What is the class of differentiable n-dimensional manifolds that allow a differential structure, in which all transition maps are isometric? Note that isometric must be overlapping pieces of charts only; however, mapping from charts to manifold can be arbitrary non-linear function.

I know such manifolds exist. For example:

  1. Trivial example: a graph of a differentiable function is such manifold.

  2. A circle: I can construct an atlas of two charts: upper and lower parts that overlap by $\varepsilon$. Since charts overlap by intervals of equal length $\varepsilon$, I can build isometric transition map.

  3. A 2D torus ($\mathbb{T}^2$): I can build 4 charts that overlap by small rectangles. These rectangles can be made isometric to one another. Also, places where all 4 charts overlap can be made isometric as well.

This construction seems to fail for a 2D sphere. So, I'm wondering which class of n-dimensional differentiable manifolds satisfies given constraints on differential structure?


In mathematical terms: Which differentiable manifolds allow differential structure $\big\{\big(\phi_\alpha, U_\alpha\big)\big\}_{\alpha \in \mathcal{A}}$ such that: $\forall \alpha, \beta \in \mathcal{A}$, if $U_\alpha \bigcap U_\beta \not= \emptyset$, then the restriction $\phi_\alpha \circ \phi^{-1}_\beta\big|_{\phi_\beta (U_\alpha \cap U_\beta)}$ is an isometric map between sets $\phi_\beta (U_\alpha \cap U_\beta)$ and $\phi_\alpha (U_\alpha \cap U_\beta)$ in $\mathbb{R}^n$?

Note that neither $\phi_\alpha$ nor $\phi_\beta$ need not be isometric maps. Also note that $U_\alpha$ and $U_\beta$ need not be isometric to each other as well.

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    $\begingroup$ I don't understand what you mean by «transition maps are isometric». $\endgroup$ Dec 17, 2014 at 16:46
  • $\begingroup$ The transition map between any two charts that have overlapping open cover of a manifold is defined as: $$ \phi_1 \circ \phi^{-1}_2 : \mathbb{R}^n \to \mathbb{R}^n$$, where $\phi_2$ and $\phi_2$ are the corresponding chart diffeomorphisms. So, I want this transition maps be an isomorphism between images $\phi_1(U_1 \cap U_2)$ and $\phi_2(U_1 \cap U_2)$. $\endgroup$ Dec 17, 2014 at 17:30
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    $\begingroup$ IIn your comment and in your edit you are using the word «isomorphic» but you probably mean something else. $\endgroup$ Dec 17, 2014 at 17:50
  • $\begingroup$ Alright, I should not have rushed. I instead of isomorphisms I meant isometirc maps. My mistake. $\endgroup$ Dec 17, 2014 at 17:53

1 Answer 1


Do you mean: The transition maps as maps between subsets of $\mathbb{R}^n$ are isometries with respect to the standard Euclidean metric on $\mathbb{R}^n$?

Then the answer is manifolds admitting a flat metric. If the metric is complete, the manifold has universal cover $\mathbb{R}^n$ with the Euclidean metric and the manifold is $\mathbb{R}^n$ modulo some group of isometries. All the compact ones are covered by tori.

One way to prove this is to build a map from $\mathbb{R}^n$ to the universal cover of your manifold by using the exponential map (from Riemannian geometry) at a point and use that there are unique geodesics in a given homotopy class in non-positively curved spaces. The map then must be an isometry, and the original manifold a quotient of this by isometries.

In two dimensions, the compact ones are the torus and Klein bottle. In three dimensions, there are more, given by Euclidean Seifert fibered spaces.

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    $\begingroup$ Just an addendum: If the flat metric is not assumed to be complete, then there are many more examples. For instance, each smooth connected noncompact manifold with trivial tangent bundle admits such a metric. (E.g., every noncompact connected oriented 3-dimensional manifold.) $\endgroup$ Dec 17, 2014 at 20:27

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