Equivalence classes and orientation

Question

From Elements of Algebraic Topology by Munkres,

Let $\sigma$ be a simplex. Define two orderings of its vertex set to be equivalent if they differ from one another by an even permutation. If $\dim \sigma >0$, the orderings of the vertices of $\sigma $ then fall into two equivalence classes.

I do not understand the highlighted part. There could be some that repeat but how is it 2?

Answer 1

Take two possible orderings $A, B$. What the statement means is that either: - There exists an even permutation $\sigma$ such that $A=\sigma B$. - There exists an odd permutation $\sigma$ such that $A=\sigma B$.

In the first case, $A$ and $B$ are in the first equivalence class (i.e. they are equivalent). In the second case, they are in different equivalence classes (they are not equivalent).

So we can see that there are not less and at least two equivalence classes. Let $C$ be a third ordering. Then: - There exists an even permutation $\omega$ such that $A=\omega C$. - There exists an odd permutation $\omega$ such that $A=\omega C$.

This means that, respectively, $C$ is in the equivalence class of $A$ or in another equivalence class. In the second case, and if $B$ is a representative of the second equivalence class as in the second case above, then either: $\omega C = \sigma B\Rightarrow C = \omega^{-1}\sigma B$.

$\omega^{-1}\sigma$ is even, so $C$ and $B$ are in the same equivalence class.

In conclusion, there are two equivalence classes which partition your set of ordering into 'equivalent bits' via your equivalence relation.

Answer 2

If you fix an ordering $x$, then any other ordering $y$ corresponds to a unique permutation $\tau_y\in \mathfrak{S}_n$, such that $\tau_y\cdot x = y$ (where permutations act in the obvious way on orderings). Now the equivalence relation is basically: $y\sim x$ iff $\tau_y\in \mathfrak{A}_n$ (the alternating group).

You can then see that half of the orderings will be in the same class as $x$, and the other half will be in their own class (corresponding to odd permutations).

More generally, when a group $G$ (here $\mathfrak{S}_n$) acts on a set $X$ (here the orderings of your simplex), for any subgroup $H$ you can define an equivalence relation by: $x\sim y$ iff $x$ and $y$ are in the same $H$-orbit. When the action is simple and transitive (which is the case in our example), then the set of equivalence classes is in bijection with $G/H$. Here it is easy to see that $\mathfrak{S}_n/ \mathfrak{A}_n$ is a set of cardinality 2.