Let $S^n$ be the standard unit $n$-sphere, embedded in Euclidian space as $S^n = \{ x \in {\mathbb{R}}^{n+1} | \| x \| = 1 \}$. Define geodesic distance as $d(x, y) = \arccos (x \cdot y)$, where $\cdot$ is the Euclidian dot product.

My lecture notes introduce geodesic distance on page 11, but they're quite hand-wavy about the proof that distance is subadditive.

To establish the triangle inequality, let $x, y, z ∈ S^{n}$, $\theta = \arccos(\langle x, y \rangle)$ and $\varphi = \arccos(\langle y, z \rangle)$. Then we can write $x = y \cos \theta + u \sin \theta$ and $z = y \cos \varphi + v \sin \varphi$ where $u$ and $v$ are unit vectors orthogonal to $y$. An easy calculation now gives $\langle x, z \rangle = \cos \theta \cos \varphi + \langle u, v \rangle \sin \theta \sin \varphi$.

  • Concretely, how do I find $u$ and $v$?

  • What is this "easy calculation"?

  • 1
    $\begingroup$ Geodetics on $S^n$ are arcs of maximal circles, whose length just depends on the subtended angle at the centre of the sphere. If we define a distance as "the length of the shortest path", it trivially fulfills the triangle inequality $d(X,Y)+d(Y,Z)\geq d(X,Z)$. $\endgroup$ – Jack D'Aurizio Aug 8 '16 at 19:37

You are projecting $x, z$ onto $y$ respectively: Consider the plane spanned by $x, y$ (let's say $x\neq \pm y$). Then one can find a unit vector in this plane which is orthogonal to $y$. Call this $u$. Then one has

$$ x = \langle x, y\rangle y + \langle x, u\rangle u.$$

By definition, $\arccos \langle x, y\rangle = \theta$, so $\cos\theta = \langle x, y\rangle$. Since$x, y, u$ are orthonormal to each other, this force $\langle x, u\rangle = \sin \theta$. Thus

$$ x = y\cos \theta + u\sin\theta .$$

(Similar for $z,v$).

(2) is really easy calculations, starting at

$$ \langle x,z\rangle = \langle y \cos\theta + u\sin\theta , y\cos\varphi + v\sin\varphi \rangle$$

and then expanding the right hand side.


Given three points on the $n$-sphere, there is a $3$-plane containing them and the origin; the intersection of this with the sphere is totally geodesic.

It suffices to prove things for $S^2,$ by whatever means you like.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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