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I am working my way through John G. Ratcliffes 'Foundations of Hyperbolic Manifolds', and while doing this, I need some help proving a theorem;

Theorem 2.1.5 A function $\phi: R \rightarrow S^n$ is a geodesic line if and only if there are orthogonal vectors $x,y\in S^n$ such that $\phi (t) = (\cos t)x+(\sin t)y$

An easy corollary of this is obviously that the geodesics of $S^n$ are its great circles, seeing as the geodesics are by definition the images of geodesic lines, and the above is clearly parametrizing a great circle on a sphere.

I have seen a lot of questions in here that are somewhat similar, but those questions are usually about proving it for the 'special case' of $n=2$, using knowledge about Riemannian Manifolds, which is something I have never heard about. So far, in this book, all I have been introduced to is the definitions;

Definition: A geodesic line in a metric space $X$ is a locally distance preserving function $\phi : R\rightarrow X$.

Definition: A great circle of $S^n$ is the intersection of $S^n$ with a 2-dimensional vector subspace of $R^{n+1}$

And then I have proved the following two lemmas which might be helpful;

Lemma 1: If $x,y,z\in S^n$ and $\theta(x,y)+\theta(y,z)=\theta(x,z)$ then $x,y,z$ are spherically collinear (Meaning that there is a great circle containing them). Here, $\theta$ is the new spherical metric given as the angle between two points that we know from our usual euclidean metric given by

$x\cdot y = |x| |y|\cos \theta (x,y)$

Lemma 2: A curve $\alpha: [a,b] \rightarrow S^n$ is a geodesic arc (ie. a distance preserving function) if and only if there are orthogonal vectors $x,y\in S^n$ such that $$ \alpha(t)=(\cos(t-a))x+(\sin(t-a))y,\quad b-a\leq \pi$$

My original idea to prove the Theorem was to take any three points in the image of $\phi$ and then try to show that they must be spherically collinear, ie. they lie on the same great circle. Thus, the image of $\phi(t)$ must be a great circle, which is parametrized as above. But I am not sure this will even work. At the same time, I have a feeling I need to use the two lemmas, combining them. Any help would be greatly appreciated.

  • https://math.stackexchange.com/questions/845828/geodesic-on-n-dimensional-sphere – ned grekerzberg Feb 19 '19 at 07:53
  • Once you have Lemma 2, you are pretty much done. But how did you prove it? The simplest proof that I know (given your definition of geodesics) requires one to know that geodesics are differentiable almost everywhere and, moreover, are absolutely continuous functions. This holds because geodesics are Lipschitz (but one one needs to invoke Rademacher-Stepanov theorem). – Moishe Kohan Feb 20 '19 at 01:57

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Well, there are plenty of proofs. If $P$ is a $2-$plane through the origin, the reflexion $\sigma _P$ through $P$ (ie the restriction to the sphere to the symmetry whose restriction to $P$ is $Id$ and $-id$ to $P^{\perp}$) is an isometry. let us prove that the fixed points (great circles) are therefore geodesics. Consider two points $p,q$ on this circle, which are closed enough, so that the shortest geodesics $[p,q]$ between them is unique. Then $\sigma _P [p,q]$ is another geodesic so we have equality, and $[p,q]$ is therefore contained in this circle. It is therefore a totaly geodesics submanifold of dimension 1, and therefore a geodesic.

Thomas
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    You are taking it for granted that minimizing geodesics between points which are sufficiently close to each other are unique. This is true but the proof is rather long (if you start with the definition of geodesics as local length-minimizers) and requires some knowledge of Riemannian geometry (i.e. the alternative definition of geodesics as curves of zero acceleration, etc). – Moishe Kohan Feb 20 '19 at 01:50