A *finite abstract simplicial complex* is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation, e.g. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.


For $n\geq 0$ the topological space $\Delta^n=\{(x\_0,...,x\_n)\in\mathbb{R}^{n+1}\mid x\_i\geq 0, \sum x\_i =1\}$ is called the *standard $n$-simplex*. A topological space homeomorphic to the standard $n$-simplex is called an *$n$-simplex*. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.


A *finite topological simplicial complex* is a pair $(X,F)$ where $X$ is a topological space and $F=(F\_1,...,F\_m)$ is a finite sequence of embeddings $F\_k:\Delta^{i\_k}\to X$ such that


* $X=\cup\_k F\_k(\Delta^{i\_k})$
* $F\_k\neq F\_l$ if $k\neq l$
* for every $1\leq k\leq m$ with $i\_k\geq 1$ and for every face $A$ of the $i\_k$-simplex $F\_k(\Delta^{i\_k})$ there is a $1\leq l\leq m$ with $F\_l(\Delta^{i\_l})=A$
* for every $1\leq k\neq k'\leq m$ the simplex $F\_k(\Delta^{i\_k})\cap F\_{k'}(\Delta^{i\_{k'}})$ is a face of each of them.


I hope, the definitions are correct.


There is the notation of a *geometric realization* of a finite abstract simplicial complex: Let $D=(S,D)$ be a finite abstract simplicial complex. Then choose a total order on $S$, w.l.o.g. $S=\{1,...,M\}$. The colimit of the functor sending an element $\{0,...,n\}$ of the poset $D$ (considered as a category) to $\Delta^n$ is the geometric realization $|D|$ of $D$.


~~If I am not mistaken there are finite topological simplicial complexes which are not the geometric realization of a finite abstract simplicial complex. This is because the choice of the total order determines an orientation of the realization. I think the projective plane for example is not in the image of the realization functor.~~


~~My question is: Is there a reasonable notation of a geometric realization for abstract simplicial complexes which has exactly the topological simplicial complexes as its image or do I have a wrong understanding somehow?~~


I have realized that the original question does not make sense. Please let me ask if this is the right way to understand the situation:


A finite triangulation of a space is the same as a "finite topological simplicial complex". Every finite triangulation is the realization of a finite abstract simplicial complex. The realization of a finite abstract simplicial complex comes with a "direction" of each 1-simplex such that the neighbouring edges are pointing in the **same** directions (they are glued together in this way). The triangulation is orientable if and only if one can permute these "directions" of the 1-simplices such that all the neighbouring edges are pointing in **opposite** directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?



Simplicial complexes vs. geometric realization of abstract simplicial complexes
At least in the realm of topology an abstract simplicial complex is equivalent to a topological (or geometric) simplicial complex, and neither of these two notions involves anything about orienting the simplices or ordering the vertices. If one has a simplicial complex of either type, one can choose a partial ordering of the vertices that restricts to a linear ordering of the vertices of each simplex, and this gives the notion of an ordered simplicial complex. For a simplicial complex that is a manifold (or more generally a pseudo-manifold) one can define an orientation to consist of a choice of orientations of the top-dimensional simplices such that the two induced orientations on each codimension one face are opposite. For a general simplicial complex the only definition of an orientation that I have seen is the trivial one where an orientation is chosen for each simplex with no compatibility assumptions on these orientations.


