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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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 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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