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I know that if $f(t)= \cos (at)e_1+ \sin(at)e_2$, then $\ddot f(t)=-a^2 f(t)= \pm N(f(t))\perp T_{f(t)}S$, where $N(p)= \frac{\nabla f(p)}{\|\nabla f(p)\|}$ is the unit normal vector, $S= g^{-1}(1), g(x_1,x_2,...,x_{n+1})= \sum_{i=1}^{n+1} x_i^2$ and $T_{f(t)}S=$ the tangent space to $S$ at $f(t)$. Hence, it follows that $f$ is a geodesic on $S$.

I'm not sure how to show the converse: Given a geodesic $f$ on $S$, it is of the form $f(t)= \cos (at)e_1+ \sin(at)e_2; e_1,e_2$ are orthogonal vectors in $\mathbb R^{n+1}$.

Suppose that $f(t)$ is a geodesic on the sphere, then $\ddot f(t)= c f(t)$, because $N(f(t))= f(t)$. With this, one gets $f_i(t)= d_i e^{\sqrt c t}+d_i' e^{\sqrt{-c} t}$ so that $f(t)= (f_i(t))_{i=1,2,..., n+1}$. But I don't understand how to show from here that $f(t)$ is of the aforementioned form.

Koro
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    Does this answer your question? The Geodesics of a Sphere. – mathcounterexamples.net Aug 19 '23 at 19:45
  • @mathcounterexamples.net: No, it doesn't. 1) Not only their notations are drastically different, they also seem to be using some other definition of geodesic. 2) My question is about taking it forward from where I've left it off, i.e., bringing f(t) in the said form using $f_i$'s. – Koro Aug 19 '23 at 19:52
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    You now need to use the fact that $|f(t)|^2 =1$ for all $t$. This fact does imply that the exponential terms of real exponent in your solution all vanish, that is $d_i ' = 0$ since $c<0$. Now, each component is the sum of a cosine and a sin. This gives the desired representation of writing the solution. – Eric Aug 19 '23 at 20:08
  • @Eric: How do we conclude that $c<0$? And, shouldn't we only have that $|f(t)|=$ constant? How can it be shown that this constant is $=1$? – Koro Aug 19 '23 at 20:09
  • We actually don’t need that fact in the proof, since if $c>0$, then we would just get that $d_i=0$, because otherwise, one of the components would diverge, leaving the unit sphere. If c=0, the equation $f''=cf$ yields a line or a point, but a line would instantly leave the unit sphere, so the only possible case for $c=0$ is that $f(t)$ is constant. For the second question, remember that we are on the unit sphere, so $f(t)$ has length 1. – Eric Aug 19 '23 at 20:46
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    Try sin and cos, rather than exponentials. Why does $c$ have to be constant, anyhow? The right approach for $n=2$ is to consider $f\times\ddot f$. This generalizes if you know about exterior product of vectors (rather than of differential firms). – Ted Shifrin Aug 19 '23 at 21:13

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