## Notes of Davide Vittone’s lecture

Height estimate for minimal surfaces in Heisenberg groups

With R. Monti.

1. The main result

Theorem 1 In ${\mathbb{H}^n}$, ${n\geq 2}$. There exist constats ${\epsilon_0}$, ${c_0}$, ${K}$ such that if ${E\subset\mathbb{H}^n}$ is a local minimizer of perimeter containing the origin, contained in horizontal cylinder ${C_1}$, and ${E}$ has small excess in direction ${X_1}$,

$\displaystyle \begin{array}{rcl} \mathrm{Exc}(E,C_1,X_1)<\epsilon_0, \end{array}$

then the intersection of ${E}$ with a slightly smaller cylinder is nearly contained in a strip in direction ${X_1}$,

$\displaystyle \begin{array}{rcl} \sup\{|p_1|\;\,p\in \partial E\cap C_{1/K}\}\leq c_0 \mathrm{Exc}(E,C_1,X_1)^{1/(4n+2)}. \end{array}$

1.1. Cylinders

Let ${D_r}$ be the intersection of the Koranyi ${r}$-ball with the hyperplane ${\{x_1=0\}}$. Let

$\displaystyle C_r:=\{p(s,0,\ldots,0)\,;\,p\in D_r,\,|s|

1.2. Excess

The excess of ${E}$ inside a horizontal cylinder ${C_r}$ in horizontal direction ${\nu}$ is

$\displaystyle \begin{array}{rcl} \mathrm{Exc}(E,C_r,\nu):=\frac{1}{2r^{Q-1}}\int_{\partial E\cap C_r}|\nu_E-\nu|^2 \,d\mathcal{P}er. \end{array}$

The inequality we prove can be thought of as a non linear version of Poincaré inequality.

It fails for ${\mathbb{H}^1}$: there, there are sets with constant horizontal normal, and therefore vanishing excess, which are not vertical planes.

It is a first step in a regularity theory. We expect a reverse Poincaré inequality of the form

$\displaystyle \begin{array}{rcl} \mathrm{Exc}(E,C_{1/K^2},X_1)\leq C' \,\sup\{|p_1|\,;\,p\in\partial E\cap C_{1/K}\}. \end{array}$

This would imply an estimate on excess,

$\displaystyle \begin{array}{rcl} \mathrm{Exc}(E,C_r(p),\nu_E(p))\leq C\,r^\alpha, \end{array}$

and one knows that this implies that the normal ${\nu_E}$ is ${C^\alpha}$, and the boundary is ${C^{1,\alpha}_{\mathbb{H}}}$.

2. Proof

Follows Schoen-Simon 1980.

2.1. Step 0

If ${\epsilon_0}$ is small enough, ${E}$ is contained in a bounded strip. This follows from the fact that vertical hyperplanes are minimizing.

2.2. Step 1

Let ${s_0}$ be such that hyperplane ${\{x_1=s_0\}}$ divides ${\partial E}$ into two pieces of equal perimeters.

Let ${s_1>s_0}$ be such that hyperplane ${\{x_1=s_1\}}$ divides ${\partial E}$ into a piece of perimeter ${<\sqrt{\mathrm{Exc}}}$.

Let ${s_2=\sup\{|p_1|\,;\,p\in\partial E\cap C_{1/4}\}}$.

We prove that ${s_2-s_1\leq C\,\mathrm{Exc}^{1/(2Q-2)}}$. Indeed, the density estimate states that

$\displaystyle \begin{array}{rcl} (s_2-s_1)^{Q-1}&\leq&\mathcal{P}er(\partial E\cap B_{s_2-s_1}(p))\\ &\leq&\mathcal{P}er(\partial E\cap C_{1/2}\cap\{x_1>s_1\})\\ &\leq&\sqrt{\mathrm{Exc}}. \end{array}$

2.3. Step 2

This is the serious step. We show that ${s_1-s_0\leq C\,\mathrm{Exc}^{1/2Q-2}}$. By Hölder,

$\displaystyle \begin{array}{rcl} \sqrt{\mathrm{Exc}}&\geq&\int_{\partial E\cap C_{1/2}}(1-\nu_E^1)\,d\mathcal{P}er\\ &\geq&\sqrt{2}\int_{\partial E\cap C_{1/2}}\sqrt{1-\nu_E^1}\,d\mathcal{P}er\\ &\geq&\sqrt{2}\int_{\partial E\cap C_{1/2}}\sqrt{1-(\nu_E^1)^2}\,d\mathcal{P}er\\ &\geq&\int_{-1/2}^{1/2}\mu_E^s(D_{1/2}^{s})\,ds. \end{array}$

Here, we have used the coarea formula, ${D_{1/2}^{s}=C_{1/2}\{p_1=s\}}$, and ${\mu_E^s}$ is the perimeter measure for the sub-Riemannian metric of hyperplane ${\{p_1=s\}}$. This is new.

Using the isoperimetric inequality (this is the place where the argument collapses for ${\mathbb{H}^1}$), this is estimated below by

$\displaystyle \begin{array}{rcl} \int_{s_0}^{s_1}\mathcal{L}^{2n}(E\cap\{p_1=s\})^{2n/2n+1}\,ds &\geq& \int_{s_0}^{s_1}(\mathcal{P}er(\partial E\cap C_{1/2}\cap\{p_1>s\})-C\,\mathrm{Exc})^{2n/2n+1}\,ds\\ &\geq&\int_{s_0}^{s_1}\sqrt{\mathrm{Exc}}^{2n/2n+1}\,ds. \end{array}$

2.4. Step 3

Let ${s_3=\inf\{|p_1|\,;\,p\in\partial E\cap C_{1/4}\}}$. Then ${s_0-s_3\leq C\,\mathrm{Exc}^{1/2Q-2}}$. For this, one uses the isometry ${x_1\mapsto -x_1}$, ${t\mapsto -t}$.