Fundamental group of $T/mathbb Z_2setminus{text{singular points}}$
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Let $T=S^1times S^1$. There is a $mathbb Z_2$-action on $T$ defined by $xsim -x$ (considering $T$ as a quotient of $mathbb R^2$).
The quotient $X:=T/mathbb Z_2$this has four singular points, say $x_1,...,x_4$ (corresponding to the four half lattice points in $mathbb R^2$). I want to calculate the fundamental group of $Xsetminus{x_1,...,x_4}$.
So far I could think of the following:
Consider $Y:=Tsetminus{x_1,...,x_4}$ (abuse of notation). Then $Ytilde = S^1vee S^1vee S^1vee S^1vee S^1.$ Now $X=Y/mathbb Z_2$. So we need to check how $mathbb Z_2$ acts on the $S^1$-summands.
From these we can choose three summands to be small circles around a singularity. The $mathbb Z_2$-action on them is then simply the $mathbb Z_2$-action on a circle with center $0$ in $mathbb R^2$.
So for them $S^1/mathbb Z_2=mathbb RP^1=S^1$.
But what about the other two summands? They seem to be glued together in a way, such that the $mathbb Z_2$ action acts more complicatedly.
Update: Thanks to the comment by Mike Miller, I was able to look up the construction of the "pillowcase". SO this is basically answered.
However, I would be interested in generalisations of this in higher dimensions. For example $T^4/mathbb Z_2$.
geometry algebraic-topology group-actions fundamental-groups
asked Nov 17 at 15:14
mathsta
1858
|
show 1 more comment
up vote
4
down vote
favorite
Let $T=S^1times S^1$. There is a $mathbb Z_2$-action on $T$ defined by $xsim -x$ (considering $T$ as a quotient of $mathbb R^2$).
The quotient $X:=T/mathbb Z_2$this has four singular points, say $x_1,...,x_4$ (corresponding to the four half lattice points in $mathbb R^2$). I want to calculate the fundamental group of $Xsetminus{x_1,...,x_4}$.
So far I could think of the following:
Consider $Y:=Tsetminus{x_1,...,x_4}$ (abuse of notation). Then $Ytilde = S^1vee S^1vee S^1vee S^1vee S^1.$ Now $X=Y/mathbb Z_2$. So we need to check how $mathbb Z_2$ acts on the $S^1$-summands.
From these we can choose three summands to be small circles around a singularity. The $mathbb Z_2$-action on them is then simply the $mathbb Z_2$-action on a circle with center $0$ in $mathbb R^2$.
So for them $S^1/mathbb Z_2=mathbb RP^1=S^1$.
But what about the other two summands? They seem to be glued together in a way, such that the $mathbb Z_2$ action acts more complicatedly.
Update: Thanks to the comment by Mike Miller, I was able to look up the construction of the "pillowcase". SO this is basically answered.
However, I would be interested in generalisations of this in higher dimensions. For example $T^4/mathbb Z_2$.
geometry algebraic-topology group-actions fundamental-groups
asked Nov 17 at 15:14
mathsta
1858
$X-{x_1,x_2,x_3,x_4}$ shouldn't have boundary, it's an open manifold that should deformation retract onto a 1-complex.
– Neal
Nov 17 at 15:18
Ok, I will fix that. This shouldn't be important for the calculation, right?
– mathsta
Nov 17 at 15:20
3
It's easier to just identify $X$ as the sphere with four singular points (draw a fundamental domain for the action on $T^2$ and see how it glues up boundaries; the picture you get is often called a "pillowcase"). Then deleting those 4 points gives you the thrice-punctured plane, with fundamental group the free group on 3 generators.
– Mike Miller
Nov 17 at 15:29
2
Please fix your title so that it is not misleading.
– Pedro Tamaroff♦
Nov 18 at 13:12
Regarding your update, if you have something else to ask regarding the generalization to higher dimensions, that might be good for a new question if it can be formulated clearly.
– Lee Mosher
Nov 18 at 14:22
|
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $T=S^1times S^1$. There is a $mathbb Z_2$-action on $T$ defined by $xsim -x$ (considering $T$ as a quotient of $mathbb R^2$).
The quotient $X:=T/mathbb Z_2$this has four singular points, say $x_1,...,x_4$ (corresponding to the four half lattice points in $mathbb R^2$). I want to calculate the fundamental group of $Xsetminus{x_1,...,x_4}$.
So far I could think of the following:
Consider $Y:=Tsetminus{x_1,...,x_4}$ (abuse of notation). Then $Ytilde = S^1vee S^1vee S^1vee S^1vee S^1.$ Now $X=Y/mathbb Z_2$. So we need to check how $mathbb Z_2$ acts on the $S^1$-summands.
From these we can choose three summands to be small circles around a singularity. The $mathbb Z_2$-action on them is then simply the $mathbb Z_2$-action on a circle with center $0$ in $mathbb R^2$.
So for them $S^1/mathbb Z_2=mathbb RP^1=S^1$.
But what about the other two summands? They seem to be glued together in a way, such that the $mathbb Z_2$ action acts more complicatedly.
Update: Thanks to the comment by Mike Miller, I was able to look up the construction of the "pillowcase". SO this is basically answered.
However, I would be interested in generalisations of this in higher dimensions. For example $T^4/mathbb Z_2$.
geometry algebraic-topology group-actions fundamental-groups
asked Nov 17 at 15:14
mathsta
1858
Let $T=S^1times S^1$. There is a $mathbb Z_2$-action on $T$ defined by $xsim -x$ (considering $T$ as a quotient of $mathbb R^2$).
The quotient $X:=T/mathbb Z_2$this has four singular points, say $x_1,...,x_4$ (corresponding to the four half lattice points in $mathbb R^2$). I want to calculate the fundamental group of $Xsetminus{x_1,...,x_4}$.
So far I could think of the following:
Consider $Y:=Tsetminus{x_1,...,x_4}$ (abuse of notation). Then $Ytilde = S^1vee S^1vee S^1vee S^1vee S^1.$ Now $X=Y/mathbb Z_2$. So we need to check how $mathbb Z_2$ acts on the $S^1$-summands.
From these we can choose three summands to be small circles around a singularity. The $mathbb Z_2$-action on them is then simply the $mathbb Z_2$-action on a circle with center $0$ in $mathbb R^2$.
So for them $S^1/mathbb Z_2=mathbb RP^1=S^1$.
But what about the other two summands? They seem to be glued together in a way, such that the $mathbb Z_2$ action acts more complicatedly.
Update: Thanks to the comment by Mike Miller, I was able to look up the construction of the "pillowcase". SO this is basically answered.
However, I would be interested in generalisations of this in higher dimensions. For example $T^4/mathbb Z_2$.
geometry algebraic-topology group-actions fundamental-groups
geometry algebraic-topology group-actions fundamental-groups
asked Nov 17 at 15:14
mathsta
1858
asked Nov 17 at 15:14
mathsta
1858
edited Nov 18 at 20:28
asked Nov 17 at 15:14
mathsta
1858
asked Nov 17 at 15:14
mathsta
1858
asked Nov 17 at 15:14
mathsta
1858
1858
$X-{x_1,x_2,x_3,x_4}$ shouldn't have boundary, it's an open manifold that should deformation retract onto a 1-complex.
– Neal
Nov 17 at 15:18
Ok, I will fix that. This shouldn't be important for the calculation, right?
– mathsta
Nov 17 at 15:20
3
It's easier to just identify $X$ as the sphere with four singular points (draw a fundamental domain for the action on $T^2$ and see how it glues up boundaries; the picture you get is often called a "pillowcase"). Then deleting those 4 points gives you the thrice-punctured plane, with fundamental group the free group on 3 generators.
– Mike Miller
Nov 17 at 15:29
2
Please fix your title so that it is not misleading.
– Pedro Tamaroff♦
Nov 18 at 13:12
Regarding your update, if you have something else to ask regarding the generalization to higher dimensions, that might be good for a new question if it can be formulated clearly.
– Lee Mosher
Nov 18 at 14:22
|
show 1 more comment
$X-{x_1,x_2,x_3,x_4}$ shouldn't have boundary, it's an open manifold that should deformation retract onto a 1-complex.
– Neal
Nov 17 at 15:18
Ok, I will fix that. This shouldn't be important for the calculation, right?
– mathsta
Nov 17 at 15:20
3
It's easier to just identify $X$ as the sphere with four singular points (draw a fundamental domain for the action on $T^2$ and see how it glues up boundaries; the picture you get is often called a "pillowcase"). Then deleting those 4 points gives you the thrice-punctured plane, with fundamental group the free group on 3 generators.
– Mike Miller
Nov 17 at 15:29
2
Please fix your title so that it is not misleading.
– Pedro Tamaroff♦
Nov 18 at 13:12
Regarding your update, if you have something else to ask regarding the generalization to higher dimensions, that might be good for a new question if it can be formulated clearly.
– Lee Mosher
Nov 18 at 14:22
$X-{x_1,x_2,x_3,x_4}$ shouldn't have boundary, it's an open manifold that should deformation retract onto a 1-complex.
– Neal
Nov 17 at 15:18
$X-{x_1,x_2,x_3,x_4}$ shouldn't have boundary, it's an open manifold that should deformation retract onto a 1-complex.
– Neal
Nov 17 at 15:18
Ok, I will fix that. This shouldn't be important for the calculation, right?
– mathsta
Nov 17 at 15:20
Ok, I will fix that. This shouldn't be important for the calculation, right?
– mathsta
Nov 17 at 15:20
3
3
It's easier to just identify $X$ as the sphere with four singular points (draw a fundamental domain for the action on $T^2$ and see how it glues up boundaries; the picture you get is often called a "pillowcase"). Then deleting those 4 points gives you the thrice-punctured plane, with fundamental group the free group on 3 generators.
– Mike Miller
Nov 17 at 15:29
It's easier to just identify $X$ as the sphere with four singular points (draw a fundamental domain for the action on $T^2$ and see how it glues up boundaries; the picture you get is often called a "pillowcase"). Then deleting those 4 points gives you the thrice-punctured plane, with fundamental group the free group on 3 generators.
– Mike Miller
Nov 17 at 15:29
2
2
Please fix your title so that it is not misleading.
– Pedro Tamaroff♦
Nov 18 at 13:12
Please fix your title so that it is not misleading.
– Pedro Tamaroff♦
Nov 18 at 13:12
Regarding your update, if you have something else to ask regarding the generalization to higher dimensions, that might be good for a new question if it can be formulated clearly.
– Lee Mosher
Nov 18 at 14:22
Regarding your update, if you have something else to ask regarding the generalization to higher dimensions, that might be good for a new question if it can be formulated clearly.
– Lee Mosher
Nov 18 at 14:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You may think of $T^2 = Bbb R^2/Bbb Z^2$ as $[0,1]^2$ with opposite edges identified; picture this as the unit square in $Bbb R^2$. That $[0,1]^2$ is a fundamental domain means precisely that everything in $Bbb R^2$ is equivalent to some point in $[0,1]^2$, and the only remaining relations glue pieces of the boundary. In this case, they are $(0,t) sim (1,t)$ and $(t,0) sim (t,1)$.
The action of negation on the unit square in $Bbb R^2/Bbb Z^2$ is to send $$(x,y) mapsto (-x, -y) sim (1-x, 1-y).$$ So pictured on the unit square, the action is reflection across $(frac 12, frac 12)$.
After quotienting by this action, everything in the unit square is equivalent to something in $[0, 1/2] times [0, 1]$. The points in the interior are equivalent to no other points in the interior, so the only thing that remains is to understand the ways the boundary glues together. This is a systematic procedure that requires no cleverness on our part, we just look at the boundary and see how it glues under our given relations.
We know that $(t,0) sim (t, 1)$ from the usual relations on the torus; this gives us the gluing of top to bottom you see there. We now have a cylinder.
We also know from the reflection relation that $(1/2, t) sim (1/2, 1-t)$ and $(0,t) sim (1,1-t)$, and this is equivalent to $(0,1-t)$. These are the two "fold the edges" relations you see in your picture. On the cylinder, it is like gluing the boundary circles together by "folding them into a crease": this is where the pillowcase picture comes from.
There is no difficulty in carrying this out in higher dimensions: you work with $[0,1]^n$, the reflection relation reduces you to understanding the gluings on the boundary of $[0,1/2] times [0,1]^{n-1}$.
answered Nov 18 at 20:45
Mike Miller
35.1k467132
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You may think of $T^2 = Bbb R^2/Bbb Z^2$ as $[0,1]^2$ with opposite edges identified; picture this as the unit square in $Bbb R^2$. That $[0,1]^2$ is a fundamental domain means precisely that everything in $Bbb R^2$ is equivalent to some point in $[0,1]^2$, and the only remaining relations glue pieces of the boundary. In this case, they are $(0,t) sim (1,t)$ and $(t,0) sim (t,1)$.
The action of negation on the unit square in $Bbb R^2/Bbb Z^2$ is to send $$(x,y) mapsto (-x, -y) sim (1-x, 1-y).$$ So pictured on the unit square, the action is reflection across $(frac 12, frac 12)$.
After quotienting by this action, everything in the unit square is equivalent to something in $[0, 1/2] times [0, 1]$. The points in the interior are equivalent to no other points in the interior, so the only thing that remains is to understand the ways the boundary glues together. This is a systematic procedure that requires no cleverness on our part, we just look at the boundary and see how it glues under our given relations.
We know that $(t,0) sim (t, 1)$ from the usual relations on the torus; this gives us the gluing of top to bottom you see there. We now have a cylinder.
We also know from the reflection relation that $(1/2, t) sim (1/2, 1-t)$ and $(0,t) sim (1,1-t)$, and this is equivalent to $(0,1-t)$. These are the two "fold the edges" relations you see in your picture. On the cylinder, it is like gluing the boundary circles together by "folding them into a crease": this is where the pillowcase picture comes from.
There is no difficulty in carrying this out in higher dimensions: you work with $[0,1]^n$, the reflection relation reduces you to understanding the gluings on the boundary of $[0,1/2] times [0,1]^{n-1}$.
answered Nov 18 at 20:45
Mike Miller
35.1k467132
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
|
show 1 more comment
up vote
2
down vote
accepted
You may think of $T^2 = Bbb R^2/Bbb Z^2$ as $[0,1]^2$ with opposite edges identified; picture this as the unit square in $Bbb R^2$. That $[0,1]^2$ is a fundamental domain means precisely that everything in $Bbb R^2$ is equivalent to some point in $[0,1]^2$, and the only remaining relations glue pieces of the boundary. In this case, they are $(0,t) sim (1,t)$ and $(t,0) sim (t,1)$.
The action of negation on the unit square in $Bbb R^2/Bbb Z^2$ is to send $$(x,y) mapsto (-x, -y) sim (1-x, 1-y).$$ So pictured on the unit square, the action is reflection across $(frac 12, frac 12)$.
After quotienting by this action, everything in the unit square is equivalent to something in $[0, 1/2] times [0, 1]$. The points in the interior are equivalent to no other points in the interior, so the only thing that remains is to understand the ways the boundary glues together. This is a systematic procedure that requires no cleverness on our part, we just look at the boundary and see how it glues under our given relations.
We know that $(t,0) sim (t, 1)$ from the usual relations on the torus; this gives us the gluing of top to bottom you see there. We now have a cylinder.
We also know from the reflection relation that $(1/2, t) sim (1/2, 1-t)$ and $(0,t) sim (1,1-t)$, and this is equivalent to $(0,1-t)$. These are the two "fold the edges" relations you see in your picture. On the cylinder, it is like gluing the boundary circles together by "folding them into a crease": this is where the pillowcase picture comes from.
There is no difficulty in carrying this out in higher dimensions: you work with $[0,1]^n$, the reflection relation reduces you to understanding the gluings on the boundary of $[0,1/2] times [0,1]^{n-1}$.
answered Nov 18 at 20:45
Mike Miller
35.1k467132
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You may think of $T^2 = Bbb R^2/Bbb Z^2$ as $[0,1]^2$ with opposite edges identified; picture this as the unit square in $Bbb R^2$. That $[0,1]^2$ is a fundamental domain means precisely that everything in $Bbb R^2$ is equivalent to some point in $[0,1]^2$, and the only remaining relations glue pieces of the boundary. In this case, they are $(0,t) sim (1,t)$ and $(t,0) sim (t,1)$.
The action of negation on the unit square in $Bbb R^2/Bbb Z^2$ is to send $$(x,y) mapsto (-x, -y) sim (1-x, 1-y).$$ So pictured on the unit square, the action is reflection across $(frac 12, frac 12)$.
After quotienting by this action, everything in the unit square is equivalent to something in $[0, 1/2] times [0, 1]$. The points in the interior are equivalent to no other points in the interior, so the only thing that remains is to understand the ways the boundary glues together. This is a systematic procedure that requires no cleverness on our part, we just look at the boundary and see how it glues under our given relations.
We know that $(t,0) sim (t, 1)$ from the usual relations on the torus; this gives us the gluing of top to bottom you see there. We now have a cylinder.
We also know from the reflection relation that $(1/2, t) sim (1/2, 1-t)$ and $(0,t) sim (1,1-t)$, and this is equivalent to $(0,1-t)$. These are the two "fold the edges" relations you see in your picture. On the cylinder, it is like gluing the boundary circles together by "folding them into a crease": this is where the pillowcase picture comes from.
There is no difficulty in carrying this out in higher dimensions: you work with $[0,1]^n$, the reflection relation reduces you to understanding the gluings on the boundary of $[0,1/2] times [0,1]^{n-1}$.
answered Nov 18 at 20:45
Mike Miller
35.1k467132
You may think of $T^2 = Bbb R^2/Bbb Z^2$ as $[0,1]^2$ with opposite edges identified; picture this as the unit square in $Bbb R^2$. That $[0,1]^2$ is a fundamental domain means precisely that everything in $Bbb R^2$ is equivalent to some point in $[0,1]^2$, and the only remaining relations glue pieces of the boundary. In this case, they are $(0,t) sim (1,t)$ and $(t,0) sim (t,1)$.
The action of negation on the unit square in $Bbb R^2/Bbb Z^2$ is to send $$(x,y) mapsto (-x, -y) sim (1-x, 1-y).$$ So pictured on the unit square, the action is reflection across $(frac 12, frac 12)$.
After quotienting by this action, everything in the unit square is equivalent to something in $[0, 1/2] times [0, 1]$. The points in the interior are equivalent to no other points in the interior, so the only thing that remains is to understand the ways the boundary glues together. This is a systematic procedure that requires no cleverness on our part, we just look at the boundary and see how it glues under our given relations.
We know that $(t,0) sim (t, 1)$ from the usual relations on the torus; this gives us the gluing of top to bottom you see there. We now have a cylinder.
We also know from the reflection relation that $(1/2, t) sim (1/2, 1-t)$ and $(0,t) sim (1,1-t)$, and this is equivalent to $(0,1-t)$. These are the two "fold the edges" relations you see in your picture. On the cylinder, it is like gluing the boundary circles together by "folding them into a crease": this is where the pillowcase picture comes from.
There is no difficulty in carrying this out in higher dimensions: you work with $[0,1]^n$, the reflection relation reduces you to understanding the gluings on the boundary of $[0,1/2] times [0,1]^{n-1}$.
answered Nov 18 at 20:45
Mike Miller
35.1k467132
answered Nov 18 at 20:45
Mike Miller
35.1k467132
answered Nov 18 at 20:45
Mike Miller
35.1k467132
answered Nov 18 at 20:45
Mike Miller
35.1k467132
35.1k467132
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
|
show 1 more comment
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
Thanks for you answer. Regarding higher dimensions: Is the quotient still a sphere?
– mathsta
Nov 18 at 21:08
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
@mathsta The quotient is no longer a topological manifold. Neighborhoods of the fixed points in $T^n$ look like the unit ball in $Bbb R^n$ with the negation action (which you may think of as the cone on $S^{n-1}$), and quotienting by that gives you the cone on $Bbb{RP}^{n-1}$. Cones on manifolds are never manifolds unless what you started with was a sphere.
– Mike Miller
Nov 18 at 21:15
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
Ah right. I was actually aware of this when I asked the main question (hence I wanted to remove the singular points). Do you know if the quotient without the singular points is homotopically equivalent to something more familiar? Also, do you know what happens if instead of removing the singular points, I blow them up?
– mathsta
Nov 18 at 21:31
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
@mathsta You have a lot of questions! In general it would be better to add all these details to a new question instead of adding them afterwards. I will say what I can here though. In the case of $T^4 = Bbb C^2/Bbb Z^2$, there is a natural way to "blow up" the singular points, replacing neighborhoods with copies of $T^*Bbb{CP}^1$. This is called the Kummer construction and is a construction of the $K3$ surface in this dimension. There are analogues in other dimensions that one can use to construct manifolds with special holonomy, due to Joyce.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
If you delete the singular points, your $T^n$ (equivariantly) deformation retracts onto the subset given by the demand that one of your coordinates is $1/4$ or $3/4$. Picturing this in the fundamental domain, you see that the quotient is the union of $2n - 1$ tori of dimension $(n-1)$ glued together a certain way along subtori. If you could work out the combinatorics of the way these glue together you could determine the fundamental group / homology groups. I am not going to try right now.
– Mike Miller
Nov 18 at 21:41
|
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$X-{x_1,x_2,x_3,x_4}$ shouldn't have boundary, it's an open manifold that should deformation retract onto a 1-complex.
– Neal
Nov 17 at 15:18
Ok, I will fix that. This shouldn't be important for the calculation, right?
– mathsta
Nov 17 at 15:20
3
It's easier to just identify $X$ as the sphere with four singular points (draw a fundamental domain for the action on $T^2$ and see how it glues up boundaries; the picture you get is often called a "pillowcase"). Then deleting those 4 points gives you the thrice-punctured plane, with fundamental group the free group on 3 generators.
– Mike Miller
Nov 17 at 15:29
2
Please fix your title so that it is not misleading.
– Pedro Tamaroff♦
Nov 18 at 13:12
Regarding your update, if you have something else to ask regarding the generalization to higher dimensions, that might be good for a new question if it can be formulated clearly.
– Lee Mosher
Nov 18 at 14:22