Question on Jacobian determinant

[mnb96]

Hello,

it is true that linear transformations have constant Jacobian determinant.
Is the converse true? That is, if a transformation has constant Jacobian determinant, then is it necessarily linear?

 

[Vargo]

Hello,

Yes, linear transformations have constant Jacobian determinant. You can check this by manual calculation.

The converse is not true. In fact, there is an important class of transformations in physics called canonical transformations (or symplectic transformations) which preserve volume, but which are not, in general, linear.

I wish I had a nice simple example at hand, but perhaps someone else will come along with a good one.

 

[homeomorphic]

The total derivative of a linear map is the linear map itself, and in particular, it's constant. The total derivative of a map is just a linearized version of the map at each point. If it's already linear, nothing happens to it when you take its derivative at a point.

Just so we don't get confused, the baby case is 5x, whose derivative is 5. The latter 5 can be interpreted as the linear map that multiplies stuff by 5, so as a linear transformation, it's the same as the first map. For a linear map from ℝ^2 to ℝ^2, the total derivative is a constant linear map, which is represented by a 2 by 2 matrix. The determinant of that matrix is the Jacobian.

 

[Erland]

Hello,

Yes, linear transformations have constant Jacobian determinant. You can check this by manual calculation.

The converse is not true. In fact, there is an important class of transformations in physics called canonical transformations (or symplectic transformations) which preserve volume, but which are not, in general, linear.

I wish I had a nice simple example at hand, but perhaps someone else will come along with a good one.

Don't know much about symplectic transformations, but what about the following one:

$u=\ln x$, $v=xy$, for $x>0$, $y>0$.

This is certainly not linear but

$\partial u/\partial x = 1/x$, $\partial u/\partial y=0$, $\partial v/\partial x=y$, $\partial v/\partial y= x$.

The Jacobian determinant is then $1/x * x -0*y=1$ for all $x,\,y>0$.

So the Jacobian determinant is constant but the transformation is not linear.

 

[mnb96]

Hi!
thank you all for the explanations.
Very interesting replies actually!

 

[mathwonk]

nice example erland, and it shows how to construct infinitely more.