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.
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.