3
$\begingroup$

Let $F\colon \mathbb{R}^n \to S^1$ be a smooth mapping.

Then, I strongly suspect that there must be a smooth function $f\colon \mathbb{R}^n \to \mathbb{R}$ such that \begin{equation} F(x) = \exp \big( if(x) \bigr) \end{equation}

Moreover, $f$ must be unique up to integer multiple of $2\pi$.

Could anyone please tell me how to prove this or correct me if I am wrong?

$\endgroup$
1
  • 1
    $\begingroup$ I think it is about this :notice $F_{*}(\pi_1(\mathbb{R}^n,\cdot)) = \{0\}\subseteq p_{*}(\pi_1(\mathbb{R},\cdot))$ where $p$ is the exponential function, which is a covering. When this condition is met, from nice spaces as those manifolds, $F$ lifts to a map into $\mathbb{R}$. Moreover, the lift has degeneracy upto each $2\pi k$ which is degeneracy of $p$ for a fixed base point $\endgroup$ Commented Jul 10 at 12:32

1 Answer 1

7
$\begingroup$

This is true. Recall that the map $\phi\colon \mathbb{R} \to S^1$, $\phi(t) = \exp(it)$ is a smooth covering map. By the smooth covering correspondence and the fact that $\mathbb{R}^n$ is contractible, any smooth map $f\colon \mathbb{R}^n \to S^1$ lifts to the covering, i.e. there's a smooth map $\tilde{f}\colon \mathbb{R}^n \to \mathbb{R}$ such that $f = \phi \circ \tilde{f}$, which is precisely what you were asking for. The uniqueness condition then follows from the fact that $\phi(t) = \phi(t')$ holds exactly if $t = t' + 2n \pi$ for some $n \in \mathbb{Z}$.

$\endgroup$
1
  • 3
    $\begingroup$ (+1, but being a bit too picky:) The uniqueness condition then follows from the fact that $\phi(t)=\phi(t')$ holds exactly if $t=t′+2n\pi$ for some $n \in \Bbb Z$ and from the continuity of $f$ :) $\endgroup$
    – Didier
    Commented Jul 10 at 12:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .