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Cubical tangent bundle

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Cubical (aka "discrete") manifolds are still manifolds, but made up of cells. What are the tangents of cubical complexes?

They aren't differentiable manifolds. Then there should be no tangents at the corners!

So, how do we understand the tangents (and differential forms)?


A $1$-manifold looks like this:

Discrete1ManifoldTangents.png

What happens at $a$? What do we need to do instead for $T_aM$?

Observation: Just a few "bad" points, the rest is OK.

BadPoints.png

Stand them up.

StandThemUp.png

Glue them together:

GlueThemTogether.png

Tangent vectors come from parametrization:

TangentVectorsFromParametrization.png

Here the tangent turns $90^o$. That's why $T_aM$ is "thick" here. This idea comes from the smooth case, when we record how the tangents turn, as follows.

How to glue these together?

TangentSpaceGlueTogether.png

What about $2$-manifolds? What is $T_aM$ at corners?

A $2$-manifold is a surface and as a cubical complex it will look like this, compared to a $1$-manifold:

TangentSpaceGlueTogether2.png TangentSpaceGlueTogether3.png

If we are to compute the tangent bundle of this kind of corner, then, algebraically, $T_aS$ should be the closure of the complement of the following: $$\left\{ \begin{array}{} x > 0 \\ y > 0 \\ z > 0 \end{array} \right\} \cup \left\{ \begin{array}{} x < 0 \\ y < 0 \\ z < 0 \end{array} \right\}.$$

Let's recall that the tangent bundle of a simple curve is homeomorphic to a strip:

Tangent bundle of a simple curve.png

Note: In general, tangent bundles are manifolds.

We expect to have the same result for cubical curves.

Let's consider the tangent bundle of the two-edge curve above. The tangent bundles of the two edges are strips and the result should be just a simple strip.

This is not the case.

Tangent bundle of edge curve.png

So, whether we use the tangent cone or its span, the tangent bundle created this way isn't homeomorphic to the that of the corresponding smooth curve! Too little in the first: too much in the last.

We define a discrete tangent bundle as: $$T(K) = \displaystyle\bigsqcup_{A \in V(K)} T_A(K),$$ as the disjoint union of all tangent spaces with each simply the $1$-dimensional star of the vertex: $$T_A(K)=\{AB \in K\}.$$

Note: In this sense, $T(K)$ is more like a disk bundle than a vector bundle.

One can also define the tangent bundle based on the concept of tangent cone. However, it applies to the realization of the complex and relies on the concept of distance.