Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

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.

As for the expression in terms of the Riemann tensor, your thinking is backwards. You should start from the geometric definition of the sectional curvature, and derive the given relationship with the Riemann tensor.

share|improve this answer
    
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
    
Well, the 'derivation' is what I wrote; I just don't think it deserves the name! ;-) But anyway, yes, the sectional curvature is just the coefficient of that term, by definition. –  Rhys Apr 10 '13 at 13:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.