Is there a uniform definition of oriented simplex that gives two orientations in all dimensions?

Question

Define a simplex to be a finite subset of vertices: a subset with cardinality $n$ defines an $(n - 1)$-simplex. An oriented simplex is typically defined to be a simplex and an equivalence class of orderings on that simplex that agree up to an even permutation^[0][1].

However, this leaves $0$-simplices (and $(-1)$-simplices?) to only admit a single orientation. As Munkres notes, this gives a “natural” basis of $0$-chains, which is not the case for higher dimensions. Instead, some authors^[1][2][3] choose to single out oriented $0$-simplices as a special case by pairing them with a positive or negative sign.

Is there a definition of orientation that gives exactly two orientations for simplices of dimension $\geq -1$ which can be used for the same purposes as the standard definitions of simplex orientation, but which does not treat lower-dimensional simplices as a special case (as in the sources already listed)? Additionally, are there reasons to prefer one definition over another based on the number of orientations they assign to lower-dimensional simplices?

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My initial instinct would be to use the following definition:

An orientation of an $(n - 1)$-simplex $\sigma$ is an element of $(\{+, -\} \times \mathrm{Bij}([n], \sigma))/{\sim}$, where:

$$ (\varepsilon_0, \pi_0) \sim (\varepsilon_1, \pi_1) \iff \mathrm{sgn}(\pi_1^{-1} \circ \pi_0) = \varepsilon_0 \cdot \varepsilon_1 $$

Any further constructions can be restated with a sign prepended. For example, for any $n \in \mathbb{N}$: $$ \newcommand{\bd}{\partial} \require{cancel} \bd_{n - 1}(\varepsilon[v_0, \dots, v_{n - 1}]) = \sum_{i \in [n]} (-1)^i \cdot \varepsilon[v_0, \dots, \cancel{v_i}, \dots, v_{n - 1}] $$ Of course, a choice of canonical sign induces a canonical basis of $0$- and $(-1)$-chains.

Is this a suitable definition, and if so, does anyone know of a reference where a definition like this is used? Moreover, does it ultimately matter whether $0$- or $(-1)$-simplices are given two orientations?

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0. J. R. Munkres, Elements Of Algebraic Topology
1. Seifert and Threlfall: A Textbook of Topology
2. L. S. Pontryagin, Foundations Of Combinatorial Topology
3. W. Rudin, Principles of Mathematical Analysis

Answer 1

I do not know a reference in the literature, but your definition works. I suggest to modify your approach technically.

The set $\operatorname{Bij} ([n],\sigma)$ is nothing else than the set $o(\sigma)$ of ordered simplices over $\sigma$, i.e. the set of sequences of length $n+1$ which traverse all vertices of $\sigma$. Note that for $n = -1$ we have $0$ vertices, i.e. $\sigma = \emptyset$. In this case we have only the trivial (or empty) vertex-sequence of length $0$. If $\sigma = \{v\}$ is a $0$-simplex we also have only one vertex-sequence $(v)$ which has length $1$ and is non-trivial. For $n > 0$ we have $n! > 1$ such vertex-sequences.

The symmetric group $\mathfrak S_{n+1} = \operatorname{Bij} ([n],[n])$ operates in the obvious way on $o(\sigma)$ by permuting sequences. Note that $\mathfrak S_0$ and $\mathfrak S_1$ are the trivial groups. Letting denote $\mathfrak E_{n+1}$ the subgroup of even permutations, the orbits of the action of $\mathfrak E_{n+1}$ on $o(\sigma)$ are the "classical" oriented simplices associated to $\sigma$. For $n > 0$ we have exactly two of them, for $n = 0,-1$ only one.

Define $$S(\sigma) = \{+1,-1\} \times o(\sigma) .$$ Its elements can be written as sequences of length $n+2$ which have an entry $\epsilon = \pm 1$ at position $1$ followed by a vertex-sequence $\Sigma \in o(\sigma)$.

The symmetric group $\mathfrak S_{n+1}$ operates on $S(\sigma)$ by $$\pi \cdot (\epsilon,\Sigma) = (\operatorname{sgn} \pi \cdot \epsilon,\pi \cdot \Sigma ).$$ The orbits of this action are your "new" equivalence classes (which you may understand as oriented simplices over $\sigma$), and we have exactly two of them for each $\sigma$. In dimensions $0$ and $-1$ we can associate to $\sigma$ a canonical oriented simplex ($\epsilon = +1$ followed by the unique vertex-sequence over $\sigma$). For $n > 0$ this is impossible (we need to make a choice in this case).

Moreover, does it ultimately matter whether $0$- or $−1$-simplices are given two orientations?

The concept of oriented simplices is introduced for the purposes of simplicial homology. So what is the impact on simplicial homology groups if we work with the new concept of orientation?

Let $K$ be a simplicial complex. By $O_n(K)$ we denote the set of all "classically" oriented simplices over $n$-simplices in $K$ and by $O'_n(K)$ we denote the set of all oriented simplices in the new sense over $n$-simplices in $K$. Taking these sets as bases for free abelian groups $D_n(K)$ and $D'_n(K)$, we get chain complexes $D_*(K)$ and $D'_*(K)$ with the usual boundary operators. For $D'_*(K)$ you have defined this operator in your question. Next form the quotient of $D_n(K), D'_n(K)$ by the subgroups $E_n(K), E'_n(K)$ generated by all sums $s + s'$, where $(s,s')$ are pairs of oppositely oriented simplices (for $n = 0,-1$ we have $E_n(K) = 0$ because there do not exist oppositely oriented simplices). This produces chain complexes $C_*(K)$ and $C'_*(K)$ which are known as oriented simplicial chain complexes of $K$.

Now define $$\phi : D_n(K) \to D'_n(K), \phi([\Sigma]) = [+1,\Sigma] .$$ For $n > 0$ it is easily verified that this establishes a bijection between the generators, thus $\phi$ is an isomorphism for $n > 0$. It is also easy to see that the maps $\phi$ form chain map. If $s' = [\Sigma']$ has the opposite orientation as $s = [\Sigma]$ (which is possible only for $n > 0$), we have $\Sigma' = \pi \cdot \Sigma$ with some odd permutation $\pi$. We get $\phi(s') = \phi([\Sigma']) = [+1,\Sigma'] = [-1,\Sigma]$ which has the opposite orientation as $\phi(s) = [+1,\Sigma]$. Thus $\phi(E_n(K)) \subset E'_n(K)$ and $\phi(E_n(K)) = E'_n(K)$ for $n > 0$. We conclude that the maps $\phi$ induce maps $$\psi : C_n(K) \to C'_n(K)$$ which clearly constitute a chain map.

$\psi : C_n(K) \to C'_n(K)$ is by construction an isomorphism for $n > 0$.

In dimensions $0,-1$ define $$\chi : D'_n(K) \to D_n(K), \chi([\epsilon,\Sigma]) = \epsilon[\Sigma] .$$ This is well-defined because for $n = 0,-1$ each oriented simplex $[\epsilon,\Sigma] \in O'_n(K)$ has a unique representative $(\epsilon,\Sigma)$. We see that $\chi(E'_n(K)) = 0 = E_n(K)$. Since $\chi$ is clearly onto, it induces an isomorphism $\chi^* : C'_n(K) \to C_n(K)$ which is an inverse for $\psi$.

This shows that $\psi : C_*(K) \to C'_*(K)$ is an isomorphism of chain complexes. In other words, old and new concepts of orientation yield the same simplicial homology groups.