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Given a triangle $(A_1,A_2,A_3)$ in $\mathbb{R}^2$, it is easy to determine whether the order $A_1,A_2,A_3$ is "clockwise" or "counter-clockwise", by imagining a real clock sitting on top of the triangle.

Moreover, if we triangulate a triangle to smaller triangles and travel the vertices of each triangle in a clockwise order, then each internal edge is travelled exactly twice, in two opposite directions, while each boundary edge is travelled exactly once, in a clockwise direction; see below:

enter image description here

Is there a generalization of this fact to $n$-dimensional simplexes? I.e, if $(A_1,\ldots,A_{n+1})$ is a simplex in $\mathbb{R}^n$:

  • Is there a way to determine whether the order $A_1,\ldots,A_{n+1}$ is "clockwise" or "counter-clockwise"?
  • If an n-dimensional simplex is triangulated to smaller $n$-dimensional simplexes, and the vertices of each sub-simplex are enumerated in a clockwise order, is it still true that each internal sub-simplex of dimension $n-1$ is enumerated twice in two opposite directions?
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  • $\begingroup$ I notice you chose the tag "orientation". One answer to your question is that the $n$-dimensional generalization of "clockwise-counterclockwise" is the concept of orientation of a simplex, which you will find discussed here: en.wikipedia.org/wiki/Simplicial_homology#Definition $\endgroup$ – Lee Mosher Mar 24 '17 at 13:04
  • $\begingroup$ @LeeMosher I saw this page, but, it implies that an "orientation" of a simplex is just an arbitrary ordering of its vertices. I do not understand how to select the orientations of two adjacent simplices in a consistent way (maybe I just do not understand the sentence: "choose an ordering of all the vertices and give each simplex the orientation corresponding to the induced ordering of its vertices"). $\endgroup$ – Erel Segal-Halevi Mar 24 '17 at 14:36
  • $\begingroup$ No, an "orientation" of a simplex is an equivalence class of orderings of its vertices, two being equivalent if they differ by an even permutation. In your picture, the equivalence class depicted is $$\{(A_1,A_2,A_3),(A_2,A_3,A_1),(A_3,A_1,A_2)\}$$ $\endgroup$ – Lee Mosher Mar 24 '17 at 14:51
  • $\begingroup$ @LeeMosher OK, but how do I pick an equivalence class to several different simplices simultaneously in a consistent way? E.g, in my picture, how do I verify that all 4 small simplices have the same orientation? $\endgroup$ – Erel Segal-Halevi Mar 25 '17 at 17:15
  • $\begingroup$ This brings to mind differential forms and exterior algebra. I'm kind of rusty on those, though, so maybe someone else can explain it. (I suppose it's possible Lee Mosher's answer has already done that but that I am so rusty I didn't recognize it.) $\endgroup$ – David K Mar 25 '17 at 19:17
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Here's a brief account of how this works.

Let me use $\Delta = [v_0,v_1,...,v_n] \subset \mathbb{R}^m$ as the notation for an oriented $n$-simplex whose vertices are $v_0,v_1,...,v_n$, and whose orientation is the equivalence class of the ordering $(v_0,v_1,...,v_n)$.

The $n$-simplex $\Delta=[v_0,...,v_n]$ has $n+1$ faces each of dimension $n-1$. These faces can be listed and assigned orientations by cyclically permuting $\Delta$ and deleting the last vertex of the list: $$[v_0,...,v_{n-1}], \, [v_1,...,v_n], \, [v_2,...,v_n,v_0], ... , [v_n,v_0,...,v_{n-1}] $$ So in your example the three oriented faces of dimension $1$ are $$[A_1,A_2], [A_2,A_3], [A_3,A_1] $$ I'll refer to the orientation on each of the faces of $\Delta$ as the boundary orientation relative to $\Delta$.

Any $n$-simplex $\Delta$ can be subdivided, in many different ways, into a union of simplices such that the intersection of any two is either empty or a smaller dimensional simplex in the subdivision. In your example the $2$-simplex $[A_1,A_2,A_3]$ is subdivided into four $2$-simplices, nine $1$-simplices, and six $0$-simplices. When $\Delta$ is subdivided in this manner, you can prove that each subsimplex $\Delta'$ of $\Delta$ has a unique orientation that is "induced" from the orientation on $\Delta$. One way to state what this means is to proceed inductively, something like this:

  1. For any sub $n$-simplex $\Delta' \subset \Delta$, if there exist $n-1$ dimensional faces $\Gamma' \subset \Delta'$ and $\Gamma < \Delta$ such that $\Gamma'$ is a sub $n-1$-simplex of $\Gamma$, then the orientation on $\Gamma'$ is the orientation induced from the orientation on $\Gamma$ (notice the inductive structure of this statement: $\Gamma$ and $\Gamma'$ have one dimension lower than $\Delta$ and $\Delta'$).
  2. For any two sub $n$-simplexes $\Delta',\Delta'' \subset \Delta$, if $\Gamma = \Delta' \cap \Delta''$ is an $n-1$ simplex then the boundary orientation on $\Gamma$ relative to $\Delta'$ is the opposite of the boundary orientation on $\Gamma$ relative to $\Delta''$.

In your example, you can see item 2 in action by looking at each interior $1$-simplex $\tau$: the arrows on opposite sides of $\tau$ point in the opposite direction.

If you want to read about these things in detail, I would recommend one of the older algebraic topology books. Some of these concepts are required in any algebraic topology book including the current textbook of Hatcher, but older books generally contain more details regarding things like subdivision of simplices.

Also, these issues of oriented simplices are very closely connected with the theory of oriented manifolds which is covered in most differential topology books.

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  • $\begingroup$ Regarding the inductive process - what is the induction base? $\endgroup$ – Erel Segal-Halevi Mar 25 '17 at 20:19
  • $\begingroup$ A $0$-simplex is just a point and it has no faces. A $1$-simplex $\Delta = [v_0,v_1]$ is an oriented line segment with initial endpoint $v_0$ and terminal endpoint $v_1$. $\endgroup$ – Lee Mosher Mar 25 '17 at 22:04

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