coherence

Floor 2, Les sutures · introduced in Chapter 1, Un objet, plusieurs noms

Do seams chosen one at a time agree around every route?

A family of seams is coherent when every route between two incarnations gives the same map; automatic for rigid objects, and otherwise the same as coming from one choice of alignments.

Plate 2.4A coherent cycle on the object of size 24: a cyclic labelling

\ell

goes by its Singer map into

7B

, by inversion into

7A

, back to the labelling

-\ell

, and by reversal home to

\ell

Definition

A family of seams is coherent when passing from one incarnation to another along different routes gives the same map. For all the seams of a family this says that the seam groupoid is the pair groupoid; for a seam system, a chosen set of seams, it says that every cycle has trivial monodromy.

Corollary(Coherence)

Let $(Y_i)_{i\in I}$ be a family of incarnations of a rigid object, and $s_{ij}$ the unique seam $Y_i\to Y_j$ . Then $s_{ii}=\id$ and $s_{jk}\circ s_{ij}=s_{ik}$ for all $i,j,k$ .

Proof

$s_{jk}\circ s_{ij}$ and $s_{ik}$ are both seams $Y_i\to Y_k$ , and there is only one.

Proposition

Let $X$ be an object. (a) If $X$ is rigid, every seam system for $X$ is coherent. (b) A seam system whose graph is connected is coherent if and only if there are alignments $\varphi_Y\colon X\to Y$ , one for each member, with $s=\varphi_{Y'}\circ\varphi_Y^{-1}$ for every seam $s\colon Y\to Y'$ of the system.

So monodromy measures exactly how far a family of seams is from being induced by one choice of alignments.

Proof

(b) If the alignments exist, every cycle composes to $\varphi_Y\varphi_Y^{-1}=\id$ . Conversely, fix a spanning tree and an alignment at one vertex, and define the others along the tree by $\varphi_{Y'}=s\circ\varphi_Y$ . A seam outside the tree closes a cycle whose monodromy is $\varphi_{Y'}^{-1}\circ s\circ\varphi_Y$ , conjugated back to the base; it is trivial by coherence, so $s=\varphi_{Y'}\varphi_Y^{-1}$ .

Example

On the object of size 24 the cycle

\begin{aligned}\{0,1,3\}\text{-labellings}&\xrightarrow{\text{Singer}}7B\xrightarrow{\text{inversion}}7A\xrightarrow{\text{Singer}^{-1}}\{0,4,6\}\text{-labellings}\\&\xrightarrow{\text{reversal}^{-1}}\{0,1,3\}\text{-labellings}\end{aligned}

is coherent: for a cyclic labelling $\ell$ of the Fano plane, the Singer map of $-\ell$ is $(-\ell)^{-1}\circ(x\mapsto x+1)\circ(-\ell)=\ell^{-1}\circ(x\mapsto x-1)\circ\ell$ , the inverse of the Singer map of $\ell$ . Seams fixed by the conventions of their theories (inversion, transvection, rotation, Singer, reversal) are consistent wherever they meet, and on the projective line the three natural seams of the object of size 56, given by the multiplier 2, the square class of $[a,b][b,c][c,a]$ and the cyclic order of a three-subset, form a coherent triangle.

Remark

In a gauge, a coherent seam system over a graph is pure gauge: every holonomy is trivial, and the system is gauge equivalent to the one with every link variable 1. An incoherent system becomes coherent on a covering of its graph: on the covering whose fundamental group is the kernel $N$ of its holonomy the pulled-back system is coherent, and every connected covering on which it becomes coherent covers that one. For the connection on the complex of stars of the twenty-eight pairs, $N$ has index 12 and the covering is the graph on the 336 elements of $\SL(2,7)$ .

Built from: seam groupoidrigid objectalignment
Builds: seam monodromy
Objects: the twenty-eightthe object of size 24the object of size 56
In the Esquisse: 4La monodromie des sutures10La table en deux, en sept et à l’infini11Le revêtement double et le miroir12Où se rencontrent les deux parents
In the volume: XIRulial RelativityXXThe Commuting Squares