Do I have this right?

 

A completely antisymmetric tensor or type (0,k) on a k dimensional manifold in orthonormal coordinates may be expressed as a scalar times the Levi-Civita symbol, [math]\epsilon_{\mu \nu \rho \sigma}[/math].

 

Under a general linear transformation, is the result also a scalar times the Levi-Civita symbol?

Apparenty, for A, antisymmetric, [math]A = (1/n!) A_{[\mu \nu \rho \sigma]}[/math].

 

[math]A = a \epsilon[/math]

and

[math] A' = |J| A [/math]

where |J| is the determinant of the Jacobian.

 

But I can't find a reference to verify.

For sure you pick up the sign of Jacobian when you look at the Levi-Civita symbol. It is a pseudo tensor. But antisymmetric tensors are antisymmetric tensors and transform accordingly. That was my worry with your first question, but as long as you consider only orientation preserving transformations then you are probably okay.