Often we can describe a particular compact manifold as (homeomorphic to) some simplicial complex. This always works in 3 or fewer dimensions. Is there a graceful way to extend this to noncompact manifolds? Two approaches I can think of would be:

  1. Use an infinite number of simplices.

  2. Use a finite number of simplices, but remove the boundary.

Does either or both of these work? Is one nicer than the other? Can anyone point me to a treatment that works out the details?


Are all Infinite Simplicial Complexes non-compact?

manifold as simplicial complex


1 Answer 1


Infinite simplicial complexes are standard and useful things. I'm not sure what kinds of detail you want to know, but there's not too much to say, other than that finite simplicial complexes are compact and infinite simplicial complexes are noncompact. There is nothing in the definition of a simplicial complex that requires finiteness, so any reference about general simplicial complexes applies just as well to the infinite case as to the finite case.

Infinite simplicial complexes occur naturally, for example, in the proof of the triangulation theorems for surfaces and for 3-manifolds --- see the book of E. E. Moise, or Hatcher's article. Those theorems, by the way, apply to noncompact manifolds as well as compact manifolds. As said above, in the noncompact case the complex is infinite; in the compact case it is finite. In fact, the noncompact case of these theorems is an ineluctable part of the proof of the compact case.

I am also aware of a few special places in the mathematical literature where one studies a space $X$ by showing that it is homeomorphic to a space of the form $K-L$ where $K$ is a simplicial complex and $L$ is a subcomplex. The example closest to my own work is the outer space of a finite rank free group. A very, very important special case, familiar to many different branches of mathematics, is the modular diagram for $SL(2,\mathbb{Z})$ (aka the modular group). That diagram is a tiling of the hyperbolic plane of the form $K-L$ where $K$ is a certain infinite simplicial 2-complex and $L$ is the 0-simplices of $K$.

  • $\begingroup$ Hi Lee -- +1 for "ineluctable"! $\endgroup$
    – Jack Lee
    May 4, 2017 at 3:38
  • $\begingroup$ Heh. That word has been stuck in my brain for the last 40-odd years since I first read Joyce's Ulysses. It finally found a way out. $\endgroup$
    – Lee Mosher
    May 4, 2017 at 13:31
  • $\begingroup$ Thanks for the answer. Is there anything specifically wrong with using a finite simplicial complex and deleting the boundary? $\endgroup$
    – user13618
    May 4, 2017 at 14:50
  • $\begingroup$ The boundary of what? If you mean the "boundary of the given finite simplicial complex", the trouble is that the concept of a "boundary" is not defined for a finite simplicial complex. Note that the examples I gave were of a simplicial complex minus a specified subcomplex; "boundary" was not mentioned in those examples. $\endgroup$
    – Lee Mosher
    May 4, 2017 at 17:40
  • 1
    $\begingroup$ So, now I understand that you want to know what happens when you take a finite simplicial complex $K$ which is a manifold-with-boundary, and you then remove its boundary $\partial K$, leaving its interior $\text{int}(K) = K - \partial K$. It is straightforward to prove, using an invariance of domain type of argument, that $\partial K$ is a subcomplex of $K$, and so $\text{int}(K)$ will be an example of what I wrote in the last paragraph of my answer, namely a simplicial complex minus a subcomplex. So in that sense there are no objections, and I don't have any other objections in mind. $\endgroup$
    – Lee Mosher
    May 4, 2017 at 23:12

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