Universal Kernel

seam monodromy

What does going around a loop of natural identifications do?

The composite of seams around a closed walk, an automorphism of the incarnation; it measures how far a family of seams is from one choice of alignments.

the other seven flex triangles turn with ity = 0x = 0z = 0(1:0:0)(0:0:1)(0:1:0)τ124powerτ
Plate 2.6The monodromy τ\tau on one flex triangle of the Klein quartic: each flex goes along its tangent to the other flex on it, and three steps return.
Definition(Seam system, monodromy)

The monodromy of a cycle γ=(s1ϵ1,…,smϵm)\gamma=(s_1^{\epsilon_1},\dots,s_m^{\epsilon_m}) of a seam system, starting and ending at YY, is

mon(γ)=smϵm∘⋯∘s1ϵ1∈Aut⁡G(Y).\mathrm{mon}(\gamma)=s_m^{\epsilon_m}\circ\cdots\circ s_1^{\epsilon_1}\in\Aut_G(Y).

The word is used as for coverings: going around a loop of identifications returns a permutation of the fibre. Here the fibre is an incarnation and the permutation is an automorphism of the object; through an alignment it is an element of NG(H)/HN_G(H)/H, well defined up to conjugation.

Proposition

If NG(H)/HN_G(H)/H is abelian, then for each incarnation YY the isomorphism Aut⁡G(Y)≅Aut⁡G(X)\Aut_G(Y)\cong\Aut_G(X) given by an alignment does not depend on the alignment, and monodromy is a homomorphism from the fundamental group of the graph of the system to Aut⁡G(X)≅NG(H)/H\Aut_G(X)\cong N_G(H)/H.

Proof

Two alignments differ by an automorphism aa of XX, and the two isomorphisms differ by conjugation by aa, which is trivial in an abelian group. Concatenating cycles composes monodromies.

Theorem(Monodromy of the object of size 24)

Let τ\tau be the composite of the tangent and residual-point seams: a flex of the Klein quartic goes to the other flex on its tangent. Then τ\tau is an automorphism of the flexes of power 4, so the cycle flexes →\to flex tangents →\to flexes, along the two natural seams, has monodromy of order 3. It permutes each flex triangle cyclically:

τ ⁣: (1:0:0)↦(0:0:1)↦(0:1:0)↦(1:0:0).\tau\colon\ (1:0:0)\mapsto(0:0:1)\mapsto(0:1:0)\mapsto(1:0:0).
Proof

The rotation of (0:0:1)(0:0:1) is gg, and the rotation of (0:1:0)(0:1:0) is the element acting there by ζ\zeta, which is g4g^4, since ρ(g)\rho(g) acts there by ζ2\zeta^2 and so ρ(g)4\rho(g)^4 by ζ8=ζ\zeta^8=\zeta. As τ(0:0:1)=(0:1:0)\tau(0:0:1)=(0:1:0), the rotation seam carries τ\tau to a map sending gg to g4g^4, which is the fourth-power map.

Remark

So the answer for non-rigid objects is mixed. Seams fixed by the conventions of their theories are consistent wherever they meet. But a single theory may supply two natural seams between the same two incarnations, and then a cycle of length two already has nontrivial monodromy: the contact point and the residual point of a flex tangent, or the roles of a point in its line, whose three seams have relative powers 1, 2 and 4. The monodromy is then an invariant of the theory; here it is the cyclic order that the tangents put on each flex triangle, a fact of the projective geometry of the quartic.

Theorem(The Coxeter edges and the marking)

Let the Coxeter graph be in its antiflag model, with GG acting through a marking μ\mu. Each edge has the form {(p,B),(q,B′)}\{(p,B),(q,B')\}, with BB and B′B' meeting in the third point cc of the line pqpq.

(a) The point rule, which goes from each point of BB off pqpq to the third point of its line with pp, and from each point of B′B' off pqpq to the third point of its line with qq, traces a directed 4-cycle on the quadrangle complementary to pqpq. The line rule traces, dually, a directed 4-cycle on the four lines missing cc. Each rule, followed by the element of order 4 that advances its cycle one step, is a seam from the edges to 4A4A, and the two rules give mutually inverse elements.

(b) The vertex seam sends an antiflag to the pair of points of P1(F7)\Proj^1(\F_7) with the same stabilizer, and an edge to a harmonic pair of disjoint pairs {{a,b},{c,d}}\{\{a,b\},\{c,d\}\}. Of the two directed 4-cycles a→c→b→d→aa\to c\to b\to d\to a and a→d→b→c→aa\to d\to b\to c\to a, exactly one has [a,c][c,b][b,a][a,c][c,b][b,a] a nonzero square, and the bracket rule sends the edge to the element of order 4 advancing that cycle one step.

(c) If μ\mu differs from μA\mu_A by an inner automorphism, the bracket rule agrees with the point rule on every edge; if by an outer one, it agrees with the line rule.

Consequently the seam system for G/C4G/C_4 formed by the Coxeter edges, the harmonic pairs of pairs and the class 4A4A, with the vertex seam, the bracket rule and the point rule, is coherent when the marking is in the class of μA\mu_A, and its monodromy is the nontrivial automorphism otherwise.

Proof

(a) The rules use only incidence and treat the two antiflags of an edge alike, so they are GG-maps; that they give inverse elements was checked by machine. (b) In [a,c][c,b][b,a][a,c][c,b][b,a] each point occurs twice, so its square class does not depend on the coordinate vectors, and it is invariant under SL⁡(2,7)\SL(2,7). With a=0a=0, b=∞b=\infty, harmonicity gives d=−cd=-c, and the products for the cycle a→c→b→da\to c\to b\to d are all in the square class of cc, while the reverse cycle gives that of −c-c; as −1-1 is not a square modulo 7, exactly one cycle has a square product. (c) For μA\mu_A the agreement was checked on all 42 edges. An inner change of marking is induced by a collineation, which commutes with all the constructions. An outer change, by conjugation with a Möbius map of non-square determinant, multiplies every bracket by a non-square, so it reverses the bracket rule.

Proposition(Monodromy as the kernel of holonomy)

Let a seam system over a connected graph G\mathcal G have holonomy hol ⁣:π1(G,v)→A\mathrm{hol}\colon\pi_1(\mathcal G,v)\to A, read as a description; its kernel NN is the group of loops around which the seams close up. The system is coherent if and only if N=π1(G,v)N=\pi_1(\mathcal G,v), and the group of monodromies is π1(G,v)/N\pi_1(\mathcal G,v)/N. So monodromy is what remains of the loops once the kernel of the holonomy is divided out.

Proof

Holonomy is a homomorphism on the fundamental group, and the system is coherent exactly when every holonomy is trivial.

Examplecomputed

Monodromy can be the spinor sign. In the lattice E8E_8 preserved by SL⁡(2,7)\SL(2,7) for one class of tetrahedra of Thurston’s manifold, the 224 half-roots form two copies O1O_1 and O2O_2 of the new object of size 112, whose automorphism group is C4C_4. The reflection seam, changing the sign of a half-root at the point of its support fixed by its stabilizer, is a seam from O1O_1 to O2O_2 and back, and the cycle it forms has trivial monodromy, since its square is the identity. The sign seam, the sign pattern of CrCr on the support of rr, with CC the conference matrix of the Weil representation, equals the reflection seam on O1O_1 and its negative on O2O_2: the cycle it forms has monodromy −I-I. Half of CrCr off the support is an automorphism of O2O_2 of order 4 with square −1-1, the integral shadow of multiplication by ii, and it generates the automorphism group.

Open questionopen

On the object G/V4bG/V_4^b the automorphism group is S3S_3, not abelian, so monodromy is defined only up to conjugation. A natural seam system with non-abelian monodromy is not known: the natural seams found there, the elation and centre seams, carry swaps to swaps and rotations to rotations, and are coherent.

In the volume
XIRulial Relativity