Smooth covering map and smooth embedding

  • Context: Graduate 
  • Thread starter Thread starter huyichen
  • Start date Start date
  • Tags Tags
    Map Smooth
Click For Summary
SUMMARY

The discussion centers on the smooth covering map F: S^2 -> R^4 defined by F(x,y) = (x^2 - y^2, xy, xz, yz) and its relationship with the smooth embedding of RP^2 into R^4. Participants clarify that the task is to demonstrate the existence of a map f: RP^2 -> R^4 such that f o p = F, where p is the smooth covering map from S^2 to RP^2. The key point is that F must be constant on the fibers of p, which simplifies the process of verifying that f is a smooth embedding.

PREREQUISITES
  • Understanding of smooth manifolds and embeddings
  • Familiarity with covering maps and quotient spaces
  • Knowledge of Jacobian matrices and their role in determining smoothness
  • Basic concepts of projective spaces, specifically RP^2
NEXT STEPS
  • Study the properties of smooth embeddings in differential topology
  • Learn about the construction and implications of covering maps
  • Investigate the role of the Jacobian matrix in smooth mappings
  • Explore the relationship between S^2 and RP^2 in the context of topology
USEFUL FOR

Mathematicians, particularly those specializing in differential geometry and topology, as well as students seeking to understand the concepts of smooth maps and embeddings in higher-dimensional spaces.

huyichen
Messages
28
Reaction score
0
Now F:S^2->R^4 is a map of the following form:
F(x,y)=(x^2-y^2,xy,xz,yz)
now using the smooth covering map p:S^2->RP^2, p is the composition of inclusion map i:S^2->R^3 and the quotient map q:R^3\{0}->RP^2. show that F descends to a smooth embedding of RP^2 into R^4.

Is the problem asked to show that F。p^(-1) is a smooth embedding? I am confused, and if it is the case, then how should we compute the Jacobian matrix for F。p^(-1)?
 
Physics news on Phys.org
Essentially, yes, but notice that p^-1 is ill defined. Instead, you must show that there exists a map f: RP^2-->R^4 such that f o p = F (and that it is a smooth embedding). Observe that this only means checking that F is constant on the fibers of p.
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
6K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K