Stable Bernstein Theorems | Gaoming Wang

Stable Bernstein in Dimension Two

Mon, 29 Jun 2026 00:00:00 +0000

Theorem 5.1.1

Let \(M\) be a complete, stable minimal surface in \(\mathbb{R}^3\). Then \(M\) is a plane.

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

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

\begin{equation} \tag{5.1.1} \label{eq:variational} \lambda_{1}(D)=\inf \left\{\int_{D}\left(|\nabla f|_{g}^{2}+q f^{2}\right) \mathrm{dvol}_{g}: \mathrm{spt}\, f \subset D, \int_{D} f^{2} \mathrm{dvol}_{g}=1\right\}, \end{equation}

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:

Lemma 5.1.2

If \(D, D'\) are connected domains in \(M\) with \(D \subset D'\), then \(\lambda_{1}(D) \geq \lambda_{1}(D')\). Moreover, if \(D' \setminus \bar{D} \neq \emptyset\), then \(\lambda_{1}(D)\gt{}\lambda_{1}(D')\).

We now state the main result of this section.

Theorem 5.1.3

The following conditions are equivalent:

\(\lambda_{1}(D) \geq 0\) for every bounded domain \(D \subset M\);
\(\lambda_{1}(D)\gt{}0\) for every bounded domain \(D \subset M\);
there exists a positive function \(g\) satisfying the equation \(\Delta g-q g=0\) on \(M\).

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

\[ \begin{cases} \Delta u-q u=0 & \text{on } B_{R}(x_{0}), \\ u=1 & \text{on } \partial B_{R}(x_{0}). \end{cases} \]

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

\[ \Delta g_{R}-q g_{R}=0 \text{ on } B_{R}(x_{0}), \quad g_{R}(x_{0})=1, \, g_{R}\gt{}0 \text{ on } B_{R}(x_{0}). \]

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 \(R\gt{}2\sigma\)

\[ g_{R} \leq C \text{ on } B_{\sigma}(x_{0}). \]

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

\begin{equation} \tag{5.1.2} \label{eq:w} \Delta w=q-|\nabla w|^{2}. \end{equation}

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

\[ -\int_{M} q f^{2} \,\mathrm{dvol}+\int_{M}|\nabla w|^{2} f^{2} \,\mathrm{dvol}=2 \int_{M} f\langle \nabla f, \nabla w\rangle \,\mathrm{dvol}. \]

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

\[ 2|f\langle \nabla f, \nabla w\rangle| \leq 2|f||\nabla f||\nabla w| \leq f^{2}|\nabla w|^{2}+|\nabla f|^{2}. \]

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

\[ -\int_{M} q f^{2} \,\mathrm{dvol}\leq \int_{M}|\nabla f|^{2} \,\mathrm{dvol}. \]

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

\[ \int_{D}\left(|\nabla f|^{2}+q f^{2}\right) \,\mathrm{dvol}\geq 0. \]

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

Corollary 5.1.4

If \(D \subset M\) is any bounded domain, and if there is a function \(g\gt{}0\) in \(D\) satisfying \(\Delta g-q g=0\), then \(\lambda_{1}(D) \geq 0\).

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.

Theorem 5.1.5

Assume \(\mathrm{d}s^{2}\) is complete. For \(a \gt{} \frac{1}{2}\) there is no positive solution \(g\) of \(\Delta g-a K g=0\) on \(M\).

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.,

\[ \frac{\Delta h}{h}-\frac{|\nabla h|^{2}}{h^{2}}=K. \]

In particular, \(h\) satisfies

\begin{equation} \tag{5.1.3} \label{eq:h} h\Delta h=K h^2+|\nabla h|^{2}. \end{equation}

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

\begin{equation} \tag{5.1.4} \label{eq:calc1} \begin{aligned} \int_{M}\left(|\nabla(\zeta h)|^{2}+aK(\zeta h)^{2}\right) &=\int_{M} |\nabla\zeta|^{2} h^{2} + \frac{1}{2}\langle \nabla\zeta^2, \nabla h^2\rangle + \zeta^2 |\nabla h|^2+aK(\zeta h)^{2}\\ &=\int_{M}|\nabla\zeta|^{2} h^{2}+\frac{1-a}{2}\langle \nabla\zeta^2, \nabla h^2\rangle+\frac{1-a}{2}\zeta^2|\nabla h|^2\\ &\quad - \frac{a}{2}\zeta^2 h \Delta h +aK(\zeta h)^{2} \,\mathrm{dvol}\\ &=\int_{M}\left(|\nabla\zeta|^{2} h^{2} +\frac{1-2a}{2} \zeta^2 |\nabla h|^{2}\right) + \frac{1-a}{2} \langle \nabla\zeta^2, \nabla h^2\rangle\\ &\leq \int_{M}C|\nabla\zeta|^{2} h^{2} -\varepsilon\int_{M}|\nabla h|^{2} \zeta^{2} \,\mathrm{dvol}. \end{aligned} \end{equation}

So we have

\begin{equation} \tag{5.1.5} \label{eq:calc3} \lambda_{1}(D) \int_{M}(\zeta h)^{2} \,\mathrm{dvol}\leq \int_{M}|\nabla\zeta|^{2} h^{2} \,\mathrm{dvol}-\int_{M}|\nabla h|^{2} \zeta^{2} \,\mathrm{dvol}. \end{equation}

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

\begin{equation} \tag{5.1.6} \label{eq:zeta} \begin{gathered} \zeta(r)=1 \text{ for } r \leq \frac{1}{2} R, \, \zeta(r)=0 \text{ for } r \geq R, \\ \zeta \geq 0 \text{ for all } r, \quad |\zeta'| \leq \frac{3}{R} \text{ for all } r. \end{gathered} \end{equation}

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

\[ \int_{M}|\nabla\zeta|^{2} h^{2} \,\mathrm{dvol}\leq \frac{9}{R^{2}} \int_{M} \mathrm{d}x \mathrm{d}y=\frac{9\pi}{R^{2}}. \]

Putting this into (5.1.5) we have

\begin{equation} \tag{5.1.7} \label{eq:calc4} \lambda_{1}\left(B_{R}(0)\right) \int_{M}(\zeta h)^{2} \,\mathrm{dvol}\leq \frac{9\pi}{R^{2}}-\int_{M}|\nabla h|^{2} \zeta^{2} \,\mathrm{dvol}. \end{equation}

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).

Corollary 5.1.6

Let \(\mathrm{d}s^{2}=\mu(z)|d z|^{2}\) be a complete metric on the disc. If \(a \geq 1\) and \(P\) is a non-negative function, then there is no positive solution \(g\) of \(\Delta g-a K g+P g=0\) on \(M\).

Complete Stable Minimal Surfaces in 3-Manifolds

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

\begin{equation} \tag{5.1.8} \label{eq:stability1} \int_{M}\left[|\nabla f|^{2}-\left(\mathrm{Ric}(e_{3})+\sum_{i,j=1}^{2} h_{ij}^{2}\right) f^{2}\right] \,\mathrm{dvol}\geq 0, \end{equation}

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

\[ K=K_{12}+h_{11}h_{22}-h_{12}^{2}, \]

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

\[ K=K_{12}-\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2}. \]

Inequality (5.1.8) may then be written in the form

\begin{equation} \tag{5.1.9} \label{eq:stability2} \int_{M}\left[|\nabla f|^{2}-\left(S-K+\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2}\right) f^{2}\right] \,\mathrm{dvol}\geq 0, \end{equation}

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

Set

\[ P:=S-K+\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2}. \]

According to (5.1.1), this inequality is equivalent to

\[ \lambda_1(D;-\Delta-P)\geq0 \]

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

\begin{equation} \tag{5.1.10} \label{eq:operator} \Delta+\left(S-K+\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2}\right). \end{equation}

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

Theorem 5.1.7

Let \(N\) be a complete oriented 3-manifold of non-negative scalar curvature. Let \(M\) be an oriented complete stable minimal surface in \(N\). There are two possibilities:

If \(M\) is compact, then \(M\) is conformally equivalent to the sphere \(S^{2}\) or \(M\) is a totally geodesic flat torus \(T^{2}\). If \(S\gt{}0\) on \(N\), then \(M\) is conformally equivalent to \(S^{2}\).
If \(M\) is not compact, then \(M\) is conformally equivalent to the complex plane \(\mathbb{C}\) or the cylinder \(\Lambda\). If \(M\) is a cylinder and the absolute total curvature of \(M\) is finite, then \(M\) is flat and totally geodesic. If the scalar curvature of \(N\) is everywhere positive, then \(M\) cannot be a cylinder with finite total curvature.

If the Ricci curvature of \(N\) is non-negative, then the assumption of finite total curvature in (ii) can be removed.

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 \(\mathbb{R}^{3}\).

Corollary 5.1.8

The only complete stable oriented minimal surface in \(\mathbb{R}^{3}\) is the plane.

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

\[ \int_{M}\left(S+\frac{1}{2} \sum h_{ij}^{2}\right) \,\mathrm{dvol}\leq \int_{M} K \,\mathrm{dvol}. \]

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

\[ \lambda_{1} \equiv \inf \left\{\int_{M}\left(|\nabla f|^{2}+K f^{2}\right) \,\mathrm{dvol}: \int_{M} f^{2}=1\right\}. \]

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

\[ \Delta g-K g+\left(S+\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2}\right) g=0. \]

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

\[ \frac{9}{R^{2}} \int_{B_{R}(z_{0})} \mathrm{d}x \mathrm{d}y-\int_{M}\left(S-K+\frac{1}{2} \sum h_{ij}^{2}\right) f^{2} \,\mathrm{dvol}\geq 0, \]

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

\[ \int_{M}\left(S+\frac{1}{2} \sum_{i,j=1}^{2} h_{ij}^{2}\right) \,\mathrm{dvol}\leq \int_{M} K \,\mathrm{dvol}. \]

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

\[ \Delta w=K-|\nabla w|^{2}. \]

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

\[ \int_{M}|\nabla w|^{2} \zeta^{2} \,\mathrm{dvol}=\int_{M} \zeta^{2} K \,\mathrm{dvol}+2 \int_{M} \zeta\langle \nabla\zeta, \nabla w\rangle \leq \int_{M} \zeta^{2} K +\frac{1}{4}|\nabla w|^{2} \zeta^{2} \,\mathrm{dvol}+4 |\nabla\zeta|^{2}. \]

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

\[ 2|\zeta||\langle \nabla\zeta, \nabla w\rangle| \leq \frac{1}{4} \zeta^{2}|\nabla w|^{2}+4|\nabla\zeta|^{2}. \]

Therefore,

\[ \frac{3}{4} \int_{M}|\nabla w|^{2} \zeta^{2} \,\mathrm{dvol}\leq \int_{M} \zeta^{2} K \,\mathrm{dvol}+4 \int_{M}|\nabla\zeta|^{2} \,\mathrm{dvol}. \]

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

\[ \frac{3}{4} \int_{M}|\nabla w|^{2} \,\mathrm{dvol}\leq \int_{M} K \,\mathrm{dvol}. \]

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

\[ \mathrm{Ric}(e_{3})=K_{13}+K_{23}=0 \text{ on } M. \]

Since

\[ \mathrm{Ric}(e_{1})=K_{12}+K_{13} \geq 0, \quad \mathrm{Ric}(e_{2})=K_{12}+K_{23} \geq 0, \]

we have

\[ \mathrm{Ric}(e_{1})+\mathrm{Ric}(e_{2})=2 K_{12}=2 K \geq 0. \]

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. ◻

Higher-Dimensional Stable Bernstein Theorems

Mon, 29 Jun 2026 00:00:00 +0000

We outline the strategy for the unconditional stable Bernstein theorem in dimensions \(n=3,4,5\). The argument has three main steps: a conformal change producing positive spectral bi-Ricci curvature, the construction of \(\mu\)-bubbles, and uniform area bounds for these bubbles via spectral Ricci estimates. The conformal and spectral curvature estimates are proved in §5.2.1; the \(\mu\)-bubble reduction is treated in §5.2.2, and the spectral Ricci estimates are recorded in §5.2.6.

Step 1: Conformal change and spectral bi-Ricci curvature. Let \(M^n\hookrightarrow \mathbb{R}^{n+1}\) be a complete, two-sided, stable minimal immersion. Fix \(0\in M\) and write \(r(x)=|x|\) for the Euclidean distance from the origin. Following Schoen–Simon–Yau and the recent stable Bernstein literature, we pass to the conformal metric

\[ \tilde{g}=r^{-2}g, \qquad g=\varphi^*\delta, \]

where \(g\) is the induced metric. Stability of \(M\) yields a spectral inequality for the conformal Laplacian with a curvature potential. To state it uniformly in dimension, we introduce the bi-Ricci curvature.

Definition 5.2.1

Let \((N,g)\) be a Riemannian manifold of dimension \(m\geq 2\). For orthonormal \(v,w\in T_pN\), define

\[ \mathrm{BiRic}(v,w):=\mathrm{Ric}(v,v)+\mathrm{Ric}(w,w)-R(v,w,v,w). \]

As functions on \(N\), we write \(\mathrm{BiRic}\) and \(\mathrm{Ric}\) for the minimum over unit directions (with \(v\perp w\) for \(\mathrm{BiRic}\)):

\[ \mathrm{BiRic}(p):=\inf_{\substack{v,w\in T_pN\\ |v|=|w|=1,\ \langle v,w\rangle=0}} \mathrm{BiRic}(v,w), \qquad \mathrm{Ric}(p):=\inf_{\substack{v\in T_pN\\ |v|=1}} \mathrm{Ric}(v,v). \]

In dimension \(m=3\), the bi-Ricci curvature is independent of the chosen orthonormal pair:

\[ \mathrm{BiRic}(v,w)=\frac12 R. \]

This is why the three-dimensional part of the argument is often stated as a spectral scalar-curvature condition.

We use spectral curvature condition to mean a lower bound for the first eigenvalue of a Schrödinger operator whose potential is built from one of these curvature quantities. The coefficient depends on the dimension and on the normalization used in the \(\mu\)-bubble argument.

In the conformal metric \(\tilde g=r^{-2}g\), stability gives the concrete inputs needed below: for \(n=3\),

\[ \lambda_1\left(-\tilde{\Delta}+\frac12\tilde R\right)\geq \frac32 \]

by the conformal scalar-curvature computation in §5.2.1, while for \(n=4\),

\[ \lambda_1\left(-\tilde{\Delta}-\widetilde{\mathrm{BiRic}}\right)\geq 1 \]

after the corresponding conformal bi-Ricci normalization.

Thus, after conformal reparametrization, the problem is reduced to studying a manifold with non-negative spectral bi-Ricci curvature (in dimension three this is precisely non-negative spectral scalar curvature). This replaces the extrinsic curvature input of the minimal hypersurface by an intrinsic spectral curvature hypothesis on \((M,\tilde{g})\).

Step 2: Construction of \(\mu\)-bubbles. The second step is to produce separating hypersurfaces inside a collar of \((M,\tilde{g})\) with controlled geometry. These are \(\mu\)-bubbles.

Classically, a soap bubble is a hypersurface of constant mean curvature (CMC) enclosing a prescribed volume—the archetypal prescribed mean curvature problem. More generally, one studies hypersurfaces whose mean curvature is a prescribed function of position and geometry; Gromov introduced and systematically used such constructions in his work on scalar curvature, naming them \(\mu\)-bubbles (the name reflects the measure \(\mu\) specifying the prescription). In the present setting, one considers a Riemannian manifold \((N,g)\) with boundary \(\partial N=\partial_+N\cup\partial_-N\) and seeks a domain \(\Omega\subset N\) whose boundary \(\Sigma=\partial\Omega\setminus\partial_-N\) (the \(\mu\)-bubble) solves a variational problem of prescribed mean curvature type: schematically, one minimizes a functional of the form

\[ \mathcal{A}(\Omega)=\int_{\Sigma} w\,d\mathcal{H}^{m-1}-\int_{\Omega} w\,h\,d\mathcal{H}^m \]

for a weight \(w\) (often the first eigenfunction of a suitable operator on \(N\)) and a carefully chosen function \(h\) determined by the curvature of \(N\). Under the spectral bi-Ricci condition from Step 1, such a minimizer exists and provides a \(\mu\)-bubble \(\Sigma\) separating \(\partial_-N\) from \(\partial_+N\) (Theorem 5.2.8).

In recent years, \(\mu\)-bubbles have become a central tool in positive scalar curvature geometry and have led to breakthroughs in the classification of manifolds with positive scalar curvature, the resolution of the Riemannian positive mass theorem in many settings, and the stable Bernstein program in dimensions three through five. The method is flexible: by choosing the prescription (the \(\mu\)-data and the function \(h\)), one encodes the desired curvature inequality into the Euler–Lagrange equation of the \(\mu\)-bubble.

Step 3: Volume and diameter estimates for \(\mu\)-bubbles. The third step controls the size of each \(\mu\)-bubble \(\Sigma\). This is analogous in spirit to the dimension reduction in the Schoen–Yau proof of the positive mass theorem: one does not work directly on the full ambient manifold, but on a hypersurface of one dimension lower whose geometry is more rigid.

After Step 2, each \(\mu\)-bubble \(\Sigma\) is a closed hypersurface in \(N\) (of dimension \(m-1\)) lying in a thin collar between \(\partial_-N\) and \(\partial_+N\). Using the construction and the spectral bi-Ricci hypothesis on \(N\), one shows that \(\Sigma\) itself carries a spectral non-negative Ricci condition: the first eigenvalue of an operator of the form \(-\Delta^\Sigma+\alpha\,\mathrm{Ric}\) is non-negative for some \(\alpha\in(0,2)\) depending on \(n\). One then studies manifolds with spectral non-negative Ricci curvature by methods similar to Bray’s proof of the Bishop–Gromov comparison: spectral bounds on \(-\Delta+\mathrm{Ric}\) imply diameter and volume upper bounds (Theorem 5.2.13 and the discussion in §5.2.6). In particular, each component of \(\partial\Omega\) has area and intrinsic diameter bounded by constants depending only on the spectral bound, not on the size of the collar.

Conclusion of the proof. Combining Steps 1–3, one obtains separating \(\mu\)-bubbles in long conformal annuli with uniform \(\tilde g\)-area and diameter bounds. As explained in the proof of the Euclidean volume-growth estimate in §5.2.6, these conformal estimates imply

\[ |B_\rho^M(0)|_g\leq C\rho^n \]

for every \(\rho\gt{}0\). Thus the \(\mu\)-bubble argument supplies the area-growth hypothesis needed for the stable Bernstein theorems with area growth, and the cited classification results force \(M\) to be an affine hyperplane. This completes the proof sketch; the conformal and spectral bi-Ricci estimates used in Steps 1–3 are proved in §5.2.1.

Conformal change of the metric on minimal hypersurfaces

Let \(M^n\hookrightarrow \mathbb{R}^{n+1}\) be a complete, two-sided, stable minimal immersion with induced metric \(g=\varphi^*\delta\). Fix \(0\in M\) and write \(r(x)=|x|\). Throughout we use the conformal metric

\[ \tilde{g}=r^{-2}g, \qquad d\tilde{\mu}=r^{-n}\,d\mu, \qquad \tilde{\nabla}=r^2\nabla^M,\quad e_i=r \tilde{e}_i. \]

We use the bi-Ricci convention fixed above: \(\mathrm{BiRic}\) denotes the minimum over orthonormal two-frames, and in dimension three \(\mathrm{BiRic}=\frac12 R\).

Theorem 5.2.2

Let \(M^n\hookrightarrow \mathbb{R}^{n+1}\) be a complete, two-sided, stable minimal immersion with induced metric \(g=\varphi^*\delta\). Then, the conformal metric \(\tilde{g}=r^{-2}g\) satisfies

\[ -\tilde{\Delta}+\frac{2}{n-2}\widetilde{\mathrm{BiRic}}\geq \frac{2(-n^3+6n^{2}-4n-8)}{8(n-2)} \]

where \(\tilde{\Delta}\) is the conformal Laplacian and \(\widetilde{\mathrm{BiRic}}\) is the conformal bi-Ricci curvature.

Under a general conformal change \(\tilde{g}=f^{-2}g\), we first relate the intrinsic Hessian on \(M\) to the ambient Euclidean Hessian.

Lemma 5.2.3

Let \(M^n\subset\mathbb{R}^{n+1}\) be a hypersurface with unit normal \(\nu\), second fundamental form \(A(X,Y)=\langle D_X Y,\nu\rangle\), and ambient connection \(D\). For \(f\in C^\infty(M)\) and tangent vector fields \(X,Y\) on \(M\),

\[ \mathrm{Hess}^M f(X,Y):=\langle \nabla^M_X\nabla^M f,Y\rangle = D^2 f(X,Y)+A(X,Y)\langle Df,\nu\rangle, \]

where \(D^2 f(X,Y):=D_X(D_Y f)\) and \(\nabla^M\) is the Levi-Civita connection of the induced metric.

Proof. Write the ambient gradient \(Df=\nabla^M f+\langle Df,\nu\rangle\,\nu\), so \(\nabla^M f\) is the tangential part. By direct computation,

\[ \begin{aligned} \nabla^M_X\nabla^M_Y f ={}& X Y (f)-\nabla_{X}^MY f = XY(f)-D_XY(f)+\langle D_X Y,\nu\rangle \langle Df,\nu\rangle\\ ={}& D^{2} f(X,Y)+A(X,Y)\langle Df,\nu\rangle. \end{aligned} \]

◻

For the conformal metric \(\tilde{g}=e^{2\varphi}g\), the change of Riemannian curvatures is given by

\[ \begin{aligned} \tilde{R}_{ijij} &= e^{2\varphi} R_{ijij}-e^{2\varphi}(T_{ii}+T_{jj}) \end{aligned} \]

where

\[ T_{ij} = \nabla_i \nabla_j \varphi-\nabla_i \varphi \nabla_j \varphi+\frac{1}{2}|\nabla \varphi|^2 g_{ij}. \]

For \(\varphi=-\log r\), we have

\[ T_{ii}=-\nabla^{2}_{ii}\log r - |r_i|^{2}/r^{2}+\frac{1}{2}|\nabla r|^{2}/r^{2}=-D^{2}_{ii}\log r-\frac{|r_i|^{2}}{r^{2}} +\frac{1}{2}|\nabla r|^{2}/r^{2}-A_{ii}\langle Dr,\nu\rangle/r \]

\[ T_{ii}=-\frac{1}{r^{2}}+|r_i|^{2}/r^{2}+\frac{1}{2}|\nabla r|^{2}/r^{2}-A_{ii}\langle Dr,\nu\rangle/r \]

For the conformal metric \(\tilde{g}=r^{-2}g\) with \(r(x)=|x|\), \(\varphi=-\log r\), the conformal curvatures satisfy

\[ \begin{aligned} \tilde{R}_{\tilde{i}\tilde{j}\tilde{i}\tilde{j}} &= r^4\tilde{R}_{ijij}\\ &= r^2 R_{ijij}-r^{2}(T_{ii}+T_{jj})\\ &= r^{2}R_{ijij}-(-2+|r_i|^{2}+|r_j|^{2}+|\nabla r|^{2}-A_{ii}\langle rDr,\nu\rangle-A_{jj}\langle rDr,\nu\rangle)\\ \widetilde{\mathrm{BiRic}}(\tilde e_1,\tilde e_2) &= r^2\mathrm{BiRic}(e_1,e_2)+2(n-3)-(2n-1)|\nabla r|^{2} - (n-3)(|r_i|^{2}+|r_j|^{2})\\ &\quad +(n-3)(A_{11}+A_{22})\langle rDr,\nu\rangle, \end{aligned} \]

We compute \(\mathrm{BiRic}\) on \(M\) using the Gauss equation.

Proposition 5.2.4

Let \(\{e_1,\ldots,e_n\}\) be a local orthonormal frame on a minimal hypersurface \(M^n\subset\mathbb{R}^{n+1}\). Then

\[ \mathrm{BiRic}(e_1,e_2) =-\sum_{i=1}^n A_{1i}^2-\sum_{j=2}^n A_{2j}^2-A_{11}A_{22}. \]

Proof. Using the Gauss equation, we compute

\[ \begin{aligned} \mathrm{BiRic}(e_1,e_2) &=\sum_{i=2}^n R_{1i1i}+\sum_{j=3}^n R_{2j2j}\\ &=\sum_{i=2}^n (A_{11}A_{ii}-A_{1i}^2)+\sum_{j=3}^n (A_{22}A_{jj}-A_{2j}^2)\\ &=-\sum_{i=1}^n A_{1i}^2-\sum_{j=2}^n A_{2j}^2-A_{11}A_{22}, \end{aligned} \]

where the last line follows since \(\mathrm{tr}\,A=0\) for a minimal hypersurface. ◻

Now, we choose \(\tilde{e}_1\) and \(\tilde{e}_2\) such that \(\widetilde{\mathrm{BiRic}}(\tilde{e}_1,\tilde{e}_2)\) takes the minimum value. Then, we have

\[ \begin{aligned} \widetilde{\mathrm{BiRic}}={}& -\sum_{i=1}^n A_{1i}^2-\sum_{j=2}^n A_{2j}^2-A_{11}A_{22}+2(n-3)-(2n-1)|\nabla r|^{2} - (n-3)(|r_i|^{2}+|r_j|^{2})\\ &+(n-3)(A_{11}+A_{22})\langle rDr,\nu\rangle. \end{aligned} \]

Proposition 5.2.5

For \(n \geq 3\), we have

\[ r^2 |A|^2 \geq \frac{2}{n-2} \left( (3n-3) - (2n-1)|dr|^2 - \widetilde{\mathrm{BiRic}} \right). \]

Proof. Recall that

\begin{equation} \tag{5.2.1} \label{eq:3.1} \begin{aligned} & r^2 \left( \sum_{i=1}^n A_{1i}^2 + \sum_{j=2}^n A_{2j}^2 + A_{11}A_{22} \right) + (n-3)\langle \vec x,\nu\rangle (A_{11}+A_{22}) \\ &\qquad = (4n-6) - (2n-1)|dr|^2 - (n-3)\bigl(dr(e_1)^2+dr(e_2)^2\bigr) - \widetilde{\mathrm{BiRic}} . \end{aligned} \end{equation}

Since \(\langle \vec x,\nu\rangle = r\,dr(\nu)\), we use Young’s inequality to obtain

\[ \left| (n-3)\langle \vec x,\nu\rangle (A_{11}+A_{22}) \right| \leq (n-3)dr(\nu)^2 + \frac{n-3}{4}r^2(A_{11}+A_{22})^2 . \]

Combined with (5.2.1) and the fact that

\[ dr(e_1)^2+dr(e_2)^2+dr(\nu)^2 \leq 1, \]

we have

\[ \begin{aligned} & r^2 \left( \sum_{i=1}^n A_{1i}^2 + \sum_{j=2}^n A_{2j}^2 + A_{11}A_{22} + \frac{n-3}{4}(A_{11}+A_{22})^2 \right) \\ &\qquad \geq (3n-3) - (2n-1)|dr|^2 - \widetilde{\mathrm{BiRic}} . \end{aligned} \]

Now we compute, using the fact that \(\operatorname{Tr} A=0\),

\[ \begin{aligned} & A_{11}^2+A_{22}^2+A_{11}A_{22} + \frac{n-3}{4}(A_{11}+A_{22})^2 \\ &\qquad = \frac12(A_{11}^2+A_{22}^2) + \frac{n-1}{4}(A_{11}+A_{22})^2 \\ &\qquad = \frac12(A_{11}^2+A_{22}^2) + \frac{n-1}{4}\sigma(A_{11}+A_{22})^2 + \frac{n-1}{4}(1-\sigma)(A_{33}+\cdots+A_{nn})^2 \\ &\qquad \leq \left( \frac12+\frac{n-1}{2}\sigma \right) (A_{11}^2+A_{22}^2) + \frac{(n-1)(n-2)}{4}(1-\sigma) (A_{33}^2+\cdots+A_{nn}^2) \\ &\qquad = \frac{n-2}{2} (A_{11}^2+\cdots+A_{nn}^2), \end{aligned} \]

where we took

\[ \sigma = \frac{n-3}{n-1} \]

in the last line. Hence, for \(n\geq 3\), we have

\[ \begin{aligned} \frac{n-2}{2}r^2|A|^2 &\geq r^2 \left( \frac{n-2}{2}\sum_{i=1}^n A_{ii}^2 + \sum_{i=2}^n A_{1i}^2 + \sum_{j=3}^n A_{2j}^2 \right) \\ &\geq r^2 \left( \sum_{i=1}^n A_{1i}^2 + \sum_{j=2}^n A_{2j}^2 + A_{11}A_{22} + \frac{n-3}{4}(A_{11}+A_{22})^2 \right) \\ &\geq (3n-3) - (2n-1)|dr|^2 - \widetilde{\mathrm{BiRic}} . \end{aligned} \]

Therefore,

\[ r^2 |A|^2 \geq \frac{2}{n-2} \left( (3n-3) - (2n-1)|dr|^2 - \widetilde{\mathrm{BiRic}} \right), \]

and the proposition follows. ◻

Proposition 5.2.6

For any \(\psi \in C_c^{0,1}(N,\tilde g)\), we have

\[ \int_N |\tilde \nabla \psi|_{\tilde g}^2 \, d\tilde \mu \geq \int_N \left( r^2 |A|^2 - \frac{n(n-2)}{2} + \left( \frac{n(n-2)}{2} - \frac{(n-2)^2}{4} \right) |dr|^2 \right) \psi^2 \, d\tilde \mu . \]

Proof. Recall that in the conformal metric, we have

\[ d\tilde \mu = r^{-n} d\mu \qquad\text{and}\qquad |\tilde \nabla f|_{\tilde g}^2 = r^2 |\nabla f|^2 . \]

Then the stability inequality for \(M\) implies

\[ \int_N r^{n-2} |\tilde \nabla f|_{\tilde g}^2 \, d\tilde \mu \geq \int_N r^{n-2} (r^2 |A|^2) f^2 \, d\tilde \mu \]

for any \(f \in C_c^{0,1}(N,\tilde g)\). We take

\[ f = r^{\frac{2-n}{2}} \psi \]

for \(\psi \in C_c^{0,1}(N,\tilde g)\). Then

\[ \tilde \nabla f = r^{\frac{2-n}{2}} \tilde \nabla \psi - \frac{n-2}{2} r^{-\frac n2} \psi \tilde \nabla r , \]

\[ \begin{aligned} |\tilde \nabla f|_{\tilde g}^2 &= r^{2-n} |\tilde \nabla \psi|_{\tilde g}^2 + \frac{(n-2)^2}{4} r^{-n} \psi^2 |\tilde \nabla r|_{\tilde g}^2 \\ &\quad - (n-2) r^{1-n} \psi \langle \tilde \nabla \psi, \tilde \nabla r \rangle_{\tilde g} \\ &:= a+b+c . \end{aligned} \]

We have

\[ \int_N r^{n-2} a \, d\tilde \mu = \int_N |\tilde \nabla \psi|_{\tilde g}^2 \, d\tilde \mu . \]

Since

\[ r^{-2} |\tilde \nabla r|_{\tilde g}^2 = |dr|^2 , \]

we have

\[ \int_N r^{n-2} b \, d\tilde \mu = \int_N \frac{(n-2)^2}{4} |dr|^2 \psi^2 \, d\tilde \mu . \]

Finally, we use integration by parts and the displayed formula for \(\tilde\Delta\log r\) below to compute

\[ \begin{aligned} \int_N r^{n-2} c \, d\tilde \mu &= - \int_N \frac{n-2}{2} \left\langle \tilde \nabla(\psi^2), \tilde \nabla(\log r) \right\rangle_{\tilde g} d\tilde \mu \\ &= \int_N \frac{n-2}{2} \tilde \Delta(\log r) \psi^2 \, d\tilde \mu \\ &= \int_N \left( \frac{n(n-2)}{2} - \frac{n(n-2)}{2} |dr|^2 \right) \psi^2 \, d\tilde \mu . \end{aligned} \]

Recall that we have the following

\[ \begin{aligned} \tilde{\Delta} \log r = r^{2}(\Delta \log r -(n-2)r^{-2}|\nabla r|^{2}) = n-nr^{-2}|\nabla r|^{2} \end{aligned} \]

Altogether, we obtain

as desired. ◻

Proposition 5.2.7

For \(3\leq n\leq5\), we have

\[ -\tilde{\Delta}+\frac{2}{n-2}\widetilde{\mathrm{BiRic}}\geq \frac{2(-n^3+6n^{2}-4n-8)}{8(n-2)} \]

in the spectral sense.

Proof. Combining Proposition 5.2.6 with Proposition 5.2.5, for every \(\psi\in C_c^{0,1}(N,\tilde g)\) we get

\[ \begin{aligned} \int_N \left( |\tilde\nabla\psi|_{\tilde g}^2 +\frac{2}{n-2}\widetilde{\mathrm{BiRic}}\psi^2 \right)d\tilde\mu \geq \int_N \left(C_0+C_1|dr|^2\right)\psi^2\,d\tilde\mu, \end{aligned} \]

where

\[ C_0=\frac{2(3n-3)}{n-2}-\frac{n(n-2)}{2}, \qquad C_1=-\frac{2(2n-1)}{n-2} +\frac{n(n-2)}{2} -\frac{(n-2)^2}{4}. \]

Since \(0\leq |dr|^2\leq1\) and \(C_1\leq0\) for \(3\leq n\leq5\), the right-hand side is bounded from below by

\[ C_0+C_1 = \frac{2(-n^3+6n^{2}-4n-8)}{8(n-2)}. \]

This proves the claimed spectral lower bound. ◻

Construction of the \(\mu\)-Bubble

Theorem 5.2.8

Suppose \((N^n,g)\), \(3\leq n\leq4\), is a compact manifold with boundary \(\partial N=\partial_+N\cup\partial_-N\). Assume that there are \(0\lt{}a\leq 2\) and a positive smooth function \(w\) such that

\[ -a\Delta^N w+\mathrm{BiRic}_N\,w\geq w. \]

Suppose that the distance between \(\partial_-N\) and \(\partial_+N\) is bounded below by \(5\pi\). Then one can find a smooth \(\mu\)-bubble \(\Sigma\) separating \(\partial_-N\) from \(\partial_+N\) such that, for every \(\psi\in C_c^\infty(\Sigma)\),

\[ \frac{4}{4-a}\int_\Sigma |\nabla^\Sigma\psi|^2 \geq \int_\Sigma \left(\frac12-\mathrm{Ric}_\Sigma\right)\psi^2. \]

Equivalently, in the spectral sense,

\[ -\frac{4}{4-a}\Delta^\Sigma+\mathrm{Ric}_\Sigma\geq \frac12. \]

The constant \(1\) is only a normalization and can be replaced by any positive constant after rescaling the metric.

Second variation of weighted area functional

Lemma 5.2.9

The minimizer to the following functional

\[ \mathcal{A}(\Omega)=\int_{\partial \Omega} w d\mathcal{H}^{n-1}-\int_{\Omega} h w d\mathcal{H}^{n}, \]

satisfies the following equation

\[ \begin{aligned} {} & \int_{ \Sigma} w \left|\nabla^\Sigma \phi\right|^2+\frac{1}{2}R_\Sigma\phi^2w-\phi\Delta^\Sigma w-\frac{1}{2}w^{-1}\left< \nabla^N w, \nu \right> ^2\phi^2-\int_{ \Sigma} \phi^2(-\Delta^Nw+\frac{1}{2}R_Nw) \\ {} & -\int_{ \Sigma}\left[ \phi^2 w \left< \nabla^Nh,\nu \right> + \frac{1}{2}h^2\phi^2w \right]\geq 0. \end{aligned} \]

for any smooth function \(\phi\) on \(\Sigma\).

Given a domain \(\Omega\) in \(N\), we do a variation \(\Omega_t\) under vector field \(V\) and denote \(\Sigma_t=\partial\Omega_t\backslash \partial_-N\) and \(\Sigma=\Sigma_0\). We choose \(V\) such that \(\nabla_{V}V=0\) and \(V=\phi \nu\) along \(\Sigma\). (Note that in general, we do not have \(V\bot \Sigma_t\) for every \(t\).)

Lemma 5.2.10

The following first and second variation formulas hold:

\[ \begin{aligned} \mathcal{A}'={} & \int_{ \Sigma} \phi\left[ \left< \nabla^Nw, \nu\right>+w H-wh \right], \text{ Recall that }H=\sum_{i =1}^{n}\left< \nabla_{e_i}e_i, \nu \right>\\ \mathcal{A}''={} & \int_{ \Sigma} \phi^2 (\Delta^N w-\Delta^\Sigma w)+\int_{ \Sigma} w\left( \left|\nabla^\Sigma \phi\right|^2-\left( |A_\Sigma|^2+\mathrm{Ric}(\nu,\nu) \right)\phi^2 \right) \\ {}& - \int_{ \Sigma}\left(\phi^2 h\left< \nabla^N w,\nu \right>+\phi^2 w\left< \nabla^Nh,\nu \right>\right). \end{aligned} \]

The first variation is straightforward.

\[ \begin{aligned} \mathcal{A}'={} & \int_{ \Sigma} \left< \nabla^Nw, V \right>-w \mathrm{div}^\Sigma(V)-hw \left< V, \nu \right> d\mathcal{H}^3 \\ ={} & \int_{ \Sigma} \left< \nabla^Nw, V \right>+w \mathrm{div}^\Sigma(V^T)+w H \left< V, \nu \right> -hw \left< V,\nu \right> d\mathcal{H}^{n-1} \end{aligned} \]

Hence, the stationary gives us the formula

\[ H=h-w^{-1}\left< \nabla^N w,\nu \right> \]

For the derivative of \(\int_{\Sigma} w\,\mathrm{div}^\Sigma(V^T)\,d\mathcal{H}^{n-1}\), we rewrite it using integration by parts as

\[ -\int_{ \Sigma} \left< \nabla^\Sigma w, V^T \right>. \]

After differentiating,

\[ -\int_{ \Sigma} \left< \nabla^\Sigma w, \nabla_{V}(V^T) \right> =\int_{ \Sigma} \left< \nabla^\Sigma w, \phi \nabla_{V}\nu \right> \]

where we have used \(\nabla_{V}V=0\). Note that we have

\[ \left< \nabla_{V}\nu,e_i \right> = -\left<\nu, \nabla_{V}e_i \right> = -\left< \nu, \nabla_{e_i}(\phi\nu) \right> =-e_i(\phi) \implies \nabla_{V}\nu=-\nabla^\Sigma \phi \]

Then, the derivative of \(\int_{ \Sigma} \left< \nabla^\Sigma w, \phi \nabla_{V}\nu \right>\) is

\[ \int_{ \Sigma} \left< \nabla^\Sigma w, \phi \nabla_{V}\nu \right> = \int_{ \Sigma} -\left< \nabla^\Sigma w, \nabla^\Sigma\phi \right> \phi \]

Now, we can compute the second variation. Note that we can use \(V\left< V, \nu \right> =0\).

\[ \begin{aligned} \mathcal{A}''={}& \int_{ \Sigma} \phi^2 \mathrm{Hess}^Nw(\nu,\nu)-\phi \left< \nabla^\Sigma w, \nabla^\Sigma \phi \right> -\phi^2 \left< \nabla^Nw, \nu \right> H \\ {}& -\int_{ \Sigma} w\phi \left( \Delta^\Sigma \phi +(|A_\Sigma|^2+\mathrm{Ric}(\nu,\nu))\phi \right) -\int_{ \Sigma}\left(\phi^2 h\left< \nabla^N w,\nu \right>+\phi^2w\left< \nabla^Nh,\nu \right>\right). \end{aligned} \]

For the Hessian, we have

\[ \begin{aligned} \mathrm{Hess}^Nw(\nu,\nu)={}&\mathrm{div}^N(\nabla^Nw)-\mathrm{div}^\Sigma(\nabla^Nw)=\Delta^Nw -\Delta^\Sigma w - \mathrm{div}^\Sigma(\left< \nabla^Nw, \nu \right> \nu)\\ ={}&\Delta^Nw-\Delta^\Sigma w+\left< \nabla^Nw,\nu \right>H \end{aligned} \]

using integration by parts, we have

\[ \begin{aligned} &\mathcal{A}''=\int_{ \Sigma} \phi^{2}(\Delta^N w-\Delta^\Sigma w)+ w\left( |\nabla^\Sigma \phi|^{2}-(|A|^{2}+\mathrm{Ric}_N(\nu,\nu))\phi^{2} \right)\\ &-\int_{ \Sigma}\left(\phi^{2}h\left< \nabla^N w,\nu \right>+\phi^{2} w \left< \nabla^Nh,\nu \right>\right). \end{aligned} \]

This is the weighted-area second variation formula used below.

Properties of the minimizer

Proposition 5.2.11

Suppose \(\Omega\) is the minimizer to \(\mathcal{A}\). Assume that the weight is \(W=w^a\), where \(0\lt{}a\leq 2\), \(w\gt{}0\), and

\[ -a\Delta^Nw+\mathrm{BiRic}_N\,w\geq w. \]

We also assume that the function \(h\) satisfies the following condition:

\[ 1+h^{2}-2|\nabla^N h|\geq 0. \]

Then, for any \(\psi\in C_c^\infty(\Sigma)\), we have

\[ \begin{aligned} \frac{4}{4-a}\int_{ \Sigma} |\nabla^\Sigma \psi|^{2}\geq{}& \int_{ \Sigma} \psi^{2}\left( \frac{1}{2}-Ric_\Sigma \right) \end{aligned} \]

We can also write it as

\[ -\frac{4}{4-a}\Delta^\Sigma+Ric_\Sigma\geq \frac{1}{2}. \]

Proof. Suppose \(\Omega\) is the minimizer to \(\mathcal{A}\). In the spectral bi-Ricci application the weight in the functional is

\[ W=w^a, \]

and we consider

\[ \phi=W^{-\frac12}\psi=w^{-\frac a2}\psi. \]

Then

\[ \nabla^\Sigma\phi = W^{-\frac12}\nabla^\Sigma\psi -\frac12 W^{-\frac12}\psi\nabla^\Sigma\log W. \]

Hence

\[ \begin{aligned} \int_{\Sigma} W|\nabla^\Sigma\phi|^2 = \int_{\Sigma}|\nabla^\Sigma\psi|^2 +\frac14\psi^2|\nabla^\Sigma\log W|^2 -\psi\left<\nabla^\Sigma\psi,\nabla^\Sigma\log W\right>. \end{aligned} \]

The first variation gives

\[ H=h-\left<\nabla^N\log W,\nu\right> =h-a\left<\nabla^N\log w,\nu\right>. \]

Put

\[ v=\left<\nabla^N\log w,\nu\right>. \]

Then \(H=h-av\), and hence

\[ ahv=\frac12h^2+\frac12a^2v^2-\frac12H^2. \]

We also need to compare the ambient and intrinsic Laplacians of the weight \(W=w^a\). Along \(\Sigma\),

\[ \begin{aligned} \frac{\Delta^\Sigma W}{W} ={}& a\frac{\Delta^\Sigma w}{w} +a(a-1)|\nabla^\Sigma\log w|^2,\\ \frac{\Delta^N W}{W} ={}& a\frac{\Delta^N w}{w} +a(a-1)\left(|\nabla^\Sigma\log w|^2+v^2\right). \end{aligned} \]

Thus

\[ \frac{\Delta^N W}{W}-\frac{\Delta^\Sigma W}{W} = a\left(\frac{\Delta^N w}{w}-\frac{\Delta^\Sigma w}{w}\right) +a(a-1)v^2. \]

Since \(\Sigma\) is stable for the weighted functional, the second variation formula with weight \(W\) gives, after moving the potential terms to the right-hand side,

\begin{equation} \tag{5.2.2} \begin{aligned} &\int_\Sigma W\left(|\nabla^\Sigma\phi|^2 -a w^{-1}\Delta^\Sigma w\,\phi^2\right) \notag\\ \geq{}& \int_\Sigma \psi^2\left( -a\frac{\Delta^N w}{w} +|A|^2+\mathrm{Ric}_N(\nu,\nu)-\frac12H^2 +\frac12h^2+\left<\nabla^Nh,\nu\right> \right) \notag\\ &+\frac{a(2-a)}2\int_\Sigma \psi^2 v^2. \label{eq:mububble-weighted-stability-with-normal-term} \end{aligned} \end{equation}

The final term is obtained by combining the Laplacian comparison contribution \(-a(a-1)v^2\) with the \(ahv\) term. Since \(0\lt{}a\leq2\), it is nonnegative and can be discarded.

It remains to estimate the left-hand side of (5.2.2). After substituting \(\phi=w^{-a/2}\psi\), it becomes

\[ \begin{aligned} \int_\Sigma |\nabla^\Sigma\psi|^2 +a\psi\left<\nabla^\Sigma\psi,\nabla^\Sigma\log w\right> -\left(a-\frac{a^2}{4}\right)\psi^2|\nabla^\Sigma\log w|^2, \end{aligned} \]

and Young’s inequality with \(\varepsilon=(4-a)^{-1}\) gives the estimate. Therefore,

\[ \begin{aligned} \int_{\Sigma} W\left(|\nabla^\Sigma\phi|^2-a w^{-1}\Delta^\Sigma w\,\phi^2\right) \leq \frac{4}{4-a}\int_{\Sigma}|\nabla^\Sigma\psi|^2. \end{aligned} \]

Combining this estimate with (5.2.2) and the spectral inequality \(-a\Delta^Nw+\mathrm{BiRic}_Nw\geq w\) gives

\[ \begin{aligned} \frac{4}{4-a}\int_{\Sigma}|\nabla^\Sigma\psi|^2 \geq{}& \int_\Sigma \psi^2\left( 1-\mathrm{BiRic}_N +|A|^2+\mathrm{Ric}_N(\nu,\nu)-\frac12H^2 \right) \\ &+\int_\Sigma \psi^2\left(\frac12h^2+\left<\nabla^Nh,\nu\right>\right). \end{aligned} \]

Now, we need to use \(|A|^{2}+\mathrm{Ric}_N(\nu,\nu)\) to bound \(\mathrm{BiRic}_N\).

Lemma 5.2.12

We have the following inequality:

\[ |A|^{2}+\mathrm{Ric}_N(\nu,\nu)\geq \mathrm{BiRic}_N-\mathrm{Ric}_\Sigma + \frac{6-n}{4}H^{2}. \]

This is the extension of the usual Schoen–Yau trick. Recall that we have (for \(n=3\))

\[ |A|^{2}+\mathrm{Ric}_N(\nu,\nu)=\frac{1}{2}R_N-\frac{1}{2}R_\Sigma+\frac{1}{2}|A|^2+\frac{1}{2}H^2\geq \frac{1}{2}R_N-\frac{1}{2}R_\Sigma+\frac{3}{4}H^{2}. \]

This is exactly the same as the inequality in the lemma.

Proof. Suppose \(e_1\in T_p\Sigma\) is a unit direction where \(Ric_\Sigma\) attains its minimum. Using the Gauss equation, we have

\[ \begin{aligned} Ric_\Sigma(e_1,e_1) ={}&\sum_{i=2}^{n-1}R^\Sigma_{1i1i} \\ ={}&\sum_{i=2}^{n-1}R^N_{1i1i} +A_{11}\sum_{i=2}^{n-1}A_{ii} -\sum_{i=2}^{n-1}A_{1i}^{2}. \end{aligned} \]

Note that we have

\[ \begin{aligned} A_{11}\sum_{i=2}^{n-1}A_{ii}={}& -A_{11}^{2}+A_{11}H=-A_{11}^{2}-H\sum_{i=2}^{n-1}A_{ii}+H^{2}\\ \geq{}& -A_{11}^{2}-\frac{1}{n-2}\left( \sum_{i=2}^{n-1}A_{ii} \right)^{2}+\left( 1-\frac{n-2}{4} \right)H^{2}\\ \geq{}& -\sum_{i=1}^{n-1}A_{ii}^{2}+\frac{6-n}{4}H^{2} \end{aligned} \]

By the choice of \(e_1\), \(\mathrm{Ric}_\Sigma=\mathrm{Ric}_\Sigma(e_1,e_1)\). Hence, we have

\[ \begin{aligned} \mathrm{Ric}_\Sigma+\mathrm{Ric}_N(\nu,\nu) \geq{}& \sum_{i=2}^{n-1}R^N_{1i1i} +\mathrm{Ric}_N(\nu,\nu)-|A|^{2}+\frac{6-n}{4}H^{2}\\ \geq{}& \mathrm{BiRic}_N-|A|^{2} + \frac{6-n}{4}H^{2}. \end{aligned} \]

Equivalently,

\[ |A|^{2}+\mathrm{Ric}_N(\nu,\nu) \geq \mathrm{BiRic}_N-\mathrm{Ric}_\Sigma+\frac{6-n}{4}H^{2}, \]

which is the desired inequality. ◻

Now, we go back to our inequality. When \(3\leq n\leq 4\), we have \(\frac{6-n}{4}\geq \frac{1}{2}\) and therefore

\[ \begin{aligned} \frac{4}{4-a}\int_{ \Sigma} |\nabla^\Sigma \psi|^{2}\geq{}& \int_{ \Sigma} \psi^{2}(1-Ric_\Sigma) +\int_\Sigma\psi^2\left(\frac{1}{2}h^{2}+\left< \nabla^Nh,\nu \right>\right). \end{aligned} \]

Together with the condition for \(h\), \(1+h^{2}-2|\nabla^N h|\geq0\), we have

\[ 1+\frac12h^2+\left<\nabla^Nh,\nu\right> \geq \frac12+\frac12(1+h^2-2|\nabla^Nh|) \geq \frac12. \]

Hence

\[ \frac{4}{4-a}\int_\Sigma|\nabla^\Sigma\psi|^2 \geq \int_{ \Sigma} \psi^{2}\left( \frac{1}{2}-Ric_\Sigma \right). \]

This is the desired inequality. ◻

Completion of the \(\mu\)-Bubble Construction

To prove Theorem 5.2.8, choose \(h\) by

\[ h(x)=-\frac{1}{2}\tan(\frac{\tilde{d}(x,\partial_-N)}{4}-\frac{\pi}{2}) \]

where \(\tilde{d}(x,\partial_-N)\) is a smoothing of \(d_N(x,\partial_-N)\) with \(\mathrm{Lip}\,\tilde{d}\leq 2\). We denote \(\varphi(x)=\frac{\tilde{d}(x,\partial_-N)}{4}-\frac{\pi}{2}\). Then

\[ |\nabla^N h| \leq \frac12(1+\tan^2\varphi), \qquad 2|\nabla^N h|\leq 1+h^2 . \]

To show the existence of the minimizer, fix a domain \(\Omega_0\) containing \(\partial_-N\) but not \(\partial_+N\) such that \(\partial \Omega_0\) lies in \(\tilde{d}(x,\partial_-N)\lt{} 4\pi\). We consider minimizing the following (relative) energy functional

\[ \mathcal{A}(\Omega)=\int_{ \partial \Omega} w^a d\mathcal{H}^{n-1}-\int_{\Omega\backslash \Omega_0} h w^a d\mathcal{H}^{n}+\int_{ \Omega_0\backslash \Omega}hw^a d\mathcal{H}^n \]

The direct method gives a minimizer in the corresponding relative homology class. Since \(3\leq n\leq4\), the free boundary part \(\Sigma=\partial\Omega\setminus\partial_-N\) is smooth after the usual regularity theory for prescribed-mean-curvature hypersurfaces. The choice \(d_N(\partial_+N,\partial_-N)\geq5\pi\) gives enough room for the smoothing \(\tilde d\) and hence for the above choice of \(h\). Applying the stability inequality from the preceding proposition proves Theorem 5.2.8.

Diameter estimate and volume estimate under spectral Ricci curvature bound

The following result of Antonelli–Xu [AX24] gives the radius/diameter and volume estimates under a spectral Ricci lower bound. In this subsection \(\mathrm{Ric}_N(v,v)\) denotes the Ricci tensor on a unit vector; when \(\mathrm{Ric}_N\) appears without arguments in a spectral inequality, it means the smallest eigenvalue of the Ricci tensor. The bi-Ricci notation is the one fixed above. In the low-dimensional range considered here, the minimizers appearing in the \(\mu\)-bubble arguments are smooth. The parameter \(\alpha\) below denotes the coefficient of the spectral Laplacian term.

Theorem 5.2.13

Let \((N^n,g)\) be a compact smooth Riemannian manifold, \(3\leq n\leq 5\), and let

\[ 0\leq \alpha\leq \frac{n-1}{n-2},\qquad \lambda\gt{}0 . \]

Suppose that there is a positive smooth function \(u\) such that

\begin{equation} \tag{5.2.3} \label{eq:AX-spectral-Ric} -\alpha\Delta u+\mathrm{Ric}_N\,u\geq (n-1)\lambda u . \end{equation}

Let \((\widetilde N,\tilde g)\) be the universal cover and let \(\tilde u\) be the lift of \(u\). Then the diameter upper bound is

\begin{equation} \tag{5.2.4} \label{eq:AX-diam-bound} \mathrm{diam}(\widetilde N,\tilde g) \leq \frac{\pi}{\sqrt{\lambda}} \left(\frac{\max_N u}{\min_N u}\right)^{\frac{n-3}{n-1}\alpha}. \end{equation}

Moreover the sharp volume upper bound is

\begin{equation} \tag{5.2.5} \label{eq:AX-volume-bound} |\widetilde N| \leq \lambda^{-\frac n2}|S^n|, \end{equation}

where \(S^n\) denotes the unit round sphere. In particular \(\pi_1(N)\) is finite. If equality holds in the volume estimate, then \(\tilde u\) is constant and \(\widetilde N\) is the round sphere of radius \(\lambda^{-1/2}\).

Corollary 5.2.14

Let \((N^n,g)\) be complete, not assumed compact, with \(3\leq n\leq 5\). If \(n\gt{}3\) and \(0\leq\alpha\lt{}\frac{4}{n-1}\), or if \(n=3\) and \(0\leq\alpha\leq2\), and if (5.2.3) holds for some positive smooth \(u\), then \(N\) is compact. Moreover \(\pi_1(N)\) is finite and, for the universal cover \(\widetilde N\),

\[ \mathrm{diam}(\widetilde N,\tilde g) \leq \frac{\pi}{\sqrt{\lambda}} \left(\frac{\max_N u}{\min_N u}\right)^{\frac{n-3}{n-1}\alpha}, \qquad |\widetilde N|\leq \lambda^{-\frac n2}|S^n|. \]

In particular \(|N|\leq \lambda^{-n/2}|S^n|\). In addition there is a constant \(C=C(n,\alpha)\) such that

\[ \mathrm{diam}(N)\leq C\lambda^{-1/2}. \]

Proof. This is precisely the complete-case corollary in [AX24], based on Xu’s subcritical spectral Bonnet–Myers theorem. The latter gives compactness and the uniform bound \(\mathrm{diam}(N)\leq C(n,\alpha)\lambda^{-1/2}\) in the stated range. Once compactness is known, Theorem 5.2.13 applies and gives (5.2.4) and (5.2.5). Since \(N\) is the quotient of \(\widetilde N\) by a finite group of deck transformations, the same volume upper bound also holds for \(N\). ◻

Even in the larger sharp range \(0\leq\alpha\leq (n-1)/(n-2)\), if \(N\) is complete and the positive function \(u\) in (5.2.3) satisfies \(0\lt{}\inf_N u\leq\sup_Nu\lt{}\infty\), then Corollary 5.2.17 below proves compactness directly. Thus noncompact examples in the super-subcritical range must have \(u\) degenerating at infinity.

Weighted geodesic proof of the diameter estimates.

We also record the weighted-geodesic calculation, since it gives another way to see the Bonnet–Myers part.

For a positive function \(u\) on \((M^m,g)\), we define the weighted geodesic distance from \(p\) to \(q\) by

\[ L_u^\alpha(p,q) = \inf_\eta\int_\eta u^\alpha\,ds_g , \]

where the infimum is taken over piecewise smooth curves from \(p\) to \(q\). As above, \(\mathrm{Ric}_M\) in a spectral inequality denotes the smallest eigenvalue of the Ricci tensor, while \(\mathrm{Ric}_M(T,T)\) denotes the tensor on the vector \(T\).

Lemma 5.2.15

Let \((M^m,g)\) be complete, \(m\geq3\), and suppose that \(u\gt{}0\) satisfies

\begin{equation} \tag{5.2.6} \label{eq:AX-alpha-spectral-positive} -\alpha\Delta u+\mathrm{Ric}_M\,u\geq (m-1)\lambda u, \qquad \lambda\gt{}0 . \end{equation}

Let \(\alpha:[0,l]\to M\) be an \(L_u^\alpha\)-minimizing curve from \(p\) to \(q\), parametrized by \(g\)-arclength, and assume first that \(q\) is not a weighted cut point along \(\alpha\). Then the following hold in the barrier sense at \(q\).

If \(0\leq\alpha\lt{}4/(m-1)\), then for every \(C^1\) function \(\psi\) on \([0,l]\) with

\[ \psi(0)=0,\qquad \psi(l)=u(q)^{\alpha/2}, \]

we have

\begin{equation} \tag{5.2.7} \label{eq:AX-subcritical-weighted-lap} \Delta_q L_u^\alpha(p,q) \leq \int_0^l \left(C_{m,\alpha}\psi_s^2-(m-1)\lambda\psi^2\right)\,ds, \end{equation}

where

\[ C_{m,\alpha} = (m-1)+ \frac{\alpha(m-3)^2}{4\left(1-\frac{m-1}{4}\alpha\right)} . \]

In particular, choosing

\[ \psi(s)=u(q)^{\alpha/2}\sin\left(\frac{\pi s}{2l}\right) \]

gives

\begin{equation} \tag{5.2.8} \label{eq:AX-subcritical-weighted-lap-sine} \Delta_q L_u^\alpha(p,q) \leq u(q)^\alpha\left( \frac{C_{m,\alpha}\pi^2}{8l} -\frac{(m-1)\lambda l}{2} \right). \end{equation}

If \(0\leq\alpha\leq (m-1)/(m-2)\), then for every \(C^1\) function \(\psi\) with

\[ \psi(0)=0,\qquad \psi(l)=u(q)^{\alpha/(m-1)}, \]

we have

\begin{equation} \tag{5.2.9} \label{eq:AX-bounded-weighted-lap} \Delta_q L_u^\alpha(p,q) \leq \int_0^l u^{\frac{m-3}{m-1}\alpha} \left((m-1)\psi_s^2-(m-1)\lambda\psi^2\right)\,ds . \end{equation}

Proof. Let \(T=\alpha'\) and put

\[ \tilde g=u^{2\alpha}g . \]

Then \(L_u^\alpha\) is exactly the distance function of the conformal metric \(\tilde g\). Write

\[ \rho=L_u^\alpha(p,\cdot). \]

Along \(\alpha\) we have, by the definition of \(\tilde g\) and of the distance function \(\rho\),

\[ d\tilde s=u^\alpha\,ds,\qquad \tilde T=u^{-\alpha}T, \qquad \nabla^g L_u^\alpha=u^\alpha T . \]

We first compute \(\tilde\Delta\rho\) in the conformal metric. Choose a \(\tilde g\)-parallel orthonormal frame along \(\alpha\),

\[ \tilde E_1,\ldots,\tilde E_{m-1},\qquad \tilde E_m=\tilde T, \]

and write \(\tilde E_i=u^{-\alpha}e_i\). Thus \(e_1,\ldots,e_{m-1},T\) is \(g\)-orthonormal and \(e_i\perp T\). Let \(\phi\) be a test function with \(\phi(0)=0\) and \(\phi(l)=1\), and set

\[ \tilde V_i=\phi\tilde E_i . \]

These fields vanish at \(p\) and equal \(\tilde E_i(q)\) at \(q\). Hence the ordinary second variation formula for the \(\tilde g\)-distance gives, for \(1\leq i\leq m-1\),

\[ \tilde\nabla^2\rho(\tilde E_i,\tilde E_i)(q) \leq \int_0^{\tilde l} \left( (\tilde T\phi)^2 -\phi^2\widetilde R(\tilde T,\tilde E_i,\tilde T,\tilde E_i) \right)\,d\tilde s . \]

The missing \(m\)-th direction is radial and contributes nothing, since \(\tilde\nabla^2\rho(\tilde T,\tilde T)=0\) away from the cut locus. Summing the preceding inequality over the transverse directions gives

\begin{equation} \tag{5.2.10} \label{eq:AX-tilde-index-simple} \tilde\Delta\rho(q) \leq \int_0^l u^{-\alpha} \left((m-1)\phi_s^2-\phi^2\widetilde{\mathrm{Ric}}(T,T)\right)\,ds . \end{equation}

Here we used \(d\tilde s=u^\alpha ds\), \(\tilde T\phi=u^{-\alpha}\phi_s\), and \(\widetilde{\mathrm{Ric}}(\tilde T,\tilde T)=u^{-2\alpha}\widetilde{\mathrm{Ric}}(T,T)\).

Now return to the original metric. The conformal Laplacian formula for \(\tilde g=e^{2\log u^\alpha}g\) gives, at \(q\),

\begin{equation} \tag{5.2.11} \label{eq:AX-lap-conformal-weighted} \Delta^g\rho = u(q)^{2\alpha}\tilde\Delta\rho-(m-2)u(q)^\alpha(\log u^\alpha)_s(l). \end{equation}

Combining (5.2.10) and (5.2.11), we get

\begin{equation} \tag{5.2.12} \begin{aligned} \Delta^g\rho(q) \leq{}& u^{2\alpha}(q)\int_0^l u^{-\alpha} \left((m-1)\phi_s^2-\phi^2\widetilde{\mathrm{Ric}}(T,T)\right)\,ds \notag\\ &\qquad -(m-2)u(q)^\alpha(\log u^\alpha)_s(l). \label{eq:AX-before-test-change} \end{aligned} \end{equation}

Now rewrite the curvature term in (5.2.12) in the original metric. The conformal Ricci formula is

\[ \begin{aligned} \widetilde{\mathrm{Ric}} ={}& \mathrm{Ric} -(m-2)\left( \nabla^2\log u^\alpha -d\log u^\alpha\otimes d\log u^\alpha \right) \\ &-\left( \Delta\log u^\alpha +(m-2)|\nabla\log u^\alpha|^2 \right)g . \end{aligned} \]

The conformal connection formula is

\[ \tilde\nabla_XY = \nabla_XY+X(\log u^\alpha)Y+Y(\log u^\alpha)X -g(X,Y)\nabla\log u^\alpha . \]

Applying it to the \(\tilde g\)-geodesic equation \(\tilde\nabla_{\tilde T}\tilde T=0\) gives

\[ 0=u^{2\alpha}\tilde\nabla_{\tilde T}\tilde T = \nabla_TT+(\log u^\alpha)_sT-\nabla\log u^\alpha, \]

and therefore

\[ \nabla_TT=\nabla^\perp\log u^\alpha,\qquad \nabla^2\log u^\alpha(T,T) =(\log u^\alpha)_{ss}-|\nabla^\perp\log u^\alpha|^2 . \]

Using this in the Ricci formula gives the pointwise identity along \(\alpha\)

\begin{equation} \tag{5.2.13} \begin{aligned} \widetilde{\mathrm{Ric}}(T,T) ={}& \mathrm{Ric}_M(T,T) -(m-2)\left( (\log u^\alpha)_{ss}-\nabla_TT \cdot \nabla \log u^\alpha -(\log u^\alpha)_s^2 \right) \notag\\ &-\left( \Delta\log u^\alpha +(m-2)|\nabla\log u^\alpha|^2 \right) \notag\\ ={}& \mathrm{Ric}_M(T,T) -(m-2)\left( (\log u^\alpha)_{ss}-|\nabla^\perp\log u^\alpha|^2 -(\log u^\alpha)_s^2 \right) \notag\\ &-\left(\Delta\log u^\alpha +(m-2)|\nabla\log u^\alpha|^2\right) \notag\\ ={}& \mathrm{Ric}_M(T,T) -\Delta\log u^\alpha -(m-2)(\log u^\alpha)_{ss}. \label{eq:AX-tilde-ric-to-g} \end{aligned} \end{equation}

The endpoint term in (5.2.12) is written in the same \(u(q)^{2\alpha}\int u^{-\alpha}(\cdots)\,ds\) scale as

\[ -(m-2)u(q)^\alpha(\log u^\alpha)_s(l) = -u(q)^{2\alpha}\int_0^l \left((m-2)\phi^2u^{-\alpha}(\log u^\alpha)_s\right)_s\,ds, \]

because \(\phi(0)=0\) and \(\phi(l)=1\). Substituting (5.2.13) into (5.2.12) and expanding this total derivative gives the original-metric formula

\begin{equation} \tag{5.2.14} \begin{aligned} \Delta^g\rho(q) \leq{}& u(q)^{2\alpha}\int_0^l u^{-\alpha}\Big[ (m-1)\phi_s^2 -2(m-2)\phi\phi_s(\log u^\alpha)_s \notag\\ &\qquad +(m-2)\phi^2(\log u^\alpha)_s^2 -\mathrm{Ric}_M(T,T)\phi^2 +\phi^2\Delta\log u^\alpha \Big]\,ds .\label{eq:AX-original-metric-before-change} \end{aligned} \end{equation}

Using (5.2.6) and \(\mathrm{Ric}_M(T,T)\geq \mathrm{Ric}_M\), along the curve

\[ -\mathrm{Ric}_M(T,T)+\alpha\frac{\Delta u}{u} \leq -(m-1)\lambda . \]

Together with

\[ \Delta\log u^\alpha = \alpha\frac{\Delta u}{u} -\alpha\frac{|\nabla u|^2}{u^2}, \]

this gives

\begin{equation} \tag{5.2.15} \begin{aligned} \Delta^g\rho(q) \leq{}& u(q)^{2\alpha}\int_0^l u^{-\alpha}\Big[ (m-1)\phi_s^2 -2(m-2)\phi\phi_s(\log u^\alpha)_s \notag\\ &\qquad +(m-2)\phi^2(\log u^\alpha)_s^2 -(m-1)\lambda\phi^2 -\alpha\phi^2\frac{|\nabla u|^2}{u^2} \Big]\,ds .\label{eq:AX-original-metric-lambda-before-change} \end{aligned} \end{equation}

Only at this point do we change the test function. The subcritical estimate uses the substitution

\[ \phi=\frac{u^{\alpha/2}}{u(q)^\alpha}\psi,\qquad \psi(0)=0,\qquad \psi(l)=u(q)^{\alpha/2}. \]

Thus

\[ \phi_s=\frac{u^{\alpha/2}}{u(q)^\alpha} \left(\psi_s+\frac{\alpha}{2}\psi\frac{u_s}{u}\right), \qquad u_s=T(u). \]

Inserting this into (5.2.15) cancels the weight \(u^\alpha\) and gives

\begin{equation} \tag{5.2.16} \begin{aligned} \Delta_q L_u^\alpha(p,q) \leq{}& \int_0^l\Big[ (m-1)\psi_s^2 +\alpha(3-m)\psi\psi_s\frac{u_s}{u} +\frac{m-1}{4}\alpha^2\psi^2\frac{u_s^2}{u^2} \notag\\ &\qquad -\alpha\psi^2\frac{|\nabla u|^2}{u^2} -(m-1)\lambda\psi^2 \Big]\,ds .\label{eq:AX-half-power-change} \end{aligned} \end{equation}

Since \(|\nabla u|^2\geq u_s^2\), this implies

\[ \begin{aligned} \Delta_q L_u^\alpha(p,q) \leq{}& \int_0^l\Big[ (m-1)\psi_s^2 +\alpha(3-m)\psi\psi_s\frac{u_s}{u} \notag\\ &\qquad -\alpha\left(1-\frac{m-1}{4}\alpha\right) \psi^2\frac{u_s^2}{u^2} -(m-1)\lambda\psi^2 \Big]\,ds . \end{aligned} \]

When \(\alpha\lt{}4/(m-1)\), Cauchy’s inequality gives

\[ \alpha(3-m)\psi\psi_s\frac{u_s}{u} -\alpha\left(1-\frac{m-1}{4}\alpha\right)\psi^2\frac{u_s^2}{u^2} \leq \frac{\alpha(m-3)^2}{4\left(1-\frac{m-1}{4}\alpha\right)}\psi_s^2 . \]

This proves (5.2.7); the sine choice gives (5.2.8).

For the bounded-\(u\) diameter estimate one uses a different exponent. Put

\[ \phi = \frac{u^{\frac{m-2}{m-1}\alpha}}{u(q)^\alpha}\psi, \qquad \psi(0)=0,\qquad \psi(l)=u(q)^{\alpha/(m-1)} . \]

Then

\[ \phi_s = \frac{u^{\frac{m-2}{m-1}\alpha}}{u(q)^\alpha} \left( \psi_s+\frac{m-2}{m-1}\alpha\psi\frac{u_s}{u} \right). \]

Substituting this into (5.2.15), the cross term cancels exactly and we obtain

\[ \begin{aligned} \Delta_q L_u^\alpha(p,q) \leq{}& \int_0^l u^{\frac{m-3}{m-1}\alpha}\Big[ (m-1)\psi_s^2 +\alpha\left(\frac{m-2}{m-1}\alpha-1\right) \psi^2\frac{u_s^2}{u^2} \notag\\ &\qquad -\alpha\psi^2\frac{|\nabla u|^2-u_s^2}{u^2} -(m-1)\lambda\psi^2 \Big]\,ds . \end{aligned} \]

If \(\alpha\leq (m-1)/(m-2)\), the two terms involving derivatives of \(u\) are nonpositive. Dropping them gives (5.2.9). ◻

Corollary 5.2.16

Let \((M^m,g)\) be complete, \(m\geq3\), and suppose \(u\gt{}0\) satisfies (5.2.6). If

\[ 0\leq\alpha\lt{}\frac{4}{m-1}, \]

then

\begin{equation} \tag{5.2.17} \label{eq:AX-subcritical-diam-from-geodesic} \mathrm{diam}(M) \leq \pi\sqrt{\frac{C_{m,\alpha}}{(m-1)\lambda}} . \end{equation}

In particular \(M\) is compact.

Proof. Assume first that two points \(p,q\) are joined by an \(L_u^\alpha\)-minimizing curve \(\alpha:[0,l]\to M\) parametrized by \(g\)-arclength. Let \(x=\alpha(l/2)\). The weighted excess

\[ e(y)=L_u^\alpha(p,y)+L_u^\alpha(y,q)-L_u^\alpha(p,q) \]

has a local minimum at \(x\). Using the two subsegments of \(\alpha\) as barriers and applying (5.2.8) to each half, we get, in the viscosity sense,

\[ 0\leq\Delta e(x) \leq u(x)^\alpha \left( \frac{C_{m,\alpha}\pi^2}{2l} -\frac{(m-1)\lambda l}{2} \right). \]

Therefore \(l\leq \pi\sqrt{C_{m,\alpha}/((m-1)\lambda)}\). Since \(d_g(p,q)\leq l\), this gives (5.2.17).

If the weighted minimizer is not realized, take a minimizing sequence. The usual Calabi limiting argument gives a broken minimizing object made of finite minimizing segments, weighted minimizing rays, and weighted minimizing lines. At the point where the corresponding excess vanishes, replace any ray by the point at parameter \(t\) on that ray and use the finite-segment barrier above. The right hand side contains the term \(-(m-1)\lambda t/2\) for that ray and is negative for \(t\) sufficiently large, contradicting the local minimum of the excess. Thus the non-realized case cannot occur, and the same diameter bound holds for all pairs of points. Hopf–Rinow then implies compactness. ◻

Corollary 5.2.17

Let \((M^m,g)\) be complete, \(m\geq3\), and suppose \(u\gt{}0\) satisfies (5.2.6) with

\[ 0\leq\alpha\leq\frac{m-1}{m-2}. \]

\[ 0\lt{}u_-:=\inf_Mu\leq \sup_Mu=:u_+\lt{}\infty, \]

then

\begin{equation} \tag{5.2.18} \label{eq:AX-bounded-u-diam-from-geodesic} \mathrm{diam}(M) \leq \frac{\pi}{\sqrt\lambda} \left(\frac{u_+}{u_-}\right)^{\frac{m-3}{2(m-1)}\alpha}. \end{equation}

In particular \(M\) is compact.

Proof. Because \(u\) is bounded above and below, the weighted metric \(u^{2\alpha}g\) is complete, so \(L_u^\alpha\)-minimizers exist. Let \(\alpha:[0,l]\to M\) be an \(L_u^\alpha\)-minimizer from \(p\) to \(q\), parametrized by \(g\)-arclength, and let \(x=\alpha(l/2)\). As above, the weighted excess has a local minimum at \(x\).

Use (5.2.9) on each half of \(\alpha\) with

\[ \psi(s)=u(x)^{\alpha/(m-1)}\sin\left(\frac{\pi s}{2(l/2)}\right). \]

Since \(\alpha(m-3)/(m-1)\geq0\), we estimate the positive term by \(u_+^{\frac{m-3}{m-1}\alpha}\) and the negative term by \(u_-^{\frac{m-3}{m-1}\alpha}\). Thus

\[ 0\leq\Delta e(x) \leq (m-1)u(x)^{\frac{2\alpha}{m-1}} \left( \frac{u_+^{\frac{m-3}{m-1}\alpha}\pi^2}{2l} -\frac{\lambda u_-^{\frac{m-3}{m-1}\alpha} l}{2} \right). \]

Hence

\[ l\leq \frac{\pi}{\sqrt\lambda} \left(\frac{u_+}{u_-}\right)^{\frac{m-3}{2(m-1)}\alpha}. \]

Since \(d_g(p,q)\leq l\) and \(p,q\) were arbitrary, this proves (5.2.18). Compactness follows from Hopf–Rinow. ◻

Volume estimate.

Lemma 5.2.18

Assume the hypotheses of Theorem 5.2.13 and normalize \(\min_{\widetilde N}\tilde u=1\). Set \(\theta=2\alpha/(n-1)\) and define, on \(\widetilde N\),

\begin{equation} \tag{5.2.19} \label{eq:AX-weighted-profile} I(v)=\inf\left\{ \int_{\partial E}\tilde u^\alpha: E\Subset\widetilde N,\quad \int_E \tilde u^\theta=v \right\}. \end{equation}

Then \(I\) satisfies

\begin{equation} \tag{5.2.20} \label{eq:AX-profile-ODE-ineq} I''I\leq -\frac{(I')^2}{n-1}-(n-1)\lambda \end{equation}

in the viscosity sense.

Proof. By the diameter part, \(\widetilde N\) is compact; hence the minimizer \(E\) exists for every \(v_0\in(0,\int_{\widetilde N}\tilde u^\theta)\). We suppress tildes. Put

\[ V(E)=\int_Eu^\theta,\qquad A(E)=\int_{\partial E}u^\alpha . \]

Let \(\Sigma=\partial E\) and take a smooth normal variation with variational field \(\varphi\nu\). We extend \(\varphi\) to a neighborhood of \(\Sigma\) and use the same symbol for the extension. With the convention used here,

\[ \partial_t d\mu_t=H\varphi\,d\mu,\qquad \partial_t\nu=-\nabla^\Sigma\varphi,\qquad \partial_t H=-\Delta^\Sigma\varphi -\bigl(|A_\Sigma|^2+\mathrm{Ric}_N(\nu,\nu)\bigr)\varphi . \]

Also

\[ \partial_t(u^{-1}u_\nu) = u^{-1}\mathrm{Hess}^N u(\nu,\nu)\varphi -u^{-1}\langle\nabla^\Sigma u,\nabla^\Sigma\varphi\rangle -u^{-2}u_\nu^2\varphi . \]

Thus

\[ V'(0)=\int_\Sigma u^\theta\varphi,\qquad V''(0)=\int_\Sigma (H+\theta u^{-1}u_\nu)u^\theta\varphi^2 +u^\theta\varphi\varphi_\nu, \]

\[ A'(0)=\int_\Sigma u^\alpha\varphi(H+\alpha u^{-1}u_\nu), \]

and differentiating this last expression gives the following formula. In the following displays, the same color marks the terms with the same origin: magenta terms cancel by integration by parts or by the normal derivative of \(\varphi=u^{-\alpha}\); green terms are the spectral-Ricci pair; blue terms come from \(|A_\Sigma|^2\); red terms come from \(u_\nu^2\); orange terms come from \(Hu_\nu\); and purple terms come from the surviving part of the last line.

\begin{equation} \tag{5.2.21} \begin{aligned} A''(0) ={}& \int_\Sigma \left( \begin{aligned} &\class{note-ax-cancel}{-\Delta^\Sigma\varphi} \class{note-ax-spec}{-\mathrm{Ric}_N(\nu,\nu)\varphi} \class{note-ax-geom}{-|A_\Sigma|^2\varphi} \end{aligned} \right) u^\alpha\varphi \notag\\ &+\int_\Sigma \left( \begin{aligned} &\class{note-ax-square}{-\alpha u^{-2}u_\nu^2\varphi} \class{note-ax-spec}{+\alpha u^{-1}\Delta^Nu\,\varphi} \class{note-ax-cancel}{-\alpha u^{-1}\Delta^\Sigma u\,\varphi}\\ &\class{note-ax-normal}{-\alpha u^{-1}Hu_\nu\varphi} \class{note-ax-cancel}{-\alpha u^{-1}\langle\nabla^\Sigma u,\nabla^\Sigma\varphi\rangle} \end{aligned} \right) u^\alpha\varphi \notag\\ &+\int_\Sigma \left( \begin{aligned} &\class{note-ax-cancel}{\alpha u^{\theta-1}u_\nu\varphi^2} \class{note-ax-cancel}{+u^\theta\varphi\varphi_\nu}\\ &\qquad \class{note-ax-last}{+Hu^\theta\varphi^2} \end{aligned} \right) u^{\alpha-\theta}\class{note-ax-last}{(H+\alpha u^{-1}u_\nu)}. \label{eq:AX-area-second-profile} \end{aligned} \end{equation}

The last line is just the derivative of the factor \(u^\alpha\varphi\,d\mu_t\) in \(A'(t)\), rewritten with a factor \(u^\theta\):

\[ \partial_t(u^\alpha\varphi\,d\mu_t) = \left( \begin{aligned} &\class{note-ax-cancel}{\alpha u^{\alpha-1}u_\nu\varphi^2} \class{note-ax-cancel}{+u^\alpha\varphi\varphi_\nu}\\ &\qquad \class{note-ax-last}{+Hu^\alpha\varphi^2} \end{aligned} \right)d\mu . \]

The volume constraint implies that

\[ u^{\alpha-\theta}(H+\alpha u^{-1}u_\nu)=A_v'(v_0), \]

where \(A_v\) is the area of this variation written as a function of \(V\). Choose \(\varphi=u^{-\alpha}\) and set

\[ Q=\int_\Sigma u^{\theta-\alpha},\qquad X=u^{\theta-\alpha}A_v'(v_0),\qquad Y=u^{-1}u_\nu . \]

Then \(H=X-\alpha Y\). For \(\varphi=u^{-\alpha}\), the tangential \(\Delta^\Sigma u\) term cancels after integration by parts with the tangential gradient term

\[ \class{note-ax-cancel}{-\alpha u^{-1}\langle\nabla^\Sigma u,\nabla^\Sigma\varphi\rangle}. \]

Hence the first two lines of (5.2.21), together with (5.2.3) and \(|A_\Sigma|^2\geq H^2/(n-1)\), are bounded above by

\[ \int_\Sigma u^{-\alpha} \left[ \class{note-ax-geom}{-\frac{H^2}{n-1}} \class{note-ax-normal}{-\alpha HY} \class{note-ax-square}{-\alpha Y^2} \class{note-ax-spec}{-(n-1)\lambda} \right]. \]

The last line of (5.2.21) is

\[ \int_\Sigma u^{-\alpha}\class{note-ax-last}{XH}, \]

because \(\class{note-ax-cancel}{\varphi_\nu=-\alpha u^{-\alpha}Y}\) and \(\class{note-ax-last}{H+\alpha Y=X}\). Therefore

\[ A''(0) \leq \int_\Sigma u^{-\alpha} \left[ \class{note-ax-geom}{-\frac{H^2}{n-1}} \class{note-ax-normal}{-\alpha HY} \class{note-ax-square}{-\alpha Y^2} \class{note-ax-spec}{-(n-1)\lambda} \class{note-ax-last}{+XH} \right]. \]

Expanding \(H=X-\alpha Y\) gives

\[ \begin{aligned} A''(0) \leq{}& \int_\Sigma u^{-\alpha}\Big[ \class{note-ax-geom}{-\frac{X^2}{n-1}} \class{note-ax-geom}{+\frac{2\alpha}{n-1}XY} \class{note-ax-geom}{-\frac{\alpha^2}{n-1}Y^2} \class{note-ax-square}{-\alpha Y^2} \class{note-ax-normal}{-\alpha XY} \class{note-ax-normal}{+\alpha^2Y^2} \notag\\ &\qquad \class{note-ax-last}{+X^2} \class{note-ax-last}{-\alpha XY} \class{note-ax-spec}{-(n-1)\lambda} \Big]. \end{aligned} \]

The inverse-function chain rule gives

\[ A_v''(v_0) = (V'(0))^{-2}A''(0)-(V'(0))^{-3}A'(0)V''(0). \]

Substituting the formulas for \(V',V''\) and using \(\theta=2\alpha/(n-1)\) yields

\[ \begin{aligned} Q^2A_v''(v_0) \leq{}& \int_\Sigma u^{-\alpha}\left[ -\frac{X^2}{n-1} +\left(\frac{2\alpha}{n-1}-\theta\right)XY +\left(\frac{n-2}{n-1}\alpha^2-\alpha\right)Y^2 -(n-1)\lambda \right]\notag\\ \leq{}& -\left(\frac{A_v'(v_0)^2}{n-1}+(n-1)\lambda\right) \int_\Sigma u^{2\theta-3\alpha}. \end{aligned} \]

Here we used \(\alpha\leq(n-1)/(n-2)\) and \(u\geq1\). Holder’s inequality gives

\[ A_v(v_0)\int_\Sigma u^{2\theta-3\alpha} \geq \left(\int_\Sigma u^{\theta-\alpha}\right)^2=Q^2. \]

Thus

\[ A_v(v_0)A_v''(v_0) \leq -\frac{A_v'(v_0)^2}{n-1}-(n-1)\lambda . \]

Since \(A_v\) is an upper barrier for \(I\) at \(v_0\), this is exactly the viscosity inequality (5.2.20). ◻

Lemma 5.2.19

Let \(V\in(0,\infty]\) and let \(I:[0,V)\to\mathbb R\) be continuous with \(I(0)=0\) and \(I(v)\gt{}0\) for \(v\in(0,V)\). Suppose

\[ I''I\leq -\frac{(I')^2}{n-1}-(n-1)\lambda \]

in the viscosity sense on \((0,V)\), and suppose

\[ \limsup_{v\to0^+}v^{-\frac{n-1}{n}}I(v) \leq n|B^n|^{1/n}. \]

Then

\[ V\leq \lambda^{-n/2}|S^n|. \]

Proof. Set

\[ \psi=I^{\frac n{n-1}}. \]

Applying the chain rule to positive upper test functions for \(I\), the preceding viscosity inequality is equivalent to

\begin{equation} \tag{5.2.22} \label{eq:AX-psi-ODE} \psi''\leq -n\lambda \psi^{\frac{2-n}{n}} \end{equation}

in the viscosity sense. The small-volume assumption gives

\[ \psi'_+(0):=\limsup_{v\to0^+}\frac{\psi(v)}{v} \leq n^{\frac n{n-1}}|B^n|^{\frac1{n-1}} =n|S^{n-1}|^{\frac1{n-1}}. \]

For \(\zeta\gt{}0\) define

\[ \mu(r)=\frac{\sin(\sqrt\lambda r)}{\sqrt\lambda}, \qquad v_\zeta(r)=\zeta\int_0^r\mu(s)^{n-1}\,ds, \qquad 0\leq r\leq\frac{\pi}{\sqrt\lambda}, \]

and define the model profile \(I_\zeta\) by

\[ I_\zeta(v_\zeta(r))=\zeta\mu(r)^{n-1}. \]

Let \(\psi_\zeta=I_\zeta^{n/(n-1)}\). Since

\[ \psi_\zeta(v_\zeta(r))=\zeta^{\frac n{n-1}}\mu(r)^n, \qquad \frac{dv_\zeta}{dr}=\zeta\mu(r)^{n-1}, \]

direct differentiation gives

\[ \psi_\zeta''=-n\lambda\psi_\zeta^{\frac{2-n}{n}}, \qquad (\psi_\zeta)'_+(0)=n\zeta^{\frac1{n-1}}. \]

The existence interval of \(I_\zeta\) has length

\[ V_\zeta = \zeta\int_0^{\pi/\sqrt\lambda}\mu(s)^{n-1}\,ds. \]

For \(\zeta=|S^{n-1}|\), this is exactly \(\lambda^{-n/2}|S^n|\).

Assume by contradiction that \(V\gt{}\lambda^{-n/2}|S^n|\). Choose \(\zeta\gt{}|S^{n-1}|\) so close to \(|S^{n-1}|\) that \(V_\zeta\lt{}V\). Then \((\psi_\zeta)'_+(0)\gt{}\psi'_+(0)\), so \(\psi\lt{}\psi_\zeta\) on \((0,\delta)\) for some \(\delta\gt{}0\). If the two functions first meet at \(a\in(0,V_\zeta)\), then on \((0,a)\) we have \(\psi\lt{}\psi_\zeta\). Since the function \(s\mapsto -n\lambda s^{(2-n)/n}\) is increasing on \((0,\infty)\), the difference \(w=\psi-\psi_\zeta\) satisfies \(w''\lt{}0\) in the viscosity sense on \((0,a)\). Thus \(w\) is strictly concave there. But \(w(0)=w(a)=0\) and \(w\lt{}0\) on \((0,a)\), which is impossible for a concave function. Hence \(\psi\lt{}\psi_\zeta\) on \((0,V_\zeta)\). Since \(\psi_\zeta(V_\zeta)=0\) while \(V_\zeta\lt{}V\) and \(I\gt{}0\) on \((0,V)\), continuity gives a contradiction at \(V_\zeta\). Therefore \(V\leq \lambda^{-n/2}|S^n|\). ◻

Proof. Normalize \(\min_{\widetilde N}\tilde u=1\) and set

\[ V_0=\int_{\widetilde N}\tilde u^\theta . \]

The profile \(I\) satisfies (5.2.20). Since \(\tilde u\) attains its minimum at some point \(\tilde p\), small geodesic balls centered at \(\tilde p\) satisfy

\[ \int_{B_r(\tilde p)}\tilde u^\theta=|B^n|r^n+O(r^{n+1}), \qquad \int_{\partial B_r(\tilde p)}\tilde u^\alpha=n|B^n|r^{n-1}+O(r^n). \]

Since these balls are admissible competitors for \(I\), it follows that

\[ \limsup_{v\to0}v^{-\frac{n-1}{n}}I(v)\leq n|B^n|^{1/n}. \]

Lemma 5.2.19 applied with \(V=V_0\) therefore gives

\[ V_0\leq \lambda^{-n/2}|S^n|. \]

Since \(\tilde u\geq1\), we have

\[ |\widetilde N|\leq\int_{\widetilde N}\tilde u^\theta=V_0 \leq \lambda^{-n/2}|S^n|. \]

If equality holds, then \(\int_{\widetilde N}\tilde u^\theta=|\widetilde N|\) and \(\tilde u\geq1\), hence \(\tilde u\equiv1\). The spectral inequality becomes the pointwise Ricci bound \(\mathrm{Ric}_{\widetilde N}\geq(n-1)\lambda\), and the equality case in the classical Bishop–Gromov theorem gives that \(\widetilde N\) is the round sphere of radius \(\lambda^{-1/2}\). ◻

We need a comparison of distance.

Lemma 5.2.20

Let \(\varphi:M\to \mathbb{R}^{n+1}\) be the immersion and \(\tilde{g}=r^{-2}g\) and \(g=\varphi^*(\delta)\) with \(0 \in M\). Given two points \(p,q\in M\) with \(d_{\tilde{g}}(p,q)\leq D\), we have \(r(p)\leq e^{D}r(q)\).

Proof. Let \(\alpha\) be a curve joining \(p\) and \(q\) with length \(D+\varepsilon\). Then, we have

\[ \begin{aligned} \log r(p)-\log r(q)={} & \int_{ } \frac{\left< \nabla r, \alpha'(t) \right> }{r(\alpha(t))}dt\leq \int_{ } \frac{|\alpha'(t)|_g}{r}dt \\ ={} & \int_{ } |\alpha'(t)|_{\tilde{g}}d\tilde{t}=D+\varepsilon. \end{aligned} \]

◻

Proof. Let \(M^n\hookrightarrow\mathbb{R}^{n+1}\) be as in the theorem, where \(n=3\) or \(n=4\). We fix any point \(x_0\in M\) and suppose \(x_0=0\) after a translation. Let \(\tilde{g}=r^{-2}g\), where \(r(x)=|x|\).

The preceding spectral estimates and \(\mu\)-bubble constructions give the following dimension-dependent constants. There exist \(L_n,A_n,D_n\lt{}\infty\) such that, whenever \(N\) is a compact collar in \((M,\tilde{g})\) with \(d_{\tilde{g}}(\partial_-N,\partial_+N)\geq L_n\), one can find a smooth \(\mu\)-bubble component \(\Sigma\) which separates \(\partial_-N\) from \(\partial_+N\), lies in \(\tilde B_{L_n}(\partial_-N)\), and satisfies

\[ |\Sigma|_{\tilde{g}}\leq A_n,\qquad \mathrm{diam}_{\tilde{g}}(\Sigma)\leq D_n . \]

For \(n=3\) and \(n=4\) this follows from the \(\mu\)-bubble reduction in Theorem 5.2.8 together with the spectral Ricci estimates in §5.2.6.

For any \(\rho\gt{}0\), choose \(R\) large such that

\[ d_{\tilde{g}}(\partial B_\rho^M,\partial B_R^M)\geq L_n. \]

Set \(N=B_R^M\setminus B_\rho^M\), with \(\partial_-N=\partial B_\rho^M\) and \(\partial_+N=\partial B_R^M\), and apply the preceding paragraph. If \(x\in\Sigma\), then there is \(y\in\partial B_\rho^M\) with \(d_{\tilde{g}}(x,y)\leq L_n\). By the distance comparison lemma,

\[ r(x)\leq e^{L_n}r(y)\leq e^{L_n}\rho . \]

Therefore \(\Sigma\subset B_{C_n\rho}\) in the Euclidean metric. Since \(g=r^2\tilde{g}\), the induced measures on the \((n-1)\)-dimensional hypersurface \(\Sigma\) satisfy \(d\mu_g=r^{n-1}d\mu_{\tilde{g}}\), and hence

\[ |\Sigma|_g \leq (C_n\rho)^{n-1}|\Sigma|_{\tilde{g}} \leq C_n\rho^{n-1}. \]

Since \(M\) is simply connected and has one end [CSZ97], the side of \(\Sigma\) containing \(B_\rho^M(x_0)\) is a compact region. Let \(\Omega_\rho\) denote this region; then \(B_\rho^M(x_0)\subset\Omega_\rho\) and \(\partial\Omega_\rho=\Sigma\). The Michael–Simon isoperimetric inequality for minimal submanifolds [MS73] gives

\[ |\Omega_\rho|_g\leq C_n|\Sigma|_g^{\frac{n}{n-1}}. \]

Consequently,

\[ |B_\rho^M(x_0)| \leq |\Omega_\rho|_g \leq C_n|\Sigma|_g^{\frac{n}{n-1}} \leq C_n\rho^n. \]

◻

Mazet’s proof in the ambient space \(\mathbb{R}^6\)

Here \(\mathbb{R}^6\) means the hypersurface dimension is \(5\). Mazet [Maz24] proves that every complete, connected, two-sided stable minimal immersion

\[ M^5\hookrightarrow \mathbb{R}^6 \]

is flat. The proof follows the Chodosh–Li–Minter–Stryker strategy [CLMS25], but with one extra parameter in the curvature quantity. This extra parameter is the weighted bi-Ricci curvature.

Let \((N^m,g)\) be a Riemannian manifold, and let \(\{e_1,\ldots,e_m\}\) be an orthonormal basis. For \(\alpha\in\mathbb{R}\), Mazet defines the \(\alpha\)-weighted bi-Ricci curvature by

\[ \operatorname{BRic}_{\alpha}(e_1,e_2) := \sum_{i=2}^{m}R(e_1,e_i,e_i,e_1) + \alpha \sum_{j=3}^{m}R(e_2,e_j,e_j,e_2). \]

Its pointwise minimum is

\[ \Lambda_\alpha(p) := \min_{\substack{e,f\in T_pN\\ |e|=|f|=1,\ \langle e,f\rangle=0}} \operatorname{BRic}_{\alpha}(e,f). \]

When \(\alpha=1\), this is the usual bi-Ricci curvature:

\[ \operatorname{BRic}_{1}(e_1,e_2) = \operatorname{Ric}(e_1,e_1) + \operatorname{Ric}(e_2,e_2) - R(e_1,e_2,e_2,e_1). \]

Thus \(\alpha\) lets one change the relative weight of the curvature directions which are tangent to the eventual \(\mu\)-bubble.

The first step is the same conformal change as before. Remove the points where the immersion hits the origin and set

\[ \tilde g=r^{-2}g, \qquad N:=M\setminus F^{-1}(0). \]

The Gulliver–Lawson observation is that \((N,\tilde g)\) is complete. Mazet’s main spectral estimate says that, in dimension \(5\), one can choose

\[ a=\frac{11}{10}, \qquad \alpha=\frac{40}{43}, \qquad \delta=\frac{3}{10}, \]

so that the stability inequality implies the following weighted bi-Ricci spectral lower bound:

\[ \int_N |\tilde\nabla\varphi|^2\,d\tilde\mu \geq \frac1a\int_N V\varphi^2\,d\tilde\mu, \qquad V\geq \delta-\widetilde{\Lambda}_\alpha, \qquad \varphi\in C_c^1(N). \]

Equivalently,

\[ \int_N \left( a|\tilde\nabla\varphi|^2 +(\widetilde{\Lambda}_\alpha-\delta)\varphi^2 \right)d\tilde\mu \geq0. \]

So \((N,\tilde g)\) does not have a pointwise lower bound \(\widetilde{\Lambda}_\alpha\geq\delta\), but it has the corresponding spectral lower bound.

The second step is to build a weighted \(\mu\)-bubble in a long annulus of \((N,\tilde g)\). The bubble is a hypersurface

\[ \Sigma^4=\partial\Omega \]

which separates the inner and outer boundary of the annulus. The second variation of the weighted \(\mu\)-bubble turns the spectral \(\operatorname{BRic}_\alpha\) bound on \(N\) into a spectral Ricci bound on \(\Sigma\). More precisely, if

\[ \lambda_\Sigma(x):=\min_{|v|=1}\operatorname{Ric}_\Sigma(v,v), \]

then the induced metric on \(\Sigma\) satisfies an inequality of the form

\[ \int_\Sigma \left( \frac{4}{(4-a)\alpha}|\nabla\phi|^2 + \lambda_\Sigma\phi^2 \right)d\mu_\Sigma \geq \frac{\delta}{2\alpha}\int_\Sigma \phi^2\,d\mu_\Sigma. \]

For Mazet’s parameters,

\[ \frac{4}{(4-a)\alpha}=\frac{43}{29}\lt{}\frac32 = \frac{k-1}{k-2}, \qquad k=\dim\Sigma=4. \]

This is the numerical point that allows the Antonelli–Xu spectral Bishop–Gromov theorem [AX24] to be applied to the \(4\)-dimensional bubble. It gives a uniform \(\tilde g\)-volume bound for \(\Sigma\).

Finally one returns to the original metric \(g\). If \(\Sigma\) is chosen around \(B_\rho^M(p_0)\), the conformal collar bound gives

\[ r\leq C\rho \qquad\text{on }\Sigma, \]

so the \(\tilde g\)-volume bound for \(\Sigma\) becomes a \(g\)-area bound

\[ |\Sigma|_g\leq C\rho^4. \]

Since \(M\) has one end, one can choose the relevant component of \(\Sigma\) to enclose \(B_\rho^M(p_0)\). The Michael–Simon–Brendle isoperimetric inequality [MS73, Bre21] on minimal hypersurfaces then gives

\[ |B_\rho^M(p_0)|\leq C|\Sigma|_g^{5/4}\leq C\rho^5. \]

This is the Euclidean volume growth needed by the Schoen–Simon–Yau stable Bernstein theorem. Hence \(M^5\) is flat. In short, the new feature in \(\mathbb{R}^6\) is that the weighted curvature \(\operatorname{BRic}_{40/43}\) creates just enough spectral Ricci positivity on the \(4\)-dimensional \(\mu\)-bubble for Antonelli–Xu’s volume estimate to close.

Green-kernel proof of the stable Bernstein theorem in R^4

Mon, 29 Jun 2026 00:00:00 +0000

The Green-kernel proof of the stable Bernstein theorem in \(\mathbb{R}^4\) is due to Cabré–Catino–Mari–Mastrolia–Roncoroni [CCM+26]. The key point is that this proof does not use \(\mu\)-bubbles. Instead, it combines:

the positive supersolution furnished by stability;
the minimal positive Green kernel \(\mathscr G\) of \(-\Delta^M\);
a sharp pointwise estimate for \(|\nabla\mathscr G|\) obtained from Bochner–Kato and a simple convexity lemma;
a weighted Schoen–Simon–Yau inequality;
a logarithmic cut-off in the variable \(\mathscr G\).

Theorem 5.3.1

Let \(M^3\hookrightarrow\mathbb{R}^4\) be a connected, complete, two-sided, stable minimal immersion. Then \(M\) is an affine hyperplane.

Reduction to bounded curvature. It is enough to prove the theorem under the auxiliary assumption

\[ |A|\in L^\infty(M). \]

Indeed, if the theorem were false, then \(|A|\) is nonzero somewhere. If \(|A|\) is globally bounded, we are already in the bounded-curvature case. If not, one performs the standard point-picking argument on intrinsic balls \(B_j^M(p_0)\): choose \(q_j\) and a scale \(\lambda_j=|A(q_j)|\to\infty\) so that, after replacing the immersion by

\[ F_j(x)=\lambda_j\bigl(F(x)-F(q_j)\bigr), \]

the rescaled hypersurfaces have \(|A_j|(q_j)=1\) and uniformly bounded curvature on larger and larger intrinsic balls around \(q_j\). Stability and minimality are scale invariant, so a smooth local compactness theorem gives a complete, two-sided, stable minimal limit in \(\mathbb{R}^4\) with bounded second fundamental form and \(|A|(0)=1\). The bounded-curvature case below would force this limit to be flat, a contradiction.

Green kernel and stability supersolution. Let now \(M^n\hookrightarrow\mathbb{R}^{n+1}\) be complete, two-sided, stable, and minimal, with \(n\geq3\), and assume \(|A|\in L^\infty(M)\). We keep \(n\) arbitrary until the last step. Stability means

\begin{equation} \tag{5.3.1} \label{eq:green-stability} \int_M |A|^2\phi^2\,d\mu \leq \int_M |\nabla\phi|^2\,d\mu, \qquad \phi\in \operatorname{Lip}_c(M). \end{equation}

Equivalently, the operator \(-\Delta-|A|^2\) is nonnegative. By the positive solution characterization of nonnegative Schrödinger operators, there exists a positive \(C^2\) function \(u\) such that

\begin{equation} \tag{5.3.2} \label{eq:green-u-supersolution} \Delta u+|A|^2u\leq0 \qquad\text{on }M. \end{equation}

We next explain why the Green kernel exists. The needed potential-theoretic input is the following.

Definition 5.3.2

Let \((N^m,g)\) be a connected, complete, noncompact Riemannian manifold. We say that \(N\) is non-parabolic if the Laplacian \(-\Delta^N\) admits a positive minimal Green kernel: for some, equivalently for every, point \(o\in N\) there is a function \(G(o,\cdot)\gt{}0\) on \(N\setminus\{o\}\) such that

\[ \Delta^N G(o,\cdot)=-\delta_o, \]

in the distributional sense, and \(G\) is minimal among all positive fundamental solutions. If no such positive Green kernel exists, \(N\) is called parabolic.

Equivalently, \(N\) is parabolic if every positive superharmonic function on \(N\) is constant; equivalently, the Brownian motion on \(N\) is recurrent. Another useful equivalent criterion is the capacity criterion below.

Definition 5.3.3

For a compact set \(K\Subset N\), define its \(2\)-capacity by

\[ \operatorname{Cap}_2(K) := \inf\left\{ \int_N |\nabla\phi|^2\,d\mu: \phi\in C_c^\infty(N),\quad \phi\geq1 \text{ on a neighborhood of }K \right\}. \]

Then \(N\) is non-parabolic if and only if \(\operatorname{Cap}_2(K)\gt{}0\) for some nonempty compact set \(K\); it is parabolic if and only if \(\operatorname{Cap}_2(K)=0\) for every compact \(K\).

Proposition 5.3.4

Let \((N^m,g)\) be complete and noncompact, with \(m\gt{}2\). Suppose that there is a constant \(S\gt{}0\) such that

\begin{equation} \tag{5.3.3} \label{eq:sobolev-nonparabolic} \left(\int_N |f|^{\frac{2m}{m-2}}\,d\mu\right)^{\frac{m-2}{m}} \leq S\int_N |\nabla f|^2\,d\mu \qquad \forall f\in C_c^\infty(N). \end{equation}

Then \(N\) is non-parabolic.

Proof. Choose a compact set \(K\Subset N\) with \(|K|\gt{}0\). If \(\phi\in C_c^\infty(N)\) and \(\phi\geq1\) near \(K\), then

\[ |K| \leq \int_K |\phi|^{\frac{2m}{m-2}}\,d\mu \leq \int_N |\phi|^{\frac{2m}{m-2}}\,d\mu. \]

Raising both sides to the power \((m-2)/m\) and using (5.3.3), we get

\[ |K|^{\frac{m-2}{m}} \leq \left(\int_N |\phi|^{\frac{2m}{m-2}}\,d\mu\right)^{\frac{m-2}{m}} \leq S\int_N |\nabla\phi|^2\,d\mu. \]

Taking the infimum over all admissible \(\phi\) gives

\[ \operatorname{Cap}_2(K) \geq S^{-1}|K|^{\frac{m-2}{m}}\gt{}0. \]

By the capacity criterion, \(N\) is non-parabolic. ◻

Proposition 5.3.5

Let \(M^n\hookrightarrow\mathbb{R}^{n+1}\) be a complete minimal hypersurface with \(n\gt{}2\). Then \(M\) is non-parabolic.

Proof. We use the sharp Michael–Simon Sobolev inequality of Brendle [Bre21]; in the present minimal hypersurface case it gives the following \(W^{1,2}\) Sobolev consequence:

\begin{equation} \tag{5.3.4} \label{eq:minimal-W12-sobolev} \left(\int_M |f|^{\frac{2n}{n-2}}\,d\mu\right)^{\frac{n-2}{n}} \leq C_n \int_M |\nabla f|^2\,d\mu . \end{equation}

Since \(n\gt{}2\), Proposition 5.3.4 applies and \(M\) is non-parabolic. ◻

For the theorem, Proposition 5.3.5 applies to our \(M^n\), since \(n\geq3\). Hence there exists a minimal positive Green kernel \(\mathscr G\) of \(-\Delta\) with pole \(o\in M\):

\[ \Delta\mathscr G=-\delta_o,\qquad \mathscr G\gt{}0,\qquad \mathscr G(x)\to0\quad\text{as }x\to\infty. \]

For a.e. \(s\gt{}0\), the level set

\[ \Sigma_s:=\{\mathscr G=s\}, \]

is smooth, and the Green flux identity gives

\begin{equation} \tag{5.3.5} \label{eq:green-flux} \int_{\Sigma_s}|\nabla\mathscr G|\,d\sigma=1. \end{equation}

Indeed, applying the divergence theorem to \(\{\mathscr G\gt{}s\}\setminus B_\varepsilon(o)\) and letting \(\varepsilon\to0\) gives exactly the mass of the pole.

Ricci lower bound from the Gauss equation. Let \(\{e_i\}_{i=1}^n\) diagonalize \(A\) at a point, with principal curvatures \(\kappa_i\) and \(\sum_i\kappa_i=0\). The traced Gauss equation gives

\[ \mathrm{Ric}(e_i,e_i) = \sum_{j\neq i}A_{ii}A_{jj} = -\kappa_i^2. \]

Thus, for every unit vector \(v\),

\[ \mathrm{Ric}(v,v)=-|A(v,\cdot)|^2. \]

Since \(A\) is trace-free,

\[ \kappa_i^2 = \left(\sum_{j\neq i}\kappa_j\right)^2 \leq (n-1)\sum_{j\neq i}\kappa_j^2 =(n-1)(|A|^2-\kappa_i^2), \]

and hence \(\kappa_i^2\leq\frac{n-1}{n}|A|^2\). Therefore

\begin{equation} \tag{5.3.6} \label{eq:green-ricci-lower} \mathrm{Ric}\geq -\frac{n-1}{n}|A|^2g. \end{equation}

The key convexity lemma. We shall use the following elementary computation several times.

Lemma 5.3.6

Let \(w,v\gt{}0\) be \(C^2\) functions on a domain \(\Omega\subset M\) satisfying

\[ \Delta w+Ww\geq0,\qquad \Delta v+Vv\leq0. \]

For \(\delta\geq0\), set

\[ \xi_\delta=w^{1+\delta}v^{-\delta}. \]

Then

\begin{equation} \tag{5.3.7} \label{eq:green-magic} \Delta\xi_\delta \geq \bigl(\delta V-(1+\delta)W\bigr)\xi_\delta +\delta(1+\delta) \left|\nabla\log\frac{w}{v}\right|^2\xi_\delta . \end{equation}

Proof. Since \(\xi_\delta=e^{\log\xi_\delta}\),

\[ \Delta\xi_\delta = \xi_\delta\left(\Delta\log\xi_\delta+ |\nabla\log\xi_\delta|^2\right). \]

Now

\[ \log\xi_\delta=(1+\delta)\log w-\delta\log v. \]

Using

\[ \Delta\log w=\frac{\Delta w}{w}-|\nabla\log w|^2, \qquad \Delta\log v=\frac{\Delta v}{v}-|\nabla\log v|^2, \]

we get

\[ \begin{aligned} \Delta\log\xi_\delta &=(1+\delta)\Delta\log w-\delta\Delta\log v \\ &\geq \delta V-(1+\delta)W -(1+\delta)|\nabla\log w|^2 +\delta|\nabla\log v|^2 . \end{aligned} \]

Also,

\[ \begin{aligned} |\nabla\log\xi_\delta|^2 &=(1+\delta)^2|\nabla\log w|^2 +\delta^2|\nabla\log v|^2 \\ &\quad -2\delta(1+\delta) \left\langle\nabla\log w,\nabla\log v\right\rangle . \end{aligned} \]

Adding the last two displays gives

\[ \Delta\log\xi_\delta+|\nabla\log\xi_\delta|^2 \geq \delta V-(1+\delta)W +\delta(1+\delta) \left|\nabla\log w-\nabla\log v\right|^2, \]

which is (5.3.7). ◻

Pointwise gradient estimate for the Green kernel. We first record the local differential inequality for the Green kernel.

Proposition 5.3.7

Let

\[ q=|\nabla\mathscr G|, \qquad \alpha=\frac{n-2}{n-1}, \qquad w=q^\alpha . \]

On the open set \(\{q\gt{}0\}\setminus\{o\}\), the function \(w\) satisfies

\begin{equation} \tag{5.3.8} \label{eq:green-gradient-subsolution} \Delta w+\frac{n-2}{n}|A|^2w\geq0. \end{equation}

Proof. Recall first the standard Bochner formula: for every smooth function \(f\) on a Riemannian manifold,

\[ \frac12\Delta |\nabla f|^2 = |\nabla^2 f|^2 +\left\langle\nabla f,\nabla\Delta f\right\rangle +\mathrm{Ric}(\nabla f,\nabla f). \]

We apply this to \(f=\mathscr G\) on \(M\setminus\{o\}\). Since \(\Delta\mathscr G=0\) away from the pole, the middle term vanishes. Hence

\[ \frac12\Delta q^2 = |\nabla^2\mathscr G|^2+\mathrm{Ric}(\nabla\mathscr G,\nabla\mathscr G). \]

The refined Kato inequality for harmonic functions gives

\[ |\nabla^2\mathscr G|^2\geq \frac{n}{n-1}|\nabla q|^2. \]

Together with the Ricci lower bound (5.3.6), this gives

\[ \frac12\Delta q^2 \geq \frac{n}{n-1}|\nabla q|^2 -\frac{n-1}{n}|A|^2q^2. \]

Since

\[ \frac12\Delta q^2=q\Delta q+|\nabla q|^2, \]

we subtract \(|\nabla q|^2\) and obtain

\begin{equation} \tag{5.3.9} \label{eq:green-q-ineq} q\Delta q \geq \frac{1}{n-1}|\nabla q|^2 -\frac{n-1}{n}|A|^2q^2. \end{equation}

Now compute the Laplacian of \(w=q^\alpha\):

\[ \begin{aligned} \Delta w &=\alpha q^{\alpha-1}\Delta q +\alpha(\alpha-1)q^{\alpha-2}|\nabla q|^2 \\ &\geq \alpha q^{\alpha-2} \left[ \frac{1}{n-1}|\nabla q|^2 -\frac{n-1}{n}|A|^2q^2 \right] +\alpha(\alpha-1)q^{\alpha-2}|\nabla q|^2, \end{aligned} \]

where we used (5.3.9) in the second line. The coefficient of \(|\nabla q|^2\) is

\[ \alpha\left(\frac1{n-1}+\alpha-1\right) = \alpha\left(\frac1{n-1}+\frac{n-2}{n-1}-1\right)=0. \]

Hence

\[ \Delta w \geq -\alpha\frac{n-1}{n}|A|^2q^\alpha = -\frac{n-2}{n}|A|^2w, \]

which is exactly (5.3.8). ◻

Apply Lemma 5.3.6 to

\[ W=\frac{n-2}{n}|A|^2,\qquad V=|A|^2,\qquad v=u, \qquad \delta\geq0. \]

For

\[ \xi_\delta=w^{1+\delta}u^{-\delta}, \]

the coefficient of the zeroth-order \(|A|^2\xi_\delta\) term in Lemma 5.3.6 is

\[ \delta V-(1+\delta)W = \left[ \delta-(1+\delta)\frac{n-2}{n} \right]|A|^2 = \frac{2\delta-(n-2)}{n}|A|^2 . \]

Thus this coefficient is nonnegative when \(\delta\geq\frac{n-2}{2}\), and it vanishes exactly at the critical choice

\[ \delta=\frac{n-2}{2}. \]

With this choice, \(1+\delta=\frac n2\), and

\[ \xi:=w^{n/2}u^{-(n-2)/2} = |\nabla\mathscr G|^{\frac{n(n-2)}{2(n-1)}}u^{-\frac{n-2}{2}} \]

is subharmonic on \(\{q\gt{}0\}\setminus U\), where

\[ U:=\{\mathscr G\gt{}1\}. \]

Because \(|A|\) is bounded, (5.3.6) gives a lower Ricci bound, and the Cheng–Yau gradient estimate [CY75] applied outside \(U\) gives

\[ |\nabla\mathscr G|\leq C\mathscr G. \]

We package the remaining two comparison-principle arguments into one lemma. First \(\mathscr G\) is compared with the stability supersolution \(u\); this auxiliary estimate is used only to remove the negative power of \(u\) and prove \(\xi\to0\) at infinity. Then \(\xi\) is compared directly with the harmonic barrier \(\mathscr G\).

Lemma 5.3.8

There is a constant \(C_0\gt{}0\) such that

\[ \xi\leq C_0\mathscr G \qquad\text{on }M\setminus U. \]

Proof. We first prove the auxiliary comparison

\[ \mathscr G\leq Cu \qquad\text{on }M\setminus U. \]

Since \(\mathscr G\to0\) at infinity, the set \(\bar U=\{\mathscr G\geq1\}\) is compact. Choose \(C\) so large that

\[ \mathscr G\leq Cu \qquad\text{on }\partial U. \]

This is possible because \(\mathscr G=1\) on \(\partial U\) and \(u\gt{}0\) there. Let \(\Omega_R\) be a smooth exhaustion of \(M\) with \(\bar U\Subset\Omega_R\), and let \(\mathscr G_R\) be the Dirichlet Green function on \(\Omega_R\) with pole \(o\). On the annular region \(\Omega_R\setminus \bar U\), the function \(\mathscr G_R\) is harmonic and vanishes on \(\partial\Omega_R\). Also,

\[ \Delta(Cu)=C\Delta u\leq -C|A|^2u\leq0, \]

so \(Cu\) is superharmonic. On the boundary of \(\Omega_R\setminus \bar U\) we have

\[ \mathscr G_R\leq \mathscr G=1\leq Cu \quad\text{on }\partial U, \qquad \mathscr G_R=0\leq Cu \quad\text{on }\partial\Omega_R. \]

The comparison principle on the bounded annulus therefore gives

\[ \mathscr G_R\leq Cu \qquad\text{on }\Omega_R\setminus U. \]

Letting \(R\to\infty\) and using \(\mathscr G_R\uparrow\mathscr G\), we obtain

\[ \mathscr G\leq Cu \qquad\text{on }M\setminus U. \]

This is where the first comparison is used. Combining it with \(|\nabla\mathscr G|\leq C\mathscr G\), we get

\[ \xi \leq C\mathscr G^{\frac{n(n-2)}{2(n-1)}}u^{-\frac{n-2}{2}} \leq C\mathscr G^{\frac{n-2}{2(n-1)}}\to0 \qquad\text{as }x\to\infty. \]

Now we perform the second comparison, this time between the subharmonic quantity \(\xi\) and a multiple of the harmonic Green kernel \(\mathscr G\). Since \(\bar U\) is compact and \(\xi\) is continuous up to \(\partial U\), while \(\mathscr G=1\) on \(\partial U\), we can choose \(C_0\) so large that

\[ \xi\leq C_0\mathscr G \qquad\text{on }\partial U. \]

Set

\[ h:=\xi-C_0\mathscr G . \]

On \(M\setminus U\), the pole \(o\) is not present, so \(\mathscr G\) is harmonic. Also \(\xi\) is subharmonic there, in the weak sense obtained from the preceding pointwise computation by the standard regularization \(q_\tau=(q^2+\tau)^{1/2}\) and then letting \(\tau\downarrow0\). Hence

\[ \Delta h=\Delta\xi-C_0\Delta\mathscr G\geq0 \qquad\text{on }M\setminus U. \]

Thus \(h\) is subharmonic. Moreover,

\[ h\leq0\qquad\text{on }\partial U. \]

We now explain how the boundary condition at infinity is used. Since \(\xi(x)\to0\) and \(\mathscr G(x)\to0\) as \(x\to\infty\), for every \(\varepsilon\gt{}0\) there is a compact set \(K_\varepsilon\supset \bar U\) such that

\[ h(x)=\xi(x)-C_0\mathscr G(x)\leq \xi(x)\leq\varepsilon \qquad\text{on }M\setminus K_\varepsilon . \]

Choose a smooth bounded domain \(\Omega_\varepsilon\) with \(K_\varepsilon\Subset\Omega_\varepsilon\). On the bounded annular domain \(\Omega_\varepsilon\setminus\bar U\), the maximum principle gives

\[ h\leq\max\{0,\varepsilon\}=\varepsilon. \]

Indeed, the inner boundary gives \(h\leq0\), and the outer boundary lies in \(M\setminus K_\varepsilon\), where \(h\leq\varepsilon\). Letting \(\varepsilon\downarrow0\), we obtain \(h\leq0\) on \(M\setminus U\), namely \(\xi\leq C_0\mathscr G\). ◻

Taking the power \(2/n\) yields the Green-kernel gradient estimate

\begin{equation} \tag{5.3.10} \label{eq:green-gradient-estimate} |\nabla\mathscr G|^{\frac{n-2}{n-1}} \leq C u^{\frac{n-2}{n}}\mathscr G^{\frac2n} \qquad\text{on }M\setminus U. \end{equation}

Weighted Schoen–Simon–Yau inequality. Set

\[ \beta=\frac{n-2}{n}. \]

Simons’ identity and the refined Kato inequality for \(A\) give, on \(\{|A|\gt{}0\}\),

\begin{equation} \tag{5.3.11} \label{eq:green-simons-kato} |A|\Delta |A| \geq \frac2n|\nabla|A||^2-|A|^4. \end{equation}

Let \(\widetilde w=|A|^\beta\). Then

\[ \begin{aligned} \Delta\widetilde w &=\beta |A|^{\beta-1}\Delta|A| +\beta(\beta-1)|A|^{\beta-2}|\nabla|A||^2 \\ &\geq \beta |A|^{\beta-2} \left[ \frac2n|\nabla|A||^2-|A|^4 \right] +\beta(\beta-1)|A|^{\beta-2}|\nabla|A||^2 . \end{aligned} \]

Again the gradient coefficient vanishes because

\[ \frac2n+\beta-1 = \frac2n+\frac{n-2}{n}-1=0. \]

Hence

\begin{equation} \tag{5.3.12} \label{eq:green-A-subsolution} \Delta\widetilde w+\frac{n-2}{n}|A|^2\widetilde w\geq0. \end{equation}

For \(0\lt{}t\leq1\), (5.3.2) gives

\[ \Delta u^t = t u^{t-1}\Delta u+t(t-1)u^{t-2}|\nabla u|^2 \leq -t|A|^2u^t. \]

Apply Lemma 5.3.6 to

\[ w=\widetilde w,\qquad v=u^t,\qquad W=\frac{n-2}{n}|A|^2,\qquad V=t|A|^2. \]

For \(\delta\geq0\), define

\[ z=|A|^{\beta(1+\delta)}u^{-t\delta}. \]

Then

\[ \Delta z \geq \left(t\delta-(1+\delta)\frac{n-2}{n}\right)|A|^2z. \]

Adding \(|A|^2z\) to both sides,

\begin{equation} \tag{5.3.13} \label{eq:green-z-subsolution} \Delta z+|A|^2z \geq \gamma |A|^2z, \qquad \gamma:= \frac2n+\left(t-\frac{n-2}{n}\right)\delta . \end{equation}

Assume \(\gamma\gt{}0\). Using (5.3.1) with test function \(z\psi\) and integrating by parts gives

\[ \begin{aligned} 0 &\leq \int_M |\nabla(z\psi)|^2-|A|^2z^2\psi^2 \\ &= \int_M z^2|\nabla\psi|^2 -\int_M z\psi^2(\Delta z+|A|^2z). \end{aligned} \]

Together with (5.3.13),

\[ \gamma\int_M |A|^2z^2\psi^2 \leq \int_M z^2|\nabla\psi|^2. \]

Put

\[ m:=1+\beta(1+\delta). \]

Taking \(\psi=\varphi^m\) and using Young’s inequality,

\begin{equation} \tag{5.3.14} \label{eq:green-weighted-ssy} \int_M |A|^{2m}u^{-2t\delta}\varphi^{2m} \leq C\int_M u^{-2t\delta}|\nabla\varphi|^{2m}, \qquad \varphi\in\operatorname{Lip}_c(M). \end{equation}

Indeed, the right-hand side before Young is

\[ C\int_M |A|^{2(m-1)}u^{-2t\delta}\varphi^{2m-2}|\nabla\varphi|^2, \]

and this is bounded by

\[ \frac12\int_M |A|^{2m}u^{-2t\delta}\varphi^{2m} + C\int_M u^{-2t\delta}|\nabla\varphi|^{2m}. \]

The parameter choice and the logarithmic cut-off. We now choose \(\varphi=\eta(\mathscr G)\). By the coarea formula,

\begin{equation} \tag{5.3.15} \label{eq:green-coarea} \begin{aligned} \int_M u^{-2t\delta}|\nabla\varphi|^{2m} &= \int_M u^{-2t\delta}|\eta'(\mathscr G)|^{2m} |\nabla\mathscr G|^{2m} \\ &= \int_0^\infty |\eta'(s)|^{2m} \left[ \int_{\Sigma_s} u^{-2t\delta}|\nabla\mathscr G|^{2m-1}\,d\sigma \right]ds . \end{aligned} \end{equation}

The estimate (5.3.10) gives

\[ u^{-t\delta}|\nabla\mathscr G|^{\beta(1+\delta)} \leq C u^{-t\delta+ \frac{(n-1)(n-2)}{n^2}(1+\delta)} \mathscr G^{\frac{2(n-1)}{n^2}(1+\delta)}. \]

We therefore impose

\begin{equation} \tag{5.3.16} \label{eq:green-choice-1} t\delta= \frac{(n-1)(n-2)}{n^2}(1+\delta). \end{equation}

Then

\[ u^{-2t\delta}|\nabla\mathscr G|^{2m-1} = |\nabla\mathscr G| \left( u^{-t\delta}|\nabla\mathscr G|^{\beta(1+\delta)} \right)^2 \leq C|\nabla\mathscr G| \mathscr G^{\frac{4(n-1)}{n^2}(1+\delta)}. \]

On \(\Sigma_s\) this becomes

\[ u^{-2t\delta}|\nabla\mathscr G|^{2m-1} \leq C|\nabla\mathscr G|s^{\frac{4(n-1)}{n^2}(1+\delta)}. \]

Using the flux identity (5.3.5), (5.3.15) becomes

\[ \int_M u^{-2t\delta}|\nabla\eta(\mathscr G)|^{2m} \leq C\int_0^\infty |\eta'(s)|^{2m} s^{\frac{4(n-1)}{n^2}(1+\delta)}\,ds. \]

For a logarithmic cut-off to close, we need

\begin{equation} \tag{5.3.17} \label{eq:green-choice-2} \frac{4(n-1)}{n^2}(1+\delta)\geq 2m-1 = 1+\frac{2(n-2)}{n}(1+\delta). \end{equation}

This inequality has a nonnegative solution \(\delta\) only for \(n=3\). In that case,

\[ \beta=\frac13,\qquad \delta=\frac72,\qquad t=\frac27, \qquad m=\frac52, \qquad \gamma=\frac12. \]

Then (5.3.14) reads

\begin{equation} \tag{5.3.18} \label{eq:green-final-ssy-n3} \int_M |A|^5u^{-2}\varphi^5 \leq C\int_M u^{-2}|\nabla\varphi|^5. \end{equation}

Also, the Green gradient estimate becomes

\[ |\nabla\mathscr G|^{1/2} \leq Cu^{1/3}\mathscr G^{2/3}. \]

Therefore

\[ u^{-1}|\nabla\mathscr G|^{3/2} \leq C\mathscr G^2, \]

and, on \(\Sigma_s\),

\[ u^{-2}|\nabla\mathscr G|^4 = |\nabla\mathscr G| \left(u^{-1}|\nabla\mathscr G|^{3/2}\right)^2 \leq C|\nabla\mathscr G|s^4. \]

Hence

\[ \int_M u^{-2}|\nabla\eta(\mathscr G)|^5 \leq C\int_0^\infty |\eta'(s)|^5s^4\,ds. \]

For \(R\gt{}1\), choose the logarithmic cut-off

\[ \eta_R(s)= \begin{cases} 0,&0\leq s\leq R^{-2},\\[2mm] 2+\dfrac{\log s}{\log R},&R^{-2}\lt{}s\lt{}R^{-1},\\[3mm] 1,&s\geq R^{-1}. \end{cases} \]

Then \(|\eta_R'(s)|=(s\log R)^{-1}\) on \((R^{-2},R^{-1})\) and \(0\) elsewhere. Putting \(\varphi=\eta_R(\mathscr G)\) in (5.3.18),

\[ \begin{aligned} \int_{\{\mathscr G\geq R^{-1}\}} |A|^5u^{-2} &\leq C\int_M u^{-2}|\nabla\eta_R(\mathscr G)|^5 \\ &\leq C\int_{R^{-2}}^{R^{-1}} \frac{s^4}{s^5(\log R)^5}\,ds \\ &= \frac{C}{(\log R)^5}\int_{R^{-2}}^{R^{-1}}\frac{ds}{s} = \frac{C}{(\log R)^4}. \end{aligned} \]

Letting \(R\to\infty\), the sets \(\{\mathscr G\geq R^{-1}\}\) exhaust \(M\) up to the end, and the right-hand side tends to zero. Hence

\[ |A|^5u^{-2}\equiv0. \]

Since \(u\gt{}0\), we conclude \(A\equiv0\). Thus the immersion is totally geodesic, and the connected complete image is an affine hyperplane in \(\mathbb{R}^4\). This proves Theorem 5.3.1.