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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?

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3 Answers 3

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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.)

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  • $\begingroup$ Thanks! Looking at Theorem 4.1 of the Richter-Gebert/Ziegler paper to which Benjamin linked on his blog, I see that knowing $M$ is essentially equivalent to knowing both $N$ and the oriented matroid $P$ obtained from $M$ by deleting the distinguished element. I can't put my finger on a precise statement yet, but I feel as if the passage from $M$ to $P$ involves forgetting much of the orientation data. As you noted, the passage from $M$ to $L(M)$ also involves forgetting much of the orientation data. So I hold out some hope that $M$ is determined by $L(M)$ and $N$. $\endgroup$
    – Nicholas Proudfoot
    Commented Jun 25, 2014 at 17:33
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$L(M)$ 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.

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  • $\begingroup$ Thanks! I'm very glad to have learned about the Bohne-Dress theorem from your blog post. I was not aware of this theorem or even of the term "single-element lifting", though this is in fact exactly what I am interested in. Let me revise my question accordingly.... $\endgroup$
    – Nicholas Proudfoot
    Commented Jun 25, 2014 at 14:23
  • $\begingroup$ Suppose that I know that $M$ is a single-element lifting of an oriented matroid $N$, which comes with an explicit representation. That is, I am given $N$, a representation of $N$, the poset $L(M)$, and the negation operation, but not the composition operation. Surely the knowledge of $N$ buys me something. What more do I need to recover the composition operation on $L(M)$, and therefore $M$ itself? $\endgroup$
    – Nicholas Proudfoot
    Commented Jun 25, 2014 at 14:34
  • $\begingroup$ I have to put single element liftings back in my brain to give a good answer. I am not an oriented matroid person. I have to think what single element liftings means in monoid terms. In general the poset is just the poset of principal right ideals under reverse inclusion. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 25, 2014 at 16:35
  • $\begingroup$ I think if I remember correctly that N is a contraction of M and a fairly big one. So you should get a fair amount of info, but most likely not everything. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 25, 2014 at 16:39
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I am aware, this is a quite old post, but I still would like to put my answer here.

In my opinion, the topological point of view (i.e. the topological representation theorem due to Falkman and Lawrence) is the most suitable way to view your questions. The covector poset is the face poset of a shellable regular CW-decomposition of the sphere. So it contains all the geometric information about the representation of the oriented matroid by an arrangement of pseudospheres (See Chapter 4 and 5 in the big red book by Björner et al). The corresponding pseudospheres correspond to certain codimension 1 subposets and the coatoms correspond to the tope-cells. So choosing one positive tope you can reconstruct all the sign-vectors (up to reorientation). It is even enough to only know the poset from corank 2 to the top, since this is basically the tope-graph of the oriented matroid, and it is shown in Chapter 4 of the aforementioned book that this contains all the information to reconstruct the whole covector poset.

One more comment: the covector poset does not see any loops or parallel elements, so in any case, it is only possible to reconstruct simple OMs or the simplification of the OM.

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  • $\begingroup$ Thanks! Looking back at this, I'm struggling to remember what my motivation was for asking the question in the first place. But it seems interesting! $\endgroup$
    – Nicholas Proudfoot
    Commented May 8 at 23:20

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