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simplified the argument
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Ian Agol
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This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $p>1$$m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $p$$m$ must move a point of $M$ at least a distance $d$.

Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\tilde{f}$ of $f$ to anythe universal cover of $T^n$$\mathbb{R}^n$ which hasfixes a fixed point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). Pass to the $m^n$-fold cover The homotopy of $T^n$ (coming from $ker\{\mathbb{Z}^n\to (\mathbb{Z}/m\mathbb{Z})^n\}$), and rescale the metric on this cover by $1/m$, every point in this cover will be distance at most $D/m$ from its image under$f$ to the liftidentity lifts to a homotopy of $f$$\tilde{f}$ to this cover with a fixed pointthe identity, whichso $\tilde{f}$ also has period $p$ ifthe property that it moves points at most $f$ does. Taking$D$ and has order $m$ large enough (so that. Let $D/m <d$$d$ be the constant from Newman's theorem), Newman's theorem implies thatfor the lift ofmanifold $f$ to this cover will be the identity if$\mathbb{R}^n$ with Euclidean metric and $f$ has period$m$, then rescale the metric by $p$$d/D$, we see that $\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity.

Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.

This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a manifold $M$ with a metric and an integer $p>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $p$ must move a point of $M$ at least a distance $d$.

Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$ and fixes a point, there is a lift of $f$ to any cover of $T^n$ which has a fixed point. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). Pass to the $m^n$-fold cover of $T^n$ (coming from $ker\{\mathbb{Z}^n\to (\mathbb{Z}/m\mathbb{Z})^n\}$), and rescale the metric on this cover by $1/m$, every point in this cover will be distance at most $D/m$ from its image under the lift of $f$ to this cover with a fixed point, which also has period $p$ if $f$ does. Taking $m$ large enough (so that $D/m <d$ from Newman's theorem), Newman's theorem implies that the lift of $f$ to this cover will be the identity if $f$ has period $p$, and hence $f$ is the identity.

Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.

This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $m$ must move a point of $M$ at least a distance $d$.

Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\tilde{f}$ of $f$ to the universal cover $\mathbb{R}^n$ which fixes a point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). The homotopy of $f$ to the identity lifts to a homotopy of $\tilde{f}$ to the identity, so $\tilde{f}$ also has the property that it moves points at most $D$ and has order $m$. Let $d$ be the constant from Newman's theorem for the manifold $\mathbb{R}^n$ with Euclidean metric and $m$, then rescale the metric by $d/D$, we see that $\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity.

Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.

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Ian Agol
  • 68.9k
  • 3
  • 194
  • 358

This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a manifold $M$ with a metric and an integer $p>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $p$ must move a point of $M$ at least a distance $d$.

Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$ and fixes a point, there is a lift of $f$ to any cover of $T^n$ which has a fixed point. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). Pass to the $m^n$-fold cover of $T^n$ (coming from $ker\{\mathbb{Z}^n\to (\mathbb{Z}/m\mathbb{Z})^n\}$), and rescale the metric on this cover by $1/m$, every point in this cover will be distance at most $D/m$ from its image under the lift of $f$ to this cover with a fixed point, which also has period $p$ if $f$ does. Taking $m$ large enough (so that $D/m <d$ from Newman's theorem), Newman's theorem implies that the lift of $f$ to this cover will be the identity if $f$ has period $p$, and hence $f$ is the identity.

Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.