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.