# Two questions on isometric embedding

According to the answer of the following question, I try a new version:

An special isometric embedding

1. Let $M$ be a Riemannian manifold and $\gamma$ be a small part of a geodesic. Is there an isometric embedding of $M$ into some $\mathbb{R}^{n}$ which sends $\gamma$ into a small part of a straight line?

2. Assume we have a foliation of the upper half plane $\mathbb{H}^{2}$ by geodesics. Is there an embedding of $\mathbb{H}^{2}$ into some $\mathbb{R}^{n}$ which sends the family of geodesics to a family of straight lines?

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By Nash's Theorem there is an isometric embedding of M into some euclidean space. It can be chosen $C^2$ (even $C^\infty$) so geodesics must be sent to geodesics in euclidean space, that is straight lines. –  ThiKu Mar 9 '14 at 14:26
But ... how does this fit with Degtyarev's example? –  ThiKu Mar 9 '14 at 14:29
@user39082 an isometric embedding does not necessarily send geodesic to geodesic. –  Ali Taghavi Mar 9 '14 at 14:31
Right, I just noticed immediately after sending. –  ThiKu Mar 9 '14 at 14:32

To show (2) is impossible, suppose one geodesic in the foliation is sent to a straight line in $\mathbb R^n$. Then consider the Jacobi field on this geodesic corresponding to "differentiating the foliation" (i.e. looking at nearby geodesics) and look at its image in $\mathbb R^n$. Since Jacobi fields in $\mathbb H^2$ have exponential growth and the embedding $\mathbb H^2\to\mathbb R^n$ is isometric, this image vector field along the line in $\mathbb R^n$ also has exponential growth. It follows that the nearby geodesics in the folation are not sent to straight lines.

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I don't know the answer to question #1, especially if you want a global embedding of the manifold. But if you'll settle for an isometric embedding in a neighborhood of a geodesic segment, here is one possible way to do it (I don't know if it works or not):

1) First, construct an embedding that is isometric up to infinite order along the geodesic and maps it into a straight line. This is straightforward, if you know the requisite theorems or tricks. One is to use the Cartan-Kähler theorem. The other is to use induction over the dimension of the submanifold and use Cauchy-Kovalevski to extend the near-isometric embedding one dimension at a time.

2) Now you want to use the Nash-Moser implicit function theorem to deform this embedding that is approximately isometric near the geodesic into one that is isometric near the geodesic. To do this, you need to invert the linearized operator and apply a smoothing operator and verify that you can do this without moving the geodesic.

3) If the Euclidean space has enough dimensions, the linearized operator is just a pointwise underdetermined system of linear algebraic equations (no PDE's). So what you want to do is concoct some additional linear constraints on the linearized solution that force the solution to be zero along the geodesic.

4) Finally, you need to show that there exists a 1-parameter family of smoothing operators on functions that converge to the identity and that are equal to the identity along the geodesic.

I don't have a copy handy, but I vaguely remember that maybe Gromov does things similar to this in his book Partial Differential Relations.

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For the second question the answer is obviously negative since distinct Euclidean lines cannot converge to each other (in the sense that the distance function goes to zero). For the first question, I am not sure, you may have to take a look at a proof of Nash theorem to see if it can be relativized.

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Is not possible that the foliating geodesics would be mapped to a familly of half straight line approaching to the (deleted) origion? –  Ali Taghavi Mar 9 '14 at 18:21
Yes, this is impossible, since an isometric embedding preserves arclength of curves. –  Misha Mar 9 '14 at 18:23
What if the foliating family of geodesics doesn't have any pair which share a point on the boundary? –  John Pardon Mar 9 '14 at 18:24
John: Then argue that Euclidean rays cannot diverge exponentially fast. –  Misha Mar 9 '14 at 18:30
John: I see, this is the same as your argument. Incidentally, I was confusing a and the, not for the first or last time. –  Misha Mar 9 '14 at 18:37