Universal Kernel

twisting element

What does a symmetry that normalizes the group, rather than commuting with it, give?

The element c by which a symmetry conjugates the marking; correcting the symmetry by c gives an automorphism, its equivariant twist, whose power is inverse to that of c.

(1:0:0)(0:0:1)(0:1:0)τthe flexes of the klein quarticw0w1w2w3w4w5w6F8 = F2[w]/(w3 + w + 1)on its 24 coordinatizations, post-composition: τp ↦ (f ↦ f(p))124twisting element124equivariant twistΦ(y) = c−1y
Plate 2.11The Frobenius at 2, twice: as a twisting element it has power 2 (blue); made equivariant it is τ\tau, of power 4 (gold), on the flexes of the Klein quartic and on the coordinatizations of the Fano plane by F8\F_8.
Definition

Let YμY_\mu be a marked set and σ\sigma a bijection of YY with σ∘μ(x)∘σ−1=μ(cxc−1)\sigma\circ\mu(x)\circ\sigma^{-1}=\mu(cxc^{-1}) for all x∈Gx\in G and some c∈Gc\in G. The element cc is the twisting element of σ\sigma, and Φ=μ(c)−1∘σ\Phi=\mu(c)^{-1}\circ\sigma is its equivariant twist.

So a symmetry that normalizes the group gives two objects: an element of GG, and an automorphism of the incarnation.

Lemma(Twisted symmetries)

Φ\Phi is an automorphism of YμY_\mu. If GG has trivial centre, cc is unique. If σ\sigma fixes a point yy, then Φ(y)=c−1y\Phi(y)=c^{-1}y.

In the language of powers: if cc normalizes a cyclic stabilizer ⟨x⟩\langle x\rangle with power kk, then Φ\Phi has power k−1k^{-1} on the corresponding incarnation.

Proof

Φμ(x)=μ(c)−1μ(cxc−1)σ=μ(x)μ(c)−1σ=μ(x)Φ\Phi\mu(x)=\mu(c)^{-1}\mu(cxc^{-1})\sigma=\mu(x)\mu(c)^{-1}\sigma=\mu(x)\Phi. If c′c' also works, c−1c′c^{-1}c' commutes with every xx.

Theorem(The Frobenius at two on the Klein quartic)

Let σ2\sigma_2 be the automorphism ζ↦ζ2\zeta\mapsto\zeta^2 of Q(ζ7)\Q(\zeta_7), the Frobenius at 2, acting on points of P2\Proj^2 coordinatewise.

(a) σ2(ρ(x))=ρ(h−1xh)\sigma_2(\rho(x))=\rho(h^{-1}xh) for every x∈Gx\in G; the twisting element c=h−1c=h^{-1} satisfies cgc−1=g2cgc^{-1}=g^2, so it has power 2.

(b) Φ=ρ(h)∘σ2\Phi=\rho(h)\circ\sigma_2 is an automorphism of every incarnation in the quartic that σ2\sigma_2 preserves. On the flexes Φ=τ\Phi=\tau, the flex-tangent map, of power 4; on the bitangents, the centres and the flex triangles Φ\Phi is the identity.

(c) At a prime above 2 the 24 flexes reduce to the 24 points of the quartic over F8\F_8, and the arithmetic Frobenius v↦v2v\mapsto v^2, twisted by ρ(h)\rho(h), is again τ\tau.

(d) The vectors fixed by this twisted Frobenius form a three-dimensional F2\F_2-space V0⊂F83V_0\subset\F_8^3, stable under GG. The two primes above 2 give two Fano planes P(V0)\Proj(V_0), whose points have stabilizers of class S4bS_4^b and S4aS_4^a respectively.

(e) For each point pp of the quartic over F8\F_8, f↦f(p)f\mapsto f(p) is an F8\F_8-coordinatization of the dual plane P(V0∗)\Proj(V_0^*); sending pp to it is a seam onto the 24 coordinatizations, and it carries τ\tau to post-composition with the Frobenius of F8\F_8.

Proof

(a) σ2(ρ(g))=diag(ζ8,ζ4,ζ2)=ρ(g)2=ρ(h−1gh)\sigma_2(\rho(g))=\mathrm{diag}(\zeta^8,\zeta^4,\zeta^2)=\rho(g)^2=\rho(h^{-1}gh) and σ2(ρ(h))=ρ(h)\sigma_2(\rho(h))=\rho(h); for ρ(s)\rho(s), and then for all xx, by machine. (b) The flex (0:0:1)(0:0:1) is rational, so Φ(0:0:1)=ρ(h)(0:0:1)=(0:1:0)=τ(0:0:1)\Phi(0:0:1)=\rho(h)(0:0:1)=(0:1:0)=\tau(0:0:1), and two GG-maps that agree at one point of a transitive GG-set agree everywhere. The objects of sizes 28, 21 and 8 are rigid, so Φ\Phi is the identity on their incarnations. (c) The reduction of the flexes is in Elkies; reduction commutes with ρ\rho and with tangent lines. (d), (e) The fixed vectors were computed; in a basis of V0V_0 the coordinates of Φˉ(p)\bar\Phi(p) are the squares of those of pp.

Theorem(The Frobenius at two is coherent)

Under every seam between incarnations of the object of size 24 in the Klein quartic and in the Fano plane coordinatized by F8\F_8, the vertex link of the octonion completion, the GG-equivariant Frobenius automorphisms correspond: both equal τ\tau, of power 4. The twisting elements, h−1h^{-1} for the quartic and the Frobenius for the completion, both have power 2.

So there is no incoherence. The relation between τ\tau, of power 4, and a Frobenius of power 2 is the general relation Φ(y)=c−1y\Phi(y)=c^{-1}y between a twisting element and its equivariant twist, and it takes the same form in both theories. The flex-tangent map τ\tau, a fact of the projective geometry of the quartic, is the Frobenius at 2 made equivariant.

Proof

For the completion, with Mx(v)=wxv2M_x(v)=w^xv^2 on F8=F2[w]/(w3+w+1)\F_8=\F_2[w]/(w^3+w+1), conjugation by the Frobenius v↦v2v\mapsto v^2 sends MxM_x to M2xM_{2x}, so the Frobenius conjugates the Singer cycle to its square, and post-composition with it on the 24 coordinatizations has power 4. The power does not depend on the seam.

Remark

Twisted symmetries are the inner case of seams over an automorphism: a symmetry σ\sigma with σμ(x)σ−1=μ(cxc−1)\sigma\mu(x)\sigma^{-1}=\mu(cxc^{-1}) is a seam over the inner automorphism x↦cxc−1x\mapsto cxc^{-1}, and correcting it by μ(c)−1\mu(c)^{-1} gives a seam.

Proposition(Lifting the Frobenius)

On the double cover G~=SL⁡(2,7)\tilde G=\SL(2,7), acting through its Weil representation, the Galois automorphism ζ↦ζ2\zeta\mapsto\zeta^2 carries W(x)W(x) to W(cxc−1)W(cxc^{-1}) both for c=d4c=d_4 and for c=d3=−d4c=d_3=-d_4, the two lifts of the twisting element h−1h^{-1}; d4d_4 has order 3 and is the odd lift, and d3d_3 has order 6. The twist is the identity on the sixteen vectors for d4d_4 and −1-1 for d3d_3. On the new object of size 48, the nonzero vectors of F72\F_7^2, the automorphisms lying over τ\tau are the scalings by 2, of order 3, and by 5=−25=-2, of order 6, whose cube is −I-I.

So τ3=−I\tau^3=-I is not forced. The twisting element is determined only up to the centre {±I}\{\pm I\} of G~\tilde G, which is the deck transformation; the odd lift, the only lift of the same order as h−1h^{-1}, gives the lift of τ\tau of order 3, and the lift of τ\tau of order 6, with τ3=−I\tau^3=-I, is the choice of the twisting element of even order.

Proof

Scaling by λ\lambda has power λ2\lambda^2 on the vectors up to sign, so the automorphisms over τ\tau are the scalings with λ2=4\lambda^2=4. The rest was checked by machine, in Q(ζ)\Q(\zeta).

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