Extending a smooth map

Question

When can I extend a smooth map $f:\mathbb{R^2}-\lbrace 0 \rbrace \to S^1$ to a smooth map $\tilde{f}:\mathbb{R^2} \to S^1$. For instance, consider $g(x,y)=(x,y)/\sqrt{x^2+y^2}$? Am I able to extend that to a smooth map $g$? My end goal is having the smooth map from $\mathbb{R^2}$ to $S^1$. It can be any map, this is just how I am trying to do it. Thanks!

Answer 1

If you are looking for a specific smooth function from $\mathbb R^2$ to $S^1$ you can use for example : $$f:\mathbb R^2\rightarrow S^1$$ $$f(x,y)=(\sin x, \cos x)$$ There are some other ways to construct smooth maps between manifolds. One way is to use the so called bump functions: http://en.wikipedia.org/wiki/Bump_function

These functions are smooth and have a compact support. If you extend the function to be zero in the rest of the manifold and nonzero inside its compact support you have a smooth map, from the manifold to $\mathbb R$ in this case.

Answer 2

Of course a necessary condition is that $f$ extend continuously to $\mathbb{R}^2$. Your map does not extend continuously: it takes a constant nonzero value along each ray through the origin, and this value is different for different rays, so $\lim_{P \rightarrow 0} f(P)$ does not exist.

Here is a criterion for continuous extension over a point:

$\bullet$ Theorem: Suppose $f: \mathbb{R}^n \setminus \{0\} \rightarrow \mathbb{R}$ is continuous. Then $f$ extends continuously to $0$ iff it is uniformly continuous on each bounded subset of $\mathbb{R}^n \setminus \{0\}$.

See $\S$ 10.11 of these notes for a proof, especially Corollary 10.48. The text takes $n = 1$ but the arguments work verbatim in the general case.

This adapts immediately to continuity of vector-valued functions, because a function $f = (f_1,\ldots,f_m): \mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous iff each of its coordinate functions $f_i: \mathbb{R}^n \rightarrow \mathbb{R}$ is continuous.

It is harder to give sufficient conditions for differentiability. I can think of one in one variable:

Theorem: Let $I$ be an open interval, let $a \in I$, and let $f: I \setminus \{a\} \rightarrow \mathbb{R}$. Suppose:
(i) $f$ extends continuously to $a$.
(ii) $f$ is differentiable at each $x \in I \setminus \{a\}$, and
(iii) $\lim_{x \rightarrow a} f'(x) = L$ exists.
Then $f$ is differentiable at $a$ and $f'(a) = L$.

I learned this result from Spivak's Calculus. A proof is given in $\S$ 5.5.5 loc. cit..

Off the top of my head, I'm not sure whether this result extends to differentiable functions of several variables. (Someone around here must know -- please tell me.)