Pullback on a manifold

  • A
  • Thread starter Rico1990
  • Start date
  • #1
Rico1990
3
0
Hey,
we had in the lecures the following:
Let M and N be smooth manifolds, and dim(M)=dim(N)=n, while $$x^i$$ and $$ y^i$$ are coordinate functions around $$p\in M$$ respective $$F(p) \in N$$, then we get for the pullback of F
Untitled01.jpg

Which entries has the matrix we take the determinant of? I thaught of partial derivatives but am not sure.
 

Attachments

  • Untitled01.jpg
    Untitled01.jpg
    6.3 KB · Views: 773

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,843
10,357
Yes it is the partial derivatives of ##y^j \circ F## with respect to ##x^i##, which is what the image you pasted says. This is just the Jacobian of the transformation ##x^i \to y^j## as subsets of ##\mathbb R^n##.
 
  • #3
Rico1990
3
0
Ok, thank you for your answer. But answer me please two last questions that arose. I deduce that these partial derivatives are defined, but they are vague in the sense, that $$x^i , y^i$$ are both functions the derivatives depend on. Is it the interest to leave them this vague or does one insert certain values so that the coordinate functions give „real" coordinates.
What is the use of this formula?

Best wishes Rico
 

Suggested for: Pullback on a manifold

Replies
4
Views
166
  • Last Post
Replies
4
Views
1K
Replies
2
Views
292
Replies
4
Views
785
  • Last Post
Replies
6
Views
2K
Replies
9
Views
2K
  • Last Post
Replies
1
Views
782
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
1
Views
1K
Top