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.
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.n_cube(3).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 QQ^5 defined as the convex hull of 10 vertices
The characteristic polynomial is a basic invariant of a hyperplane arrangement. It is defined as
where the sum is \(P\) is the intersection_poset() of the arrangement and \(\mu\) is the Moebius 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.
TESTS:
sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: h = H([(1, 106), 106266], [(83, 101), 157866], [(111, 110), 186150], [(453, 221), 532686],
....: [(407, 237), 516882], [(55, 32), 75620], [(221, 114), 289346], [(452, 115), 474217],
....: [(406, 131), 453521], [(28, 9), 32446], [(287, 19), 271774], [(241, 35), 244022],
....: [(231, 1), 210984], [(185, 17), 181508], [(23, -8), 16609])
sage: h.n_regions()
85
sage: H()
Empty hyperplane arrangement of dimension 2
sage: Zero = HyperplaneArrangements(QQ)
sage: Zero
Hyperplane arrangements in 0-dimensional linear space over Rational Field with coordinate
sage: Zero()
Empty hyperplane arrangement of dimension 0
sage: Zero.an_element()
Empty hyperplane arrangement of dimension 0
AUTHORS:
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 [RS] on hyperplane arrangements.
REFERENCES:
[RS] | Stanley, Richard: Hyperplane Arrangements, Geometric Combinatorics (E. Miller, V. Reiner, and B. Sturmfels, eds.), IAS/Park City Mathematics Series, vol. 13, American Mathematical Society, Providence, RI, 2007, pp. 389-496. |
Bases: sage.structure.element.Element
An element in a hyperplane arrangement.
Warning
You should never create HyperplaneArrangementElement instances directly, always use the parent.
The union of self with other.
INPUT:
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
sage: A.add_hyperplane(x+y-1) # alias
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
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.
See also
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
Return hyperplane arrangement over the new base ring.
INPUT:
OUTPUT:
The hyperplane arrangement obtained by changing the base field, as a new hyperplane arrangement.
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>
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
TESTS:
sage: H.<s,t,u,v> = HyperplaneArrangements(QQ)
sage: m = matrix([(0, -1, 0, 1, -1), (0, -1, 1, -1, 0), (0, -1, 1, 0, -1),
....: (0, 1, 0, 0, 0), (0, 1, 0, 1, -1), (0, 1, 1, -1, 0), (0, 1, 1, 0, -1)])
sage: R.<x> = QQ[]
sage: expected_charpoly = (x - 1) * x * (x^2 - 6*x + 12)
sage: for s in SymmetricGroup(4): # long time (about a second on a Core i7)
....: m_perm = [m.column(i) for i in [0, s(1), s(2), s(3), s(4)]]
....: m_perm = matrix(m_perm).transpose()
....: charpoly = H(m_perm.rows()).characteristic_polynomial()
....: assert charpoly == expected_charpoly
Return the cone over the hyperplane arrangement.
INPUT:
OUTPUT:
A new yperplane 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 \([0, 1, 0, \ldots, 0]\).
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)
Return the hyperplane arrangement obtained by removing h.
INPUT:
OUTPUT:
A new hyperplane arrangement with the given hyperplane(s) h removed.
See also
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>
TESTS:
sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([0,1,0], [1,0,1], [-1,0,1], [0,1,-1], [0,1,1])
sage: h = H([0,4,0])
sage: A.deletion(h)
Arrangement <y - 1 | y + 1 | x - y | x + y>
sage: l = H([1,2,3])
sage: A.deletion(l)
Traceback (most recent call last):
...
ValueError: hyperplane is not in the arrangement
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
Return the number of hyperplanes separating the two regions.
INPUT:
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
Return the generating function for the number of hyperplanes at given distance.
INPUT:
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
Return the \(i,j\)-th doubly-indexed Whitney number.
If kind=1, this number is obtained by adding the Moebius 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:
OUTPUT:
Integer. The \((i,j)\)-th entry of the kind Whitney number.
See also
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:
[GZ] | Greene; Zaslavsky “On the Interpretation of Whitney Numbers Through Arrangements of Hyperplanes, Zonotopes, Non-Radon Partitions, and Orientations of Graphs” Transactions of the American Mathematical Society, Vol. 280, No. 1. (Nov., 1983), pp. 97-126. |
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
TESTS:
sage: b = hyperplane_arrangements.coordinate(2)
sage: b.is_essential()
True
sage: b.essentialization() is b
True
Return the face vector.
OUTPUT:
A vector of integers.
The \(d\)-th entry is the number of faces of dimension \(d\). A face is is the intersection of a region with a hyperplane of the arrangehment.
EXAMPLES:
sage: A = hyperplane_arrangements.Shi(3)
sage: A.face_vector()
(0, 6, 21, 16)
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:
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
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
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
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
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.
See also
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
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
Test whether the hyperplane separates the given regions.
INPUT:
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
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
TESTS:
sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([(1,1),0], [(2,3),-1], [(4,5),3])
sage: B = A.change_ring(FiniteField(7))
sage: B.n_bounded_regions()
Traceback (most recent call last):
...
TypeError: base field must have characteristic zero
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
The number of regions of the hyperplane arrangement.
OUTPUT:
An integer.
EXAMPLES:
sage: A = hyperplane_arrangements.semiorder(3)
sage: A.n_regions()
19
TESTS:
sage: H.<x,y> = HyperplaneArrangements(QQ)
sage: A = H([(1,1), 0], [(2,3), -1], [(4,5), 3])
sage: B = A.change_ring(FiniteField(7))
sage: B.n_regions()
Traceback (most recent call last):
...
TypeError: base field must have characteristic zero
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
Return the Poincare polynomial of the hyperplane arrangement.
OUTPUT:
The Poincare polynomial in \(\QQ[x]\).
EXAMPLES:
sage: a = hyperplane_arrangements.coordinate(2)
sage: a.poincare_polynomial()
x^2 + 2*x + 1
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.n_simplex(5)
sage: H = p.hyperplane_arrangement()
sage: H.rank()
5
The region in the hyperplane arrangement containing a given point.
The base field must have characteristic zero.
INPUT:
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
TESTS:
sage: A = H([(1,1),0], [(2,3),-1], [(4,5),3])
sage: B = A.change_ring(FiniteField(7))
sage: B.region_containing_point((1,2))
Traceback (most recent call last):
...
ValueError: base field must have characteristic zero
sage: A = H([(1,1),0], [(2,3),-1], [(4,5),3])
sage: A.region_containing_point((1,-1))
Traceback (most recent call last):
...
ValueError: point sits on a hyperplane
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, y0 in CartesianProduct(range(N+1), 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
Return the restriction to a hyperplane.
INPUT:
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
See also
Indicates on which side of each hyperplane the given point \(p\) lies.
The base field must have characteristic zero.
INPUT:
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)
TESTS:
sage: H.<x,y> = HyperplaneArrangements(GF(3))
sage: A = H(x, y)
sage: A.sign_vector([1, 2])
Traceback (most recent call last):
...
ValueError: characteristic must be zero
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.
See also
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)
The union of self with other.
INPUT:
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
sage: A.add_hyperplane(x+y-1) # alias
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
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:
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)
Return the vertices.
The vertices are the zero-dimensional faces, see face_vector().
INPUT:
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, y0 in CartesianProduct(range(N+1), 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))
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]
)
Return the k-th Whitney number.
If kind=1, this number is obtained by summing the Moebius 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 [GZ] for more details.
INPUT:
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
Bases: sage.structure.parent.Parent, sage.structure.unique_representation.UniqueRepresentation
Hyperplane arrangements.
For more information on hyperplane arrangements, see sage.geometry.hyperplane_arrangements.arrangement.
INPUT:
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>
alias of HyperplaneArrangementElement
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
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
Return hyperplane arrangements over a different base ring.
INPUT:
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
TESTS:
sage: L.change_ring(QQ) is L
True
Return the \(i\)-th coordinate hyperplane.
INPUT:
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
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)
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