0

I'm new to algebraic topology -- sorry if this is a silly question or covered in some obvious place. I didn't find it in Hatcher or other questions on this site.

What I get so far: Consider an $n$-dimensional CW complex $X$ with skeletons $\{X^{k}\}_{k = 0}^n$, a $k$-cell $\sigma$ of $X$, and a $k - 1$-cell $\tau$. The incidence number between $\sigma$ and $\tau$, which I'll write as $[\sigma, \tau]$, is defined as the topological degree of the attaching map $\varphi : \tau \to \partial\sigma$. Let $\{\mathscr{C}_k\}_{k = 0}^n$ be the chain modules on $X$, i.e. the formal integer linear combinations of $k$-cells; if we agree to also write "$\sigma$" for the $k$-chain $1\cdot\sigma$, then we can write

$$\partial\sigma = \sum_{\tau \in X^{k - 1}}[\sigma, \tau]\tau.$$

We can then extend this by $\mathbb{Z}$-linearity to define the boundary operators $\partial : \mathscr{C}_k \to \mathscr{C}_{k - 1}$. Assuming the complex has a finite number of cells, we can represent the boundary operators as sparse matrices with integer entries, which makes it possible to compute homology algorithmically.

The incidence numbers between cells can, in principle, take arbitrary integer values. Nonetheless, most examples that you might come up with by hand will have all incidence numbers be either -1, 0, or +1.

My question: Given a CW complex $X$ with incidence numbers of arbitrary magnitude, is there an equivalent complex with all incidence numbers between -1 and +1? Does the answer change if by "equivalent" I mean homeomorphic vs homotopy-equivalent? Alternatively, are there reasonable, finite complexes that you can only represent using attaching maps of higher degree?

Reason for asking: if you compute homology or generally mess about with CW complexes algorithmically, it's pretty nice if you can assume all the matrix entries are between -1 and +1. For starters, it means you can do some compression on the data type; you can use 2-bit integers instead of 32-bit.

Daniel Shapero
  • 503
  • 1
    A very useful example is to have a single $0$-cell, a single $n$-cell, and a single $(n+1)$-cell, attached to the $n$-cell by a map of degree $d$. The homology is then cyclic of order $d$ in dimension $n$. (I'm taking issue with "Nonetheless, most examples that you might come up with by hand will have all incidence numbers be either -1, 0, or +1.") – John Palmieri Nov 10 '22 at 18:55
  • If the CW complex is "regular" (see for example Hatcher, p. 533), then it can be subdivided to produce a simplicial complex, and a simplicial complex has the incidence numbers you want. – John Palmieri Nov 10 '22 at 18:57
  • Ah I missed the appendices -- if you post this as an answer I'll accept. I'm just wrapping my head around simplicial stuff and I guess I pictured CW complexes like a bunch of polygons glued together nicely instead of simplexes, which definitely misses examples like yours. – Daniel Shapero Nov 10 '22 at 19:26
  • This question gets at the issue some. – Daniel Shapero Nov 10 '22 at 19:29
  • See also https://math.stackexchange.com/questions/1528005/simplicial-complex-vs-delta-complex-vs-cw-complex – John Palmieri Nov 10 '22 at 19:46

1 Answers1

1

If the CW complex is "regular" (see for example Hatcher, p. 533), then it can be subdivided to produce a simplicial complex, and a simplicial complex has the incidence numbers you want.

In terms of algorithms, a standard way to compute homology groups is to put the boundary maps, as matrices, into Smith normal form. That procedure can end up dealing with quite large integers — see https://doc.sagemath.org/html/en/reference/topology/sage/topology/simplicial_complex_examples.html#sage.topology.simplicial_complex_examples.SumComplex, for example. So converting to small incidence numbers may help initially, but it may not provide any benefit in the end.

John Palmieri
  • 9,136