How many charts are needed to cover a 2-torus?
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Can you please answer this question with explanation ? I just learned about the charts needed to cover 1 and 2 spheres but got confused for the case of torus. It would be great if you guys could help.
manifolds
asked Jan 28 '14 at 1:48
Mush
383
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up vote
7
down vote
favorite
Can you please answer this question with explanation ? I just learned about the charts needed to cover 1 and 2 spheres but got confused for the case of torus. It would be great if you guys could help.
manifolds
asked Jan 28 '14 at 1:48
Mush
383
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Can you please answer this question with explanation ? I just learned about the charts needed to cover 1 and 2 spheres but got confused for the case of torus. It would be great if you guys could help.
manifolds
asked Jan 28 '14 at 1:48
Mush
383
Can you please answer this question with explanation ? I just learned about the charts needed to cover 1 and 2 spheres but got confused for the case of torus. It would be great if you guys could help.
manifolds
asked Jan 28 '14 at 1:48
Mush
383
asked Jan 28 '14 at 1:48
Mush
383
asked Jan 28 '14 at 1:48
Mush
383
asked Jan 28 '14 at 1:48
Mush
383
383
add a comment |Â
add a comment |Â
3 Answers
3
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5
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Four will do it. Wrap one around the outside of the doughnut and another on the inside. Let them overlap a little. Both of these are diffeomorphic to hollow cylinders, which require two patches to cover.
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
1
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
add a comment |Â
up vote
4
down vote
Hint $mathbb T^2=mathbb S^1timesmathbb S^1$
It is easily to see that $mathbb S^1$ can be charted by two covers, then $mathbb T^2$ is four.
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
add a comment |Â
up vote
2
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If the images of the charts are not restricted to be simply connected (which they are not according to the Wikipedia definition of "chart"), then there is a two-chart-system covering the torus.
And even with simply connected open sets, there is a way to do it with only three charts. Here is a picture showing how these charts look like on the flat torus (the right one shows that they indeed cover everything).
If I have time I may compose one on the embedded torus.
answered Aug 29 at 8:41
M. Winter
18.1k62764
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Four will do it. Wrap one around the outside of the doughnut and another on the inside. Let them overlap a little. Both of these are diffeomorphic to hollow cylinders, which require two patches to cover.
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
1
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
add a comment |Â
up vote
5
down vote
accepted
Four will do it. Wrap one around the outside of the doughnut and another on the inside. Let them overlap a little. Both of these are diffeomorphic to hollow cylinders, which require two patches to cover.
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
1
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Four will do it. Wrap one around the outside of the doughnut and another on the inside. Let them overlap a little. Both of these are diffeomorphic to hollow cylinders, which require two patches to cover.
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
Four will do it. Wrap one around the outside of the doughnut and another on the inside. Let them overlap a little. Both of these are diffeomorphic to hollow cylinders, which require two patches to cover.
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
answered Jan 28 '14 at 1:56
ncmathsadist
41.4k257100
41.4k257100
1
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
add a comment |Â
1
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
1
1
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
I know this is an old answer, but I have a question. Do you assume that the images of your charts $phi:Uto Bbb R^2$ are simply connected? As far as I can tell (e.g. from the definition from Wikipedia), this is no part of the definition of chart. Because if you do not assume this, then the cylinders are diffeomorphic to annuli in $Bbb R^2$ and these two maps suffice to cover the torus.
– M. Winter
May 4 at 12:33
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@M.Winter This deserves a seperate answer. Please turn your comment into an answer so I can upvote it.
– Babelfish
Aug 28 at 16:28
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
@Babelfish Thanks for motivating me to answer. I added one below.
– M. Winter
Aug 29 at 8:42
add a comment |Â
up vote
4
down vote
Hint $mathbb T^2=mathbb S^1timesmathbb S^1$
It is easily to see that $mathbb S^1$ can be charted by two covers, then $mathbb T^2$ is four.
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
add a comment |Â
up vote
4
down vote
Hint $mathbb T^2=mathbb S^1timesmathbb S^1$
It is easily to see that $mathbb S^1$ can be charted by two covers, then $mathbb T^2$ is four.
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Hint $mathbb T^2=mathbb S^1timesmathbb S^1$
It is easily to see that $mathbb S^1$ can be charted by two covers, then $mathbb T^2$ is four.
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
Hint $mathbb T^2=mathbb S^1timesmathbb S^1$
It is easily to see that $mathbb S^1$ can be charted by two covers, then $mathbb T^2$ is four.
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
answered Jan 28 '14 at 2:11
gaoxinge
2,555728
2,555728
add a comment |Â
add a comment |Â
up vote
2
down vote
If the images of the charts are not restricted to be simply connected (which they are not according to the Wikipedia definition of "chart"), then there is a two-chart-system covering the torus.
And even with simply connected open sets, there is a way to do it with only three charts. Here is a picture showing how these charts look like on the flat torus (the right one shows that they indeed cover everything).
If I have time I may compose one on the embedded torus.
answered Aug 29 at 8:41
M. Winter
18.1k62764
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
add a comment |Â
up vote
2
down vote
If the images of the charts are not restricted to be simply connected (which they are not according to the Wikipedia definition of "chart"), then there is a two-chart-system covering the torus.
And even with simply connected open sets, there is a way to do it with only three charts. Here is a picture showing how these charts look like on the flat torus (the right one shows that they indeed cover everything).
If I have time I may compose one on the embedded torus.
answered Aug 29 at 8:41
M. Winter
18.1k62764
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If the images of the charts are not restricted to be simply connected (which they are not according to the Wikipedia definition of "chart"), then there is a two-chart-system covering the torus.
And even with simply connected open sets, there is a way to do it with only three charts. Here is a picture showing how these charts look like on the flat torus (the right one shows that they indeed cover everything).
If I have time I may compose one on the embedded torus.
answered Aug 29 at 8:41
M. Winter
18.1k62764
If the images of the charts are not restricted to be simply connected (which they are not according to the Wikipedia definition of "chart"), then there is a two-chart-system covering the torus.
And even with simply connected open sets, there is a way to do it with only three charts. Here is a picture showing how these charts look like on the flat torus (the right one shows that they indeed cover everything).
If I have time I may compose one on the embedded torus.
answered Aug 29 at 8:41
M. Winter
18.1k62764
edited Aug 29 at 16:14
answered Aug 29 at 8:41
M. Winter
18.1k62764
answered Aug 29 at 8:41
M. Winter
18.1k62764
answered Aug 29 at 8:41
M. Winter
18.1k62764
18.1k62764
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
add a comment |Â
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
This torus looks yummy. What a nice picture.
– Babelfish
Aug 29 at 9:59
add a comment |Â
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