Universit`
a degli Studi di Pisa
FACOLT `
A DI SCIENZE MATEMATICHE, FISICHE E NATURALI
Corso di Laurea Magistrale in Matematica
Tesi di laurea magistrale
On convex PL-manifolds as regular images of
R
n
Candidato:
Antonio Carbone
Relatori:
Prof. Jos´
e F. Fernando
Prof. Jos´
e M. Gamboa
Correlatore:
Abstract
Modern real algebraic geometry was born with Tarski's article [26], where it is proved that the image of a semialgebraic set under a polynomial map is a semialgebraic set. We are interested in studying what might be the inverse problem to Tarski's result, that is to represent semialgebraic sets as polynomial or regular images of Euclidean spaces. In June 1990, during the Oberwolfach reelle algebraische geometrie week [20], Gamboa proposed:
Problem 1. To characterize those semialgebraic subsets of Rm that are either
polynomial or regular images of Euclidean spaces.
As speci c example of open questions he stated in Oberwolfach: 1. Is the set {x2+ y2>1}a polynomial image of R2?
2. Is the open quadrant {x > 0, y > 0} a regular image of R2?
In 2002 Fernando and Gamboa answered both these questions in [8]. It con-stituted the starting point of the systematic study of the problem of representing semialgebraic sets as polynomial or regular images of Euclidean spaces.
Them, joinly with Ueno, have attempted to understand better polynomial and regular images of Rn, in the last two decades, with the following targets:
• To nd obstructions to be either polynomial or regular images [8], [9], [11], [17] and [28].
• To prove (costructively) that large families of semialgebraic sets with piecewise linear boundary (convex polyhedra, their interiors, their com-plements and the comcom-plements of their interiors) are either polynomial or regular images of Euclidean spaces [10], [14], [15], [16], [19] and [27]. Given a semialgebraic set S ⊂ Rm, we introduce the following two invariants:
p(S) := inf
n≥1{there exists a polynomial mapf : R n→
Rmsuch that f(Rn) = S},
r(S) := inf
n≥1{there exists a regular mapf : R
n →Rmsuch that f(Rn) = S}.
If S is not representable as a polynomial image (resp. a regular image) of any Rn, then we set p(S) := +∞ (resp. r(S) := +∞).
In this dissertation we will focus on the regular case, and we will provide a complete computation, following the works of Fernando, Gamboa and Ueno;
ment of their interiors.
Given a d-dimensional convex polyhedron K ⊂ Rn, with d ≥ 1, denote Int K
its interior as topological manifold with boundary. The values of the invariants r(K), r(Int K), r(Rn\ K)and r(Rn\Int K) are collected in the following table
K bounded Kunbounded r(·) d=1 d>1 d=1 d>1 K 1 d 1 d Int K 2 d 2 d Rn\ K +∞ n 2 n,+∞∗ Rn\Int K +∞ n 1 n,+∞∗
(∗) We have r(Rn\ K) = r(Rn\Int K) = +∞ if and only if K disconnects Rn.
The closed ball Bn := Bn(0, 1) = {x ∈ Rn : kxk ≤ 1} ⊂ Rn can be
seen as the limit of a suitable family of bounded convex polyhedra (consider triangulations of the closed ball). It is thus natural to pose the same questions we approached for convex polyhedra. We will provide a complete answer for the closed ball Bn, the open ball Bn:= {x ∈Rn: kxk < 1}and their complements.
We will prove that for each n ≥ 2, the following equalities hold r(Bn) = r(Bn) = r(Rn\ Bn) = r(Rn\ Bn) = n.
Moreover we will compute the invariant r for the spheres Sn−1:= ∂B
n, showing
that in this case we have
r(Sn−1) = n − 1.
The computation of the invariants r(Sn−1)and r(Rn\ B
n)is made for the rst
time in this dissertation, completing the computation of the invariant r in the limit case of balls and sets derived by them.
Contents
1 Real algebraic geometry 1
1.1 Real algebraic geometry . . . 1
1.2 Semialgebraic geometry . . . 3
1.3 Dimension of semialgebraic sets . . . 5
2 Presentation of the problem 9 2.1 The invariants p and r . . . 10
2.2 The purpose of this dissertation . . . 10
3 Polynomial and regular images of Rn 13 3.1 Obstructions for the representation . . . 13
3.2 Obstruction for dimension two . . . 15
3.3 A list of examples . . . 17
3.4 Semialgebraic subsets of R . . . 18
4 Open quadrant 21 4.1 Polynomial case . . . 21
4.2 Regular case . . . 23
5 Generalities on convex polyhedra 25 5.1 Convex sets in Euclidean space . . . 25
5.2 Convex polyhedra . . . 27
6 Interior of convex polyhedra as regular images of Rn 31 6.1 Reduction to bounded polyhedra . . . 31
6.2 ABT -partition of the boundary of a convex polyhedron . . . 34
6.3 Interior of a convex polyhedron . . . 41
7 Convex polyhedra as regular images of Rn 47 7.1 Convex polyhedron . . . 47
7.2 Proof of the technical lemma . . . 49
8 Complements of convex polyhedra as regular images of Rn 57 8.1 Complement of basic nowhere dense semialgebraic sets . . . 57
8.2 Trimming tools of type I . . . 60
8.3 Technical results on trimming of type I . . . 60
9 Complement of the interior of convex polyhedra as regular
im-ages of Rn 73
9.1 Trimming tools of type II . . . 73 9.2 Technical results on trimming of type II . . . 74 9.3 Complement of the interior of a convex polyhedron . . . 74
10 Limit cases: balls 79
10.1 The ball Bn . . . 79
10.2 An extra example: the cylinder Cn . . . 83
Chapter 1
Real algebraic geometry
In this chapter, we will provide a brief introduction to real algebraic and semial-gebraic geometry, focusing on the results needed in order to keep the exposition self contained. Since the purpose of this dissertation is to investigate regular im-ages of Euclidean spaces Rn, we will restrict the introduction to real algebraic
geometry with coe cients in the eld R of real numbers, instead of an arbi-trary real closed eld. Standard references for real algebraic and semialgebraic geometry are [2] and [5].
1.1 Real algebraic geometry
The main goal of algebraic geometry is the study of those subsets of Kn de ned
as common zero sets of polynomials in K[x1, . . . , xn], where K is a eld. The
easiest situation corresponds to choose K algebraically closed. In particular, the case K := C has deserved major attention, and nowadays complex algebraic geometry is a very one of the most sophisticated parts of mathematics. As shown afterwards, the non algebraically closed case often leads to surprising results (at least for classical algebraic geometers). In particular, when we work with real coe cients we can no longer follow blindly our standard C-geometrical intuition, and some care is needed. The reason is that the dictionary: algebraic geometry-commutative algebra does not work so nicely in the non algebraically closed case. We will begin with some standard de nitions and examples to show the weird behaviour that can appear working with real coe cients. In what follows, K is either the eld R or C.
De nition 2. A subset V ⊂ Knis called an algebraic set if it can be represented
as
V = V(S) := {x ∈Kn: p(x) = 0 ∀p ∈ S}, where S ⊂ K[x] := K[x1, . . . , xn]is a non-empty subset.
If I is the ideal generated by a non-empty subset S ⊂ K[x], it is straight-forward to see that V(I) = V(S). Hilbert's basis theorem (see [1] as standard reference) asserts that K[x] is a Noetherian ring; in particular each ideal I ⊂ K[x] is nitely generated, and every algebraic set V ⊂ Kn can be represented as the
Remark 3. Recall some standard facts about algebraic sets.
1. The ideal of a subset V ⊂ Kn is I(V ) := {f ∈ K[x] : f(x) = 0 ∀x ∈ V }.
2. Every real algebraic set V ⊂ Rn can be represented as the zero set of a
single polynomial. In fact, if V := {p1= 0, . . . , pk = 0}then V = {P = 0},
where P := p2
1+ . . . + p2k.
3. Algebraic sets in Kn are the closed sets for a topology on Kn, called the
Zariski topology.
4. Every Zariski closed set in Kn is also closed in the standard Euclidean
topology because polynomial functions are continuous with respect to the Euclidean topology.
5. The Zariski topology of Kn is not Hausdor, but the points are closed.
De nition 4. An algebraic set V ⊂ Knis called reducible if there exist algebraic
subsets V1( V and V2( V such that V = V1∪ V2. Otherwise it is said that V
is irreducible.
Remark 5. The algebraic set V ⊂ Knis irreducible if and only if I(V ) ⊂ K[x]
is a prime ideal. In fact V = V1∪ V2 with V1, V26= V if and only if there exist
f1∈ I(V1)and f2∈ I(V2)such that f1, f2∈ I(V )/ and f1f2∈ I(V ).
Proposition 6 (Decomposition into irreducible components). Every algebraic set V ⊂ Kn can be decomposed into a nite union of irreducible algebraic sets
V = V1∪ . . . ∪ Vk,
where Vi6⊂ Vj if i 6= j. In addition, up to reordering the indices, this
decompo-sition is unique. Proof. See [2].
Examples 7. The following examples show that real algebraic sets present a wilder behaviour than complex algebraic sets.
1. The ideal I := (x2(x − 1)2+ y2) ⊂R[x, y] is prime because x2(x − 1)2+ y2
is an irreducible polynomial, but the algebraic set V(I) = {(0, 0), (1, 0)} ⊂R2
is reducible.
2. I := ((xy − 1)2+ x2) ⊂R[x, y] is a proper prime ideal, but V(I) = ∅.
3. The irreducible algebraic set V(y2−x3+x2) ⊂R2is not connected. Indeed
one of its connected components is an isolated point.
4. The set V(y2− x3+ x) ⊂ R2 is an irreducible algebraic set with two
connected components. One of them is a bounded set.
5. The circle S1:= {x2+ y2− 1 = 0} ⊂R2is a bounded irreducible algebraic
1.2. Semialgebraic geometry
Figure 1.1: The cubic curves y2− x3+ x2= 0(left) and y2− x3+ x = 0(right)
1.2 Semialgebraic geometry
It is well known that C does not admit any order structure compatible with its eld structure. On the other hand R has a (unique) standard order structure. So, to understand the topology of the subsets of Rn it is natural to use this
order structure. We will provide three examples to explain it, and to see why this is the (best) way to investigate real objects (see [24] for a more complete explanation).
Examples 8. 1. Consider the family of polynomials Fa,b:= {t2+ at + b} ⊂R[t].
Then, V(t2+ at + b) ⊂R is nonempty if and only if a2− 4b ≥ 0.
2. A nonsingular elliptic curve {y2 = x3+ ax + b} ⊂R2 is connected if and
only if its discriminant ∆ := −16(4a3+ 27b2)is negative (see [25]).
3. Consider the sphere S2:= {x2+ y2+ z2− 1 = 0} ⊂R3and the projection
π:R3→R2,(x, y, z) 7→ (x, y). Then, π(S2) = {(x, y) ∈R2: x2+ y2≤ 1}.
De nition 9. A subset S ⊂ Rn is called semialgebraic if it admits a
represen-tation of the form
S := k [ i=1 ri \ j=1 {x ∈Rn: pij(x) ?ij0},
where each pij∈R[x] is a polynomial, and ?ij is one of the symbols <, = or >,
Observe that the semialgebraic subsets of Rn form the smallest family of
subsets containing all the sets of the form
{x ∈Rn: p(x) > 0}, where p ∈ R[x],
and it is closed under nite intersections, nite unions and complements. Alter-natively, if we call the conditions p(x) > 0, p(x) < 0 or p(x) = 0, sign conditions on the polynomial p, then a semialgebraic subset of Rn is de ned by boolean
combinations of sign conditions involving a nite number of polynomials. Theorem 10 (Tarski-Seidenberg). Let π : Rn+1→Rn be the projection
(x1, . . . , xn+1) 7→ (x1, . . . , xn),
and let S ⊂ Rn+1 be a semialgebraic set. Then π(S) is a semialgebraic subset
of Rn.
Proof. See [2] or [5].
Corollary 11. The closure and the interior (with respect to the Euclidean topol-ogy)of a semialgebraic set are semialgebraic.
Proof. Let S ⊂ Rn be a semialgebraic set. The closure of S in the Euclidean
topology is the set, ClRn(S) = {x ∈R
n: ∀t ∈
R ∃y ∈ S (ky − xk2< t2 or t = 0)},
and it can be written as Rn
\ π2(Rn+1\ π1(A)),
where
A := {(x, y, t) ∈R2n+1: y ∈ S and (ky − xk2< t2 or t = 0)},
π1: R2n+1 →Rn+1 is the projection given by π1(x, y, t) = (x, t)and the map
π2 : Rn+1 → Rn the one given by π2(x, t) = x. Hence, by Tarski-Seidenberg
theorem, ClRn(S)is a semialgebraic set.
As the interior IntRn(S)can be written as
IntRn(S) =R n\Cl
Rn(R n\ S),
it follows that IntRn(S)is semialgebraic too.
Remark 12. It would be wrong to belive that the closure of a semialgebraic set is always obtained just by relaxing the strict inequalities describing it. Consider the cubic curve of Example 3(3)
y2− x3+ x2= 0.
The closure of the semialgebraic set
S := {(x, y) ∈R2: x3− x2− y2>0},
is not the set
T := {(x, y) ∈R2: x3− x2− y2≥ 0}.
The closure of S is obtained from T removing the point (0, 0) ∈ R2, and may
be descriced as
ClRn(S) = {(x, y) ∈R
1.3. Dimension of semialgebraic sets
De nition 13. A map f : S → T between the semialgebraic sets S ⊂ Rmand
T ⊂Rn is a semialgebraic map if its graph
Γf := {(x, y) ∈Rm×Rn: y = f (x)}
is a semialgebraic set of Rm+n.
De nition 14. A semialgebraic homeomorphism between the semialgebraic sets S ⊂ Rmand T ⊂ Rn is a bijective semialgebraic map f : S → T that is a
homeomorphism when S and T are endowed, respectively, with the Euclidean topology inherited from Rm and Rn. In such a case it is said that S and T are
semialgebraically homeomorphic.
Recall that a regular function f : Rn →R is a rational function f ∈ R(x) :=
R(x1, . . . , xn)of the form f := g/h, where g, h ∈ R[x1, . . . , xn]and h(x) 6= 0 for
every point x ∈ Rn.
A regular map is a map F : Rn → Rm, whose components are all regular
functions.
Corollary 15. Let F : Rn → Rm be a regular map and let S ⊂ Rn be a
semialgebraic subset. Then F (S) is a semialgebraic subset of Rm.
Proof. The graph of the regular map F : Rn →Rmis the set
ΓF = {(x, y) ∈Rn×Rm: y − F (x) = 0}
or, in other words, if each coordinate of F is given by gi/hi,
ΓF = {(x, y) ∈Rn×Rm: hi(x)yi− gi(x) = 0, i = 1, . . . , m}.
This shows that ΓF is a semialgebraic set and, by Tarski-Seidenberg theorem,
the image
F(S) = π((S ×Rm) ∩ ΓF),
where π : Rn×Rm → Rm is the canonical projection, is a semialgebraic set
too.
1.3 Dimension of semialgebraic sets
In order to introduce the de nition of dimension of a semialgebraic set, we recall a decomposition theorem for semialgebraic sets. The proof can be found in [5]. Theorem 16 (Decomposition for semialgebraic sets). Every semialgebraic sub-set of Rn is a disjoint union of a nite number of semialgebraic sets, each of
them semialgebraically homeomorphic to an open hypercube (0, 1)d ⊂ Rd, for
some d ∈ N (with (0, 1)0 being a point).
This theorem clearly shows what the dimension of a semialgebraic set should be. If S is a semialgebraic set homeomorphic to `k
i=1(0, 1)
dk, then the dimension
of S should be the maximum of the di's. Note that this de nition is well posed
due to the invariance of dimension theorem (see [21]). We want to give an equivalent algebraic de nition, whose advantage is that it is intrinsic, that is, it does not depend on the decomposition above, and it coincides with the classical de nition for complex and real algebraic sets. To do that, we introduce more notation. Given a semialgebraic set S ⊂ Rn, denote P(S) := R[x]/I(S) the ring
De nition 17. Let S ⊂ Rn be a semialgebraic set. The dimension of S is the
Krull dimension of the ring P(S), i.e. the maximal length of chains of prime ideals of P(S).
As P(S) is a quotient of R[x] its Krull dimension is not bigger than the Krull dimension of R[x], which equals n. Indeed,
(0)( (x1)( . . . ( (x1, . . . , xn)
is a chain of prime ideals of R[x] of maximal length n.
We introduce now some standard results on dimension whose proofs can be found in [5, 2.8].
Proposition 18. Let S ⊂ Rn be a semialgebraic set. Then
dim S = dimClRn(S) = dimClzar
Rn(S),
where Clzar
Rn(S) = V(I(S))is the closure of S in the Zariski topology of R n.
Theorem 19. If V = V1∪ . . . ∪ Vk is the decomposition of the algebraic set
V ⊂Rn into its irreducible components, then dim V = max
idim Vi.
Proposition 20. Let U be a nonempty open semialgebraic subset of Rn. Then
dim U = n.
Theorem 21. Let S ⊂ Rn be a semialgebraic set, and let f : S → Rm be a
semialgebraic map. Then
dim f (S) ≤ dim S.
In addition, if f is a bijection from S onto f(S), then dim S = dim f(S). Corollary 22. Let S ⊂ Rn be a semialgebraic set semialgebrically
homeomor-phic to `k i=1(0, 1)
dk. Then dim S = max{d
1, . . . , dk}.
The next result allows us to have a well posed de nition of local dimension. There exist several dierent and equivalent de nitions of the local dimension of a semialgebraic set at each of its points. We choose a geometrical one.
Proposition 23. Let S ⊂ Rn be a semialgebric set and let x ∈ S. There exists
a semialgebraic neighbourhood U of x in S such that, for every semialgebraic neighbourhood V of x in S contained in U, the equality dim U = dim V holds. De nition 24. Let S ⊂ Rn be a semialgebric set, and let x ∈ S. The local
dimension of S at x, denoted dim Sx, is dim U where U is a semialgebraic
neighbourhood U of x in S as in Proposition 23. Moreover, if dim Sx = dim S
for each x ∈ S, it is said that S is pure dimensional. Example 25. The Whitney umbrella is the algebraic set
V := {(x, y, z) ∈R3: zx2− y2= 0}.
It is an irreducible algebraic set but it is not pure dimensional. In fact at each point p on the z-axis with z < 0 its local dimension is dim Vp = 1 because
1.3. Dimension of semialgebraic sets
Figure 1.2: Whitney umbrella
Proposition 26. Let S ⊂ Rn be a semialgebraic set of dimension d. Then, the
set of points of S of local dimension d, that is, S(d):= {x ∈ S : dim Sx= d},
is a nonempty closed semialgebraic subset of S.
Proposition 27. Let S ⊂ Rn be a semialgebraic set. Then
dim(ClRn(S) \ S) < dim S.
Theorem 28. Let S ⊂ Rnbe a semialgebraic set which is a smooth submanifold
Chapter 2
Presentation of the problem
Modern real algebraic geometry was born with Tarski's article [26], where it is proved that the image of a semialgebraic set under a polynomial map is a semialgebraic set. We are interested in studying what might be the inverse problem to Tarski's result, that is to represent semialgebraic sets as polynomial or regular images of Euclidean spaces. In June 1990, during the Oberwolfach reelle algebraische Geometrie Woche [20], Gamboa proposed:
Problem 29. To characterize those semialgebraic subsets of Rmthat are either
polynomial or regular images of Euclidean spaces.
As speci c example of open questions he stated in Oberwolfach: 1. Is the set {x2+ y2>1}a polynomial image of R2?
2. Is the open quadrant {x > 0, y > 0} a regular image of R2?
In 2002 Fernando and Gamboa answered both these questions in [8]. It con-stituted the starting point of the systematic study of the problem of representing semialgebraic sets as polynomial or regular images of Euclidean spaces.
Them, joinly with Ueno, have attempted to understand better polynomial and regular images of Rn, in the last two decades, with the following targets:
• To nd obstructions to be either polynomial or regular images [8], [9], [11], [17] and [28].
• To prove (costructively) that large families of semialgebraic sets with piecewise linear boundary (convex polyhedra, their interiors, their com-plements and the comcom-plements of their interiors) are either polynomial or regular images of Euclidean spaces [10], [14], [15], [16], [19] and [27]. In [6] appears a complete solution to Problem 29 for one dimensional semi-algebraic sets. The general case seems very far to be solved. Currently there are no general principles to approach directly Problem 29 in its full generality. We point out here some of the obstacles that quickly arise in the solution of the general case:
• The rigidity of polynomial and regular maps hinders their manipulation in order to realize a semialgebraic set satisfying the known obstructions, as a polynomial or regular image of an Euclidean space.
• It is di cult to compute the image of an arbitrary polynomial or regular map.
• The family of polynomial or regular images do not behave nicely with respect to the usual set-theoretic operations or geometric constructions (closures, interiors, suspensions over a polynomial or regular image, ...) For these reasons the arguments used until now in the existential proofs have been developed ad hoc. So (constructive) proofs are developed using the, generally hard, strategy:
(i) Given a semialgebraic set S ⊂ Rm satisfying the known obstructions to
be an either polynomial or regular image, to invent a suitable either poly-nomial or regular map f : Rn →Rm whose image is contained in S and
it seems likely to have S as image. (ii) To prove by hand that f(Rn) = S.
2.1 The invariants p and r
Given a semialgebraic set S ⊂ Rm, we introduce the following two invariants:
p(S) := inf
n≥1{there exists a polynomial mapf : R n→
Rm such that f(Rn) = S},
r(S) := inf
n≥1{there exists a regular mapf : R
n→Rmsuch that f(Rn) = S}.
If S is not representable as a polynomial image (resp. a regular image) of any Rn, then we set p(S) := +∞ (resp. r(S) := +∞). By Theorem 21 we have
dim S ≤ r(S) ≤ p(S).
These inequalities may be strict, although it is not known if there exists a semialgebric set S ⊂ Rmsuch that
dim S < r(S) < p(S) < +∞.
At the moment it is only known the existence of semialgebraic sets S ⊂ Rm
such that
p(S), r(S) ∈ {dim S, dim S + 1, +∞}.
2.2 The purpose of this dissertation
In this dissertation we will focus on the regular case, and we will provide a complete computation, following the works of Fernando, Gamboa and Ueno, of the invariant r for large families of semialgebraic sets with piecewise linear boundary:
• Convex polyhedra (Chapter 7), • their interiors (Chapter 6), • their complements (Chapter8),
2.2. The purpose of this dissertation
• the complement of their interiors (Chapter 9).
Given a d-dimensional convex polyhedron K ⊂ Rn, with d ≥ 1, we denote by
Int K its interior as topological manifold with boundary (see Remark 51 for more details). The values of the invariants r(K), r(Int K), r(Rn\ K)and r(Rn\Int K)
are collected in the following table
K bounded Kunbounded r(·) d=1 d>1 d=1 d>1 K 1 d 1 d Int K 2 d 2 d Rn\ K +∞ n 2 n,+∞∗ Rn\Int K +∞ n 1 n,+∞∗
(∗) We have r(Rn\ K) = r(Rn\Int K) = +∞ if and only if K disconnects Rn.
The closed ball Bn := Bn(0, 1) = {x ∈ Rn : kxk ≤ 1} ⊂ Rn can be
seen as the limit of a suitable family of bounded convex polyhedra (consider triangulations of the closed ball). It is thus natural to pose the same questions we approached for convex polyhedra. We will provide a complete answer for the closed ball Bn, the open ball Bn := {x ∈Rn: kxk < 1}and their complements.
We will prove, in Chapter 10, that for each n ≥ 2, the following equalities hold r(Bn) = r(Bn) = r(Rn\ Bn) = r(Rn\ Bn) = n.
Moreover we will compute the invariant r for the spheres Sn−1:= ∂B
n, showing
that in this case we have
r(Sn−1) = n − 1.
The computation of the invariants r(Sn−1)and r(Rn\ B
n)is made for the rst
time in this dissertation, completing the computation of the invariant r in the limit case of balls and sets derived by them.
Chapter 3
Polynomial and regular
images of R
n
In this chapter we will brie y present some of the obstructions for a semialge-braic set S ⊂ Rmto be a polynomial or regular image of some Euclidean space
Rn. We will also furnish several examples and counterexamples, both in the
polynomial and regular case. In the last section we will compute the invariant r(S)for each semialgebraic set S ⊂ R, a result which will be useful in the next chapters during the discussion on convex polyhedra and sets derived by them.
3.1 Obstructions for the representation
We will present some elementary obstructions for the niteness of p(S) and r(S). Some other obstructions are known (see Section 2.2 for a complete list of references). The most general obstruction, at the moment, for a semialgebraic set S ⊂ Rm to be a polynomial image of Rn is the following result from [17].
The proof is involved and requires deep knowledge of resolution of singularities, complex algebraic geometry and algebraic topology and it is beyond the purpose of this dissertation.
Denote the hyperplane at in nity of the real projective space RPm with
H∞:= {x0= 0}, and consider the natural embedding
Rm
→RPm\H∞, (x1, . . . , xm) 7→ [1 : x1: . . . : xm].
We can state the following theorem.
Theorem 30. Let S ⊂ Rmbe a semialgebraic set that is not a point and such
that p(S) < +∞. Then the set S∞:=ClRPm(S) ∩H∞ of points at in nity of S
is nonempty and connected.
This theorem provides also a new evidence of the dierence between regular and polynomial images of Rn. In fact we can consider the following examples.
Examples 31. (1) Let S1 := {(x, y) ∈R2 : x2+ y2 = 1}. As it is bounded,
p(S1) = +∞but S1 is the image of the regular map
f :R → R2, t 7→ t 2− 1 t2+ 1 2 − 2t t2+ 1 2 ,2 t 2− 1 t2+ 1 2t t2+ 1 ! .
This map is the composition of the inverse of the stereographic projection of S1
with respect to the point (1, 0) with the polynomial map g:C ≡ R2→
C ≡ R2, z:= x + iy 7→ z2≡ (x2− y2,2xy).
(2) The 1-dimensional semialgebraic set S := {x > 0, xy = 1} is the image of the regular map
f :R2→R2, (x, y) 7→ (xy − 1)2+ x2, 1 (xy − 1)2+ x2 .
This map is the composition of the map g : R2 →R, (x, y) 7→ (xy − 1)2+ x2
whose image is the open interval (0, +∞) and the rational map h:R2 99KR2, t 7→ t,1 t .
With the notations of Theorem 30 the set S∞ := {[0 : 1 : 0], [0 : 0 : 1]} is
disconnected, thus p(S) = +∞. Instead, r(S) = 2. In fact r(S) ≤ 2 because f(R2) = Sand r(S) 6= 1. Otherwise there would exist a regular map F : R → R2
with F (R) = S.
Composing this map with the projection π : R2→R, (x, y) 7→ x we would
have r((0, +∞)) = 1, that is false (see Proposition 43).
The very rst obstruction to be a polynomial or regular image of Rn has
been presented in [8] for the polynomial case. We provide a proof in the regular case, that contains the polynomial one as well.
Proposition 32. Let S ⊂ Rm be a semialgebraic set with r(S) < +∞ (resp.
p(S) < +∞). Then, S is connected, pure dimensional and its Zariski closure is irreducible.
Proof. Since r(S) = n < +∞ there exists a regular map f : Rn → Rm such
that f(Rn) = S. Regular maps are in particular continuous, hence S = f(Rn)
must be connected.
Consider now the ideal a := I(S) ⊂ R[x1, . . . , xn] consisting of all
polyno-mials that vanish on S. Suppose, by the way of contradiction, that the Zariski closure Clzar
Rn(S) of S is reducible. Then a is not a prime ideal (see Remark
5), and we pick two polynomials p1, p2 ∈ a/ such that p1p2 ∈ a. Note that
(p1p2) ◦ f = 0, that is, (p1◦ f ) · (p2◦ f ) ≡ 0 on Rn. By the identity principle
for polynomials applied after a manipulation on nowhere zero denominators it follows that either p1◦ f = 0or p2◦ f = 0. We suppose, without loss of
gen-erality, that p1◦ f = 0. Then S ⊂ {p1 = 0}, that is, p1 ∈ a, which is absurd.
Thus a is prime and ClzarRn(S)is irreducible.
Suppose to nish that S is not pure dimensional. This means that there exists a point x ∈ S such that the local dimension dim Sx of S at x is smaller
than dim S. Let U be an open semialgebraic neighbourhood of x in Rm such
that dim (S ∩ U) < dim S, and let q ∈ R[x1, . . . , xm]such that
X:=ClzarRn(U ∩ S) = {x ∈Rm: q(x) = 0}.
Since the composition q ◦ f vanishes identically on the nonempty open subset f−1(U ) of Rn then q ◦ f is identically zero on Rn (using again the identity
principle for polynomials after a manipulation on denominators). We obtain S ⊂ X, that is absurd because dim(X) = dim (S ∩ U) < dim(S).
3.2. Obstruction for dimension two
Examples 33. 1. The Whitney umbrella V := {zx2− y2 = 0} ⊂ R3 has
r(V ) = p(V ) = +∞ because V is not pure dimensional.
2. Let ∂P be the boundary of a nondegenerate polygon P. Then, the equal-ities r(∂P) = p(∂P) = +∞ hold because the Zariski closure of ∂P is not irreducible. The same argument holds for the boundary of any nondegen-erate polyhedron.
3. The complement {|x1| > 1} ⊂Rn of a layer (a closed interval if n = 1) is
neither a regular nor a polynomial image of any Euclidean space because it is disconnected.
We now present an obstruction that occurs only for polynomial images. Let S be a polynomial image of some polynomial map f : Rm→Rnand let q : Rn →R
be a polynomial which is nonconstant on S. Then q(S) ⊂ R is unbounded. Indeed, for each v ∈ Rm the image q(S) would contain the image ϕ
v(R) of the
polynomial map ϕv(t) := q(f (tv)). If ϕv(R) were bounded for each v ∈ Rm
then ϕv(t) would be a constant map, say ϕv(R) := rv ∈ R. Therefore, given
two vectors v, u ∈ Rmwe would have
q(f (v)) = rv = ϕv(0) = ϕu(0) = ru= q(f (u)),
so, q would be constant on S, a contraddiction. Thus, the linear projections of S are either unbounded or a point. In particular S itself is either unbounded or a point. We have proved the following
Proposition 34. Let S ⊂ Rmbe a semialgebraic set and suppose p(S) < +∞.
For each polynomial function f : Rm→R the image f(S) is a singleton or an
unbounded set. In particular, S is unbounded or a singleton.
Examples 35. 1. Let K ⊂ Rm be a bounded convex polyhedron. Then
p(K) = +∞. By contrast, we will see that r(K) = dim K.
2. Consider the closed ball B2:= {x2+ y2≤ 1} ⊂R2. It holds p(B2) = +∞
since it is a bounded set. Instead we have r(B2) = 2 since it is the image
of the regular map
F :R2→R2, (x, y) 7→ 2x x2+ y2+ 1, 2y x2+ y2+ 1 .
3.2 Obstruction for dimension two
One of the open questions posed by Gamboa in 1990 was to determine whether the complement of the closed ball {x2+ y2>1} ⊂R2 is a polynomial image of
R2. The question was solved by Fernando and Gamboa in [8] using a key result
of Jelonek. In [22] Jelonek called parametric semilines the nontrivial polynomial images of R. Recall that a continuous map f : Rn →Rm is proper at a point
p ∈Rmif there exists a compact neighborhood K ⊂ Rmof p such that f−1(K)
is compact.
Recall in addition that the exterior boundary of a subset T ⊂ Rm is the
subset ∂T := ClRm(T ) \ T, where the closure is taken with respect to the
Remark 36. We claim that the exterior boundary ∂S of S := f(Rn), where
f : Rn → Rm is any continuous map, is a subset of the set N (f) of points
p ∈Rmat which f is not proper.
Indeed, if p ∈ ∂S \ N (f) ⊂ ClRm(S) there exists a compact neighborhood
K of p in Rm such that f−1(K) is compact. Thus, S ∩ K = f(f−1(K)) is a
closed subset of K. As K is a closed subset of Rm it follows that S ∩ K is a
closed subset of Rm. Hence, p ∈ K ∩ Cl
Rm(S) = K ∩ S, which is a contradiction
because p ∈ ∂S. Consequentely ∂S ⊂ N (f).
The next result, which is speci c of the bidimensional case, describes accu-rately the exterior boundary of the image of a polynomial map R2→R2.
Proposition 37. Let S ⊂ R2 be a semialgebraic set with p(S) = 2. Then,
ClzarR2(∂S)is a nite union of parametric semilines.
Proof. Let f : R2 → R2 be a polynomial map such that f(R2) = S. The set
N (f ) ⊂ ClR2(S) of points in R2 at which f is not proper is, by Jelonek [23],
either empty or a nite union of parametric semilines L1, . . . , Ls. Let L be an
irreducible component of Clzar
R2(∂S). Then, using Remark 36,
L ⊂ClzarR2(∂S) ⊂ClzarR2(N (f )) =ClzarR2(L1) ∪ . . . ∪ClzarR2(Ls).
Since L and the Clzar
R2(Li)'s are irreducible, we deduce that L = Cl zar
R2(Li) for
some i = 1, . . . , s, as claimed.
In particular the answer to the question posed by Gamboa is negative. The complement S := {x2+ y2 > 1} ⊂ R2 of the closed ball is not a polynomial
image of R2 because its exterior boundary is the unit circle, which is not a
nite union of parametric semilines since it is a bounded set. Consequentely, p(S) > 2. Actually it holds p(S) = 3 (see [9]).
Example 38. Both S := {xy < 1} and T := {x > 0, xy > 1} are semialgebraic subsets of R2such that p(S) > 2 and p(T ) > 2. This is so because the common
Zariski closure of their exterior boundary is the hyperbola {xy = 1}, which is not a nite union of parametric semilines.
The next example, suggested to us by Gamboa, shows that Proposition 36 is false in the regular case.
Example 39. The semialgebraic set S := {x ≥ 0, y ≥ 0, xy < 1} ⊂ R2 is a
regular image of R2. In fact S is the image of the regular map
f :R2→R2, (x, y) 7→ x2 y2+ 1, y2 x2+ 1 .
The Zariski closure of its exterior boundary is the algebraic set: Clzar
R2(∂S) = {xy = 1},
which is not a union of parametric semilines.
The previous example shows that Theorem 30 is not longer true if we replace polynomial by regular. In fact the set S∞ := {[0 : 1 : 0], [0 : 0 : 1]}of points at
3.3. A list of examples
3.3 A list of examples
We provide some basic examples of polynomial and regular images of Euclidean spaces.
1. The upper half-plane H := {y > 0} ⊂ R2 is the image of the polynomial
map
R2→
R2, (x, y) 7→ (y(xy − 1), (xy − 1)2+ x2).
2. The punctured plane S := R2\ {0} has p(S) = 2 because it is the image
of the polynomial map R2→R2, (x, y) 7→ (xy − 1, (xy − 1)x2− y).
3. The open half-band B := {x > 0, 0 < y < 1} has p(B) = +∞ by Propo-sition 34. However, r(B) = 2 because B = h(Q), where Q is the open quadrant Q := {x > 0, y > 0}, which has r(Q) = 2 (see Section 4.2), and h:R2
99KR2 is the rational map
(x, y) 7→ x, 1 1 + y .
Even if polynomial and regular images do not behave nicely with respect to most of the usual topological operations, it is possible to construct cones over polynomial or regular images.
Remark 40 (Cones). Let f : Rm→Rnbe either a polynomial or regular map,
and let S := f(Rm). If p 6∈ S is a point of Rn, the cone of S on p is the image
of the map
G:Rm+1→
Rn, (x, t) 7→ (1 − t2)p + t2f(x).
By Proposition 34 we can not obtain bounded cones in the polynomial case, but we can obtain them in the regular one. Since [0, 1] ⊂ R is the image of the regular map
h:R → R, t 7→ t 1 + t2 +
1 2,
in order to obtain the (bounded) cone of the regular image S = f(Rm) on the
point p 6∈ S, it is su cient to consider the regular map H:Rm+1→
Rn, (x, t) 7→ h(t)f (x) + (1 − h(t))p.
The next example shows that sometimes the regular case is easier to deal with. The polynomial case were presented by Ueno in [28]. To that end he found a suitable ad hoc polynomial map f : R2 → R2 whose image is clearly
contained in S and the hardest part of the proof is to show that f(R2) = S. We
present right now an original proof for regular maps with a geometrical avor. To do that we bene t of the fact that the regular case is less rigid than the polynomial one.
Example 41. The semialgebraic set S := {0 < y < x2+ 1} ⊂R2 is a regular
image of R2. We start with a regular map f : R2 → R2 whose image is the
open half-plane H := {y > 0} ⊂ R2 as in the previous example. If we compose
f with the regular map
g: H →R2, (x, y) 7→ x, 1 1 + y ,
we obtain g ◦ f(R2) = R × (0, 1). Since both the line ` := {y = 0} and the
parabola C := {y = x2+ 1} are regular images of R, there exist regular maps
h1:R → R2 and h2:R → R2 such that ` = h1(R) and C = h2(R). Consider a
convex combination of both maps. Namely, h:R × (0, 1) → R2, (x, λ) 7→ λh
1(x) + (1 − λ)h2(x).
It is straightforward to check that the composition h ◦ g ◦ f sati es (h ◦ g ◦ f )(R2) = S.
If we are only interested to understand if a particular convex semialgebraic set is a regular image of some Rn, without looking for the optimal map realizing
this, convex combinations are a useful tool. The next example shows, using con-vex combinations, that every bounded concon-vex polyhedron K ⊂ Rn is a regular
image of some Euclidean space. The purpouse of the example is only to show the power of this tool in the regular case, since nowadays we already know the precise value of the invariant r(K) for any convex polyhedron K (bounded or not) of any dimension.
Example 42. Any bounded convex polyhedron K ⊂ Rn is regular image of some
Euclidean space.
Since K is convex and bounded it is the convex hull of its vertices. Suppose K has m ≥ 3 vertices v1, . . . , vm. We want to show that there exists a regular
map Rm−1→Rn whose image is K. The image of the regular map
f :R → R, t 7→ t 1 + t2 +
1 2,
is the closed interval [0, 1], so the image of the map F : Rm−1 →Rm−1 given
by (x1, . . . , xm−1) 7→ (f (x1), . . . , f (xm−1))is the cube [0, 1]m−1. Consider next
the polynomial map G : Rm−1→Rn, given by
(λ1, . . . , λm−1) 7→ λ1v1+ . . . + λm−1vm−1+ 1 − m−1 X j=1 λj vm.
Then, the image of the composition G ◦ F is the convex hull of the points v1, . . . , vm, that is, (G ◦ F )(Rm−1) = K.
3.4 Semialgebraic subsets of R
In [6] Fernando provided a complete characterization of the one dimensional polynomial and regular images of Rn. For the purpouse of this dissertation
it is su cient to characterize those semialgebraic sets S ⊂ R that are regular image of some Rn. If a semialgebraic set S ⊂ R is the image of a regular map
f :Rn →R, then S is connected. The only connected, nonempty subsets of R
are the intervals. Since the points and R itself are trivially regular images of R we focus on proper intervals. Then, up to precomposing with an a nity, we have to consider the following cases
I1:= [0, 1], I2:= [0, 1), I3:= (0, 1), I4:= [0, +∞), I5:= (0, +∞).
3.4. Semialgebraic subsets of R
(i) I1= f1(R), where f1:R → R is the regular map
t 7→ t 1 + t2 +
1 2, (ii) I2= f2(R), where f2:R → R is the regular map
t 7→1 − 1 1 + t2,
(iii) I3= f3(R2), where f3:R2→R is the regular map
(x, y) 7→ (xy − 1)
2+ x2
1 + (xy − 1)2+ x2,
(iv) I4= f4(R), where f4:R → R is the polynomial map
t 7→ t2,
(v) I5= f5(R2), where f5:R2→R is the polynomial map
(x, y) 7→ (xy − 1)2+ x2.
To complete our characterization we have to prove that I3 and I5 are not
regular images of R. To that end we use a result of Fernando, presented in [6]. Proposition 43. Let I ⊂ R be a proper interval. Then r(I) ≤ 2 and r(I) = 2 if and only if I ( R is open.
Proof. By our previous discussion it is su cient to show that (0, +∞) and (0, 1) are not regular images of R. Observe that a regular map f : R → R can be always extended to a regular map F : RP1 → RP1, where RP1 is the real
projective line. Indeed, the regular map f can be written as f(t) = g(t)/h(t) with g, h ∈ R[t] polynomials and h nowhere zero. Since R[t] is an Euclidean domain we can divide g by h if deg h ≤ deg g. So we can always suppose that f is of the form
f(t) = q(t) +p(t)
h(t) ∈R(t),
with q, p, h ∈ R[t], h nowhere zero and n := deg p < m := deg h. Next we consider the homogenized of the map f. If d := deg q there exist homogeneous polynomialsq,epeand eh in R[x0, x1]such that the only zero of eh is (0, 0),
deg(eq) = d, deg(p) = n, deg(ee h) = m and fx1 x0 = eq(x0, x1) xd0 + x m−n 0 · e p(x0, x1) e h(x0, x1)
Since (0, 0) is the unique zero of the homogeneous polynomial eh it is also the unique zero of the product v(x0, x1) := xd0· eh(x0, x1)and this is a homogeneous
polynomial of degree m + d. Also the sum u(x0, x1) :=eq(x0, x1)eh(x0, x1) + x
m−n+d
is a homogeneous polynomial of degree m + d. Thus the map F :RP1→RP1,[x
0: x1] 7→ [v(x0, x1) : u(x0, x1)]
is a well de ned polynomial map; it is called the homogenized of the map f. Notice that F coincides with f on the chart U0:= {x06= 0} ⊂RP1.
Since RP1 is compact and connected, the image F (RP1)is either RP1 or a
proper closed interval J ⊂ RP1. If F ([0 : 1]) = [0 : 1] then I := J \ {[0 : 1]} is an
unbounded closed interval of R. On the other hand, if F ([0 : 1]) = c ∈ R then J = [a, b]is a bounded closed interval of R and either I = J, if F−1(c)is not a
singleton, or I = J \ {c} if F−1(c)is a singleton. As I is connected, it coincides
with either [a, b] or one of the half-open bounded intervals [a, b) or (a, b]. Thus, if I ( R is open, then r(I) ≥ 2, and we are done.
Chapter 4
Open quadrant
During an Oberwolfach week in June 1990 [20], Gamboa proposed the following problem: which are the semialgebraic sets of Rn that are either polynomial or
regular images of some Euclidean space Rm? One of the open questions proposed
by Gamboa was the so called open quadrant problem. He asked whether the open quadrant Q := {x > 0, y > 0} ⊂ R2 is a polynomial or regular image
of the plane R2. In 2002 Fernando and Gamboa solved this problem showing
that Q is a polynomial (hence regular) image of the plane, publishing their results in [8]. The authors were not completely satis ed with this proof, as it requires computer assistance and the total degree of the solution mapping is very high. Next they have provided, in joint works with Ueno, see [12] and [18], two alternative and elegant proofs.
The relevance of this result is due to the fact that it constitutes the starting point of the systematic study of the problem of representing some semialgebraic sets as polynomial and regular images of Euclidean spaces. Moreover, the repre-sentation of the open quadrant as polynomial or regular image of the plane has been used in the solution of most of the problems solved until nowadays: either as base step in some inductive arguments, or because most of the maps known until today factorize through the open quadrant Q and this factorization makes these problems aordable. Because of the importance of the problem, even if in this dissertation we focus on the regular case, we present in the rst section an overview without proofs, following [13], of the polynomial case. In the second section we present a proof for the regular case following [7].
4.1 Polynomial case
As we have seen in Proposition 36, for an open 2-dimensional semialgebraic set S ⊂R2the geometry of the exterior boundary ∂S := Cl
R2(S)\S, which is empty
if S is closed, provides some obstructions for S to be a polynomial image of R2.
This partially explains why to represent open subsets as either polynomial or regular images of Euclidean spaces is in general more di cult than to represent the closed ones. In the case of the open quadrant Q we nd this phenomenon too. In fact it is much more di cult to represent Q as a polynomial image of R2 than to represent Q ∪ {(0, 0)} and Cl
computation shows that Q ∪ {(0, 0)} = f(R2)and Cl R2(Q) = g(R 2)where: f(x, y) := (x4 y2, x2y4) and g(x, y) := (x2 , y2).
Three dierent solutions of the open quadrant problem in the polynomial case are known, see [8], [12] and [18], and we present them brie y.
4.1.A Computational solution. This is the rst solution of the open quad-rant problem. It used computer assistance in order to apply Sturm's algo-ritm to decide the existence of real roots of an univariate polynomial of a very high degree. In the proof it is employed the polynomial map h1 : R2 → R2,
(x, y) 7→ (h11(x, y), h12(x, y))where
h11(x, y) := (1−x3y+y−xy2)2+(x2y)2, h12(x, y) := (1−xy+x−x4y)2+(x2y)2.
It holds that h1(R2) = Q ∪ K where K := {(1, 0), (0, 1)}. As
h−11 (K) = {(−1, 0), (0, −1)},
is a nite set, by Lemma 105, there exists a polynomial map h0:R2→R2such
that h0(R2) =R2\ h1−1(K)and consequentely, h1◦ h0(R2) = Q.
The total degree, i.e. the sum of the degrees of its components, of this polynomial map is 56 and the total number of its monomials is 168.
4.1.B Algebraic proof. This proof is less technical and does not involve the aid of computers. It is the shortest known proof. The polynomial map is f3◦ f2◦ f1where
f1:R2→R2, (x, y) 7→ ((xy − 1)2+ x2,(xy − 1)2+ y2),
f2:R2→R2, (x, y) 7→ (x, y(xy − 2)2+ x(xy − 1)2),
f3:R2→R2, (x, y) 7→ (x(xy − 2)2+ xy2/2, y).
Figure 4.1: Sketch of the algebraic proof
Although the polynomial maps f1, f2 and f3 have small degree, the total
degree of the composition is 72 and the total number of monomials is 350. 4.1.C Topological proof. This proof is particularly nice, elegant and bril-liant (at least according to the personal preferences of the author of this disserta-tion). The proof involves arguments of algebraic topology. The map considered is g := (g1, g2) :R2→R2where
g1(x, y) := (x2y4+ x4y2− y2− 1)2+ x6y4,
g2(x, y) := (x6y2+ x2y2− x2− 1)2+ x6y4.
The polynomial map g has total degree 28 and its total number of monomials is 22.
4.2. Regular case
4.1.D A new map. In an unpublished work, Ueno has recently found a polynomial map f : R2 → R2 of total degree 8 such that f(R2) = Q. In
addition, he has proved that deg(h) ≥ 6 for every polynomial map h : R2→R2
with h(R2) = Q. The polynomial map F := (f
1, f2) :R2→R2given by
F(x, y) = (x2y2+ (x2y+ xy2− 1)2(x2+ 1), x2y2+ (x2y+ xy2− 1)2(y2+ 1)), satis es F (R2) = Q.
4.2 Regular case
Consider the regular map g := (g1, g2) :R2→R2given by
(x, y) 7→ x2(xy − 1)2+ (xy + 1) 2 1 + (x + y)2, y 2(xy − 1)2+ (xy + 1) 2 1 + (x + y)2 .
Its image is the open quadrant Q.
Proof. Observe rst that g1, g2 are strictly positive on R2, so g(R2) ⊂ Q. To
prove the converse inclusion, we begin by proving that the half-line `+:= {x − y = 0, x > 0},
is contained in the image of g. Note that g(t, t) = t2(t2− 1)2+(t 2+ 1)2 1 + 4t2 , t 2(t2− 1)2+(t 2+ 1)2 1 + 4t2 . Since g(1, 1) = 4 5, 4 5 , and lim
t→+∞g1(t, t) = +∞, the image of the line {x = y}
contains the half-line `+
1 :=(s, s) : 4
5 ≤ s < +∞
. Furthermore, observe that g t,1 t = 4t2 t2+ (t2+ 1)2, 4t2 t2+ (t2+ 1)2 and lim
t→+∞gi(t, 1/t) = 0 for i = 1, 2. Thus the image of g contains the segment
(s, s) : 0 < s ≤4 5
. Consequently, `+ ⊂ g(R2).
Let us prove now that the points (a, b) ∈ Q with a 6= b belong to g(R2). We
may assume without lost of generality that a > b because g(y, x) = (g2(x, y), g1(x, y)).
We are looking for two real numbers x, y ∈ R such that g(x, y) = (a, b). In order to do that, we can focus on a suitably chosen family of hyperbolas that cover the plane R2 except, at most, those points in the bers of g of the form (a, a).
Let us write then y := λ/x for some λ ∈ R. We want to solve the equation g1(x, λ/x) = a, that is, (λ − 1)2x2+ (λ + 1) 2x2 x2+ (x2+ λ)2 = a. (4.2.1) Observe that a − b= g1 x,λ x − g2 x,λ x = (λ − 1)2 x2−λ 2 x2 .
In particular, for λ 6= 1, x4− a − b (λ − 1)2x 2− λ2= 0, and so, x2:= r(λ) = 1 2 a − b (λ − 1)2 + s (a − b)2 (λ − 1)4+ 4λ 2 ! .
The assumption λ 6= 1 is not restrictive, as the points of the form (x, 1/x) ∈ R2
are mapped into points of the form (a, a) ∈ Q, and we have proved that g(R2)
contains `+.
Now we can replace x2 by r(λ) in equation (4.2.1), obtaining
ϕ(λ) := (λ − 1)2r(λ) + (λ + 1) 2r(λ) r(λ) + (r(λ) + λ)2 = a. We have lim λ→+∞ϕ(λ) = +∞ and λ→1lim+ϕ(λ) = a − b.
Since (a, b) ∈ Q we have a > a−b; in addition a > b by our previous assumption. Thus, there exists λ?>1such that ϕ(λ?) = a. In particular we have
g ϕ(λ?), λ? ϕ(λ?) = (a, b), as required.
Chapter 5
Generalities on convex
polyhedra
In this chapter we will brie y introduce those results on convex sets and convex polyhedra, in order to keep the exposition as much self contained as possible. In the rst section we introduce the de nition and some general properties of convex sets in Rn. In the second section we present some of the basic properties
of convex polyhedra. The main references will be the books [3] and [4].
5.1 Convex sets in Euclidean space
De nition 44. A non-empty subset S ⊂ Rnis called convex if, for any x, y ∈ S,
we have [x, y] ⊂ S, where
[x, y] := {λx + (1 − λ)y : λ ∈ [0, 1]}. Examples 45. (1) Rn itself is a convex set.
(2) Every, nite or in nite, intersection of convex sets is a convex set. Recall that the a nity group of Rn is de ned as the semidirect product
A(Rn) :=
Rn
o GL(Rn),
where GL(Rn)is the general linear group and its natural action on Rn is given
by linear trasformations (or matrix multiplication of a vector if we work in coordinates). Since a nities map lines into lines and segments into segments, the next proposition is straightforword.
Proposition 46. Let f : Rn →Rn be an a nity and let S, T ⊂ Rn be convex
sets. Then f(S) and f−1(T )are convex sets.
Since a ne maps are polynomial bijections of Rn that trasform convex sets
into convex sets, we will usually employ a ne bijections f : Rn→Rn, that will
be called change of coordinates, to place convex polyhedra, or more generally a semialgebraic set S ⊂ Rn, in convenient positions. As usual we say that S and
It follows from Example 45, that given any non-empty subset A ⊂ Rn, there
exists the smallest convex set containing A. So the following de nition is well posed.
De nition 47. Given any non-empty set A ⊂ Rn, the smallest convex set
containing A is called the convex hull of A, and denoted by E(A).
Convex sets behave nicely with respect to some standard topological opera-tions in Rn endowed with the Euclidean topology.
Proposition 48. If S ⊂ Rn is convex, so is Int
Rn(S), and one has
IntRn(S) =IntRn(ClRn(S)).
Proof. See [3, 11.2.4].
Of course, the equality IntRn(S) =IntRn(ClRn(S))is false for arbitrary sets
S ⊂Rn, as shown in the following gure. The equality Cl
Rn(S) =ClRn(IntRn(S))
may not hold even if S is convex. To that end it is su cient to consider a hy-perplane H ⊂ Rn.
Figure 5.1: A counterexample to the equality IntRn(S) =IntRn(ClRn(S))
Proposition 49. If S ⊂ Rn is a convex set, so is its closure Cl Rn(S).
Proof. Pick two points x, y ∈ ClRn(S) and let z := λx + (1 − λ)y for some
λ ∈[0, 1]. We must prove that z ∈ S. Let {xn}∞n=1 and {yn}∞n=1 be sequences
of points in S such that x := limn→∞xn and y := limn→∞yn. As S is convex,
each point zn := λxn+ (1 − λ)yn∈ S. Consequently, z:= λx + (1 − λ)y = λ lim n→∞xn + (1 − λ) lim n→∞yn = lim n→∞(λxn+ (1 − λ)yn) = limn→∞zn ∈ClR n(S), as claimed.
5.2. Convex polyhedra
On the other hand the fact that A ⊂ Rn is closed does not su ces to
guarantee that its convex hull E(A) is also closed. Consider, for example, the closed semialgebric set
A = {xy ≥ 1 , x > 0} ∪ {(0, 0)} ⊂R2,
whose convex hull is the semialgebraic set
E(A) = {x > 0, y > 0} ∪ {(0, 0)}, which is not closed.
5.2 Convex polyhedra
Recall that a closed half-space H+⊂Rn is a semialgebraic subset of Rn of the
form
H+:= {`(x) ≥ 0} ⊂Rn,
where ` ∈ R[x] is a polynomial of degree one.
De nition 50. A convex polyhedron K ⊂ Rn is a non-empty subset of Rn
obtained as a nite intersection of closed half-spaces.
Note that a convex polyhedron K ( Rnis a topological manifold with
bound-ary, and denote by ∂K its boundary and by Int(K) := K \ ∂K its interior, that is, the largest topological manifold (without boundary) contained in K. In the extremal case, where K = {v} is a point, we use the usual convention and write Int(K) = K and ∂K = ∅. The dimension dim K = dim Int(K) of K is its dimen-sion as topological manifold with boundary. Observe that the dimendimen-sion of K de ned above coincides with the dimension of K as semialgebraic set as de ned in Section 1.3.
Remark 51. Let K ⊂ Rn be a convex polyhedron. Its interior Int(K) as
topological manifold, and its relative interior IntRnK, as a topological subspace
of Rn, are dierent in general. It holds that Int(K) = Int
X(K) where X is
the a ne subspace of Rn generated by K (i.e. the smallest a ne subspace of
Rn containing K). Both notions coindide if dim K = n. For the same reason
the exterior boundary ∂K := ClRn(K) \IntRn(K) is in general dierent to its
boundary as topological manifold, although we will denote both with the same symbol ∂K. As before, the boundary of K as topological manifold coincides with its exterior boundary partialXK := ClX(K) \IntX(K), where X is the a ne
space generated by K.
Note that in the de nition of convex polyhedra, there may appear super u-ous half-spaces, and even if we write the polyhedron as the intersection of as few half-spaces as possible, there may still be many ways of doing it. In the fol-lowing gure it is shown a convex polyhedron where two of the four half-spaces used in de ning it can be changed at will.
Theorem 52 (Structure of polyhedra). Let K ⊂ Rn be a convex polyhedron of
maximal dimension dim K = n, de ned as K := Tm i=1H
+
i , where H +
i are closed
Figure 5.2: We can change the hyperplanes A and B at will (i) the H+
i 's are well-determined, up to renaming the indices,
(ii) if Hi is the hyperplane ∂Hi+ de ning H +
i , the intersection Hi∩ K is a
convex polyhedron of dimension n − 1, (iii) the boundary ∂K of K is the union Sm
i=1Hi∩ K.
Proof. See [4, 12.1.5].
From this theorem, if K ⊂ Rn is a convex polyhedron of dimension n, there
exists a unique family H := {H1, . . . , Hm} of a ne hyperplanes of Rn, which is
empty just in case K = Rn, such that K := Tm i=1H
+
i . We refer to this family
as the minimal presentation of K.
Another consequence of the Structure theorem, part (iii), is that if K has minimal presentation {H1, . . . , Hm}, then
Int(K) = K \ ∂K = K \ m [ i=1 (K ∩ Hi) = m \ i=1 Hi+∩ m \ i=1 (Rn\ H i) = m \ i=1 (H+ i \ Hi).
De nition 53. Given K ⊂ Rn a convex polyhedron of dimension n with
mini-mal presentation {H1, . . . , Hm}, the facets, or (n − 1)-faces, of K are the
inter-sections Fi := K ∩ Hi for 1 ≤ i ≤ m. For 0 ≤ j ≤ n − 2 we de ne inductively
the j-faces of K as the facets of the (j + 1)-faces of K. As usual the 0-faces are the vertices of K and the 1-faces are the edges of K.
Remark 54. Let K ⊂ Rn be a convex polyhedron of dimension n with minimal
presentation {H1, . . . , Hm}. If K has at least one vertex v, then m ≥ n. In fact
v is the intersection of all those hyperplanes that contains it (see [4, 12.1.9]) {v} = \
v∈Hi
Hi.
De nition 55. A convex polyhedron K of Rnis nondegenerate if it has at least
one vertex. Otherwise, the convex polyhedron is said to be degenerate.
Lemma 56. Let K ⊂ Rnbe an n-dimensional convex polyhedron. The following
assertions are equivalent: (i) K is nondegenerate,
5.2. Convex polyhedra
(ii) either K = Rnor there exists a nondegenerate, (n−k)-dimensional, convex
polyhedron P ⊂ Rn−k, where 1 ≤ k < n, such that, after a change of
coordinates, K = P × Rk,
(iii) K contains a line `.
Proof. (i) → (ii) : Suppose K 6= Rn. Let E be a face of K of minimal dimension.
Since K 6= Rn and K has no vertices, we have 1 ≤ dim E = k < n. Observe
that since the facets of E, if any, are also faces of K whose dimension is strictly smaller than the dimension of E, it follows that E has no facets, and so it is a nely equivalent to Rk. Hence after a change of coordinates, we may assume
that
E:= {x ∈Rn : xk+1= 0, . . . , xn= 0} =Rk× {0}.
Let {H1, . . . , Hm} be the minimal presentation of K and let
`i(x) := ai1x1+ . . . + ainxn+ ai0∈R[x],
be a polynomial of degree 1 such that H+
i = {`i(x) ≥ 0}. Since E ⊂ K,
ai1y1+ . . . + aikyk+ ai0= `i(y, 0) ≥ 0,
for all y ∈ Rk. Thus a
i1= 0, . . . , aik= 0 for 1 ≤ i ≤ m, that is,
`i(x) = ai(k+1)xk+1+ . . . + ainxn+ ai0.
Hence K = P × Rn−k, where
P := {z := (zk+1, . . . , zn) ∈Rn−k : `1(0, z) ≥ 0, . . . , `m(0, z) ≥ 0}
is a convex polyhedron of Rn−k. Note that there exists a face E? of P such
that E = Rk× E? and, comparing their dimensions, k = dim E = k + dim E?.
Therefore, dim E?= 0, that is, E? is a vertex of P, so P is nondegenerate.
(ii) → (iii): It is straightforward since k > 0.
(iii) → (i): We claim that each face E of K contains a line `E parallel to `. We
may assume K 6= Rn and let H := {H
1, . . . , Hm} be its minimal presentation.
Each hyperplane Hi∈ His parallel to `; otherwise Hi∩ ` ∈ Kis a unique point,
and ` 6⊂ H+ i .
Next we prove the result for each facet of K. Fix a facet Fi := K ∩ Hi and
a point pi ∈ Fi. Let us prove that Fi contains the line `i parallel to ` and
passing through pi. As ` is parallel to each Hj ∈ Hthen, either `i ⊂ Hj or `i
is parallel to Hj. In particular `i ⊂ Hi because pi∈ `i∩ Hi. Observe also that
pi∈ K ∩ `i⊂ Hj+∩ `i. Therefore `i⊂ Hj+and `i⊂ Hi∩ m \ j=1 Hj+= Hi∩ K = Fi.
Now given an arbitrary face E of K, there exists, by [4, 12.1.9], some facets F1, . . . , Fs of K such that E = T
s
j=1Fj. Pick a point p ∈ E, and note that,
since the line `E parallel to ` and passing through p is contained in each facet
Fj for 1 ≤ j ≤ s, it is also contained in E. To complete the proof, suppose K
has a vertex v. Since v is a face of K, it should contain the line `v, which is
We conclude this section recalling a result (see [3, 11.3.4] for a proof) that characterizes topologically bounded convex polyhedra.
Lemma 57. Let K ⊂ Rn be a nondegenerate, bounded, convex polyhedron of
dimension d. Then K is semialgebraically homeomorphic to the d-dimensional closed ball
Chapter 6
Interior of convex polyhedra
as regular images of R
n
In this chapter we will prove that the interior of an n-dimensional convex poly-hedron K ⊂ Rn is a regular image of Rn. We want to prove the following
Theorem 58 (Interior of convex polyhedra). Let K ⊂ Rn be an n-dimensional
convex polyhedron, with n ≥ 2. Then, Int(K) is a regular image of Rn.
Of course, if K ⊂ Rn is a d-dimensional polyhedron with 2 ≤ d < n, then K
is contained in some d-dimensional a ne subspace of Rn and, up to a change
of coordinates, K can be identi ed with a polyhedron of Rd× {0}. Thus, it
follows from Theorem 58 that Int(K) is a regular image of Rd. This is why
we will consider only n-dimensional convex polyhedra K ⊂ Rn. To prove this
theorem some preparation is needed. First we will see how to place a convex polyhedron in a suitable position, and how to reduce the problem to the case of bounded polyhedra (via a rational map through the projective space). After this reduction, in order to prove that the interior Int(K) of an n-dimensional bounded convex polyhedron K ⊂ Rn is a regular image of Rn, it will be crucial
to nd a partition of its boundary ∂K := K \ Int(K). Such a partition will be determined by the choice of an exterior point p ∈ Rn\ Kand it is called ABT
-partition. This construction has interest by its own, and will be the crucial ingredient to obtain Theorem 58. We will follow the article [10].
Remark 59. From Proposition 34 we know that the projection onto any line of Rn of a polynomial image of an Euclidean space is either a point or an
unbounded set. Thus, neither bounded convex polyhedra nor their interiors are polynomial images of any Euclidean space.
6.1 Reduction to bounded polyhedra
To begin with we will show that every convex polyhedron can be placed in a suitable position. A nondegenerate n-dimensional convex polyhedron K ⊂ Rn
with minimal presentation {H1, . . . , Hm}is facing upwards if there exists a
sub-family {Hi1, . . . , Hin} whose common intersection is a vertex v := (v1, . . . , vn)
of K such that Tn j=1H
+
vertex of K with minimum xn-coordinate. We want to show that after a change
of coordinates, we can always suppose that every nondegenerate n-dimensional convex polyhedron K is facing upwards.
Lemma 60. Let K ⊂ Rn be an n-dimensional, nondegenerate, convex
poly-hedron. Thus, after a change of coordinates, we can always assume that K is facing upwards and it does not intersect the hyperplane {xn= 0}.
Proof. Let {H1. . . , Hm} be the minimal presentation of K. Since K is
nonde-generate, we have m ≥ n (see Remark 54). We can assume, after a change of coordinates and up to reordering the indices i = 1, . . . , m, that Tn
j=1Hj = {0}
is a vertex of K and H+
i = {xi≥ 0}for 1 ≤ i ≤ n. In particular
K ⊂ {x1≥ 0, . . . , xn≥ 0} ⊂ {x1+ . . . + xn≥ 0}.
Thus, it su ces to compose the change of coordinates above with a second one that transforms the half-space {x1 + . . . + xn ≥ 0} into the half-space
{xn ≥ 1}.
We can show now that for a nondegenerate unbounded n-dimensional convex polyhedron K ⊂ Rn, it is always possible to nd a rational map that transforms
it into a bounded polyhedron. The original proof of this fact can be found in [10]. We provide an alternative topological proof.
Lemma 61. Let K ⊂ Rn be an n-dimensional, nondegenerate, unbounded,
con-vex polyhedron facing upwards that does not intersect the hyperplane {xn= 0}.
Consider the rational map f :Rn 99KRn, (x1, . . . , xn) 7→ x1 xn , . . . ,xn−1 xn , 1 xn . Then ClRn(f (K)) ⊂R
n is a bounded convex polyhedron.
Proof. The rational map f can be interpreted as a transition map between two charts of the real projective space RPn
f :Rn g
99KRPn 99Kh Rn.
In particular it preserves a ne subspaces and the convexity of those subsets that do not intersect the hyperplane {xn = 0}. Hence, f(K) is a convex subset
of Rn and, by Proposition 49, its closure Cl
Rn(f (K)) is a convex polyhedron
of Rn. Now we have to show that f(K) is bounded. Since, by hypothesis, K
is placed in a suitable position, ClRPn(g(K)) does not intersect the projective
hyperplane
Un:= {[x1: . . . : xn+1] ∈RPn: xn= 0}.
Henceforth, the restriction h
ClRPn(g(K)):ClRP
n(g(K)) →Rn
is a regular map. Notice that ClRPn(g(K))is a compact space because it is a
closed subset of the compact space RPn. Thus, h(Cl
RPn(g(K))) is a compact
subset of Rnand, in particular, it is bounded. Hence f(K) is bounded too since
