\) having a stable \(\mathbb{C}\). Similarly \(\Lambda \times \mathbb{R}\) has a flat metric with a stable \(\Lambda\).

Proof. Case (i) was observed by Schoen–Yau [SY79a] and follows by choosing \(f\) identically equal to one in inequality (5.1.9) to obtain

The Gauss-Bonnet theorem now implies that \(M\) is the sphere or the torus. In the torus case \(S \equiv 0\) on \(M\). The stability operator reduces to \(\Delta-K\) and its first eigenvalue is

Since \(\lambda_{1} \geq 0\) by stability and \(\int_{M} K \,\mathrm{dvol}=0\) we conclude that \(\lambda_{1}=0\) and the constant function \(f \equiv 1\) satisfies \(\Delta f-K f=0\) which implies that \(K \equiv 0\).

To prove case (ii), we first show that the universal covering of \(M\) is conformally equivalent to \(\mathbb{C}\). If this is not true, then \(M\) is covered by the disc. Using stability and Theorem 5.1.3 we have a positive function \(g\) on \(M\) satisfying

Lifting \(g\) to the disc we obtain a positive solution of this equation on the disc endowed with a complete metric. Since \(S+\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2} \geq 0\), this yields a contradiction by Corollary 5.1.6. Thus we have shown that \(M\) is conformally covered by \(\mathbb{C}\) and hence \(M\) is either conformally equivalent to \(\mathbb{C}\) or \(M\) is conformal to a cylinder \(\Lambda\).

If \(M\) is a cylinder, let \(z=x+i y\) be a complex coordinate for \(M\) so that \(|d z|^{2}\) is the flat metric on \(M\), and the given metric on \(M\) is \(\mathrm{d}s^{2}=\mu(z)|d z|^{2}\). Fix a point \(z_{0} \in M\) and let \(r\) be the distance from \(z_{0}\) taken with respect to the flat metric. For any \(R\gt{}0\), choose \(\zeta(r)\) satisfying (5.1.6). Substituting \(\zeta\) for \(f\) in (5.1.9) and using (5.1.6) and the conformal invariance of the Dirichlet integral we have

where \(B_{R}(z_{0})\) is the ball taken with respect to the flat metric. Since \(\int_{B_{R}(z_{0})} \mathrm{d}x \mathrm{d}y\) has growth bounded by a constant times \(R\) and we are assuming \(\int_{M}|K| \,\mathrm{dvol}\lt{}\infty\) we can use the dominated convergence theorem to let \(R\) tend to infinity to achieve

Recall that the Cohn–Vossen inequality states that if \(M\) is a complete non-compact surface with finite total curvature, then \(\int_{M} K \,\mathrm{dvol}\leq 2\pi \chi(M)\), where \(\chi(M)\) is the Euler characteristic of \(M\). Since \(M\) is topologically a cylinder, we have \(\chi(M)=0\). Thus the Cohn–Vossen inequality gives \(\int_{M} K \,\mathrm{dvol}\leq 0\). Since \(S+\frac{1}{2} \sum h_{ij}^{2} \geq 0\) we conclude that \(M\) is totally geodesic and \(S \equiv 0\) on \(M\).

Hence the stability operator reduces to \(\Delta-K\). By Theorem 5.1.3 there is a positive function \(g\) on \(M\) satisfying \(\Delta g-K g=0\). Set \(w=\log g\). Computing we have

Choosing \(\zeta\) as above, we multiply by \(\zeta^{2}\) and integrate by parts to get

The Cauchy-Schwarz inequality and the arithmetic-geometric mean inequality give

Therefore,

Letting \(R \to \infty\) as above, we conclude that

Thus \(w\) is constant, so \(g\) is constant and we have \(K \equiv 0\).

In case \(N\) has non-negative Ricci curvature, we write the stability operator as \(\Delta+\mathrm{Ric}(e_{3})+\sum_{i,j=1}^{2} h_{ij}^{2}\) and note that the proof that \(M\) is totally geodesic now follows as in the previous paragraph (without the assumption of finite total curvature). From the previous proof we also get that

Since

we have

Thus the Gauss curvature of \(M\) is non-negative and, since \(M\) is a cylinder, we have \(K \equiv 0\). This completes the proof of Theorem 5.1.7. ◻