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Let $K$ be a simplicial complex and let $L$ be a subcomplex of $K$.

Questions:

  • Is it possible to define an operation on (some) simplicial complexes so that $K/L$ is a simplicial complex for which $|K/L|\cong |K|/|L|$?
  • Is it the case that $|K|/|L|$ is always triangulable, i.e., that there's a simplicial complex $Q$ such that $|Q|\cong |K|/|L|$?

For example, if $K=\{a,b,ab\}$ and $L=\{a,b\}$ then $K/L$ would be a vertex with a self-loop. That's not a simplicial complex, but it is a cellular complex, and its homology is isomorphic to the homology of $|K|/|L|$. (In this case, to get $K/L$ I identified all simplices in $L$ with a point, preserving "connections" between them.)

Eric Wofsey
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user12344567
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1 Answers1

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For simplicial sets rather than simplicial complexes, you can define quotients perfectly fine: if $K$ is a simplicial set and $L\subset K$ is a sub-simplicial set, there is a simplicial set $K/L$ defined by $(K/L)_n=K_n/L_n$, and the canonical map $|K|/|L|\to |K/L|$ is a homeomorphism. If $K$ and $L$ are simplicial complexes, then you can consider them as simplicial sets, and take the simplicial set $K/L$; if the nondegenerate simplices of $K/L$ happen to form a simplicial complex, then you've found a natural simplicial complex whose geometric realization is $|K|/|L|$. In general, the nondegenerate simplices of a simplicial set form a simplicial complex iff the vertices of any nondegenerate simplex are all distinct and no two nondegenerate simplices have the same vertices. In the case of $K/L$, this translates to the following (quite strong!) conditions on $L$: if two vertices of a simplex of $K$ are in $L$, then the entire simplex must be in $L$, and for any simplex of $K$ not containing any vertices in $L$, there is at most one vertex in $L$ that can be added to it to give a simplex of $K$.

If you want an operation stated purely in terms of simplicial complexes without mentioning simplicial sets, you can do this as follows. Let $K$ be a simplicial complex with vertex set $V$ and $L$ be a subcomplex of $K$ with vertex set $W$ such that if two vertices of a simplex of $K$ are in $W$, then the entire simplex is in $L$, and for any simplex of $K$ not containing any vertices in $W$, there is at most one vertex in $W$ that can be added to it to give a simplex of $K$. Then you can define a simplicial complex $K/L$ as follows: the vertex set of $K/L$ is $V/W$, and a subset $S$ of $V/W$ is a simplex of $K/L$ iff it is the image of a simplex of $K$ under the quotient map $V\to V/W$. There is then a canonical homeomorphism $|V|/|W|\cong|V/W|$.

As for your second question, the answer is yes, since the geometric realization of any simplicial set is triangulable. The proof is nontrivial; see this answer on MO for an indication of some of the ideas involved and references to more detailed proofs. In particular, while there does exist a simplicial complex $Q$ such that $|Q|\cong|K|/|L|$, there is not (as far as I know) any canonical choice of such a $Q$.

Eric Wofsey
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  • Thanks for answering. Yes, I want an operation stated in terms of simplicial complexes only. I don't quite understand the construction, however. (1) So what does $V/W$ stand for ($V$ and $W$ are sets)? (2) What simplex are you referring to in "If two vertices of $K$ are in $W$ then the entire simplex is in $L$"? (3) What exactly do you mean by "For any simplex of $K$ not containing any vertices in $W$ there is at most one vertex in $W$ that can be added to it to give a simplex of $K$"? (I am accepting the answer as soon as I understand how and why the construction works.) – user12344567 Dec 10 '15 at 14:27
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    (1) $V/W$ is the quotient of the set $V$ by the equivalence relation that make all the points of $W$ equivalent to each other. (2) I said "two vertices _of a simplex_ of $K$", so you have some particular simplex of $K$ you're looking at. (3) I mean that if $S$ is the set of vertices of a simplex in $K$ and $S\cap W=\emptyset$, then there cannot be two distinct vertices $x,y\in W$ such that $S\cup\{x\}$ and $S\cup\{y\}$ are both the vertex sets of simplices of $K$. (The point is that if there were, then $S\cup\{x\}$ and $S\cup\{y\}$ would have the same vertices in the quotient.) – Eric Wofsey Dec 10 '15 at 19:23
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    As for why it works, this should be straightforward to see once you understand it. The point is that the hypotheses on $L$ are extremely restrictive, such that when you just glue together all the vertices of $L$ and throw away all the higher-dimensional simplices of $L$ (or equivalently, collapse them together down to a single point), you still get a simplicial complex (no simplex you had before has had two of its vertices glued together, and simplices are still uniquely determined by their vertex sets). – Eric Wofsey Dec 10 '15 at 19:41
  • Yes, okay, I see. These assumptions are pretty restrictive. – user12344567 Dec 14 '15 at 12:25
  • Could you elaborate on a canonical homeomorphism $f:|V/W|\to|V|/|W|$? I understand that once you have $f$ you can extend it to a homeomorphism $g:|K/L|\to|K|/|L|$. For instance, given $f$ for a suitable geometric realization, take $x\in|K/L|$ for $x=\lambda_1v_1+\ldots+\lambda_rv_r$, where $\lambda_i$'s are barycentric coordinates. Now define $g(x)=\lambda_1f(v_1)+\ldots+\lambda_rf(v_r)$, which is a homeomorphism between $|K/L|$ and $|K|/|L|$. – user12344567 Dec 14 '15 at 15:03
  • Ok, I see, it's obvious there's a homeomorphism $f:|V/W|\to|V|/|W|$. – user12344567 Dec 14 '15 at 15:21