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Adjoint operators and diffeomorphism

  1. May 9, 2013 #1
    Hello,

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

    My question is: can we give an analogous definition of adjoint operator when [itex]\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n[/itex] is a diffeomorphism of ℝn?
     
  2. jcsd
  3. May 10, 2013 #2
    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 [itex]\phi[/itex] is linear. Then we can write [itex]\phi(\mathbf{x})=A\mathbf{x}[/itex], where A is a n×n matrix. We have the well-known result that: [tex]\left\langle x,\; Ay \right\rangle = \left\langle A^T x, \; y \right\rangle[/tex] Obviously in this case the adjoint of [itex]\phi[/itex] is simply given by the transpose of the corresponding matrix A.
    Note however that we can write: [tex]\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[/tex]

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

    Does this result sill hold for "curvilinear" deformations, where [itex]\phi[/itex] is a transformation of curvilinear coordinates, and G is the metric tensor expressed in terms of the Jacobian matrix?
     
    Last edited: May 10, 2013
  4. May 10, 2013 #3

    WannabeNewton

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    Yes ##g = J^{T}J## where ##J## is the jacobian associated with ##\phi##.
     
  5. May 10, 2013 #4
    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 [itex]\phi:U\rightarrow \mathbb{R}^n[/itex], e.g. [itex]<\phi(u), \; \phi(v)> \; = \; ?[/itex]
     
  6. May 10, 2013 #5

    WannabeNewton

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