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(Originally posted at m.se without answers.)

Let $T$ be a set of triangles in an abstract simplicial complex, with orientation of the triangles chosen such that

$$\partial \left( \sum \limits_{t \in T} t \right) = 0$$

where the boundary operator is as usual, and we consider coefficients in $\mathbb{Z}$. Let $S$ be the support of $T$, i.e. the set of nodes in the complex that appear in some $t \in T$.

Does there necessarily exist a $3$-dimensional geometric realization of $T$, in the following sense:

Is there a polyhedron $P$ with vertices $N$ and all triangular faces $F$, as well as a map $\phi : N \to S$ such that $\phi(F) = T$?

(where $\phi(F) = \bigcup \limits_{(a, b, c) \in F} (\phi(a), \phi(b), \phi(c))$)

I mean "polyhedron" here in the classical sense: a 3d solid with finitely many faces, all of which are "flat," such that only adjacent faces intersect and they do so only at a boundary point or a line. The polyhedron can be non-convex (although I suspect that this question is equivalent if we restrict to the convex case).

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  • $\begingroup$ What do you mean by a polyhedron? A geometric realization of the complex $T$ such that two triangles intersect only along a side or a vertex and only if they do intersect in the abstract simplicial complex? Also, what do you mean by "possibly with repeats"? $\endgroup$ – Ivan Izmestiev Jun 6 '17 at 11:16
  • $\begingroup$ @IvanIzmestiev Thanks -- I edited the post with clarifications/more precise language that will hopefully clear up these points. $\endgroup$ – GMB Jun 6 '17 at 14:54
  • $\begingroup$ Any $2D$ homology class can be realized by a map of a compact surface into the space. (This is not true in all dimensions.) Then, if you want a 3D polyhedron, just think of your surface as the boundary of a cube with several holes drilled through the middle. $\endgroup$ – Jim Conant Jun 6 '17 at 16:05
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    $\begingroup$ A special case of your question is: given an abstract triangulation of a surface, can it be realized as a polyhedral surface in $\mathbb{R}^3$? The answer is "no", although counterexamples are complicated (in genus 6 if I remember correctly). $\endgroup$ – Ivan Izmestiev Jun 6 '17 at 16:24
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    $\begingroup$ If I understand your question correctly, then counterexamples are discussed on the top of page 4 of this article. $\endgroup$ – Ivan Izmestiev Jun 16 '17 at 6:23

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