What maps descend to homeomorphisms
$begingroup$
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 $2times 2$ 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.
algebraic-topology covering-spaces mapping-class-group
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
$endgroup$
add a comment |
$begingroup$
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 $2times 2$ 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.
algebraic-topology covering-spaces mapping-class-group
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
$endgroup$
1
$begingroup$
I'd say let $T^x = left[ {begin{array}{cc} 1 & x \ 0&1 end{array} } right]$ and $G = {T^x,x inmathbb{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 $A = H setminus E ={ Hv,vin E}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hvmapsto HT^xv$ is an homeomorphism $A to A$. And the kernel of $G subset Aut(E)to Gsubset Aut(A) $ is $H$ so $G/H$ acts faithfully on $Hsetminus E$.
$endgroup$
– reuns
Dec 28 '18 at 22:27
1
$begingroup$
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
add a comment |
$begingroup$
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 $2times 2$ 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.
algebraic-topology covering-spaces mapping-class-group
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
$endgroup$
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 $2times 2$ 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.
algebraic-topology covering-spaces mapping-class-group
algebraic-topology covering-spaces mapping-class-group
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
asked Dec 27 '18 at 17:54
Vidit DVidit D
10211
10211
1
$begingroup$
I'd say let $T^x = left[ {begin{array}{cc} 1 & x \ 0&1 end{array} } right]$ and $G = {T^x,x inmathbb{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 $A = H setminus E ={ Hv,vin E}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hvmapsto HT^xv$ is an homeomorphism $A to A$. And the kernel of $G subset Aut(E)to Gsubset Aut(A) $ is $H$ so $G/H$ acts faithfully on $Hsetminus E$.
$endgroup$
– reuns
Dec 28 '18 at 22:27
1
$begingroup$
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
add a comment |
1
$begingroup$
I'd say let $T^x = left[ {begin{array}{cc} 1 & x \ 0&1 end{array} } right]$ and $G = {T^x,x inmathbb{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 $A = H setminus E ={ Hv,vin E}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hvmapsto HT^xv$ is an homeomorphism $A to A$. And the kernel of $G subset Aut(E)to Gsubset Aut(A) $ is $H$ so $G/H$ acts faithfully on $Hsetminus E$.
$endgroup$
– reuns
Dec 28 '18 at 22:27
1
$begingroup$
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
1
1
$begingroup$
I'd say let $T^x = left[ {begin{array}{cc} 1 & x \ 0&1 end{array} } right]$ and $G = {T^x,x inmathbb{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 $A = H setminus E ={ Hv,vin E}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hvmapsto HT^xv$ is an homeomorphism $A to A$. And the kernel of $G subset Aut(E)to Gsubset Aut(A) $ is $H$ so $G/H$ acts faithfully on $Hsetminus E$.
$endgroup$
– reuns
Dec 28 '18 at 22:27
$begingroup$
I'd say let $T^x = left[ {begin{array}{cc} 1 & x \ 0&1 end{array} } right]$ and $G = {T^x,x inmathbb{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 $A = H setminus E ={ Hv,vin E}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hvmapsto HT^xv$ is an homeomorphism $A to A$. And the kernel of $G subset Aut(E)to Gsubset Aut(A) $ is $H$ so $G/H$ acts faithfully on $Hsetminus E$.
$endgroup$
– reuns
Dec 28 '18 at 22:27
1
1
$begingroup$
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
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $pi:mathbf Rtimes [0,1]to A$ be the covering map. If $xin A$, you can pick a $yin pi^{-1}(x)$. 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).
answered Dec 28 '18 at 21:23
TomTom
27118
$endgroup$
$begingroup$
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 $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
add a comment |
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$begingroup$
Let $pi:mathbf Rtimes [0,1]to A$ be the covering map. If $xin A$, you can pick a $yin pi^{-1}(x)$. 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).
answered Dec 28 '18 at 21:23
TomTom
27118
$endgroup$
$begingroup$
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 $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
add a comment |
$begingroup$
Let $pi:mathbf Rtimes [0,1]to A$ be the covering map. If $xin A$, you can pick a $yin pi^{-1}(x)$. 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).
answered Dec 28 '18 at 21:23
TomTom
27118
$endgroup$
$begingroup$
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 $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
add a comment |
$begingroup$
Let $pi:mathbf Rtimes [0,1]to A$ be the covering map. If $xin A$, you can pick a $yin pi^{-1}(x)$. 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).
answered Dec 28 '18 at 21:23
TomTom
27118
$endgroup$
Let $pi:mathbf Rtimes [0,1]to A$ be the covering map. If $xin A$, you can pick a $yin pi^{-1}(x)$. 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).
answered Dec 28 '18 at 21:23
TomTom
27118
answered Dec 28 '18 at 21:23
TomTom
27118
answered Dec 28 '18 at 21:23
TomTom
27118
answered Dec 28 '18 at 21:23
TomTom
27118
27118
$begingroup$
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 $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
add a comment |
$begingroup$
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 $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
$begingroup$
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
$begingroup$
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 $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
$begingroup$
well if $g$ is the one coming from $M^{-1}$, then $fcirc g(x) = f(pi(M^{-1} y)) = pi(MM^{-1}y) = x$ where $yin pi^{-1}(x)$. And the same holds for $gcirc f$
$endgroup$
– Tom
Dec 29 '18 at 12:15
add a comment |
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1
$begingroup$
I'd say let $T^x = left[ {begin{array}{cc} 1 & x \ 0&1 end{array} } right]$ and $G = {T^x,x inmathbb{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 $A = H setminus E ={ Hv,vin E}$. Then $H$ is a normal subgroup of $G$ implies that $T^xHv =HT^xv$ so $Hvmapsto HT^xv$ is an homeomorphism $A to A$. And the kernel of $G subset Aut(E)to Gsubset Aut(A) $ is $H$ so $G/H$ acts faithfully on $Hsetminus E$.
$endgroup$
– reuns
Dec 28 '18 at 22:27
1
$begingroup$
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