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Lipschitz orthonormal frames on submanifolds of $\mathbf{R}^n$ ?

Suppose we are given a $d-$dimensional submanifold of $\mathbf{R}^n$ with a trivial normal bundle, whose $d-$dimensional volume is $V$ and has a non-self-intersecting tube of radius $r$ around it. Can one obtain an explicit upper bound $L$ such that there must exist a frame of orthonormal vector fields that is $L-$Lipschitz (with respect to the Euclidean norm on both the frame and the manifold)?

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 Have you worked this out when $d = 1$? That seems like a somewhat easier initial case to understand. – Deane Yang Jul 22 2010 at 0:38

Sorry, I meant an orthonormal frame for the normal bundle that is Lipschitz. Supposing the manifold is closed, When $d=1$ the manifold is $S^1$, so can't one just do parallel transport along the curve starting from a point $x$ all the way back and then use a constant-rate rotation to match up at $x$?

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 You might possibly be having a problem of browser losing the cookies. Thus you are shown as multiple users. You might want to contact the moderators to merge them. – Anweshi Jul 25 2010 at 6:53 And then normally you can comment on questions for additional remarks, without needing to add them as answer. – Anweshi Jul 25 2010 at 6:55 Thanks a lot, but I think I won't be shown as the first user any more, so it should be fine. – Hari Jul 25 2010 at 7:41

You will probably need to require $C^2$ smoothness of your submanifold. Take a simple example of $d=1$ and the manifold the graph of the function $f(x) = x^\alpha$ for $x > 0$ and $f(x) = 0$ if $x \le 0$. Then for $1 < \alpha < 2$, the normal bundle is only Hölder continuous, so no $L$ exists at $x = 0$. Or would you exclude this counterexample for the reason that the exponential map from the normal bundle is not a diffeomorphism onto any ball of radius $r > 0$ at $x = 0$?

Generally, you will probably need to have a bound on the extrinsic curvature, for which $C^2$ and compactness would suffice.

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