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I was reading "A primer on Mapping Class Groups" by B. Farb and D. Margalit and I'm stuck at a point in the proof of Mod$(A)\cong \mathbb{Z}$ where $A$ is the annulus, where the restriction of $M$ on $\mathbb{R}\times\{1\}$ descends to a homeomorphism. I can't figure out how.

The question is as follows, Let $M$ be a $2\times 2$ real matrix defining a linear transformation from $\mathbb{R}^2$ to itself. If we restrict it to the universal cover of the annulus $(\mathbb{R}\times [0,1])$ such that it is equivariant under the group of deck transformations $(\mathbb{Z})$ then this restriction descends to a homeomorphism of the annulus.

In this case the matrix is$$ M= \left[ {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right] $$ I'm unable to prove this fact.

Thanks in advance.

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    $\begingroup$ I'd say let $T^x = \left[ {\begin{array}{cc} 1 & x \\ 0&1 \end{array} } \right]$ and $G = \{T^x,x \in\mathbb{R}\} $ and $E = \{ \left[ {\begin{array}{cc} y \\ z \\ \end{array} } \right], (y,z) \in \mathbb{R} \times [0,1]\}$. Left multiplication by $T^x$ is an homeomorphism $E \to E$. Let $H = \{ T^n, n \in \mathbb{Z}\}$ and $A = H \setminus E =\{ Hv,v\in E\}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hv\mapsto HT^xv$ is an homeomorphism $A \to A$. And the kernel of $G \subset Aut(E)\to G\subset Aut(A) $ is $H$ so $G/H$ acts faithfully on $H\setminus E$. $\endgroup$ – reuns Dec 28 '18 at 22:27
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    $\begingroup$ If you replace $T^x$ by some $M \in GL_2(\mathbb{R})$ which sends $E$ to itself then $M = \left[ {\begin{array}{cc} t & x \\ 0&1 \end{array} } \right]$ and "equivariant" means $M Hv = HMv$ so $MH = HM$ and $M = \left[ {\begin{array}{cc} \pm 1 & x \\ 0&1 \end{array} } \right]$. $\endgroup$ – reuns Dec 28 '18 at 22:34
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Let $\pi:\mathbf R\times [0,1]\to A$ be the covering map. If $x\in A$, you can pick a $y\in \pi^{-1}(x)$. Thanks to the equivariance property, $f(x) = \pi(My)$ does not depend on the choice of $y$. Local triviality gives you the continuity of $f$.

In our case we can do the same things with $M^{-1}$, so this is an homeomorphism (I don't know what is needed in general).

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  • $\begingroup$ oh yes, this does it. But what guarantees that $M^{-1}$ descends to $f^{-1}$, I mean there should be some kind of uniqueness right? $\endgroup$ – Vidit D Dec 29 '18 at 7:08
  • $\begingroup$ well if $g$ is the one coming from $M^{-1}$, then $f\circ g(x) = f(\pi(M^{-1} y)) = \pi(MM^{-1}y) = x$ where $y\in \pi^{-1}(x)$. And the same holds for $g\circ f$ $\endgroup$ – Tom Dec 29 '18 at 12:15

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