- #1

ayan849

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There cannot be a single chart for the manifold n-sphere.

I have been trying a proof for long but ...

Can you help me out?

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- Thread starter ayan849
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- #1

ayan849

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There cannot be a single chart for the manifold n-sphere.

I have been trying a proof for long but ...

Can you help me out?

- #2

lavinia

Science Advisor

Gold Member

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There cannot be a single chart for the manifold n-sphere.

I have been trying a proof for long but ...

Can you help me out?

What attempts at proof have you tried?

- #3

ayan849

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- #4

ayan849

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From this example, we were trying to prove the existence of single charts for S^1: {(cos t, sin t): t ε [0,2π)}.

- #5

Alesak

- 111

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There are problems with this "chart". For example, it's range is not open. And even if you wanted to remedy this situation by saying that circle is a manifold with boundary (so that [0,2π) is open in HFrom this example, we were trying to prove the existence of single charts for S^1: {(cos t, sin t): t ε [0,2π)}.

- #6

Alesak

- 111

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What happened was I was reading a book where they have discussed the construction of charts for the figure-8: {(sin t, sin 2t): t ε (0,2π)}. Using the subspace topology of R^2, you cannot make it into a manifold. But there are topologies that admit its manifold structure, and there is a topology that has a single chart for figure-8.

Could you please share that topology? I frankly don't see how to do it, unless you have different definition of (topological) manifold.

- #7

ayan849

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a map is a chart if it is ahomeomorphismfrom open subset of manifold towholeℝ^{n}. So if there was a single chart for a sphere or for a circle, it would have to be homeomorphic to ℝ^{n}.

There are problems with this "chart". For example, it's range is not open. And even if you wanted to remedy this situation by saying that circle is a manifold with boundary (so that [0,2π) is open in H^{1}), it still doesn't satisfy one other crucial condition to count as a chart.

I agree. But is there a proof that there CANNOT be ANY single chart for S

- #8

ayan849

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Could you please share that topology? I frankly don't see how to do it, unless you have different definition of (topological) manifold.

In stead of taking the subspace topology of ℝ^2, consider open intervals in (0,2π) to get mapped to the figure-8. This is indeed a homeomorphism, as in case of subspace topology of ℝ^2, it was not (because of the presence of the "cross" at the middle which was making our life hard).

- #9

Alesak

- 111

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I agree. But is there a proof that there CANNOT be ANY single chart for S^{n}?

Well, if you want it formaly, I would do proof by contradiction.

Suppose there is a single chart for all S

- #10

Alesak

- 111

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In stead of taking the subspace topology of ℝ^2, consider open intervals in (0,2π) to get mapped to the figure-8. This is indeed a homeomorphism, as in case of subspace topology of ℝ^2, it was not (because of the presence of the "cross" at the middle which was making our life hard).

Sorry, still don't see it. How exactly do you map open intervals in (0,2π) to figure 8?

There still have to be open set containing the middle point of figure 8 homeomorphic to, say, (0, 1), but IMHO that's impossible.

Also, every connected 1-manifold is homeomorphic to either circle or real line, so it can be embedded into ℝ^2, so it has to have inherited topology from R^2.

- #11

ayan849

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Sorry, still don't see it. How exactly do you map open intervals in (0,2π) to figure 8?

There still have to be open set containing the middle point of figure 8 homeomorphic to, say, (0, 1), but IMHO that's impossible.

Also, every connected 1-manifold is homeomorphic to either circle or real line, so it can be embedded into ℝ^2, so it has to have inherited topology from R^2.

map the nbd of π in (0, 2π) to one branch at the middle of fig8, right nbd of 0 to one portion of the remaining branch near the middle of fig8 and left nbd of 2π to the remaining portion of the remaining branch near the middle of fig8. Thats how u can cover the whole of fig8.

- #12

Alesak

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map the nbd of π in (0, 2π) to one branch at the middle of fig8, right nbd of 0 to one portion of the remaining branch near the middle of fig8 and left nbd of 2π to the remaining portion of the remaining branch near the middle of fig8. Thats how u can cover the whole of fig8.

Well, but is inverse of this function really continuous? Homomorphism is bijection that is continuous and has continuous inverse.

- #13

ayan849

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- #14

Alesak

- 111

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R^2 doesn't really need to be considered. For this map to qualify as a chart, it would need to be homeomorphic to (0, 1) or (0, 2π) or R with usual euclidean topology (by definition of chart). What book are you reading?

- #15

Alesak

- 111

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It is also possible to to define manifold using only coordinate charts without any reference to underlying topology, with that the topology can be inherited. So with manifolds, your are really restricted when it comes to choosing underlying topology. See this example of nonmanifold.

- #16

ayan849

- 22

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R^2 doesn't really need to be considered. For this map to qualify as a chart, it would need to be homeomorphic to (0, 1) or (0, 2π) or R with usual euclidean topology (by definition of chart). What book are you reading?

A course in differential geometry-s.kumaresan.

- #17

ayan849

- 22

- 0

Sorry, still don't see it. How exactly do you map open intervals in (0,2π) to figure 8?

There still have to be open set containing the middle point of figure 8 homeomorphic to, say, (0, 1), but IMHO that's impossible.

Also, every connected 1-manifold is homeomorphic to either circle or real line, so it can be embedded into ℝ^2, so it has to have inherited topology from R^2.

Try to make fig-8 with a piece of open string. The map is trivial.

- #18

wisvuze

- 372

- 1

Well, if you want it formaly, I would do proof by contradiction.

Suppose there is a single chart for all S^{n}. Then, by definition, there is a homeomorphism between R^{n}and S^{n}. But R^{n}and S^{n}are not homeomorphic. So contradiction.

this is your answer, one is compact, the other is not.

- #19

Alesak

- 111

- 0

Try to make fig-8 with a piece of open string. The map is trivial.

Ok, to convince you 8 is not a manifold, try this: since you say it is a manifold, there has to be open nbd around every point homeomorphic to (0, 1). So find me homeomorphism between figure X, which is a ndb of middle point of 8, and (0, 1). The (0, 1) interval is with usual euclidean topology.

- #20

ayan849

- 22

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note carefully my figure 8: {(sin t, sin 2t): t ε (0, 2π)}. It is really not a 8. The middle portion is not a "cross". If t ε [0, 2π], then it is certainly not a manifold. By my "figure 8" IS a manifold and I am sure you know how. :)

- #21

Alesak

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That's a really sneaky example;)

- #22

ayan849

- 22

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please see my another thread regarding connection and give some light on it if possible.

- #23

Alesak

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I would, but I don't know single iota about connections!

- #24

lavinia

Science Advisor

Gold Member

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From this example, we were trying to prove the existence of single charts for S^1: {(cos t, sin t): t ε [0,2π)}.

Manifolds that can be covered by a single coordinate chart are homoemorphic to Euclidean space.

There are many ways to see that Euclidean space can not be homeomorphic to a sphere.

For one thing a sphere is compact and Euclidean space is not.

Also,'Euclidean space can be continuously shrunk to a single point. The map (X,t) -> tX for t between 0 and 1 does the trick.

A sphere can not be continuously shrunk to a point. For otherwise its top homology group would be zero.

Can you think of any other arguments? For starters think about whether Euclidean space minus a point can be homeomorphic to all of Euclidean space.

- #25

ayan849

- 22

- 0

many thanks for the discussion. I think I have my answer now.

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