Recall that all inner product spaces of the same dimension are isometric.

For example, if $(M,\mathrm{g})$ and $(N,\mathrm{h})$ are Riemannian manifolds of the same dimension, then $(T_pM,\mathrm{g}_p)$ and $(T_qN,\mathrm{h}_q)$ are isometric inner product spaces. Let $\beta:T_pM\to T_qN$ be a (linear) isometry.

Question: Why isn't $\exp_q\!\circ\,\beta\circ\exp_p^{-1}$ a local isometry wherever it is defined?

To clarify, I know that it isn't a local isometry, at least in general. If it was, then I'd have a local isometry between neighborhoods of every point on each equidimensional Riemannian manifold, which is clearly absurd. I'm looking for an explanation as to why this composition is not a local isometry. I'm not looking for a counterexample.

I think my mental impasse comes from Gauss' lemma, but I'm not sure. Using the chain rule, shouldn't $\exp_{q*}\!\circ\,\beta\circ\exp_{p*}^{-1}$, as a composition of linear isometries, itself be a linear isometry?


2 Answers 2


The differential is an isometry of vector spaces. The local map between Manifolds is a radial isometry, but in general it will fail to map geodesic distance spheres from the base point to isometric distance spheres. Simply because these need not admit an isometry. So a path orthogonal to geodesic rays starting in the base points will be mapped onto a path with different length.

Just map a hyperbolic geodesic ball around some point to a Euclidean on in standard models, or to a spheric one. (Recall spheric trigonometry)...

Edit: actually this can be made much more precise and to some extent even quantified. If you want to learn more about the topic search for 'Comparison theorems in Riemannian Geometry'. The classic is Cheeger, Ebin (with exactly that keyword as title), which was quite an eye opener for me. Another key word is 'Jacobi fields', these are the vector fields which describe the tangent distortions (and are the objects which are investigate in the study of comparison theorems).


Exponential maps are in general not isometries (otherwise every Riemannian manifold would be flat), and the term "linear" does not make sense since general Riemannian manifolds do not have a linear structure.

  • $\begingroup$ It makes sense for vector spaces, though. My linear maps are between vector spaces. $\endgroup$ Jun 28, 2015 at 5:24
  • $\begingroup$ You are talking about the composition of the linear map $\beta$ with two maps between a tangent space and the manifold itself. Only $\beta$ is a linear isometry, the exponential maps are not. (Or maybe, since you are mentioning the chain rule, you are only thinking of the differential at the base point which would indeed be an isometry. However, this argument does not work at any other point.) $\endgroup$ Jun 28, 2015 at 5:32

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