## question on Jacobian with function composition and inverse functions

Hello,

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

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

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

What can we say about $J_f(u)$, the Jacobian of f ?

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 Quote by mnb96 What can we say about $J_f(u)$, the Jacobian of f ?
hello!
you go first!

 Quote by tiny-tim you go first!
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:
$$J_f(u) = J_{\phi^-1}(T(\phi(u))) \; J_{T}(\phi(u)) \; J_\phi(u)$$

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:

$$|J_f(u)| = \lambda \cdot |J_{\phi^-1}(T(\phi(u))) |\cdot|J_\phi(u)|$$

...and here I get stuck.
I told you I wasn't going to go very far...

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## question on Jacobian with function composition and inverse functions

well, that's quite a long way!

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 ?

 Quote by tiny-tim now what would you like to be able to say about Jφ-1Jφ ?
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 $(T \circ \phi)(u)$ while the term Jφ depends on u.

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 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?
 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.