# In cell-decomposed manifolds, how easy is it to arrange for the tubular neighborhood of a diagonal to contract onto the diagonal?

Suppose that you have decomposed a manifold $M$ into cells (I care most, if it matters, about compact oriented smooth manifolds; but if my question can be solved in the PL category, all the better). By this, I have in mind some sort of combinatorial description, so there probably should be various words like "regular" or whatnot. For various reasons, I cannot restrict my attention to simplicial decompositions — I must allow cubes, for example. In any case, if you have done so, then on the set of cells in $M$ there is a metric, generated by declaring that $\operatorname{distance}(a,b) = 1$ if $a$ is a facet of $b$ (i.e. if the cell $a$ lies in the closure of the cell $b$). Then given any collection $Y$ of cells and any $\ell \in \mathbb N = \lbrace 0,1,2,\dots\rbrace$, I can define the set $\mathrm{B}_\ell(Y)$ to consist of the closure of the union of all cells at distance at most $\ell$ from some element of $Y$. It is a sub-cell-complex of $M$. Suppose that the cell decomposition is very fine compared to the topologies of $Y$ and $M$: then one should expect that $\mathrm{B}_\ell(Y)$ contracts onto the closure of $Y$. Note that the closure of $Y$ is precisely $\mathrm{B}_0(Y)$.

My specific situation is as follows. I have a manifold (compact, oriented, etc., if it matters) $X$. If I choose a cell decomposition of $X$, then I can induce a cell decomposition of $M = X^n$ by declaring that a cell of $M$ is an $n$-tuple of cells in $X$. (Certainly, if my cells were simplices with ordered vertices, then I could make other choices, but for my application this is most natural.) The diagonal map $X \hookrightarrow X^n$ is not a map of cell complexes, but still for each cell in $X$, there is a corresponding diagonal cell of $M$, and I will define $Y$ to be this "diagonal" copy of $X$.

My question is the following:

For fixed $\ell$, but letting $n$ vary, how can one find a cell decomposition of $X$ such that, with the notation above, $\mathrm B_\ell(Y)$ has the homotopy type of $X$? Or at least the same rational homology?

Almost surely, the $\ell$-fold barycentric subdivision of any cell decomposition of $X$ will do the trick — probably the $\lceil \log_2\ell \rceil$-fold barycentric subdivision would work — but I find myself unable to prove this, even after talking to various friends who know more topology than I do. Or perhaps I'm supposed to find a Riemannian metric for which I would have the appropriate result, and then choose a cell decomposition in which all vertices are at distance roughly $1$ from each other. Or something. In any case, I know that my intuition for high-dimensional manifolds is poor.

I do know how to prove that after one barycentric subdivision, $\mathrm{B}_0(Y)$ has the rational homology of $X$.

• Your question is very close in spirit to Abrams and Ghrist's work on discretized configuration spaces. It's not quite the same but it's close. – Ryan Budney May 18 '13 at 20:28
• @Ryan: Thanks for the comment! I was unaware of their work, but I'll look it up. – Theo Johnson-Freyd May 18 '13 at 21:57