Pullback of F on Manifolds: What Matrix Do We Take Determinant Of?

  • A
  • Thread starter Rico1990
  • Start date
  • Tags
    Manifold
In summary, the conversation discusses the pullback of a smooth manifold, with the participants mentioning coordinate functions and partial derivatives. They clarify that the partial derivatives are defined and depend on the coordinate functions, and that the formula has a specific purpose.
  • #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: 893
Physics news on Phys.org
  • #2
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
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
 

1. What is the "pullback" of F on manifolds?

The pullback of F on manifolds refers to a mathematical operation that allows us to transfer the properties of a function F from one manifold to another. It involves using a function to map points from one manifold to another, and then using this mapping to pull back the properties of F onto the second manifold.

2. Why do we need to take the determinant of the matrix in the pullback of F on manifolds?

Taking the determinant of the matrix in the pullback of F on manifolds is necessary because it allows us to understand how the function F behaves under changes of coordinates on the manifold. The determinant of this matrix gives us important information about the change in volume and orientation of the manifold under the mapping.

3. How is the matrix chosen for the pullback of F on manifolds?

The matrix for the pullback of F on manifolds is chosen based on the specific properties of the function F and the manifolds involved. In general, the matrix is constructed using the partial derivatives of F with respect to the coordinates of the two manifolds, and it is chosen to ensure that the pullback operation is well-defined and consistent.

4. Can the pullback of F on manifolds be extended to higher dimensions?

Yes, the pullback of F on manifolds can be extended to higher dimensions. In fact, the concept of pullback can be applied to any smooth map between manifolds, regardless of their dimension. However, the specific matrix used in the pullback operation may vary depending on the dimensionality of the manifolds involved.

5. What are some applications of the pullback of F on manifolds?

The pullback of F on manifolds has various applications in mathematics and physics. It is used in differential geometry, where it helps study the properties of smooth manifolds and their transformations. It also has applications in the field of mechanics, where it is used to analyze the behavior of physical systems under different coordinate systems and transformations.

Similar threads

  • Differential Geometry
Replies
20
Views
2K
  • Differential Geometry
Replies
2
Views
588
Replies
9
Views
3K
Replies
6
Views
352
Replies
17
Views
3K
  • Differential Geometry
Replies
10
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
  • Differential Geometry
Replies
1
Views
2K
  • Topology and Analysis
Replies
6
Views
912
Back
Top