In a paper by Yann Ollivier:

Let $x$ be a point in $X$, $v$ a small tangent vector at $x$, $y$ the endpoint of $v$, $w_x$ a small tangent vector at $x$, and $w_y$ the parallel transport of $w_x$ from $x$ to $y$ along $v$. If, instead of a Riemannian manifold, we were working in ordinary Euclidean space, the endpoints $x_0$ and $y_0$ of $w_x$ and $w_y$ would constitute a rectangle with $x$ and $y$. But in a manifold, generally these four points do not constitute a rectangle any more.

Indeed, because of curvature, the two geodesics starting along $w_x$ and $w_y$ may diverge from or converge towards each other. Thus, on a sphere (positive curvature), two meridians starting at two points on the equator have parallel initial velocities, yet they converge at the North (and South) pole. Since the initial velocities $w_x$ and $w_y$ are parallel to each other, this effect is at second order in the distance along the geodesics (Fig.).

Thus, let us consider the points lying at distance $\varepsilon$ from $x$ and $y$ on the geodesics starting along $w_x$ and $w_y$, respectively. In a Euclidean setting, the distance between those two points would be $|v|$, the same as the distance between $x$ and $y$. The discrepancy from this Euclidean case is used as a definition of a curvature.

Definition(Sectional curvature). Let $(X, d)$ be a Riemannian manifold. Let $v$ and $w_x$ be two unit-length tangent vectors at some point $x \in X$. Let $\varepsilon, \delta > 0$. Let $y$ be the endpoint of $v$ and let $w_y$ be obtained by parallel transport of $w_x$ from $x$ to $y$. Then $$d(exp_{x} \varepsilon w_x, exp_{y} \varepsilon w_y) = \delta (1 −\frac{\varepsilon^2}{2} K(v,w)+ O(\varepsilon^3 +\delta \varepsilon^2)) $$

when $\delta , \varepsilon \to 0$. This defines a quantity $K(v,w)$, which is the sectional curvature at $x$ in the directions $(v,w)$.

Question1 How can I derive the formula $d(exp_{x} \varepsilon w_x, exp_{y} \varepsilon w_y) = \delta (1 −\frac{\varepsilon^2}{2} K(v,w)+ O(\varepsilon^3 +\delta \varepsilon^2)) $ from figure?

Question2 How this definition of sectional curvature can be derived from its usual definition ($K(v,w)=\frac{\langle R(v,w)w, v\rangle}{\langle v,v\rangle \langle w,w \rangle - \langle v,w \rangle ^2}$)

The first equation doesn't really need to be 'derived' -- it is a definition. The point $y$ is defined by travelling a small distance $\delta$ in the direction of $v$. So $$\delta = d(x,y) = d\big(\exp_x(0),\exp_y(0)\big)~,$$ and this gives you the term of zeroth order in $\epsilon$. As explained in the text, there is no first-order term, and we then simply define $K(v,w)$ via the equation given.
• thanks for your answer But I think the formula is not definition and can be derived from figure what is definition of sectional curvature is coefficient of $-\frac{\varepsilon^2}{2}$ in this formula. – Sepideh Bakhoda Apr 10 '13 at 13:29
Define a variation $$d(x,y)=\delta,\ c(t):=\exp_x \ tv,\ t\in [0,\delta ]$$
$$f(s,t):=\exp_{c(t)}\ sw(t),\ s\in [0,\varepsilon ]$$ where $w$ is a parallel vector field of $w$ along $c$ So $l(s):={\rm length}\ f(s,\ ),\ l(0)=\delta$ so that $l(\varepsilon )=d(\exp_x \varepsilon w_x,\exp_y \varepsilon w_y)$ Hence $$l(s) =\int_0^\delta |f_t| dt$$ so that $$l' =0,\ l'' = \int - K (c',w)$$ by variation formulas
That is, $$l(s)=l(0) +\frac{s^2}{2} (\int -K) + O(s^3)$$