The lattice of covectors of an oriented matroid Let $M$ be an oriented matroid on the ground set $E$, and let $L(M)$ be its ranked poset of covectors.  By definition, $L(M)$ is a sub-poset of the poset $\{0, \pm 1\}^E$, ordered by putting $0 < \pm 1$, with the rank being the number of nonzero coordinates.  However, I want to consider $L(M)$ as an abstract ranked poset, rather than as a subposet of $\{0, \pm 1\}^E$.
Question 1:  Is the oriented matroid $M$ determined, up to isomorphism, by $L(M)$?
Question 2:  Is there a nice characterization of which ranked posets arise in this way?
 A: This is an answer to question 1 (not the modified version in the comments to Ben's answer).  
If you consider only L(M), you have lost all the information the orientation provides (beyond the bare fact that an orientation exists).  Since there can be more than one, non-isomorphic oriented matroid on the same matroid, it can't be possible to reconstruct the oriented matroid from L(M).  (I don't actually have an explicit example of multiple oriented structures on one matroid in mind, but counts of the orientable and oriented matroids imply that there are orientable matroids with more than one orientation starting in rank 3 with 6 points.)
A: $L(M)$ is a monoid with covector composition and has a unary operation, its negation. Two oriented matroids are isomorphic iff their associated unary monoids are isomorphic, see my blogpost 
http://bensteinberg.wordpress.com/2013/08/13/equivalence-or-oriented-matroids-from-a-semigroup-viewpoint/
I would be surprised if you can recover the negation and product from just the order. But I don't have an example. 
