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|<r\}.

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}.

About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
This entry was posted in seminar and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s