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Suppose $(M,g_M),(N,g_N)$ are two Riemann manifolds of the same dimension, and $f$ is a totally geodesic diffeomorphism between them, is it true that $M,N$ must be isometric (probably not through $f$, but another isometric map $h$; and probably scaling by some constant, so $h^*g_N=cg_M$ for some constant c)?

I would guess the answer to be negative, but the only examples of totally geodesic diffeomorphism I have in mind are isometries and linear transformations in $\mathbb{R}^n$.

Richard
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  • What is a totally geodesic diffeomorphism? A constant map is not a diffeomorphism. – Deane Aug 03 '23 at 11:49
  • Thanks for the correction. – Richard Aug 03 '23 at 12:02
  • Could you also provide a definition of a totally geodesic map? – Deane Aug 03 '23 at 12:14
  • A totally geodesic map $f$ is such that its hessian $\nabla df$ vanishes everywhere. Equivalently, a map is totally geodesic if the image of every geodesic is again a geodesic. So if $f$ is both totally geodesic and diffeomorphic, then its inverse should also be totally geodesic. – Richard Aug 03 '23 at 12:38
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    A linear transformation on $\mathbb{R}^n$ has zero Hessian but is not necessarily an isometry. – Deane Aug 03 '23 at 12:41
  • What you call totally geodesic is usually called "affine." Indeed, flat tori provide many counter examples to your conjecture. – Moishe Kohan Aug 03 '23 at 12:52
  • Yes but it is a map from $\mathbb{R}^n$ to itself, and $\mathbb{R}^n$ is isometric to itself via the identity map. As I have mentioned, the totally geodesic diffeomorphism $f$ need not be isometric, but there might exist another isometry $h$ between $M,N$. – Richard Aug 03 '23 at 12:52
  • What would be a counterexample from flat tori? @MoisheKohan – Richard Aug 03 '23 at 12:54
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    Any two flat tori of the same dimension are affine-equivalent (see Deane's comment). But most of them are not isometric. (Start by working out 1-dimensional examples.) See also my answer here: https://math.stackexchange.com/questions/2301858/could-a-riemannian-metric-be-uniquely-determined-by-its-exponential-map/3614249#3614249 – Moishe Kohan Aug 03 '23 at 13:06
  • Ah I see. Could there be an example that is not locally isometric (to be more precise, there exist $x$ that has not neighborhood locally isometric to some neighborhood of $f(x)$)? It just seems to me that affine maps are too close to (local) isometry from the examples I have, but I guess there should be more gaps between them. – Richard Aug 03 '23 at 13:17
  • As I said, read my linked answer... – Moishe Kohan Aug 03 '23 at 13:22
  • I believe that a vanishing Hessian implies that constant speed geodesics map to constant speed geodesics, but the speed is not 1. If the speed is 1, then the map is necessarily an isometry. Affine maps between flat manifolds are totally geodesic maps that are not necessarily isometries. What I do not know is if there are any other examples. @MoisheKohan, do you know the answer to this? – Deane Aug 03 '23 at 15:40
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    @Deane: There is one more construction which locally amounts to rescaling the product metric by different factors, i.e. $a_1g_1+ a_2 g_2+...+a_kg_k$. But (locally) nothing else is possible. See my answer to http://math.stackexchange.com/questions/2301858. – Moishe Kohan Aug 03 '23 at 16:12

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