Disconnected coverings for oriented matroids via simultaneous mutations
Let \({\mathcal U}_{n,r}\) be a uniform oriented matroid of rank \(r\) on the set \(E=\{1,2,\ldots,n\}\). Hence, as (unsigned) sets, the circuits of \({\mathcal U}_{n,r}\) are exactly the subsets of \(E\) with \(r+1\) elements. \({\mathcal U}_{n,r}\) is defined given the set of signed circuits, \({\mathcal C}\), where a \(C\subset E\) is signed by choosing \(A\) and \(B\) with \(A\cup B=C\) and \(A\cap B=\emptyset\) (the parts of \(C\) signed differently). If the signatures in \({\mathcal C}_1 \subset {\mathcal C}\) already determine \({\mathcal U}_{n,r}\), then every \(r\)-set (or basis) must be contained in an element of \({\mathcal C}_1\) (seen as a set). On the other hand, if this condition is true and, by joining two circuits \(A\) and \(B\) of \({\mathcal C}_1\) by an edge if \(A\cap B\) has \(r\) elements, one obtains a connected graph, then \({\mathcal C}_1\) determines \({\mathcal U}_{n,r}\) completely [\textit{D. Forge} and \textit{J. L. Ramírez Alfonsín}, Discrete Math. 187, 109-121 (1998; Zbl 0958.05018)]. In this note, an example is given where every basis is contained in an element of \({\mathcal C}_1\) but the graph above is disconnected, and yet \({\mathcal C}_1\) determines completely the orientation in \({\mathcal U}_{n,r}\).
scientific article; zbMATH DE number 1839042
scientific article