# Contradictory Orientations of Faces in Simplicial Complexes

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?

• Actually, they are supposed to be contradictory. Otherwise you wouldn't get a chain complex. – Zhen Lin Dec 12 '14 at 8:27
• Can you please elaborate? I don't understand why we would want an element of the complex to have contradicting orientations. – user153312 Dec 12 '14 at 10:35
• Have you not seen the construction of simplicial homology? – Zhen Lin Dec 12 '14 at 10:47
• I have, and I am confused by orientation. Also, in Hatcher, edges are oriented from according to ordering of vertices, but following Rotman (using $(-1)^i[v_0,\dots,\hat {v_i},\dots,v_n]$), I would have differently oriented edges. – user153312 Dec 12 '14 at 10:51
• The only purpose in defining the orientation of simplices is to define the boundary of a chain, and in order to have a chain complex, we need $\partial \circ \partial = 0$. For this to happen, we need a lot of cancellation to happen, so it is a good thing that the signs disagree. – Zhen Lin Dec 12 '14 at 10:53

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.
• It's not obvious that a general procedure should exist, even for $2$-simplices. Hence the notion of a nonorientable surface, and more generally a nonorientable manifold. – Gyu Eun Lee Dec 12 '14 at 10:34