What does modular space $mathbb{H}/ mathrm{SL}_2(mathbb{Z})$ mean?
Juts a quick question. In Freitag's Complex Analysis as an example for The Quotient Topology it comes:
The "modular space" $mathbb{H}/mathrm{SL}_2(mathbb{Z}).$
Every element in $mathbb{H}$ can be mapped by a linear fractional transformation in $mathbb{H}/ mathrm{SL}_2(mathbb{Z})$ to some fixed element ${{tau_0}}$ in $mathbb{H}$ so is it true to say $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong {{tau_0}}$? So basically a modular space is just any single point in $mathbb{H}$?
I have a little background in Modular Forms so much appreciated a simple explanation.
general-topology complex-analysis quotient-spaces
asked Nov 26 at 23:59
72D
564116
add a comment |
Juts a quick question. In Freitag's Complex Analysis as an example for The Quotient Topology it comes:
The "modular space" $mathbb{H}/mathrm{SL}_2(mathbb{Z}).$
Every element in $mathbb{H}$ can be mapped by a linear fractional transformation in $mathbb{H}/ mathrm{SL}_2(mathbb{Z})$ to some fixed element ${{tau_0}}$ in $mathbb{H}$ so is it true to say $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong {{tau_0}}$? So basically a modular space is just any single point in $mathbb{H}$?
I have a little background in Modular Forms so much appreciated a simple explanation.
general-topology complex-analysis quotient-spaces
asked Nov 26 at 23:59
72D
564116
$Gamma = mathrm{SL}_2(mathbb{Z})$ is a group of biholomorphisms $mathbb{H}to mathbb{H}$. Let $ Gamma tau = { gamma tau, gamma in Gamma}$ a subset of $mathbb{H}$. Then $Gamma setminus mathbb{H} = { Gamma tau, tau in mathbb{H}}$ is a topological space whose points are subsets of $mathbb{H}$. The topology is the complex topology inherited from $mathbb{H}$ so it is a Riemann surface. For any $Gamma tau$ there is a representative with $gammatau in mathcal{F}$ the fundamental domain mentioned by carmichael.
– reuns
Nov 27 at 0:20
add a comment |
Juts a quick question. In Freitag's Complex Analysis as an example for The Quotient Topology it comes:
The "modular space" $mathbb{H}/mathrm{SL}_2(mathbb{Z}).$
Every element in $mathbb{H}$ can be mapped by a linear fractional transformation in $mathbb{H}/ mathrm{SL}_2(mathbb{Z})$ to some fixed element ${{tau_0}}$ in $mathbb{H}$ so is it true to say $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong {{tau_0}}$? So basically a modular space is just any single point in $mathbb{H}$?
I have a little background in Modular Forms so much appreciated a simple explanation.
general-topology complex-analysis quotient-spaces
asked Nov 26 at 23:59
72D
564116
Juts a quick question. In Freitag's Complex Analysis as an example for The Quotient Topology it comes:
The "modular space" $mathbb{H}/mathrm{SL}_2(mathbb{Z}).$
Every element in $mathbb{H}$ can be mapped by a linear fractional transformation in $mathbb{H}/ mathrm{SL}_2(mathbb{Z})$ to some fixed element ${{tau_0}}$ in $mathbb{H}$ so is it true to say $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong {{tau_0}}$? So basically a modular space is just any single point in $mathbb{H}$?
I have a little background in Modular Forms so much appreciated a simple explanation.
general-topology complex-analysis quotient-spaces
general-topology complex-analysis quotient-spaces
asked Nov 26 at 23:59
72D
564116
asked Nov 26 at 23:59
72D
564116
edited Nov 27 at 0:15
asked Nov 26 at 23:59
72D
564116
asked Nov 26 at 23:59
72D
564116
asked Nov 26 at 23:59
72D
564116
564116
$Gamma = mathrm{SL}_2(mathbb{Z})$ is a group of biholomorphisms $mathbb{H}to mathbb{H}$. Let $ Gamma tau = { gamma tau, gamma in Gamma}$ a subset of $mathbb{H}$. Then $Gamma setminus mathbb{H} = { Gamma tau, tau in mathbb{H}}$ is a topological space whose points are subsets of $mathbb{H}$. The topology is the complex topology inherited from $mathbb{H}$ so it is a Riemann surface. For any $Gamma tau$ there is a representative with $gammatau in mathcal{F}$ the fundamental domain mentioned by carmichael.
– reuns
Nov 27 at 0:20
add a comment |
$Gamma = mathrm{SL}_2(mathbb{Z})$ is a group of biholomorphisms $mathbb{H}to mathbb{H}$. Let $ Gamma tau = { gamma tau, gamma in Gamma}$ a subset of $mathbb{H}$. Then $Gamma setminus mathbb{H} = { Gamma tau, tau in mathbb{H}}$ is a topological space whose points are subsets of $mathbb{H}$. The topology is the complex topology inherited from $mathbb{H}$ so it is a Riemann surface. For any $Gamma tau$ there is a representative with $gammatau in mathcal{F}$ the fundamental domain mentioned by carmichael.
– reuns
Nov 27 at 0:20
$Gamma = mathrm{SL}_2(mathbb{Z})$ is a group of biholomorphisms $mathbb{H}to mathbb{H}$. Let $ Gamma tau = { gamma tau, gamma in Gamma}$ a subset of $mathbb{H}$. Then $Gamma setminus mathbb{H} = { Gamma tau, tau in mathbb{H}}$ is a topological space whose points are subsets of $mathbb{H}$. The topology is the complex topology inherited from $mathbb{H}$ so it is a Riemann surface. For any $Gamma tau$ there is a representative with $gammatau in mathcal{F}$ the fundamental domain mentioned by carmichael.
– reuns
Nov 27 at 0:20
$Gamma = mathrm{SL}_2(mathbb{Z})$ is a group of biholomorphisms $mathbb{H}to mathbb{H}$. Let $ Gamma tau = { gamma tau, gamma in Gamma}$ a subset of $mathbb{H}$. Then $Gamma setminus mathbb{H} = { Gamma tau, tau in mathbb{H}}$ is a topological space whose points are subsets of $mathbb{H}$. The topology is the complex topology inherited from $mathbb{H}$ so it is a Riemann surface. For any $Gamma tau$ there is a representative with $gammatau in mathcal{F}$ the fundamental domain mentioned by carmichael.
– reuns
Nov 27 at 0:20
add a comment |
1 Answer
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While all elements of $mathbb{H}$ are $mathrm{SL}_2(mathbb{R})$-equivalent, this is no longer the case if $mathbb{R}$ is replaced by $mathbb{Z}$. If (as you say) you have some background in modular forms, you've likely seen a picture of the standard fundamental domain for the modular group:
$$ mathcal{F}={z=x+iy:y>0,-frac{1}{2}leq xleq frac{1}{2},|z|geq 1}$$
You can visualize the quotient space by gluing together the boundary of $mathcal{F}$: the points $-frac{1}{2}+y$ and $frac{1}{2}+y$ are identified, as are $z$ and $-frac{1}{z}$ if $zinmathcal{F},|z|=1$.
answered Nov 27 at 0:09
carmichael561
46.9k54382
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
|
show 1 more comment
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1 Answer
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votes
While all elements of $mathbb{H}$ are $mathrm{SL}_2(mathbb{R})$-equivalent, this is no longer the case if $mathbb{R}$ is replaced by $mathbb{Z}$. If (as you say) you have some background in modular forms, you've likely seen a picture of the standard fundamental domain for the modular group:
$$ mathcal{F}={z=x+iy:y>0,-frac{1}{2}leq xleq frac{1}{2},|z|geq 1}$$
You can visualize the quotient space by gluing together the boundary of $mathcal{F}$: the points $-frac{1}{2}+y$ and $frac{1}{2}+y$ are identified, as are $z$ and $-frac{1}{z}$ if $zinmathcal{F},|z|=1$.
answered Nov 27 at 0:09
carmichael561
46.9k54382
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
|
show 1 more comment
While all elements of $mathbb{H}$ are $mathrm{SL}_2(mathbb{R})$-equivalent, this is no longer the case if $mathbb{R}$ is replaced by $mathbb{Z}$. If (as you say) you have some background in modular forms, you've likely seen a picture of the standard fundamental domain for the modular group:
$$ mathcal{F}={z=x+iy:y>0,-frac{1}{2}leq xleq frac{1}{2},|z|geq 1}$$
You can visualize the quotient space by gluing together the boundary of $mathcal{F}$: the points $-frac{1}{2}+y$ and $frac{1}{2}+y$ are identified, as are $z$ and $-frac{1}{z}$ if $zinmathcal{F},|z|=1$.
answered Nov 27 at 0:09
carmichael561
46.9k54382
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
|
show 1 more comment
While all elements of $mathbb{H}$ are $mathrm{SL}_2(mathbb{R})$-equivalent, this is no longer the case if $mathbb{R}$ is replaced by $mathbb{Z}$. If (as you say) you have some background in modular forms, you've likely seen a picture of the standard fundamental domain for the modular group:
$$ mathcal{F}={z=x+iy:y>0,-frac{1}{2}leq xleq frac{1}{2},|z|geq 1}$$
You can visualize the quotient space by gluing together the boundary of $mathcal{F}$: the points $-frac{1}{2}+y$ and $frac{1}{2}+y$ are identified, as are $z$ and $-frac{1}{z}$ if $zinmathcal{F},|z|=1$.
answered Nov 27 at 0:09
carmichael561
46.9k54382
While all elements of $mathbb{H}$ are $mathrm{SL}_2(mathbb{R})$-equivalent, this is no longer the case if $mathbb{R}$ is replaced by $mathbb{Z}$. If (as you say) you have some background in modular forms, you've likely seen a picture of the standard fundamental domain for the modular group:
$$ mathcal{F}={z=x+iy:y>0,-frac{1}{2}leq xleq frac{1}{2},|z|geq 1}$$
You can visualize the quotient space by gluing together the boundary of $mathcal{F}$: the points $-frac{1}{2}+y$ and $frac{1}{2}+y$ are identified, as are $z$ and $-frac{1}{z}$ if $zinmathcal{F},|z|=1$.
answered Nov 27 at 0:09
carmichael561
46.9k54382
answered Nov 27 at 0:09
carmichael561
46.9k54382
answered Nov 27 at 0:09
carmichael561
46.9k54382
answered Nov 27 at 0:09
carmichael561
46.9k54382
46.9k54382
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
|
show 1 more comment
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes I made a mistake about $mathrm{SL}_2(mathbb{R})$ with $mathrm{SL}_2(mathbb{Z})$ and I know the $mathcal{F}$. But for clarification: so is it true that $mathbb{H}/ mathrm{SL}_2(mathbb{R}) cong$ any single point in $mathbb{H}$; and $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}$?
– 72D
Nov 27 at 0:18
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
Yes, the quotient $mathbb{H}/mathrm{SL}_2(mathbb{R})$ can be identified with a single point
– carmichael561
Nov 27 at 0:19
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
And $mathbb{H}/ mathrm{SL}_2(mathbb{Z}) cong mathcal{F}?$
– 72D
Nov 27 at 0:20
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
Not sure what $cong$ means in this context. And don't forget that the boundary of $mathcal{F}$ must be handled correctly.
– carmichael561
Nov 27 at 0:21
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
≅ means that the two spaces are homemorphic.
– 72D
Nov 27 at 0:22
|
show 1 more comment
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$Gamma = mathrm{SL}_2(mathbb{Z})$ is a group of biholomorphisms $mathbb{H}to mathbb{H}$. Let $ Gamma tau = { gamma tau, gamma in Gamma}$ a subset of $mathbb{H}$. Then $Gamma setminus mathbb{H} = { Gamma tau, tau in mathbb{H}}$ is a topological space whose points are subsets of $mathbb{H}$. The topology is the complex topology inherited from $mathbb{H}$ so it is a Riemann surface. For any $Gamma tau$ there is a representative with $gammatau in mathcal{F}$ the fundamental domain mentioned by carmichael.
– reuns
Nov 27 at 0:20