Curvdimension and curvature of a metric profile III

I continue from the previous post “Curvdimension and curvature of a metric profile II“.

Let’s see what is happening for $(X,g)$, a sufficiently smooth ( $C^{4}$ for example),  complete, connected  riemannian manifold.  The letter “ $g$” denotes the metric (scalar product on the tangent space) and the letter “ $d$” will denote the riemannian distance, that is for any two points $x,y \in X$ the distance $d(x,y)$ between them is the infimum of the length of absolutely continuous curves which start from $x$ and end in $y$. The length of curves is computed with the help of the metric $g$.

Notations.   In this example $X$ is a differential manifold, therefore it has tangent spaces at every point, in the differential geometric sense. Further on, when I write “tangent space” it will mean tangent space in this sense. Otherwise I shall write “metric tangent space” for the metric notion of tangent space.

Let $u,v$ be vectors in the tangent space at $x \in X$. When the basepoint $x$ is fixed by the context then I may renounce to mention it in the various notations. For example $\|u\|$ means the norm of the vector $u$ with respect to the scalar product $g_{x}$ on the tangent space $T_{x} X$  at the point $x$. Likewise, $\langle u,v \rangle$ may be used instead of $g_{x}(u,v)$;  the riemannian curvature tensor at $x$  may be denoted by $R$ and not by $R_{x}$, and so on …

Remark 2. The smoothness of the riemannian manifold $(X,g)$ should be just enough such that the curvature tensor is $C^{1}$ and such that for any compact subset $C \subset X$ of $X$, possibly by rescaling $g$, the geodesic exponential $exp_{x} u$ makes sense (exists and it is uniquely defined) for any $x \in C$ and for any $u \in T_{x} X$ with $\|u\| \leq 2$.

Let us fix such a compact set $C$ and let’s take a  point $x \in C$.

Definition 5. For any $\varepsilon \in (0,1)$ we define on the closed ball of radius $1$ centered at $x$ (with respect to the distance $d$) the following distance: for any $u,v \in T_{x} X$ with $\|u\| \leq 1$, $\| v\| \leq 1$ $d^{x}_{\varepsilon} (exp_{x} \, u, exp_{x} v) \, = \, \frac{1}{\varepsilon} d((exp_{x} \, \varepsilon u, exp_{x} \varepsilon v)$.

(The notation used here is in line with the one used in  dilation structures.)

Recall that the sectional curvature $K_{x}(u,v)$ is defined for any pair of vectors $u,v \in T_{x} X$ which are linearly independent (i.e. non collinear).

Proposition 1. Let $M > 0$ be greater or equal than $\mid K_{x}(u,v)\mid$, for any $x \in C$ and any non-collinear pair of vectors $u,v \in T_{x} X$ with $\|u\| \leq 1$, $\| v\| \leq 1$.  Then for any $\varepsilon \in (0,1)$ and any $x \in C$ $u,v \in T_{x} X$ with $\|u\| \leq 1$, $\| v\| \leq 1$ we have $\mid d^{x}_{\varepsilon} (exp_{x} \, u, exp_{x} v) - \|u-v\|_{x} \mid \leq \frac{1}{3} M \varepsilon^{2} \|u-v\|_{x} \|u\|_{x} \|v\|_{x} + \varepsilon^{2} \|u-v\|_{x} O(\varepsilon)$.

Corollary 1. For any $x \in X$ the metric space $(X,d)$ has a metric tangent space at $x$, which is the isometry class of the unit ball in $T_{x}X$ with the distance $d^{x}_{0}(u,v) = \|u - v\|_{x}$.

Corollary 2. If the sectional curvature at $x \in X$ is non trivial then the metric profile at $x$ has curvdimension 2 and moreover $d_{GH}(P^{m}(\varepsilon, [X,d,x]), P^{m}(0, [X,d,x]) \leq \frac{2}{3} M \varepsilon^{2} + \varepsilon^{2} O(\varepsilon)$.

Proofs next time, but you may figure them out by looking at the section 1.3 of these notes on comparison geometry , available from the page of  Vitali Kapovitch.

Sub-riemannian geometry from intrinsic viewpoint, course notes

Here are the course notes prepared   for a course at CIMPA research school on sub-riemannian geometry (2012):

Sub-riemannian geometry from intrinsic viewpoint ( 27.02.2012) (14.06.2012)

I want to express my thanks for the invitation and also my excuses for not being abble to attend the school (due to very bad weather conditions in this part of Europe, I had to cancel my plane travel).