The Determinant, Tensors, and OrientationI am a bit confused about orientation and tensors as exemplified by the determinant.If we have an inner product or a metric and transform a vector from one coordinate system to another the magnitude of that vector is unchanged as the coordinate representation of the metric also changes. Therefore it is natural to think of a vector as a tensor and it's length doesn't change under orientation reversing transformations.In contrast, we can consider the determinant as an alternating tensor from the exterior algebra, for instance in R 3 , d e t ( v 1 , v 2 , v 3 ) = e 1 ∧ e 2 ∧ e 3 ( v 1 , v 2 , v 3 )But if we perform a reflection, A then we get, d e t ( A v 1 , A v 2 , A v 3 ) = d e t ( A ) d e t ( v 1 , v 2 , v 3 ) = − d e t ( v 1 , v 2 , v 3 )Does that mean that the 3-covector e 1 ∧ e 2 ∧ e 3 when contracted with 3 vectors is not invariant, which would seem to violate the idea of contracting tensors to a scalar produces an invariant under transformations.Or do we instead by transforming our 3-covector and the input vectors get an invariant oriented volume measure? e ′ 1 ∧ e ′ 2 ∧ e ′ 3 ( A v 1 , A v 2 , A v 3 ) = e 1 ∧ e 2 ∧ e 3 ( v 1 , v 2 , v 3 ) ?That would seem to mean that the determinant includes an arbitrary sign convention and that the definition of the determinant is different in left and right handed oriented coordinate systems.

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The Determinant, Tensors, and OrientationI am a bit confused about orientation and tensors as exemplified by the determinant.If we have an inner product or a metric and transform a vector from one coordinate system to another the magnitude of that vector is unchanged as the coordinate representation of the metric also changes. Therefore it is natural to think of a vector as a tensor and it&#039;s length doesn&#039;t change under orientation reversing transformations.In contrast, we can consider the determinant as an alternating tensor from the exterior algebra, for instance in             R        3  ,  d  e  t  (      v    1    ,      v    2    ,      v    3    )  =      e    1    ∧      e    2    ∧      e    3    (      v    1    ,      v    2    ,      v    3    )But if we perform a reflection, A then we get,  d  e  t  (  A      v    1    ,  A      v    2    ,  A      v    3    )  =  d  e  t  (  A  )  d  e  t  (      v    1    ,      v    2    ,      v    3    )  =  −  d  e  t  (      v    1    ,      v    2    ,      v    3    )Does that mean that the 3-covector       e    1    ∧      e    2    ∧      e    3   when contracted with 3 vectors is not invariant, which would seem to violate the idea of contracting tensors to a scalar produces an invariant under transformations.Or do we instead by transforming our 3-covector and the input vectors get an invariant oriented volume measure?      e          ′      1        ∧      e          ′      2        ∧      e          ′      3        (  A      v    1    ,  A      v    2    ,  A      v    3    )  =      e    1    ∧      e    2    ∧      e    3    (      v    1    ,      v    2    ,      v    3    )  ?That would seem to mean that the determinant includes an arbitrary sign convention and that the definition of the determinant is different in left and right handed oriented coordinate systems.

Bryce Obrien · Accepted Answer

Tu&#039;s book, Differential Geometry, answers this question in it&#039;s discussion of the Volume Form on a Riemannian Manifold.If we an   A  ∈  G  L  (  n  ,      R    )is a transformation between coordinate systems with:            v        ′    =  A      v  in the new basis. Then we similarly have a relation on the dual bases.            ω        ′    =      A          −      1            ω  So then the volume form transforms as,  d  e  t  (      A          −      1        )      e    1    ∧      e    2    ∧      e    3    =      e          ′      1        ∧      e          ′      2        ∧      e          ′      3      And then if we transform both the volume form and the vectors we are measuring we get the following.                              e          1                ∧                  e          2                ∧                  e          3                (        u        ,        v        ,        w        )                            =        d        e        t        (                  A                      −            1                          )        d        e        t        (        A        )                  e          1                ∧                  e          2                ∧                  e          3                (        u        ,        v        ,        w        )                                          =        d        e        t        (                  A                      −            1                          )                  e          1                ∧                  e          2                ∧                  e          3                (        A        u        ,        A        v        ,        A        w        )                                          =                  e                      ′            1                          ∧                  e                      ′            2                          ∧                  e                      ′            3                          (        A        u        ,        A        v        ,        A        w        )                                          =                  e                      ′            1                          ∧                  e                      ′            2                          ∧                  e                      ′            3                          (                  u          ′                ,                  v          ′                ,                  w          ′                )            So the volume form or top form is an invariant volume measure.

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