Intrinsic curvature of curves and surfaces and a Gauss–Bonnet theorem in the Heisenberg group

Abstract

We use a Riemannnian approximation scheme to define a notion of intrinsic Gaussian curvature for a Euclidean $C^{2}$-smooth surface in the Heisenberg group $\mathbb {H}$ away from characteristic points, and a notion of intrinsic signed geodesic curvature for Euclidean $C^{2}$-smooth curves on surfaces. These results are then used to prove a Heisenberg version of the Gauss–Bonnet theorem. An application to Steiner’s formula for the Carnot–Carathéodory distance in $\mathbb {H}$ is provided.

Appendix: Examples

We want to collect here a list of explicit examples where we compute the intrinsic Gaussian curvature explicitly.

Example 8.6

Any vertically ruled surface Euclidean $C^{2}$-smooth surface $\Sigma$ has vanishing intrinsic Gaussian curvature, i.e. if

$$\begin{aligned} \Sigma = \{ (x_1, x_2, x_3)\in \mathbb {H}: f(x_1,x_2)=0\}, \end{aligned}$$

for $f \in C^{2}(\mathbb {R}^2)$, then ${\mathcal {K}}_0 =0$. In particular, every vertical plane has constant intrinsic Gaussian curvature ${\mathcal {K}}_0 = 0$.

Example 8.7

The horizontal plane through the origin, $\Sigma = \{ (x_1,x_2,x_3) \in \mathbb {H}: x_3=0\},$ has

$$\begin{aligned} {\mathcal {K}}_0 = -\dfrac{2}{(x_{1}^{2} + x_{2}^{2})}. \end{aligned}$$

Example 8.8

The Korányi sphere, $\Sigma = \{ (x_1,x_2,x_3) \in \mathbb {H}: (x_{1}^2 + x_{2}^2)^2 + 16 x_{3}^2 -1 =0 \},$ has

$$\begin{aligned} {\mathcal {K}}_0 = -\dfrac{2}{(x_{1}^2 + x_{2}^2)} + 6 (x_{1}^2 + x_{2}^2). \end{aligned}$$

Example 8.9

Let $\alpha >0$. The paraboloid, $\Sigma = \{ (x_1, x_2, x_3)\in \mathbb {H}: x_3 = \alpha (x_{1}^2 + x_{2}^2)\},$ has

$$\begin{aligned} {\mathcal {K}}_0 = - \dfrac{1}{1+ 16 \alpha ^2} \, \dfrac{2}{(x_{1}^2 + x_{2}^2)}. \end{aligned}$$

Example 8.10

Every surface $\Sigma$ given as a $x_3$-graph, $\Sigma = \{ (x_1,x_2,x_3)\in \mathbb {H}: x_3 = f(x_1,x_2)\}$, with $f\in C^{2}(\mathbb {R}^2)$, has

$$\begin{aligned} {\mathcal {K}}_0 = -\dfrac{2}{\Vert \nabla _{ H }u \Vert _{ H }^{2}}+ \dfrac{1}{\Vert \nabla _{ H }u \Vert _{ H }^{4}} (\mathrm {Hess}f) \left( \nabla _{ H }u, J \nabla _{ H }u\right) , \end{aligned}$$

where $u(x_1, x_2, x_3):= x_3 - f(x_1,x_2)$.

For $x_3$-graphs, we have another useful result that provides a sufficient condition for the intrinsic Gaussian curvature ${\mathcal {K}}_0$ to vanish.

Lemma 8.11

Let $\Sigma \subset \mathbb {H}$ be as before. Let $g \in \Sigma$ and suppose that $\Sigma$ is a Euclidean $C^{2}$-smooth $x_3$-graph and $X_1u$ and $X_2u$ are linearly dependent in a neighborhood of g. Then ${\mathcal {K}}_0(g)=0$.

Proof

In the case of a $x_3$-graph, $X_3 u =1$ and we have

$$\begin{aligned} {\mathcal {K}}_0 = -\frac{1}{|\nabla _0 u|^2} + \frac{X_2 u X_1(\tfrac{1}{2}|\nabla _0 u|^2) - X_1 u X_2(\tfrac{1}{2}|\nabla _0 u|^2)}{|\nabla _0 u|^4}. \end{aligned}$$

Assume that $aX_1 u + bX_2 u =$ in a neighborhood of g. Let us suppose that $b \ne 0$; the case $a \ne 0$ is similar. Without loss of generality, assume that $b=1$. We expand

$$\begin{aligned} {\mathcal {K}}_0 = \frac{-(X_1u)^2-(X_2u)^2 + X_1uX_2uX_1X_1u+(X_2u)^2X_1X_2u-(X_1u)^2X_2X_1u-X_1uX_2uX_2X_2u}{|\nabla _0 u|^4} \end{aligned}$$

(8.4)

and use the identities $X_2u=aX_1u$,

$$\begin{aligned} X_1X_2u=aX_1X_1u,\\ X_2X_1u=X_1X_2u-X_3u=aX_1X_1u-1 \end{aligned}$$

and

$$\begin{aligned} X_2X_2u=aX_2X_1u=a-a^2X_1X_1u \end{aligned}$$

to rewrite the numerator of (8.4) entirely in terms of $X_1u$ and $X_1X_1u$. A straightforward computation shows that the expression for ${\mathcal {K}}_0$ vanishes. The case when $a\ne 0$ is similar. $\square$

Example 8.12

Every surface $\Sigma$ given as a $x_1$-graph, $\Sigma = \{ (x_1,x_2,x_3)\in \mathbb {H}: x_1 = f(x_2,x_3) \}$ with $f\in C^{2}(\mathbb {R}^2)$, has

$$\begin{aligned} {\mathcal {K}}_0= & {} -\dfrac{f_3^2}{\Vert \nabla _{ H }u \Vert _{ H }^2} + \dfrac{(x_1^2-x_2^2) f_{33}}{8\, \Vert \nabla _{ H }u \Vert _{ H }^4} \left( x_1 \, f_3 + \dfrac{x_1 x_2 \, f_3^2}{2}\right) -\dfrac{f_{23} (1+ \tfrac{x_2}{2}f_3)}{\Vert \nabla _{ H }u \Vert _{ H }^2} \\&+ \dfrac{1}{\Vert \nabla _{ H }u \Vert _{ H }^4}\left( \dfrac{x_1^2 \, f_3^3}{8} + \dfrac{(1+\tfrac{x_2}{2}f_3)}{2}f_3\right) , \end{aligned}$$

where $u(x_1, x_2, x_3):= x_1 - f(x_2,x_3)$. A similar result holds for $x_2$-graphs.