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From this it is immediate that L \ e is the circuitspace of G \ e, and, consequently, its dual Ll/e is the cocircuitspace of G \ e. Similarily, the circuitspace of Gle is easily seen to be Lle and hence the cocircuitspace of G/e is Ll\e. Perhaps the duality between the two operations deletion and contraction is best understood in the case of planar graphs. Assume G is CHAPTER 5. ORIENTED MATROIDS 62 planar and G1 is its dual. Thus L and Ll are the circuitspaces of G and G1, resp. Assume that e E E is contained in the boundary of two regions Rl and R2 in some embedding of G.

Then there is a bounded face whose boundary obviously is a circuit X C E of G. Hence deleting an edge e E E leaves the graph connected and decreases both f and m by 1. Thus the formula follows by induction. The most important thing about planar graphs in our context is the concept of dual graphs. This is as follows. Suppose G = (V, E) is a plane graph. Then we may define a new graph by considering each country as a vertex and by joining two vertices by an edge, provided the corresponding countries are neighbouring.

E. whose negative (positive) part equals the positive (negative) part of X. Clearly this notion also applies to sets of sign vectors, thus for S C_ 2±E, -S denotes the set of opposite sign vectors. e. x- fl A = 0), we say that X is nonnegative on A, denoted by XA > 0. XA < 0 and XA = 0 are defined similarily. We will usually write X _> 0 and X < 0 instead of XE > 0 and XE < 0, resp. 1. SIGN VECTORS 59 or b) 3YES',eEsuppYandY>0, but not both. 2 Let a : 1KE -, 2±E denote the "forgetting magnitudesmap", which associates to every x E 1KE the corresponding sign vector X = a(x), and let S = a(L) and S'= a(Ll) for some pair of orthogonal subspaces.