), it may or may not be the case that the N*1 vertices of G define a permutation of the integers 0,…,N. When the N*1 vertices (x_{i},y_{i}) of G define a permutation of the integers 0,…,N, G will be termed “canonical”, and Wagner[1] has extended Hart’s results on N-free dim 2 posets so as to delimit precisely the set of possible canonical ordered DAGs (oDAGs). (These include ordered trees with sets of vertices such as {(0,0),(1,3),(2,1),(3,2)}, ordered forests with sets of vertices such as {(0.2),(1.3),(2,0)(3,1)}, and oDAGS which are neither trees and forests, such as those with sets of vertices like {(0,0),(1,2),(2,1),(3,3)}.

In addition to Wagner’s demonstration that the “orderability” of a canonical oDAG arises from its “N-freeness”, Jamison has investigated the conditions under which a canonical oDAG will possess a novel kind of symmetry when plotted on a torus instead of in a plane. To see intutively the nature of this symmetry, consider the canonical oDAG G with vertices {(0,0),(1,4),(2,1),(3,3),(4,2)}. If these vertices are plotted in the plane in the usual way,, one can cut the plane along lines x = -1/2, x = 9/2, y = -1/2, and y = 9/2, and after doing so, form a torus on which one line corresponds to the two lines x = -1/2 and x = 9/2, and one line corresponds to the two lines y = -1/2 and y = 9/2. Then, one can find a way to recut the torus back into the plane so that the original vertices of G now appear in the plane as the symmetrical set of vertices {(0,1),(1,0),(2,2),(3,4),(4,3)} (or {(-2,-1),(-1,-2),(0,0),(1,2),(2,1)} after translation of origin.)

For Wagner’s and Jamison’s purposes, it is immaterial that in any canonical oDAG G with N+1 vertices, any vertex (x_{i},y_{i}) of G can also be uniquely identified with an integer-valued point (p_{i},t_{i},d_{i},s_{i}) in E4 lying in a hyperplane that satisfies the equation p_{i} + t_{i} + d_{i} + s_{i} = N, where: i) p_{i} is the number of vertices (x_{j},y_{j}) in G such that x_{j} < x_{i} and y_{j}> y_{i}; ii) t_{i} is the number of vertices (x_{j},y_{j}) in G such that x_{j} > x_{i} and y_{j}< y_{i}; iii) d_{i} is the number of vertices (x_{j},y_{j}) in G such that x_{j} < x_{i} and y_{j}< y_{i}; iv) s_{i} is the number of vertices (x_{j},y_{j}) in G such that x_{j} > x_{i} and y_{j}> y_{j}; v) x_{i} = p_{i}

*d<sub>i</sub>; vi) y<sub>i</sub> = t<sub>i</sub>*d_{i}.

But inasmuch as the equation p_{i} + t_{i} + d_{i} + s_{i}

= N implies that any oDAG will naturally generate a variety of *hyperplane arrangements* in the sense of Stanley , this equation is actually quite useful in the study of empirical phenomena whose formal characterization requires the postulation of hierarchically-organized structures that contain both +opaque +and *transparent* substructures, e.g. ordered trees that not only contain *opaque* subtrees from which one cannot remove any proper subtree, but also *transparent* subtrees from which one can remove a proper subtree at will. (As first observed by Chomsky , examples of such opacity abound in natural languages, as evidenced, for example, by the fact that speakers of English can say “*Whose picture did you see?*” but not “*Whose did you see picture?*“, even though they can and often do say “*Whom did you see a picture of?*” instead of “*Of whom did you see a picture?*“)

To see why the equation p + t + d + s = N implicitly generates hyperplane arrangements of potential interest to students of opacity in hierarchical structures, consider first the trivial hyperplane arrangement that arises because any non-trivial subtree of an ordered tree G is itself an ordered tree. For example, consider the values of p, t, d, and s for the vertices of the “toroidally symmetric” ordered tree T mentioned above:

(x,y) p t d s

(0,0) 0 0 0 4

(1,4) 0 3 1 0

(2,1) 1 0 1 2

(3,3) 1 1 2 0

(4,2) 2 0 2 0

Just as all the vertices of T lie in a hyperplane normal to and passing through the point (N/4, N/4,N/4, N/4 ) = (1,1,1,1), so also do the vertices {(2,1), (3,3),(4,2)} of the single subtree T’ of T lie in a second hyperplane also normal to but passing through the point (1*2/4, 0*2/4, 1*2/4, 2*2/4) (because the vertices of T’ can be recoordinatized as {(0,0,0,2), (0,1,1,0,(1,0,1,0)} relative to an origin at (1,0,1,0) ) So T and T’ together define an hyperplane arrangement consisting of two hyperplanes that are parallel in the sense of being both normal to the same vector .

More importantly, T and T’ are each bounded in E4 by an additional hyperplane arrangement that may readily be seen by considering what happens to the equation p + t + d + s = N when p, t, d, and s each assume the minimum possible value of 0: (p-N/3)*(t-N/3)*(d-N/3) = 0, (p-N/3)*(t-N/3)*(s-N/3) = 0, (p-N/3)*(d-N/3)*(s-N/3) = 0, and (t-N/3)*(d-N/3)*(s-N/3) = 0. These equations imply that: i) T is itself bounded by a tetrahedron in E4; ii) we may consider any proper non-trivial subtree T’ of T to be bounded by a tetrahedron, once we have recoordinatized the vertices of T’ relative to a new origin at the root of T’. And any such tetrahedron of course defines a hyperplane arrangement consisting of four planes in E4.</p>