Removing a variable is not the way to get the equation for a 0D point in 2D, or for a 1D line in 3D, or for our 2D intersection-of-two-3D-hyperplanes in 4D. For every hypergraph onnvertices there is an associated subspace arrangement in Rncalled a hypergraph arrangement. Even the simple equation y=1 is a 3D hyperplane if we are in 4D. y=5x+2 is a line, y=x is a line, x=6 is a line, y=0 is a line. It takes 2 equations to represent a 2D object in 4D (the intersection of two 3D hyperplanes is a 2D object.) For analogy, tell me the single "equation" that maps as a point in 2D. Here, we take a different approach, which makes direct use of Zaslavsky's formula relating the intersection. Chapter 1: Basic Definitions, the Intersection Poset and the Characteristic Polynomial. Perhaps someday these notes will be expanded into a textbook on arrangements. These numbers count a class of permutations known as Dumont derangements. After going through these notes a student should be ready to study the deeper algebraic and topological aspects of the theory of hyperplane arrangements. Variable replacement does not work in 3D, we do the cross-product as outlined, and it does not work in 4D. Hetyei recently introduced a hyperplane arrangement (called the homogenized Linial arrangement) and used the finite field method of Athanasiadis to show that its number of regions is a median Genocchi number. And z+y=5 is still a 2d plane in 3d, just like x+y+z=5 is. The result is a vector parallel to the intersection line. The essentialization of a hyperplane arrangement whose base field has characteristic 0 is obtained by intersecting the hyperplanes by the space spanned by their. Where (x, y, z) is the normal vector of each plane. This cross product is simply taking the determinant of matrix: i j k x1 y1 z1 x2 y2 z2. The fact that the equation has one less variable does not decrease the dimensionality of the object.įor analogy: y=7 is still a one-dimensional line in 2d, just like y=x+7 is. computable from the intersection lattice), and freeness of the module behaves well under certain operations on hyperplane arrangements which are used to. In the case of finding the line at which two planes intersect, you need to take the cross product of the normal of the two planes. 3x + 4y + 2z - 7(2 - 2x + 3y - 2z) = 10 is itself a three-dimensional hyperplane in four-dimensional space and it does not represent the intersection of the two given three-dimensional hyperplanes in four-dimensional space. The answer with a simple variable replacement is incorrect.
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