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Simplicial complexes can be defined in two different way, i.e. either abstractly as purely combinatorial objects, or embedded in Euclidean space. Let me briefly mention which definitions I use exactly:

Definition 1:

Let $\mathcal{V}$ be a finite set. An "abstract simplicial complex" is a collection of non-empty subsets $\Delta$ of $\mathcal{V}$ such that it contains all the singelton sets and such that for any non-empty $\tau\subset\sigma$ for $\sigma\in\Delta$ it holds that $\tau\in\Delta$

Definition 2:

A Euclidean simplicial complex is a collection $\Delta$ of Euclidean simplices (=the convex hull of a bunch of affinely-independent vectors in $\mathbb{R}^{n}$) such that every face of a simplex is again contained in $\Delta$, such that the intersection of any two simplices is either empty or the face of both simplices and such that $\Delta$ is locally finite.

Obviously, to every Euclidean simplicial complex we can associate an abstract one by defining the abstract simplices to be the sets containing all the vertices of the corresponding Euclidean simplices. The abstract simplicial complex corresponding to a Euclidean one is usually called "vertex scheme". Last but not least, we need a notion of equivalence:

Let $(\mathcal{V}_{1},\Delta_{1})$ and $(\mathcal{V}_{2},\Delta_{2})$ be two abstract simplicial complexes. A pair of maps $(\varphi:\Delta_{1}\to\Delta_{2},\varphi_{0}:\mathcal{V}_{1}\to\mathcal{V}_{2})$ is called "simplicial isomorphism", if it is of the form $\varphi(\{v_{0},\dots,v_{k}\})=\{\varphi_{0}(v_{0}),\dots,\varphi_{0}(v_{k})\}$ and if both $\varphi$ and $\varphi_{0}$ are bijective.

In other words, two simplicial isomorphic abstract simplicial complexes admit the same number and the same gluing pattern of simplices.

Now, it is a well-known fact that every finite abstract simplicial complex admits a "geometric realization", which is a Euclidean simplicial complex whose vertx scheme is simplicial isomorphic to our given complex. This is exactly the point which I do not quite understand. Why does there always exists an isomorphic simplicial complex which has the property that the intersection of two simplices is either a face or empty? It seems that this property in the definition of Euclidean simplicial complexes is much more restrictive, since for an abstract simplicial complex, the intersection of two simplices must not necessarily be again a simplex. For example, consider the following "triangulation" of the 2-torus:

enter image description here

where we glue together the top and bottom as well as the right and left. Now, this is a well-defined abstract simplicial complex, since the non-empty subset of any simplex is again a simplex. However, it is certainly not an Euclidean simplicial complex, since the intersection of the two triangles is the union of two edges and not again a simplex. So, how does this triangulation, viewed as an abstract simplicial complex, has a geometric realization?

Obviously, I just have a thinking error and misunderstand some of the definitions above, but I can't figure out which part I miss exactly...

Thank you all!

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    $\begingroup$ By gluing the top to the bottom and the left to the right in your square picture, you've identified all four corners. The resulting abstract simplicial complex would have just one vertex, say $v$. Each of the edges in your picture and each of the triangles would become just $\{v\}$. So your abstract simplicial complex would be unexpectedly simple, and its geometric realization would be just a single point. $\endgroup$ Mar 1, 2022 at 16:00
  • $\begingroup$ In Definition 2 do you require that all Euclidean simplices lie in a fixed $\mathbb R^n$? This would b very restrictive. $\endgroup$
    – Paul Frost
    Mar 1, 2022 at 16:58
  • $\begingroup$ @AndreasBlass : Indeed, that was not the best example. But the argument also works with the example drawn in this question: math.stackexchange.com/questions/821380/…. This is also clearly an abstract simplicial complex, but not an Euclidean one $\endgroup$
    – B.Hueber
    Mar 1, 2022 at 17:36
  • $\begingroup$ @PaulFrost : Yes, this is the usual definition, as for example stated in Lee's "Introduction to Topological Manifolds" on page 149 $\endgroup$
    – B.Hueber
    Mar 1, 2022 at 17:38
  • $\begingroup$ In Lee's book, you also find the Theorem about the fact that every finite abstract simplicial complex admits a geometric realization (Proposition 5.41) $\endgroup$
    – B.Hueber
    Mar 1, 2022 at 17:40

1 Answer 1

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In the comments, where you rephrased your question, you expressed your concern regarding this statement:

... it could in principle happen that the intersection of two simplices is a union of several simplices.

But that statement is wrong: In an abstract simplicial complex, the intersection of two simplices is always either the empty set or a single simplex.

Here's the proof. Consider two simplices $\sigma_1,\sigma_2 \in \Delta$. Consider their intersection $\tau = \sigma_1 \cap \sigma_2$. Supposing that $\tau = \emptyset$, we are done. Supposing on the other hand that $\tau \ne \emptyset$, from the definition of intersection it follows that $\tau \subset \sigma_1$ (and that $\tau \subset \sigma_2$). From the definition of a simplicial complex, since $\sigma_1 \in \Delta$, and since $\tau \subset \sigma_1$, it follows that $\tau \in \Delta$. Therefore, $\sigma_1 \cap \sigma_2 = \tau \in \Delta$, that is to say, $\sigma_1 \cap \sigma_2$ is a simplex.

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  • $\begingroup$ Thank you very much. But what is if, for example, in some 3d complex I have two tetrahedra which are glued together on two pairs of faces. In this case, the intersection of them two is just the set of all four vertices of each of the tetrahedra and hence it is again a simplex. However, still, the intersection of them two is the union of two faces, when I think about them geometrically... $\endgroup$
    – B.Hueber
    Mar 3, 2022 at 9:01
  • $\begingroup$ Well, that's not a simplicial complex, but there are other types of complexes that we use in topology which that one fits in to. For instance, it is a CW complex (which is explained in most algebraic topology books), and it is also a $\Delta$ complex (which is explained in Hatcher's algebraic topology book). $\endgroup$
    – Lee Mosher
    Mar 3, 2022 at 14:46

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