SU(3)-fixed chiral spinor lines in six dimensions

For $x,y\in V$ and $\omega \in {\mathrm{\Lambda }}^{\bullet }V$, wedge and contraction satisfy ${\iota }_{\bar{x}}(y\wedge \omega )+y\wedge {\iota }_{\bar{x}}\omega =⟨y,x⟩\omega$, while two wedge operators and two contractions anticommute. Expanding with $c(x)=x\wedge -{\iota }_{\bar{x}}$ therefore gives $c(x)c(y)+c(y)c(x)=-(⟨x,y⟩+⟨y,x⟩)I=-2\mathrm{Re}⟨x,y⟩I\mathrm{.}$ Thus $S_6$ is a complex Clifford module for the underlying Euclidean six-space, with even and odd exterior degree as its two chiral modules. For $U_g={\mathrm{\Lambda }}^{\bullet }g$, preservation of wedge products gives $U_g(x\wedge )U_g^{-1}=(gx)\wedge$. Preservation of the Hermitian form gives $⟨gv,gx⟩=⟨v,x⟩$; applying this equality to each term of the displayed contraction formula in [A2] yields $U_g{\iota }_{\bar{x}}{U}_g^{-1}={\iota }_{\overline{gx}}$. Subtracting these identities proves $U_{g}c(x)U_g^{-1}=c(gx)$, the required Clifford covariance.

For the displayed Koszul formula, write its right-hand side as $K(X,Y,Z)$. Substituting $fX$, $fY$, and $fZ$ and using $[X,fY]=f[X,Y]+X(f)Y$ gives $K(fX,Y,Z)=fK(X,Y,Z)$, $K(X,Y,fZ)=fK(X,Y,Z)$, and $K(X,fY,Z)=fK(X,Y,Z)+2X(f)g(Y,Z)$. Thus nondegeneracy of $g$ makes the formula a smooth affine connection. Directly exchanging $X,Y$ gives $K(X,Y,Z)-K(Y,X,Z)=2g([X,Y],Z)$, so it is torsion-free; adding the expressions with $Y,Z$ exchanged gives $K(X,Y,Z)+K(X,Z,Y)=2Xg(Y,Z)$, so it is metric-compatible. These identities also force the formula for any torsion-free metric-compatible connection, proving uniqueness.

The local spin-frame change $s^{\mathrm{\prime }}=sa$, with $A={\lambda }_6(a)$, changes the orthonormal-frame connection form by ${\omega }^{\mathrm{\prime }}=A^{-1}\omega A+A^{-1}dA$. Applying $d{\lambda }_6$ gives $A_{s^{\mathrm{\prime }}}=\mathrm{Ad}(a^{-1})A_s+a^{-1}da$. Associated-bundle coordinates change by ${\psi }^{\mathrm{\prime }}={\rho }_6^{\pm }(a{)}^{-1}\psi$, so differentiating this identity gives $d{\psi }^{\mathrm{\prime }}+d{\rho }_6^{\pm }(A_{s^{\mathrm{\prime }}}){\psi }^{\mathrm{\prime }}={\rho }_6^{\pm }(a{)}^{-1}(d\psi +d{\rho }_6^{\pm }(A_s)\psi )$. Thus the displayed local covariant derivatives agree on overlaps. Along each smooth path segment in a local frame the transport equation is $\dot{v}(t)=-B(t)v(t)$ with continuous matrix $B(t)=d{\rho }_6^{\pm }(A_s(\dot{\gamma }(t)))$. On a compact interval, its fundamental matrix is the uniformly convergent Picard series whose $m$th term has norm at most $(T\sup \mathrm{\parallel }B\mathrm{\parallel }{)}^m\mathrm{/}m!$; substitution into the integral equation proves existence, and iterating that equation for a difference proves uniqueness. Reversing the path gives the inverse map. For a smooth family of straight coordinate paths in a chart, the same series and its parameter derivatives converge uniformly on compact parameter sets, so transport depends smoothly on the endpoint. Composing finitely many chart-segment transports defines the piecewise-$C^1$ transport used in the Statement.

The degree-zero line ${\mathrm{\Lambda }}^{0}V$ is fixed by every $g\in SU(3)$, and the degree-three line ${\mathrm{\Lambda }}^{3}V$ is fixed because $g$ acts on it by $\det g=1$. Neither $V={\mathrm{\Lambda }}^{1}V$ nor $V^{\ast }={\mathrm{\Lambda }}^{2}V$ has a nonzero invariant vector: $D_1=\mathrm{diag}(i,-i,1)$ and $D_2=\mathrm{diag}(1,i,-i)$ lie in $SU(3)$, and their fixed subspaces in $V$ are respectively $\mathbb{C}{e}_3$ and $\mathbb{C}{e}_1$, whose intersection is zero; their dual actions have fixed subspaces $\mathbb{C}{e}^3$ and $\mathbb{C}{e}^1$, also with zero intersection. The map $u\wedge v\mapsto (w\mapsto \mathrm{\Omega }(u,v,w))$, for $\mathrm{\Omega }=e^1\wedge e^2\wedge e^3$, is an isomorphism ${\mathrm{\Lambda }}^{2}V\to V^{\ast }$ and is $SU(3)$-equivariant because $\mathrm{\Omega }$ is invariant. Hence $(S_6^{+}{)}^{SU(3)}={\mathrm{\Lambda }}^{0}V$ and $(S_6^{-}{)}^{SU(3)}={\mathrm{\Lambda }}^{3}V$, each of complex dimension $1$.

The operators in [A3] satisfy the ten-dimensional Clifford relations: the same-factor anticommutators are those of ${\mathrm{\Delta }}_4$ and $S_6$, while the mixed anticommutators vanish because ${\mathrm{\Gamma }}_6$ anticommutes with each ${\rho }_a$. Their module has complex dimension $4\cdot 8=32$. The product chirality operator therefore splits its positive eigenspace into $({\mathrm{\Delta }}_4^{+}\otimes S_6^{+})\oplus ({\mathrm{\Delta }}_4^{-}\otimes S_6^{-})$, up to swapping both internal signs if the orientation sign is reversed; the negative eigenspace has the opposite pairing. Either ten-dimensional chirality uses one four-dimensional Weyl module with each internal chirality.

Define $y\sim z$ when a piecewise-smooth path in $Y$ joins them. Constant paths, reversal, and concatenation make this an equivalence relation. Each class is open: in a coordinate ball about any point of the class, coordinate straight segments join that point to every other point of the ball. If there were more than one class, one class and the union of the others would be disjoint nonempty open sets covering $Y$, contrary to connectedness. Thus a piecewise-smooth path from $y_0$ to every $y\in Y$ exists. Evaluation at $y_0$ injects global parallel sections into the holonomy-fixed fiber vectors, since a parallel section is recovered along every path by its transport. Conversely, for a holonomy-fixed vector $v$ in a chiral fiber at $y_0$, set $\psi (y):=P_{\gamma }v$ for any piecewise-smooth path $\gamma$ from $y_0$ to $y$. This is path-independent because two such paths compose with the reverse of one to a based loop, whose transport fixes $v$. The smooth endpoint-dependence established in step 1.3 makes $\psi$ smooth locally, and the transport ODE makes it parallel. Hence evaluation is an isomorphism from global parallel sections to the $SU(3)$-fixed vectors in the chiral fiber. By step 2.1 each such invariant space is one-dimensional over $\mathbb{C}$.

Let $L_{+}$ and $L_{-}$ be the one-dimensional $SU(3)$-fixed lines in $S_6^{+}$ and $S_6^{-}$ from step 2.1. By steps 2.1, 3.1, and 3.2, the holonomy-fixed part of either ten-dimensional chirality is $U\oplus V$, where $U=W\otimes L_{+}$ and $V=\bar{W}\otimes L_{-}$, or with the two internal lines interchanged. The real structure $R$ in [A4] preserves this subspace and commutes with $H$. There is no nonzero $H$-equivariant antilinear map $W\to W$: such a map has the form $z\mapsto C\bar{z}$, and for $g_t=\mathrm{diag}(t,t^{-1})\in S{L}_2(\mathbb{C})$ equivariance gives $C\mathrm{diag}(\bar{t},{\bar{t}}^{-1})=\mathrm{diag}(t,t^{-1})C$; $t=2$ forces the off-diagonal entries of $C$ to vanish, and $t=i$ forces the diagonal entries to vanish. The same calculation applies to $\bar{W}$. Hence $R$ exchanges $U$ and $V$. Choose $e_{+}\mathrm{\ne }0$ in $L_{+}$. The map $W_{\mathbb{R}}\to \mathrm{Fix}(R)$, $w\mapsto w\otimes e_{+}+R(w\otimes e_{+})$, is a real-linear $H$-equivariant bijection, with inverse given by projection to $U$ followed by the identification $w\otimes e_{+}\mapsto w$. Likewise $w\mapsto w+J_{4}w$ identifies the underlying real representation $W_{\mathbb{R}}$ with $S_4^{\mathrm{M}}$, with inverse given by projection to $W$. Thus $\mathrm{Fix}(R)\cong S_4^{\mathrm{M}}$ as real $H$-modules, so it has real dimension $4$. The same argument applies when the ten-dimensional chirality or orientation interchanges the two internal lines.