Compact Smooth Manifolds in n-dimensional Euclidean Space

  • Thread starter Thread starter gotjrgkr
  • Start date Start date
  • Tags Tags
    Manifolds Smooth
Click For Summary
SUMMARY

In the discussion on compact smooth manifolds in n-dimensional Euclidean space, it is established that not all smooth manifolds are compact. Specifically, while R itself is a smooth manifold, it is not compact due to its unbounded nature. Compact smooth manifolds can exist if they include boundaries, as closed and bounded segments of R are compact. However, closed smooth manifolds in R^n without boundaries are not compact.

PREREQUISITES
  • Understanding of smooth manifolds
  • Familiarity with Euclidean space concepts
  • Knowledge of compactness in topology
  • Basic principles of manifolds with boundaries
NEXT STEPS
  • Study the properties of smooth manifolds in differential geometry
  • Explore the concept of compactness in topology
  • Learn about manifolds with boundaries and their implications
  • Investigate examples of compact smooth manifolds in R^n
USEFUL FOR

Mathematicians, students of differential geometry, and anyone interested in the properties of smooth manifolds and their compactness in Euclidean spaces.

gotjrgkr
Messages
84
Reaction score
0
Hi!
I want to know if any smooth manifold in n-dimensional euclidean space can be compact or not.
If it is possible, then could you give me an example about that?

I also want to comfirm whether a cylinder having finite volume in 3-dimensional euclidean space can be a smooth manifold.

I hope to receive your reply soon. Thank you . Have a nice day!:smile:
 
Physics news on Phys.org
I'm not sure what you mean by "any smooth manifold in n-dimensional euclidean space can be compact". If you mean "is every smooth manifold compact?", the answer is trivially "no". R itself is a smooth manifold but is not compact because it is not bounded.

If you mean "do there exist compact smooth manifolds?", then the answer depends upon whether you are including "manifolds with boundary" in "smooth manifolds". If you are then any closed and bounded segement or R is compact so the answer is yes. If not then the answer is "no" again. The only closed smooth manifolds in [itex]R^n[/itex] without boundaries are the Rn themselves which are not bounded.
 
I actually haven't studied the subject related with manifolds,yet. But I will study it, sooner or later.
I don't know whay you mean "manifolds with boundary".
Do you mean that it implies an union of a set called "manifolds" and its boundary?
Could you explain about it more??
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
10K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 20 ·
Replies
20
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 44 ·
2
Replies
44
Views
7K
  • · Replies 5 ·
Replies
5
Views
3K