Adjoint operators and diffeomorphism

[mnb96]

Hello,

we known that for each linear operator \phi:\mathbb{R}^n\rightarrow \mathbb{R}^n there exists an adjoint operator \overline{\phi} such that: <\phi(\mathbf{x}),\mathbf{y}>=<\mathbf{x},\overline{\phi}(\mathbf{y})> for all x,y in ℝⁿ, and where <\cdot,\cdot> is the inner product.

My question is: can we give an analogous definition of adjoint operator when \phi:\mathbb{R}^n\rightarrow \mathbb{R}^n is a diffeomorphism of ℝⁿ?

 

[mnb96]

It's a pity that nobody answered to this question. Perhaps I did not formulate my question properly ... anyways, I will show a partial solution to the problem, and finally I will ask you if the result can be generalized.

Let's assume the diffeomorphism \phi is linear. Then we can write \phi(\mathbf{x})=A\mathbf{x}, where A is a n×n matrix. We have the well-known result that: \left\langle x,\; Ay \right\rangle = \left\langle A^T x, \; y \right\rangle Obviously in this case the adjoint of \phi is simply given by the transpose of the corresponding matrix A.
Note however that we can write: \left\langle A^T x, \; y \right\rangle = \left\langle (A^T A)A^{-1} x, \; y \right\rangle = \left\langle G A^{-1} x, \; y \right\rangle

I don't know if this is a useful step, but notice that G=A^T A is the metric tensor. The question now reduces to:

Does this result sill hold for "curvilinear" deformations, where \phi is a transformation of curvilinear coordinates, and G is the metric tensor expressed in terms of the Jacobian matrix?

 

[WannabeNewton]

Yes $g = J^{T}J$ where $J$ is the jacobian associated with $\phi$.

 

[mnb96]

Hi WannabeNewton,

thanks for your reply, and good to hear that the result holds for admissible change of coordinates.
Do you have any hint on what strategy I could use in order to prove this result?

I suppose that the result can be deduced from the knowledge of how the inner product in ℝⁿ changes under a diffeomorphism given by a change of coordinates \phi:U\rightarrow \mathbb{R}^n, e.g. <\phi(u), \; \phi(v)> \; = \; ?

 

[WannabeNewton]

It is related to that yes. See here: https://www.physicsforums.com/showthread.php?t=649422
The result is quite geometric even if it might not come off as such.