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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 $n$-dimensional space $\mathcal{G}_n$ forms a one-dimensional subspace. Its members are called $n$-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 $\mathcal{G}_4$ 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
    \[\left| {{{\mathbf{e}}_0} \wedge {{\mathbf{e}}_1} \wedge {{\mathbf{e}}_2} \wedge {{\mathbf{e}}_3}} \right| = {\text{ }}{\left| {{\text{det}} {{[\mathbf{e}}_i} \cdot {{\mathbf{e}}_j}]} \right|^{1/2}}\]
  4. The unit volume has an inverse ${\text{I}}^{4}$. If the basis is orthonormal:
    \[{\text{I}}^{4} = {\text{ }}{{\mathbf{e}}^0} \wedge {{\mathbf{e}}^1} \wedge {{\mathbf{e}}^2} \wedge {{\mathbf{e}}^3}\qquad {\text{I}}^{4} = - {\operatorname{I} _{4}}{\qquad}(\operatorname{I} _{4})^2 = - 1\]

   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.