A (challenging?) question around the Jacobian matrix

[Aleph-0]

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 $$g^j$$ are the components of g.
Question : Is $$(H_{i,j}(x))_{i,j}$$ (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 ?

 

[ObsessiveMathsFreak]

I believe you may have to move into three dimensions to obtains a counterexample. It is possible that this theorem is in fact true.

Edit:
Actually, I believe that a counterexample is provided by;

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

Can anyone else check this?

 

[Aleph-0]

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).

 

[Haelfix]

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.

 

[ObsessiveMathsFreak]

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.

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 $$\mathbf{h}_i$$ are linearly dependent. So there exist non zero $$\alpha_i$$ such that

$$\Sigma \alpha_i \mathbf{h}_i = \mathbf{0}$$

Now by the definition

$$\mathbf{h}_i = \int_0^1 \frac{\partial \mathbf{g}}{\partial x_i}(t \mathbf{x}) dt$$
Therefore

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

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

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

$$\int_0^1 \mathbf{v}(t) dt = \int_0^1 \frac{d \mathbf{z}}{dt}(t) dt = \mathbf{z}(1)-\mathbf{z}(0) = \mathbf{0}$$

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

We now make a change of variables by letting
$$\mathbf{x}(\mathbf{u}) = \left( \begin{array}{cc}<br /> \alpha_1 & -\alpha_2 \\<br /> \alpha_2 & \alpha_1 <br /> \end{array} \right) \mathbf{u}$$

Then it can be shown that $$\mathbf{v}(t) = \frac{d \mathbf{g}}{d u_1}(t(1,0))$$, i.e. $$v$$ is the tangent vector to the $$u_2=0$$ 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 $$\alpha_i$$ 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.