Suppose $h_{g}: \mathbb{R}^n \to \mathbb{R}^{n-1}$ be a coboundary crossed homomorphism with action $g$ as a cyclic permutation of coordinates on $\mathbb{R}^n$ vectors. So, the acting group is a cyclic group $C_n$ .
The problem is to found vectors $v \in \mathbb{R}^n$ which under every action $g$ mapped to vectors $h_g(v)$ with only nonnegative or nonpositive components/coordinates. Formally, with index notation for vector components:
$$\forall\, g \in C_n \, ,\;\forall\, i \in \overline{1, n-1} : (h_g(v))_i \geq 0 \, \vee (h_g(v))_i \leq 0$$
It means, that every vector have it's own "sign" independently. (Kind of "orientation")
If I read properly about such map, $h_g$ can be computed in the next way: $h_g(v)=g\,p-p$ , where $$p=p(v)=(v_1, v_1+v_2, \dots, v_1+ \dots +v_{n-1})$$
(we also can use $p(g\,v-v)$ becouse $p$ is linear and $g$ commute with $p$ , it's easy to see)
So.. I wrote a program to found solutions "symbolically" : if here only 2 values$(x_1, x_2)$ among all $v_i$-'s we can easily check the sign, becouse $\,x_1-x_2 \geq 0\,$ iff $\,x_1 \geq x_2$. But for more values this approach is bad. Probably, we can use some linear order on $x_i$'s linear combinations . Lexicographic order is quite good, so I used it. And here something strange happen - every (non-"binary") solution I found$(n<10)$ had only "positive" or "negative" coboundaries, for any $C_n$-permutation.
But I found a way to explain, why this happen. It was made possible by a better way of working with this problem. It's simpler than previous one! Spoiler - in lexicographic order only $0$-vector is $0$.
At first, fix some small $n$ and explicitly compute every $h_g$ with $v_i=x_i$ variables :
let $n=6$ \begin{align*} &h_{1}(v)=(x_2 - x_1, x_3 - x_1, x_4 - x_1, x_5 - x_1, x_6 - x_1)\\ &h_{2}(v)=(x_3 - x_1, x_3 + x_4 - x_1 - x_2, x_4 + x_5 - x_1 - x_2, x_5 + x_6 - x_1 - x_2, x_6 - x_2)\\ &h_{3}(v)=(x_4 - x_1, x_4 + x_5 - x_1 - x_2, x_4 + x_5 + x_6 - x_1 - x_2 - x_3, x_5 + x_6 - x_2 - x_3, x_6 - x_3)\\ &h_{4}(v)=(x_5 - x_1, x_5 + x_6 - x_1 - x_2, x_5+x_6 - x_2 - x_3, x_5 + x_6 - x_3 - x_4, x_6 - x_4)\\ &h_{5}(v)=(x_6 - x_1, x_6 - x_2, x_6 - x_3, x_6 - x_4, x_6 - x_5)\\ \end{align*} I use integer notation for $g\,$, $h_0(v)$ is every time $0$-vector, so we can omit it.
Can you see something? - every vector has one common component with some other one. It must be true for other dimentions but I not prove that...
Seems really like a complete graph (in fact intersection graph). It says, that if one common coordinate is not zero, it's vertices($h_g$'s) must have the same sign, and all connected edges also, and so one... The "color" spreads quickly as soon as new node has at least one "uncolored" edge.(generated by $\neq 0$ component)
It figures out that if I want solutions with opposite sign at some permutation(-s) $\sigma \in C_n \,$, the graph must be not connected. To do that - remove some edges, is the same as make them $0$.
With lex. ordering every non-empty linear combination of $x_i$ is not zero. Well, that would explain, why solutions based on it were totally positive/negative.(for most cases)
But if you do so, you add a "global" constraints, so if you remove one edge, you probably, remove some other edge. Clearly, in this way you will reduce the solution set more and more and eventually only trivial solutions will remain - just vectors from the kernel of $h_g$. I even don't know this critical amount of removes. Practice has shown that this number can be $n-3$, but it depends on which components you nullify to split the graph. Some nodes are more “connected” than others.
Finally, I can get to the fun part.
Every time I've tried to found such "2-sign" solutions, - every time $v$ has components with maximum 2 different values!
Why? I don't know how to work here... so I need a help.
My first hypotesis - low dimention. Maybe, in some higher dimension, we can found at least "ternary" vectors with non-connected and "2-colored" intersection graph between coboundaries for $C_n$. But I not believe in that.
Maybe it's becouse this map lives between cochains of order 0 and 2 ?
Or, we need to rewrite this problem with more group theory related objects, for example - represent a sign as a map to $C_2$ ... like here: https://mathoverflow.net/a/417719/547498
Perhaps, the answer is not hard, even easy... Who knows?
Some "multiple-sign" solutions: $(x, y, y, x, y),\,$ $(x, x, y, x, x),\,$ $(x, y, x, x, y, x),\,$ $(x, x, x, y, x, x),\,$ $(x, x, x, y, x, y),\,$ $(x, y, x, y, x, y, y, x, y)$ .
- This problem arose as an applied problem, and if anyone is interested, it relates to music theory.
