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I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for $n=1$, all Riemannian manifolds are modeled on $\mathbb R$. When $n=2$, I believe that it suffices for the scalar curvature to vanish everywhere (this is certainly necessary). But my intuition is poor for higher-dimensional structures.

Put another way: given a Riemannian structure $g$ on a smooth manifold, when can I find coordinates $x^1,\dots,x^n$ so that $g_{ij}(x) = \delta_{ij}$?

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4 Answers 4

up vote 11 down vote accepted

If the Riemannian metric is twice differentiable in some co-ordinate system, then this holds in any dimension if and only if the Riemann curvature tensor vanishes identically.

In dimension 2, it suffices for the scalar curvature to vanish. In dimension 3, it suffices for the Ricci curvature to vanish. In higher dimensions, you need to have the full Riemann curvature vanish.

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It's probably worth saying a word or two about how to prove this. My suggested proof: 1) Show that parallel translation from one point to another one is independent of the path joining the two points, as long as the paths are homotopic 2) Fix an orthonormal basis of tangent vectors at a point and parallel translate them to everywhere in a neighborhood. 3) By the torsion-free property of the connection, these vector fields commute. By the Frobenius theorem, they form an orthonormal foliation. The flat co-ordinates are now easily constructed. –  Deane Yang Dec 6 '09 at 20:07
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If remember correctly, there is a more general result due to somebody, that any two Riemannian manifolds are locally isometric if and only if their curvature tensors are locally the "same". –  Greg Kuperberg Dec 6 '09 at 20:24
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In detail, the proof is at deltaepsilons.wordpress.com/2009/11/12/…. –  Akhil Mathew Dec 6 '09 at 21:09
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Greg and Deane: This is a special case of the local equivalence problem in riemannian geometry, which is something which the General Relativity community is very interested in. Namely, you have two solutions of Einstein equations and would like to know whether they are locally equivalent; that is, whether they are describing the same spacetime but in different coordinates. (The special case in this question is when one of the metrics is the flat metric in flat coordinates.) [continued below] –  José Figueroa-O'Farrill Dec 6 '09 at 21:31
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Deane: "I don't believe I've ever seen this result used for anything." If you're referencing Cartan's Theorem (the local version of Cartan-Ambrose-Hicks), then I believe it can be used to prove that any Riemannian manifold with $\nabla R = 0$ is locally isometric to a symmetric space. –  Jesse Madnick Dec 28 '13 at 8:11

Greg's comment on Deane's answer is sort of correct (given suitable hypotheses), but maybe a bit misleading in the context of this discussion. Since the character count doesn't allow it, I'm adding this comment as an "answer" (though it is not an answer to the original question).

There are non-isometric 2-spheres $S_1,S_2$ for which there is a diffeomorphism $f$ from $S_1$ to $S_2$ so that the curvature at each point $p \in S_1$ is equal to the curvature of $f(p)$ in $S_2$. For example, let $O$ be a curve in the plane with dihedral $D_2$ symmetry whose curvature has 4 critical points (is this called an "oval"? I forget). If $S$ is a surface of revolution of $O$, then $S$ is foliated (in the complement of two "poles") by "latitude" circles of constant curvature, and the value of the curvature moves monotonically between two extreme values as one moves from the "poles" to the "equator". One can easily produce nonisometric surfaces with "the same" curvature function. The length of the circle with a given curvature value is an invariant of the isometry type which is not captured by the curvature itself (thought of as smooth function on $S$).

I think this example (and more discussion) is in Berger's book "A panoramic view of Riemannian geometry".

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Deane already answered the question. I just want to add that knowing the existence of local flat coordinates (by the vanishing of the curvature) and actually finding the flat coordinates are two very different things. I've had "fun" in the past finding explicit flat coordinates for flat metrics and it can be nontrivial, albeit highly satisfying when you work them out!

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In principle the exponential map from a point would always give it to you, right? –  Greg Kuperberg Dec 6 '09 at 20:26
    
That's correct, but explicitly integrating the geodesic equations is often nontrivial, at least for me. In some cases the only I could it was by cooking up enough 'translations'. –  José Figueroa-O'Farrill Dec 6 '09 at 20:35
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I'm in full agreement with José here. It is much easier to build flat co-ordinates using mutually orthogonal vector fields and/or families of hypersurfaces, rather than radial geodesics and/or geodesic spheres. –  Deane Yang Dec 6 '09 at 21:53

This is just a rephrasing of Deane's answer, but let me add one general comment. To any Riemannian metric (or pseudo-Riemannian metric) $g$ on a manifold $M$, you can associate a Levi-Civita connection $\nabla : T_M \to T_M \otimes \Omega_M$, where $T_M$ is the cotangent sheaf and $\Omega_M$ is the cotangent sheaf. Just like how the de Rham $d : \mathcal{O}_M \to \Omega_M$ can be extended to $d : \Omega_M^i \to \Omega_M^{i+1}$, the connection $\nabla : T_M \to T_M \otimes \Omega_M$ can be extended to $\nabla : T_M \otimes \Omega_M^i \to T_m \otimes \Omega_M^{i+1}$.

Then being able to find local coordinates $x_i$ such that $g_{ij} = \delta_{ij}$ is equivalent to $\nabla^2 = \nabla \circ \nabla : T_M \to T_M \otimes \Omega_M^2$, which corresponds to the Riemann curvature tensor that Deane mentions, being zero. This should be reminiscent of the $d^2 = 0$ of de Rham cohomology, or homological algebra in general... ;-)

It is perhaps not-so-standard (or not as standard as I would like) to talk about connections in terms of sheaves, but this point of view is better because it generalizes better. What I said above works in all of the "standard" geometric categories: $C^\infty$, real analytic, complex analytic, super-, etc. (Maybe it even works in the algebraic category? But I'm not sure, as the notion of local coordinates is more complicated in algebraic geometry.)

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Wait, you can define connections on sheaves? Do you have a reference for that? –  Akhil Mathew Dec 7 '09 at 3:03
    
I'm not sure I see any difficulty in a formal definition of what a connection on a sheaf is (but please correct me if I'm wrong about this). The issue, I think, is whether any such connection exists or not on a given sheaf. But in the specific setting of the tangent and/or cotangent sheaves of a smooth manifold, this is not in question. On the other hand, I'm not sure that there is anything to be gained in using the language of sheaves here. I'd be interested in learning if there is. –  Deane Yang Dec 7 '09 at 3:52
    
Most standard treatments define a connection on a bundle $E$ as a map $\Gamma(E) \to \Gamma(E \otimes \Omega^1)$. If we work in the $C^\infty$ category, and if $\Gamma$ here denotes $C^\infty$ global sections, then I think this definition is the same as the sheaf definition because we have partitions of unity. On the other hand, if we work in algebraic or analytic categories, this definition in terms of global sections won't work. For example, in these categories we may not even have any global sections at all. But I haven't thought too much about this, someone should correct me if I'm wrong. –  Kevin H. Lin Dec 7 '09 at 13:05
    
Connections can be defined in a fairly general context (including sheaves). Keywords: "Grothendieck connection", "crystal", "infinitesimal site" –  S. Carnahan Dec 9 '09 at 6:33
    
What I'm wondering about, though, is whether Levi-Civita connections can be defined in a general algebraic context. –  Kevin H. Lin Dec 9 '09 at 14:05

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