Mixing
Homology group of $SL(2,mathbb{R})/SL(2,mathbb{Z})$
$begingroup$
My friend asks me how to compute the homology group of $X=SL(2,mathbb{R})/SL(2,mathbb{Z})$. It is not hard to see that $H_0(X)=mathbb{Z}$, and for $qge 4$ we have $H_q(X)=0$.
But I don't know how to compute $H_q(X)$ for $q=1,2,3$, since the action of $SL(2,mathbb{Z})$ seems 'strange' to me, and I can't make $X$ into a CW complex. Can anyone help me?
algebraic-topology manifolds homology-cohomology
asked Jan 16 at 5:52
BonbonBonbon
47118
$endgroup$
add a comment |
$begingroup$
My friend asks me how to compute the homology group of $X=SL(2,mathbb{R})/SL(2,mathbb{Z})$. It is not hard to see that $H_0(X)=mathbb{Z}$, and for $qge 4$ we have $H_q(X)=0$.
But I don't know how to compute $H_q(X)$ for $q=1,2,3$, since the action of $SL(2,mathbb{Z})$ seems 'strange' to me, and I can't make $X$ into a CW complex. Can anyone help me?
algebraic-topology manifolds homology-cohomology
asked Jan 16 at 5:52
BonbonBonbon
47118
$endgroup$
2
$begingroup$
For $q geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold.
$endgroup$
– Mike Miller
Jan 16 at 6:32
$begingroup$
@MikeMiller Thanks, my mistake.
$endgroup$
– Bonbon
Jan 16 at 6:38
1
$begingroup$
It's a well-known fact that $SL(2,mathbb Z)$ is not cocompact in $SL(2,mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :)
$endgroup$
– Cheerful Parsnip
Jan 16 at 6:51
add a comment |
$begingroup$
My friend asks me how to compute the homology group of $X=SL(2,mathbb{R})/SL(2,mathbb{Z})$. It is not hard to see that $H_0(X)=mathbb{Z}$, and for $qge 4$ we have $H_q(X)=0$.
But I don't know how to compute $H_q(X)$ for $q=1,2,3$, since the action of $SL(2,mathbb{Z})$ seems 'strange' to me, and I can't make $X$ into a CW complex. Can anyone help me?
algebraic-topology manifolds homology-cohomology
asked Jan 16 at 5:52
BonbonBonbon
47118
$endgroup$
My friend asks me how to compute the homology group of $X=SL(2,mathbb{R})/SL(2,mathbb{Z})$. It is not hard to see that $H_0(X)=mathbb{Z}$, and for $qge 4$ we have $H_q(X)=0$.
But I don't know how to compute $H_q(X)$ for $q=1,2,3$, since the action of $SL(2,mathbb{Z})$ seems 'strange' to me, and I can't make $X$ into a CW complex. Can anyone help me?
algebraic-topology manifolds homology-cohomology
algebraic-topology manifolds homology-cohomology
asked Jan 16 at 5:52
BonbonBonbon
47118
asked Jan 16 at 5:52
BonbonBonbon
47118
edited Jan 16 at 7:36
Bonbon
asked Jan 16 at 5:52
BonbonBonbon
47118
asked Jan 16 at 5:52
BonbonBonbon
47118
asked Jan 16 at 5:52
BonbonBonbon
47118
47118
2
$begingroup$
For $q geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold.
$endgroup$
– Mike Miller
Jan 16 at 6:32
$begingroup$
@MikeMiller Thanks, my mistake.
$endgroup$
– Bonbon
Jan 16 at 6:38
1
$begingroup$
It's a well-known fact that $SL(2,mathbb Z)$ is not cocompact in $SL(2,mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :)
$endgroup$
– Cheerful Parsnip
Jan 16 at 6:51
add a comment |
2
$begingroup$
For $q geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold.
$endgroup$
– Mike Miller
Jan 16 at 6:32
$begingroup$
@MikeMiller Thanks, my mistake.
$endgroup$
– Bonbon
Jan 16 at 6:38
1
$begingroup$
It's a well-known fact that $SL(2,mathbb Z)$ is not cocompact in $SL(2,mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :)
$endgroup$
– Cheerful Parsnip
Jan 16 at 6:51
2
2
$begingroup$
For $q geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold.
$endgroup$
– Mike Miller
Jan 16 at 6:32
$begingroup$
For $q geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold.
$endgroup$
– Mike Miller
Jan 16 at 6:32
$begingroup$
@MikeMiller Thanks, my mistake.
$endgroup$
– Bonbon
Jan 16 at 6:38
$begingroup$
@MikeMiller Thanks, my mistake.
$endgroup$
– Bonbon
Jan 16 at 6:38
1
1
$begingroup$
It's a well-known fact that $SL(2,mathbb Z)$ is not cocompact in $SL(2,mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :)
$endgroup$
– Cheerful Parsnip
Jan 16 at 6:51
$begingroup$
It's a well-known fact that $SL(2,mathbb Z)$ is not cocompact in $SL(2,mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :)
$endgroup$
– Cheerful Parsnip
Jan 16 at 6:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We may write just as well $X = widetilde{SL_2 Bbb R}/widetilde{SL_2 Bbb Z}$, where the latter term is the preimage of $SL_2 Bbb Z$ under the covering map $widetilde{SL_2 Bbb R} to SL_2 Bbb R$. This identifes $X = B(widetilde{SL_2 Bbb Z}, 1)$, because its universal cover is the contractible space $widetilde{SL_2 Bbb R}$.
You are thus asking for a group homology computation. This is especially fortuitous because as a particularly emotive parsnip observed, $H_k X = 0$ for $k geq 3$. So all we need to know is $H_k X = H_k(widetilde{SL_2 Bbb Z}; Bbb Z)$ for $k = 1, 2$, where this last term means group homology (or equivalently, the homology of any space with contractible universal cover and $pi_1 X = widetilde{SL_2 Bbb Z}$).
To facilitate things, observe that as here this is the fundamental group of the complement of the trefoil knot, with presentation $langle x, y mid x^2 = y^3 rangle.$ Knot complements always have aspherical universal cover, so we have $X simeq S^3 setminus T_{2,3}$, where $T_{2,3}$ is the trefoil knot. Knot complements always have $H_1 X = Bbb Z$ and $H_k X = 0$ for $k > 1$ by Alexander duality or a Mayer-Vietoris calculation.
Likely there is an explicit homeomorphism, which I do not know but the clever 3-dimensional geometers do.
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
$endgroup$
1
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
add a comment |
$begingroup$
The space of unimodular lattices in $mathbb{R}^2$ is acted upon transitively by $mathrm{SL}(2,mathbb{R})$, using the underlying action of $mathrm{SL}(2,mathbb{R})$ on $mathbb{R}^2$. One can view such a matrix as containing a basis for a unimodular lattice. The stabilizer of the ${(1,0),(0,1)}$ lattice is $mathrm{SL}(2,mathbb{Z})$, and so $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is in bijective correspondence with the space of unimodular lattices.
(Warning: what follows is my attempt to expand out a story someone once told me. I don't know the theory of elliptic functions well enough to know I haven't made any mistakes.)
The Weierstrass elliptic function $wp(z;Lambda)$ for a complex number $z$ and a lattice $Lambda$ for $mathbb{C}$ gives a point on the Riemann sphere $mathbb{C}_infty$. The quotient of $mathbb{C}$ by a lattice is a flat torus, and $wp(-;Lambda)$ can be thought of as a function from a particular flat torus to the Riemann sphere with four branch points of order two, via a hyperelliptic involution. The four branch points are the images of $0,omega_1/2,omega_2/2,(omega_1+omega_2)/2$, where $omega_1,omega_2inmathbb{C}$ generate $Lambda$. Let
begin{align*}
e_1&=wp(omega_1/2;Lambda)\
e_2&=wp(omega_2/2;Lambda)\
e_3&=wp((omega_1+omega_2)/2;Lambda)
end{align*}
which are distinct elements of $mathbb{C}$ satisfying $e_1+e_2+e_3=0$. These three numbers determine $Lambda$ in that $wp$ is the unique solution to the differential equation $(frac{dy}{dz})^2=4(y-e_1)(y-e_2)(y-e_3)$ with a pole at $0$. Expanded, the right-hand-side polynomial is $4y^3-g_2y-g_3$. Hence, polynomials of this form parameterize lattices, if we exclude those that have double roots. The discriminant of this polynomial is $(g_2^3-27 g_3^2)/16$, and $g_2^3-27 g_3^2$ is called the modular discriminant.
Every lattice has a unimodular representative by scaling, and in particular there is a deformation retract onto unimodular lattices. So, if we take all lattices parameterized using $(g_2,g_3)$ in $mathbb{C}^2$, remove those for which the discriminant is zero, then intersect with the unit $3$-sphere ${(z,w):lvert zrvert^2+lvert wrvert ^2=1}$, we have a parameterization of unimodular lattices after isotopy through the deformation retract. It is known that the link of the singularity of $g_2^3-27 g_3^2$ (i.e., the intersection between the zero-set and $S^3$) is a trefoil knot. (There should be a way to say the condition on $g_2$ and $g_3$ so that the corresponding lattice is unimodular, but I couldn't find it.)
So: $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is homeomorphic to $S^3$ minus a trefoil knot, which can concretely be written as
$${(z,w)inmathbb{C}^2:lvert zrvert^2+lvert wrvert ^2=1text{ and }z^3neq 27w^3}.$$
Knot complements have trivial reduced homology except $tilde{H}_1(S^3-K)cong mathbb{Z}$ due to Alexander duality.
Stepping back a bit: if all we wanted was the fundamental group, then consider the following. Recall that each lattice is parameterized by three distinct points in $mathbb{C}$ which sum to $0$. Unimodularity can be achieved after some positive real scaling factor of these points. Loops in $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ correspond to loops in the configuration space of triples of distinct points. Except for the small complication of rescaling which doesn't change anything homotopically, this is Artin's definition of the braid group.
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
$endgroup$
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075357%2fhomology-group-of-sl2-mathbbr-sl2-mathbbz%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We may write just as well $X = widetilde{SL_2 Bbb R}/widetilde{SL_2 Bbb Z}$, where the latter term is the preimage of $SL_2 Bbb Z$ under the covering map $widetilde{SL_2 Bbb R} to SL_2 Bbb R$. This identifes $X = B(widetilde{SL_2 Bbb Z}, 1)$, because its universal cover is the contractible space $widetilde{SL_2 Bbb R}$.
You are thus asking for a group homology computation. This is especially fortuitous because as a particularly emotive parsnip observed, $H_k X = 0$ for $k geq 3$. So all we need to know is $H_k X = H_k(widetilde{SL_2 Bbb Z}; Bbb Z)$ for $k = 1, 2$, where this last term means group homology (or equivalently, the homology of any space with contractible universal cover and $pi_1 X = widetilde{SL_2 Bbb Z}$).
To facilitate things, observe that as here this is the fundamental group of the complement of the trefoil knot, with presentation $langle x, y mid x^2 = y^3 rangle.$ Knot complements always have aspherical universal cover, so we have $X simeq S^3 setminus T_{2,3}$, where $T_{2,3}$ is the trefoil knot. Knot complements always have $H_1 X = Bbb Z$ and $H_k X = 0$ for $k > 1$ by Alexander duality or a Mayer-Vietoris calculation.
Likely there is an explicit homeomorphism, which I do not know but the clever 3-dimensional geometers do.
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
$endgroup$
1
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
add a comment |
$begingroup$
We may write just as well $X = widetilde{SL_2 Bbb R}/widetilde{SL_2 Bbb Z}$, where the latter term is the preimage of $SL_2 Bbb Z$ under the covering map $widetilde{SL_2 Bbb R} to SL_2 Bbb R$. This identifes $X = B(widetilde{SL_2 Bbb Z}, 1)$, because its universal cover is the contractible space $widetilde{SL_2 Bbb R}$.
You are thus asking for a group homology computation. This is especially fortuitous because as a particularly emotive parsnip observed, $H_k X = 0$ for $k geq 3$. So all we need to know is $H_k X = H_k(widetilde{SL_2 Bbb Z}; Bbb Z)$ for $k = 1, 2$, where this last term means group homology (or equivalently, the homology of any space with contractible universal cover and $pi_1 X = widetilde{SL_2 Bbb Z}$).
To facilitate things, observe that as here this is the fundamental group of the complement of the trefoil knot, with presentation $langle x, y mid x^2 = y^3 rangle.$ Knot complements always have aspherical universal cover, so we have $X simeq S^3 setminus T_{2,3}$, where $T_{2,3}$ is the trefoil knot. Knot complements always have $H_1 X = Bbb Z$ and $H_k X = 0$ for $k > 1$ by Alexander duality or a Mayer-Vietoris calculation.
Likely there is an explicit homeomorphism, which I do not know but the clever 3-dimensional geometers do.
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
$endgroup$
1
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
add a comment |
$begingroup$
We may write just as well $X = widetilde{SL_2 Bbb R}/widetilde{SL_2 Bbb Z}$, where the latter term is the preimage of $SL_2 Bbb Z$ under the covering map $widetilde{SL_2 Bbb R} to SL_2 Bbb R$. This identifes $X = B(widetilde{SL_2 Bbb Z}, 1)$, because its universal cover is the contractible space $widetilde{SL_2 Bbb R}$.
You are thus asking for a group homology computation. This is especially fortuitous because as a particularly emotive parsnip observed, $H_k X = 0$ for $k geq 3$. So all we need to know is $H_k X = H_k(widetilde{SL_2 Bbb Z}; Bbb Z)$ for $k = 1, 2$, where this last term means group homology (or equivalently, the homology of any space with contractible universal cover and $pi_1 X = widetilde{SL_2 Bbb Z}$).
To facilitate things, observe that as here this is the fundamental group of the complement of the trefoil knot, with presentation $langle x, y mid x^2 = y^3 rangle.$ Knot complements always have aspherical universal cover, so we have $X simeq S^3 setminus T_{2,3}$, where $T_{2,3}$ is the trefoil knot. Knot complements always have $H_1 X = Bbb Z$ and $H_k X = 0$ for $k > 1$ by Alexander duality or a Mayer-Vietoris calculation.
Likely there is an explicit homeomorphism, which I do not know but the clever 3-dimensional geometers do.
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
$endgroup$
We may write just as well $X = widetilde{SL_2 Bbb R}/widetilde{SL_2 Bbb Z}$, where the latter term is the preimage of $SL_2 Bbb Z$ under the covering map $widetilde{SL_2 Bbb R} to SL_2 Bbb R$. This identifes $X = B(widetilde{SL_2 Bbb Z}, 1)$, because its universal cover is the contractible space $widetilde{SL_2 Bbb R}$.
You are thus asking for a group homology computation. This is especially fortuitous because as a particularly emotive parsnip observed, $H_k X = 0$ for $k geq 3$. So all we need to know is $H_k X = H_k(widetilde{SL_2 Bbb Z}; Bbb Z)$ for $k = 1, 2$, where this last term means group homology (or equivalently, the homology of any space with contractible universal cover and $pi_1 X = widetilde{SL_2 Bbb Z}$).
To facilitate things, observe that as here this is the fundamental group of the complement of the trefoil knot, with presentation $langle x, y mid x^2 = y^3 rangle.$ Knot complements always have aspherical universal cover, so we have $X simeq S^3 setminus T_{2,3}$, where $T_{2,3}$ is the trefoil knot. Knot complements always have $H_1 X = Bbb Z$ and $H_k X = 0$ for $k > 1$ by Alexander duality or a Mayer-Vietoris calculation.
Likely there is an explicit homeomorphism, which I do not know but the clever 3-dimensional geometers do.
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
edited Jan 16 at 15:25
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
answered Jan 16 at 8:00
Mike MillerMike Miller
37.6k473140
37.6k473140
1
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
add a comment |
1
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
1
1
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
$begingroup$
(I asked one of those clever 3-dimensional geometers and tried writing it up.)
$endgroup$
– Kyle Miller
Feb 17 at 4:32
add a comment |
$begingroup$
The space of unimodular lattices in $mathbb{R}^2$ is acted upon transitively by $mathrm{SL}(2,mathbb{R})$, using the underlying action of $mathrm{SL}(2,mathbb{R})$ on $mathbb{R}^2$. One can view such a matrix as containing a basis for a unimodular lattice. The stabilizer of the ${(1,0),(0,1)}$ lattice is $mathrm{SL}(2,mathbb{Z})$, and so $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is in bijective correspondence with the space of unimodular lattices.
(Warning: what follows is my attempt to expand out a story someone once told me. I don't know the theory of elliptic functions well enough to know I haven't made any mistakes.)
The Weierstrass elliptic function $wp(z;Lambda)$ for a complex number $z$ and a lattice $Lambda$ for $mathbb{C}$ gives a point on the Riemann sphere $mathbb{C}_infty$. The quotient of $mathbb{C}$ by a lattice is a flat torus, and $wp(-;Lambda)$ can be thought of as a function from a particular flat torus to the Riemann sphere with four branch points of order two, via a hyperelliptic involution. The four branch points are the images of $0,omega_1/2,omega_2/2,(omega_1+omega_2)/2$, where $omega_1,omega_2inmathbb{C}$ generate $Lambda$. Let
begin{align*}
e_1&=wp(omega_1/2;Lambda)\
e_2&=wp(omega_2/2;Lambda)\
e_3&=wp((omega_1+omega_2)/2;Lambda)
end{align*}
which are distinct elements of $mathbb{C}$ satisfying $e_1+e_2+e_3=0$. These three numbers determine $Lambda$ in that $wp$ is the unique solution to the differential equation $(frac{dy}{dz})^2=4(y-e_1)(y-e_2)(y-e_3)$ with a pole at $0$. Expanded, the right-hand-side polynomial is $4y^3-g_2y-g_3$. Hence, polynomials of this form parameterize lattices, if we exclude those that have double roots. The discriminant of this polynomial is $(g_2^3-27 g_3^2)/16$, and $g_2^3-27 g_3^2$ is called the modular discriminant.
Every lattice has a unimodular representative by scaling, and in particular there is a deformation retract onto unimodular lattices. So, if we take all lattices parameterized using $(g_2,g_3)$ in $mathbb{C}^2$, remove those for which the discriminant is zero, then intersect with the unit $3$-sphere ${(z,w):lvert zrvert^2+lvert wrvert ^2=1}$, we have a parameterization of unimodular lattices after isotopy through the deformation retract. It is known that the link of the singularity of $g_2^3-27 g_3^2$ (i.e., the intersection between the zero-set and $S^3$) is a trefoil knot. (There should be a way to say the condition on $g_2$ and $g_3$ so that the corresponding lattice is unimodular, but I couldn't find it.)
So: $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is homeomorphic to $S^3$ minus a trefoil knot, which can concretely be written as
$${(z,w)inmathbb{C}^2:lvert zrvert^2+lvert wrvert ^2=1text{ and }z^3neq 27w^3}.$$
Knot complements have trivial reduced homology except $tilde{H}_1(S^3-K)cong mathbb{Z}$ due to Alexander duality.
Stepping back a bit: if all we wanted was the fundamental group, then consider the following. Recall that each lattice is parameterized by three distinct points in $mathbb{C}$ which sum to $0$. Unimodularity can be achieved after some positive real scaling factor of these points. Loops in $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ correspond to loops in the configuration space of triples of distinct points. Except for the small complication of rescaling which doesn't change anything homotopically, this is Artin's definition of the braid group.
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
$endgroup$
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
add a comment |
$begingroup$
The space of unimodular lattices in $mathbb{R}^2$ is acted upon transitively by $mathrm{SL}(2,mathbb{R})$, using the underlying action of $mathrm{SL}(2,mathbb{R})$ on $mathbb{R}^2$. One can view such a matrix as containing a basis for a unimodular lattice. The stabilizer of the ${(1,0),(0,1)}$ lattice is $mathrm{SL}(2,mathbb{Z})$, and so $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is in bijective correspondence with the space of unimodular lattices.
(Warning: what follows is my attempt to expand out a story someone once told me. I don't know the theory of elliptic functions well enough to know I haven't made any mistakes.)
The Weierstrass elliptic function $wp(z;Lambda)$ for a complex number $z$ and a lattice $Lambda$ for $mathbb{C}$ gives a point on the Riemann sphere $mathbb{C}_infty$. The quotient of $mathbb{C}$ by a lattice is a flat torus, and $wp(-;Lambda)$ can be thought of as a function from a particular flat torus to the Riemann sphere with four branch points of order two, via a hyperelliptic involution. The four branch points are the images of $0,omega_1/2,omega_2/2,(omega_1+omega_2)/2$, where $omega_1,omega_2inmathbb{C}$ generate $Lambda$. Let
begin{align*}
e_1&=wp(omega_1/2;Lambda)\
e_2&=wp(omega_2/2;Lambda)\
e_3&=wp((omega_1+omega_2)/2;Lambda)
end{align*}
which are distinct elements of $mathbb{C}$ satisfying $e_1+e_2+e_3=0$. These three numbers determine $Lambda$ in that $wp$ is the unique solution to the differential equation $(frac{dy}{dz})^2=4(y-e_1)(y-e_2)(y-e_3)$ with a pole at $0$. Expanded, the right-hand-side polynomial is $4y^3-g_2y-g_3$. Hence, polynomials of this form parameterize lattices, if we exclude those that have double roots. The discriminant of this polynomial is $(g_2^3-27 g_3^2)/16$, and $g_2^3-27 g_3^2$ is called the modular discriminant.
Every lattice has a unimodular representative by scaling, and in particular there is a deformation retract onto unimodular lattices. So, if we take all lattices parameterized using $(g_2,g_3)$ in $mathbb{C}^2$, remove those for which the discriminant is zero, then intersect with the unit $3$-sphere ${(z,w):lvert zrvert^2+lvert wrvert ^2=1}$, we have a parameterization of unimodular lattices after isotopy through the deformation retract. It is known that the link of the singularity of $g_2^3-27 g_3^2$ (i.e., the intersection between the zero-set and $S^3$) is a trefoil knot. (There should be a way to say the condition on $g_2$ and $g_3$ so that the corresponding lattice is unimodular, but I couldn't find it.)
So: $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is homeomorphic to $S^3$ minus a trefoil knot, which can concretely be written as
$${(z,w)inmathbb{C}^2:lvert zrvert^2+lvert wrvert ^2=1text{ and }z^3neq 27w^3}.$$
Knot complements have trivial reduced homology except $tilde{H}_1(S^3-K)cong mathbb{Z}$ due to Alexander duality.
Stepping back a bit: if all we wanted was the fundamental group, then consider the following. Recall that each lattice is parameterized by three distinct points in $mathbb{C}$ which sum to $0$. Unimodularity can be achieved after some positive real scaling factor of these points. Loops in $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ correspond to loops in the configuration space of triples of distinct points. Except for the small complication of rescaling which doesn't change anything homotopically, this is Artin's definition of the braid group.
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
$endgroup$
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
add a comment |
$begingroup$
The space of unimodular lattices in $mathbb{R}^2$ is acted upon transitively by $mathrm{SL}(2,mathbb{R})$, using the underlying action of $mathrm{SL}(2,mathbb{R})$ on $mathbb{R}^2$. One can view such a matrix as containing a basis for a unimodular lattice. The stabilizer of the ${(1,0),(0,1)}$ lattice is $mathrm{SL}(2,mathbb{Z})$, and so $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is in bijective correspondence with the space of unimodular lattices.
(Warning: what follows is my attempt to expand out a story someone once told me. I don't know the theory of elliptic functions well enough to know I haven't made any mistakes.)
The Weierstrass elliptic function $wp(z;Lambda)$ for a complex number $z$ and a lattice $Lambda$ for $mathbb{C}$ gives a point on the Riemann sphere $mathbb{C}_infty$. The quotient of $mathbb{C}$ by a lattice is a flat torus, and $wp(-;Lambda)$ can be thought of as a function from a particular flat torus to the Riemann sphere with four branch points of order two, via a hyperelliptic involution. The four branch points are the images of $0,omega_1/2,omega_2/2,(omega_1+omega_2)/2$, where $omega_1,omega_2inmathbb{C}$ generate $Lambda$. Let
begin{align*}
e_1&=wp(omega_1/2;Lambda)\
e_2&=wp(omega_2/2;Lambda)\
e_3&=wp((omega_1+omega_2)/2;Lambda)
end{align*}
which are distinct elements of $mathbb{C}$ satisfying $e_1+e_2+e_3=0$. These three numbers determine $Lambda$ in that $wp$ is the unique solution to the differential equation $(frac{dy}{dz})^2=4(y-e_1)(y-e_2)(y-e_3)$ with a pole at $0$. Expanded, the right-hand-side polynomial is $4y^3-g_2y-g_3$. Hence, polynomials of this form parameterize lattices, if we exclude those that have double roots. The discriminant of this polynomial is $(g_2^3-27 g_3^2)/16$, and $g_2^3-27 g_3^2$ is called the modular discriminant.
Every lattice has a unimodular representative by scaling, and in particular there is a deformation retract onto unimodular lattices. So, if we take all lattices parameterized using $(g_2,g_3)$ in $mathbb{C}^2$, remove those for which the discriminant is zero, then intersect with the unit $3$-sphere ${(z,w):lvert zrvert^2+lvert wrvert ^2=1}$, we have a parameterization of unimodular lattices after isotopy through the deformation retract. It is known that the link of the singularity of $g_2^3-27 g_3^2$ (i.e., the intersection between the zero-set and $S^3$) is a trefoil knot. (There should be a way to say the condition on $g_2$ and $g_3$ so that the corresponding lattice is unimodular, but I couldn't find it.)
So: $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is homeomorphic to $S^3$ minus a trefoil knot, which can concretely be written as
$${(z,w)inmathbb{C}^2:lvert zrvert^2+lvert wrvert ^2=1text{ and }z^3neq 27w^3}.$$
Knot complements have trivial reduced homology except $tilde{H}_1(S^3-K)cong mathbb{Z}$ due to Alexander duality.
Stepping back a bit: if all we wanted was the fundamental group, then consider the following. Recall that each lattice is parameterized by three distinct points in $mathbb{C}$ which sum to $0$. Unimodularity can be achieved after some positive real scaling factor of these points. Loops in $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ correspond to loops in the configuration space of triples of distinct points. Except for the small complication of rescaling which doesn't change anything homotopically, this is Artin's definition of the braid group.
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
$endgroup$
The space of unimodular lattices in $mathbb{R}^2$ is acted upon transitively by $mathrm{SL}(2,mathbb{R})$, using the underlying action of $mathrm{SL}(2,mathbb{R})$ on $mathbb{R}^2$. One can view such a matrix as containing a basis for a unimodular lattice. The stabilizer of the ${(1,0),(0,1)}$ lattice is $mathrm{SL}(2,mathbb{Z})$, and so $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is in bijective correspondence with the space of unimodular lattices.
(Warning: what follows is my attempt to expand out a story someone once told me. I don't know the theory of elliptic functions well enough to know I haven't made any mistakes.)
The Weierstrass elliptic function $wp(z;Lambda)$ for a complex number $z$ and a lattice $Lambda$ for $mathbb{C}$ gives a point on the Riemann sphere $mathbb{C}_infty$. The quotient of $mathbb{C}$ by a lattice is a flat torus, and $wp(-;Lambda)$ can be thought of as a function from a particular flat torus to the Riemann sphere with four branch points of order two, via a hyperelliptic involution. The four branch points are the images of $0,omega_1/2,omega_2/2,(omega_1+omega_2)/2$, where $omega_1,omega_2inmathbb{C}$ generate $Lambda$. Let
begin{align*}
e_1&=wp(omega_1/2;Lambda)\
e_2&=wp(omega_2/2;Lambda)\
e_3&=wp((omega_1+omega_2)/2;Lambda)
end{align*}
which are distinct elements of $mathbb{C}$ satisfying $e_1+e_2+e_3=0$. These three numbers determine $Lambda$ in that $wp$ is the unique solution to the differential equation $(frac{dy}{dz})^2=4(y-e_1)(y-e_2)(y-e_3)$ with a pole at $0$. Expanded, the right-hand-side polynomial is $4y^3-g_2y-g_3$. Hence, polynomials of this form parameterize lattices, if we exclude those that have double roots. The discriminant of this polynomial is $(g_2^3-27 g_3^2)/16$, and $g_2^3-27 g_3^2$ is called the modular discriminant.
Every lattice has a unimodular representative by scaling, and in particular there is a deformation retract onto unimodular lattices. So, if we take all lattices parameterized using $(g_2,g_3)$ in $mathbb{C}^2$, remove those for which the discriminant is zero, then intersect with the unit $3$-sphere ${(z,w):lvert zrvert^2+lvert wrvert ^2=1}$, we have a parameterization of unimodular lattices after isotopy through the deformation retract. It is known that the link of the singularity of $g_2^3-27 g_3^2$ (i.e., the intersection between the zero-set and $S^3$) is a trefoil knot. (There should be a way to say the condition on $g_2$ and $g_3$ so that the corresponding lattice is unimodular, but I couldn't find it.)
So: $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ is homeomorphic to $S^3$ minus a trefoil knot, which can concretely be written as
$${(z,w)inmathbb{C}^2:lvert zrvert^2+lvert wrvert ^2=1text{ and }z^3neq 27w^3}.$$
Knot complements have trivial reduced homology except $tilde{H}_1(S^3-K)cong mathbb{Z}$ due to Alexander duality.
Stepping back a bit: if all we wanted was the fundamental group, then consider the following. Recall that each lattice is parameterized by three distinct points in $mathbb{C}$ which sum to $0$. Unimodularity can be achieved after some positive real scaling factor of these points. Loops in $mathrm{SL}(2,mathbb{R})/mathrm{SL}(2,mathbb{Z})$ correspond to loops in the configuration space of triples of distinct points. Except for the small complication of rescaling which doesn't change anything homotopically, this is Artin's definition of the braid group.
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
edited Feb 17 at 4:37
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
answered Feb 17 at 4:27
Kyle MillerKyle Miller
9,023929
9,023929
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
add a comment |
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
$begingroup$
Very nice - this makes the picture eminently clear.
$endgroup$
– Mike Miller
Feb 17 at 6:17
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075357%2fhomology-group-of-sl2-mathbbr-sl2-mathbbz%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
For $q geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold.
$endgroup$
– Mike Miller
Jan 16 at 6:32
$begingroup$
@MikeMiller Thanks, my mistake.
$endgroup$
– Bonbon
Jan 16 at 6:38
1
$begingroup$
It's a well-known fact that $SL(2,mathbb Z)$ is not cocompact in $SL(2,mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :)
$endgroup$
– Cheerful Parsnip
Jan 16 at 6:51