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Let $\mathbb{S}^n$ be the $n$-dimensional unit sphere, equipped with the standard round Riemannian metric.

Let $f:\mathbb{S}^n \to \mathbb{S}^n$ be a diffeomorphism and suppose that for every (parametrized) geodesic $\gamma$, $f \circ \gamma$ is also a (parametrized) geodesic.

Must $f$ be an isometry? (that is the restriction of an orthogonal matrix on $\text{O}(n+1)$).

An equivalent condition on $f$ is that $\nabla df=0$ where $\nabla=\nabla^{(T\mathbb{S}^n)^*} $ $ \otimes \nabla^{f^*T\mathbb{S}^n}$ is the relevant tensor product connection.

Asaf Shachar
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    Is there any counterexample if we consider diffeomorphism $f:M\to M$ with same properties? ($M$ is closed arbitrary manifold). A related old post – C.F.G Nov 05 '20 at 20:30
  • Thanks, this was my next natural question. I asked about it here: https://math.stackexchange.com/questions/3896934/is-every-geodesic-preserving-diffeomorphism-an-isometry – Asaf Shachar Nov 06 '20 at 20:07

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Yes, such an $f$ must be an isometry.

Let $p\in S^n$, $v\in T_p S^n$ with $v\neq 0$. By polarization, it is enough to show that that $\|d_p f(v)\|= \|v\|$.

So, let's establish this equality. To do so, first note that $\gamma:\mathbb{R}\rightarrow S^n$ defined by $\gamma(t) = \exp_p(tv)$ is injective on $\left[0,\frac{2\pi}{\|v\|}\right)$.

Because $f$ is a diffeomorphism, $f\circ\gamma$ is injective on the same interval. But since $f$ maps geodesics to geodesics, we know $f\circ \gamma = \exp_{f(p)}(td_p f(v))$. This implies that $\|d_p f(v)\| \leq \|v\|.$

Further, because $\gamma$ fails to be injective on $\left[0, \frac{2\pi}{\|v\|}\right]$, the same must be true of $f\circ \gamma$. This implies that $\|d_pf(v)\| \geq \|v\|$. The two inequalities together now give the result.

Jason DeVito - on hiatus
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  • Thanks! Just to be sure I understand: Are you in fact showing that for every $v \in T_pM$ which corresponds to a periodic geodesic, $||df_p(v)||=||v||$? (not just on the sphere.) – Asaf Shachar Nov 06 '20 at 15:14
  • Actually, on a second thought, I think that your argument uses the fact that all the "closed geodesic loops" have the same length, right? you are using that $\gamma$ and $f \circ \gamma$ stop minimizing after the same time/length. – Asaf Shachar Nov 06 '20 at 15:30
  • Yes, to the second point. – Jason DeVito - on hiatus Nov 06 '20 at 15:33
  • @JasonDeVito The key argument, as noticed Asaf Shachar, seems to be that all geodesics are closed and have the same length. Would this mean that if $(M,g)$ is a Zoll manifold with all geodesics of the same length (I once read it as "Besse manifold"), then this result is also true? – Didier Nov 11 '20 at 16:22
  • @DIdier_: That certainly seems to be the case. – Jason DeVito - on hiatus Nov 11 '20 at 16:27
  • @DIdier_: Just a comment: "Besse" has another book with this title: Manifolds all of whose Geodesics are Closed. and Zoll’s Surfaces is in chapter 4. – C.F.G Nov 11 '20 at 16:34
  • @C.F.G Now that I remember it, I heard the name "Besse manifold" in a talk in which the orator said he called these manifolds Besse as a tribute to this book. Thank you for the reference. – Didier Nov 11 '20 at 16:36