Question on Jacobian with function composition and inverse functions

In summary, we have two functions, phi and T, that are both diffeomorphisms and T is also linear. We consider the function f(u) = (phi^-1 ∘ T ∘ phi)(u). Since T is linear, the Jacobian determinant of T is constant. We can use the chain rule to find the Jacobian of f, which is a multiple of Jφ-1Jφ. However, since the two Jacobians are evaluated at different points, we cannot simplify this expression any further.
  • #1
mnb96
715
5
Hello,

let's suppose I have two functions [itex]\phi:U\rightarrow V[/itex], and [itex]T:V\rightarrow V[/itex] that are both diffeomorphisms having inverse.
Furthermore [itex]T[/itex] is linear.

I consider the function [itex]f(u) = (\phi^{-1}\circ T \circ \phi)(u)[/itex], where [itex]\circ[/itex] is the composition of functions.

Since [itex]T[/itex] is linear, we already know that the Jacobian determinant is constant: [itex]J_T(v)=\lambda[/itex].

What can we say about [itex]J_f(u)[/itex], the Jacobian of f ?
 
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  • #2
mnb96 said:
What can we say about [itex]J_f(u)[/itex], the Jacobian of f ?

hello! :smile:

you go first! :wink:
 
  • #3
tiny-tim said:
you go first! :wink:
Ok, but you'll see I won't go very far :)
Let's try, though...

If I am not wrong the Jacobian obeys an analogous rule of the chain rule for derivatives, so we can write:
[tex]J_f(u) = J_{\phi^-1}(T(\phi(u))) \; J_{T}(\phi(u)) \; J_\phi(u)[/tex]

The Jacobian determinant of a product of Jacobians is given by the product of Jacobian determinants. We also know that the Jacobian determinant of T is constant, thus:

[tex]|J_f(u)| = \lambda \cdot |J_{\phi^-1}(T(\phi(u))) |\cdot|J_\phi(u)|[/tex]

...and here I get stuck.
I told you I wasn't going to go very far... :smile:
 
  • #4
well, that's quite a long way! :smile:

ok, you've proved that Jf is a multiple of Jφ-1Jφ

now what would you like to be able to say about Jφ-1Jφ ? :wink:
 
  • #5
tiny-tim said:
now what would you like to be able to say about Jφ-1Jφ ? :wink:

Basically, I would like to know if that term can be somehow simplified.
My guess is that we can't say much more than that because Jφ-1 is a function of [itex](T \circ \phi)(u)[/itex] while the term Jφ depends on u.
 
  • #7
I think the difficulty here though is that the Jacobian of the inverse map is not being evaluated at the image point of the forward map phi. Rather it is being evaluated at T(phi(u)). I think that is why the OP is stuck. Am I missing a way to simplify this?
 
  • #8
Hi Vargo,

the problem you mentioned in your post describes well why I get stuck.
Basically I can't see a way to simplify that expression, because the two Jacobians are evaluated at different points.
 

1. What is a Jacobian in relation to function composition and inverse functions?

A Jacobian is a mathematical concept that represents the rate of change of one set of variables with respect to another set of variables. In the context of function composition and inverse functions, the Jacobian is used to calculate the change in the output of a composite function or inverse function based on changes in the input variables.

2. How is the Jacobian used in function composition?

In function composition, the Jacobian is used to calculate the derivative of the composite function. This is done by taking the product of the individual derivatives of each function in the composition. It is a useful tool for finding the rate of change of complex functions.

3. Can the Jacobian be used to find the inverse of a function?

Yes, the Jacobian can be used to find the inverse of a function. In order for a function to have an inverse, the Jacobian must be non-zero at all points in the domain of the function. The inverse function can then be found by solving for the original input variables in terms of the output variables.

4. What is the relationship between the Jacobian and the determinant?

The Jacobian and the determinant are closely related, as the Jacobian of a function is equal to the determinant of the matrix of partial derivatives of that function. This relationship is particularly useful in finding the Jacobian for functions with multiple input variables.

5. How is the Jacobian used in multivariable calculus?

In multivariable calculus, the Jacobian is used to calculate the derivative of a multivariable function. It is also used in change of variables, where it is used to convert integrals from one coordinate system to another. The Jacobian is an important tool in understanding and solving problems in multivariable calculus.

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