Download MATH 890 HOMEWORK 2 (1) 1. Background Suppose P is a

MATH 890
HOMEWORK 2
DAVID MEREDITH
(1) 1. Background
Suppose P is a polytope with vertex set V. The combinatorial structure
of P , the structure captured by the face graph of P , is determined by the
subsets of V consisting of vertex sets of facets of P . Each of these subsets
is called a facet of V. The faces of V, subsets of V consisting of vertex sets
of faces of P , are all possible non-empty intersections of the facets of V.
Conversely, given a set V and a collection of subsets covering V, we
can ask if there is a polytope whose vertices can be put into one-to-one
correspondence with the elements of V such that the facets of the polytope
correspond to the subsets. In general there is no such polytope, and it is
not obvious in general how to define a subset structure of V to guarantee
that the subsets can be realized as the facets of a polytope.
Definition 1. Suppose P and Q are polytopes with vertex sets V and W.A
bijection φ : V → W is a combinatorial equivalence P → Q if F ⊂ V is a
facet of V implies φ(F ) is a facet of W.
Proposition 1. Using the notation of the definition, suppose φ : P → Q
is a combinatorial equivalence. If G is a facet of Q, there exists a facet F
of P such that φ(F ) = G.
Proof. A combinatorial equivalence P → Q, which is a map sending vertices
of P to vertices of Q, can be extended to a homeomorphism P → Q that
maps each facet F of P homeomorphically to the facet φ(F ). Since both P
and Q are homeomorphic to a sphere with the facets covering the sphere,
under a combinatorial equivalence P → Q every facet of Q is the image of
a facet of P .
Suppose V is the set of vertices for a full-dimensional polytope in Rd .
The points of V are vectors in Rd that we will consider as column vectors.
A subset F ⊂ V is a facet if and only if
(a) F spans an affine set H of dimension d − 1.
(b) All vertices not in F are on the same side of H
2. Realization Spaces
Let P be a full-dimensional polytope in Rd and let F be a flag of P .
That is, F = (u0 , . . . , ud ) is an ordered set of vertices of P such that each
leading subset (u0 , . . . , ue ) spans an e-dimensional face of P . After an affine
coordinate change, we may assume u0 = 0 and ui = ei , a standard basis
Date: March 9, 2012.
1
A set is covered by a
collection of subsets
if the union of the
subsets is the set.
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DAVID MEREDITH
vector. Suppose the remaining vertices of P are v1 , . . . , vn . The realization
space R(P, F) ⊂ Rnd is the set of points (w1 , . . . , wn ) ∈ Rnd such that
(a) each segment wi ∈ Rd
(b) the points (u0 , . . . , ud , w1 , . . . , wn ) are the vertices of a polytope Q
(c) the assignment ui → ui and vj → wj is a combinatorial equivalence
P → Q. 
x1i
 .. 
Let wi =  . .
xdi
Theorem 1. The set of points u1 , . . . , ud , w1 , . . . , wn are the vertices of a
polytope Q combinatorially equivalent to P (under the assignment ui → ui
and vj → wj ) if and only if the variables xji satisfy certain polynomial
equalities and inequalities with integer coefficients.
Proof. Let V = (u0 , . . . , ud , v1 , . . . , vn ) be the vertices of P and W =
(u0 , . . . , ud , w1 , . . . , wn ) be the vertices of Q. We must show that W are
the vertices of a polytope Q and F 0 = φ(F) is a facet of Q for all facets F
of P if and only if the variables xji satisfy certain polynomial equalities and
inequalities with integer coefficients. It suffices to show for each facet F of
P that F 0 spans a hyperplane; and all points of W not in F 0 are on the
same side of F 0 if and only if the variables xji satisfy certain polynomial
equalities and inequalities with integer coefficients.
(a) Let F = (x0 , . . . , xm ) be a facet of.P , and let x0i = φ(xi ). To show
let yi0 = x0i − x00 and
that F 0 = (x00 , . . . , x0n ) spans
0a hyperplane,
0
define a d × n matrix M = y1 · · · yn . The points x0i span a d − 1
dimensional subspace if and only if every d × d submatrix of M has
zero determinant and (at least) one d − 1 × d − 1 submatrix has a
non-zero determinant. One d − 1 × d − 1 subdeterminant is non-zero if
and only if the sum of the squares of the d − 1 × d − 1 subdeterminants
is positive. Since these determinants are polynomials in the xji with
integer coefficients (because the entries in the vectors xi are either
variables xji or integers), the x0i span a d − 1 dimensional subspace if
and only if some polynomial equalities and inequalities are satisfied.
(b) To show all the points x0 ∈ W\F 0 are on the same side of the hyperplane spanned by F 0 if and only if some polynomial equations
and inequalities with integer coefficients are satisfied, start by fixing
a flag x0 , . . . , xd−1 in the facet F. Let x0i = φ(xi ), yi = xi − x0 and
yi0 = x0i − x00 . For any point x0 ∈ W\F 0 . let x0 =
φ(x). Necessar
0
0
0
y1 · · · yd−1 y and
ily x ∈
V\F.
Let
y
=
x
−
x
,
y
=
x
−
x
,
M
=
0
0
0
0
y0 . Since every point x ∈ V\F is on the same side of
M = y1 · · · yd−1
F, all the determinants det(M ) are non-zero with the same sign. Since
P and Q share a flag {ui } (so Q is not a mirror image of P ), all x0 are
on the same side of F 0 if and only if sign(det(M )) = sign(det(M 0 )).
Since det(M 0 ) is a polynomial with integer coefficients in xji , we conclude that all the points x0 ∈ W\F 0 are on the same side of the hyperplane spanned by F 0 if and only if some polynomial equations and
inequalities with integer coefficients are satisfied.
MATH 890—HW 2
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(2) Let x0 = (1, 0, 0), x1 = (0, 1, 0), x2 = (−1, −1, 0), x3 = (0, 0, 1) and
x4 = (0, 0, −1). These points are the vertices of a bisimplex polytope P
in R3 , and the points V = (x0 , . . . ,x3) are a flag. The realization space
x
R(P, V) consists of all points x = y  where (x0 , . . . , x3 , x) also form a
z
bisimplex. This is the set of points x on the same side of the xy-plane
as x4 and below the three planes spanned by the three sides of P that lie
above the x-axis. The equations defining this set are
z<0
x+y+z <1
x − 2y + z < 1
−2x + y + z < 1
(3) Use the same points as above (I think the problem is slightly misstated on
the homework sheet). Let V 0 = (x1 , . . . , x4 ), which is 
also a flag of P . The
x
realization space R(P, V 0 ) consists of points x = y  where x is on the
z
same side of the planes spanned by x1 , x2 , x3 and x1 , x2 , x4 as x0 . The
equations are
2x − y − z > −1
2x − y + z > −1
Since R(P, V 0 ) is bounded by two planes while R(P, V) is bounded by four,
the two representation spaces are not affinely isomorphic.
(4) Let P be a full-dimensional polytope in a real vector space V , and let
H1 , . . . , Hn be affine forms on V defining distinct hyperplanes that meet the
interior of P . (Note: the intersections of the hyperplanes need not meet inside P. Some of them could even be parallel.) The union of the hyperplanes
divides P into a polytopal complex P consisting of d-dimensional polytopes
Ps1 ...sn = P ∩H1s1 ∩· · ·∩Hnsn , where si ∈ {+, −}. Thus there are potentially
2n polytopes in P, although some of them may be empty.
Pn The polytopal
complex P is regular. It has a convex function f (x) = i=1 |Hi (x)| whose
domains of linearity are the non-empty polytopes Ps1 ...sn .
Proof. On each polytope Ps1 ...sn each linear form Hi is either non-negative
or non-positive. Thus |Hi | = ±Hi on Ps1 ...sn , so f is linear on Ps1 ...sn .
It remains to show that f is convex and the domains of linearity do not
extend beyond the Ps1 ...sn .
To show that f is convex on P = |P|, it suffices to show that each
component |Hi (x) is convex (the sum of convex functions is convex). Let
x, y ∈ P and let a, b > 0, a + b = 1. Then
|Hi (ax + by)| = |aHi (x) + bHi (y)|
≤ |aHi (x)| + |bHi (y)|
= a|Hi (x)| + b|Hi (y)|
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DAVID MEREDITH
Finally, we must show that the Ps1 ...sn are the domains of linearity of f .
Suppose x, y ∈ P are not contained in the same polytopal subdivision.
Then for one index i we may assume Hi (x)P< 0 and Hi (y) > 0. By
the previous paragraph, the function g(x) = j6=i |Hj (x)| is convex, and
f (x) = g(x)+|Hi (x)|. We will show that f (x+y) < f (x)+f (y), thus showing that f is not linear on a domain not contained in one of the polytopal
components of P.
f (x + y) = g(x + y) + |Hi (x + y)|
≤ g(x) + g(y) + |Hi (x) + Hi (y)|
< g(x) + g(y) + |Hi (x)| + |Hi (y)|
= f (x) + f (y)
(5) Let P ⊂ V be a full-dimensional polytope in a vector space V with vertices
V. For each v ∈ V define Cv = R+ (P − v), a cone in V . Define
Cv∗ = {α ∈ V ∗ : α(x) ≥ 0, ∀x ∈ Cv }
= {α ∈ V ∗ : α(w − v) ≥ 0, ∀w ∈ P }
= {α ∈ V ∗ : α(w − v) ≥ 0, ∀w ∈ V\{v}}
= {α ∈ V ∗ : α(w − v) ≥ 0, ∀w ∈ V which are neighbors of v}
We will introduce a family of linear functionals on V ∗ and related subsets of V ∗ used in our proofs. Define the linear functional avw ∈ V ∗∗ by
avw (α) = α(w − v). Then define the hyperplane Hvw = ker(avw ) and the
+
= {α ∈ V ∗ : avw (α) ≥ 0}.
half-space Hvw
+
+
Lemma 1. Hvw
∩ Hwv
= Hvw = Hwv
Proof. Obvious
Proposition 2.
Cv∗
=
T
w∈V
w6=v
+
Hvw
is a polyhedral cone.
Proof of Proposition. Clearly Cv∗ is the stated intersection, from which it
follows that Cv∗ is a polyhedral cone.
Proposition 3. The collection Cv∗ for v ∈ V is a fan in V ∗ .
Proof. We must show that the intersection F = Cv∗1 ∩ · · · ∩ Cv∗n is a face of
Cv∗1 . But
F = Cv∗1 ∩ Hv1 v2 ∩ · · · ∩ Hv1 vn
∗
+
Since Cv1 ⊂ Hv1 vi , Cv∗1 ∩ Hv1 vi is a face of Cv∗1 . Thus the intersection F
is a face of Cv∗1 .
Proposition 4. F is complete.
Proof. Suppose α ∈ V ∗ . We must find v ∈ V such that α ∈ Cv∗ . Chose
v ∈ V such that α(v) ≤ α(w), all w ∈ V. Then α ∈ Cv∗ .
∗
Lemma 2. Suppose α ∈ Cv∗ ∩ Cw
. Then α(v) = α(w).
Proof. Since w ∈ Cv∗ , α(w) ≥ α(v). Similarly α(v) ≥ α(w) so α(w) =
α(v).
Proposition 5. F is projective.
MATH 890—HW 2
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Proof. We must produce a convex function f : V → R whose domains of
linearity are the cones Cv∗ . If α ∈ Cv∗ let f (α) = −α(v). The completeness
of F and the lemma above show that f is well-defined on V . It is obvious
that f is linear on Cv∗ , and since all the v ∈ V are distinct the domains
of linearity of f are precisely the cones Cv∗ . It remains to show that f is
convex.
∗
. Choose a, b ≥ 0 such that a + b = 1 and
Suppose α ∈ Cv∗ and β ∈ Cw
define γ = aα + bβ. Suppose γ ∈ Cu∗ . Then α(v) ≤ α(u) and β(w) ≤ β(u).
Thus f is convex:
f (aα + bβ) = −γ(u)
= a(−α(u)) + b(−β(u))
≤ a(−α(v)) + b(−β(w))
= af (α) + bf (β)
Definition 2. The projective fan F is the normal fan of the polytope P .
Proposition 6. Let F be a projective fan in V ∗ . Then there exists a
polytope P ⊂ V such that F is the normal fan of P .
Proof. Let f : V ∗ → R be the convex function associated with the projective fan F. For each cone C ∈ F, let −vC ∈ V ∗∗ = V be the unique vector
such that −vC |C = f |C. That is, α ∈ C if and only if f (α) = −α(vC ).
(The “if” statement follows from the fact that C is a domain of linearity
for F .) We must show that the vectors vC , C ∈ F, are the vertices of a
polytope P whose normal fan is F.
For each cone C ∈ F, define another cone:
Cv∗C = {α ∈ V ∗ : α(vC ) ≤ α(vC 0 ), ∀C 0 ∈ F}
It suffices to show for each cone C ∈ F that
(2.1)
Cv∗C ⊂ C
First we explain why 2.1 suffices to prove the proposition, then we prove
2.1. Suppose the vectors W = {vC : C ∈ F} span a polytope P with
vertices V. Then V ⊂ W. By the first part, the cones Cv∗C for vC ∈ V
form a projective fan F 0 in V ∗ , and by 2.1 each cone of F 0 is a subset of a
different cone of F. Since F 0 is complete, W = V and F 0 = F.
Finally we show that 2.1 holds. Suppose α ∈ Cv∗C . We must show α ∈ C
or, equivalently, f (α) = −α(vC ). First we show f (α) + α(vC ) ≤ 0.
Since α ∈ Cv∗C , α(vC ) ≤ α(vC 0 ) for all C 0 ∈ F. Since F is complete,
there exists C 0 ∈ F such that α ∈ C 0 . Then f (α) = −α(vC 0 ) and f (α) +
α(vC ) ≤ 0.
We conclude the proof by showing f (α) + α(vC ) ≥ 0. Suppose α ∈
C 0 . Since dim(C) = d, α is a linear combination of linear functionals
in C. We may choose linear functionals αP
i and βj in
P C together with
positive coefficients ai and bj such that α + i ai αi = j bj βj . Note that
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DAVID MEREDITH
α+
P
i
ai αi ∈ C. Thus:
−α(vC ) −
X
ai αi (vC ) = (α +
X
i
ai αi )(−vC )
i
= f (α +
X
ai αi )
i
≤ f (α) + f (
X
ai αi )
X
= −α(vC 0 ) −
ai αi (vC )
Thus −α(vC ) ≤ −α(vC 0 ) or α(vC ) + f (α) ≥ 0.
(6) Example showing that not every complete fan in R3 projective. (This example is similar to Bruns and Gubeladze, Polytopes, Rings and K-Theory,
pp. 35.)
Let A = (1, 0, 1), B = (0, 1, 1), C = (−1, −1, 1), A0 = (1, 0, −1),
0
B = (0, 1, −1) and C 0 = (−1, −1, −1). Let P be the simplical complex
consisting of triangles ABC, ABB 0 , AA0 B 0 , BCC 0 , BB 0 C 0 , CAA0 , CC 0 A0
and A0 B 0 C 0 . This complex is a cylinder with rectangular sides divided into
two triangles and a triangular top and bottom. Let F be the fan consisting
of cones generated by the triangles in P. The cone generated by the triangle ABC will be denoted CABC , and the other cones will be denoted in a
similar fashion. Since P surrounds the origin of R3 , F is a complete fan. It
remains to show that F is not projective. We must show that there cannot
exist a convex function f : R3 → R such that f is linear on each cone and
the cones are the maximal domains of linearity for f . That is, there cannot
exist a convex function f such that for each triangle T ∈ P,
(a) there exists a linear functional αT on R3 such that f |CT = αT |CT ; and
(b) If T and T 0 are different triangles then αT 6= αT 0
Suppose such a function f existed. We may assume f |A0 B 0 C 0 = 0. Let
D = (1/2, 1/2, 0) = 21 A + 12 B 0 = 12 A0 + 12 B. Since the edge AB 0 lies in
a domain of linearity for f , f (D) = 12 f (A). Since f is convex, f (D) =
f ( 21 A0 + 21 B) ≤ 12 f (B). Thus f (A) ≤ f (B). By similar arguments we have
f (A) ≤ f (B) ≤ f (C) ≤ f (A), so f (A) = f (B) = f (C).
f (A)
(x + y + z). We have
Consider the linear functional α(x, y, z) =
2
α(A) = f (A) = αABB 0 (A) = αAA0 B 0 (A)
α(B) = f (B) = αABB 0 (B)
α(B 0 ) = f (B 0 ) = αABB 0 (B 0 ) = αAA0 B 0 (B)
α(A0 ) = f (A0 ) = αAA0 B 0 (A0 )
Therefore α and αABB 0 agree on A, B and B 0 , so α = αABB 0 . Similarly
α = αAA0 B 0 . Thus αABB 0 = αAA0 B 0 , contradicting condition (b) above.
MATH 890—HW 2
Department of Mathematics
San Francisco State University
San Francisco, CA 94132
E-mail address: [email protected]
URL: http://online.sfsu.edu/~meredith
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