In the following I will use definitions and constructions of simplicial complexes and $\Delta$-complexes/sets from Greg Friedman's An elementary illustrated Introduction to simplicial sets.

I'm looking for an example of a $\Delta$-complex which cannot be endowed with structure of a simplicial complex or a proof that it always works. In this case one should say this space is "triangulable".

Note that of course it's easy to construct a $\Delta$-complex which isn't a simplicial complex with respect the choices of simplices comming from the $\Delta$-complex datum: see example on page 11 in Friedman's notes. And this should be not surprising at all because $\Delta$-complexes allow much more flexibility in gluing boudaries than simplicial complex. But that's not what I'm asking about: if we subdivide the $2$-cell in tree new $2$-cells, then we can endow this $\Delta$-complex with another simplex structure which turns it into a honest simplicial complex.

And so my question is if every $\Delta$-complex "triangulable" in the sense that one can endow it with structure of a simplicial complex (ie decompose it into simplices satisfying glueing relations allowed for simplicial complex) which might have nothing to do with original $\Delta$-complex structure?

Supplemental picture:

enter image description here

  • $\begingroup$ The word you are looking for is "triangulable", as written on the page you cite. The geometric realisation of any simplicial set – hence also any $\Delta$-set – is triangulable, by subdividing twice. $\endgroup$
    – Zhen Lin
    Commented Jan 22, 2023 at 22:19

1 Answer 1


Here is an exercise: The 2nd barycentric subdivision of every $\Delta$-complex is a simplicial complex.

To follow up on our discussion in the comments, let me address the example edited in to your post.

Let me denote the large blue $2$-simplex as $\Sigma$, which we think of as the domain of one of the simplices in the given $\Delta$-complex $X$, with corresponding characteristic map denoted $f : \Sigma \to X$.

Let me also denote the highest vertex of $\Sigma$ as $A$, the lower right vertex of $\Sigma$ as $E$, and all the other vertices of the 2nd barycentric subdivision along $\overline{AE}$ as $A,B,C,D,E$.

A key observation here is that while it is possible that $f(A)=f(E)$, there is no other possibility of two points on $\overline{AE}$ being mapped to the same image by $f$: the function $f$ is injective on the half-open edge $[A,E)$. This follows from the definition of a $\Delta$-complex, together with the assumption that $f$ is one of the characteristic maps of the given $\Delta$-complex.

Consider also the two red outlined 2-simplices in your diagram, each of which is a simplex of the 2nd barycentric subdivision of $\Sigma$; let me denote them as $\sigma$, which has $\overline{AB}$ as one of its edges, and $\tau$ which has $\overline{BC}$ as one of its edges. Notice that $\sigma \cap \tau$ is a common $1$-simplex face of each of $\sigma$ and $\tau$.

The thing to observe is that $f$ is injective on $\sigma$, also $f$ is injective on $\tau$, and finally $f(\sigma) \cap f(\tau) = f(\sigma \cap \tau)$. This holds because $f$ is injective on $[A,C)$.

  • $\begingroup$ When we perform the first subdivision we obtain a $\Delta$-complex such that in every simplex the vertices are different. (because by def of $\Delta$-complex the intersection of the inner $\overset{\circ}{\Delta_k }$ of any $k$-simplex $\Delta_k$ with any $(k-1)$-simplex must be trivial) What I not understand why after second subdivision we obtain a simplicial complex? One of the central features of´simplicial complexes is that every simplex is uniquely determined by it's vertices. $\endgroup$
    – user267839
    Commented Feb 11, 2023 at 21:49
  • $\begingroup$ But I not see why after second subdivision we cannot obtain a complex containing two distinct simplices having identical vertices. Could you give a hint how to resolve this problem? $\endgroup$
    – user267839
    Commented Feb 11, 2023 at 21:49
  • 1
    $\begingroup$ Here's a hint: Do a proof by induction on the dimension of the skeleta $X^{(k)}$ of the given $\Delta$ complex $X$. First prove that the second barycentric subdivision of $X^{(0)}$ is a simplicial complex (this is obvious). Then, assuming that the second barycentric subdivision of $X^{(k)}$ is a simplicial complex, prove that the second barycentric subdivision of $X^{(k+1)}$ is a simplicial complex. $\endgroup$
    – Lee Mosher
    Commented Feb 12, 2023 at 0:21
  • $\begingroup$ There is one point I'm not sure about how to resolve it. Pick a $(k+1)$-simplex $\Delta_{k+1}$ and perform two barycentric subdivisions on it and assume that we already know that the $k$ skeleton $X^k$ is already simplicial complex. Then we obtain $(k+1)^2$ new $(k+1)$-simpleces. $\endgroup$
    – user267839
    Commented Mar 7, 2023 at 0:48
  • $\begingroup$ The difference between a $\Delta$- and a simplicial complex is that of the latter there are two additional glueing rules: 1) all vertices of a simplex are different 2) any two simplices share at most one common lower simplex as intersection. Clearly 1) is satisfied because if there is a new $(k+1)$-simplex having two or more identified vertices by construction it's $k$-face which contains these vertices is contained in $k$-face of initial $(k+1)$-simplex $\Delta_{k+1}$, so we are inside $k$-skeleton $X^k$. We conclude by induction that all vertices of $k$-face are different. $\endgroup$
    – user267839
    Commented Mar 7, 2023 at 0:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .