I have found a very nice paper which generalizes much of vector calculus to discrete lattices through the use of simplicial complexes,
https://journals.aps.org/pre/abstract/10.1103/PhysRevE.59.1217
alternative link (should be free to access):
Following the authors, we define a 0-, 1-, 2-, and 3-fields according to $$\phi = \sum_{[i]} \phi_i \,(i)$$ $$\alpha = \sum_{[i,j]} \alpha_{ij} \, (i,j)$$ $$\beta = \sum_{[i,j,k]} \beta_{ijk} \, (i,j,k)$$ $$\gamma = \sum_{[i,j,k,l]} \gamma_{ijkl} \, (i,j,k,l)$$ where $(i_0, \cdots, i_p)$ is a $p$-simplex. We take the usual convention that a simplex changes sign under an odd permutation of its indices, which in the authors' notation is written as $$(\mathcal{P}(i_0), \cdots \mathcal{P}(i_p)) = \mathrm{sgn}(\mathcal{P}) (i_0, \cdots, i_p)$$
If I understand correctly, each field should be invariant under reordering of the indices, and thus we conclude that each of the field "tensor" components should be fully anti-symmetric, "The summations include each simplex only once. Thus the square bracket $[\cdots]$ denotes one representative of an equivalence class related by exchange of indices."
Now this is the part I am confused about, the authors state:
Renumbering the vertex set of $\mathcal{G}$ [(the underlying graph)] is a formal symmetry which cannot have physical consequences. Reversing the vertex ordering changes the sign of $p$-simplices with odd $p$, so it changes the sign of the components for $p$-fields with $p=1$ and 3. Thus by analogy with the parities of continuum fields with respect to coordinate inversion, we classify the basic $p$-field quantities as 0-fields (scalars), 1-fields (polar vectors), 2-fields (axial vectors), and 3-fields (pseudoscalars).
I am not able to figure out precisely what this means since they do not give examples. My understanding is this: Assume we label every vertex with an index $\{1,2,\cdots\}$, then we can take for each representative of the equivalence classes the one in which the indices are in ascending order, i.e. we choose the representative 2-simplex of the equivalence class $[i_0,i_1,i_2]$ to be $(i_{n_0}, i_{n_1}, i_{n_2})$ where $i_{n_0} < i_{n_1} < i_{n_2}$ and $\{n_0,n_1,n_2\}$ is a permutation of $\{0,1,2\}$ (apologies, I realize this notation is rather cumbersome but I hope I made my point).
Now, when the authors say "reversing the vertex ordering", I am not sure what they mean. If the graph is finite, and the list of labels ends, $\{0,1,\cdots, N\}$, then I would assume they mean relabel each vertex $i \rightarrow N-i$, but I cannot figure out how to recover their statement that this changes the sign of $p$-simplices with odd $p$. Every simplex now has an ordering which is descending rather than ascending, but then I find
$$(1) \rightarrow (1) = +(1)$$ $$(1,2) \rightarrow (2,1) = -(1,2)$$ $$(1,2,3) \rightarrow (3,2,1) = -(1,2,3)$$ $$(1,2,3,4) \rightarrow (4,3,2,1) = +(1,2,3,4)$$
which is not the statement the authors make, I am expecting to find that the signs alternate as $+,-,+,-$.
I seem to have thoroughly confused myself, and am doubting that I really understand what they mean by reversing the ordering, how this is akin to spatial inversions in vector calculus, and generally think I am failing to grasp something here. Any help appreciated parsing the quoted statement.
For $n\in \Bbb N$ let $\tau$ be the permutation of $\{1,2,\ldots,n\}$ which reverses its order, that is $\tau(j)=n+1-j$. The sign of the permutation depends on $n$ modulo $4$; it is $+1$ (an even permutation) if $n\equiv0$ or $1\pmod 4$, and it is $-1$ (an odd permutation) if $n\equiv2$ or $3\pmod 4$.
The reason: $\tau$ consists of $\lfloor n/2\rfloor$ cycles of length $2$ (and one fixed point when $n$ is odd), so $\tau$ shares the parity of $\lfloor n/2\rfloor$.