# Local Isomorphism on Topological Groups

I'm currently studying Lie Groups, and reading "Theory of Lie Groups I" by C. Chevalley. He talks about topological groups in chapter two. To be more precise, on page 38 he presents two examples in order to show that it's not always possible to extend a local isomorphism to the whole group. He says:

"Let $\phi$ be the mapping which assigns to every real number $x$ its residue class $\phi(x)$ modulo 1; let $f$ be the contraction of $\phi$ to the interval $]-1/4,+1/4[$. It is clear that f is a local isomorphism of $R$ into $T$, but these groups are not isomorphic."

Well, I've got a few doubts... Does it make sense to consider a residue class of a real number modulo 1? I don't understand that, since residue classes are taken to integers (sorry for my lack of knowledge if I'm wrong). Second, does "contraction" mean the same as in metric spaces? $T$ is the 1-dimensional torus (in general, the factor group $R^{n}/H$, where $H$ is the subgroup of $R^{n}$ where the elements have integers coordinates, is the $n$-dimensional torus $T^{n}$). Well, I didn't understand neither why this local isomorphism happens.

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Modulu 1 also means looking at fractional parts, e.g. the fractional part of $x$ is defined to be $x - \floor{x}$, where $\floor{x}$ is the floor function. –  user38268 Sep 12 '12 at 2:18
$\mathbb{Z}$ is a subgroup of $\mathbb{R}$ and you're considering the quotient map $\phi\colon \mathbb{R \to R/Z}$ –  t.b. Sep 12 '12 at 2:24
It appears that contraction here means restriction: $f=\varphi\upharpoonright(-1/4,1/4)$, which maps an interval around $0_{\Bbb R}$ isomorphically to an interval around $0_{T}$. –  Brian M. Scott Sep 12 '12 at 2:31
Oh, sorry, I didn't make it clear. When I started to write this post I first put the title of the topic. He talks about generators of a group later in the same topic. –  Br09 Sep 12 '12 at 2:42
@tomasz: Chevalley notes at the beginning of the section that every zero neighborhood of a connected group generates the group. I suspect that's the reason for the title, which would probably more appropriately contain something about local isomorphisms of groups. –  t.b. Sep 12 '12 at 2:44

I do not understand why the author use "contraction" (did Chevalley define it earlier? I could not find it in Google Books), it is not widely used in the modern Lie group literature. But suppose you are given with the local quotient map $$\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}$$ It is clear that at the open interval $(-\frac{1}{4},\frac{1}{4})$ the quotient map's image is homeomorphic to itself. Now the original additive group structure is also inherited from the quotient structure (since $\mathbb{Z}$ is also an additive subgroup). To be precise for any $x\in (-\frac{1}{4},\frac{1}{4})$, let $\epsilon\in (-\frac{1}{4},\frac{1}{4})$ be small enough such that $x+\epsilon\in (-\frac{1}{4},\frac{1}{4})$. Then we can claim $f(x)+f(\epsilon)\in f(-\frac{1}{4},\frac{1}{4})$ as well. So $f$ is well defined in terms of group structure. I guess restricting to an open neighborhood there exists a group isomorphism is what "locally isomorphic" means.
From a cursory glance Chevalley doesn't define it, he just uses some obsolete terminology: contraction of $f$ to $A$* means *restriction of a function $f\colon X \to Y$ to a subset $A \subseteq X$. –  t.b. Sep 12 '12 at 2:45
Modulo 1 means taking the fractional part. If $x$ is a real number and $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$, then $x-\lfloor x\rfloor \equiv x \bmod 1.$
Because $im(\phi)$ is interval $(0,1)$ my guess is that the contraction $f:(0,1) \rightarrow (-1/4. 1/4)$. Here torus means the circle $S^{1} = \mathbb{R} / \mathbb{Z}$.