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Let $M$ be a smooth manifold with boundary, endowed with a smooth Riemannian metric $g$.

Suppose $g$ is flat, and let $p \in \partial M$.

Is there an open neighbourhood of $p$ which is isometric to an open subset of the standard half-space $\mathbb{H}^n=\{ (x_1,...,x_n)| x_n \ge 0\}$, endowed with the standard Euclidean metric?

One approach would to be to use the exponential map at $p$ (which is a local isometry in the case with no boundary). However, I am not sure where $\exp_p$ is defined in this case.

I guess its domain will be the subset of inward-pointing vectors in $T_pM$.

Edit:

As seen by the example $M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \} $, the suggetsion to use the exponential map does not work. The problem is that to cover the boundary, we need to conside geodesics with initial velocities in $T_p\partial M$. But, if we consider these geodesics in the ambient space $\mathbb{R}^2$, then they are standard straight lines (which are tangent to the unit circle), hence do not lie in $M$ for any positive time.

So, indeed, the boundary "cannot be straightened out".

Chill2Macht
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Asaf Shachar
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2 Answers2

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I would say no, because you cannot assure that the boundary will be straight. Consider the disk $$ M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \}\,. $$ It is clearly flat, but because the only isometries in $\mathbb R^2$ are translations and rotations you cannot straighten out the boundary, not even locally.

I would suggest the following amendment: every flat Riemannian manifold with boundary is locally isometric to an open subset together with parts of its smooth boundary.

Martins Bruveris
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  • Thanks for the observation! I am not entirely sure about the validity of your last suggestion, though. Indeed, it seems to me very plausible that what you are saying is true, i.e that every flat manifold is locally isometric to an open subset together with parts of its smooth boundary. However, I am interested to see a more rigorous argument for this. Do you have such an argument? – Asaf Shachar Mar 11 '17 at 16:04
  • what is the definition of the boundary points? They are the ones with a homeomorphism into part a point on y-axis in the half-plane. Now you can carry over the metric to there. Now you need to show that by bending the boundary, it is possible to attain the same metric from Euclidean metric. Details? Uhh! :( – Behnam Esmayli Mar 11 '17 at 16:11
  • No, unfortunately I do not. At first I thought that since around every interiour point the manifold is locally isometric to Euclidean space, one would just extend this isometry to the boundary. But one does not of course have control over the size of this neighborhood. – Martins Bruveris Mar 11 '17 at 16:22
  • What you are saying is very interesting. I also thought on the problem of bounding the sizes of the "good" neighbourhoods, where $f$ is invertible. In fact, this was the main motivation which made me ask the question... This is relevant to another question I asked on isometric immersions between manifolds with boundary, as you may see here: http://math.stackexchange.com/questions/2171418/isometric-immersions-between-manifolds-with-boundary-are-locally-distance-preser if you are interested. (The relevant part to our discussion is after "Edit 2"). – Asaf Shachar Mar 14 '17 at 07:41
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The answer to the question is negative. The necessary and sufficient condition is that the geodesic is totally geodesic, i.e. the second fundamental form must vanish. Existence and rigidity questions for submanifolds are discussed in depth in volume 4 of Spivak's book. See also J. Bernt, S. Console and C. Enrique Olmos Submanifolds and holonomy, CRC Press, 2016.

Marc Troyanov
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