# Hyperplane Arrangements¶

Before talking about hyperplane arrangements, let us start with individual hyperplanes. This package uses certain linear expressions to represent hyperplanes, that is, a linear expression $$3x + 3y - 5z - 7$$ stands for the hyperplane with the equation $$x + 3y - 5z = 7$$. To create it in Sage, you first have to create a HyperplaneArrangements object to define the variables $$x$$, $$y$$, $$z$$:

sage: H.<x,y,z> = HyperplaneArrangements(QQ)
sage: h = 3*x + 2*y - 5*z - 7;  h
Hyperplane 3*x + 2*y - 5*z - 7
sage: h.normal()
(3, 2, -5)
sage: h.constant_term()
-7


The individual hyperplanes behave like the linear expression with regard to addition and scalar multiplication, which is why you can do linear combinations of the coordinates:

sage: -2*h
Hyperplane -6*x - 4*y + 10*z + 14
sage: x, y, z
(Hyperplane x + 0*y + 0*z + 0,
Hyperplane 0*x + y + 0*z + 0,
Hyperplane 0*x + 0*y + z + 0)


See sage.geometry.hyperplane_arrangement.hyperplane for more functionality of the individual hyperplanes.

## Arrangements¶

There are several ways to create hyperplane arrangements:

Notation (i): by passing individual hyperplanes to the HyperplaneArrangements object:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: box = x | y | x-1 | y-1;  box
Arrangement <y - 1 | y | x - 1 | x>
sage: box == H(x, y, x-1, y-1)    # alternative syntax
True


Notation (ii): by passing anything that defines a hyperplane, for example a coefficient vector and constant term:

sage: H = HyperplaneArrangements(QQ, ('x', 'y'))
sage: triangle = H([(1, 0), 0], [(0, 1), 0], [(1,1), -1]);  triangle
Arrangement <y | x | x + y - 1>

sage: H.inject_variables()
Defining x, y
sage: triangle == x | y | x+y-1
True


The default base field is $$\QQ$$, the rational numbers. Finite fields are also supported:

sage: H.<x,y,z> = HyperplaneArrangements(GF(5))
sage: a = H([(1,2,3), 4], [(5,6,7), 8]);  a
Arrangement <y + 2*z + 3 | x + 2*y + 3*z + 4>


Notation (iii): a list or tuple of hyperplanes:

sage: H.<x,y,z> = HyperplaneArrangements(GF(5))
sage: k = [x+i for i in range(4)];  k
[Hyperplane x + 0*y + 0*z + 0, Hyperplane x + 0*y + 0*z + 1,
Hyperplane x + 0*y + 0*z + 2, Hyperplane x + 0*y + 0*z + 3]
sage: H(k)
Arrangement <x | x + 1 | x + 2 | x + 3>


Notation (iv): using the library of arrangements:

sage: hyperplane_arrangements.braid(4)
Arrangement of 6 hyperplanes of dimension 4 and rank 3
sage: hyperplane_arrangements.semiorder(3)
Arrangement of 6 hyperplanes of dimension 3 and rank 2
sage: hyperplane_arrangements.graphical(graphs.PetersenGraph())
Arrangement of 15 hyperplanes of dimension 10 and rank 9
sage: hyperplane_arrangements.Ish(5)
Arrangement of 20 hyperplanes of dimension 5 and rank 4


Notation (v): from the bounding hyperplanes of a polyhedron:

sage: a = polytopes.cube().hyperplane_arrangement();  a
Arrangement of 6 hyperplanes of dimension 3 and rank 3
sage: a.n_regions()
27


New arrangements from old:

sage: a = hyperplane_arrangements.braid(3)
sage: b = a.add_hyperplane([4, 1, 2, 3])
sage: b
Arrangement <t1 - t2 | t0 - t1 | t0 - t2 | t0 + 2*t1 + 3*t2 + 4>
sage: c = b.deletion([4, 1, 2, 3])
sage: a == c
True

sage: a = hyperplane_arrangements.braid(3)
sage: b = a.union(hyperplane_arrangements.semiorder(3))
sage: b == a | hyperplane_arrangements.semiorder(3)    # alternate syntax
True
sage: b == hyperplane_arrangements.Catalan(3)
True

sage: a
Arrangement <t1 - t2 | t0 - t1 | t0 - t2>
sage: a = hyperplane_arrangements.coordinate(4)
sage: h = a.hyperplanes()[0]
sage: b = a.restriction(h)
sage: b == hyperplane_arrangements.coordinate(3)
True


A hyperplane arrangement is essential is the normals to its hyperplane span the ambient space. Otherwise, it is inessential. The essentialization is formed by intersecting the hyperplanes by this normal space (actually, it is a bit more complicated over finite fields):

sage: a = hyperplane_arrangements.braid(4);  a
Arrangement of 6 hyperplanes of dimension 4 and rank 3
sage: a.is_essential()
False
sage: a.rank() < a.dimension()  # double-check
True
sage: a.essentialization()
Arrangement of 6 hyperplanes of dimension 3 and rank 3


The connected components of the complement of the hyperplanes of an arrangement in $$\RR^n$$ are called the regions of the arrangement:

sage: a = hyperplane_arrangements.semiorder(3)
sage: b = a.essentialization();   b
Arrangement of 6 hyperplanes of dimension 2 and rank 2
sage: b.n_regions()
19
sage: b.regions()
(A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 6 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays)
sage: b.bounded_regions()
(A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 6 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices)
sage: b.n_bounded_regions()
7
sage: a.unbounded_regions()
(A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices, 1 ray, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices, 1 ray, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices, 1 ray, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices, 1 ray, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices, 1 ray, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices, 1 ray, 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line)


The distance between regions is defined as the number of hyperplanes separating them. For example:

sage: r1 = b.regions()[0]
sage: r2 = b.regions()[1]
sage: b.distance_between_regions(r1, r2)
1
sage: [hyp for hyp in b if b.is_separating_hyperplane(r1, r2, hyp)]
[Hyperplane 2*t1 + t2 + 1]
sage: b.distance_enumerator(r1)  # generating function for distances from r1
6*x^3 + 6*x^2 + 6*x + 1


Note

bounded region really mean relatively bounded here. A region is relatively bounded if its intersection with space spanned by the normals of the hyperplanes in the arrangement is bounded.

The intersection poset of a hyperplane arrangement is the collection of all nonempty intersections of hyperplanes in the arrangement, ordered by reverse inclusion. It includes the ambient space of the arrangement (as the intersection over the empty set):

sage: a = hyperplane_arrangements.braid(3)
sage: p = a.intersection_poset()
sage: p.is_ranked()
True
sage: p.order_polytope()
A 5-dimensional polyhedron in ZZ^5 defined as the convex hull of 10 vertices


The characteristic polynomial is a basic invariant of a hyperplane arrangement. It is defined as

$\chi(x) := \sum_{w\in P} \mu(w) x^{dim(w)}$

where the sum is $$P$$ is the intersection_poset() of the arrangement and $$\mu$$ is the Möbius function of $$P$$:

sage: a = hyperplane_arrangements.semiorder(5)
sage: a.characteristic_polynomial()               # long time (about a second on Core i7)
x^5 - 20*x^4 + 180*x^3 - 790*x^2 + 1380*x
sage: a.poincare_polynomial()                     # long time
1380*x^4 + 790*x^3 + 180*x^2 + 20*x + 1
sage: a.n_regions()                               # long time
2371
sage: charpoly = a.characteristic_polynomial()    # long time
sage: charpoly(-1)                                # long time
-2371
sage: a.n_bounded_regions()                       # long time
751
sage: charpoly(1)                                 # long time
751


For finer invariants derived from the intersection poset, see whitney_number() and doubly_indexed_whitney_number().

Miscellaneous methods (see documentation for an explanation):

sage: a = hyperplane_arrangements.semiorder(3)
sage: a.has_good_reduction(5)
True
sage: b = a.change_ring(GF(5))
sage: pa = a.intersection_poset()
sage: pb = b.intersection_poset()
sage: pa.is_isomorphic(pb)
True
sage: a.face_vector()
(0, 12, 30, 19)
sage: a.face_vector()
(0, 12, 30, 19)
sage: a.is_central()
False
sage: a.is_linear()
False
sage: a.sign_vector((1,1,1))
(-1, 1, -1, 1, -1, 1)
sage: a.varchenko_matrix()
[          1          h2       h2*h4       h2*h3    h2*h3*h4 h2*h3*h4*h5]
[         h2           1          h4          h3       h3*h4    h3*h4*h5]
[      h2*h4          h4           1       h3*h4          h3       h3*h5]
[      h2*h3          h3       h3*h4           1          h4       h4*h5]
[   h2*h3*h4       h3*h4          h3          h4           1          h5]
[h2*h3*h4*h5    h3*h4*h5       h3*h5       h4*h5          h5           1]


There are extensive methods for visualizing hyperplane arrangements in low dimensions. See plot() for details.

AUTHORS:

• David Perkinson (2013-06): initial version
• Qiaoyu Yang (2013-07)
• Kuai Yu (2013-07)
• Volker Braun (2013-10): Better Sage integration, major code refactoring.

This module implements hyperplane arrangements defined over the rationals or over finite fields. The original motivation was to make a companion to Richard Stanley’s notes [Sta2007] on hyperplane arrangements.

class sage.geometry.hyperplane_arrangement.arrangement.HyperplaneArrangementElement(parent, hyperplanes, check=True)

A hyperplane arrangement.

Warning

You should never create HyperplaneArrangementElement instances directly, always use the parent.

add_hyperplane(other)

The union of self with other.

INPUT:

• other – a hyperplane arrangement or something that can be converted into a hyperplane arrangement

OUTPUT:

A new hyperplane arrangement.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([1,2,3], [0,1,1], [0,1,-1], [1,-1,0], [1,1,0])
sage: B = H([1,1,1], [1,-1,1], [1,0,-1])
sage: A.union(B)
Arrangement of 8 hyperplanes of dimension 2 and rank 2
sage: A | B   # syntactic sugar
Arrangement of 8 hyperplanes of dimension 2 and rank 2


A single hyperplane is coerced into a hyperplane arrangement if necessary:

sage: A.union(x+y-1)
Arrangement of 6 hyperplanes of dimension 2 and rank 2
Arrangement of 6 hyperplanes of dimension 2 and rank 2

sage: P.<x,y> = HyperplaneArrangements(RR)
sage: C = P(2*x + 4*y + 5)
sage: C.union(A)
Arrangement of 6 hyperplanes of dimension 2 and rank 2

bounded_regions()

Return the relatively bounded regions of the arrangement.

A region is relatively bounded if its intersection with the space spanned by the normals to the hyperplanes is bounded. This is the same as being bounded in the case that the hyperplane arrangement is essential. It is assumed that the arrangement is defined over the rationals.

OUTPUT:

Tuple of polyhedra. The relatively bounded regions of the arrangement.

EXAMPLES:

sage: A = hyperplane_arrangements.semiorder(3)
sage: A.bounded_regions()
(A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices and 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices and 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices and 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 6 vertices and 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices and 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices and 1 line,
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 3 vertices and 1 line)
sage: A.bounded_regions()[0].is_compact()    # the regions are only *relatively* bounded
False
sage: A.is_essential()
False

change_ring(base_ring)

Return hyperplane arrangement over the new base ring.

INPUT:

• base_ring – the new base ring; must be a field for hyperplane arrangements

OUTPUT:

The hyperplane arrangement obtained by changing the base field, as a new hyperplane arrangement.

Warning

While there is often a one-to-one correspondence between the hyperplanes of self and those of self.change_ring(base_ring), there is no guarantee that the order in which they appear in self.hyperplanes() will match the order in which their counterparts in self.cone() will appear in self.change_ring(base_ring).hyperplanes()!

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([(1,1), 0], [(2,3), -1])
sage: A.change_ring(FiniteField(2))
Arrangement <y + 1 | x + y>

characteristic_polynomial()

Return the characteristic polynomial of the hyperplane arrangement.

OUTPUT:

The characteristic polynomial in $$\QQ[x]$$.

EXAMPLES:

sage: a = hyperplane_arrangements.coordinate(2)
sage: a.characteristic_polynomial()
x^2 - 2*x + 1

closed_faces(labelled=True)

Return the closed faces of the hyperplane arrangement self (provided that self is defined over a totally ordered field).

Let $$\mathcal{A}$$ be a hyperplane arrangement in the vector space $$K^n$$, whose hyperplanes are the zero sets of the affine-linear functions $$u_1, u_2, \ldots, u_N$$. (We consider these functions $$u_1, u_2, \ldots, u_N$$, and not just the hyperplanes, as given. We also assume the field $$K$$ to be totally ordered.) For any point $$x \in K^n$$, we define the sign vector of $$x$$ to be the vector $$(v_1, v_2, \ldots, v_N) \in \{-1, 0, 1\}^N$$ such that (for each $$i$$) the number $$v_i$$ is the sign of $$u_i(x)$$. For any $$v \in \{-1, 0, 1\}^N$$, we let $$F_v$$ be the set of all $$x \in K^n$$ which have sign vector $$v$$. The nonempty ones among all these subsets $$F_v$$ are called the open faces of $$\mathcal{A}$$. They form a partition of the set $$K^n$$.

Furthermore, for any $$v = (v_1, v_2, \ldots, v_N) \in \{-1, 0, 1\}^N$$, we let $$G_v$$ be the set of all $$x \in K^n$$ such that, for every $$i$$, the sign of $$u_i(x)$$ is either $$0$$ or $$v_i$$. Then, $$G_v$$ is a polyhedron. The nonempty ones among all these polyhedra $$G_v$$ are called the closed faces of $$\mathcal{A}$$. While several sign vectors $$v$$ can lead to one and the same closed face $$G_v$$, we can assign to every closed face a canonical choice of a sign vector: Namely, if $$G$$ is a closed face of $$\mathcal{A}$$, then the sign vector of $$G$$ is defined to be the vector $$(v_1, v_2, \ldots, v_N) \in \{-1, 0, 1\}^N$$ where $$x$$ is any point in the relative interior of $$G$$ and where, for each $$i$$, the number $$v_i$$ is the sign of $$u_i(x)$$. (This does not depend on the choice of $$x$$.)

There is a one-to-one correspondence between the closed faces and the open faces of $$\mathcal{A}$$. It sends a closed face $$G$$ to the open face $$F_v$$, where $$v$$ is the sign vector of $$G$$; this $$F_v$$ is also the relative interior of $$G_v$$. The inverse map sends any open face $$O$$ to the closure of $$O$$.

INPUT:

• labelled – boolean (default: True); if True, then this method returns not the faces itself but rather pairs $$(v, F)$$ where $$F$$ is a closed face and $$v$$ is its sign vector (here, the order and the orientation of the $$u_1, u_2, \ldots, u_N$$ is as given by self.hyperplanes()).

OUTPUT:

A tuple containing the closed faces as polyhedra, or (if labelled is set to True) the pairs of sign vectors and corresponding closed faces.

Todo

Should the output rather be a dictionary where the keys are the sign vectors and the values are the faces?

EXAMPLES:

sage: a = hyperplane_arrangements.braid(2)
sage: a.hyperplanes()
(Hyperplane t0 - t1 + 0,)
sage: a.closed_faces()
(((0,),
A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 line),
((1,),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line),
((-1,),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line))
sage: a.closed_faces(labelled=False)
(A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 line,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line)
sage: [(v, F, F.representative_point()) for v, F in a.closed_faces()]
[((0,),
A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 line,
(0, 0)),
((1,),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(0, -1)),
((-1,),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(-1, 0))]

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: a = H(x, y+1)
sage: a.hyperplanes()
(Hyperplane 0*x + y + 1, Hyperplane x + 0*y + 0)
sage: [(v, F, F.representative_point()) for v, F in a.closed_faces()]
[((0, 0),
A 0-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex,
(0, -1)),
((0, 1),
A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 ray,
(1, -1)),
((0, -1),
A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 ray,
(-1, -1)),
((1, 0),
A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 ray,
(0, 0)),
((1, 1),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
(1, 0)),
((1, -1),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
(-1, 0)),
((-1, 0),
A 1-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 1 ray,
(0, -2)),
((-1, 1),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
(1, -2)),
((-1, -1),
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
(-1, -2))]

sage: a = hyperplane_arrangements.braid(3)
sage: a.hyperplanes()
(Hyperplane 0*t0 + t1 - t2 + 0,
Hyperplane t0 - t1 + 0*t2 + 0,
Hyperplane t0 + 0*t1 - t2 + 0)
sage: [(v, F, F.representative_point()) for v, F in a.closed_faces()]
[((0, 0, 0),
A 1-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex and 1 line,
(0, 0, 0)),
((0, 1, 1),
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(0, -1, -1)),
((0, -1, -1),
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(-1, 0, 0)),
((1, 0, 1),
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(1, 1, 0)),
((1, 1, 1),
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
(0, -1, -2)),
((1, -1, 0),
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(-1, 0, -1)),
((1, -1, 1),
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
(1, 2, 0)),
((1, -1, -1),
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
(-2, 0, -1)),
((-1, 0, -1),
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(0, 0, 1)),
((-1, 1, 0),
A 2-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 1 ray, 1 line,
(1, 0, 1)),
((-1, 1, 1),
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
(0, -2, -1)),
((-1, 1, -1),
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
(1, 0, 2)),
((-1, -1, -1),
A 3-dimensional polyhedron in QQ^3 defined as the convex hull of 1 vertex, 2 rays, 1 line,
(-1, 0, 1))]


Let us check that the number of closed faces with a given dimension computed using self.closed_faces() equals the one computed using face_vector():

sage: def test_number(a):
....:     Qx = PolynomialRing(QQ, 'x'); x = Qx.gen()
....:     RHS = Qx.sum(vi * x ** i for i, vi in enumerate(a.face_vector()))
....:     LHS = Qx.sum(x ** F[1].dim() for F in a.closed_faces())
....:     return LHS == RHS
sage: a = hyperplane_arrangements.Catalan(2)
sage: test_number(a)
True
sage: a = hyperplane_arrangements.Shi(3)
sage: test_number(a) # long time
True

cone(variable='t')

Return the cone over the hyperplane arrangement.

INPUT:

• variable – string; the name of the additional variable

OUTPUT:

A new hyperplane arrangement. Its equations consist of $$[0, -d, a_1, \ldots, a_n]$$ for each $$[d, a_1, \ldots, a_n]$$ in the original arrangement and the equation $$[0, 1, 0, \ldots, 0]$$.

Warning

While there is an almost-one-to-one correspondence between the hyperplanes of self and those of self.cone(), there is no guarantee that the order in which they appear in self.hyperplanes() will match the order in which their counterparts in self.cone() will appear in self.cone().hyperplanes()!

EXAMPLES:

sage: a.<x,y,z> = hyperplane_arrangements.semiorder(3)
sage: b = a.cone()
sage: a.characteristic_polynomial().factor()
x * (x^2 - 6*x + 12)
sage: b.characteristic_polynomial().factor()
(x - 1) * x * (x^2 - 6*x + 12)
sage: a.hyperplanes()
(Hyperplane 0*x + y - z - 1,
Hyperplane 0*x + y - z + 1,
Hyperplane x - y + 0*z - 1,
Hyperplane x - y + 0*z + 1,
Hyperplane x + 0*y - z - 1,
Hyperplane x + 0*y - z + 1)
sage: b.hyperplanes()
(Hyperplane -t + 0*x + y - z + 0,
Hyperplane -t + x - y + 0*z + 0,
Hyperplane -t + x + 0*y - z + 0,
Hyperplane t + 0*x + 0*y + 0*z + 0,
Hyperplane t + 0*x + y - z + 0,
Hyperplane t + x - y + 0*z + 0,
Hyperplane t + x + 0*y - z + 0)

defining_polynomial()

Return the defining polynomial of A.

Let $$A = (H_i)_i$$ be a hyperplane arrangement in a vector space $$V$$ corresponding to the null spaces of $$\alpha_{H_i} \in V^*$$. Then the defining polynomial of $$A$$ is given by

$Q(A) = \prod_i \alpha_{H_i} \in S(V^*).$

EXAMPLES:

sage: H.<x,y,z> = HyperplaneArrangements(QQ)
sage: A = H([2*x + y - z, -x - 2*y + z])
sage: p = A.defining_polynomial(); p
-2*x^2 - 5*x*y - 2*y^2 + 3*x*z + 3*y*z - z^2
sage: p.factor()
(-1) * (x + 2*y - z) * (2*x + y - z)

deletion(hyperplanes)

Return the hyperplane arrangement obtained by removing h.

INPUT:

• h – a hyperplane or hyperplane arrangement

OUTPUT:

A new hyperplane arrangement with the given hyperplane(s) h removed.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([0,1,0], [1,0,1], [-1,0,1], [0,1,-1], [0,1,1]);  A
Arrangement of 5 hyperplanes of dimension 2 and rank 2
sage: A.deletion(x)
Arrangement <y - 1 | y + 1 | x - y | x + y>
sage: h = H([0,1,0], [0,1,1])
sage: A.deletion(h)
Arrangement <y - 1 | y + 1 | x - y>

derivation_module_basis(algorithm='singular')

Return a basis for the derivation module of self if one exists, otherwise return None.

INPUT:

• algorithm – (default: "singular") can be one of the following:
• "singular" – use Singular’s minimal free resolution
• "BC" – use the algorithm given by Barakat and Cuntz in [BC2012] (much slower than using Singular)

OUTPUT:

A basis for the derivation module (over $$S$$, the symmetric space) as vectors of a free module over $$S$$.

ALGORITHM:

Singular

This gets the reduced syzygy module of the Jacobian ideal of the defining polynomial $$f$$ of self. It then checks Saito’s criterion that the determinant of the basis matrix is a scalar multiple of $$f$$. If the basis matrix is not square or it fails Saito’s criterion, then we check if the arrangement is free. If it is free, then we fall back to the Barakat-Cuntz algorithm.

BC

Return the product of the derivation module free chain matrices. See Section 6 of [BC2012].

EXAMPLES:

sage: W = WeylGroup(['A',2], prefix='s')
sage: A = W.long_element().inversion_arrangement()
sage: A.derivation_module_basis()
[(a1, a2), (0, a1*a2 + a2^2)]

derivation_module_free_chain()

Return a free chain for the derivation module if one exists, otherwise return None.

EXAMPLES:

sage: W = WeylGroup(['A',3], prefix='s')
sage: A = W.long_element().inversion_arrangement()
sage: for M in A.derivation_module_free_chain(): print("%s\n"%M)
[ 1  0  0]
[ 0  1  0]
[ 0  0 a3]

[ 1  0  0]
[ 0  0  1]
[ 0 a2  0]

[  1   0   0]
[  0  -1  -1]
[  0  a2 -a3]

[ 0  1  0]
[ 0  0  1]
[a1  0  0]

[ 1  0 -1]
[a3 -1  0]
[a1  0 a2]

[       1        0        0]
[      a3       -1       -1]
[       0       a1 -a2 - a3]

dimension()

Return the ambient space dimension of the arrangement.

OUTPUT:

An integer.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: (x | x-1 | x+1).dimension()
2
sage: H(x).dimension()
2

distance_between_regions(region1, region2)

Return the number of hyperplanes separating the two regions.

INPUT:

• region1, region2 – regions of the arrangement or representative points of regions

OUTPUT:

An integer. The number of hyperplanes separating the two regions.

EXAMPLES:

sage: c = hyperplane_arrangements.coordinate(2)
sage: r = c.region_containing_point([-1, -1])
sage: s = c.region_containing_point([1, 1])
sage: c.distance_between_regions(r, s)
2
sage: c.distance_between_regions(s, s)
0

distance_enumerator(base_region)

Return the generating function for the number of hyperplanes at given distance.

INPUT:

• base_region – region of arrangement or point in region

OUTPUT:

A polynomial $$f(x)$$ for which the coefficient of $$x^i$$ is the number of hyperplanes of distance $$i$$ from base_region, i.e., the number of hyperplanes separated by $$i$$ hyperplanes from base_region.

EXAMPLES:

sage: c = hyperplane_arrangements.coordinate(3)
sage: c.distance_enumerator(c.region_containing_point([1,1,1]))
x^3 + 3*x^2 + 3*x + 1

doubly_indexed_whitney_number(i, j, kind=1)

Return the $$i,j$$-th doubly-indexed Whitney number.

If kind=1, this number is obtained by adding the Möbius function values $$mu(x,y)$$ over all $$x, y$$ in the intersection poset with $$\mathrm{rank}(x) = i$$ and $$\mathrm{rank}(y) = j$$.

If $$kind=2$$, this number is the number of elements $$x,y$$ in the intersection poset such that $$x \leq y$$ with ranks $$i$$ and $$j$$, respectively.

INPUT:

• i, j – integers
• kind – (default: 1) 1 or 2

OUTPUT:

Integer. The $$(i,j)$$-th entry of the kind Whitney number.

EXAMPLES:

sage: A = hyperplane_arrangements.Shi(3)
sage: A.doubly_indexed_whitney_number(0, 2)
9
sage: A.whitney_number(2)
9
sage: A.doubly_indexed_whitney_number(1, 2)
-15


REFERENCES:

essentialization()

Return the essentialization of the hyperplane arrangement.

The essentialization of a hyperplane arrangement whose base field has characteristic 0 is obtained by intersecting the hyperplanes by the space spanned by their normal vectors.

OUTPUT:

The essentialization as a new hyperplane arrangement.

EXAMPLES:

sage: a = hyperplane_arrangements.braid(3)
sage: a.is_essential()
False
sage: a.essentialization()
Arrangement <t1 - t2 | t1 + 2*t2 | 2*t1 + t2>

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: B = H([(1,0),1], [(1,0),-1])
sage: B.is_essential()
False
sage: B.essentialization()
Arrangement <-x + 1 | x + 1>
sage: B.essentialization().parent()
Hyperplane arrangements in 1-dimensional linear space over
Rational Field with coordinate x

sage: H.<x,y> = HyperplaneArrangements(GF(2))
sage: C = H([(1,1),1], [(1,1),0])
sage: C.essentialization()
Arrangement <y | y + 1>

sage: h = hyperplane_arrangements.semiorder(4)
sage: h.essentialization()
Arrangement of 12 hyperplanes of dimension 3 and rank 3

face_product(F, G, normalize=True)

Return the product $$FG$$ in the face semigroup of self, where $$F$$ and $$G$$ are two closed faces of self.

The face semigroup of a hyperplane arrangement $$\mathcal{A}$$ is defined as follows: As a set, it is the set of all open faces of self (see closed_faces()). Its product is defined by the following rule: If $$F$$ and $$G$$ are two open faces of $$\mathcal{A}$$, then $$FG$$ is an open face of $$\mathcal{A}$$, and for every hyperplane $$H \in \mathcal{A}$$, the open face $$FG$$ lies on the same side of $$H$$ as $$F$$ unless $$F \subseteq H$$, in which case $$FG$$ lies on the same side of $$H$$ as $$G$$. Alternatively, $$FG$$ can be defined as follows: If $$f$$ and $$g$$ are two points in $$F$$ and $$G$$, respectively, then $$FG$$ is the face that contains the point $$(f + \varepsilon g) / (1 + \varepsilon)$$ for any sufficiently small positive $$\varepsilon$$.

In our implementation, the face semigroup consists of closed faces rather than open faces (thanks to the 1-to-1 correspondence between open faces and closed faces, this is not really a different semigroup); these closed faces are given as polyhedra.

The face semigroup of a hyperplane arrangement is always a left-regular band (i.e., a semigroup satisfying the identities $$x^2 = x$$ and $$xyx = xy$$). When the arrangement is central, then this semigroup is a monoid. See [Br2000] (Appendix A in particular) for further properties.

INPUT:

• F, G – two faces of self (as polyhedra)
• normalize – Boolean (default: True); if True, then this method returns the precise instance of $$FG$$ in the list returned by self.closed_faces(), rather than creating a new instance

EXAMPLES:

sage: a = hyperplane_arrangements.braid(3)
sage: a.hyperplanes()
(Hyperplane 0*t0 + t1 - t2 + 0,
Hyperplane t0 - t1 + 0*t2 + 0,
Hyperplane t0 + 0*t1 - t2 + 0)
sage: faces = {F0: F1 for F0, F1 in a.closed_faces()}
sage: xGyEz = faces[(0, 1, 1)] # closed face x >= y = z
sage: xGyEz.representative_point()
(0, -1, -1)
sage: xGyEz = faces[(0, 1, 1)] # closed face x >= y = z
sage: xGyEz.representative_point()
(0, -1, -1)
sage: yGxGz = faces[(1, -1, 1)] # closed face y >= x >= z
sage: xGyGz = faces[(1, 1, 1)] # closed face x >= y >= z
sage: a.face_product(xGyEz, yGxGz) == xGyGz
True
sage: a.face_product(yGxGz, xGyEz) == yGxGz
True
sage: xEzGy = faces[(-1, 1, 0)] # closed face x = z >= y
sage: xGzGy = faces[(-1, 1, 1)] # closed face x >= z >= y
sage: a.face_product(xEzGy, yGxGz) == xGzGy
True

face_semigroup_algebra(field=None, names='e')

Return the face semigroup algebra of self.

This is the semigroup algebra of the face semigroup of self (see face_product() for the definition of the semigroup).

Due to limitations of the current Sage codebase (e.g., semigroup algebras do not profit from the functionality of the FiniteDimensionalAlgebra class), this is implemented not as a semigroup algebra, but as a FiniteDimensionalAlgebra. The closed faces of self (in the order in which the closed_faces() method outputs them) are identified with the vectors $$(0, 0, \ldots, 0, 1, 0, 0, \ldots, 0)$$ (with the $$1$$ moving from left to right).

INPUT:

• field – a field (default: $$\QQ$$), to be used as the base ring for the algebra (can also be a commutative ring, but then certain representation-theoretical methods might misbehave)
• names – (default: 'e') string; names for the basis elements of the algebra

Todo

Also implement it as an actual semigroup algebra?

EXAMPLES:

sage: a = hyperplane_arrangements.braid(3)
sage: [(i, F[0]) for i, F in enumerate(a.closed_faces())]
[(0, (0, 0, 0)),
(1, (0, 1, 1)),
(2, (0, -1, -1)),
(3, (1, 0, 1)),
(4, (1, 1, 1)),
(5, (1, -1, 0)),
(6, (1, -1, 1)),
(7, (1, -1, -1)),
(8, (-1, 0, -1)),
(9, (-1, 1, 0)),
(10, (-1, 1, 1)),
(11, (-1, 1, -1)),
(12, (-1, -1, -1))]
sage: U = a.face_semigroup_algebra(); U
Finite-dimensional algebra of degree 13 over Rational Field
sage: e0, e1, e2, e3, e4, e5, e6, e7, e8, e9, e10, e11, e12 = U.basis()
sage: e0 * e1
e1
sage: e0 * e5
e5
sage: e5 * e0
e5
sage: e3 * e2
e6
sage: e7 * e12
e7
sage: e3 * e12
e6
sage: e4 * e8
e4
sage: e8 * e4
e11
sage: e8 * e1
e11
sage: e5 * e12
e7
sage: (e3 + 2*e4) * (e1 - e7)
e4 - e6

sage: U3 = a.face_semigroup_algebra(field=GF(3)); U3
Finite-dimensional algebra of degree 13 over Finite Field of size 3

face_vector()

Return the face vector.

OUTPUT:

A vector of integers.

The $$d$$-th entry is the number of faces of dimension $$d$$. A face is the intersection of a region with a hyperplane of the arrangement.

EXAMPLES:

sage: A = hyperplane_arrangements.Shi(3)
sage: A.face_vector()
(0, 6, 21, 16)

has_good_reduction(p)

Return whether the hyperplane arrangement has good reduction mod $$p$$.

Let $$A$$ be a hyperplane arrangement with equations defined over the integers, and let $$B$$ be the hyperplane arrangement defined by reducing these equations modulo a prime $$p$$. Then $$A$$ has good reduction modulo $$p$$ if the intersection posets of $$A$$ and $$B$$ are isomorphic.

INPUT:

• p – prime number

OUTPUT:

A boolean.

EXAMPLES:

sage: a = hyperplane_arrangements.semiorder(3)
sage: a.has_good_reduction(5)
True
sage: a.has_good_reduction(3)
False
sage: b = a.change_ring(GF(3))
sage: a.characteristic_polynomial()
x^3 - 6*x^2 + 12*x
sage: b.characteristic_polynomial()  # not equal to that for a
x^3 - 6*x^2 + 10*x

hyperplanes()

Return the number of hyperplanes in the arrangement.

OUTPUT:

An integer.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([1,1,0], [2,3,-1], [4,5,3])
sage: A.hyperplanes()
(Hyperplane x + 0*y + 1, Hyperplane 3*x - y + 2, Hyperplane 5*x + 3*y + 4)


Note that the hyperplanes can be indexed as if they were a list:

sage: A[0]
Hyperplane x + 0*y + 1

intersection_poset()

Return the intersection poset of the hyperplane arrangement.

OUTPUT:

The poset of non-empty intersections of hyperplanes.

EXAMPLES:

sage: a = hyperplane_arrangements.coordinate(2)
sage: a.intersection_poset()
Finite poset containing 4 elements

sage: A = hyperplane_arrangements.semiorder(3)
sage: A.intersection_poset()
Finite poset containing 19 elements

is_central()

Test whether the intersection of all the hyperplanes is nonempty.

OUTPUT:

A boolean whether the hyperplane arrangement is such that the intersection of all the hyperplanes in the arrangement is nonempty.

EXAMPLES:

sage: a = hyperplane_arrangements.braid(2)
sage: a.is_central()
True

is_essential()

Test whether the hyperplane arrangement is essential.

A hyperplane arrangement is essential if the span of the normals of its hyperplanes spans the ambient space.

OUTPUT:

A boolean indicating whether the hyperplane arrangement is essential.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: H(x, x+1).is_essential()
False
sage: H(x, y).is_essential()
True

is_formal()

Return if self is formal.

A hyperplane arrangement is formal if it is 3-generated [Yuz1993], where $$k$$-generated is defined in minimal_generated_number().

EXAMPLES:

sage: P.<x,y,z> = HyperplaneArrangements(QQ)
sage: A = P(x, y, z, x+y+z, 2*x+y+z, 2*x+3*y+z, 2*x+3*y+4*z, 3*x+5*z, 3*x+4*y+5*z)
sage: B = P(x, y, z, x+y+z, 2*x+y+z, 2*x+3*y+z, 2*x+3*y+4*z, x+3*z, x+2*y+3*z)
sage: A.is_formal()
True
sage: B.is_formal()
False

is_free(algorithm='singular')

Return if self is free.

A hyperplane arrangement $$A$$ is free if the module of derivations $$\operatorname{Der}(A)$$ is a free $$S$$-module, where $$S$$ is the corresponding symmetric space.

INPUT:

• algorithm – (default: "singular") can be one of the following:
• "singular" – use Singular’s minimal free resolution
• "BC" – use the algorithm given by Barakat and Cuntz in [BC2012] (much slower than using Singular)

ALGORITHM:

singular

Check that the minimal free resolution has length at most 2 by using Singular.

BC

This implementation follows [BC2012] by constructing a chain of free modules

$D(A) = D(A_n) < D(A_{n-1}) < \cdots < D(A_1) < D(A_0)$

corresponding to some ordering of the arrangements $$A_0 \subset A_1 \subset \cdots \subset A_{n-1} \subset A_n = A$$. Such a chain is found by using a backtracking algorithm.

EXAMPLES:

For type $$A$$ arrangements, chordality is equivalent to freeness. We verify that in type $$A_3$$:

sage: W = WeylGroup(['A',3], prefix='s')
sage: for x in W:
....:    A = x.inversion_arrangement()
....:    assert A.matroid().is_chordal() == A.is_free()

is_linear()

Test whether all hyperplanes pass through the origin.

OUTPUT:

A boolean. Whether all the hyperplanes pass through the origin.

EXAMPLES:

sage: a = hyperplane_arrangements.semiorder(3)
sage: a.is_linear()
False
sage: b = hyperplane_arrangements.braid(3)
sage: b.is_linear()
True

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: c = H(x+1, y+1)
sage: c.is_linear()
False
sage: c.is_central()
True

is_separating_hyperplane(region1, region2, hyperplane)

Test whether the hyperplane separates the given regions.

INPUT:

• region1, region2 – polyhedra or list/tuple/iterable of coordinates which are regions of the arrangement or an interior point of a region
• hyperplane – a hyperplane

OUTPUT:

A boolean. Whether the hyperplane hyperplane separate the given regions.

EXAMPLES:

sage: A.<x,y> = hyperplane_arrangements.coordinate(2)
sage: A.is_separating_hyperplane([1,1], [2,1], y)
False
sage: A.is_separating_hyperplane([1,1], [-1,1], x)
True
sage: r = A.region_containing_point([1,1])
sage: s = A.region_containing_point([-1,1])
sage: A.is_separating_hyperplane(r, s, x)
True

matroid()

Return the matroid associated to self.

Let $$A$$ denote a central hyperplane arrangement and $$n_H$$ the normal vector of some hyperplane $$H \in A$$. We define a matroid $$M_A$$ as the linear matroid spanned by $$\{ n_H | H \in A \}$$. The matroid $$M_A$$ is such that the lattice of flats of $$M$$ is isomorphic to the intersection lattice of $$A$$ (Proposition 3.6 in [Sta2007]).

EXAMPLES:

sage: P.<x,y,z> = HyperplaneArrangements(QQ)
sage: A = P(x, y, z, x+y+z, 2*x+y+z, 2*x+3*y+z, 2*x+3*y+4*z)
sage: M = A.matroid(); M
Linear matroid of rank 3 on 7 elements represented over the Rational Field


We check the lattice of flats is isomorphic to the intersection lattice:

sage: f = sum([list(M.flats(i)) for i in range(M.rank()+1)], [])
sage: PF = Poset([f, lambda x,y: x < y])
sage: PF.is_isomorphic(A.intersection_poset())
True

minimal_generated_number()

Return the minimum $$k$$ such that self is $$k$$-generated.

Let $$A$$ be a central hyperplane arrangement. Let $$W_k$$ denote the solution space of the linear system corresponding to the linear dependencies among the hyperplanes of $$A$$ of length at most $$k$$. We say $$A$$ is $$k$$-generated if $$\dim W_k = \operatorname{rank} A$$.

Equivalently this says all dependencies forming the Orlik-Terao ideal are generated by at most $$k$$ hyperplanes.

EXAMPLES:

We construct Example 2.2 from [Yuz1993]:

sage: P.<x,y,z> = HyperplaneArrangements(QQ)
sage: A = P(x, y, z, x+y+z, 2*x+y+z, 2*x+3*y+z, 2*x+3*y+4*z, 3*x+5*z, 3*x+4*y+5*z)
sage: B = P(x, y, z, x+y+z, 2*x+y+z, 2*x+3*y+z, 2*x+3*y+4*z, x+3*z, x+2*y+3*z)
sage: A.minimal_generated_number()
3
sage: B.minimal_generated_number()
4

n_bounded_regions()

Return the number of (relatively) bounded regions.

OUTPUT:

An integer. The number of relatively bounded regions of the hyperplane arrangement.

EXAMPLES:

sage: A = hyperplane_arrangements.semiorder(3)
sage: A.n_bounded_regions()
7

n_hyperplanes()

Return the number of hyperplanes in the arrangement.

OUTPUT:

An integer.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([1,1,0], [2,3,-1], [4,5,3])
sage: A.n_hyperplanes()
3
sage: len(A)    # equivalent
3

n_regions()

The number of regions of the hyperplane arrangement.

OUTPUT:

An integer.

EXAMPLES:

sage: A = hyperplane_arrangements.semiorder(3)
sage: A.n_regions()
19

orlik_solomon_algebra(base_ring=None, ordering=None)

Return the Orlik-Solomon algebra of self.

INPUT:

• base_ring – (default: the base field of self) the ring over which the Orlik-Solomon algebra will be defined
• ordering – (optional) an ordering of the ground set

EXAMPLES:

sage: P.<x,y,z> = HyperplaneArrangements(QQ)
sage: A = P(x, y, z, x+y+z, 2*x+y+z, 2*x+3*y+z, 2*x+3*y+4*z)
sage: A.orlik_solomon_algebra()
Orlik-Solomon algebra of Linear matroid of rank 3 on 7 elements
represented over the Rational Field
sage: A.orlik_solomon_algebra(base_ring=ZZ)
Orlik-Solomon algebra of Linear matroid of rank 3 on 7 elements
represented over the Rational Field

plot(**kwds)

Plot the hyperplane arrangement.

OUTPUT:

A graphics object.

EXAMPLES:

sage: L.<x, y> = HyperplaneArrangements(QQ)
sage: L(x, y, x+y-2).plot()
Graphics object consisting of 3 graphics primitives

poincare_polynomial()

Return the Poincaré polynomial of the hyperplane arrangement.

OUTPUT:

The Poincaré polynomial in $$\QQ[x]$$.

EXAMPLES:

sage: a = hyperplane_arrangements.coordinate(2)
sage: a.poincare_polynomial()
x^2 + 2*x + 1

rank()

Return the rank.

OUTPUT:

The dimension of the span of the normals to the hyperplanes in the arrangement.

EXAMPLES:

sage: H.<x,y,z> = HyperplaneArrangements(QQ)
sage: A = H([[0, 1, 2, 3],[-3, 4, 5, 6]])
sage: A.dimension()
3
sage: A.rank()
2

sage: B = hyperplane_arrangements.braid(3)
sage: B.hyperplanes()
(Hyperplane 0*t0 + t1 - t2 + 0,
Hyperplane t0 - t1 + 0*t2 + 0,
Hyperplane t0 + 0*t1 - t2 + 0)
sage: B.dimension()
3
sage: B.rank()
2

sage: p = polytopes.simplex(5, project=True)
sage: H = p.hyperplane_arrangement()
sage: H.rank()
5

region_containing_point(p)

The region in the hyperplane arrangement containing a given point.

The base field must have characteristic zero.

INPUT:

• p – point

OUTPUT:

A polyhedron. A ValueError is raised if the point is not interior to a region, that is, sits on a hyperplane.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([(1,0), 0], [(0,1), 1], [(0,1), -1], [(1,-1), 0], [(1,1), 0])
sage: A.region_containing_point([1,2])
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 2 vertices and 2 rays

regions()

Return the regions of the hyperplane arrangement.

The base field must have characteristic zero.

OUTPUT:

A tuple containing the regions as polyhedra.

The regions are the connected components of the complement of the union of the hyperplanes as a subset of $$\RR^n$$.

EXAMPLES:

sage: a = hyperplane_arrangements.braid(2)
sage: a.regions()
(A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex, 1 ray, 1 line)

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H(x, y+1)
sage: A.regions()
(A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays)

sage: chessboard = []
sage: N = 8
sage: for x0 in range(N+1):
....:     for y0 in range(N+1):
....:         chessboard.extend([x-x0, y-y0])
sage: chessboard = H(chessboard)
sage: len(chessboard.bounded_regions())   # long time, 359 ms on a Core i7
64

restriction(hyperplane)

Return the restriction to a hyperplane.

INPUT:

• hyperplane – a hyperplane of the hyperplane arrangement

OUTPUT:

The restriction of the hyperplane arrangement to the given hyperplane.

EXAMPLES:

sage: A.<u,x,y,z> = hyperplane_arrangements.braid(4);  A
Arrangement of 6 hyperplanes of dimension 4 and rank 3
sage: H = A[0];  H
Hyperplane 0*u + 0*x + y - z + 0
sage: R = A.restriction(H);  R
Arrangement <x - z | u - x | u - z>
sage: D = A.deletion(H);  D
Arrangement of 5 hyperplanes of dimension 4 and rank 3
sage: ca = A.characteristic_polynomial()
sage: cr = R.characteristic_polynomial()
sage: cd = D.characteristic_polynomial()
sage: ca
x^4 - 6*x^3 + 11*x^2 - 6*x
sage: cd - cr
x^4 - 6*x^3 + 11*x^2 - 6*x

sign_vector(p)

Indicates on which side of each hyperplane the given point $$p$$ lies.

The base field must have characteristic zero.

INPUT:

• p – point as a list/tuple/iterable

OUTPUT:

A vector whose entries are in $$[-1, 0, +1]$$.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([(1,0), 0], [(0,1), 1]);  A
Arrangement <y + 1 | x>
sage: A.sign_vector([2, -2])
(-1, 1)
sage: A.sign_vector((-1, -1))
(0, -1)

unbounded_regions()

Return the relatively bounded regions of the arrangement.

OUTPUT:

Tuple of polyhedra. The regions of the arrangement that are not relatively bounded. It is assumed that the arrangement is defined over the rationals.

EXAMPLES:

sage: A = hyperplane_arrangements.semiorder(3)
sage: B = A.essentialization()
sage: B.n_regions() - B.n_bounded_regions()
12
sage: B.unbounded_regions()
(A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 3 vertices and 1 ray,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays,
A 2-dimensional polyhedron in QQ^2 defined as the convex hull of 1 vertex and 2 rays)

union(other)

The union of self with other.

INPUT:

• other – a hyperplane arrangement or something that can be converted into a hyperplane arrangement

OUTPUT:

A new hyperplane arrangement.

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([1,2,3], [0,1,1], [0,1,-1], [1,-1,0], [1,1,0])
sage: B = H([1,1,1], [1,-1,1], [1,0,-1])
sage: A.union(B)
Arrangement of 8 hyperplanes of dimension 2 and rank 2
sage: A | B   # syntactic sugar
Arrangement of 8 hyperplanes of dimension 2 and rank 2


A single hyperplane is coerced into a hyperplane arrangement if necessary:

sage: A.union(x+y-1)
Arrangement of 6 hyperplanes of dimension 2 and rank 2
Arrangement of 6 hyperplanes of dimension 2 and rank 2

sage: P.<x,y> = HyperplaneArrangements(RR)
sage: C = P(2*x + 4*y + 5)
sage: C.union(A)
Arrangement of 6 hyperplanes of dimension 2 and rank 2

varchenko_matrix(names='h')

Return the Varchenko matrix of the arrangement.

Let $$H_1, \ldots, H_s$$ and $$R_1, \ldots, R_t$$ denote the hyperplanes and regions, respectively, of the arrangement. Let $$S = \QQ[h_1, \ldots, h_s]$$, a polynomial ring with indeterminate $$h_i$$ corresponding to hyperplane $$H_i$$. The Varchenko matrix is the $$t \times t$$ matrix with $$i,j$$-th entry the product of those $$h_k$$ such that $$H_k$$ separates $$R_i$$ and $$R_j$$.

INPUT:

• names – string or list/tuple/iterable of strings. The variable names for the polynomial ring $$S$$.

OUTPUT:

The Varchenko matrix.

EXAMPLES:

sage: a = hyperplane_arrangements.coordinate(3)
sage: v = a.varchenko_matrix();  v
[    1    h2    h1]
[   h2     1 h1*h2]
[   h1 h1*h2     1]
sage: factor(det(v))
(h2 - 1) * (h2 + 1) * (h1 - 1) * (h1 + 1)

vertices(exclude_sandwiched=False)

Return the vertices.

The vertices are the zero-dimensional faces, see face_vector().

INPUT:

• exclude_sandwiched – boolean (default: False). Whether to exclude hyperplanes that are sandwiched between parallel hyperplanes. Useful if you only need the convex hull.

OUTPUT:

The vertices in a sorted tuple. Each vertex is returned as a vector in the ambient vector space.

EXAMPLES:

sage: A = hyperplane_arrangements.Shi(3).essentialization()
sage: A.dimension()
2
sage: A.face_vector()
(6, 21, 16)
sage: A.vertices()
((-2/3, 1/3), (-1/3, -1/3), (0, -1), (0, 0), (1/3, -2/3), (2/3, -1/3))
sage: point2d(A.vertices(), size=20) + A.plot()
Graphics object consisting of 7 graphics primitives

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: chessboard = []
sage: N = 8
sage: for x0 in range(N+1):
....:     for y0 in range(N+1):
....:         chessboard.extend([x-x0, y-y0])
sage: chessboard = H(chessboard)
sage: len(chessboard.vertices())
81
sage: chessboard.vertices(exclude_sandwiched=True)
((0, 0), (0, 8), (8, 0), (8, 8))

whitney_data()

Return the Whitney numbers.

OUTPUT:

A pair of integer matrices. The two matrices are the doubly-indexed Whitney numbers of the first or second kind, respectively. The $$i,j$$-th entry is the $$i,j$$-th doubly-indexed Whitney number.

EXAMPLES:

sage: A = hyperplane_arrangements.Shi(3)
sage: A.whitney_data()
(
[  1  -6   9]  [ 1  6  6]
[  0   6 -15]  [ 0  6 15]
[  0   0   6], [ 0  0  6]
)

whitney_number(k, kind=1)

Return the k-th Whitney number.

If kind=1, this number is obtained by summing the Möbius function values $$mu(0, x)$$ over all $$x$$ in the intersection poset with $$\mathrm{rank}(x) = k$$.

If kind=2, this number is the number of elements $$x, y$$ in the intersection poset such that $$x \leq y$$ with ranks $$i$$ and $$j$$, respectively.

See [GZ1983] for more details.

INPUT:

• k – integer
• kind – 1 or 2 (default: 1)

OUTPUT:

Integer. The k-th Whitney number.

EXAMPLES:

sage: A = hyperplane_arrangements.Shi(3)
sage: A.whitney_number(0)
1
sage: A.whitney_number(1)
-6
sage: A.whitney_number(2)
9
sage: A.characteristic_polynomial()
x^3 - 6*x^2 + 9*x
sage: A.whitney_number(1,kind=2)
6
sage: p = A.intersection_poset()
sage: r = p.rank_function()
sage: len([i for i in p if r(i) == 1])
6

class sage.geometry.hyperplane_arrangement.arrangement.HyperplaneArrangements(base_ring, names=())

Hyperplane arrangements.

For more information on hyperplane arrangements, see sage.geometry.hyperplane_arrangement.arrangement.

INPUT:

• base_ring – ring; the base ring
• names – tuple of strings; the variable names

EXAMPLES:

sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: x
Hyperplane x + 0*y + 0
sage: x + y
Hyperplane x + y + 0
sage: H(x, y, x-1, y-1)
Arrangement <y - 1 | y | x - 1 | x>

Element
ambient_space()

Return the ambient space.

The ambient space is the parent of hyperplanes. That is, new hyperplanes are always constructed internally from the ambient space instance.

EXAMPLES:

sage: L.<x, y> = HyperplaneArrangements(QQ)
sage: L.ambient_space()([(1,0), 0])
Hyperplane x + 0*y + 0
sage: L.ambient_space()([(1,0), 0]) == x
True

base_ring()

Return the base ring.

OUTPUT:

The base ring of the hyperplane arrangement.

EXAMPLES:

sage: L.<x,y> = HyperplaneArrangements(QQ)
sage: L.base_ring()
Rational Field

change_ring(base_ring)

Return hyperplane arrangements over a different base ring.

INPUT:

• base_ring – a ring; the new base ring.

OUTPUT:

A new HyperplaneArrangements instance over the new base ring.

EXAMPLES:

sage: L.<x,y> = HyperplaneArrangements(QQ)
sage: L.gen(0)
Hyperplane x + 0*y + 0
sage: L.change_ring(RR).gen(0)
Hyperplane 1.00000000000000*x + 0.000000000000000*y + 0.000000000000000

gen(i)

Return the $$i$$-th coordinate hyperplane.

INPUT:

• i – integer

OUTPUT:

A linear expression.

EXAMPLES:

sage: L.<x, y, z> = HyperplaneArrangements(QQ);  L
Hyperplane arrangements in 3-dimensional linear space over Rational Field with coordinates x, y, z
sage: L.gen(0)
Hyperplane x + 0*y + 0*z + 0

gens()

Return the coordinate hyperplanes.

OUTPUT:

A tuple of linear expressions, one for each linear variable.

EXAMPLES:

sage: L = HyperplaneArrangements(QQ, ('x', 'y', 'z'))
sage: L.gens()
(Hyperplane x + 0*y + 0*z + 0,
Hyperplane 0*x + y + 0*z + 0,
Hyperplane 0*x + 0*y + z + 0)

ngens()

Return the number of linear variables.

OUTPUT:

An integer.

EXAMPLES:

sage: L.<x, y, z> = HyperplaneArrangements(QQ);  L
Hyperplane arrangements in 3-dimensional linear space over Rational Field with coordinates x, y, z
sage: L.ngens()
3