decraig Posted January 18, 2014 Share 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? Link to comment Share on other sites More sharing options...
decraig Posted January 21, 2014 Author Share 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. Link to comment Share on other sites More sharing options...
ajb Posted January 21, 2014 Share 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. Link to comment Share on other sites More sharing options...
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