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A (challenging?) question around the Jacobian matrix

  1. Apr 5, 2008 #1
    Here is the problem:

    Suppose that g is a diffeomorphism on R^n. Then we know that its jacobian matrix is everywhere invertible.
    Let us define the following matrix valued function on R^n
    H_{i,j} (x) = \int_0^1 \partial_i g^j(tx) dt
    where [tex]g^j[/tex] are the components of g.
    Question : Is [tex](H_{i,j}(x))_{i,j} [/tex] (which could be interpreted as a mean of the Jacobian matrix of g) invertible for any x ?

    My guess is that the answer is negative, but I find no counter-examples.
    Any Help ?
  2. jcsd
  3. Apr 7, 2008 #2
    I believe you may have to move into three dimensions to obtains a counterexample. It is possible that this theorem is in fact true.

    Actually, I believe that a counterexample is provided by;

    [tex]\mathbf{g}((x,y,z))=(x \cos(2\pi z)-y \sin(2\pi z),x \sin(2\pi z)+y \cos(2\pi z),z)[/tex]
    [tex]H((0,0,1))[/tex] being singular.

    Can anyone else check this?
    Last edited: Apr 7, 2008
  4. Apr 7, 2008 #3
    Indeed, we even have H(x,y,1) singular for any x,y !
    Thank you !
    I wonder if the theorem is true in dimension 2...
    (it is trivially true in dimension 1).
  5. Apr 7, 2008 #4


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    Interesting problem. I thought about this for awhile, and couldn't come up with a counter in D =2. Can anyone prove this statement? Its not clear to me why it should be true, and there seems to be something interesting (maybe topological) lurking behind it.
    Last edited: Apr 7, 2008
  6. Apr 14, 2008 #5
    I believe it is true because of the nature of coordinate lines and their tangents in 2D.

    Firstly, assume that we can find a H in 2D that is singular. Therefore the columns of H, which we'll denote [tex]\mathbf{h}_i[/tex] are linearly dependant. So there exist non zero [tex]\alpha_i[/tex] such that

    [tex]\Sigma \alpha_i \mathbf{h}_i = \mathbf{0}[/tex]

    Now by the definition

    [tex]\mathbf{h}_i = \int_0^1 \frac{\partial \mathbf{g}}{\partial x_i}(t \mathbf{x}) dt[/tex]

    [tex]\Sigma \alpha_i \mathbf{h}_i =\int_0^1 \Sigma \alpha_i \frac{\partial \mathbf{g}}{\partial x_i}(t \mathbf{x}) dt = \mathbf{0} [/tex]

    Now looking at the integral along this curve. We let
    [tex]\mathbf{v}(t)=\Sigma \alpha_i \frac{\partial \mathbf{g}}{\partial x_i}(t(1,0)) = \alpha_1 \frac{\partial \mathbf{g}}{\partial x_1}(t(1,0)) + \alpha_2 \frac{\partial \mathbf{g}}{\partial x_2}(t(1,0))[/tex]
    And so we have
    [tex]\int_0^1 \mathbf{v}(t) dt = \mathbf{0} [/tex]
    We cannot have [tex]\mathbf{v}=\mathbf{0}[/tex], because the jacobian is always fully ranked, and hence any linear combination of its columns cannot be zero.

    So [tex]\mathbf{v}(t)[/tex] is non zero everywhere, and its integral is zero. But if we let [tex]\mathbf{v}(t)=\frac{d \mathbf{z}}{dt}(t)[/tex], we can see that

    [tex]\int_0^1 \mathbf{v}(t) dt = \int_0^1 \frac{d \mathbf{z}}{dt}(t) dt = \mathbf{z}(1)-\mathbf{z}(0) = \mathbf{0}[/tex]

    Which means that [tex]\mathbf{v}[/tex] is the tangent vector to some closed curve [tex]\mathbf{z}(t)[/tex]

    We now make a change of variables by letting
    [tex]\mathbf{x}(\mathbf{u}) = \left( \begin{array}{cc}
    \alpha_1 & -\alpha_2 \\
    \alpha_2 & \alpha_1
    \end{array} \right) \mathbf{u}

    Then it can be shown that [tex]\mathbf{v}(t) = \frac{d \mathbf{g}}{d u_1}(t(1,0))[/tex], i.e. [tex]v[/tex] is the tangent vector to the [tex]u_2=0[/tex] coordinate curve. But this means that this coordinate curve is closed. Since the mapping from u to x is a diffeomorphism, and the function g is a diffeomorphism, the function from u to g is a diffeomorphism on R^2. This means that it cannot have any closed coordinate curves. Therefore we have a contridiction, and so H must be non singular.

    I hope this is OK. I'm not being especially rigorous here.

    In higher dimensions, some of the [tex]\alpha_i[/tex] can be zero, and hence the mapping from u to x may not be a diffeomorphism. Essentially you have some wiggle room once you move into higher dimensions.
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