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Extensions of functions on manifolds

Anonymous Sun 08 Sep 23:04:29 2019 No.10957120 View Reply Original Report

Quoted By: >>10957149 >>10957172 >>10957711 >>10958938

Let $M \subseteq \mathbb{R}^n$ be a compact manifold. If $f:M \to \mathbb{R}$ is differentiable, does a differentiable $F: \mathbb{R}^n \to \mathbb{R}$ with $F_{|M} \equiv f$ exist?

Anonymous

Anonymous Sun 08 Sep 2019 23:15:04 No.10957149 Report

Quoted By: >>10957170

>>10957120
what does $F_{|M}$ mean?

Anonymous

Anonymous Sun 08 Sep 2019 23:20:43 No.10957170 Report

Quoted By: >>10957174 >>10957177

>>10957149

it means the restriction, that is, $F_{|M}: M \to \mathbb{R}, x \mapsto F(x).$

Anonymous

Anonymous Sun 08 Sep 2019 23:21:48 No.10957172 Report

Quoted By: >>10957194

>>10957120
Your manifold is already given as embedded on R^n, so yes? Just take F = f, right?

Anonymous

Anonymous Sun 08 Sep 2019 23:22:23 No.10957174 Report

Quoted By: >>10957238

>>10957170
you know how to check your TeX here right? there is a thing on the top left of the post box that say $\LaTeX$ that you click, then copypaste your post in there to check that the TeX parses correctly

Anonymous

Anonymous Sun 08 Sep 2019 23:22:59 No.10957177 Report

Quoted By:

>>10957170
Meant to post: it means the restriction, that is, $F_{|M}: M \to \mathbb{R}, x \mapsto F(x)$ .

Anonymous

Anonymous Sun 08 Sep 2019 23:25:28 No.10957184 Report

Quoted By: >>10957190 >>10957196

how can M map to R if M is allowed to be N-dimensional?

Anonymous

Anonymous Sun 08 Sep 2019 23:28:36 No.10957190 Report

Quoted By:

>>10957184

For example by $x \mapsto 0$ . This is a trivial example, that is clearly extensible.

Anonymous

Anonymous Sun 08 Sep 2019 23:30:10 No.10957194 Report

Quoted By:

>>10957172
That manifold might be smooth curve or a sphere. Extending it is not trivial.

Anonymous

Anonymous Sun 08 Sep 2019 23:30:31 No.10957196 Report

Quoted By:

>>10957184
Imagine the manifold is a surface on which you can messure a temperature. So you could define f: M->R as function that maps a point of the manifold to it's temperature, to give a example.

Anonymous

Anonymous Sun 08 Sep 2019 23:35:31 No.10957217 Report

Quoted By: >>10957256

Get it as a boundary of a manifold with boundary, apply tubular neighborhood and smooth bump functions.

Anonymous

Anonymous Sun 08 Sep 2019 23:39:34 No.10957238 Report

Quoted By:

>>10957174
Not the same anon. But cool, I have never noticed it there. I thought it was just a logo, that tells you this board stupors latex.

Anonymous

Anonymous Sun 08 Sep 2019 23:44:57 No.10957256 Report

Quoted By:

>>10957217

could you be more elaborate

Anonymous

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Anonymous Mon 09 Sep 2019 02:39:01 No.10957711 Report

Quoted By: >>10958660 >>10958956

>>10957120
https://en.wikipedia.org/wiki/Tietze_extension_theorem

Anonymous

Anonymous Mon 09 Sep 2019 14:07:57 No.10958660 Report

Quoted By: >>10958951

>>10957711
It says continuous, not differentiable. Even in a real analysis where mostly smooth stuff is considered it was only listed in this form.

Anonymous

Anonymous Mon 09 Sep 2019 16:06:34 No.10958938 Report

Quoted By: >>10959026

>>10957120
This is a handwavy physicist way of thinking about this but I think it's on the right track. Imagine $F_0$ is initially defined to be zero everywhere except on $M$ , and then use this as initial conditions for the heat equation. After any finite time you will have a differentiable $F$ and the value on $M$ will not have changed much.

Anonymous

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Anonymous Mon 09 Sep 2019 16:13:17 No.10958951 Report

Quoted By: >>10958956 >>10959035

>>10958660
Smooth maps are homotopic to (piece-wise) continuous maps for closed $M$ . Without additional structure on $M$ (namely being a locally convex Banach) we cannot use resolution of unity and Hahn-Banach to glue extensions together.

Anonymous

Anonymous Mon 09 Sep 2019 16:15:00 No.10958956 Report

Quoted By:

>>10957711
>>10958951
anime loving math student =
socially retarded wage slave
Have fun getting cucked by the whole system, stemlet.

Anonymous

Anonymous Mon 09 Sep 2019 16:42:01 No.10959026 Report

Quoted By: >>10959193

>>10958938
I will pick up on the handwaviness: My M could be null-set, so "energetically" - what ever that means - it should not really be able to heat up things. If it would be, then I could use some reasonable smoothing kernel and could convolve things. also it "not changing much" is not good enough for what I need.

Anonymous

Anonymous Mon 09 Sep 2019 16:44:55 No.10959035 Report

Quoted By: >>10959116

>>10958951
what are you talking about. functions are always homotopic to each other by t f + (1-t) g. Are sure you understand the question?

Anonymous

Anonymous Mon 09 Sep 2019 17:11:22 No.10959116 Report

Quoted By: >>10959159

>>10959035
You're kidding, right? No one can be this stupid.

Anonymous

Anonymous Mon 09 Sep 2019 17:32:17 No.10959159 Report

Quoted By: >>10959208

>>10959116
i am referring to maps to $\mathbb{R}$ , which is obviously the case here.

Anonymous

Anonymous Mon 09 Sep 2019 17:42:14 No.10959193 Report

Quoted By:

>>10959026
Consider F_0 a delta function-like generalized function with support on M. Or just use the piecewise continuous function implied by that extension theorem. The whole point is using heat kernels to make the functions smooth.

"Not changing much" is for you to figure out the picky details of. There is no doubt a whole family of F_0 that will convolve to the F you want

Anonymous

Anonymous Mon 09 Sep 2019 17:46:50 No.10959208 Report

Quoted By:

>>10959159
>possibly disconnected $M$
>no singular values for Morse function
Regardless, Hahn-Banach would still be needed for paracompact $M$ .

Anonymous

Anonymous Mon 09 Sep 2019 19:09:17 No.10959378 Report

Quoted By:

https://math.stackexchange.com/questions/1216520/extending-a-function-on-a-submanifold-to-the-ambient-manifold-proof-of-a-prope

And then use a bump function to extend it from the neighbourhood of M to R?

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