Simons’ Inequality and Generalized Bernstein Theorems | Gaoming Wang

Simons’ Identity and Inequality

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

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

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

◻

Classification of Stable Minimal Cones for n<= 6

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

Proposition 2.2.1

Suppose \(\mathbf C\) is a minimal cone in \(\mathbb{R}^{n+1}\), smooth away from the origin. Then we have the following Simons-type inequality on \(\mathbf C\setminus\{0\}\):

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

Proof. Choose \(\{ e_i \}\) to be a local orthonormal frame on \(\mathbf C\setminus\{0\}\) such that \(e_n\) is the radial direction \(\partial_r\). Recall the Simons identity for minimal hypersurfaces in \(\mathbb{R}^{n+1}\):

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

It suffices to show

\[ |\nabla A|^2 \geq 2 \frac{|A|^2}{r^2}+|\nabla |A||^2. \]

Since \(\mathbf C\) is a cone, we have \(A_{in}=0\) and \(A_{ij,n}=-\frac{1}{r}A_{ij}\) for \(i,j\lt{}n\). Indeed,

\[ A(rp)=\frac{A(p)}{r}. \]

Taking derivative in the radial direction gives

\[ A_{ij,n} (rp)= - \frac{A_{ij}(p)}{r^{2}}= -\frac{1}{r}A_{ij}(rp). \]

So we have

\[ \sum_{i,j,k}^{}A_{ij,k}^{2}=\sum_{\alpha,\beta=1}^{n-1}3 A_{\alpha\beta,n}^{2} + \sum_{\alpha,\beta,\gamma=1}^{n-1}A_{\alpha\beta,\gamma}^{2} = 2 \frac{|A|^2}{r^2} + \sum_{\alpha,\beta=1}^{n-1}A_{\alpha\beta,n}^{2}+ \sum_{\alpha,\beta,\gamma=1}^{n-1}A_{\alpha\beta,\gamma}^{2}. \]

At a fixed point, choose \(\{e_\alpha\}_{\alpha=1}^{n-1}\) so that

\[ A_{\alpha\beta}=\lambda_\alpha\delta_{\alpha\beta}. \]

Then

\[ |\nabla |A||^2 =\frac{1}{|A|^2}\sum_{k=1}^{n}\Big(\sum_{\alpha=1}^{n-1}\lambda_\alpha A_{\alpha\alpha,k}\Big)^2 =\frac{1}{|A|^2}\sum_{\beta=1}^{n-1}\Big(\sum_{\alpha=1}^{n-1}\lambda_\alpha A_{\alpha\alpha,\beta}\Big)^2 +\frac{1}{|A|^2}\Big(\sum_{\alpha=1}^{n-1}\lambda_\alpha A_{\alpha\alpha,n}\Big)^2. \]

By Cauchy–Schwarz and \(A_{\alpha\alpha,n}=-\frac1rA_{\alpha\alpha}\),

\[ |\nabla |A||^2 \leq \sum_{\alpha,\beta=1}^{n-1}A_{\alpha\alpha,\beta}^2+\frac{|A|^2}{r^2} \leq \sum_{\alpha,\beta,\gamma=1}^{n-1}A_{\alpha\beta,\gamma}^2+\sum_{\alpha,\beta=1}^{n-1}A_{\alpha\beta,n}^2. \]

Therefore,

\[ |\nabla A|^2 =2\frac{|A|^2}{r^2} +\sum_{\alpha,\beta=1}^{n-1}A_{\alpha\beta,n}^2 +\sum_{\alpha,\beta,\gamma=1}^{n-1}A_{\alpha\beta,\gamma}^2 \geq 2\frac{|A|^2}{r^2}+|\nabla |A||^2. \]

This completes the proof. ◻

A similar computation yields

\[ \frac{1}{2}\Delta|A|^{2}\geq p |\nabla |A||^{2} + (3-p)\frac{|A|^{2}}{r^{2}} - |A|^4, \]

for any \(p\leq 1+\frac{2}{n-1}\).

Theorem 2.2.2

Let \(\mathbf C^n \subset \mathbb{R}^{n+1}\) be a minimal nonflat cone, smooth away from the origin. If \(n \leq 6\), \(\mathbf C\) is unstable. If \(n \geq 7\), there exists a stable minimal nonflat cone.

Proof. Suppose \(\mathbf C\) is stable. Testing the stability inequality with \(|A|\varphi\), we get

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

Hence, we obtain

\[ 2 \int_{ } \frac{|A|^{2}}{r^{2}}\varphi^{2} \leq \int_{ } |A|^{2}|\nabla \varphi|^2. \]

Now choose

\[ \varphi= \max\{ 1,r \}^{1-\frac{n}{2}-2\varepsilon} r^{1+\varepsilon}. \]

We need to verify that this function is admissible. We compute

\[ \begin{aligned} \int_{ } |A|^{2}|\nabla \varphi|^2={}&(1+\varepsilon)^{2}\int_{ \{ r\lt{}1 \}}|A|^{2} r^{2\varepsilon}+(2-\frac{n}{2}-\varepsilon)^{2}\int_{ \{ r\geq 1 \}} |A|^{2} r^{-n+2-\varepsilon}\\ ={}&(1+\varepsilon)^{2}\int_{ 0}^1dr \int_{ \Sigma} |A_\Sigma|^{2} r^{2\varepsilon-2+n-1}d\Sigma\\ &+(2-\frac{n}{2}-\varepsilon)^{2}\int_{ 1}^\infty dr \int_{ \Sigma} |A_\Sigma|^{2} r^{-n-2\varepsilon+n-1}d\Sigma\\ ={}&(\int_{ 0}^1 r^{2\varepsilon+n-3}dr + (2-\frac{n}{2}-\varepsilon)^{2}\int_{ 1}^\infty r^{-2\varepsilon-1}dr)\int_{ \Sigma} |A_\Sigma|^{2} d\Sigma\lt{}+\infty \end{aligned} \]

Here, \(\Sigma=\mathbf C\cap S^n\) is the link of the cone, which is a smooth closed minimal hypersurface in \(S^n\). Hence, \(\varphi\) is admissible. On the other hand, we have

\[ \begin{aligned} 2\int_{ } \frac{|A|^{2}}{r^{2}}\varphi^{2}={}& 2\int_{ \{ r\lt{}1 \}} |A|^{2} r^{2\varepsilon} + 2\int_{ \{ r\geq 1 \}} |A|^{2} r^{-n+2-4\varepsilon}\\ \end{aligned} \]

When \(n\leq 6\), we can choose \(\varepsilon\gt{}0\) small enough such that \(2\gt{}(1+\varepsilon)^{2}\) and \(2\gt{}(2-\frac{n}{2}-\varepsilon)^{2}\). Hence \(|A|^2\equiv 0\), so \(\mathbf C\) is flat, a contradiction. Therefore, \(\mathbf C\) is unstable.

It remains to construct a stable minimal nonflat cone in \(\mathbb{R}^8\).

Define the Simons-type cone by

\[ \mathbf C_{p,q} =\left\{(x,y)\in \mathbb{R}^{p+1}\times \mathbb{R}^{q+1}:\ q|x|^2=p|y|^2\right\}. \]

Lemma 2.2.3

\(\mathbf C_{p,q}\) is a minimal cone in \(\mathbb{R}^{p+q+2}\), smooth away from the origin, and it is stable if and only if \(p+q\geq 6\).

One can verify directly that \(\mathbf C_{p,q}\) is minimal by computing its mean curvature. In particular, its principal curvatures are

\[ \kappa_1=\cdots=\kappa_p=\sqrt{\frac{q}{p}}, \qquad \kappa_{p+1}=\cdots=\kappa_{p+q}=-\sqrt{\frac{p}{q}},\quad \text{and}\quad \kappa_{p+q+1}=0. \]

Its second fundamental form satisfies \(|A|^{2}=\frac{n-1}{r^{2}}\). Choose \(X= \frac{\varphi^{2}}{r^{2}}x\) and insert it into the first variation formula. Then

\[ 0=\int_{ } \mathrm{div}^{\mathbf C_{p,q}}(X)=\int_{ } n\frac{\varphi^{2}}{r^{2}}-2\frac{\varphi^{2}}{r^{4}}|x|^2+2 \frac{\varphi \nabla^{\mathbf C_{p,q}} \varphi \cdot x}{r^{2}}=\int_{ } (n-2)\frac{\varphi^{2}}{r^{2}}+2 \frac{\varphi \nabla^{\mathbf C_{p,q}} \varphi \cdot x}{r^{2}}. \]

Hence

\[ \int_{ } (n-2) \frac{\varphi^{2}}{r^{2}}\leq 2 \sqrt{\int_{ } \frac{\varphi^{2}}{r^{2}} \int_{ } |\nabla^{\mathbf C_{p,q}} \varphi|^2} \implies \int_{ } \frac{(n-2)^{2}}{4} \frac{\varphi^{2}}{r^{2}} \leq \int_{ } |\nabla^{\mathbf C_{p,q}} \varphi|^2. \]

Since \(|A|^2=\frac{n-1}{r^2}\), this can be rewritten as

\[ \int_{ } \frac{(n-2)^{2}}{4(n-1)}|A|^{2} \varphi^{2} \leq \int_{ } |\nabla^{\mathbf C_{p,q}} \varphi|^2. \]

In particular, if \(n\geq 7\), then \(\frac{(n-2)^2}{4(n-1)}\geq 1\), so \(\mathbf C_{p,q}\) is stable. ◻

In fact, one has the stronger result:

Theorem 2.2.4

Each \(\mathbf C_{p,q}\) is area-minimizing if \(p+q\geq 6\) except for the case \(p,q= (1,5)\) or \((5,1)\).

The case \(\mathbf C_{p,p}\) with \(p\geq 3\) was proved by Bombieri–De Giorgi–Giusti [BDGG69]. Lawson [Law72] later proved the result for \(p+q\gt{}6\), and Simões [Sim74] handled the remaining cases \(p,q=(2,4)\) or \((4,2)\).

The original proof is quite involved and relies on calibrations. One seeks a vector field \(\xi\) such that \(\operatorname{div}^{\mathbf C} \xi=0\) and \(\xi=\nu_{\mathbf C}\) on \(\mathbf C\setminus\{0\}\), which implies that \(\mathbf C\) is area-minimizing. Finding such a calibration for \(\mathbf C_{p,q}\) is difficult and requires solving an ODE.

Here we present a different proof due to De Philippis–Paolini [DPP09], which uses a sub-calibration argument together with an explicit construction of a sub-calibration for \(\mathbf C_{p,p}\).

Sub-minimal Sets and Sub-calibrations

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

Let \(\Omega\subset\mathbb{R}^n\) be an open set. For a set \(E\) with smooth boundary, the perimeter \(P(E,A)\) in a smooth bounded open set \(A\subset\Omega\) is the \((n-1)\)-dimensional Hausdorff measure of \(\partial E\cap A\).

Definition 2.3.1

A set \(E\) with smooth boundary is sub-minimal in \(\Omega\) if for every smooth bounded open set \(A\subset\Omega\) and every \(F\subset E\) with \(E\setminus F\subset\subset A\),

\[ P(E,A)\leq P(F,A). \]

Proposition 2.3.2

If \(E\) and \(E^c=\Omega\setminus E\) are both sub-minimal in \(\Omega\), then \(E\) is minimal in \(\Omega\).

Proof. Let \(A\subset\Omega\) be a smooth bounded open set, and let \(F\) satisfy \(E\triangle F\subset\subset A\). Define \(F'=E\cap F\subset E\) and \(F''=(E\cup F)^c\subset E^c\). Then \(E\setminus F'\subset\subset A\) and \(E^c\setminus F''\subset\subset A\). By sub-minimality:

\[ P(E,A)\leq P(F',A),\quad P(E^c,A)\leq P(F'',A). \]

Since \(P(E^c,A)=P(E,A)\) and \(P(F'',A)=P(E\cup F,A)\), we have

\[ 2P(E,A)\leq P(E\cap F,A)+P(E\cup F,A). \]

Using the identity

\[ P(E\cap F,A)+P(E\cup F,A)\leq P(E,A)+P(F,A), \]

we conclude \(P(E,A)\leq P(F,A)\), so \(E\) is minimal. ◻

Definition 2.3.3

Let \(E\) have smooth boundary. A \(C^1\) vector field \(\xi\) on \(\Omega\) is a sub-calibration of \(E\) if:

\(\xi(x)=\nu_E(x)\) (exterior unit normal) for all \(x\in\partial E\);
\(\operatorname{div}\xi(x)\leq0\) for all \(x\in E\);
\(\lvert\xi(x)\rvert\leq1\) for all \(x\in\Omega\).

Theorem 2.3.4

If \(E\) admits a sub-calibration \(\xi\), then \(E\) is sub-minimal.

Proof. Let \(A\) be a bounded open set, and let \(F\subset E\) with \(E\setminus F\subset\subset A\). Choose \(\eta_j\in C_c^1(A)\) with \(\eta_j=1\) on \(E\setminus F\), \(0\leq\eta_j\leq1\), and \(\bigcup_j\{x:\eta_j(x)=1\}=A\). Let \(\xi_j=\eta_j\xi\). Then

\[ \int_{E\cap A}\operatorname{div}\xi_j - \int_{F\cap A}\operatorname{div}\xi_j = \int_{E\setminus F}\operatorname{div}\xi \leq 0, \]

\[ \int_{E\cap A}\operatorname{div}\xi_j \leq \int_{F\cap A}\operatorname{div}\xi_j \leq P(F,A). \]

By the divergence theorem:

\[ \int_{E\cap A}\operatorname{div}\xi_j = \int_{\partial E\cap A}\langle\xi_j,\nu_E\rangle d\mathcal{H}^{n-1} = \int_{\partial E\cap A}\eta_j d\mathcal{H}^{n-1} \geq \mathcal{H}^{n-1}\bigl(\partial E\cap\{\eta_j=1\}\bigr). \]

Taking \(j\to\infty\) yields \(P(E,A)\leq P(F,A)\), so \(E\) is sub-minimal. ◻

Let \(n=2m\). The Simons cone is

\[ \mathbf C=\bigl\{(x_1,\dots,x_m,y_1,\dots,y_m)\in\mathbb{R}^{2m}: x_1^2+\dots+x_m^2=y_1^2+\dots+y_m^2\bigr\}. \]

Define

\[ \mathcal{C}=\bigl\{(x,y)\in\mathbb{R}^m\times\mathbb{R}^m: |x|\leq|y|\bigr\},\quad f(x,y)=\frac{1}{4}\bigl(|x|^4-|y|^4\bigr). \]

Define the vector field

\[ \xi=\frac{Df}{|Df|}. \]

Proposition 2.3.5

\(\xi\) is a sub-calibration for \(E\) in \(\mathbb{R}^{2m}\setminus\{0\}\), where \(E:=\{(x,y):|x|\leq |y|\}\), and \(-\xi\) is a sub-calibration for \(E^c\) in \(\mathbb{R}^{2m}\setminus\{0\}\).

Proof.

\(\lvert\xi\rvert=1\) everywhere, and \(\xi\) is the exterior normal to \(E\) on \(\partial E\). We compute

\[ Df=(|x|^2 x,\,-|y|^2 y), \qquad |Df|^2 = |x|^6 + |y|^6. \]

Hence

\[ \Delta f = (m+2)|x|^{2} - (m+2)|y|^{2},\quad D|Df|^{2}= 6\bigl(|x|^{4} x,\, |y|^{4} y\bigr). \]

Thus

\[ \begin{aligned} |Df|^3\mathrm{div}\xi ={}& |Df|^2 \Delta f - \frac{1}{2}\langle D|Df|^2, Df\rangle\\ ={}& \bigl((m+2)|x|^{2} - (m+2)|y|^{2}\bigr)(|x|^6 + |y|^6) - 3\bigl(|x|^{8} + |y|^{8}\bigr)\\ ={}& (m-1)|x|^8 - (m-1)|y|^8 + (m+2)|x|^{2}|y|^{6} - (m+2)|x|^{6}|y|^{2}\\ ={}& (m-1)(|x|^4-|y|^4)(|x|^4+|y|^4) + (m+2)|x|^{2}|y|^{2}(|y|^4 - |x|^4)\\ ={}& (|x|^4 - |y|^4)\bigl((m-1)(|x|^4 + |y|^4) - (m+2)|x|^2 |y|^2\bigr). \end{aligned} \]

Note that

\[ (m-1)a^{2} - (m+2)ab + (m-1)b^{2} \]

is nonnegative if and only if

\[ \mathrm{det}\,\begin{pmatrix} m-1 & -\frac{m+2}{2}\\ -\frac{m+2}{2} & m-1 \end{pmatrix} = (m-1)^2 - \frac{(m+2)^2}{4} = \frac{3m(m-4)}{4} \geq 0. \]

◻

Theorem 2.3.6

The Simons cone \(\mathbf C\) is area-minimizing in \(\mathbb{R}^8\).

Proof. By the preceding theorem, \(E\) and \(E^c\) are sub-minimal in \(\mathbb{R}^8\setminus\{0\}\). Since the origin has codimension \(\gt{}1\), the perimeter is unchanged, and hence \(E\) and \(E^c\) are sub-minimal in \(\mathbb{R}^8\). By the preceding proposition, \(E\) is minimal, so its boundary \(\mathbf C\) is area-minimizing. ◻

One may consider the gradient of the function

\[ \frac{1}{4}\left( q^{2}|x|^4- p^{2}|y|^4 \right) \]

in the case of the minimizing cone \(\mathbf C_{p,q}\).