Existence and Five-Form Properties of the Lorentzian Hodge Star

At each $x\in M$, [F2] supplies a pseudo-orthonormal basis. If it has negative orientation, reversing one spacelike basis vector preserves its metric matrix and changes the orientation sign, so a positive pseudo-orthonormal coframe exists. If two such positive coframes are related by $A$, then $A^T\eta A=\eta$ for $\eta =\mathrm{diag}(-1,+1,\dots ,+1)$; taking determinants gives $(\det A{)}^2=1$, and orientation preservation gives $\det A>0$, hence $\det A=1$. Therefore their top wedges agree. In a positive coordinate change $x^{\mathrm{\prime }}=f(x)$, $\det (g^{\mathrm{\prime }})=\det (g)\mathrm{/}(\det Df{)}^2$ while $d{x}^{\mathrm{\prime }0}\wedge \cdots \wedge d{x}^{\mathrm{\prime }9}=(\det Df)d{x}^0\wedge \cdots \wedge d{x}^9$; because $\det Df>0$, the displayed coordinate volume forms agree. The coefficient $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$ is smooth since $\det g$ never vanishes. Thus ${\mathrm{vol}}_g$ is a well-defined smooth top form.

In the coframe of step 1.1, for a subset $I\subset \{0,\dots ,9\}$ write $e^I$ for its increasing wedge and $I^c$ for its complement. By [F3], the $e^I$ form a basis in degree $p$ and the $e^{I^c}$ form a basis in degree $10-p$. Their metric pairing is diagonal: $⟨e^I,e^J{⟩}_g=0$ for $I\mathrm{\ne }J$ and $⟨e^I,e^I{⟩}_g={\eta }_I:={\prod }_{i\in I}{\eta }_{ii}\in \{-1,+1\}$. Also $e^I\wedge e^{I^c}={ϵ}_I{\mathrm{vol}}_g$, where ${ϵ}_I\in \{-1,+1\}$ is the permutation sign of the concatenated indices. Consequently the unique map satisfying the defining equation on basis pairs is ${\ast }_g{e}^I={\eta }_I{ϵ}_I{e}^{I^c}$; bilinearity extends the equation to all forms, and the complementary-basis pairing proves uniqueness. This map sends a basis to nonzero multiples of a basis, so it is an isomorphism. Its coordinate coefficients are algebraic expressions in $g^{-1}$ and $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$, hence are smooth.

For $\mathrm{\mid }I\mathrm{\mid }=5$, applying the formula twice gives ${\ast }_g^2{e}^I={\eta }_I{\eta }_{I^c}{ϵ}_I{ϵ}_{I^c}{e}^I$. There is exactly one negative diagonal entry, so ${\eta }_I{\eta }_{I^c}={\prod }_{i=0}^9{\eta }_{ii}=-1$. Interchanging the two blocks of five indices uses $5\cdot 5=25$ transpositions, so ${ϵ}_I{ϵ}_{I^c}=(-1{)}^{25}=-1$ by [F4]. Their product is $+1$, and linearity proves ${\ast }_g^2=\mathrm{id}$ on every $5$-form.

No $5$-element subset equals its complement, so the $252=\left(\genfrac{}{}{0px}{}{10}{5}\right)$ basis wedges split into $126$ disjoint pairs $\{e^I,e^{I^c}\}$. On each pair ${\ast }_g$ exchanges the two basis lines with nonzero coefficients, and step 3.1 makes their product $1$. The vectors $e^I+{\ast }_g{e}^I$ and $e^I-{\ast }_g{e}^I$ are nonzero eigenvectors with eigenvalues $+1$ and $-1$, respectively; each pair therefore contributes one dimension to each eigenspace. Both eigenspaces have dimension $126$.