[Oriented matroids](http://www2.lirmm.fr/~sol/Rapports/References/orientedMatroids.pdf) are abstractions of hyperplane arrangements, or equivalently vector configurations. Let me recall the definition in terms of covectors.


Let $R=\lbrace 0,+,-\rbrace$ with the monoid structure given by the table
$$\begin{array}{c|ccc}\circ& 0 &+ &-\\ \hline 0 & 0 &+&-\\ + & + &+ &+\\ - & -&-&-\end{array}$$
$R$ is the oriented matroid corresponding to the unique central arrangement in $\mathbb R$.
More generally, $R^n$ corresponds to the arrangement of coordinate hyperplanes $x\_i=0$ in $\mathbb R^n$. Elements of $R^n$ are called *covectors*.


We put $0=(0,\ldots,0)$ and $-(x\_1,\ldots, x\_n) = (-x\_1,\ldots,-x\_n)$ where negation is the automorphism of $R$ fixing $0$ and switching $+,-$. We view the structure $(R^n,\circ,x\mapsto -x)$ as a unary monoid (i.e., monoid with a distinguished unary operation).


If $x,y\in R^E$, then the *separation set* is $$S(x,y)=\lbrace e\in E\mid x\_e=-y\_e\neq 0\rbrace.$$


An *oriented matroid* with (finite) ground set *E* is a collection $\mathcal L\subseteq R^E$ of covectors such that


1. $0\in \mathcal L$
2. $x\in \mathcal L\iff -x\in \mathcal L$
3. $x,y\in \mathcal L\implies x\circ y\in \mathcal L$
4. If $x,y\in \mathcal L$ and $e\in S(x,y)$, then there exists $z\in \mathcal L$ such that $z\_e=0$ and $z\_f=(x\circ y)\_f=(y\circ x)\_f$ for $f\notin S(x,y)$.


Axioms 1-3 simply state that $\mathcal L$ is a unary submonoid of $R^E$. Axiom 4 is a statement about the geometry of hyperplane arrangements.


The motivating example is if $\mathcal A=\lbrace H\_1,\ldots, H\_n\rbrace$ is a hyperplane arrangement in $\mathbb R^d$ given by forms $f\_i(x)=0$ for $i=1,\ldots,n$ then the set $\mathcal L$ of covectors encoding the signs of the entries of $(f\_1(x),\ldots, f\_n(x))$ with $x\in \mathbb R^d$ form an oriented matroid. The covectors are in bijection with the faces of the zonotope $Z=[-f\_1,f\_1]+\cdots+[-f\_n,f\_n]$ whose normal fan is $\mathcal A$ and the face poset of $Z$ is isomorphic to $\mathcal L$ with the ordering $x\leq y$ if $y\circ x=x$.


In particular, $R$ is the oriented matroid corresponding to the hyperplane $x=0$ in $\mathbb R$. Thus it makes sense to think of a homomorphism of unary monoids $\varphi\colon \mathcal L\to R$ as a *functional* on $\mathcal L$. 



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> > **Question 1.** If $\mathcal L\subseteq R^E$ is an oriented matroid, is it true that every non-zero functional is of the form $x\mapsto x\_e$ for some $e\in E$?
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If the answer to Question 1 is no, then I am interested in the following question.



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> > **Question 2.** Let $M$ be a unary monoid. We can define $\widehat M$ to be the set of non-zero homomorphisms of unary monoids $\varphi\colon M\to R$. We can then define a *Gelfand transformation* $\gamma\colon M\to R^{\widehat M}$ by $\gamma(m)\_e = e(m)$. Is it true that $M$ is isomorphic as a unary monoid to the set of covectors of an oriented matroid if and only if $\gamma(M)$ is the set of covectors of an oriented matroid on with ground set $\widehat M$? 
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If the answer to Question 1 is positive, then the answer to Question 2 is positive. Note that if $M$ is the set of covectors of an oriented matroid, then $\gamma$ is injective. I think it is not too difficult to describe unary monoids for which $\gamma$ is injective. (If we forget the unary operation, then I know which monoids have enough homomorphisms to *R* to separate points.)



Functionals on oriented matroids
I think the answer to your first question is "yes". 


Oriented matroids can be realized topologically, and I am going to use that language. (This means is that I can pretend the oriented matroid is a hyperplane arrangement, provided I am a little careful.)


Let $\phi$ be a functional. Let the rank of the oriented matroid be $n$. I will show that if $\phi$ is zero on any $n$-dimensional cell, then it is zero everywhere, and otherwise there is some $e \in E$ such that $\phi=\phi\_e$. The proof is by induction on $n$. 


I claim first that there is an $(n-1)$-dimensional face on which $\phi$ is zero. The proof is by continuity: we can find a path starting in an n-dimensional cell on which $\phi$ is positive, and ending in a cell where $\phi$ is negative, going through only cells of dimension $n$ and $n-1$. If $X$ is an $n$-dimensional cell and $Y$ is an adjacent $n-1$-dimensional cell, then they cannot have opposite signs, since $Y\circ X=X$, but $- \circ + = -$. The same kind of argument shows that we can't have $\phi(X)=0$ but 
$\phi(Y)\ne 0$. Thus there is some $n-1$-dimensional cell, say $Y$, on which $\phi=0$.


Now consider some 1-dimensional cell, say $L$, incident with $Y$. We can consider the oriented matroid on the link of $L$ (this is generally called the "contraction"). $\phi$ is well-defined here. Induction now shows that $\phi$ is either zero on every cell incident to $L$, or else there is some $e$ such that $\phi/L=\phi\_e/L$. 


In the former case, we can work our way around the whole oriented matroid,establishing that $\phi=0$. 


In the latter case, we will find that there is a consistent choice of $e$, such that for any cell adjacent to the $n-1$-cells of $e$, we have that $\phi$ and $\phi\_e$ agree. If $\phi$ were zero anywhere else, then by induction it would be zero on some other $(n-1)$-dimensional cell, and thus also on all the cells of that hyperplane, but these two hyperplanes would meet, and that then would force $\phi$ to be zero everywhere (again by induction). 


