From what I've read, an orientation of a simplex is an equivalence class of total orders on its vertices, where equivalence means up to even permutation. An orientation on an $n$-simplex induces an orientation on its $i^{\text{th}}$ face (the one obtained by omitting the $i^{\text{th}}$ vertex): it is $(-1)^i[v_0,\dots,\hat {v_i},\dots,v_n]$ where the minus sign means the opposite orientation. This is all nice and good, but I don't understand how to make the story work if we're dealing with complexes.
Suppose we have the following complex, where the triangles are included:
There are two ways of orienting each triangle – with the "suborder", or analogously to before with $(-1)^i[v_0,\dots,\hat {v_i},\dots,v_n]$.
If I use the "suborder", I get two contradicting induced orientations on the edge $[v_0,v_2]$ – positive (from $v_0 $ to $v_2$) from the right triangle (since we're omitting the $2^{\text{nd}}$ vertex), and negative from the left triangle.
Using $(-1)^i[v_0,\dots,\hat {v_i},\dots,v_n]$ does not help either as it merely reverses the orientations of both triangles (each time we omit an odd vertex) and I stay with the same contradiction.
How do we orient the elements of a simplicial complex?
Best Answer
Let me use $(v_0,\ldots,v_n)$ to denote the unoriented $n$-simplex given by the vertices $v_0,\ldots,v_n$, and $[v_0,\ldots,v_n]$ the oriented $n$-simplex with the orientation determined by the given ordering.
For this triangulation, there exists a way to orient the $2$-simplices in the complex so that the resulting edge orientations do not contradict each other. Simply take the $2$-simplex spanned by $(v_0,v_1,v_2)$ and orient it counterclockwise; I will denote this as $[v_0,v_2,v_1]$, but any cyclic permutation of the vertices gives the same oriented $2$-simplex. (In general cyclic permutations might not result in the same orientation. It works here because $3$-cycles are even permutations.)
Then the edge spanned by $(v_0,v_2)$ has the orientation $[v_0,v_2]$ (going tail to head; note that a cyclic permutation for this oriented $1$-simplex doesn't give the same orientation, because $2$-cycles are odd permutations). Now orient the $2$-simplex spanned by $(v_0,v_2,v_3)$ so that the induced orientation on $(v_0,v_2)$ is $[v_0,v_2]$. Namely, if we give $(v_0,v_2,v_3)$ the clockwise orientation $[v_0,v_2,v_3]$ then this induces the orientation $[v_0,v_2]$ on $(v_0,v_2)$ as desired.