10

Let $\gamma$ be a regular curve on the sphere. In a lecture, the following result was used

$$L(\gamma)=\frac 14 \int_{S^2} \sharp (\gamma \cap \xi ^\perp)d\xi$$

$\xi^\perp$ is the plane with normal $\xi$ going through the origin. $\sharp(\gamma\cap \xi^\perp)$ is the number of points in the intersection of the curve $\gamma$ and the plane $\xi^\perp$.

We're saying we can get the length of $\gamma$ by integrating the function $\xi\mapsto \sharp(\gamma\cap \xi^\perp)$ which counts intersections of $\gamma$ with moving planes. By symmetry this function is even, which means we count every point twice, and I can see why there should be a factor of $\frac 12$. I'm not sure about the $\frac 14$.

I'm having a hard time finding a proof for this result. What is a simple method of proving it?

  • 2
    I'm upvoting just because of awesome symbols used in the equation. Would be eager to learn what does it exactly mean. – SBF Nov 14 '15 at 22:08
  • 1
    On that note, perhaps define what your symbols mean. Different classes across the globe may use different symbols – theREALyumdub Nov 14 '15 at 22:17
  • 3
    If you know the formula has this form then you can see $1/4$ is the right factor by letting $\gamma$ be a great circle: in this case almost every plane intersects $\gamma$ in two points, so the integral over the the sphere is $2 \cdot 4 \pi$, which you must divide by $4$ to get $L(\gamma)=2 \pi$. As for the proof, you can probably prove it easily for geodesic segments and then use an approximation argument, since both sides send concatenation to addition. – Anthony Carapetis Nov 15 '15 at 01:27
  • @AnthonyCarapetis could you explain how to prove for geodesics and what the approximation argument should be? –  Nov 15 '15 at 08:28
  • You should be able to prove it for geodesic segments by just computing - you can assume by isometry invariance that the segment is in some canonical position (probably either around the equator or starting at a pole) and then perform the integral. Use piecewise geodesic approximation - you just need to establish rigorously that as the length of the segments shrinks to zero, the approximation will eventually cut each great circle the same number of times as the exact curve. There might be a little bit of subtlety here (might need better and better approximations for different great circles). – Anthony Carapetis Nov 15 '15 at 08:34
  • The argument in your link looks like it's probably more direct. Is there any particular part of it you need clarified? I don't have time to read and explicate the whole thing. – Anthony Carapetis Nov 15 '15 at 08:35
  • @AnthonyCarapetis I don't understand where the "structure equations for $SO(3)$" come from. –  Nov 15 '15 at 08:36
  • 1
    The matrix $(\gamma, e_2, e_3)^T$ is special orthogonal for each $s$, and thus its derivative lies in the Lie algebra $\mathfrak so(3)$; i.e. is skew-symmetric. You can conclude this by differentiating the condition $M(s) M^T(s) = I$. – Anthony Carapetis Nov 15 '15 at 08:42
  • @AnthonyCarapetis thank you! Lastly, how does the author conclude from the expression for $dp_{s,\tau}$ that $F^\ast (dv_{S^2})=\cos(\tau-\phi)d\tau ds$? –  Nov 15 '15 at 10:01

2 Answers2

2

Consider first a great circle. Each plane meet it in exactly 2 points, and the result follows. For a small arc of length $2\pi \over q$, the result follows because you need $q$ such arcs to cover the circle. The additivity give you the result for arcs of length ${p\over q}\times 2\pi$, and by continuity, for every arc of circle. Now, the result follows by writing a Riemann sum of the rhs as the Crofton formula for a broken geodesic approximating the curve : let $\alpha$ be fixed. Note that for a sufficiently small arc of a $C^1$ curve between two points $a,b$, the number of intersection with a given circle is the same (0 or 1) as its intersection with the geodesic $[a,b]$ if the angle with the direction $\vec { ab}$ is $>\alpha$. Now the total area of part of the sphere made of vector which make an angle $\leq \alpha$ with a given direction is $C\alpha$. If teh curve is nice, you can compute the RHS of Crofton formula, the number $\sharp (\gamma \cup \xi ^\perp)$ is bounded, and the result follows.

Thomas
  • 8,588
0

For convenience we let $\xi^\perp =\{v\in \mathbb{S}^2| \langle v,x\rangle =0\}$

If $f: \mathbb{S}^2\rightarrow \mathbb{R}$ by $ f(\xi ) =\sharp (\gamma\cap \xi^\perp)$, then note that the restriction $$f|\gamma(t)^\perp \geq 1$$ by $\gamma(t)$ : If $\xi\in \gamma(t)\perp$, then $\xi^\perp $ contains $ \gamma(t)$

We are sufficient to consider the case where $$\gamma(t) = (0,\cos\ t,\sin\ t),\ 0\leq t\leq \theta$$

Then $ f((\pm 1,0,0))=\infty$ and note that $A=\{v\in \mathbb{S}^2| \langle v,\gamma (t)\rangle \geq 0\ {\rm for\ some}\ t\}$ is a union of two lunes whose central angle is $t$.

Note that $f$ has value $1$ on $A-\{(1,0,0),(-1,0,0)\}$ and $f$ has value $0$ on the compliment of $A$. Here $$ \frac{1}{4}\int_{\mathbb{S}^2} f(\xi)d\xi = \frac{1}{4}\int_{A} f(\xi)d\xi = \frac{1}{4} \frac{2t}{2\pi} {\rm area}\ \mathbb{S}^2= t ={\rm length} (\gamma )$$ which completes the proof.

HK Lee
  • 20,532