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I am asked to show that every automorphism of the fundamental group of a torus $T=S^1\times{}S^1$ is induced by a homeomorphism $h:T\rightarrow{}T$, which fixes the base point.

What I was thinking is composing rotations in $R^3$ and the antipodal map on $T$, which will give me base-point preserving homeomorphisms of $T$. These homeormorphisms will obviously induce automorphisms of the group $\pi_1(T)$.

But how do I show that all the automorphisms are induced this way? I can see that the order of the group $Aut(\pi_1(T))$ is $8$ and so I have to produce $8$ such homeomorphism! Is there any other elegant way?

Any help is appreciated. Thanks!

Added: As pointed out in the comments, I miscalculated the order of the group $Aut(\pi_1(T))$. But then my original argument is not working at all. Any help?

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  • $\begingroup$ Does $\mathbb Z\times\mathbb Z$ really only have $8$ automorphisms? $\endgroup$
    – Nishant
    Oct 13, 2014 at 3:24
  • $\begingroup$ @Nishant, I was thinking $(m,n)\rightarrow{}(m,\pm{}n)$, $(m,n)\rightarrow{}(\pm{}m,n)$, $(m,n)\rightarrow{}(n,\pm{}m)$ and $(m,n)\rightarrow{}(\pm{}n,m)$. Am I missing something? $\endgroup$
    – ChesterX
    Oct 13, 2014 at 3:30
  • $\begingroup$ I think $(m, n)\mapsto(m, m+n)$ also works. $\endgroup$
    – Nishant
    Oct 13, 2014 at 4:13
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    $\begingroup$ an automorphism of $Z^2$ corresponds to a $2\times 2$-matrix which is invertible, i.e. the determinant is $\pm 1$. $\endgroup$ Oct 13, 2014 at 6:45
  • $\begingroup$ @Dan, Thanks for that! I've changed the question. $\endgroup$
    – ChesterX
    Oct 13, 2014 at 6:51

1 Answer 1

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Think of the torus as $\Bbb{R}^2 / \Bbb{Z}^2$. The map $$ \begin{align} \Bbb{R}^2 &\overset{\normalsize\tilde{\varphi}^{}_A}{\to} \Bbb{R}^2 \\ \begin{bmatrix} s \\ t \end{bmatrix} &\mapsto \begin{bmatrix} as+bt \\ cs+dt \end{bmatrix} \end{align} $$ with $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in GL_2(\Bbb{Z})$ satisfies $\tilde{\varphi}^{}_A(\Bbb{Z}^2) \subseteq \Bbb{Z}^2$. Therefore, it descends to $$ \begin{align} T &\overset{\normalsize\varphi^{}_A}{\to} T \\ \begin{bmatrix} s \\ t \end{bmatrix} &\mapsto \begin{bmatrix} as+bt \pmod{1} \\ cs+dt \pmod{1} \end{bmatrix} \end{align} $$ For any $A \in GL_2(\Bbb{Z}) \cong \operatorname{Aut}\pi_1(T)$, this is an explicit construction of a self-homeomorphism of $\varphi^{}_A$ of $T$ such that $$ \bigl( \varphi^{}_A \bigr)_* = A. $$

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  • $\begingroup$ I know this is an old post, but is it necessary to say anything about the base points being preserved? $\endgroup$ Jun 3, 2018 at 4:08

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