# 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.