In other words, I’m hosed.

I’ve been thinking about a category of ‘abstract locally complete partitions’. Fix M=m\times \mathbb{R}^n, a disjoint union of copies of \mathbb{R}^n indexed by elements of the finite set m. My category is then supposed to consist of pairs (\rho,f) where \rho is a finite collection of affine spaces, A_{\rho}=\coprod_{i\in s}A_i, along with a partition, \Lambda, of the indexing set s, and f:A_{\rho}\rightarrow M an affine (on each component) map which is non-locally-constant. I interchangeably let \rho refer to the collection of spaces, or the partition on that collection of spaces.

To describe non-locally-constant, I must first remind you that I let my partition \Lambda on s induce a partition on A_{\rho} where x\sim y iff the component containing x is equivalent to the component containing y, mod \Lambda. That is, all points in a component A_i are considered equivalent, and two components are equivalent as determined by \Lambda. Now f being non-locally-constant means that there exists x,y equivalent mod \Lambda but with f(x)\neq f(y).

Now, given such an object (\rho,f) in my category, I would like to reduce it to a more restricted type of complete locally affine partition. In particular, I would like to reduce it to the case where

  • There are no more than m zero-dimensional affine spaces in A_{\rho}, and no more than m non-zero-dimensional affine spaces in A_{\rho}.
  • If A_i is a component of A_{\rho} that is not zero-dimensional, then the \Lambda-equivalence class of i is just i itself.
  • My map f takes distinct zero-dimensional spaces in A_{\rho} to distinct components of M. Similarly for non-zero-dimensional spaces.

To complete the description of my big category, I need to describe the morphisms. A map \alpha from (\rho,f) to (\rho',f') will be an affine map \alpha:A_{\rho}\rightarrow A_{\rho'} such that f'\circ \alpha=f and \overline{\alpha(\rho)}\leq \rho' – a property I will now further clarify. My \rho consists of a partition of the space A_{\rho}. By \alpha(\rho), I mean the transitive closure of the relation on A_{\rho'} where whenever x\sim y in A_{\rho} then f(x)\sim f(y) in A_{\rho'}. This process gives me an equivalence relation \alpha(\rho) on A_{\rho'}. Any time I have an equivalence relation \sigma on a (disjoint union of) affine space(s), I let \overline{\sigma} denote the finest coarsening of \sigma that is “locally affine” – meaning equivalence classes are (disjoint unions of) affine subspaces, and all equivalence classes in a given component are parallel (technically, there’s probably a little more than that, but it’s good enough for now I guess). So now we have the meaning of \overline{\alpha(\rho)}, and by \overline{\alpha(\rho)}\leq \rho', I simply mean that \overline{\alpha(\rho)} is coarser than \rho' (by which I mean the equivalence relation on A_{\rho'}.

So allow me to recap. My \alpha is an affine map with a property saying that an appropriate triangle commutes (f'\circ \alpha=f), and the affine closure of the image of the partition for \rho is coarser than the partition for \rho'. Since \rho' is a “complete” locally affine partition (any two points in the same component are in the same equivalence class), this also forces \overline{\alpha(\rho)} to be “complete”.

Of course, I’m not mentioning that this category is really “a category object in the category of topological spaces”. So really I have a space of objects, a space of morphisms, and enough maps between them to make sense of things. I’ll continue not mentioning that, saving it for another day.

Now, like I said, I want to be reducing any (\rho,f) in my big category to get it down to a particularly nice form. One of the main steps I had been relying on turns out to not be allowed, because it isn’t functorial. I hardly knew one could write down “obvious” maps that weren’t functorial, but I’ve apparently done so rather frequently lately.

So, what is this construction? Given (\rho,f), one step I want to take is to replace any subset of the components of A_{\rho} that are (1) all related by the partition, and (2) all map to the same component of M. I want to replace such a subset by the affine span (direct sum in the category of affine spaces) of the spaces in it. This seems entirely reasonable. Given a bunch of affine spaces, and a map to an affine space, I get, for free, a map from the direct sum of the original affine spaces. That’s what direct sums do.

However, since I have disjoint unions as targets of maps \alpha, I run into trouble. Consider, for example, (\rho,f) where the space A_{\rho} consists of three disjoint points, the equivalence relation has them all equivalent, and f sends them all to the same component. Consider (\rho',f') the same three points, the same f, but only two of the points are equivalent. The obvious map \alpha has \overline{\alpha(\rho)}\leq \rho', and the triangle commutes by construction. Now, when I take affine spans as mentioned, I end up with a single space in \rho (a plane, the affine span of 3 points), and two spaces in \rho' (a line (the span of two points), and a point). That’s a problem, because I no longer know what to do with \alpha. As I mentioned before, given a map from a bunch of affine spaces to a single affine space, I get a map from the affine span to the single space. However, given a map from a bunch of affine spaces to a bunch of affine spaces, I no longer get a map from the affine span to the same bunch of spaces (the single affine span, being connected, can only end up in one of the target spaces).

So that’s upsetting. It’s not even the only thing I’ve written down recently that wasn’t a functor. Back to the drawing board, as a fella says.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: