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Mapping Class Group acts properly discontinuous; Alexander method
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Let $S$ be a closed surface of genus $g$. The Alexander Method in Farb and Maraglit's "A Primer on Mapping Class Groups" (p.59) (roughly) states that if $c_1,c_2$ are two filling curves in minimal position and $phi$ is an orientation preserving homeomorphism of $S$ such that $phi(c_i)$ is isotopic to either $c_1$ or $c_2$, then a power of $phi$ is isotopic to the identity.
Later on in the book, they use this result to prove Fricke's theorem, namely that the action of the Mapping Class Group on Teichmüller space is properly discontinuous. They argue as follows:
Suppose $B$ is a compact subset of Teichmüller space and $f$ is a mapping class with $f cdot B cap B ne emptyset$. Given two curves as above, an auxiliary lemma tells them that $f^-1(c_1)$ cand $f^-1(c_2)$ can only be two of finitely many isotopy classes of curves. Then they caim "By the Alexander method, there are [only] finitely many [possible] choices for $f^-1$ once the isotopy classes $f^-1(c_i)$ are determined."
Unfortunately, I do not see how this follows. Suppose $g$ is a mapping class such that $g^-1(c_i)$ is in the same isotopy class as $f^-1(c_i)$. We need to show that there are only fintely many possibilities for $g$. The composition $f circ g^-1$ maps the isotopy class of $c_i$ to $c_i$. Thus, the Alexander method tells us that some power $(f circ g^-1)^n$ is isotopic to the identity. Furthermore, by investigating the Alexander mehtod, one can see that the power is bounded by a constant depending only on $c_1$ and $c_2$ (namely, the number of intersections factorial). However, this does not tell me that there are only finitely many options for $g$. Can I not have infinitely many non-isotopic homeomorphisms $g_k$ such that $(f circ g_k^-1)^n$ is isotopic to the identity (for some fixed $n$)? How do I disprove it?
Cheers
group-actions hyperbolic-geometry teichmueller-theory mapping-class-group
asked Jul 18 at 15:48
Florian R
3408
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1
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Let $S$ be a closed surface of genus $g$. The Alexander Method in Farb and Maraglit's "A Primer on Mapping Class Groups" (p.59) (roughly) states that if $c_1,c_2$ are two filling curves in minimal position and $phi$ is an orientation preserving homeomorphism of $S$ such that $phi(c_i)$ is isotopic to either $c_1$ or $c_2$, then a power of $phi$ is isotopic to the identity.
Later on in the book, they use this result to prove Fricke's theorem, namely that the action of the Mapping Class Group on Teichmüller space is properly discontinuous. They argue as follows:
Suppose $B$ is a compact subset of Teichmüller space and $f$ is a mapping class with $f cdot B cap B ne emptyset$. Given two curves as above, an auxiliary lemma tells them that $f^-1(c_1)$ cand $f^-1(c_2)$ can only be two of finitely many isotopy classes of curves. Then they caim "By the Alexander method, there are [only] finitely many [possible] choices for $f^-1$ once the isotopy classes $f^-1(c_i)$ are determined."
Unfortunately, I do not see how this follows. Suppose $g$ is a mapping class such that $g^-1(c_i)$ is in the same isotopy class as $f^-1(c_i)$. We need to show that there are only fintely many possibilities for $g$. The composition $f circ g^-1$ maps the isotopy class of $c_i$ to $c_i$. Thus, the Alexander method tells us that some power $(f circ g^-1)^n$ is isotopic to the identity. Furthermore, by investigating the Alexander mehtod, one can see that the power is bounded by a constant depending only on $c_1$ and $c_2$ (namely, the number of intersections factorial). However, this does not tell me that there are only finitely many options for $g$. Can I not have infinitely many non-isotopic homeomorphisms $g_k$ such that $(f circ g_k^-1)^n$ is isotopic to the identity (for some fixed $n$)? How do I disprove it?
Cheers
group-actions hyperbolic-geometry teichmueller-theory mapping-class-group
asked Jul 18 at 15:48
Florian R
3408
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $S$ be a closed surface of genus $g$. The Alexander Method in Farb and Maraglit's "A Primer on Mapping Class Groups" (p.59) (roughly) states that if $c_1,c_2$ are two filling curves in minimal position and $phi$ is an orientation preserving homeomorphism of $S$ such that $phi(c_i)$ is isotopic to either $c_1$ or $c_2$, then a power of $phi$ is isotopic to the identity.
Later on in the book, they use this result to prove Fricke's theorem, namely that the action of the Mapping Class Group on Teichmüller space is properly discontinuous. They argue as follows:
Suppose $B$ is a compact subset of Teichmüller space and $f$ is a mapping class with $f cdot B cap B ne emptyset$. Given two curves as above, an auxiliary lemma tells them that $f^-1(c_1)$ cand $f^-1(c_2)$ can only be two of finitely many isotopy classes of curves. Then they caim "By the Alexander method, there are [only] finitely many [possible] choices for $f^-1$ once the isotopy classes $f^-1(c_i)$ are determined."
Unfortunately, I do not see how this follows. Suppose $g$ is a mapping class such that $g^-1(c_i)$ is in the same isotopy class as $f^-1(c_i)$. We need to show that there are only fintely many possibilities for $g$. The composition $f circ g^-1$ maps the isotopy class of $c_i$ to $c_i$. Thus, the Alexander method tells us that some power $(f circ g^-1)^n$ is isotopic to the identity. Furthermore, by investigating the Alexander mehtod, one can see that the power is bounded by a constant depending only on $c_1$ and $c_2$ (namely, the number of intersections factorial). However, this does not tell me that there are only finitely many options for $g$. Can I not have infinitely many non-isotopic homeomorphisms $g_k$ such that $(f circ g_k^-1)^n$ is isotopic to the identity (for some fixed $n$)? How do I disprove it?
Cheers
group-actions hyperbolic-geometry teichmueller-theory mapping-class-group
asked Jul 18 at 15:48
Florian R
3408
Let $S$ be a closed surface of genus $g$. The Alexander Method in Farb and Maraglit's "A Primer on Mapping Class Groups" (p.59) (roughly) states that if $c_1,c_2$ are two filling curves in minimal position and $phi$ is an orientation preserving homeomorphism of $S$ such that $phi(c_i)$ is isotopic to either $c_1$ or $c_2$, then a power of $phi$ is isotopic to the identity.
Later on in the book, they use this result to prove Fricke's theorem, namely that the action of the Mapping Class Group on Teichmüller space is properly discontinuous. They argue as follows:
Suppose $B$ is a compact subset of Teichmüller space and $f$ is a mapping class with $f cdot B cap B ne emptyset$. Given two curves as above, an auxiliary lemma tells them that $f^-1(c_1)$ cand $f^-1(c_2)$ can only be two of finitely many isotopy classes of curves. Then they caim "By the Alexander method, there are [only] finitely many [possible] choices for $f^-1$ once the isotopy classes $f^-1(c_i)$ are determined."
Unfortunately, I do not see how this follows. Suppose $g$ is a mapping class such that $g^-1(c_i)$ is in the same isotopy class as $f^-1(c_i)$. We need to show that there are only fintely many possibilities for $g$. The composition $f circ g^-1$ maps the isotopy class of $c_i$ to $c_i$. Thus, the Alexander method tells us that some power $(f circ g^-1)^n$ is isotopic to the identity. Furthermore, by investigating the Alexander mehtod, one can see that the power is bounded by a constant depending only on $c_1$ and $c_2$ (namely, the number of intersections factorial). However, this does not tell me that there are only finitely many options for $g$. Can I not have infinitely many non-isotopic homeomorphisms $g_k$ such that $(f circ g_k^-1)^n$ is isotopic to the identity (for some fixed $n$)? How do I disprove it?
Cheers
group-actions hyperbolic-geometry teichmueller-theory mapping-class-group
asked Jul 18 at 15:48
Florian R
3408
asked Jul 18 at 15:48
Florian R
3408
asked Jul 18 at 15:48
Florian R
3408
asked Jul 18 at 15:48
Florian R
3408
3408
add a comment |Â
add a comment |Â
1 Answer
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A stronger statement is true: if $c_1$, $c_2$ are two filling curves in minimal position then the subset of the mapping class group that preserves the unordered pair of isotopy classes $[c_1],[c_2]$ is a finite subgroup.
It follows that once the unordered pair isotopy classes $f^-1[c_1],f^-1[c_2]$ is determined, the set of values of $f^-1$ lives in one left coset of that subgroup, and hence is finite.
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
A stronger statement is true: if $c_1$, $c_2$ are two filling curves in minimal position then the subset of the mapping class group that preserves the unordered pair of isotopy classes $[c_1],[c_2]$ is a finite subgroup.
It follows that once the unordered pair isotopy classes $f^-1[c_1],f^-1[c_2]$ is determined, the set of values of $f^-1$ lives in one left coset of that subgroup, and hence is finite.
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
add a comment |Â
up vote
0
down vote
accepted
A stronger statement is true: if $c_1$, $c_2$ are two filling curves in minimal position then the subset of the mapping class group that preserves the unordered pair of isotopy classes $[c_1],[c_2]$ is a finite subgroup.
It follows that once the unordered pair isotopy classes $f^-1[c_1],f^-1[c_2]$ is determined, the set of values of $f^-1$ lives in one left coset of that subgroup, and hence is finite.
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
A stronger statement is true: if $c_1$, $c_2$ are two filling curves in minimal position then the subset of the mapping class group that preserves the unordered pair of isotopy classes $[c_1],[c_2]$ is a finite subgroup.
It follows that once the unordered pair isotopy classes $f^-1[c_1],f^-1[c_2]$ is determined, the set of values of $f^-1$ lives in one left coset of that subgroup, and hence is finite.
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
A stronger statement is true: if $c_1$, $c_2$ are two filling curves in minimal position then the subset of the mapping class group that preserves the unordered pair of isotopy classes $[c_1],[c_2]$ is a finite subgroup.
It follows that once the unordered pair isotopy classes $f^-1[c_1],f^-1[c_2]$ is determined, the set of values of $f^-1$ lives in one left coset of that subgroup, and hence is finite.
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
answered Jul 18 at 18:52
Lee Mosher
45.7k33478
45.7k33478
add a comment |Â
add a comment |Â
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