Simons’ Identity and Inequality

Lemma 2.1.1

For a minimal hypersurface \(\Sigma^n\) in \(\mathbb{R}^{n+1}\), the second fundamental form \(A\) satisfies the following identity:

\[ \frac{1}{2}\Delta |A|^{2} = -|A|^4 + |\nabla A|^2. \]

Proof. Recall that the Ricci identity states that for any \(2\)-tensor \(T_{ij}\) on \(\Sigma\), we have

\[ T_{ij,kl} - T_{ij,lk} = R_{lkim} T_{mj} + R_{lkjm} T_{im}, \]

where \(R_{ijkl}\) is the Riemann curvature tensor of \(\Sigma\). The Gauss equation in Euclidean space gives

\[ R_{ijkl}=A_{ik}A_{jl}-A_{il}A_{jk}. \]

Hence

\[ \begin{aligned} \frac{1}{2}\Delta |A|^2 ={}& A_{ij}\Delta A_{ij} + A_{ij,k}^{2} \end{aligned} \]

Moreover,

\[ \begin{aligned} A_{ij}\Delta A_{ij} ={}& A_{ij}A_{ij,kk} = A_{ij}A_{ik,jk}\quad (\text{Codazzi equation}) \\ ={}& A_{ij}A_{ik,kj} + A_{ij}(R_{kjim}A_{mk}+R_{kjkm}A_{im}) \quad (\text{Ricci identity}) \\ ={}& A_{ij}A_{kk,ij} + A_{ij}(A_{ki}A_{jm}-A_{ji}A_{km})A_{mk}\\ &+ A_{ij}(A_{kk}A_{jm}-A_{km}A_{jk})A_{im} \quad (\text{Gauss equation, Codazzi equation}) \\ ={}&-A_{ij}A_{ji}A_{km}A_{mk}+ A_{ij}A_{jm}A_{mk}A_{ki}-A_{ij}A_{jk}A_{km}A_{mi}\quad \text{minimal}\\ ={}& -|A|^4. \end{aligned} \]

This proves Simons’ identity. ◻

Lemma 2.1.2

For the second fundamental form \(A\) of a minimal hypersurface \(\Sigma^n\) in \(\mathbb{R}^{n+1}\), we have

\[ |\nabla A|^{2} \geq \left( 1+\frac{2}{n} \right)|\nabla |A||^{2}. \]

Proof. Let \(\{e_i\}\) be a local orthonormal frame on \(\Sigma\) that diagonalizes \(A\) at a point, so \(A_{ij} = \lambda_i \delta_{ij}\) with \(\sum_i \lambda_i = 0\). Then

\[ |\nabla |A|^2|^2 = \left|\nabla \sum_i \lambda_i^2\right|^2 = \left(2\sum_i \lambda_i A_{ii,k}\right)^2. \]

Hence

\[ \bigl|\nabla|A|^2\bigr|^2 = 4\sum_k\Bigl(\sum_i \lambda_i A_{ii,k}\Bigr)^2 \leq 4 \sum_i \lambda_i^2 \sum_{i,k} A_{ii,k}^2 = 4|A|^2 \sum_{i,k} A_{ii,k}^2. \]

Since \(2|A|\,|\nabla|A|| = |\nabla|A|^2|\), we have

\[ |\nabla |A||^2 = \frac{|\nabla|A|^2|^2}{4|A|^2} \leq \sum_{i,k} A_{ii,k}^2. \]

Moreover,

\[ \begin{aligned} \sum_{i,k=1}^{n}A_{ii,k}^2 ={}& \sum_{i\neq k}^{} A_{ii,k}^2 + \sum_{i=1}^{n} A_{ii,i}^2 \\ \leq{}& \sum_{i\neq k}^{} A_{ii,k}^2 + \sum_{i=1}^{n}(\sum_{j\neq i} A_{jj,i})^{2} \\ \leq{}& \sum_{i\neq k}^{} A_{ii,k}^2 + (n-1)\sum_{i=1}^{n}\sum_{j\neq i} A_{jj,i}^2 \\ ={}& n \sum_{i\neq k}^{} A_{ii,k}^2\\ ={}& \frac{n}{2} \sum_{i\neq j,k}^{} A_{ik,i}^{2}+A_{ki,i}^{2} \end{aligned} \]

Therefore,

\[ \left( 1+\frac{2}{n} \right)|\nabla |A||^{2} \leq \sum_{i,k}^{}A_{i i,k}^{2}+\sum_{i\neq j,k}^{} A_{ik,i}^{2}+A_{ki,i}^{2} \leq |\nabla A|^2. \]

◻

Corollary 2.1.3

For a minimal hypersurface \(\Sigma^n\) in \(\mathbb{R}^{n+1}\), the second fundamental form \(A\) satisfies the following differential inequality:

\[ \Delta |A|^{2} \geq -2|A|^4 + 2\left( 1+\frac{2}{n} \right)|\nabla |A||^{2}. \]

Equivalently,

\[ |A|\Delta |A| +|A|^4\geq \frac{2}{n}|\nabla |A||^2. \]

Theorem 2.1.4

Let \(\Sigma^n\) be an (immersed) stable minimal hypersurface in \(\mathbb{R}^{n+1}\) with \(n \leq 5\). Suppose \(\Sigma\) has (intrinsic) Euclidean volume growth, i.e., there exists a constant \(C\gt{}0\) such that for all \(R\gt{}0\),

\[ |B^\Sigma_R| \leq C R^n, \]

where \(B^\Sigma_R\) is the intrinsic geodesic ball of radius \(R\) in \(\Sigma\). Then \(\Sigma\) must be a hyperplane.

Proof. We test the stability inequality with \(|A|^{p-1}\varphi\). Then

\[ \begin{aligned} \int_{ } |A|^{2p}\varphi^{2} \leq{}& \int_{ } |\nabla (|A|^{p-1}\varphi)|^2 \\ ={}& \int_{ } (p-1)^{2}|A|^{2p-4}|\nabla |A||^2 \varphi^2 + |A|^{2p-2}|\nabla \varphi|^2 + (2p-2)|A|^{2p-3}\varphi \langle\nabla |A|, \nabla \varphi\rangle. \end{aligned} \]

We multiply Simons’ inequality by \(|A|^{2p-4}\varphi^2\) and integrate by parts to obtain

\[ \begin{aligned} \frac{2}{n} \int |A|^{2p-4} |\nabla |A||^2 \varphi^2 \leq \int |A|^{2p} \varphi^2-(2p-3)|\nabla |A||^{2}|A|^{2p-4}\varphi^2 - 2|A|^{2p-3}\varphi \langle\nabla |A|, \nabla \varphi\rangle. \end{aligned} \]

Cancelling the \(|A|^{2p}\varphi^{2}\) term, we obtain

\[ \left( \frac{2}{n} + (2p-3)-(p-1)^{2} \right) \int |A|^{2p-4} |\nabla |A||^2 \varphi^2 \leq \int |A|^{2p-2}|\nabla \varphi|^2 + (2p-4)|A|^{2p-3}\varphi |\nabla |A|| |\nabla \varphi|. \]

If \(\frac{2}{n} + (2p-3)-(p-1)^{2} \gt{} 0\), i.e.,

\[ (p-2)^2\lt{}\frac{2}{n}, \]

then we can apply Cauchy–Schwarz to the right-hand side to get

\[ \int_{ } |A|^{2p-4}|\nabla |A||^2 \varphi^2 \leq C \int_{ } |A|^{2p-2}|\nabla \varphi|^2, \]

for some constant \(C\) depending on \(n\) and \(p\). Using the stability inequality again, we get

\[ \int_{ } |A|^{2p} \varphi^2 \leq C \int_{ } |A|^{2p-2}|\nabla \varphi|^2\leq C \left( \int_{ } |A|^{2p} \varphi^2 \right)^{\frac{p-1}{p}} \left( \int_{ } |\nabla \varphi|^{2p} \right)^{\frac{1}{p}}. \]

Hence,

\[ \int_{ } |A|^{2p} \varphi^2 \leq C \int_{ } |\nabla \varphi|^{2p}. \]

If we choose \(\varphi\) to be a cutoff function supported in \(B^\Sigma_{2R}\) with \(|\nabla \varphi|\leq \frac{C}{R}\), and equal to \(1\) in \(B^\Sigma_R\), then

\[ \int_{B^\Sigma_R} |A|^{2p} \leq C \int_{B^\Sigma_{2R}} \left( \frac{C}{R} \right)^{2p} \leq C R^{n-2p}. \]

Here the parameter \(p\) is half of the final integrability exponent, since the estimate controls \(|A|^{2p}\). Thus the decay requires \(2p\gt{}n\), equivalently \(p\gt{}n/2\). When \(n\leq 5\), the admissible interval

\[ 2-\sqrt{\frac{2}{n}}\lt{}p\lt{}2+\sqrt{\frac{2}{n}} \]

intersects \((n/2,\infty)\). Choose such a \(p\). Then \(n-2p\lt{}0\), and letting \(R \to \infty\) gives \(A \equiv 0\) on \(\Sigma\). Hence \(\Sigma\) is a hyperplane. ◻

Another useful hypothesis is extrinsic Euclidean volume growth: there exists \(C\gt{}0\) such that for all \(R\gt{}0\),

\[ |\Sigma \cap B_R(0)| \leq C R^n. \]

This condition implies intrinsic Euclidean volume growth.

Proposition 2.1.5

The following two statements are equivalent:

Every complete stable minimal hypersurface in \(\mathbb{R}^{n+1}\) with extrinsic Euclidean volume growth is a hyperplane. (This means \(\Sigma \cap B_R(0)\) has area at most \(C R^n\) for some constant \(C\) independent of \(R\).)
We have the following curvature estimate for stable minimal hypersurfaces: there exists \(C=C(n,\Lambda)\) such that if \(\Sigma^n \subset B_{2R}(0) \subset \mathbb{R}^{n+1}\) is a stable minimal hypersurface with

\[ |\Sigma \cap B_{2R}(0)| \leq \Lambda R^n, \]

then

\[ \sup_{\Sigma \cap B_R(0)} |A| \leq \frac{C}{R}. \]

Note that (ii) is scale-invariant. If it holds for some \(R\gt{}0\), then it holds for all \(R\gt{}0\) by scaling. This is because for \(\lambda\gt{}0\), the second fundamental form of \(\lambda\Sigma\) is \(\lambda^{-1}A\), and the area of \(\lambda\Sigma\cap B_{2\lambda R}(0)\) is \(\lambda^n |\Sigma\cap B_{2R}(0)|\).

Proof. (ii) \(\Rightarrow\) (i). Let \(\Sigma^n\subset \mathbb{R}^{n+1}\) be complete, stable, and minimal, with extrinsic Euclidean volume growth:

\[ |\Sigma\cap B_R(0)|\leq C_0 R^n \quad \forall R\gt{}0. \]

Fix \(p\in\Sigma\). For \(R\geq |p|\), the ball \(B_{2R}(p)\) is contained in \(B_{3R}(0)\), and hence

\[ |\Sigma\cap B_{2R}(p)|\leq C_0(3R)^n. \]

Apply (ii) after translating \(p\) to the origin, with \(\Lambda=3^nC_0\):

\[ \sup_{\Sigma\cap B_R(p)}|A|\leq \frac{C}{R}. \]

Let \(R\to\infty\). Then \(|A|(p)=0\). Since \(p\) is arbitrary, \(A\equiv 0\), so \(\Sigma\) is a hyperplane.

(i) \(\Rightarrow\) (ii). Assume (ii) fails. Then for some \(\Lambda\gt{}0\) there exist stable minimal hypersurfaces

\[ \Sigma_j\subset B_{2R_j}(0),\qquad |\Sigma_j\cap B_{2R_j}(0)|\leq \Lambda R_j^n, \]

such that

\[ \sup_{\Sigma_j\cap B_{R_j}(0)} |A_j|\,R_j \to \infty. \]

Choose \(y_j\in\Sigma_j\cap B_{R_j}(0)\) such that, with \(K_j:=|A_j|(y_j)\),

\[ K_jR_j\to\infty. \]

We use the standard point-picking argument in the ball centered at \(y_j\). Define

\[ G_j(x):=\left(\frac{R_j}{2}-|x-y_j|\right)|A_j|(x) \]

on \(\Sigma_j\cap B_{R_j/2}(y_j)\), and choose \(q_j\) where \(G_j\) attains its maximum. Since \(G_j(y_j)=R_jK_j/2\), if

\[ Q_j:=|A_j|(q_j),\qquad \rho_j:=\frac{1}{2}\left(\frac{R_j}{2}-|q_j-y_j|\right), \]

then

\[ Q_j\rho_j=\frac{G_j(q_j)}{2}\geq \frac{R_jK_j}{4}\to\infty. \]

Moreover, for \(x\in\Sigma_j\cap B_{\rho_j}(q_j)\),

\[ \frac{R_j}{2}-|x-y_j| \geq \frac{1}{2}\left(\frac{R_j}{2}-|q_j-y_j|\right), \]

and the maximality of \(G_j\) gives

\[ |A_j|(x)\leq 2Q_j. \]

Rescale:

\[ \widetilde\Sigma_j:=Q_j(\Sigma_j-q_j). \]

Then \(0\in\widetilde\Sigma_j\), \(|\widetilde A_j|(0)=1\), and on \(B_{\widetilde\rho_j}(0)\) with \(\widetilde\rho_j:=Q_j\rho_j\to\infty\):

\[ |\widetilde A_j|\leq 2. \]

Stability is scale-invariant, so each \(\widetilde\Sigma_j\) is stable minimal.

We now get the local area bound in the rescaled sequence. Because \(q_j\in B_{R_j/2}(y_j)\) and \(y_j\in B_{R_j}(0)\), we have \(q_j\in B_{3R_j/2}(0)\). Hence

\[ B_{R_j/2}(q_j)\subset B_{2R_j}(0). \]

Fix \(\sigma\gt{}0\). Since \(\rho_j\leq R_j/4\) and \(Q_j\rho_j\to\infty\), we also have \(Q_jR_j\to\infty\). For \(j\) large, \(\sigma/Q_j\leq \rho_j\) and \(\sigma/Q_j\leq R_j/2\). By the monotonicity formula, applied with center \(q_j\) and the two radii \(\sigma/Q_j\) and \(R_j/2\),

\[ \frac{|\Sigma_j\cap B_{\sigma/Q_j}(q_j)|}{(\sigma/Q_j)^n} \leq \frac{|\Sigma_j\cap B_{R_j/2}(q_j)|}{(R_j/2)^n} \leq 2^n\Lambda. \]

Hence

\[ |\widetilde\Sigma_j\cap B_\sigma(0)| =Q_j^n |\Sigma_j\cap B_{\sigma/Q_j}(q_j)| \leq 2^n\Lambda \sigma^n. \]

Thus \(\widetilde\Sigma_j\) have uniform local area growth and curvature bounds on larger and larger balls. By the compactness theorem for minimal immersions with locally bounded curvature and area, a subsequence converges smoothly on compact sets to a complete stable minimal hypersurface \(\Sigma_\infty\subset\mathbb{R}^{n+1}\) with Euclidean volume growth. By construction,

\[ |A_{\Sigma_\infty}|(0)=1. \]

But (i) says every such \(\Sigma_\infty\) is a hyperplane, so \(A_{\Sigma_\infty}\equiv 0\), contradiction.

Therefore (ii) must hold. ◻

Corollary 2.1.6

The curvature estimate in (ii) holds for \(n\leq 5\).

Theorem 2.1.7

Let \(u:\mathbb{R}^n\to \mathbb{R}\) be an entire solution of the minimal surface equation, with \(n\leq 5\). Then \(u\) is affine.

Proof. Let

\[ \Sigma=\{(x,u(x)):x\in \mathbb{R}^n\}\subset \mathbb{R}^{n+1} \]

be the graph of \(u\). By the calibration argument above, \(\Sigma\) is area-minimizing. In particular, \(\Sigma\) is stable.

We next prove Euclidean volume growth. After a translation, we may assume \(0\in \Sigma\). For a.e. \(R\gt{}0\), the intersection

\[ \Gamma_R:=\Sigma\cap \partial B_R \]

is a smooth closed \((n-1)\)-dimensional submanifold of the sphere \(\partial B_R=S_R^n\). Since

\[ H_{n-1}(S^n)=0, \]

the cycle \(\Gamma_R\) bounds an \(n\)-dimensional region \(D_R\subset S_R^n\). Choosing the smaller of the two sides of \(S_R^n\setminus \Gamma_R\), we may assume

\[ |D_R|\leq \frac12 |S_R^n|\leq C_n R^n. \]

Since \(\Sigma\) is area-minimizing and

\[ \partial(\Sigma\cap B_R)=\Gamma_R=\partial D_R, \]

we obtain

\[ |\Sigma\cap B_R|\leq |D_R|\leq C_n R^n \]

for a.e. \(R\gt{}0\).

Hence, by the Schoen–Simon–Yau theorem, \(\Sigma\) is a hyperplane. Since \(\Sigma\) is a graph over \(\mathbb{R}^n\), that hyperplane cannot be vertical. Thus

\[ u(x_1,\ldots,x_n)=a_1 x_1 + \cdots + a_n x_n + c \]

◻

Simons’ Identity and Inequality

Gaoming Wang

Assistant Professor