Rigid String Diagrams

Goal

We aim to define a 'rigid space' to be something kind of like an $n$-category, except instead of being able to draw ordinary string diagrams where the lines are "flexible", we're going to define it in such a way that they're rather rigid. Here's an example of an ordinary string diagram in 2 dimensions:

We have objects $C, D, E$, some morphisms $F: C \to D$, $G : D \to E$, $H : C \to E$, $J : E \to C$, and some $2$-cells $\alpha : G \o F \to H$ and $\beta : 1_E \to H \o J$.

Now here's a "rigid string diagram":

The constraint we mean to impose is that every 1-cell has a fixed slope everywhere it occurs in the diagram, and every 2-cell has exactly the same 1-cells emerging from it at the same angles. In the general $n$-dimensional case, every $m$-cell would occupy a finite subset of an $(n-m)$-dimensional hyperplane, again of a fixed slope.

Colors and Diagrams

Fix a natural number $n$ for what dimension we're working in. Choose a set $C$ to be a set of 'colors', or alternatively the 'cells' of the category or space that we're describing.

A 'diagram' is a map $\R^n \to C$.

An $n$-dimensional rigid space is a set $C$ together with an action \[\wat : C \to \R^n \to C\] The action can also be thought of as a map from cells to diagrams, describing what the neighborhood of that cell must locally resemble, wherever it occurs. Three properties are required, for any $(c : C) (v : \R^n) (\lambda: \R^+)$, where $\R^+ = \{x \in \R \st x > 0\}$.

$c \wat 0 = c$
$ c \wat \lambda v = c \wat v$
$\exists U . (0 \in U \subseteq \R^n) \land U \hbox { open} \land {}$ $\forall (u : U).\ c \wat (v + u) = (c \wat v) \wat u$

The first axiom says that the required neighborhood for cell $c$ must actually have $c$ at its origin. The second axiom says that scaling a point $v$ in the domain of the diagram by any positive factor $\lambda$ doesn't change what cell is assigned to it. The third axiom says that the "neighborhood diagrams" themselves have to respect the neighborhood diagrams of all the cells that occur in them: for any point $v$, there is some neighborhood of $v$ that exactly resembles the neighborhood diagram of the cell $c \wat v$, except translated so that the origin coincides with $v$.

Dimension

Despite the fact that we have said nothing about the dimensions of the cells themselves, it turns out a notion of dimension falls out of the axioms above.

First we fix a rigid space, and define the "kernel" of a cell to be the subset of $\R^n$ as follows: \[ \ker(c) = \{v \in \R^n \st c \wat v = c\} \] We claim that this is actually a linear subspace of $\R^n$. So we need to show closure under scaling by arbitrary (not just positive) constants, and closure under addition.

Lemma 1 If $v \in \ker(c)$ and $\lambda \in \R$, then $\lambda v \in \ker(c)$.

Proof If $\lambda = 0$, appeal to axiom 1. If $\lambda > 0$, appeal to axiom 2. The remaining case is $\lambda \lt 0$.

By axiom 3, let $U$ be given. Choose $\ee > 0$ small enough such that $-\ee v \in U$ and $\ee < 1$. Then we know

$c \wat (v - \ee v) = (c \wat v) \wat (-\ee v)$	(by (3))
$c \wat (1-\ee )v = c \wat (-\ee v)$	(by assumption $v \in \ker(c)$)
$c \wat v = c \wat (-\ee v) $	(by (2) and $\ee < 1$)
$c = c \wat (-\ee v)$	(by assumption $v \in \ker(c)$)
$c = c \wat \lambda v$	(b.c. $\ee > 0, \lambda < 0$ and (2))

hence $\lambda v \in \ker(c)$. $\cqed$

Lemma 2 If $v,w \in \ker(c)$, then $v+w \in \ker(c)$.

Proof

By (3), pick an open neighborhood $U \ni 0$ such that for all $u \in U$ we have \[c \wat (v + u) = (c \wat v) \wat u\] Now pick $\ee > 0$ small enough such that $\ee (v+w) \in U$ and $(\ee /(1+\ee ))w \in U$, and observe:

$ c \wat (v + w) = c \wat (\ee v + \ee w) $	(2)
$ = (c \wat v) \wat (\ee v + \ee w) $	($v \in \ker(c)$)
$ = c \wat (v + \ee v + \ee w) $	(3)
$ = c \wat ((1+\ee )v + \ee w)$
$ = c \wat (v + (\ee /(1+\ee ))w)$	(2)
$ = (c \wat v) \wat (\ee /(1+\ee ))w)$	(3)
$ = c \wat (\ee /(1+\ee ))w $	($v \in \ker(c)$)
$ = c \wat w $	(2)
$ = c $	($w \in \ker(c)$)

$\cqed$

The kernel of any cell $c$ is a linear subspace of $\R^n$, so it has a dimension. From this we can define a notion of dimension of the cell itself. We say \[ \dim(c) = n - \dim(\ker(c))\] So that the 0-dimensional "objects" are the cells $c$ whose kernel is all of $\R^n$: the boundary-diagram consists only of $c$. The 1-dimensional "morphisms" are the cells $c$ whose boundary-diagram has a codimension-1 kernel, partitioning $\R^n$ into two halves.

We can show that every other cell occurring in the boundary diagram of $c$ must have strictly lower dimension.

Lemma 3 If $c \wat v = d$, then $\ker(c) \subseteq \ker(d)$.

Proof: Let $c, v, d$, and $w \in \ker(c)$ be given, and suppose $c \wat v = d$. We know $c \wat w = c$, and we want to show $d \wat w = d$.

Apply to (3) to $w$ to find $U_w$ such that \[ \forall (u_w : U_w).\ c \wat (w + u_w) = (c \wat w) \wat u_w\] Pick $\ee$ small enough such that $\ee v \in U_w$. Let $x = (w + \ee v)$. Note that \[ c \wat x\] \[ = c \wat (w + \ee v)\] \[ = (c \wat w) \wat \ee v\] \[ = c \wat \ee v\] \[ = c \wat v\] \[ = d\]

Now apply (3) to $x$. That is, pick $U_x$ such that \[ \forall (u_x : U_x) . c \wat (x + u_x) = (c \wat x) \wat u_x\] Now pick $\delta > 0$ small enough that $\delta w \in U_x$. Observe that on the one hand \[ c \wat (x + \delta w)\] \[ = c \wat (w + \ee v + \delta w)\] \[ = c \wat ((1 + \delta )w + \ee v)\] \[ = c \wat (w + v (\ee / (1 + \delta )))\] \[ = (c \wat w) \wat (v (\ee / (1 + \delta )))\] \[ = (c \wat w) \wat v\] \[ = c \wat v\] \[ = d\] But on the other hand \[ c \wat (x + \delta w)\] \[ = (c \wat x) \wat \delta w\] \[ = d \wat \delta w\] \[ = d \wat w\] which is as required. $\cqed$

Corollary If $c \wat v = d$ and $c \ne d$, then $\dim(d) \lt \dim(c)$.

Proof We immediately know $v \ne \ker(c)$ since $c \ne d$. Choose $U$ by (3) for $v$, and pick $\ee > 0$ small enough so that $\ee v \in U$. Then \[ d \wat v = d \wat \ee v = \] \[ = (c \wat v) \wat \ee v = c \wat (v + \ee v)\] \[= c \wat v(1+\ee) = c \wat v = d\] so $v \in \ker(d) - \ker(c)$. That is $v$, is a witness that we must have a strict containment \[ \ker(d) \supsetneq \ker(c)\] hence \[ \dim(\ker(d)) > \dim(\ker(c))\] hence \[\dim(d) \lt \dim(c)\] $\cqed$