decraig Posted January 18, 2014 Posted January 18, 2014 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?
decraig Posted January 21, 2014 Author Posted January 21, 2014 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.
ajb Posted January 21, 2014 Posted January 21, 2014 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.
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