spinor system

Let $V$ be the local system on $M$ given by the defining representation of $\SL(2,\Z[\omega])$, restricted to $\Gamma(\mathfrak p)$. At a cusp, with translations $t_b$, a cocycle of $V$ restricts to $z(b)=(sb^2/2+Ab+B\bar b,\ sb)$, with $s$ holomorphic in the parameter of the cusp and $B$ antiholomorphic. The boundary image of $H^1(M;V)$ is the graph $s=L_VB$ of an antisymmetric matrix $L_V$ with zero diagonal, indexed by the eight cusps $\infty,0,\dots,6$. Let $\Theta$ be the Paley matrix on these indices, with border $\Theta_{\infty a}=1$, $\Theta_{a\infty}=-1$ and inner entries the Legendre symbol $\bigl(\frac{b-a}7\bigr)$.

The stable $\Z[\omega]$-lattices in $\Q(\omega)^2$ are all homothetic to $\Z[\omega]^2$, and this lattice fixes the boundary scattering up to one sign, the labelling of the deck group by $\SL(2,7)$ up to inner automorphisms. In this integral framing, with the deck group labelled by reduction modulo $\mathfrak p$:

(a) $L_V=y\,\Theta$ with no further signs, where

(b) the eigenspace of $\Theta$ for $-\sqrt{-7}$ is the quartet $\mathbf4$, on which $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ has trace $(1+\sqrt{-7})/2$, so $L_V$ acts on $\mathbf4$ by $v_{\mathbf4}=-\sqrt{-7}\,y$ and on $\bar{\mathbf4}$ by $-v_{\mathbf4}$; (c) in the cusp-adapted basis $(pe_1,e_2)$, $p=3+\omega$, the constant is $-2\omega\sqrt{-3}/\bar p$, and the entries $L_{c\infty}$, $c\ne\infty$, are $2\sqrt{-3}$ times the corresponding entries $\omega/\bar p$ for the trivial local system.

The rotation $D=\mathrm{diag}(-1,1)$ acts on the cochains of $V$ by a cochain map and carries the boundary image of $H^1(M;V)$ to itself. On the coordinates $B$ it acts by a signed permutation $R_D$ over $c\mapsto-c$, on the coordinates $s$ by $-R_D$, and $R_D\Theta=-\Theta R_D$. So $D$ exchanges $\mathbf4$ and $\bar{\mathbf4}$ and, read in the conventions it transports, leaves $L_V$ and $y$ unchanged: the sign that distinguishes the quartets belongs to $\Theta$, and $y$ is invariant under the outer class.

The mirror $c$ instead carries $L_V$ to its complex conjugate in the same labels, so $y\mapsto\bar y$. Since $v_{\mathfrak p}(y)-v_{\bar{\mathfrak p}}(y)=-2$, while rescaling the coordinate of every cusp by one $\eta\in\Q(\omega)^\times$ changes this difference by $-3(v_{\mathfrak p}(\eta)-v_{\bar{\mathfrak p}}(\eta))$ and a real or purely imaginary element has equal valuations, in no $\Q(\omega)$-rational normalization is $y$ even or odd under $c$.

Let $T_0$ be the ideal tetrahedron with cusps $\infty$, 0, 1 and $1+\omega$, with primitive vectors $\xi_\infty=(1,0)$, $\xi_0=(0,1)$, $\xi_1=(1,1)$ and $\xi_{1+\omega}=(1+\omega,1)$, the cusp of $(x,y)$ being $x/y$. The four are pairwise unimodular, and at each cusp the elements $u_c(t)=I+t\,\xi_c\xi_c^{\mathsf T}J$, $t\in\Z[\omega]$, $J=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$, are parabolic for $t\ne0$ and fix exactly the line of $\xi_c$. Of the 24 permutations of the cusps of $T_0$ exactly the 12 even ones are induced by Möbius maps, all of them in $\PSL(2,\Z[\omega])$, and their lifts form the binary tetrahedral group $2T$; the odd ones are induced only by anti-Möbius maps. Modulo $\mathfrak p$, the stabilizer in $\SL(2,7)$ of the quadruple of $T_0$ and its complement has 48 elements with the element orders of $2O$; its 24 elements that exchange the quadruple with its complement act on $M$ without fixed points, since no element of $\PSL(2,\Z[\omega])$ has order divisible by 4.

(a) The line of $\xi_c$ is the unique line of $V$ fixed by the parabolic stabilizer of $c$. (b) Reduction modulo $\mathfrak p$ maps the stabilizer of $T_0$ in $\SL(2,\Z[\omega])$ isomorphically onto the even part $2T$ of $2O$, and there either of the two faithful two-dimensional representations $\rho_\pm$ of $2O$ is equivalent to $V$; the stabilizer permutes the four cusp lines as it permutes the cusps. (c) The odd part of $2O$ acts on $V$ only by transport between fibers, and $V$ does not choose between $\rho_+$ and $\rho_-$, which agree on $2T$ and differ by $\sqrt2\mapsto-\sqrt2$ on the elements of order 8. (d) The cusp lines transform as $V$ and not as $\bar V$, and in the mirror link complement $\bar M$ they transform as $\bar V$.

Give the Hermitian $2\times2$ matrices the Lorentz form with $\langle A,A\rangle=\det A$, and put $N_c=\xi_c\xi_c^\dagger$, null vectors with $\langle N_a,N_b\rangle=\tfrac12$ for $a\ne b$ that form a $\Z$-basis of $\mathrm{Herm}_2(\Z[\omega])$. The six pairs of cusps of $T_0$ give the six vectors $T_{ab}=N_a+N_b$, each a unit timelike vector, with $\langle T_{ab},T_{ac}\rangle=\tfrac32$ and $\langle T_{ab},T_{cd}\rangle=2$ when $\{a,b,c,d\}$ are the four cusps. Each of the 24 ordered moves $\{c,a\}\to\{c,b\}$ is realized, $T_{cb}=gT_{ca}g^\dagger$, by a parabolic element fixing $\xi_c$ that carries $\xi_a$ to a unit multiple of $\xi_b$, and each of the 6 ordered moves $\{a,b\}\to\{c,d\}$ by an element of order 4 of $2T$. All are holomorphic, so each preserves the class of $V$. The reflections of $T_0$ also carry pairs to pairs, and they are anti-Möbius.

For $G=\SL(2,\Z[\omega])$, $\dim(V\otimes V)^G=1$, spanned by $\varepsilon$, $\dim(V\otimes\bar V)^G=0$ and $\dim(\bar V\otimes\bar V)^G=1$; hence $\mathrm{Hom}_G(V,\bar V)=0$. No positive inner product on $V$ is $G$-invariant, since $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ is a nontrivial Jordan block. In a fermionic Fock space over a finite-dimensional space with a positive inner product, holes transform by $(g^{-1})^{\mathsf T}$, which for $g\in\SL(2,\C)$ is $\varepsilon g\varepsilon^{-1}$: the holes of $V$ carry $V$ again, not $\bar V$, whatever positive inner product is used. Holes carry $\bar V$ only on a one-particle space carrying a unitary representation of the Lorentz group in which the spinor transforms as $V$, and such a representation is infinite-dimensional, as in Wigner’s classification.

$V$ carries no $G$-invariant positive form, but the vectors that the moves pass between do. For a positive $T$ of determinant one put $\langle\xi,\eta\rangle_T=\xi^\dagger T^{-1}\eta$; every $g\in\SL(2,\C)$ is an isometry from $(V,\langle\,,\rangle_T)$ to $(V,\langle\,,\rangle_{gTg^\dagger})$, and every move of $T_0$ is an isometry of these forms.

(1) On $\ell^2(G\cdot T_0)\otimes V$ with $\langle\psi,\phi\rangle=\sum_T\psi(T)^\dagger T^{-1}\phi(T)$ the action $(g\psi)(T)=g\,\psi(g^{-1}Tg^{-\dagger})$ is unitary: it is $\operatorname{Ind}_{G_{T_0}}^{G}V$, and on $L^2(\mathbb H^3)\otimes V$ the same formula, with the invariant measure, gives the unitary representation of $\SL(2,\C)$ induced from $\mathrm{SU}(2)$. (2) On functions on the cusps with values in the cusp lines, each cusp’s vector taken primitive in $\Z[\omega]^2$, $G$ acts unitarily, by the character of each cusp’s stabilizer on its vector: $\chi$ for $V$ and $\bar\chi\ne\chi$ for $\bar V$. On the four cusp lines of $T_0$ a hole relative to the full Fock state carries the conjugate character. (3) Every $gTg^\dagger$ is positive and every $g\xi\xi^\dagger g^\dagger$ positive semidefinite: neither space contains vectors of the opposite sign.

$G$ acts on $\Lambda=\mathrm{Herm}_2(\Z[\omega])$ by $X\mapsto gXg^\dagger$, preserving $\det X$ and the cone of positive $X$; in the continuum every vector with $\det X<0$ is carried to $-X$ by some element of $\SL(2,\C)$. Let $X\in\Lambda$ have $\det X<0$ and $3\mid\det X$, and suppose $X\not\equiv0$ modulo $\pi=1-\omega$. Then $X$ modulo $\pi$ is a symmetric form of rank one over $\F_3$, $\lambda\,\ell\ell^{\mathsf T}$, and the Legendre symbol $\bigl(\tfrac{\lambda}{3}\bigr)$ is invariant under $X\mapsto gXg^\dagger$ for every $g\in\GL(2,\Z[\omega])$, and under $X\mapsto X^{\mathsf T}$. Since $\bigl(\tfrac{-1}{3}\bigr)=-1$, no such map sends $X$ to $-X$.

In the box $|x_1|,|x_2|\le2$, $z=u+v\omega$ with $|u|,|v|\le2$, the elements of $\SL(2,\Z[\omega])$ with entry coordinates in $[-2,2]$ reverse 276 of the 474 primitive vectors with $\det X<0$, and those of unit-determinant $\GL(2,\Z[\omega])$ reverse 322. The unreversed are the 148 covered by the theorem above and 4 of determinant $-11$. At a prime that is not ramified there is no invariant of this kind, since at an inert prime every scalar of the residue field is a norm from its quadratic extension and at a split prime the two factors scale independently; so the four are presumably reversed by larger elements, which were not searched. No element reverses a vector with $\det X>0$.

$\SL(2,7)$ has no two-dimensional irreducible representation. The group $2O$ has eight classes, of sizes 1,1,6,6,6,8,8,12, and its irreducible characters 1, $1'$, $\mathbf2$, $\mathbf3$, $\mathbf3'$, $E_+$, $E_-$ and $\mathbf4_s$ are all obtained from restrictions, with $E_\pm=\rho_\pm$. The quartets $\mathbf4$ and $\bar{\mathbf4}$ both restrict to $\mathbf4_s=E_+\otimes\mathbf2=E_-\otimes\mathbf2$, and $E_+\otimes E_+=1+\mathbf3$.

So no quartet has a doublet summand, and the two quartets restrict to the same irreducible. The one doublet with the pattern $\mathbf2\otimes\mathbf2=\mathbf1\oplus\mathbf3$ inside a quartet’s restriction is $E$, a tensor factor shared by $\mathbf4$ and $\bar{\mathbf4}$; the faithful doublets $E_\pm$ occur only in $\mathbf6_s$, $\mathbf6_s'$ and $\mathbf8_s$. Around a cusp the parabolic element $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$, of order 7, acts on $\mathbf4$ with eigenvalues $\zeta^k$, $k\in\{0,1,2,4\}$, and on $\bar{\mathbf4}$ with $k\in\{0,3,5,6\}$: one fixed line, together with the quadratic residues, respectively the non-residues.