\( \newcommand{\bs}{\boldsymbol}
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Appendix E
The rules for calculating the inner product of a vector with
a $p$-form and the inner product of two $p$-forms are:
\[{\mathbf{a}} \cdot
({{\mathbf{a}}_1} \wedge ... \wedge {{\mathbf{a}}_p}) =
\sum\limits_{k = 1}^p {{{( - 1)}^{k + 1}}({\mathbf{a}} \cdot
{{\mathbf{a}}_k}} ){{\mathbf{a}}_1} \wedge ... \wedge
{{\mathbf{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{a}
}}_k} \wedge ... \wedge {{\mathbf{a}}_p}\]
\[({{\mathbf{a}}_p} \wedge ...
\wedge {{\mathbf{a}}_1}) \cdot ({{\mathbf{b}}_1} \wedge ... \wedge
{{\mathbf{b}}_p}) = \left| {\begin{array}{*{20}{c}}
{{{\mathbf{a}}_1} \cdot
{{\mathbf{b}}_1}}&{..}&{{{\mathbf{a}}_1} \cdot
{{\mathbf{b}}_p}} \\ {..}&{}&{..} \\ {{{\mathbf{a}}_p} \cdot
{{\mathbf{b}}_1}}&{..}&{{{\mathbf{a}}_{}} \cdot
{{\mathbf{b}}_p}} \end{array}} \right|\]
The 'check' on $\mathbf{a}_k$ in (E.1) indicates omission. Note the
reverse order of the $\mathbf{a}$-blade in (E.2). Otherwise,
a sign-factor ${( - 1)^{p(p - 1)/2}}$ needs to be factored in
because it takes $p(p - 1)/2$ interchanges to put the
$\mathbf{a}$-vectors in opposite order. This operation of reversion
on a blade ${{\mathbf{A}}_p} = {{\mathbf{a}}_1} \wedge ... \wedge
{{\mathbf{a}}_p}$ is indicated by the tilde-symbol:
$\widetilde{\mathbf{A}}_p:= {{\mathbf{a}}_p} \wedge ... \wedge
{{\mathbf{a}}_1}$.
Reversion is included in the definition of the scalar product
of two multivectors that yields the scalar part:
\[\left\langle
{{{\mathbf{A}}_p},{{\mathbf{B}}_p}} \right\rangle : =
{{\widetilde{\mathbf{A}}_p}} \cdot {{\mathbf{B}}_p}\]
The scalar product is non-zero only when ${\mathbf{A}}_p$ and
${\mathbf{B}}_p$ are of the same grade. It is symmetrical and
reversable, and may be used to define a 'norm squared' of blades and
multivectors, e.g.: ${\left| {{{\mathbf{A}}_p}} \right|^2} := \left|
{{{{\mathbf{\widetilde A}}}_p} \cdot {{\mathbf{A}}_p}} \right| =
\left| {{\text{Det}}\left[ {{{\mathbf{a}}_i} \cdot {{\mathbf{a}}_j}}
\right]} \right|$.
Identities for wedge products involving Trautman forms:
- $\quad {{\mathbf{e}}^k} \wedge {{\bs{\eta }}_a} \; \; \,= -
\delta _a^k\eta, \quad {\mathbf{\eta }} := {\text{I}}^4$
- $\quad {{\mathbf{e}}^k} \wedge {{\bs{\eta }}_{ab}} \; \, = -
2!\delta _{[a}^k{{\bs{\eta }}_{b]}}$
- $\quad {{\mathbf{e}}^k} \wedge {{\bs{\eta }}_{abc}} \, = -
3!\delta _{[a}^k{{\bs{\eta }}_{bc]}}$
- $\quad {{\mathbf{e}}^k} \wedge {\varepsilon _{abcd}} = -
4!\delta _{[a}^k{{\bs{\eta }}_{bcd]}}$
- $ \quad {{\mathbf{e}}^k} \wedge {{\mathbf{e}}^l} \wedge
{{\bs{\eta }}_{ab}} \;= - 2!\delta _{[a}^k\delta _{b]}^l\eta$
- $\quad {{\mathbf{e}}^k} \wedge {{\mathbf{e}}^l} \wedge
{{\bs{\eta }}_{abc}} = - 3!\delta _{[a}^k\delta _b^l{{\bs{\eta
}}_{c]}}$
- $ \quad {{\mathbf{e}}^k} \wedge {{\mathbf{e}}^l}{\varepsilon
_{abcd}} \; \; \;= - 4!\delta _{[a}^k\delta _b^l{{\bs{\eta
}}_{cd]}} $
To obtain these identities one identifies the $p$-blades at the
left-hand sides, takes the dual with (C.2) and applies (B.3). Then
the inverse duality operation gives the Trautman forms at the
right-hand sides.