# Is every $\Delta$-complex realizable as Simplicial complex, ie triagulazable?

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:

• 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. Commented Jan 22, 2023 at 22:19

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

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 $$[A,C)$$.

• 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. Commented Feb 11, 2023 at 21:49
• 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? Commented Feb 11, 2023 at 21:49
• 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. Commented Feb 12, 2023 at 0:21
• 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. Commented Mar 7, 2023 at 0:48
• 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. Commented Mar 7, 2023 at 0:48