In the Kochen-Specker theorem, where does the assumption $v(-n)=v(n)$ come from? I'm trying to get a better understanding of the Kochen-Specker theorem, reading the presentation in arXiv:1708.01380.
In section (2) of the above paper, the authors introduce the two KS assumptions as:

*

*Let $M\equiv \{A_i\}_i$ be a set of observables. All these observables can be simultaneously assigned a real number. In other words, there is a mapping $v$ such that $v(A_i)\in\mathbb R$ for all $i$.

*The mapping $v$ above satisfies, for all commuting observables $A,B$ we have $v(aA+bB)=av(A)+bv(B)$ for all $a,b\in\mathbb C$, and $v(AB)=v(A)v(B)$. Moreover, there is some observable $X$ such that $v(X)\neq0$ (this is the improved form of the axioms, as referred to in the paper).

I'll refer to these assumptions with $\textbf{KS}1$ and $\textbf{KS}2$, respectively.
The authors then proceed to prove a few consequences of $\textbf{KS}2$. In particular, that $v(I)=1$, $v(P)\in\{0,1\}$ for all projectors $P$. They also observe that $\sum_i P_i=I$ implies $\sum_i v(P_i)=1$ (Eq. (2.4) in the paper).
Now this is where I don't quite follow the presentation. The authors seem to claim (around Eq. (2.5) in the paper) that the $\textbf{KS}2$ condition can be reduced to the following constraints for $v$:
$$\sum_i v(n_i)=1, \qquad v(n_i)\in\{0,1\},\qquad v(-n)=v(n),\tag{2.5}$$
for any orthonormal basis of vectors $n_i$.
Where does the third condition, $v(-n)=v(n)$ come from? I can see how the first two conditions follow from $\textbf{KS}2$, but not the third one.
 A: I figured out the answer while finishing to write the question. Figured I'd post it anyway for future reference.
The conditions in Eq. (2.5) don't actually follow from $\textbf{KS}2$ at all. In Eq. (2.5), we are instead rewriting the statement $v(P)\in\{0,1\}$ when $P$ is a projection, using a slight abuse of notation.
This is pointed out in the sentence between (2.4) and (2.5) in the paper, where the authors write

(...) It is customary to identify the projectors $P_i=|n_i\rangle\!\langle n_i|$ with the corresponding unit vectors $n_i$ (defined up to a sign), with the $n_i$ forming a basis for the Hilbert space, and in $d$ dimensions write (...)

In other words, in (2.5) we are actually referring to some other function, call it $\tilde v$, that is defined on unit vectors $n\in S^{d-1}$, rather than observables, like $v$.
We define such function so that $v(P_n)=\tilde v(n)$ for any $P_n\equiv|n\rangle\!\langle n|$ and $n\in S^{d-1}$.
Then, the condition $v(P_n)\in\{0,1\}$ becomes $\tilde v(n)\in\{0,1\}$, the condition $\sum_i v(P_{n_i})=1$ becomes $\sum_i \tilde v(n_i)=1$, and, finally, the fact that $P_{-n}=P_n$ implies that $\tilde v(-n)=\tilde v(n)$.
In other words, the latter condition is consequence of defining a function on vectors, rather than on the operators projecting onto such vectors.
