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# Class Note for MATH 796 at KU 5

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This 3 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 14 views.

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Date Created: 02/06/15
Monday 310 Projectivization and Coning Let K be a eld Denote points of K by 55 x1 i i Projective space llm lK is by de nition the set of lines through the origin in K If K R we can regard Pn llR as the unit sphere S7 1 with opposite points identi ed in particular it is an n 7 1dimensional manifold Algebraically write 55 N j if a and j are nonzero scalar multiples of each other Then N is an equivalence relation and P 1KltKn6gt Linear hyperplanes in K correspond to af ne hyperplanes in Pn lK Thus given a central arrangement A C Kquot we can construct its projectivization projA C Pm lK Projectivization supplies a nice way to draw central 3dimensional real arrangements Let S be the unit sphere so that H O S is a great circle for every H E A Regard H0 0 S as the equator and project the northern hemisphere into your piece of paper proj 3 proj Brg projessBr3 Of course a diagram of projA only shows the upper half77 of A We can recover A from projA by re ecting the interior of the disc to the exterior77 Stanley For example when A 3 In particular rproj TAi De nition Let A C Kquot not necessarily central The cone CA is the central arrangement in K7 1 de ned as follows 0 Geometrically Make a copy of A in Kn1 choose a point p not in any hyperplane of A and replace each H E A With the af ne span H ofp and H Which Will be a hyperplane in Kn1l Then toss in one more hyperplane containing p and in general position With respect to every H o Algebraically For H 9 l Li39 111 E A With L a homogeneous linear form on K and al E K construct a hyperplane H xlyu cmy l Li39 aly C K7 1 in CA Then toss in the hyperplane y 0 For example if A consists of the points x 0 x 73 and x 5 in R1 then CA consists of the lines 95 y 9573y x5yandy0ian2l Proposition 1 XCAUs k E 1XAk We ll prove this next time In particular Zaslavsky s formula for the number of regions implies that rcA 2rAl More on Regions and the Characteristic Polynomial Let A C K be a hyperplane arrangement We have seen that if K R then rA of regions ofA 71dimAXAEl bA of rel bounded regions 71mquotkAXA1l Zaslavsky s theorems and that if K qu then 1 FEWll 11 a fact rst noticed implicitly by Crapo and Rota and explicitly by Athanasiadis What if A C Cquot is a complex hyperplane arrangement Since the hyperplanes of A have codimension 2 as real vector subspaces the complement X Cquot A is connected but not simply connected Theorem 2 Brieskorn 1971 The homology groups H1X Z are free abelian and the Poincare polynomial of X is the characteristic polynomial backwards Zrankz HzXZql qleLM l Il 10 Orlik and Solomon 1980 strengthened Brieskorn s result by giving a presentation of the cohomology ring HX Z in terms of LA thereby proving that the cohomology is a combinatorial invariant of A Brieskorn s theorem says only that the additive structure of H X Z is a combinatorial invariantl The homotopy type of X is not a combinatorial invariant according to Reiner by a result of Rybnikov Graphic Arrangements De nition 1 Let G be a simple graph on vertex set The graphic arrangement AG C K consists of the hyperplanes x1 95 Where ij is an edge of G The arrangement AG is central but not essential so LAa is a geometric lattice The corresponding matroid is naturally isomorphic to the graphic matroid of G In particular 7 AG TG 2 0 equals the number of acyclic orientations of G For instance if G Kn then A Brn Which we have seen has n regions On the other hand the acyclic orientations of Kn are in bijection With total orderings of its vertices Moreover the chromatic polynomial of G equals the characteristic polynomial of LAai This last fact has a concrete combinatorial interpretation Regard a point 951 i i i 957 E F as a qcoloring of G that assigns color 951 to vertex 239 Then the proper qcolorings are precisely the points of F2 AG The number of such colorings is xG q the chromatic polynomial of G evaluated at 11 on the other hand by 1 it is also the characteristic polynomial XAGqi Since xG q XAGq for in nitely many 1 namely all integer prime powers the polynomials must be equal For some graphs such as complete graphs and trees the chromatic polynomial factors into linear terms For others it doesn t Example 1 Let G C4 a cycle With four vertices and four edges and let A Aa Then LA is the lattice of ats of the matroid U34 iiei LF 4 W 3 With 7 F minlFl 3 Since the Mobius function of an element of L depends only on its rank it is easy to check that XLk k3 74k 6k 7 3 k 71k2 73k 13 Multiplying by kdim AL mnk lL k4 3 gives the characteristic polynomial of AL Which is the chromatic polynomial of C4 mas 7 Mk 71 7 3k k So the question arises For Which graphs does the chromatic polynomial factor into linear terms More generally for Which arrangements A does the characteristic polynomial XAUs factor

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