Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Assume we have a Riemannian manifold $M$ embedded isometrically in $\mathbb{R}^n$ (It may not be closed or open). Let $V(i)$ be a smooth orthonormal vector field on $M$ (which can be extended to a basis in $T(\mathbb{R}^n)$ at every point). Given that a smooth function $f$ on $M$ can be extended to $\mathbb{R}^n$ smoothly (say, the extended function is $g$), can it be extended in a way such that gradient of $g$ lies in $TM$ for all points in $M$?

share|improve this question
1  
Why do you need to know? Which examples of M do you have in mind? –  Yemon Choi Jun 27 '13 at 7:02
1  
For M, I am taking the space of positive definite matrices. I have the Laplacian(Laplace beltrami operator) on it but it cant be extended to all of R^k(k=n^2). I want to know if it can be extended atleast for functions on M which can be extended to R^k. –  user35911 Jun 27 '13 at 7:13
3  
A tubular neighborhood of $M$ in $\mathbb{R}^n$ can be identified with its normal bundle. Extend to the normal bundle by making $f$ constant on the fibres. Take a smooth cut-off some distance away from $M$. –  Willie Wong Jun 27 '13 at 8:07
1  
@willie wong I dont think this should work since by this argument, any smooth function can be extended to R^n. It will work only for nicer manifolds. –  user35911 Jun 27 '13 at 8:35
2  
What do you mean by "embedded"? As long as $M\subset \mathbb{R}^N$ is a closed set, by Whitney extension theorem any smooth $f$ on $M$ extends to a smooth $\tilde{f}$ on $\mathbb{R}^n$. The manifold structure is not even necessary. –  Willie Wong Jun 27 '13 at 14:57

1 Answer 1

up vote 2 down vote accepted

No: Take a spiral in $\mathbb R^2$ like $\mathbb R_{>0}\ni t\mapsto t.e^{it}$ with the induced metric, $g(x,y)=x$. You cannot change $g$ such that its gradient is tangent to the spiral: You have problems at 0.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.