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# arrangement(List,Ring) -- make a hyperplane arrangement

## Synopsis

• Function: arrangement
• Usage:
arrangement(L, R)
arrangement L
• Inputs:
• L, a list, of affine-linear equations in the ring $R$ or that is a product of linear forms
• R, a ring, a polynomial ring or linear quotient of a polynomial ring
• Outputs:
• , determined by the input data

## Description

A hyperplane is an affine-linear subspace of codimension one. An arrangement is a finite set of hyperplanes. When each hyperplane contains the origin, the arrangement is central.

Probably the best-known hyperplane arrangement is the braid arrangement consisting of all the diagonal hyperplanes. In $4$-space, it is constructed as follows.

 i1 : S = QQ[w,x,y,z]; i2 : A3 = arrangement {w-x, w-y, w-z, x-y, x-z, y-z} o2 = {w - x, w - y, w - z, x - y, x - z, y - z} o2 : Hyperplane Arrangement  i3 : assert isCentral A3

When a hyperplane arrangement is created from a product of linear forms, the order of the factors is not preserved.

 i4 : A3' = arrangement ((w-x)*(w-y)*(w-z)*(x-y)*(x-z)*(y-z)) o4 = {y - z, x - z, x - y, w - z, w - y, w - x} o4 : Hyperplane Arrangement  i5 : assert(A3 != A3') i6 : arrangement (x^2*y^2*(x^2-y^2)*(x^2-z^2)) o6 = {y, y, x, x, x - z, x + z, x - y, x + y} o6 : Hyperplane Arrangement 

The package can recognize that a polynomial splits into linear forms over the base field.

 i7 : kk = toField(QQ[p]/(p^2+p+1)) -- toField is necessary so that M2 treats this as a field o7 = kk o7 : PolynomialRing i8 : R = kk[s,t] o8 = R o8 : PolynomialRing i9 : arrangement (s^3-t^3) o9 = {s - t, s - p*t, s + (p + 1)t} o9 : Hyperplane Arrangement 

If we project onto a linear subspace, then we obtain an essential arrangement, meaning that the rank of the arrangement is equal to the dimension of its ambient vector space.

 i10 : R = S/ideal(w+x+y+z); i11 : A3'' = arrangement({w-x,w-y,w-z,x-y,x-z,y-z}, R) o11 = {- 2x - y - z, - x - 2y - z, - x - y - 2z, x - y, x - z, y - z} o11 : Hyperplane Arrangement  i12 : ring A3'' o12 = R o12 : QuotientRing i13 : assert(rank A3'' === dim ring A3'')

The trivial arrangement has no equations.

 i14 : trivial = arrangement({},S) o14 = {} o14 : Hyperplane Arrangement  i15 : ring trivial o15 = S o15 : PolynomialRing i16 : assert isCentral trivial

## Caveat

If the entries in $L$ are not ring elements in $R$, then the induced identity map is used to map them from the ring of first element in $L$ into $R$.