# On Gromov's proof of the systolic inequality $\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$

In the page 10 of the paper "Filling Riemannian manifolds" by Gromov (ProjetEuclid link), the author proves the following inequality (1.2) relating the systole and the filling radius of manifolds. $$\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$$

During the proof, he uses an argument something like this:

"$$M$$ is a closed $$n$$-dimensional manifold and $$[M]$$ is the fundamental class of it. In some ambient space $$X$$, the fundamental class $$[M]$$ is null-homologous. Therefore, one can find $$n+1$$-dimensional singular chain $$c$$ inside $$X$$ such that $$\partial c=[M]$$. Moreover, using a $$\textbf{piecewise linear approximation}$$ of $$c$$, one can construct a polyhedron $$P$$ such that $$M\subset P \subset X$$ and the fundamental class $$[M]$$ is null-homologous inside of this $$P$$."

So, somehow he is constructing a triangulation of realization of the singular chain $$c$$ (i.e., union of continuous images of $$\Delta_{n+1}$$) using "piecewise linear approximation". But I don't understand this step. How do we guarantee this realization of singular chain is triangulable? It is not necessarily manifold. What is precise meaning of his "piecewise linear approximation" of singular chain?

Edit: I changed the title of this question from "When a topological space (not necessarily manifold) has a triangulation?" to "A question on the Gromov's proof of $$\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$$". Even though the original question is interesting in its own right, but I want to focus on one specific topic: Understanding Gromov's argument.

• Do you seek an answer to the precise question you ask, or an explanation of why Gromov's argument is correct? These likely have different answers. – John Pardon Feb 8 at 23:58
• @John Pardon: Mainly, explanation of Gromov's argument. – S.Lim Feb 9 at 0:03
• @KGEO Then you should probably change your title accordingly. Both are very interresting though. – M. Dus Feb 9 at 0:16
• @M. Dus Makes sense. I will edit my question to focus on one topic, soon. – S.Lim Feb 9 at 0:31
• About the question in the title (at this time), namely "When does a topological space (not necessarily manifold) have a triangulation?", it is an open question whether the (compact) space of closed subgroups of $\mathbf{R}^n$ has a triangulation for $n\ge 3$ (for $n\le 2$ it's true). – YCor Feb 9 at 0:35

The following may work.

$$M$$ is a smooth manifold so it can be triangulated smoothly. Now taking a fine enough (smooth) barycentric subdivision, there is an isotopy taking $$M$$'s embedding in $$L^\infty$$ to a linearly-embedded simplicial complex $$M'$$ which doesn't move any point too far (since $$M$$ is compact and bounds within an open $$\epsilon$$-neighborhood, it bounds in some $$\epsilon'<\epsilon$$-neighborhood; take $$M'$$ such that no point moves further than $$(\epsilon-\epsilon')/2$$ or so).

Denote the image of $$c$$ under the isotopy by $$c'$$. Those faces of $$c'$$ in $$M'$$ are already piecewise-linear, and the map can be approximated by a piecewise-linear one $$c''$$ from a fine enough barycentric subdivision of the domain of $$c'$$: send the vertices to their images under $$c'$$, and extend linearly. It is an approximation in the sense that it is the same up to homotopy, and within an $$\epsilon$$-neighborhood of $$M'$$. Since the set of linear simplices in $$M' \cup c''$$ has a finite-dimensional affine span, we may now work in a vector space $$W$$ of sufficiently high dimension, and we may assume $$M'$$ bounds within an $$\epsilon$$-neighborhood inside $$W$$.

Triangulate a sufficiently small neighborhood of $$M'$$ in $$W$$ in which $$M'$$ bounds, and call the result $$P$$. A sufficiently fine derived subdivision of $$P$$ refines the complex $$M'$$ (see for instance the first 15 pages or so of Glaser's book). Since singular and simplicial homology coincide, $$M'$$ is nullhomologous (simplicially) in those simplices of this complex contained in an $$\epsilon$$-neighborhood of $$M'$$. Now one can work with this $$P$$.

(Edit: closer to the original meaning than the paragraph above: a sufficiently fine derived subdivision of $$P$$ refines $$M' \cup \text{im}(c'')$$).

Since isotopy = ambient isotopy in high codimension, we may take $$P$$ "back" through the ambient isotopy to a smooth simplicial complex containing $$M$$.

I think slight modification of the argument in Hatcher's algebraic topology book (pg 108-109) works.

Let $$[M]$$ be a $$n$$-singular cycle representing the fundamental cycle of $$M$$. Then $$[M]=\delta_1+\dots+\delta_k$$ where each $$\delta_i:\Delta_n\rightarrow M\subset B_r(M,L^\infty(M))$$ is a singular $$n$$-simplex. From the assumption, there is a singular $$(n+1)$$-chain $$c$$ such that $$\partial c= [M]$$. One can express $$c=\sum_i\varepsilon_i\sigma_i$$ with $$\varepsilon_i=\pm 1$$, allowing repetitions of singular $$(n+1)$$-simplices $$\sigma_i$$. Now, when we compute $$\partial c$$ as a sum of singular $$n$$-simplices with signs $$\pm 1$$, there may be canceling pairs consisting of two identical singular simplices with opposite sign. Choosing a maximal collection of such canceling pairs, construct an $$(n+1)$$-dimensional $$\Delta$$-complex $$K$$ from a disjoint union of $$(n+1)$$-simplices $$\Delta_{n+1}^i$$ one for each $$\sigma_i$$, by identifying the canceling pairs.

Then, we have a continuous map $$\phi:K\rightarrow B_r(M,L^\infty(M))$$ induced from $$\sigma_i$$'s. $$K$$ is $$\Delta$$-complex, not simplicial complex. $$\phi$$ is not homeomorphism, just continuous map. But I think this is enough. Let $$P:=\mathrm{Im}(\phi)$$ (then, $$M\subset P\subset B_r(M,L^\infty(M))$$). By subdividing the complex $$K$$ if necessary, one can make distance between $$\phi(u)$$ and $$\phi(v)$$ arbitrarily small (whenever $$u,v$$ are vertices). After that, just follow Gromov's argument.