(Co)homology of $S^2×S^2/ℤ_2$
3
$begingroup$
Cohomology of $S^2times S^2/mathbb{Z}_2$
I was looking at this question, the accepted answer uses the homology of the space to find the cohomology. I was wondering how one could compute the homology of an orbit space like this in the first place?
"The product of two spheres admits a diagonal $ℤ_2$ action, $(x, y)mapsto(−x,−y)$" - describe the homology groups of the orbit space of this free action.
general-topology algebraic-topology
edited Dec 9 '18 at 5:12
Andrews
1,2691421
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
$endgroup$
|
show 7 more comments
3
$begingroup$
Cohomology of $S^2times S^2/mathbb{Z}_2$
I was looking at this question, the accepted answer uses the homology of the space to find the cohomology. I was wondering how one could compute the homology of an orbit space like this in the first place?
"The product of two spheres admits a diagonal $ℤ_2$ action, $(x, y)mapsto(−x,−y)$" - describe the homology groups of the orbit space of this free action.
general-topology algebraic-topology
edited Dec 9 '18 at 5:12
Andrews
1,2691421
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
$endgroup$
2
$begingroup$
As long as you can construct a nice enough cell decomposition or triangulation of $S^2 times S^2$ has, you can derive a cellular decomposition or triangulation of the orbit, and then proceed from there directly using cellular or simplicial homology.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 0:57
$begingroup$
@RolfHoyer I am sorry, I am not sure I understand you completely. Lets say I do a cellular decomposition of $S2 x S2$ $S2xS2$ has a one $e^2$ and one $e^0$ cell for each S2. So the cell structure for $S2 times S2$ is given by the cells $e^0 times e^0$, $e^2 times e^2$, two $e^0 times e^2$.
$endgroup$
– Shelly B
Dec 9 '18 at 1:35
$begingroup$
@RolfHoyer I am unsure how I can use this to find a decomposition of the orbit
$endgroup$
– Shelly B
Dec 9 '18 at 1:37
$begingroup$
That is one decomposition, but it isn't the only one. In order to get a nice cellular structure on the orbit, you need the group action to never produce a non-identity self-map from a cell to itself. That way, when you go to the quotient it just identifies cells without needing to do any folding. I would recommend subdividing your decomposition further. It should suffice to use the decomposition of $S^2$ into two cells in each of dimensions 0,1,2, I think.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 1:37
1
$begingroup$
@RolfHoyer (3) works. I would write it up if and only if it is of interest to OP - I worry I would be writing something complicated and somewhat abstract for little gain for others. The cellular chain complex is not so impossible to calculate here (I remember doing this explicitly around the time of that original thread), but I do not have the energy to try again.
$endgroup$
– user98602
Dec 9 '18 at 3:17
|
show 7 more comments
3
3
3
0
$begingroup$
Cohomology of $S^2times S^2/mathbb{Z}_2$
I was looking at this question, the accepted answer uses the homology of the space to find the cohomology. I was wondering how one could compute the homology of an orbit space like this in the first place?
"The product of two spheres admits a diagonal $ℤ_2$ action, $(x, y)mapsto(−x,−y)$" - describe the homology groups of the orbit space of this free action.
general-topology algebraic-topology
edited Dec 9 '18 at 5:12
Andrews
1,2691421
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
$endgroup$
Cohomology of $S^2times S^2/mathbb{Z}_2$
I was looking at this question, the accepted answer uses the homology of the space to find the cohomology. I was wondering how one could compute the homology of an orbit space like this in the first place?
"The product of two spheres admits a diagonal $ℤ_2$ action, $(x, y)mapsto(−x,−y)$" - describe the homology groups of the orbit space of this free action.
general-topology algebraic-topology
general-topology algebraic-topology
edited Dec 9 '18 at 5:12
Andrews
1,2691421
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
edited Dec 9 '18 at 5:12
Andrews
1,2691421
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
edited Dec 9 '18 at 5:12
Andrews
1,2691421
edited Dec 9 '18 at 5:12
Andrews
1,2691421
edited Dec 9 '18 at 5:12
Andrews
1,2691421
1,2691421
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
asked Dec 9 '18 at 0:50
Shelly BShelly B
383
383
2
$begingroup$
As long as you can construct a nice enough cell decomposition or triangulation of $S^2 times S^2$ has, you can derive a cellular decomposition or triangulation of the orbit, and then proceed from there directly using cellular or simplicial homology.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 0:57
$begingroup$
@RolfHoyer I am sorry, I am not sure I understand you completely. Lets say I do a cellular decomposition of $S2 x S2$ $S2xS2$ has a one $e^2$ and one $e^0$ cell for each S2. So the cell structure for $S2 times S2$ is given by the cells $e^0 times e^0$, $e^2 times e^2$, two $e^0 times e^2$.
$endgroup$
– Shelly B
Dec 9 '18 at 1:35
$begingroup$
@RolfHoyer I am unsure how I can use this to find a decomposition of the orbit
$endgroup$
– Shelly B
Dec 9 '18 at 1:37
$begingroup$
That is one decomposition, but it isn't the only one. In order to get a nice cellular structure on the orbit, you need the group action to never produce a non-identity self-map from a cell to itself. That way, when you go to the quotient it just identifies cells without needing to do any folding. I would recommend subdividing your decomposition further. It should suffice to use the decomposition of $S^2$ into two cells in each of dimensions 0,1,2, I think.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 1:37
1
$begingroup$
@RolfHoyer (3) works. I would write it up if and only if it is of interest to OP - I worry I would be writing something complicated and somewhat abstract for little gain for others. The cellular chain complex is not so impossible to calculate here (I remember doing this explicitly around the time of that original thread), but I do not have the energy to try again.
$endgroup$
– user98602
Dec 9 '18 at 3:17
|
show 7 more comments
2
$begingroup$
As long as you can construct a nice enough cell decomposition or triangulation of $S^2 times S^2$ has, you can derive a cellular decomposition or triangulation of the orbit, and then proceed from there directly using cellular or simplicial homology.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 0:57
$begingroup$
@RolfHoyer I am sorry, I am not sure I understand you completely. Lets say I do a cellular decomposition of $S2 x S2$ $S2xS2$ has a one $e^2$ and one $e^0$ cell for each S2. So the cell structure for $S2 times S2$ is given by the cells $e^0 times e^0$, $e^2 times e^2$, two $e^0 times e^2$.
$endgroup$
– Shelly B
Dec 9 '18 at 1:35
$begingroup$
@RolfHoyer I am unsure how I can use this to find a decomposition of the orbit
$endgroup$
– Shelly B
Dec 9 '18 at 1:37
$begingroup$
That is one decomposition, but it isn't the only one. In order to get a nice cellular structure on the orbit, you need the group action to never produce a non-identity self-map from a cell to itself. That way, when you go to the quotient it just identifies cells without needing to do any folding. I would recommend subdividing your decomposition further. It should suffice to use the decomposition of $S^2$ into two cells in each of dimensions 0,1,2, I think.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 1:37
1
$begingroup$
@RolfHoyer (3) works. I would write it up if and only if it is of interest to OP - I worry I would be writing something complicated and somewhat abstract for little gain for others. The cellular chain complex is not so impossible to calculate here (I remember doing this explicitly around the time of that original thread), but I do not have the energy to try again.
$endgroup$
– user98602
Dec 9 '18 at 3:17
2
2
$begingroup$
As long as you can construct a nice enough cell decomposition or triangulation of $S^2 times S^2$ has, you can derive a cellular decomposition or triangulation of the orbit, and then proceed from there directly using cellular or simplicial homology.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 0:57
$begingroup$
As long as you can construct a nice enough cell decomposition or triangulation of $S^2 times S^2$ has, you can derive a cellular decomposition or triangulation of the orbit, and then proceed from there directly using cellular or simplicial homology.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 0:57
$begingroup$
@RolfHoyer I am sorry, I am not sure I understand you completely. Lets say I do a cellular decomposition of $S2 x S2$ $S2xS2$ has a one $e^2$ and one $e^0$ cell for each S2. So the cell structure for $S2 times S2$ is given by the cells $e^0 times e^0$, $e^2 times e^2$, two $e^0 times e^2$.
$endgroup$
– Shelly B
Dec 9 '18 at 1:35
$begingroup$
@RolfHoyer I am sorry, I am not sure I understand you completely. Lets say I do a cellular decomposition of $S2 x S2$ $S2xS2$ has a one $e^2$ and one $e^0$ cell for each S2. So the cell structure for $S2 times S2$ is given by the cells $e^0 times e^0$, $e^2 times e^2$, two $e^0 times e^2$.
$endgroup$
– Shelly B
Dec 9 '18 at 1:35
$begingroup$
@RolfHoyer I am unsure how I can use this to find a decomposition of the orbit
$endgroup$
– Shelly B
Dec 9 '18 at 1:37
$begingroup$
@RolfHoyer I am unsure how I can use this to find a decomposition of the orbit
$endgroup$
– Shelly B
Dec 9 '18 at 1:37
$begingroup$
That is one decomposition, but it isn't the only one. In order to get a nice cellular structure on the orbit, you need the group action to never produce a non-identity self-map from a cell to itself. That way, when you go to the quotient it just identifies cells without needing to do any folding. I would recommend subdividing your decomposition further. It should suffice to use the decomposition of $S^2$ into two cells in each of dimensions 0,1,2, I think.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 1:37
$begingroup$
That is one decomposition, but it isn't the only one. In order to get a nice cellular structure on the orbit, you need the group action to never produce a non-identity self-map from a cell to itself. That way, when you go to the quotient it just identifies cells without needing to do any folding. I would recommend subdividing your decomposition further. It should suffice to use the decomposition of $S^2$ into two cells in each of dimensions 0,1,2, I think.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 1:37
1
1
$begingroup$
@RolfHoyer (3) works. I would write it up if and only if it is of interest to OP - I worry I would be writing something complicated and somewhat abstract for little gain for others. The cellular chain complex is not so impossible to calculate here (I remember doing this explicitly around the time of that original thread), but I do not have the energy to try again.
$endgroup$
– user98602
Dec 9 '18 at 3:17
$begingroup$
@RolfHoyer (3) works. I would write it up if and only if it is of interest to OP - I worry I would be writing something complicated and somewhat abstract for little gain for others. The cellular chain complex is not so impossible to calculate here (I remember doing this explicitly around the time of that original thread), but I do not have the energy to try again.
$endgroup$
– user98602
Dec 9 '18 at 3:17
|
show 7 more comments
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2
$begingroup$
As long as you can construct a nice enough cell decomposition or triangulation of $S^2 times S^2$ has, you can derive a cellular decomposition or triangulation of the orbit, and then proceed from there directly using cellular or simplicial homology.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 0:57
$begingroup$
@RolfHoyer I am sorry, I am not sure I understand you completely. Lets say I do a cellular decomposition of $S2 x S2$ $S2xS2$ has a one $e^2$ and one $e^0$ cell for each S2. So the cell structure for $S2 times S2$ is given by the cells $e^0 times e^0$, $e^2 times e^2$, two $e^0 times e^2$.
$endgroup$
– Shelly B
Dec 9 '18 at 1:35
$begingroup$
@RolfHoyer I am unsure how I can use this to find a decomposition of the orbit
$endgroup$
– Shelly B
Dec 9 '18 at 1:37
$begingroup$
That is one decomposition, but it isn't the only one. In order to get a nice cellular structure on the orbit, you need the group action to never produce a non-identity self-map from a cell to itself. That way, when you go to the quotient it just identifies cells without needing to do any folding. I would recommend subdividing your decomposition further. It should suffice to use the decomposition of $S^2$ into two cells in each of dimensions 0,1,2, I think.
$endgroup$
– Rolf Hoyer
Dec 9 '18 at 1:37
1
$begingroup$
@RolfHoyer (3) works. I would write it up if and only if it is of interest to OP - I worry I would be writing something complicated and somewhat abstract for little gain for others. The cellular chain complex is not so impossible to calculate here (I remember doing this explicitly around the time of that original thread), but I do not have the energy to try again.
$endgroup$
– user98602
Dec 9 '18 at 3:17