Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given two Riemannian Manifolds of dimension 2, and a point on each. If the scalar curvatures are isomorphic (as functions) in some neighbourhoods of these points, are then the manifolds necessarily locally isometric?

share|improve this question

3 Answers 3

The answer is 'no' for simple reasons: If the two Riemannian surfaces $(M_1,g_1)$ and $(M_2,g_2)$ have Gauss curvatures $K_1:M_1\to\mathbb{R}$ and $K_2:M_2\to\mathbb{R}$ respectively and there happen to be points $p_i\in M_i$ such that $K_1(p_1) = K_2(p_2)$ and such that $dK_i$ is nonvanishing at $p_i$, then there will always be $p_i$-neighborhoods $U_i\subset M_i$ and a local diffeomorphism $\phi:U_1\to U_2$ with $\phi(p_1)=p_2$ such that $K_1 = K_2\circ\phi$ on $U_1$. This is just basic calculus. Of course, there is no reason for $\phi$ to be a local isometry.

However, now set $L_i = |dK_i|^2_{g_i}$ and suppose, in addition, that $L_1(p_1)=L_2(p_2)$ and that $dK_i\wedge dL_i$ is nonvanishing in a neighborhood of $p_i$ for $i=1,2$. Then there will be $p_i$-neighborhoods $U_i\subset M_i$ and a unique local diffeomorphism $\phi:U_1\to U_2$ with $\phi(p_1)=p_2$ such that $K_1 = K_2\circ\phi$ and $L_1 = L_2\circ\phi$ on $U_1$. Using $\phi$, you can now compare $g_1$ with $\phi^*(g_2)$. If these are equal on $U_1$, then the two metrics are (obviously) locally isometric. If these are not equal, then there is no local isometry between the two metrics that carries $p_1$ to $p_2$.

If you happen to have the bad luck that $dK_i\wedge dL_i$ vanishes identically on a neighborhood of $p_i$ (while, still $dK_i$ is nonvanishing), then, locally, one can write $L_i = f_i\circ K_i$ for some (essentially unique) functions $f_i$, and these will have to be equal in a neighborhood of $K_i(p_i)$ or there is no isometry. However, even if this does hold, there might still be no isometry, you have to go to higher order.

For example, if $K_1(p_1) = K_2(p_2)$, each $dK_i$ is nonvanishing on a neighborhood of $p_i$, and there exists a function $f$ on a neighborhood of $K_i(p_1)$ such that $L_i = f\circ K_i$ on some neighborhood of $p_i$, then you might consider the function $J_i = \Delta K_i$ and hope that this is independent of $K_i$, and, if so, ask whether the map $\phi$ that satisfies $(K_1,J_1) = (K_2\circ\phi, J_2\circ\phi)$ is an isometry. Etc.

The point is, though, that, after a finite number of tests of this kind (assuming that some mild nondegeneracy conditions hold), you will be able to completely determine whether the metrics are locally isometric in neighborhoods of the points $p_i$.

share|improve this answer
    
Welcome to math.SE, and thanks for the answer! It's much clearer than my own flailing attempt, anyways... –  Aaron Mazel-Gee May 10 '11 at 16:52

This is true if the curvatures are constant (Minding's theorem), but not in general. I actually don't know a counterexample, but (edit: I think that) Stillwell points out on page 344 of Mathematics and its history that Strubecker's Differentialgeometrie gives such an a related example in Volume III on page 124, but I'm not 100% sure if that is a counterexample of what you're asking, and I don't currently have access to it.


Update: Okay, here's an example I can access. In Audin's Geometry, after the Theorema egregium, there is a remark that the "converse is wrong; a classical counter-example will be found in Exercise VIII.24." Thankfully Exercise VIII.24 is available on Google Books, at least to me. There is a section with hints and solutions, too, but I can't see whether it includes that exercise.

share|improve this answer
    
The section with hints and solutions is not available to me. So how do you prove that the two surfaces in that exercise are non-isometric? –  Anirbit Dec 13 '10 at 20:01
    
@Anirbit: I'm sorry, I did not try the exercise, and I don't think I have access to the solutions either at present. If I find something I'll let you know, but I have no plans to do so soon. –  Jonas Meyer Dec 14 '10 at 0:05

I'm a little rusty on this stuff, but here's a sketch of a general answer. Basically the idea is that for any differential-geometric structure on a manifold (Riemannian metric, almost-complex structure, Finsler structure, symplectic structure, jet structure, web structure, etc.) you define the "frame bundle" to be the space of ways of identifying the structure with the canonical one on $\mathbb{R}^n$. So in the case of a Riemannian metric, point by point you choose an identification of the tangent space with $\mathbb{R}^n$ so that the inner product agrees with the usual one on $\mathbb{R}^n$, and then you allow yourself to post-compose with any element of the orthogonal group $O(n)$ in order to allow for differences in the original choices. (This is because the orthogonal group is exactly the space of linear maps that preserves the inner product.) This gives you what is known as a G-structure on the manifold (here $G=O(n)$). This is a bundle over the manifold where every fiber is a $G$-torsor, i.e. a space which is topologically the same as $G$ but there's no choice of which point is the identity because there was no canonical choice of how to identify $T_pM$ with $\mathbb{R}^n$. (So e.g. a real line is a torsor for the group $\mathbb{R}$.) The next step is to try and make a connection for this bundle, and if I remember correctly, this is the place where we tease out geometric invariants. There's a Lie-algebra-valued 1-form (with values in $Lie(G)$) that measures the failure of the connection to be trivial(izable), I think, and then you calculate something called the "Spencer cohomology". The main point is that this gives all the invariants, so for example it falls out as an immediate consequence that two manifolds are locally isomorphic if and only if there's a local diffeomorphism that lines up their Gaussian curvatures. (Something else really cool is that this perspective tells you exactly why we can have the Riemann mapping theorem in dimension 2, but then in all higher dimensions the only conformal maps are Möbius transformations. Also this can be used to prove Darboux's theorem, and the same method gives the Nijenhaus tensor.)

share|improve this answer
    
And the answer to your question is that I don't remember, but it will be yes if and only if scalar curvature determines Gaussian curvature. Embarrassingly, even though I remember the story I just told you, I don't remember whether that's true. I suspect not, though? –  Aaron Mazel-Gee Oct 16 '10 at 22:47
    
I'm rusty on my Riemannian geometry, too, but scalar curvature is a constant multiple of the Gaussian curvature in the 2-dimensional case. –  Jonas Meyer Oct 16 '10 at 23:00
2  
... the constant being exactly two. –  Willie Wong Oct 17 '10 at 0:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.