Is R mod 2pi a Compact Manifold?

[Bahram_phd]

Hi,
Why R mod 2pi is a Compact Manifold?
Isn't this like a real line which is not compact?
How should we prove it using a finite sub-cover for this manifold?
bah

 

[morphism]

R mod 2pi doesn't look like the real line; it looks like the interval [0,2pi] except the end points are identified. (So it's a circle...)

 

[mathwonk]

consider the map from R to the circle defined by t-->(cos(t), sin(t)).

 

[lavinia]

Hi,
Why R mod 2pi is a Compact Manifold?
Isn't this like a real line which is not compact?
How should we prove it using a finite sub-cover for this manifold?
bah

What is the topology that you want?
If it is R with the topology of a line modulo the discrete group of integer multiples of 2pi, then use the definition of open set in the quotient topology to show that every open cover has a finite subcover. You need to know that a closed interval is compact.

 

[Bacle2]

What is the topology that you want?
If it is R with the topology of a line modulo the discrete group of integer multiples of 2pi, then use the definition of open set in the quotient topology to show that every open cover has a finite subcover. You need to know that a closed interval is compact.

Isn't the quotient topology assumed when talking about a quotient?

It seems like the open sets here would be the sets (a,b); 0<a,b<2∏ , which lift to

the open sets U(a±2k∏, b± 2k∏ ); for k integer, in ℝ

Then, since compactness is hereditary, and this topology is smaller than the subspace topology in [0,2∏],

(which is a compact space), then the quotient is compact.

EDIT:
Correction: the quotient topology agrees with the subspace topology of S^1 in R^2. By, e.g., mathwonk's map f, which descends to

a homeo f^ between R/~ and S^1, subspace (since the map is constant in R/~) , and which can be shown to be continuous and onto.

Then, by homeo., R/~ is also compact. If you want an argument by subcovers, do the argument in S^1, and pull back to R/~ by the homeo f^

 

[Bacle2]

Actually, to show the manifold part, there are results that the quotients of manifolds by some special types of group actions are themselves manifolds--you can, e.g., show that projective space is a manifold by using this result/theorem.

 

[Bahram_phd]

Thanks guys,
It helped me a lot. I did some studies on quotient topology and now I have a better understanding of the problem.