Should sigma be in the dA formula

Should sigma be in the dA formula? PhL 2.25.12 I arrive here after a stack push from document " use of abs val J in tensor density...". The σ in question here is this one: (An)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x // en missing or An = σ (-1)n-1 e1 x e2 ... x eN // en missing σ ≡ sign[det(Sab)] = sign[det(Rab)] For example, if N = 3 we would have A3 = σ e1 x e2 σ = sign of det(S) This seems illogical to me, because this equation A3 = σ e1 x e2 is entirely stated in x-space, so how can it know about S? Well here is how. We have (en)i = Sin ! This is because the en are defined in terms of a transformation S. OK, so I accept that S at least plays a role in A3 = σ e1 x e2 . Now suppose I start with some S with σ = signdetS = +1, and I then move to S' = -S in N=3 dimensions. What happens? It is a different transformation F now. I think en → – en so e1 x e2 → e1 x e2 . Then since we start with A3 = e1 x e2 we should end up with A'3 = e1 x e2 = e'1 x e'2 . But the equation above says this is wrong, and we should have A'3 = σ' e'1 x e'2 = – e'1 x e'2. In simple terms, in N=3 imagine some signdetS = 1 situation and face vector is A3 = e1 x e2 and I can do right hand rule and find A3 and all is well, thank you. Now I take S' = -S. This changes x'-space, and that in turn causes e'n = - en. Does this change Cartesian x-space in any way? I want to say no, it is still my original right handed Cartesian coordinate system in which I am drawing whatever en vectors are determined by S. So after making this x'-space change, I should have dA'3 = same as before the change, but the sign σ is getting in the way. So where did this σ come from in the first place? The first appearance is Appendix B section (d) , but things really start in the 3-piped section there. I define A as being in the direction of E , but E as defined in Appendix A looks like this: E3 = det(R) e1 x e2 Thus, if I take S→-S', det(R) changes sign and E3 changes sign. In my 3-piped section I DEFINE my area in this way: A3 ≡ |e1 x e2| 3 so by definition, I put it in the direction of E3 here. This is why A3 is going to reverse direction under S→-S. Maybe that was a bad definition? Maybe this is a definition that does not match the physical situation using RHR cross products. I want to draw up my prototype case and argument. Both pictures are of the same x-space, only the x'-space definition is different between the two pictures. All three vectors e1, e2 and E3 change sign. Well consider the parity inversion for all points in a cube (slice shown of the cube) What has happened to this square under parity? Interior maps into interior. Both squares have an outfacing face "ab" and I show a vector representing the area of this face ab. That vector like all position vectors flips. The outfacing face ab is still outfacing in the second picture. The square has NOT been "turned inside out". If I think of the arrow as being E3, then I like the idea that E3 changes direction like this. Let's try a more 3D version of the above picture In the left picture, the cube face nearest the viewer points to the viewer as indicated by the circled dot. The cross product e1 x e2 points in this direction. After the parity transformation, that front face has become the back face. The outfacing area vector for that back face points away from the viewer. But the cross product e1 x e2 still points toward the viewer, so this new face really is A3 = – e1 x e2 and not I like the minus sign very much! In general I think a parity transformation flips the out-facing direction of every N-piped face 180 degrees, and this is what happens to En and that is why I like dAn pointing in the En direction. So this seems to explain the σ appearing in this equation, at least for N=3 : An = σ (-1)n-1 e1 x e2 ... x eN In the N=3 case, the cross product says the same, but the σ creates the minus sign we need. In the other form we write An = |det(Sab)| En and in this form, we obtain the desired sign change because En changes sign. Conclusion: I am happy with the appearance of σ in one of the An formulas: An = |det(S)| En An = σ (-1)n-1e1 x ... x eN // en missing