Spin(8) four-form chirality weights

The basis formula in [F2] gives ${\star }^2{e}_I=(-1{)}^{4(8-4)}{e}_I=e_I$ for every four-element index set $I$, so $\star$ is diagonalizable with eigenvalues $\pm 1$. Its matrix on the 70 basis vectors has zero diagonal because $I^c\mathrm{\ne }I$, so its trace is zero; the two eigenspaces therefore each have dimension $35$. The vector action ${\rho }_g$ lies in $SO(V)$ by [F1]. Its induced action preserves the Gram inner product, since each pairing $⟨{\rho }_g{v}_i,{\rho }_g{w}_j⟩$ equals $⟨v_i,w_j⟩$, and it fixes $\omega$ because its determinant is $1$. Applying it to the Hodge identity shows $g\cdot \star \beta$ satisfies the defining identity for $\star (g\cdot \beta )$; uniqueness in [F2] gives $g\cdot \star =\star g\cdot$.

Put $w_j^{+}=e_{2j-1}+i{e}_{2j}$ and $w_j^{-}=e_{2j-1}-i{e}_{2j}$. With the Cartan basis $H_j=-i{R}_j$ fixed in Spin(8) Clifford modules, real forms, and D4 weights, the vectors $w_j^{\pm }$ have weights $\pm {\xi }_j$. For signs $s_j\in \{+1,-1\}$, expansion in the ordered real basis gives $\star (w_1^{s_1}\wedge \cdots \wedge w_4^{s_4})=(s_1{s}_2{s}_3{s}_4)(w_1^{s_1}\wedge \cdots \wedge w_4^{s_4})\mathrm{.}$ Indeed, a term with $k$ even-indexed basis vectors has Hodge sign $(-1{)}^k$, and the ratio of its coefficient to that of the complementary term is $(-1{)}^k{\prod }_j{s}_j$, so every complementary pair has the displayed common eigenvalue. The vectors $w_1^{+}\wedge \cdots \wedge w_4^{+}$ and $w_1^{+}\wedge w_2^{+}\wedge w_3^{+}\wedge w_4^{-}$ therefore lie in the $+$ and $-$ eigenspaces and have weights $2{\lambda }_4={\xi }_1+{\xi }_2+{\xi }_3+{\xi }_4$ and $2{\lambda }_3={\xi }_1+{\xi }_2+{\xi }_3-{\xi }_4$, respectively. Each is killed by every positive-root operator: adding any positive root ${\xi }_i\pm {\xi }_j$ with $i<j$ gives a coordinate $2$, while every ${\mathrm{\Lambda }}^4{V}_{\mathbb{C}}$ weight is a sum of four distinct weights among $\pm {\xi }_j$ and has coordinates in $\{-1,0,1\}$. For either $\mu =2{\lambda }_4=(1,1,1,1)$ or $\mu =2{\lambda }_3=(1,1,1,-1)$, the shifted-coordinate squares $({\mu }_i+{\rho }_i{)}^2$ are $(16,9,4,1)$, where $\rho =\frac{1}{2}{\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}\alpha =(3,2,1,0)$. Pairing the positive roots ${\xi }_i\pm {\xi }_j$ in the Weyl dimension formula gives denominator ${\prod }_{i<j}({\rho }_i^2-{\rho }_j^2)=4320$ and numerator ${\prod }_{i<j}(({\mu }_i+{\rho }_i{)}^2-({\mu }_j+{\rho }_j{)}^2)=7\cdot 12\cdot 15\cdot 5\cdot 8\cdot 3=151200$ for either weight. Thus each corresponding irreducible module has dimension $151200\mathrm{/}4320=35$.

The $+$ eigenspace is Spin(8)-invariant by step 1.1, and the positive-definite Gram inner product from the Statement complexifies to a positive-definite invariant Hermitian form by step 1.1. Orthogonal complements therefore decompose this finite-dimensional complex representation into irreducible submodules, and the corresponding orthogonal projections commute with the Lie-algebra action. A nonzero projection of the highest-weight vector in step 2.1 is still killed by all positive-root operators and has weight $2{\lambda }_4$. By [F3] this weight is dominant integral; its summand is $L(2{\lambda }_4)$ by [F4], and its dimension is $35$ by the Weyl calculation in step 2.1. The whole complexified $+$ eigenspace has dimension $35$ by step 1.1, so that summand is the entire eigenspace; it is irreducible of highest weight $2{\lambda }_4$. A nonzero real invariant subspace complexifies to this whole $35$-dimensional complex module, so its real dimension is $35$ and it is the full real eigenspace. Thus the real $+$ eigenspace is irreducible.

The $-$ eigenspace is Spin(8)-invariant by step 1.1, and the positive-definite Gram inner product from the Statement complexifies to a positive-definite invariant Hermitian form by step 1.1. Orthogonal complements therefore decompose this finite-dimensional complex representation into irreducible submodules, and the corresponding orthogonal projections commute with the Lie-algebra action. A nonzero projection of the highest-weight vector in step 2.1 is still killed by all positive-root operators and has weight $2{\lambda }_3$. By [F3] this weight is dominant integral; its summand is $L(2{\lambda }_3)$ by [F4], and its dimension is $35$ by the Weyl calculation in step 2.1. Since the whole complexified $-$ eigenspace has dimension $35$ by step 1.1, it equals this summand and is irreducible of highest weight $2{\lambda }_3$. A nonzero real invariant subspace complexifies to this whole $35$-dimensional complex module, so its real dimension is $35$ and it is the full real eigenspace. Thus the real $-$ eigenspace is irreducible.