Affine structure on the circle whose atlas consists of homeomorphisms onto $\mathbb{R}$

Question

Let $S^1$ be the unit circle in the complex plane. An atlas on $S^1$ is a finite collection $\alpha = \{ (U_i, \phi_i)\}_{i=1,\ldots,n}$ of pairs, where $U_i$ is an open subset of $S^1$ and $\phi:U_i\to\mathbb{R}$ is an open embedding (a homeomorphism onto open subset $\phi_i(U_i)$).

An atlas is called affine if for any $i,j\in\{1,\ldots,n\}$ the corresponding transition map $\phi_j\circ\phi_i^{-1}:\phi_i(U_i\cap U_j) \to \phi_j(U_i\cap U_j)$ is a restriction of some affine map $g_{i,j}:\mathbb{R}\to\mathbb{R}$, $g_{i,j}(t) = at+b$, for some $a,b\in\mathbb{R}$ with $a\not=0$.

Question: do there exists an affine atlas on $S^1$ whose charts $\phi_i:U_i\to\mathbb{R}$ are homeomorphisms onto $\mathbb{R}$?

1. I suspect that the answer to the question is negative, but do not understand the obstructions for such atlases

2. The problem of existence of affine atlases with surjective charts is mentioned e.g. here.

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3. It is easy to construct an example of affine atlas with non-surjective charts.

Let $p:\mathbb{R}\to S^1$, $p(t)=e^{2\pi i t}$, be the universal cover of $S^1$, and let $\{U_i\}_{i=1,\ldots,n}$ be an open cover of $S^1$ by open intervals such that for all $i,j$ the intersection $U_i\cap U_j$ is either empty of connected. Then each interval $U_i$ is the image of some open segment $(a_i,b_i)$, defined up to a constant shift by an integer number. Let $\phi_i:U_i \to (a_i,b_i)$ be the inverse homeomorphism.

Claim. $\{ (U_i,\phi_i) \}_{i=1,\ldots,n}$ is an affine atlas.

Indeed, if $U_i\cap U_j$ is non-empty, and $\phi_i(U_i\cap U_j)=(c_i,d_i)$, then there is $k\in\mathbb{Z}$ such that $\phi_j(U_i\cap U_j)=(c_i+k,d_i+k)$ and $g_{i,j}=\phi_j\circ\phi_i^{-1}:(c_i,d_i) \to (c_i+k,d_i+k)$ is given either by

• $g_{i,j}(t) = t+k$, or by
• $g_{i,j}(t) = k + d_i + c_i - t$

so it is affine.

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4. Let $S=(0,-1)=-i$ and $N=(0,1)=+i$ be the south and north poles of the circle. Then the stereographic projections $\phi_S:S^1\setminus S \to \mathbb{R}$ and $\phi_N:S^1\setminus N \to \mathbb{R}$ constitute an atlas with surjective charts, and it is easy to see (and well known) that the transition map $g = \phi_N\circ \phi_S^{-1}: \mathbb{R}\setminus 0\to \mathbb{R}\setminus 0$ is given by $g(t) = 1/t$ and is not affine.

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I would be very grateful for any information about that question.

Answer 1

$\DeclareMathOperator\R{R}\DeclareMathOperator\Z{Z}$No.

Indeed affine circles have been classified in the 50s by Kuiper. These are

1. the standard circle $C_1=\R/\Z$ (complete)
2. exotic circles $C_t=\R_{>0}/\langle t\rangle$ (non-complete) for $t>1$.

Any interval inside one of these circles is affinely equivalent to a bounded interval in the real line, hence to the interval $\mathopen]0,\mathclose 1[$, which is not affinely equivalent to $\R$ (e.g., because the automorphism group of the former is cyclic of order 2 when that of the latter is infinite — or because the former is non-complete and the second is complete).

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Here's a plain classification-free argument. I claim that if a connected affine 1-manifold $X$ has an open subset $U$ isomorphic to $\R$, then $X=U$.

Proof: otherwise, let $x$ be an element of $\bar{U}-U$. Let $J$ be a small enough interval around $x$ and $u:\mathopen]-1,1\mathclose[\to J$ mapping $0$ to $x$. Let $v:\R\to U$ be an affine isomorphism, with $x=\lim_{+\infty}v$. So the partially defined map $w=u^{-1}\circ v$ is affine, and injective and well-defined near $+\infty$, such that $\lim_{+\infty}w=0$. But near infinity, a bounded affine map is constant. We get a contradiction.

Answer 2

The answer is no in a more general setting of $(G,X)$-structures (see https://en.wikipedia.org/wiki/(G,X)-manifold).

A couple $(G,X)$ consists of a Lie group $G$ acting on a (simply connected) manifold $X$ transitively and analytically (i.e if $g\in G$ acts as identity on a (non-empty) open subset $U\subset X$ then it acts as identity globally).

Now, a (connected) manifold $M$ is said to admit a $(G,X)$-structure if it has an atlas of charts with values in $X$ such that all transitions are restrictions of elements of $G$ acting on $X$. In this case, the universal cover $\widetilde{M}$ has a $(G,X)$-structure (by pulling back the one on $M$) and it admits a local diffeomorphism $D:\widetilde{M}\to X$ (called the developing map) which is constructed, roughly speaking, as analytically extending a local chart. If one proper charts $U\subset \widetilde{M}\to X$ is onto then there is no way to extend to a local diffeomorphism which contradicts the existence of the developing map.

In your example, $\mathbb{S}^1$ has a $(\operatorname{Aff}(\mathbb{R}), \mathbb{R})$-structure and if $U\to \mathbb{R}$ is a chart which is onto then it lifts to the universal cover to a chart which is also onto and this is impossible by the previous argument.