question on Jacobian with function composition and inverse functions

mnb96

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 ?

tiny-tim

hello!

you go first!

mnb96

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...

tiny-tim

well, that's quite a long way!

ok, you've proved that J_f is a multiple of J_φ^-1J_φ …

now what would you like to be able to say about J_φ^-1J_φ ?

mnb96

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 J_φ^-1 is a function of [itex](T \circ \phi)(u)[/itex] while the term J_φ depends on u.

tiny-tim

see http://en.wikipedia.org/wiki/Inverse_function_theorem

Vargo

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?

mnb96

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.