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Class Note for MATH 796 at KU

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This 4 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 19 views.

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Date Created: 02/06/15
Wednesday 35 Regions of Hyperplane Arrangements Let A C Rquot be a real hyperplane arrangement The regions of A are the connected components of Rquot A R UHeA H Each component is the interior of a bounded or unbounded polyhedron in particular it is homeomorphic to R We write 7 A number of regions of A We d also like to count the number of bounded regions However we must be careful because if A is not essential then every region is unbounded Accordingly call a region relatively bounded ifthe corresponding region in essA is bounded and de ne MA number of relatively bounded regions of A Note that bA 0 if and only if essA is central Example 1 Let A1 and A2 be the 2dimensional arrangements shown on the left and right of the gure below respectively Then 7 A1 6 bA1 0 TA2 10 bA2 2 Example 2 The Boolean arrangement Q7 consists of the n coordinate hyperplanes in R The complement R n is 951 957 l x y 0 for all 239 and the connected components are the open orthants speci ed by the signs of the n coordinates Therefore Mg 2 Example 3 Let A consist of m lines in R2 in general position that is no two lines are parallel and no three are coincident Draw the dual graph G the graph whose vertices are the regions of A with an edge between every two regions that share a common border A Let 7 MA 1 of vertices of G b MA 5 of edges of G f of faces of G Then 1a 1 r m m27m2 11 f 1 f because each bounded region contains exactly one point where two lines of A meet and 1c 4f71 257r7b because each unbounded face has four sides 1d Note that the number r 7 b of unbounded regions is just 2m Take a walk around a very large circle You will enter each unbounded region once and will cross each line twicei Therefore from 1c and 1b we obtain 1e em2f71 mQ Now Euler s formula for planar graphs says that v 7 e f 2 Substituting in 1a 1b and 1e and solving for r gives 7 m2m2 T 7 2 and therefore 7 2 7 m273m2 7 mil 77 7 m7 2 7 2 Example 4 The braid arrangement Brn consists of the hyperplanes x1 xj in R The complement R Br consists of all vectors in R with no two coordinates equal and the connected components of this set are speci ed by the ordering of the set of coordinates as real numbers yx zy XltZlty ZX Therefore MB n Our next goal is to prove Zaslavsky s theorems that the numbers TA and bA can be obtained as simple evaluations of the characteristic polynomial of the intersection poset LAi Deletion and Restriction Let A be a hyperplane arrangement If A is central then we know that LA is a geometric lattice I ll write MA for the corresponding matroid represented you will recall by the normal vectors 7 to the hyperplanes H E A Let x E LAi Recall that this means that x is an af ne space formed by the intersection of some subset of A De ne arrangements AxH AlH2x AEWlWH x H AAE Example 5 Let A be the 2dimensional arrangement shown on the left with the line H and point p as shown Then A17 and AH are shown on the right XX The reason for this notation is that LAx and LAx are isomorphic respectively to the principal order ideal and principal order lter generated by x in LAi LA LA H We say that A55 is the restriction of A to 95 Notice that rank A equals either rankA 7 1 or rank A according as H is or is not a coloop in the matroid of A since MA MA 7 Fly Proposition 1 Let A be a real arrangement and H E A Let A A and A AH Then 2 TM NW TM and 3 MA 7 bA MA if rankA rankA 7 0 if rankA rank A 1 Proof Consider what happens when we add H to A to obtain A Some regions of A will remain the same while others will be split into two regions The regions in the rst category each count once in both rA and rAi The regions in the second category each contribute 2 to rA but they also correspond bijectively to the regions of A This proves By the way if and only if H is a coloop then it borders every regionof A so rA 2rA in this case Now what about bounded regions If H is a coloop then A has no bounded regions 7 every region of A will contain a line so every region of A will contain a ray Otherwise the bounded regions of A come in three avors First the regions not bordered by H eg 1 below correspond bijectively to bounded regions of A through which H does not pass Second for each region R of A bordered by H the region R O H is bounded in A where R denotes the topological closure Moreover R comes from a bounded region in A if and only if walking from R across H gets you to a bounded region of A Yes in the case of the pair 2 and 3 which together contribute two to each side of 3 no in the case of 4 which contributes one to each side of D This looks a lot like a Tutte polynomial deletion contraction recurrence However we only have a matroid to work with when LA is a geometric lattice that is when A is central otherwise LA is not even a bounded poset On the other hand LA is certainly ranked by codimension for every arrangement so we can work instead with its characteristic polynomial which as you recall is de ned as 4 XAkXLAk Z amen xeLA Proposition 2 DeletionRestriction Let A be a real arrangement and H E A Let A A and A AH Then 5 MW XAk XAk Proof coming next time

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