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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.

enter image description here

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 ($K(v,w)=\frac{\langle R(v,w)w, v\rangle}{\langle v,v\rangle \langle w,w \rangle - \langle v,w \rangle ^2}$)

Thanks in advance for your time.

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closed as off-topic by Benoît Kloeckner, David White, Carlo Beenakker, Kevin P. Costello, j.c. Oct 23 '13 at 9:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benoît Kloeckner, David White, Carlo Beenakker, Kevin P. Costello, j.c.
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Hi, welcome to Mathoverflow. As there are large number of user overlaps between MO and Math.StackExchange, generally we discourage immediate cross posting a question between the two forums. For future reference, please just post your question in one of the two places. If you don't get a response after a reasonable amount of time (say, about a week), then maybe it would be worthwhile considering cross posting to the other forum. Thanks. –  Willie Wong Apr 10 '13 at 12:48
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In the interest of not duplicating efforts, a partial answer has already been given on Math.SE: math.stackexchange.com/questions/356989/… –  Willie Wong Apr 10 '13 at 12:48
    
@Willie Wong, Ok,thanks. –  Sepideh Bakhoda Apr 10 '13 at 13:19
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1 Answer 1

To supplement the answer given at SE, I would mention that the definition via the Taylor expansion of the distance function is really a slick shortcut rather than a definition. It relies on the fact that the Gaussian curvature at $x$ of the surface obtained as the image of the plane spanned by $v$ and $w$ is the same as the sectional curvature in this 2-dimensional direction. So to convince yourself of the formula, you just need to understand it for surfaces. Trying to relate it to the general formula in terms of the curvature tensor is a red herring.

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