1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: A (challenging?) question around the Jacobian matrix
  1. Questions of Jacobian (Replies: 4)

  2. Question on Jacobians (Replies: 3)

  3. The Jacobian Matrix. (Replies: 0)

  4. Challenge Question (Replies: 1)