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Tetrads in General Relativity

III. Geometric Algebra

Pseudoscalar Volume

   In geometric algebra, a $p$-vector (or $p$-blade) is the wedge product of $p$ distinct basis vectors ${{\mathbf{e}}_1} \wedge {{\mathbf{e}}_2}.... \wedge {{\mathbf{e}}_p}$, or more generally, the wedge product of $p$ linearly independent vectors. For $p = 0,1,2,3$, $p$-vectors are called, respectively, scalars, vectors, bivectors and tri-vectors; they are dual to 0-forms, 1-forms, 2-forms and 3-forms. Geometrically, a $p$-vector represents a $p$-dimensional oriented volume of space specified by the vectors that define its edges. Whenever two of the constituent vectors are exchanged, the volume changes  sign, corresponding to a volume of opposite orientation.

   The wedge product of the maximum number of independent vectors of a $d$-dimensional vector space forms a one-dimensional subspace. Its members are called $d$-volumes or pseudoscalars (top forms or volume forms in the language of differential forms). Similar to scalars, pseudoscalars do not have any direction aside from their sign.

Unit Volume

  1. As the highest-grade element in a vector space, a pseudoscalar volume is unique up to a sign and scaling. In orientable spaces, it is possible to define a unit volume.
  2. In particular, the unit volume of the tangent space ${V_4}$ spanned by a basis $\{{{\mathbf{e}}_j}\}$ is the grade-4 pseudoscalar:
    \[{\operatorname{I} _{4}}: = {{\mathbf{e}}_0} \wedge {{\mathbf{e}}_1} \wedge {{\mathbf{e}}_2}\wedge {{\mathbf{e}}_3}/\left| {{{\mathbf{e}}_0}\wedge {{\mathbf{e}}_1} \wedge {{\mathbf{e}}_2}\wedge {{\mathbf{e}}_3}} \right| \qquad \left| {{\operatorname{I} _{4}}} \right| = 1\]
    its orientation being specified by the basis vectors.
  3. The modulus in the denominator may be calculated as
    \[\left| {{{\mathbf{e}}_0} \wedge {{\mathbf{e}}_1} \wedge {{\mathbf{e}}_2} \wedge {{\mathbf{e}}_3}} \right| = {\left| {{\text{det [}}{{\mathbf{e}}_i} \cdot {{\mathbf{e}}_j}]} \right|^{1/2}}\]
  4. The unit volume has an inverse. If the basis is orthonormal:
    \[{\text{I}}_{4}^{ - 1} = {\text{ }}{{\mathbf{e}}^0} \wedge {{\mathbf{e}}^1} \wedge {{\mathbf{e}}^2} \wedge {{\mathbf{e}}^3}\qquad {\text{I}}_{4}^{ - 1} = - {\operatorname{I} _{4}}{\qquad}(\operatorname{I} _{4})^2 = - 1\]

   One usually assumes that the unit volume is continuous and differentiable over the entire manifold, that it has the same grade everywhere, and that it is single-valued. The last assumption implies that the manifold is orientable, and so rules out objects such as the Mobius strip, where the unit volume is double-valued.