I have a simplicial complex, built out of hyper-tetrahedra (5-cells) with the topology of $S_{4}$ and I would like to assign an ordering to it's vertices (some couple thousand), so that I can apply a boundary operator and co-boundary operator on it.
I have been trying to understand how to do this with a cubic lattice, but moving to a triangulated surface has thrown me, and I can't quite grasp how an ordering for the simplicies falls out from an arbitrary ordering of all the vertices.
Wikipedia says something like:
One standard way to do this is to choose an ordering of all the vertices and give each simplex the orientation corresponding to the induced ordering of its vertices.
I guess I'm not getting the "induced" part. Once I have an induced ordering for each 5-cell, I can start to apply the boundary operator and I will know the respective signs for each of the faces, but how does one determine the canonical labeling for each simplex in the complex?
Thanks
EDIT : After the first answer it was revealed to me that there are two working uses of the word "orientation" being used in homology texts. One has to do with simply labeling the complex, the other has to do with manifold orientability (which is what I was interested in).
My question should be re-phrased. How does one orient a simplicial complex so that when a couple of simplicies are acted on by the boundary operator, or co-boundary operator, the necessary sub- or super-simplices that show up appear with the correct compatible signs between the simplicies.
Thanks, and let me know if I can clarify further.
Best Answer
As Eric Wofsey says, there are two notions here: An orientation of an $n$-dimensional simplicial complex is a choice of orientation for each $n$-dimensional simplex. The easiest way to find one is to take a global ordering of all the vertices. If your simplicial complex is a manifold then your ordering will give an orientation of the manifold if and only if, whenever we have two simplices $(u,w_1, w_2, \dots, w_n)$ and $(v,w_1, w_2, \dots, w_n)$ which overlap on an $n-1$ dimensional face, then $(u,w_1, w_2, \dots, w_n)$ and $(v,w_1, w_2, \dots, w_n)$ are given opposite signs. Some manifolds are orientable and some are not. For example, I can trangulate the Mobius strip as the simplicial complex $(v_1, v_2, v_3)$, $(v_2, v_3, v_4)$, $(v_3, v_4, v_5)$, $(v_4, v_5, v_1)$, $(v_5, v_1, v_2)$ and we cannot choose the orientations of these simplices compatibly.
What I want to add to Eric's answer is that, if you have a specific simplicial complex, it is straightforward to work out if we can choose compatible orientations. Let $\Gamma$ be the graph whose vertices are the $n$-dimensional simplices of your complex and where there is an edge between vertices corresponding to adjacent vertices. Find a spanning tree $T$ of this graph. (Or a forest if it is disconnected.) Choose the orientation of one simplex arbitrarily. (If $\Gamma$ is disconnected, choose one vertex in each tree of the spanning forest.) Travel along the edges of $T$; there is a unique way to choose the orientation at each new vertex of $T$ so that the orientations are compatible along the edge of $T$ we have just traveled.
After you have oriented all the simplices to be compatible with the edges in $T$, check if they are also compatible with the other edges of $\Gamma$. If so, you have built an orientation of the manifold, if not the manifold is not orientable. If, as you say, your simplicial complex is homeomorphic to $S^4$, then you will succeed, because $S^4$ is orientable.