Differential Geometry On Cartographic Spaces

Cartographic Sets

There's a combinatorial way of describing $n$-dimensional discrete spaces that I'm quite fond of, that I've seen go by under various different names: it's very close to Abstract Polytopes, but has a more algebraic feel. I've seen them called "Maniplexes", although maybe to be a maniplex is to satisfy a slightly broader set of nondegeneracy conditions. Anyway, I'm going to refer to them here as "Cartographic Spaces", following Grothendieck's description of the Cartographic Group in Esquisse d'un Programme.

Here's the definition: an ($n$-dimensional) cartographic set is an action of the cartographic group on some set $X$. The $n$-dimensional cartographic group is given by generators and relations \[ (\sigma_0, \ldots, \sigma_n \st \sigma_i^2 = (\sigma_i\sigma_j)^2 = 1,\ (|i - j| > 1)) \] The intuition behind this definition is that a $n$-dimensional space is made up of cells: 0-dimensional vertices, 1-dimensional edges, 2-dimensional faces (which, since this isn't a simplicial complex, can be generally polygons, rather than merely triangles), 3-dimensional volumes, etc.

The elements of the set $X$ on which the group acts are not cells but flags: a flag is a choice of a 0-cell $c_0$ and a 1-cell $c_1$, and $\cdots$ and an $n$-cell $c_n$, such that all the $(c_i)_{i\in \{0,\ldots,n\}}$ are incident to one another. For example, here is a depiction of a 2-dimensional space, (it's 2-dimensional because the space itself is constituted out of at most 2-dimensional stuff, though I've depicted it as a surface embedded in 3-dimensions) with some flags indicated by short lines attached to the face and edge that they belong to, drawn nearer to the vertex they belong to:

The 10 red flags all correspond to the top, 5-sided face, and the 8 blue flags all belong to the front, 4-sided face. It is more convenient to draw flags as dots near their incident cells, like so:

so that we can draw the action of the group generators $\sigma_i$ on them as a graph:

The graph is undirected because we required that every $\sigma_i$ is an involution $\sigma_i^2 = 1$. The effect of $\sigma_i$ is an "$i$-dimensional flip": when we hit a flag with $\sigma_i$, we take as input some flag $(c_0, c_1, \ldots, c_i, \ldots, c_n)$, and we change it to $(c_0, c_1, \ldots, c_i', \ldots, c_n)$, the unique other flag that differs from the first in only its $i$-dimensional cell.

The fact that non-adjacent generators are supposed to commute ($\sigma_i\sigma_j = 1$ when $|i - j| > 2$) here means specifically that $\sigma_0$ and $\sigma_2$ commute, so we always end up with squares of blue and green edges in the above diagram, with blue and green edges alternating around each square.

Cartographic Spaces

I wanted to puzzle out for myself how one should think about discrete differential geometry on these structures, and I think there's a very pleasant answer. In order to talk about flows and forms and so on, we have to generalize cartographic sets to cartographic linear spaces. This is as easy as replacing "action on a set" with "linear action on a linear vector space".

A cartographic space is defined to be a vector space $V$ together with a homomorphism from the $n$-dimensional cartographic group to the group of linear maps $V \to V$.

For example, if we have a cartographic set $X$, we can trivially define a cartographic space from it: simply let the space $V$ be the space of all linear combinations of elements of $X$, and let the action of $\sigma_i$ on $V$ be the natural extension from its action on basis elements to general elements of $V$.

$k$-Vector Fields

We can then define what it means to be a $k$-vector field on $V$ in a nice way: if $k \in \{0,\ldots,n\}$, we say that a $k$-vector field on $V$ is an element $v \in V$ such that for any $i$, we have $\sigma_i v = -v$ if $i < k$, and $\sigma_i v = v$ if $i > k$. We can write this as a linear subspace of $V$: \[ V_k = \{ v \in V \st \forall i . (i < k \imp \sigma_i v = -v) \wedge (i > k \imp \sigma_i v = v)\}\] \[ = \bigcap_{i < k}\ker(\sigma_i + 1) \cap \bigcap_{i > k} \ker(\sigma_i - 1) \]

Here is an example of a $0$-vector field on $V$, i.e. a scalar field on the vertices of the space. The condition above requires $\sigma_1 v = v$ and $\sigma_2 v = v$; flipping the edge we're sitting on shouldn't change the coefficient of the flag we're looking at, and neither should flipping the face. All that matters is the vertex.

Here is an example of a $1$-vector field on $V$, i.e. a collection of weighted oriented edges. The condition above requires $\sigma_0 v = -v$ and $\sigma_2 v = v$. If we flip a vertex, we must flip the sign of the coefficient, but if we flip the face, we must keep the coefficient constant.

Finally, Here is an example of a $2$-vector field on $V$, i.e. a collection of weighted oriented faces. The condition above requires $\sigma_0 v = -v$ and $\sigma_1 v = -v$. If we flip either the vertex or edge of a flag, we must flip the sign of the coefficient.

Computing Boundaries

If we are to do differential geometry, we have to have some notion that plays the role of taking the boundary of a bit of space, or, dually, takes the exterior derivative of a form, bumping the dimension of a field (or form) down or up by one, respectively.

Curiously, I think both of these operations in this setting — at least when the cartographic space in question arises from a cartographic set — can be realized (when we fix the dimension $k$) as projection operators on $V$, i.e. linear idempotent maps $V \to V$. Given the involutivity of $\sigma_i$, we do know that $\mu_i = (\sigma_i + 1) / 2$ is idempotent, for \[ \mu_i \mu_i v = \mu_i (\sigma_i v + v) / 2 \] \[= (\sigma_i \sigma_i v + 2 \sigma_i v + v) / 4\] \[= ( v + 2 \sigma_i v + v) / 4\] \[= ( 2 \sigma_i v + 2v) / 4\] \[= ( \sigma_i v + v) / 2\] \[= \mu_i v\] We can also define the 'opposite' projection $\bar \mu_i = (1 - \sigma_i) / 2$. It is easy to check that $\im \mu_i = \ker \bar \mu_i$ and $\ker \mu_i = \im \bar\mu_i$ and $\mu_i + \bar\mu_i = 1$. Now I claim that there are projections $L_k, R_k : V \to V$, uniquely defined by requiring that \[\im L_k = \bigcap_{i < k} \ker \mu_i \qquad \im R_k = \bigcap_{i > k} \im \mu_i \] \[\ker L_k = \sum_{i < k} \im \mu_i \qquad \ker R_k = \sum_{i > k} \ker \mu_i \] and in fact these commute, $L_k R_k = R_k L_k$. Let's define $\pi_k = L_k R_k$. I claim moreover that the image of $\pi_k$ is exactly the set of $k$-vector fields $V_k$, since \[ V_k = \bigcap_{i < k}\ker(\sigma_i + 1) \cap \bigcap_{i > k} \ker(\sigma_i - 1) \] \[ = \bigcap_{i < k}\ker((\sigma_i + 1) / 2) \cap \bigcap_{i > k} \ker((\sigma_i - 1) / 2) \] \[ = \bigcap_{i < k}\ker(\mu_i) \cap \bigcap_{i > k} \ker(\bar\mu_i) \] \[ = \bigcap_{i < k}\ker(\mu_i) \cap \bigcap_{i > k} \im(\mu_i) \] \[ = \im L_k \cap \im R_k\]

Now the punchline is that taking a $(k+1)$-vector field that already lives $V_{k+1}$ and hitting it with $\pi_k$ is essentially the same thing as taking its boundary. In this case, if we wanted to show that the chain complex condition \[\partial \partial = 0\] holds, it would suffice to show that \[ \pi_{k-1}\pi_k\pi_{k+1} = 0 \tag{1}\] For we start with some vector $v$ that lives in $V_{k+1}$, and so we know that $\pi_{k+1} v = v$. If we take its boundary once, that's $\pi_k \pi_{k+1} v$. If we take its boundary twice, that's $\pi_{k-1}\pi_k \pi_{k+1}$. To establish (1), we need only check a few 'absorption' properties, namely that for any $k$, we have \[ i< k \imp (L_k \bar\mu_i = L_k = \bar\mu_i L_k) \quad L_k L_{k+1} = L_{k+1} \] \[ i> k \imp (R_k \mu_i = R_k = \mu_i R_k) \quad R_{k-1} R_{k} = R_{k-1}\] which means that \[ \pi_{k-1}\pi_k\pi_{k+1} = L_{k-1}R_{k-1} R_k L_k L_{k+1}R_{k+1} \] \[ = L_{k-1}R_{k-1} L_{k+1}R_{k+1} \] \[ = L_{k-1}R_{k-1}\mu_k \bar\mu_k L_{k+1}R_{k+1} \] \[ = L_{k-1}R_{k-1} 0 L_{k+1}R_{k+1} \] \[ = 0 \]

Or, in a diagram chase:

I think this is really nice! The reason that a double-boundary (or dually, double-exterior-derivative) is always zero, is really just the fact that a vector field can't simultaneously be in the image of $\mu_k$ ("$k$-even", "$k$-bosonic") and in the image of $\bar\mu_k$ ("$k$-odd", "$k$-fermionic")