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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.

kηives
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2 Answers2

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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.

David E Speyer
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  • This is great. So this is done in the dual graph, and one essentially orients the links, correct? – kηives Jan 05 '18 at 17:33
  • The computation is done in the dual graph. I'd rather not identify the $n+1$ edges out of a vertex in the dual graph with the $n+1$ vertices of the cell it is dual to. I suppose there is a canonical such bijection, but the data you need is not just in the dual graph. Given two adjacent simplices $\sigma$ and $\tau$, corresponding to two adjacent vertices $v_{\sigma}$ and $v_{\tau}$ of the dual graph, there is a bijection between the links of $v_{\sigma}$ and $v_{\tau}$, corresponding to which $n-1$ dimensional faces have which intersection with $\sigma \cap \tau$, and you need this data. – David E Speyer Jan 05 '18 at 17:43
  • You can surely figure out how to say things in terms of orienting links, but I'd find it simpler to just say that I am travelling around the dual graph, but orienting the original simplices. – David E Speyer Jan 05 '18 at 17:44
  • You mean the $n+1$ edges which are dual to the $n+1$ $n-1$-dimensional faces. That is the correct identification. – kηives Jan 05 '18 at 17:50
  • Could you say how this procedure extends to the lower dimensional simplicies? For instance, I can see immediately how this would satisfy the compatibility for the 5-cells, but what about for the lower dimensional simplicies like the tetrahedra? I don't see how this will enforce compatibility for lower dimensional simplicies. – kηives Jan 06 '18 at 18:45
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    I don't know what compatability you want for the lower dimensional simplices. I can't think of any which is true. – David E Speyer Jan 06 '18 at 19:10
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    @kηives: Such a compatibility in lower-dimensional simplices will basically never exist. In particular, if you ever have a $k$-simplex which is contained in more than two different $(k+1)$-simplices, it is impossible to orient all of them so that every pair gives opposite signs on the $k$-simplex in the boundary (since there are only two possible signs). – Eric Wofsey Jan 06 '18 at 19:12
  • @EricWofsey Thanks both of you. I have to get a better idea of this stuff. I appreciate your help a lot. – kηives Jan 06 '18 at 19:17
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Let $S$ be the set of all the vertices of your simplicial complex. If you choose a total order $<$ on the set $S$, then $<$ is also a total order on any subset of $S$. In particular, the set of vertices of each simplex is a subset of $S$, so $<$ is a total order on the vertices of each simplex.

In other words, the "induced" order for a simplex is just the exact same relation as the ordering on all the vertices, restricted to the vertices of the simplex.

Eric Wofsey
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    Thanks, I understand immediately what you mean. I don't understand how this can be a valid way to define an orientation though. Following this prescription, if I act with the boundary operator on two $p$-simplices that share a $(p-1)$-sub-simplex, that sub-simplex is not guaranteed to appear with opposite signs in each case. Musn't an orientation guarantee that? – kηives Jan 02 '18 at 22:09
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    I think you're confusing two different notions of "orientation". An "orientation" of a simplicial complex is just a choice of orientation of each simplex (or sometimes, an ordering of the set of all vertices), with no compatibility required. That is all you need to define homology. In a different context, one can speak of an "orientation" on a triangulation of a manifold, and that requires the compatibility you mention. Such an orientation need not exist in general. – Eric Wofsey Jan 02 '18 at 22:12
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    That is an extreme bummer... In all the references I have been going though in homology, their examples seem to always have their complexes oriented such that they always get that result to cancel the shared pieces. It has been a major struggle for me to understand how they have known to do that. Thanks for your time. – kηives Jan 02 '18 at 22:16