This paper explores the topological properties of general polynomial mappings from complex planes or spheres to erences, providing insights into their structure and behavior.
Contribution
It extends the Whitney theorem to complex polynomial mappings, offering a new understanding of their topological characteristics.
Findings
01
Topology of polynomial mappings characterized
02
Extension of Whitney theorem to complex case
03
Insights into polynomial mapping structures
Abstract
We describe the topology of a general polynomial mapping F=(f,g):X→C2, where X is a complex plane or a complex sphere.
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TopicsAdvanced Differential Equations and Dynamical Systems · Holomorphic and Operator Theory · Homotopy and Cohomology in Algebraic Topology
For given natural numbers d1,d2
we describe the topology of a generic polynomial mapping F=(f,g):X→C2, with deg f≤d1 and deg g≤d2. Here X is a complex plane or a complex sphere.
Key words and phrases:
polynomials, folds, cusp singularities
1991 Mathematics Subject Classification:
14 R 99, 32 A 10
The authors are partially supported by the grant of Narodowe Centrum Nauki, grant number 2015/17/B/ST1/02637, additionally the third author is partially
supported by the FAPESP grant
2014/00304-2
1. Introduction
Polynomial mappings F:Cn→Cn are the most classical objects in the complex analysis, yet their topology has not been studied up till now. To the best knowledge of the authors complex algebraic families of polynomial mappings on affine varieties have not been investigated so far. Here we describe an idea of such study. We consider the family ΩCn(d1,…,dm) of polynomial mappings F=(F1,…,Fm):Cn→Cm of degree bounded by (d1,…,dm).
For a smooth affine variety X⊂Cn we also consider the family ΩX(d1,…,dm)={F∣X:F∈ΩCn(d1,…,dm)}. In particular based on Mather Projection Theorem, we prove that a generic member of ΩX(d1,…,dm) is transversal to a given modular submanifold (in particular to a given Thom-Boardman strata ΣI) in the space of multi-jets sJk(X,Cm). Moreover, we show that a generic member of ΩX(d1,…,dm) is transversal to any smooth algebraic subvariety of the space of jets Jk(X,Cm), at least if di≥k.
Let us recall that in [11] the second author proved that if M,X,Y are affine irreducible varieties, X,Y are smooth and Φ:M×X→Y is an algebraic family of polynomial mappings such that the generic element of this family is proper then two generic members of this family are topologically equivalent. In particular if X⊂Cp is of dimension n and m≥n then any two generic members of the family ΩX(d1,…,dm) are topologically equivalent. For example, if X is a smooth surface then the numbers cX(d1,d2) and dX(d1,d2) of cusps and double folds, respectively, of a generic member of the family ΩX(d1,d2) are well-defined.
Our aim is to describe effectively the topology of such generic mappings. We consider in this paper the simplest case, when n=m=2 and X=C2 or X is the complex sphere S={(x,y,z)∈C3:x2+y2+z2=1}.
In these cases we describe the topology of the set C(F) of critical points of F and the topology of its discriminant Δ(F).
In particular we show that a generic polynomial mapping F∈ΩX(d1,d2) has only cusps, folds and double folds as singularities and we compute the number cX(d1,d2) of cusps and the number dX(d1,d2) of double folds of such generic polynomial mapping.
Our ideas work well also in higher dimensions. This paper is the first step in a study of the topology of
generic polynomial mappings F:Cn→Cn.
The problem of counting the number of cusps of a
generic perturbation of a real plane-to-plane singularity was
considered by Fukuda and Ishikawa in [3]. They proved that
the number modulo 2 of cusps of a generic perturbation F of an A
finitely determined map-germ F0:(R2,0)→(R2,0) is a topological invariant of
F0. More recently, in [14] Krzyżanowska and Szafraniec
gave an algorithm to compute the number of cusps for sufficiently
generic fixed real polynomial mapping of the real plane.
Algebraic formulas to count the number of cusps and nodes of a generic
perturbation of an A finitely determined holomorphic map-germ F0:(C2,0)→(C2,0) were given by Gaffney and
Mond in [4, 5] (see also [19]). In this case any two generic
perturbations F of F0 defined on a sufficiently small
neighborhood of [math] are topologically equivalent, so the
numbers of cusps and nodes of F are invariants of the map-germ F0.
Let us note that in some cases our result allows also to use local methods to study global mappings. Indeed, in the special case when gcd(d1,d2)=1 the numbers c(F) and d(F) can be computed by using local methods of Gaffney and Mond [5] or Ohmoto methods [18] based on Thom polynomials. Note that in this case the leading homogenous part Fh of a generic mapping F=(f,g) is A finitely determined. Moreover,
we have a deformation Ft(x)=(td1f(t−1(x)),td2g(t−1(x))).
Now we can use the fact (which is first proved in our paper) that a generic (with respect to the Zariski topology) mapping F∈ΩX(d1,d2) has only folds, cusps and double folds as singularities.
Thus for the deformation Ft∈ΩX(d1,d2) of F all Ft,t=0 are generic mappings and all cusps and nodes of Ft tend to [math] when t→0.
In this case our formulas for c(F) and d(F) coincide with formulas of Gaffney-Mond etc.
However, in the general case these approaches do not work since any homogeneous mapping is not A finitely determined if gcd(d1,d2)=1 (Gaffney-Mond, [5]). Note that even if a germ F is K finitely determined then in general the number of cusps and nodes depends on a given stable perturbation Ft of F (see Section 8).
In particular in that case the local number of cusps or nodes cannot be defined and the methods of Gaffney-Mond and Ohmoto do not work. If gcd(d1,d2)=1 our formulas do not coincide with formulas of Gaffney-Mond and Ohmoto, or rather the latter simply do not apply. Hence in general even discrete global invariants can not be obtained by local methods or methods based on Thom polynomials.
Now we will briefly describe the content of the paper.
In Section 2 we state and prove general theorems. In Section 3 we describe the topology of the set of critical points of a generic mapping
F∈ΩC2(d1,d2). Moreover we compute the number cC2(d1,d2) of cusps. In Section 4 we describe the topology of the discriminant
Δ(F) and compute the number dC2(d1,d2) of nodes of Δ(F). In Section 5 we describe the topology of the set of critical points of a generic mapping F∈ΩS(d1,d2), and compute the number cS(d1,d2), where S⊂C3 is a complex sphere. In Section 6 we describe the topology of the discriminant Δ(F) and we compute the number dS(d1,d2).
In Section 7 we introduce the notions of a generalized cusp and the index of a generalized cusp μ (see Definitions 7.1 and 7.3). We show that if F=(f,g):X→C2 is an arbitrary polynomial mapping
with degf≤d1, degg≤d2 and generalized cusps at points a1,…,ar then ∑i=1rμai≤cX(d1,d2).
We conclude the paper with Section 8 which is devoted to proper stable deformations of a given polynomial mapping F:X→Cm.
In particular we give an example of a K finitely determined polynomial mapping F:C2→C2 and its two stable deformations Ft, Gt which have different number of cusps at [math].
2. General polynomial mappings
Let Ωn(d1,…,dm) denote the space of polynomial mappings F:Cn→Cm of multi-degree bounded by d1,…,dm.
Similarly if X⊂Cp is a smooth affine variety we consider the family ΩX(d1,…,dm)={F∣X:F∈Ωp(d1,…,dm)}.
By Jq(Cn,Cm) we denote the space of q-jets of
polynomial mappings F=(f1,…,fm):Cn→Cm. We define
it exactly as in [15].
If we fix coordinates in
the domain and the target then we can identify Jq(Cn,Cm) with the
space Cn×Cm×(CNq)m, where CNq
parametrizes coefficients of polynomials of n-variables and of
degree bounded by q with zero constant term (which correspond to
suitable Taylor polynomials). In further applications, in most
cases, we treat the space Jq(Cn,Cm) in this simple way. In
particular for a given polynomial mapping F:Cn→Cm we
can define the mapping jq(F) as
[TABLE]
If Xn⊂Cp is a smooth affine variety then the space Jq(X,Cm) has the structure of a smooth algebraic manifold and can be locally represented in the same simple way as above. Indeed, locally X is a complete intersection, i.e. for every point x∈X there is an open neighborhood Ux of x such that Ux={g1=0,…,gp−n=0} (in some open set of Cp) and \operatorname{rank}\Big{[}\frac{\partial g_{i}}{\partial x_{j}}\Big{]}=p-n on Ux. We can assume that the mapping (x1,…,xn,g1,…,gp−n) is biholomorphic near x. In particular we have xi=ϕi(x1,…,xn) for i>n. Hence there exists another Zariski open neighborhood
Vx of x such that in Vx we have global local holomorphic coordinates x1,…,xn. In particular Jq(Vx,Cm) can be identified with the space Vx×Cm×(CNq)m. In local coordinates we have a mapping
[TABLE]
Now we show that the space Jq(X,Cm) has the structure of a smooth algebraic manifold. Let D be a sheaf of derivations on X.
Since D is coherent and X is affine D is generated by a finite number of global sections D1,…,Ds. For a multi-index α=(α1,…,αs) let Dα=Dα1…Dαs. Now let Q be the number of multi-indexes α
with ∣α∣≤k. Take d1=d2…=dm=k and consider the mapping
[TABLE]
It is easy to see that the mapping Ψ is algebraic and its image is exactly the space Jk(X,Cm).
By sJq(X,Cm) we denote the space of multi q-jets of polynomial mappings F=(f1,…,fn):X→Cm. We denote by Diag the set {(x1,…xs)∈Xs:xi=xj for some i=j} and for bundles πi:Wi→X we denote by DiagX the set {(w1,…ws):πi(wi)=πj(wj) for some i=j}. We have sJq(X,Cm)=(Jq(X,Cm))s∖DiagX. More generally, we define the space of (q1,…,qs)-jets to be Jq1,…,qs(X,Cm):=Jq1(X,Cm)×…×Jqs(X,Cm)∖DiagX. We call them, if there is no danger of confusion, the space of multi-jets. Again, for a given polynomial mapping F:X→Cm we have the mapping
[TABLE]
In the sequel we use the Thom-Boardman manifolds ΣI (see [1], [16]) which give stratifications of the jet space Jk(X,Cm). For a mapping F:X→Cm we denote ΣI(F):=(Jq(F))−1(ΣI). The sets Σi(F) consist of points where F has corank exactly i. Moreover, if Σ11,…,ik(F) is a manifold then Σ11,…,ik,ik+1(F)=Σik+1(F∣Σ11,…,ik(F)).
We will also use the Thom-Boardman manifolds in the space of multi-jets. For bundles πi:Wi→Y we denote by diagY the set {(w1,…ws):πi(w1)=…=πs(ws)}. We denote (ΣI1,…,ΣIs):=ΣI1×…×ΣIs∩diagY.
Let us state the following result of Mather (this is an analogue of Theorem 1 in [15], as Mather remarked after stating Theorem 6, the proof is analogous and the main change is the use of Bertini’s theorem instead of Sard’s theorem):
Theorem 2.1**.**
Let X⊂Cn be a smooth affine algebraic subvariety and let W⊂sJq(X,Cm) be a modular submanifold. There exists a Zariski open non-empty subset U in the space of all linear mappings L(Cn,Cm) such that for every L∈U the mapping L:X→Cm is transversal W.
This theorem has the following nice application (which in the real smooth case was first observed by S. Ichiki in [9]):
Corollary 2.2**.**
Let X⊂Cn be an affine smooth algebraic subvariety, let W⊂sJq(X,Cm) be a modular submanifold and let F:X→Cm be a polynomial mapping. There exists a Zariski open non-empty subset U in the space of all linear mappings
L(Cn,Cm) such that for every L∈U the mapping F+L:X→Cm is transversal to W.
Proof.
Let G:X∋x↦(x,F(x))∈X×Cm and X~=graph(G)≅X. Apply Mather’s theorem to the variety X~.
We get that for a general matrix A∈GL(m,m) and general linear mapping L∈L(Cn,Cm) the mapping H(A,L)=A(F1,…,Fm)T+L is transversal to W. Hence also the mapping A−1∘H(A,L) is transversal to W (because W is invariant with respect to action of global biholomorphisms). This means that the mapping F+A−1L is transversal to W. But we can specialize the matrix A to the identity and the mapping L to a given linear mapping L0∈L(Cn,Cm). Hence we see that there is a dense subset of linear mappings L∈L(Cn,Cm) such that the mapping F+L:X→Cm is transversal to W. However, the set of such mappings is a constructible subset of L(Cn,Cm). Since it is dense and constructible, it must contain a non-empty Zariski open subset.
∎
We have the following general result which follows directly from Corollary 2.2:
Theorem 2.3**.**
Let X⊂Cn be a smooth algebraic variety and let W⊂sJq(X,Cm) be a modular submanifold. Then there is a Zariski open subset V⊂ΩX(d1,…,dm) such that for every F∈V the mapping F is
transversal to W. In particular, there is a Zariski open subset U⊂ΩX(d1,…,dm) such that for every F∈U the mapping F is
transversal to the Thom-Boardman strata (ΣI1,…,ΣIs) in sJq(X,Cm).
Proof.
By Corollary 2.2 the set of mappings F∈Ωn(d1,…,dm) which are transversal to W is dense in Ωn(d1,…,dm). However it is also constructible. Thus it must contain a Zariski open subset. Now observe that (ΣI1,…,ΣIs) is a modular
manifold.
∎
Note that Mather’s result does not hold for every smooth subvariety in the space of jets, it requires the variety to be modular.
We prove here a result in the general direction – we omit the assumption of modularity for the price of sufficiently high degree of the mapping.
We start with the following fact:
Lemma 2.4**.**
For every sequence of pairwise different points c1,…,cs∈Cn, a number i∈{1,…,s} and sequence of numbers aα, where α ranges through multiindexes α=(α1,…,αn) with 0≤∣α∣≤qi there is a polynomial Hi of degree bounded by D≤∑j=1mqj+m−1 for i=1,…,n, such that:
(1)
for every multindex α with ∣α∣≤qi we have ∂x1α1…∂xnαn∂αHi(ci)=aα,
2. (2)
for every j=i and every multindex β with ∣β∣≤qj we have
∂x1β1…∂xnβn∂βHi(cj)=0.
Proof.
Using linear change of coordinates we can assume that ci1=cj1 for i=j. By the symmetry it is enough to construct a polynomial H1. Take
[TABLE]
We determine coefficients bα inductively. If α=(0,...,0)=0, then b0=a0/∏i=2m(c11−ci1)qi+1. Now assume that we have all bβ determined for
∣β∣=k<q1 and we show how to determine bα with ∣α∣=k+1. We have
[TABLE]
where R(c1) depends only at c1,...,cn and bγ where ∣γ∣≤k. Hence it is enough to take bα=(aα−R(c1))/α!∏i=2m(c11−ci1)qi+1.
∎
Now we can prove:
Theorem 2.5**.**
Let Xn⊂Cp be a smooth affine variety of dimension n. Let S1,…,Sk be locally closed smooth algebraic submanifolds of Jq1,…,qr(X,Cm).
Let d1,…,dm be integers such that di≥∑j=1rqj+r−1 for i=1,…,m.
Then there is a Zariski open dense subset
U⊂ΩX(d1,…,dm)
such that for every F∈U we
have
[TABLE]
Proof.
First consider the case X=Cn. For simplicity we can take m=1 (the general case is analogous).
It is enough to prove that the mapping Ωn(d1)×(∏rCn∖Diag)∋(F,x)↦Jq1,…,qr(F)(x)∈Jq1,…,qr(Cn,C) is a submersion.
Let us observe that if we have a mapping G:P×Z→P×W of the form G(p,z)=(p,g(p,z))∈P×W, then G is a submersion if the mapping Z∋z↦g(p,z)∈W is a submersion for every fixed p∈P. Now take P=∏rCn∖Diag, Z=Ωn(d1) and W in such way that P×W=Jq1,…,qm(Cn,C). Put G(p,F)=Jq1,…,qr(F)(p). From the previous statement we have that G is a submersion if G(p,⋅) is a submersion for every fixed p. But since G(p,⋅) is linear for fixed p
it is enough to prove that G(p,⋅) is surjective for fixed p. Hence our statement reduces to the Lemma 2.4.
Now assume that X is a general affine smooth variety. In generic system of linear coordinates, for a given points a1,...,ar∈X, we
can find a Zariski open subset U, which have global local coordinates
x1,…,xn, i.e., x1,…,xn are holomorphic (local) coordinates in U.
Let Ωn(d1,…,dm)(x1,…,xn)⊂Ωn(d1,…,dm) denote the set of polynomial mappings, which depend only on variables x1,…,xn. Note that we have Ωn(d1,…,dm)≅Ωn(d1,…,dm)(x1,…,xn)⊕W, where mappings in W have coefficients different from coefficients in
Ωn(d1,…,dm)(x1,…,xn). Note that
W∣U is the subset of holomorphic mappings locally depending only on variables x1,…,xn and which have coefficients independent from coefficients in
Ωn(d1,…,dm)(x1,…,xn).
Now we can prove as above that Ψ:Ωn(d1,…,dm)×U∋(F,x)↦jq1,…,qr(F∣U)(x)∈Jq1,…,qr(U,Cm) is a submersion (in the proof it is enough to use only parameters from Ωn(d1,…,dm)(x1,…,xn).
Fix 1≤i≤k. By the transversality theorem
with a parameter the set of polynomials F∈Ωn(d1,…,dn) such that jq1,…,qr(F∣X) is transversal to Si
is dense in Ωn(d1,…,dm). On the other hand this set is
constructible in Ωn(d1,…,dm).
We conclude that there is a
Zariski open dense subset Vi⊂Ωn(d1,…,dm) such
that for every F∈Vi we have jq1,…,qr(F∣X)⋔Si. Now it
is enough to take
U=⋂i=1kVi.
∎
Definition 2.6**.**
Let Σk⊂J1(X,Cn) denote the subvariety of 1-jets of
corank k. Let F∈ΩX(d1,…,dn). We say that
F is one-generic if F is proper and j1(F)⋔Σ1.
By Corollary 2.2 the subset of one-generic mappings
contains a Zariski open dense subset of
ΩX(d1,…,dn). We have the following result:
Theorem 2.7**.**
Let X be a smooth complex manifold of dimension n. Let F:X→Cn be a proper holomorphic one-generic mapping. Let C(F)
denote the set of critical points of F. Then there is an open
and dense subset U⊂C(F) such that for every a∈U the
germ Fa:(X,a)→(Cn,F(a)) is holomorphically equivalent
to a fold.
Proof.
Let Δ=F(C(F)) be the discriminant of F. Take U=C(F)∖F−1(Sing(Δ)). The set U is an open dense subset of
C(F). Take a point a∈U and consider the germ Fa:(X,a)→(Cn,F(a)). By the choice of the point a the germ
of the discriminant of Fa is smooth. Hence by [10], Corollary 1.11, the
germ Fa is biholomorphically equivalent to a k-fold:
(Cn,0)∋(x1,…,xn)↦(x1k,x2,…,xn)∈(Cn,0). In particular corank[Fa]=1.
Now note that
J1(Cn,Cn)≅Cn×Cn×M(n,n), where M(n,n)={[aij],1≤i,j≤n} is the set of n×n matrices.
In these coordinates the set Σ1 is given as {(x,y,m):det[mij]=ϕ(x,y,m)=0} on the open subset {(x,y,m):corank[mij]≤1}. Since the
mapping j1(F) is transversal to Σ1 the mapping ϕ∘j1(F)=kx1k−1 has to be a submersion at [math]. This is possible
only for k=2.
∎
3. Plane mappings
Here we will study the set Ω2(d1,d2).
Let us denote coordinates in J1(C2,C2) by
[TABLE]
For a mapping
F=(f,g)∈Ω2(d1,d2), we have
[TABLE]
which justifies our
notation. The set Σ1 is given by the equation
ϕ(x,y,f,g,fx,fy,gx,gy)=fxgy−fygx=0. Since Σ1
describes elements of rank one it is easy to see that it is a
smooth (non-closed) subvariety of J1(C2,C2).
Now we would like to describe the set Σ1,1 effectively. We restrict our
attention only to sufficiently general jets. In the space
J2(C2,C2) we introduce coordinates
[TABLE]
A generic mapping F satisfies rankdaF≥1 for every a (because
codimΣ2=4). We can assume that F=(f,g) and ∇af=0. The critical set of F is exactly the set Σ1(F)
and it has a reduced equation
∂x∂f(x,y)∂y∂g(x,y)−∂y∂f(x,y)∂x∂g(x,y)=0,
which we write for simplicity as fxgy−fygx=0. In particular
the tangent line to Σ1(F) is given as
[TABLE]
Consequently the condition for [Fa]∈Σ1,1 is:
[TABLE]
and
[TABLE]
Let us note that the last equation contains terms gxxfy2
and gyyfx2 hence for ∇f=0 these two equations
form a complete intersection. In general, if we omit the
assumption ∇f=0 the set Σ1,1 is given in J2(C2,C2)
by three equations:
[TABLE]
[TABLE]
and
[TABLE]
As above by symmetry the set Σ1,1 is smooth and locally is
given as a complete intersection of either L1,L2 or L1,L3.
We will denote by J,J1,1,J1,2 curves given by L1∘j2(F)=0, L2∘j2(F)=0 and L3∘j2(F)=0, respectively. We will also
identify these curves with their equations.
Remark 3.1**.**
These formulas give a description of Σ1,1 also in the case of a general affine surface X, however, it might be only locally in the Zariski topology of J2(X,C2).
Definition 3.2**.**
Let F:(C2,a)→(C2,F(a)) be a germ of a holomorphic mapping. We say
that F has a fold at a if F is biholomorphically
equivalent to the mapping (C2,0)∋(x,y)↦(x,y2)∈(C2,0). Moreover, we say
that F has a cusp at a if F is biholomorphically
equivalent to the mapping (C2,0)∋(x,y)↦(x,y3+xy)∈(C2,0).
Remark 3.3**.**
It is well known that F has a fold at a if j2(F)⋔Σ1 at a and j2(F)(a)∈Σ1,0, and F has a cusp if j2(F)⋔Σ1,Σ1,1 and j2(F)∈Σ1,1.
(cf. [21])
Let X⊂Cn be a smooth algebraic surface, then there is a non-empty Zariski open subset U⊂ΩX(d1,d2) such that for every F∈U the mapping F has only folds and cusps as singularities and the discriminant F(C(F)) has only cusps and nodes as singularities.
Now we compute the number of cusps of a generic polynomial mapping
F∈Ω2(d1,d2). To do this we need a series of lemmas:
Lemma 3.5**.**
Let L∞ denote the line at infinity of
C2. There is a non-empty open subset V⊂Ω2(d1,d2) such that for all (f,g)∈V:
The case d1=1 is trivial so assume d1>1. Let us note that
the set S⊂J1(C2,C2) given by {fx=fy=0} is smooth.
Hence (1) follows from Theorem 2.5. To prove (2) it is
enough to assume that f∈Hd, where Hd denotes the set of
homogenous polynomials of two variables of degree d. Let Ψ:Hd×(C×C)∖{0,0}∋(f,x,y)↦(∂x∂f(x,y),∂y∂f(x,y))∈C2. It is easy to see that Ψ is a submersion.
Indeed, if f=∑aixd−iyi then fx:=∂x∂f(x,y)=da0xd−1+…+ad−1yd−1,fy:=∂y∂f(x,y)=a1xd−1+…+dadyd−1. Since (x,y)=(0,0) we
can assume by symmetry that y=0. Now
∂ad−1∂fx=yd−1,∂ad∂fx=0,∂ad∂fy=dyd−1. Thus
∂(ad−1,ad)∂(fx,fy)=dy2(d−1)=0.
Hence for a generic polynomial f∈Hd the mapping Ψf:(C×C)∖{(0,0)}∋(x,y)↦(∂x∂f(x,y),∂y∂f(x,y))∈C2 is
transversal to the point (0,0). In particular Ψf−1(0,0)
is either zero-dimensional or the empty set. Since f is a
homogenous polynomial the first possibility is excluded. This
means that {∂x∂f=0}∩{∂y∂f=0}∩L∞=∅.
∎
Lemma 3.6**.**
Let L∞ denote the line at infinity of
C2. There is a non-empty open subset V⊂Ω2(d1,d2) such
that for all F=(f,g)∈V:
(1)
J(F)∩J1,1(F)∩L∞=∅,
2. (2)
J(F)⋔L∞.
Here J(F) denotes the projective closure of the set {J(F)=0} etc.
Proof.
Since the case d1=d2=1 is trivial we may assume that d1>1 or d2>1.
We consider the (generic) case when degf=d1 and degg=d2.
Hence J(F)∩L∞ and J1,1(F)∩L∞ depend only on the homogeneous parts of f and g of degree
d1 and d2 respectively. Let Hd denote the set of
homogeneous polynomials of degree d in two variables. It is
sufficient to show that there is an open subset V⊂Hd1,d2:=Hd1×Hd2 such that J(F)∩J1,1(F)∩L∞=∅ for all F=(f,g)∈V.
Consider the set X={(p,F)∈P1×Hd1,d2:J(F)(p)=J1,1(F)(p)=0}. Note that X is a
closed subset of P1×Hd1,d2, and if
J(F)∩J1,1(F)∩L∞=∅ then
F belongs to the image of the projection of X on
Hd1,d2. So to prove (1) it is sufficient to show that X
has dimension strictly smaller than the dimension of
Hd1,d2.
Let q=(1:0)∈P1, Y:={q}×Hd1,d2 and X0=X∩Y. Note that all fibers of the projection
X→P1 are isomorphic to X0. Thus
dim(X)=dim(X0)+dim(P1) and to prove (1) it is sufficient to
show that X0 has codimension at least 2 in Y.
Let (q,F)∈Y and let ai and bi be the parameters in Hd1,d2 giving
respectively the coefficients of f at xd1−iyi and
of g at xd2−iyi. For 0≤i+j≤d1, we have
∂xiyj∂i+jf(q)=(d1−i−j)!(d1−j)!j!aj(F) and similarly
for g and bj.
To conclude the proof of (1) we will show that the codimension of {a0b0=0}∩X0 in Y is at least 2 and ∇J and ∇J1,1 are linearly independent outside {a0b0=0}∩X0 and thus the variety X0 has
codimension 2 in Y.
Let us calculate J(p). We have J(p)=(fxgy−fygx)(q,F)=(d1a0b1−d2a1b0)(F).
Thus {a0=0}∩X0⊂{a0=a1b0=0}∩Y has codimension at least 2 and we may assume in further calculations that a0(F)=0 and similarly b0(F)=0.
Let us assume that d2>1. We have ∂b1∂J(p)=∂b1∂(d1a0b1−d2a1b0)(F)=d1a0(F) and ∂b2∂J(p)=0.
Now let us calculate ∂b2∂J1,1(p). The
coefficient b2 can only be obtained from ∂y2∂2g, which is present in J1,1 in the
summand −2∂y2∂2g(d1∂x∂f)2. Thus
∂b2∂J1,1(p)=∂b2∂(−2d12b2a02)(F)=−2(d1a0(F))2.
So det∂(b1,b2)∂(J,J1,1)(p)=−2(d1a0(F))3=0.
Similarly, if d2=1 and d1>1 then det∂(a1,a2)∂(J,J1,1)(p)=−2(d1a0(F))(d2b0(F))2=0.
To prove (2) note that \overline{\big{\{}\frac{\partial J}{\partial x}(F)=0\big{\}}}\cap\overline{\big{\{}\frac{\partial J}{\partial y}(F)=0\big{\}}}\subset\overline{J_{1,1}(F)}, hence (1) implies (2).
∎
Lemma 3.7**.**
There is a non-empty open subset V1⊂Ω2(d1,d2)
such that for all (f,g)∈V1 and every a∈C2: if
∂x∂f(a)=0 and
∂y∂f(a)=0, then
∂x∂g(a)=0 and
∂y∂g(a)=0.
Proof.
Let us consider two subsets in J1(C2,C2): R1:={(x,y,f,g,fx,fy,gx,gy):fx=0,fy=0,gx=0} and R2:={(x,y,f,g,fx,fy,gx,gy):fx=0,fy=0,gy=0}. By Theorem
2.5 there is a non-empty open subset V1⊂Ω2(d1,d2) such that for every F∈V1 the mapping
j1(F) is transversal to R1 and R2. Since these subsets
have codimension three, we see that the image of j1(F) is
disjoint with R1 and R2.
∎
Lemma 3.8**.**
There is a non-empty open subset V2⊂Ω2(d1,d2)
such that for all (f,g)∈V2 we have
\big{\{}\frac{\partial{f}}{\partial{x}}=0\big{\}}\cap\big{\{}\frac{\partial{f}}{\partial{y}}=0\big{\}}\cap J_{1,2}(f,g)=\emptyset.
Proof.
Let us consider the (non-closed) subvariety S⊂J2(2)
given by equations: fx=0, fy=0,
(fxxgy+fxgxy−fxygx−fygxx)gy−(fxygy+fxgyy−fyygx−fygxy)gx=0,
gx=0, gy=0. It is easy to check that S is a smooth
complete intersection and it has codimension three. The set of
generic mappings F which are transversal to S contains a
Zariski open dense subset V2⊂Ω2(d1,d2). By
construction for all (f,g)∈V2 we have
\big{\{}\frac{\partial{f}}{\partial{x}}=0\big{\}}\cap\big{\{}\frac{\partial{f}}{\partial{y}}=0\big{\}}\cap J_{1,2}(f,g)=\emptyset.
∎
Lemma 3.9**.**
There is a non-empty open subset V3⊂Ω2(d1,d2)
such that for all (f,g)∈V3 the curve J(f,g) is transversal
to the curve J1,1(f,g).
Proof.
There is a Zariski open subset V3 which contains only generic
mappings which satisfy hypotheses of all lemmas above. We can also
assume that the curves \big{\{}\frac{\partial{f}}{\partial{x}}=0\big{\}} and
\big{\{}\frac{\partial{f}}{\partial{y}}=0\big{\}} intersect transversally.
We have to show that the curves J(f,g) and
J1,1(f,g)
intersect transversally at every point a∈J(f,g)∩J1,1(f,g). If ∇af=0 then it follows from transversality of the mapping F
to the set S1,1. Hence we can assume
∂x∂f(a)=0 and
∂y∂f(a)=0. By Lemma 3.7
we have ∂x∂g(a)=0 and
∂y∂g(a)=0. Let us denote:
∂x∂f(x,y)=fx, ∂y∂f(x,y)=fy, etc. It is enough to prove that in
the ring Oa2 we have the
equality I=(fxgy−fygx,(fxxgy+fxgxy−fxygx−fygxx)fy−(fxygy+fxgyy−fyygx−fygxy)fx)=ma,
where ma denotes the maximal ideal of Oa2.
Put L=fxgy−fygx. Hence I=(L,Lxfy−Lyfx). Since
gx(a)=0, gy(a)=0, we have
[TABLE]
[TABLE]
By Lemma 3.8 we have [Lxgy−Lygx](a)=0, hence I=(fx,fy)=ma.
∎
Now we are in a position to prove:
Theorem 3.10**.**
There is a Zariski open, dense subset U⊂Ω2(d1,d2)
such that for every mapping F∈U the mapping F has only
folds and cusps as singularities and the number of cusps is
equal to
[TABLE]
Moreover, if d1>1 or d2>1 then the set C(F) of critical
points of F is a smooth connected curve, which is topologically
equivalent to a sphere with g=2(d1+d2−3)(d1+d2−4)
handles and d1+d2−2 points removed.
Proof.
Note that by Theorem 3.4 a generic F has only folds and cusps as singularities.
Note that every point a
of the intersection of curves J(f,g) and J1,1(f,g) with
∇af=0 is a cusp. Moreover for a generic
mapping F points with ∇af=0 are not cusps (Lemma
3.8). By Bezout Theorem we have that in J(f,g)∩J1,1(f,g) there are exactly
(d1−1)2 points with ∇f=0 and
that the number of cusps of a generic mapping is equal to
[TABLE]
Finally by Lemma 3.6 we have that C(F)=S1(F) is a
smooth affine curve which is transversal to the line at infinity.
This means that C(F) is also smooth at infinity,
hence it is a smooth projective curve of degree d=d1+d2−2.
Thus by the Riemmann-Roch Theorem the curve C(F)
has genus g=2(d−1)(d−2). This means in particular that
C(F) is homeomorphic to a sphere with
g=2(d−1)(d−2) handles. Moreover, by the Bezout Theorem
it has precisely d points at infinity.
∎
Remark 3.11**.**
The curve C(F) has d1+d2−2 (smooth) points at infinity and at each of these points it is transversal to the line at infinity.
4. The discriminant
Here we analyze the discriminant of a generic mapping from Ω(d1,d2). Let us recall that the discriminant of the mapping
F:C2→C2 is the curve Δ(F):=F(C(F)), where C(F) is the critical curve of F. From Theorem 2.3 we have:
Lemma 4.1**.**
There is a non-empty open subset U⊂Ω2(d1,d2)
such that for every mapping F∈U:
(1)
F∣C(F)* is injective outside a finite set,*
2. (2)
if p∈Δ(F) then ∣F−1(p)∩C(F)∣≤2,
3. (3)
if ∣F−1(p)∩C(F)∣=2 then the curve Δ(F) has a normal
crossing at p.
Proof.
By Theorem 2.3 can find a U such that we have the required transversality to Thom-Boardman strata. (1) and (3) follow from transversality to (Σ1,Σ1) and (2) follows from transversality to (Σ1,Σ1,Σ1).
∎
Hence for a generic F the only singularities of Δ(F) are cusps and nodes. We showed in Theorem 3.10 that there are exactly c(F)=d12+d22+3d1d2−6d1−6d2+7 cusps. Now we will compute the number d(F) of nodes of Δ(F). We will use the following theorem of Serre (see [17], p. 85):
Theorem 4.2**.**
If Γ is an irreducible curve of degree d and genus g in the complex projective plane
then
[TABLE]
where δz denotes the delta invariant of a point z.
First we compute the degree of the discriminant:
Lemma 4.3**.**
Let F=(f,g)∈Ω(d1,d2) be a generic mapping. If d1≥d2 then degΔ(F)=d1(d1+d2−2).
Proof.
Let L⊂C2 be a generic line {ax+by+c=0}. Then L intersects Δ(F) in smooth points and degΔ(F)=#L∩Δ(F).
If j:C(F)→Δ(F) is a mapping induced by F then #L∩Δ(F)=#j−1(L∩Δ(F)). The curve j−1(L)={af+bg+c=0} has no common points at infinity
with C(F). Hence by Bezout Theorem we have #j−1(L∩Δ(F))=(degj−1(L))(degC(F))=d1(d1+d2−2). Consequently degΔ(F)=d1(d1+d2−2).
∎
We have the following method of computing the delta invariant (see [17], p. 92-93):
Theorem 4.4**.**
Let V0⊂C2 be an irreducible germ of an analytic curve with the Puiseux parametrization of the form
[TABLE]
Let Dj=gcd(a0,a1,…,aj−1). Then
[TABLE]
If V=⋃i=1rVi has r branches then
[TABLE]
where V⋅W denotes the intersection product.
The main result of this section will be based on the following:
Theorem 4.5**.**
Let F∈Ω(d1,d2) be a generic mapping. Let d1≥d2 and d=gcd(d1,d2). Denote by Δ the projective closure
of the discriminant Δ. Then
[TABLE]
Proof.
Let f~(x,y,z)=zd1f(zx,zy) and g~(x,y,z)=zd2g(zx,zy) be the homogenizations of f and g, respectively, and let f(x,z)=f~(x,1,z) and g(x,z)=g~(x,1,z). For a generic mapping the curves C(F) and {f=0} have no common points at infinity (see Lemma 4.6). Moreover we may assume that (1:0:0)∈/C(F). Thus F extends to a neighborhood of C(F)∩L∞ on which it is given by the formula
[TABLE]
Let {P1,…,Pd1+d2−2}=C(F)∩L∞, fix a point P=Pi. The curve C(F) is transversal to the line at infinity so it has a local parametrization at P of the form γ(t):=(∑ieiti,t). We have the following:
Lemma 4.6**.**
If F is a generic mapping then f(P)=0, g(P)=0 and
[TABLE]
where cd=0 and d2c=d1d.
Proof.
Let J~=J~(F) be the homogenization of J(F). Obviously J~=∂x∂f~∂y∂g~−∂y∂f~∂x∂g~.
Let J(x,z)=J~(x,1,z). Since J(γ(t))=0 and ∂t∂γ(t)∣t=0=(e1,1) we have
[TABLE]
[TABLE]
[TABLE]
Consider the set
[TABLE]
Note that if f(P)=0 or c=0 then the fiber over F of the projection from X to Ω2(d1,d2) is non-empty. Hence it suffices to prove that X has codimension at least 2.
Let p=(0:1:0) and q=(a:b:0)∈L∞∖{(1:0:0)}. Let T~(x,y,z)=(bx−ay,y,z) so that T~(q)=p. Take T(x,y)=(bx−ay,y). Note that J~(F∘T)=(J~(F)∘T~)J(T~)=bJ~(F)∘T~. Furthermore
[TABLE]
Thus (p,F)↦(T−1(p),F∘T) is an isomorphism of Xp:=X∩({p}×Ω2(d1,d2)) and X∩({q}×Ω2(d1,d2)). So it is enough to show that Xp has codimension 2 in Yp:={p}×Ω2(d1,d2).
Let ai be the parameters in Ω2(d1,d2) giving the coefficients of f~ (and of f) at xd1−iyi and let bi and ci describe respectively the coefficients of g~ at xd2−iyi and xd2−i−1yiz.
The first equation of Xp is d2ad1−1bd2−d1ad1bd2−1=0 and the only summand of the second containing cd2−1 is −(ad1−1)2(d2−1)cd2−1. Clearly those equations are independent outside the set {ad1−1=0}. Moreover {ad1−1=d2ad1−1bd2−d1ad1bd2−1=0}={ad1−1=ad1=0}∪{ad1−1=bd2−1=0}, thus Xp has codimension 2 in Yp.
Finally note that if d2c=d1d then
[TABLE]
Hence we consider the set
[TABLE]
[TABLE]
Similarly as above one can show that it has codimension 2, which concludes the proof.
∎
Let Cp be the branch of C(F) at P. We find the Puiseux expansion of the branch F(CP) of Δ(F) at F(P). We have
[TABLE]
[TABLE]
If d1=d2 then by Lemma 4.6 we have d−c=0 and F(CP) is smooth at F(P). So assume d1>d2. Since the function h(t)=(g(P)f(P)f(γ(t))g(γ(t)))d1−d21=1+d1−d2d−ct+… is invertible in t=0 we can introduce a new variable T=th(t). We have F(γ(T))=(Td1−d2f(P)g(P),Td1h(t)−d1(1−ct+…)f(P)1).
Moreover
[TABLE]
By Lemma 4.6 we have d2c−d1d=0 and we can apply Theorem 4.4 to compute δ(F(CP))F(P). Since a0=d1−d2, a1=d1 and a2=d1+1, we have 2δ(F(CP))F(P)=(d1−1)(d1−d2−d)+(d1+1−1)(d−1)=(d1−1)(d1−d2−1)+(d−1), where d=gcd(d1,d2).
To proceed further we also need:
Lemma 4.7**.**
If F is a generic mapping then
[TABLE]
for i,j∈{1,2,…,d1+d2−2} and i=j.
Proof.
Consider the set X={(p,q,F)∈L∞×L∞×Ω2(d1,d2):p=q,J~(F)(p)=J~(F)(q)=f~(p)d2g~(q)d1−f~(q)d2g~(p)d1=0}. Similarly as in Lemma 4.6 we will prove that X has codimension 3, so there is a dense open subset S⊂Ω(d1,d2) such that the projection from X has empty fibers over F∈S.
Indeed, take p=(1:0:0), q=(0:1:0) and Y:={(p,q)}×Ω2(d1,d2). It suffices to show that X0=X∩Y has codimension 3 in Y. Let ai and bi be the parameters in Ω2(d1,d2) giving respectively the coefficients of f~ at xd1−iyi and of g~ at xd2−iyi.
The three equations describing X0 are w1=d1a0b1−d2a1b0=0, w2=d2ad1−1bd2−d1ad1bd2−1=0 and w3=a0d2bd2d1−ad1d2b0d1=0. Note that X0∩{a0=0}={a0=b0=w2=0}∪{a0=a1=ad1=w2=0} has codimension 3. Similarly X0∩{b0=0} and X0∩{ad1=0} have codimension 3, however outside the set {a0=b0=ad1} the three equations are obviously independent. Thus X0 has codimension 3 in X.
∎
Now we are in a position to compute ∑z∈(Δ∖Δ)δz. If d1=d2 then Δ has exactly d1+d2−2 smooth points at infinity and consequently ∑z∈(Δ∖Δ)δz=0 (see the text after the proof of Lemma 4.6). So assume d1>d2, then Δ has only one point at infinity Q=(1:0:0).
In Q the curve Δ has exactly r=d1+d2−2 branches Vi=F(CPi). We computed above that 2δ(Vi)Q=(d1−1)(d1−d2−1)+(d−1). Now we will compute Vi⋅Vj.
Let ta,b(x,y)=(x+a,y+b). By the dynamical definition of intersection there exists a neighborhood U of [math], such that for small generic a,b we have
[TABLE]
This means that Vi⋅Vj is equal to the number of solutions of the following system:
[TABLE]
[TABLE]
where a,b and S,T are sufficiently small. Take
[TABLE]
[TABLE]
Thus we have Vi⋅Vj=mult0Q. Note that by Lemma 4.7 the minimal homogenous polynomials of the two components of Q have no nontrivial common zeroes, hence Vi⋅Vj=d1(d1−d2). Consequently
[TABLE]
[TABLE]
[TABLE]
∎
We can now prove the following:
Theorem 4.8**.**
There is a Zariski open, dense subset U⊂Ω2(d1,d2)
such that for every mapping F∈U the discriminant Δ(F)=F(C(F)) has only cusps and nodes as singularities. Let d=gcd(d1,d2).
Then the number of cusps is equal to
[TABLE]
and the number of nodes is equal to
[TABLE]
Proof.
Let d1≥d2 and D=d1+d2−2. By Lemma 4.3 we have degΔ(F)=d1D. From Lemma 4.1 we know that Δ(F) has only cusps and nodes as singularities and is birational with C(F). Hence Δ(F) has genus g=21(D−1)(D−2). Thus by Theorem 4.2 we have
If d1=d2=d then the discriminant has 2d−2 smooth points at infinity and at each of these points it is tangent to the line L∞ (at infinity) with multiplicity d.
If d1>d2 then the discriminant has only one point at infinity with d1+d2−2 branches V1,…,Vd1+d2−2 and each of these branches has delta invariant
[TABLE]
and Vi⋅L∞=d1. Additionally Vi⋅Vj=d1(d1−d2). In particular the branches Vi are smooth if and only if
d1=d2 or d1=d2+1.
5. The complex sphere
In the next two sections we show that our method can be easily generalized to the case when X is a complex sphere.
Let ϕ=y2+2xz and let S be a complex sphere: S={(x,y,z):ϕ=1} (of course S is linearly equivalent with a standard sphere S′:={(x,y,z):x2+y2+z2=1}).
Here we will study the set ΩS(d1,d2). We consider on the set ΩS(d1,d2) the Zariski topology, which is the induced topology given by the mapping Θ:Ω3(d1,d2)∋F↦F∣S∈ΩS(d1,d2).
First we compute the critical set C(F) of a generic mapping F=(f,g)∈ΩS(d1,d2).
Note that x∈C(F) if rank (∇ϕ,∇f,∇g)<3, hence C(F) is the intersection of S and the surface given by
[TABLE]
In particular we have:
Corollary 5.1**.**
For a generic mapping F∈ΩS(d1,d2) we have degC(F)=2(d1+d2−1).
Now we describe cusps of a generic mapping F:S→C2. Note that a tangent line to C(F) is given by two equations:
[TABLE]
The mapping F has a cusp in a point (x,y,z) if
(1) (x,y,z)∈C(F)
(2) the line given by the kernel of d(x,y,z)F is tangent to C(F).
First let us determine the kernel of d(x,y,z)F. If rankzfxyfyxfz=2 then
the kernel is given by the vector
[TABLE]
Otherwise it is the vector
[TABLE]
Let J1,1(F):=J(F)xv1(f)+J(F)yv2(f)+J(F)zv3(f) and J1,2(F):=J(F)xv1(g)+J(F)yv2(g)+J(F)zv3(g). Let C denote the set of cusps of F, for generic F we have from the construction:
[TABLE]
Furthermore, we will show in Lemma 5.2 that S∩{J1,2(F)=0}∩{v(f)=0}=∅ which gives
[TABLE]
Lemma 5.2**.**
Let L∞ denote the plane at infinity of
C3. There is a non-empty open subset V⊂ΩS(d1,d2) such
that for all F=(f,g)∈V:
(1) The assertion can be proved locally. Consider the open set Uz={p∈S:z=0} (and similarly open sets Ux,Uy). In Uz we have globally defined local coordinates x,y. Now the proof reduces to Lemma 3.8.
(2) Similarly as in Lemma 3.6 we will show that there is an open subset V⊂Hd1,d2:=Hd1×Hd2 such that S∩{J(F)=0}∩{J1,1(F)=0}∩L∞=∅ for all F=(f,g)∈V.
Let ϕ(x,y,z)=y2+2xz and Γ:={(x,y,z)∈\tenmsyP2:ϕ(x,y,z)=0}. Obviously Γ≅\tenmsyP1.
Consider the set X={(p,F)∈Γ×Hd1,d2:ϕ(p)=J(F)(p)=J1,1(F)(p)=0}.
If {ϕ=0}∩{J(F)=0}∩{J1,1(F)=0}∩L∞=∅ then
F belongs to the image of the projection of X on
Hd1,d2. So to prove (1) it is sufficient to show that X
has dimension strictly smaller than the dimension of Hd1,d2.
Let q=(1:0:0)∈P2, Y:={q}×Hd1,d2 and X0=X∩Y. Note that all fibers of the projection
X→Γ are isomorphic to X0, because the group GL(S) of linear transformations of S acts transitively on the conic at infinity of S. Thus dim(X)=dim(X0)+dim(Γ) and to prove (1) it is sufficient to
show that X0 has codimension at least 2 in Y.
Let r=(q,F)∈Y and let ai,j and bi,j be the parameters in Hd1,d2 giving
respectively the coefficients of f at xd1−i−jyizj and
of g at xd2−i−jyizj. For 0≤i+j+k≤d1, we have
∂xiyjzk∂i+j+kf(q)=(d1−i−j−k)!(d1−j−k)!j!k!aj,k(F) and similarly
for g and bj,k.
To conclude the proof of (1) we will show that the codimension of {a1,0b1,0=0}∩X0 in Y is at least 2 and ∇J and ∇J1,1 are linearly independent outside {a1,0b1,0=0}∩X0 and thus the variety X0 has
codimension 2 in Y.
Let us calculate J(r). We have J(r)=(fxgy−fygx)(q,F)=(d1a0,0b1,0−d2a1,0b0,0)(F).
Thus {a0,0=0}∩X0⊂{a0,0=a1,0b0,0=0}∩Y has codimension at least 2 and in further calculations we may assume that a0,0(F)=0 and similarly b0,0(F)=0.
Let us assume that d2>1. We have ∂b1,0∂J(r)=∂b1,0∂(d1a0,0b1,0−d2a1,0b0,0)(F)=d1a0,0(F) and ∂b2,0∂J(r)=0.
Now let us calculate ∂b2,0∂J1,1(r). The
coefficient b2,0 can only be obtained from ∂y2∂2g, which is present in J1,1 in the
summand ∂y2∂2gzfxxfz2. Thus
∂b2,0∂J1,1(p)=∂b2,0∂(2b2,0d12a0,02)(F)=2d12a0,0(F)2.
So det∂(b1,0,b2,0)∂(J,J1,1)(p)=2d13(a0,0(F))3=0.
Similarly, if d2=1 and d1>1 then det∂(a0,1,a0,2)∂(J,J1,1)(p)=2d23(b1,0(F))3=0.
(3) Note that \overline{\big{\{}\nabla J(F)|_{S}=0\big{\}}}\subset\{\overline{J_{1,1}(F)=0\}}, hence (2) implies (3).
∎
Lemma 5.3**.**
There is a non-empty open subset V1⊂ΩS(d1,d2)
such that for all (f,g)∈V1 the curve S∩J(f,g) is transversal
to the curve S∩J1,1(f,g).
Proof.
As in Lemma 5.2 (1) we consider the sets Ux,Uy,Uz with globally defined local coordinates and reduce the proof to Lemmas 3.7 and 3.9.
∎
Lemma 5.4**.**
There is a non-empty open subset V2⊂Hd1
such that for all f∈V2 the equations:
Let Γ:={(x,y,z)∈\tenmsyP2:ϕ(x,y,z)=0}≅\tenmsyP1. Consider the set
[TABLE]
If {ϕ=0}∩{v(f)=0}=∅ then
f belongs to the image of the projection of X on
Hd1. So to prove (1) it is sufficient to show that X
has dimension strictly smaller than the dimension of
Hd1.
Let q=(1:0:0)∈P2, Y:={q}×Hd1 and X0=X∩Y. As before, all fibers of the projection
X→Γ are isomorphic to X0, so
dim(X)=dim(X0)+dim(Γ) and it is sufficient to
show that X0 has codimension at least 2 in Y.
But X0 is given by two equations: −a(1,0)=0,d1a(0,0)=0, so codimX0=2.
∎
Lemma 5.5**.**
There is a non-empty open subset V3⊂ΩS(d1,d2)
such that for all (f,g)∈V3 the equations:
(1)
y2+2xz=1,**
2. (2)
v(f)=0**
have exactly 2(d12−d1+1) common solutions.
Proof.
We have
[TABLE]
Note that generically the curve {yfyxfz=0}∩{zfxxfz=0} decomposes into {v(f)=0} and {x=fz=0}. Thus by the Bezout Theorem deg{v(f)=0}=d12−d1+1 and S∩{v(f)=0} has 2(d12−d1+1) points. We leave checking that the intersections are transversal and there are no components at infinity to the reader.
∎
Now we are in a position to prove:
Theorem 5.6**.**
There is a Zariski open, dense subset U⊂ΩS(d1,d2)
such that for every mapping F=(f,g)∈U the mapping F has only
folds and cusps as singularities and the number of cusps is
equal to
[TABLE]
Moreover the set C(F) of critical
points of F is a smooth connected curve, which is topologically
equivalent to a sphere with (d1+d2−2)2
handles and 2(d1+d2−1) points removed.
Proof.
Note that every point a
of the intersection of curves J(f,g) and J1,1(f,g) with
v(f)=0 is a cusp. Moreover for a generic
mapping F points with v(f)=0 are not cusps (Lemma
5.2). By Lemma 5.5 we have that in the set S∩{v(f)=0} there are exactly
2(d12−d1+1) points and
that the number of cusps of a generic mapping is equal to
[TABLE]
Moreover by Lemma 5.2 we have that C(F)=S1(F) is a
smooth affine curve which is transversal to the plane at infinity.
This means that J:=C(F) is also smooth at infinity,
hence it is a smooth projective curve of degree 2(d1+d2−1).
Note that Pic(S)=\tenmsyZL1⊕\tenmsyZL2, where
L1,L2 are suitable lines in S (for details see e.g. [20], Ex.2 p. 237). Moreover if H is a plane section then H∼L1+L2.
Hence in Pic(S) we have C(F)∼aL1+bL2 where a+b=2(d1+d2−1).
Take li=Li∩S and note that Pic(S) is generated freely by l1 or l2 with the relation l1+l2=0.
In particular C(F)∼(a−b)l1. But in Pic(S) we have C(F)∼(d1+d2−1)H=0.
Thus a=b=d1+d2−1.
Suppose that C(F) is not connected. Hence C(F)=Γ1+Γ2.
We have Γ1∼a1L1+b1L2 and Γ2∼a2L1+b2L2, where a1,b1,a2,b2≥0, a1+b1>0 and a2+b2>0.
Note that a1+a2=b1+b2=d1+d2−1>0 thus if a1b2=0 then a2b1>0. So Γ1.Γ2=a1b2+a2b1>0. Consequently
Γ1∩Γ2=∅ and C(F) is not smooth – a contradiction. This implies that C(F) is connected.
Let H⊂\tenmsyP3 be a hyperplane. The canonical divisor of S is
−2H=−2(L1+L2). Hence KJ=(J−2H)∣J=(d1+d2−3)(L1+L2)∣J and degKJ=2(d1+d2−3)(d1+d2−1).
By Riemmann-Roch Theorem J has genus degKJ/2+1=(d1+d2−2)2.
This means in particular that
C(F) is homeomorphic to a sphere with
(d1+d2−2)2 handles. Moreover, by the Bezout Theorem
it has precisely 2(d1+d2−1) points at infinity.
∎
Remark 5.7**.**
The curve C(F) has 2(d1+d2−1) (smooth) points at infinity and in each of these points it is transversal to the plane at infinity.
6. The complex sphere: the discriminant
Here we analyze the discriminant of a generic mapping from ΩS(d1,d2). Similarly as for the plane Theorem 2.3 implies that for a generic F the only singularities of Δ(F) are cusps and nodes. We showed in Theorem 5.6 that there are exactly c(F)=2(d12+d22+3d1d2−3d1−3d2+1) cusps. Now we will compute the number d(F) of nodes of Δ(F).
First we compute the degree of the discriminant:
Lemma 6.1**.**
Let F=(f,g)∈ΩS(d1,d2) be a generic mapping. If d1≥d2 then degΔ(F)=2d1(d1+d2−1).
Proof.
Since the proof is analogous to the proof of Lemma 4.3 we skip it.
∎
The main result of this section will be based on the following:
Theorem 6.2**.**
Let F∈ΩS(d1,d2) be a generic mapping. Let d1≥d2 and d=gcd(d1,d2). Denote by Δ the projective closure
of the discriminant Δ. Then
[TABLE]
Proof.
Let f~(x,y,z,w)=wd1f(wx,wy,wz) and g~(x,y,z,w)=wd2g(wx,wy,wz) be the homogenizations of f and g and let f(x,y,w)=f~(x,y,1,w) and g(x,y,w)=g~(x,y,1,w), respectively. For a generic mapping the curves C(F) and {f=0} have no common points at infinity (see Lemma 6.3). Moreover since F is generic, we have {z=0}∩C(F)=∅. Thus F extends to a neighborhood of C(F)∩L∞ on which it is given by the formula
[TABLE]
Let Γ=S∩L∞. Let {P1,…,P2d1+2d2−2}=C(F)∩Γ, fix a point P=Pi. The curve C(F) is transversal to the line at infinity so it has a local parametrization at P of the form γ(t):=(∑iaiti,∑ibiti,t). We have the following:
Lemma 6.3**.**
If F is a generic mapping then f(P)=0, g(P)=0 and
[TABLE]
where cd=0 and d2c=d1d.
Proof.
Let J~ be the homogenization of J. Obviously
[TABLE]
Now let J(x,y,w)=J~(x,y,1,w) and ψ(x,y,w)=2x+y2−w2=ϕ~(x,y,1,w), where ϕ~ is the homogenization of ϕ=y2+2xz−1. We have J(γ(t))=0 and ψ(γ(t))=0. Moreover,
∂t∂γ(t)∣t=0=(a1,b1,1). Thus we have
[TABLE]
[TABLE]
Consequently a1=a1δ−1 and b1=b1δ−1, where
[TABLE]
[TABLE]
Thus
[TABLE]
Take
[TABLE]
[TABLE]
Consider the set
[TABLE]
Note that if f(P)=0 or c=0 then the fiber over F of the projection from X to Ω3(d1,d2) is non-empty. Hence it suffices to prove that X has codimension at least 2.
Let p=(0:0:1:0), and q=(−a2/2:a:1:0)∈Γ. Let T~(x,y,z,w)=(x+ay−a2z/2,y−az,z,w) and T(x,y,z)=(x+ay−a2z/2,y−az,z). Thus T(S)=S and T~(q)=p. As in Lemma 4.6 we can show that (p,F)↦(T~−1(p),F∘T) is an isomorphism of Xp:=X∩({p}×Ω3(d1,d2)) and X∩({q}×Ω3(d1,d2)). So it is enough to show that Xp has codimension 2 in Yp:={p}×Ω3(d1,d2).
Let ai,j,k be the parameters in Ω3(d1,d2) giving the coefficients of f~ at xiyjzd1−i−j−kwk (i.e. of f at xiyjzd1−i−j−k) and let bi,j,k describe the analogous coefficients of g~.
The first equation of Xp is w1:=d2a0,1,0b0,0,0−d1b0,1,0a0,0,0. The second one is
[TABLE]
[TABLE]
[TABLE]
By direct computation we obtain
[TABLE]
Thus the equations w1=0 and w2=0 are independent outside the set
[TABLE]
So Xp has codimension 2 in Yp.
Finally note that if d2c=d1d then
[TABLE]
Hence we consider the set
[TABLE]
[TABLE]
Similarly as above one can show that it has codimension 2, which concludes the proof.
∎
Let Cp be the branch of C(F) at P. Exactly as in Section 4 by using the Puiseux expansion we can show that if d1=d2 then F(CP) is smooth at F(P) and if d1>d2 then 2δ(F(CP))F(P)=(d1−1)(d1−d2−d)+(d1+1−1)(d−1)=(d1−1)(d1−d2−1)+(d−1), where d=gcd(d1,d2).
To proceed further we also need:
Lemma 6.4**.**
If F is a generic mapping then
[TABLE]
for i,j∈{1,2,…,2(d1+d2−1)} and i=j.
Proof.
Consider the set X={(p,q,F)∈Γ×Γ×Ω3(d1,d2):p=q,J~(F)(p)=J~(F)(q)=f~(p)d2g~(q)d1−f~(q)d2g~(p)d1=0}. Similarly as in Lemma 6.3 we will prove that X has codimension 3, so there is a dense open subset U⊂Ω3(d1,d2) such that the projection from X has empty fibers over F∈U.
Indeed, take p=(1:0:0:0), q=(0:0:1:0) and Y:={(p,q)}×Ω3(d1,d2). It suffices to show that X0=X∩Y has codimension 3 in Y. Let aij and bij be the parameters in Ω3(d1,d2) giving respectively the coefficients of f~ at xd1−i−jyizj and of g~ at xd2−i−jyizj.
The three equations describing X0 are
[TABLE]
[TABLE]
Note that
X0∩{a0,0=0}⊂{a0,0=b0,0=w2=0}∪{a0,0=a0,1=w2=0} has codimension 3. Similarly X0∩{b0,0=0} and X0∩{a0,d1=0} have codimension 3, however outside the set {a0,0=0}∪{b0,0=0}∪{a0,d1=0} the three equations are obviously independent. Thus X0 has codimension 3 in X.
∎
Now we are in a position to compute ∑z∈(Δ∖Δ)δz. If d1=d2 then Δ has exactly 2(d1+d2−1) smooth points at infinity and consequently ∑z∈(Δ∖Δ)δz=0 (see the statement after Lemma 6.3). So assume d1>d2, then Δ has only one point at infinity Q=(1:0:0).
In Q the curve Δ has exactly r=2(d1+d2−1) branches Vi=F(CPi). We have 2δ(Vi)Q=(d1−1)(d1−d2−1)+(d−1). As in Section 4 we have Vi⋅Vj=d1(d1−d2). Consequently
[TABLE]
[TABLE]
[TABLE]
∎
We can now prove the following:
Theorem 6.5**.**
There is a Zariski open, dense subset U⊂ΩS(d1,d2)
such that for every mapping F∈U the discriminant Δ(F)=F(C(F)) has only cusps and nodes as singularities.
The number of cusps is equal to
[TABLE]
and the number of nodes is equal to
[TABLE]
where D=d1+d2−1 and d=gcd(d1,d2).
Proof.
Let d1≥d2 and D=(d1+d2−1). By Lemma 6.1 we have degΔ(F)=2d1D. Since Δ(F) is birational with C(F) it has genus g=D(D−2)+1. Moreover, Δ(F) has only cusps and nodes as singularities thus by Theorem 4.2 we have
If d1=d2=d then the discriminant has 4d−2 smooth points at infinity and in each of these points it is tangent to the line L∞ (at infinity) with multiplicity d.
If d1>d2 then the discriminant has only one point at infinity with 2(d1+d2−1) branches V1,…,V2(d1+d2−1) and each of these branches has delta invariant
[TABLE]
and Vi⋅L∞=d1. Additionally Vi⋅Vj=d1(d1−d2). In particular branches Vi are smooth if and only if
d1=d2 or d1=d2+1.
7. Generalized cusps
In this section our aim is to estimate the number of cusps of
non-generic mappings. We start from:
Definition 7.1**.**
Let F:(C2,a)→(C2,F(a)) be a germ of a holomorphic mapping. We say
that F has a generalized cusp at a if Fa is proper, the
curve J(F)=0 is reduced near a and the discriminant of Fa
is not smooth at F(a).
Remark 7.2**.**
If Fa is proper, J(F)=0 is reduced near a and J(F) is
singular at a then it follows from Theorem 1.14 from [10] that
also the discriminant of Fa is singular at F(a) and hence
F has a generalized cusp at a.
Now we introduce the index of generalized cusp:
Definition 7.3**.**
Let F=(f,g):(C2,a)→(C2,F(a)) be a germ of a holomorphic mapping.
Assume that F has a generalized cusp at a point a∈C2.
Since the curve J(F)=0 is reduced near a, we have that the set
{∇f=0}∩{∇g=0} has only isolated points near
a. For a generic linear mapping T∈GL(2), if
F′=(f′,g′)=T∘F then ∇f′ does not vanish
identically on any branch of {J(F)=0} near a. We say that
the cusp of F at a has an index μa:=dimCOa/(J(F′),J1,1(F′))−dimCOa/(fx′,fy′).
Remark 7.4**.**
We show below that the index μa is well-defined and
finite. Moreover, it is easy to see that a simple cusp has index
one.
Let X⊂Cm be a smooth surface.
Let F=(f,g)∈Ωm(d1,d2). Assume that F∣X has a
generalized cusp at a∈X. If Ua⊂X is a sufficiently
small ball around a then μa is equal to the number of
simple cusps in Ua of a mapping F′∣U where F′∈Ωm(d1′,d2′) is a generic mapping, which is sufficiently
close to F in the natural topology of Ωm(d1′,d2′). Here d1′≥d1,d2′≥d2.
Proof.
We can assume that X=C2 and ∇f does not vanish identically on any
branch of {J(F)=0} near a. In particular we have dimOa/(fx,fy)=dimOa/(J(F),fx,fy)<∞.
Let Fi=(fi,gi)∈Ω2(d1′,d2′) be a sequence of
generic mappings, which is convergent to F. Consider the
mappings Φ=(J(F),J1,1(F)), Φi=(J(Fi),J1,1(Fi)), Ψ=(∇f) and Ψi=(∇fi). Thus
Φi→Φ and Ψi→Ψ.
Since a is a cusp of F we have Φ(a)=0. Moreover
da(Φ)<∞, where da(Φ) denotes the local
topological degree of Φ at a. Indeed, if J1,1(F)=0 on
some branch B of the curve J(F)=0 then the rank of F∣B
would be zero and by Sard theorem F has to contract B, which
is a contradiction (Fa is proper). By the Rouche Theorem (see [2], p. 86), we
have that for large i the mapping Φi has exactly
da(Φ) zeroes in Ua and Ψi has exactly
da(Ψ) zeroes in Ua (counted with multiplicities, if
Ψ(a)=0 we put da(Ψ)=0). However, the mappings Fi
are generic, in particular all zeroes of Φi and Ψi are
simple. Moreover the zeroes of Φi which are not cusps of Fi are
zeroes of Ψi.
Hence μa=da(Φ)−da(Ψ) is indeed the number of
simple cusps of Fi in Ua.
∎
Corollary 7.7**.**
Let X be a smooth affine surface. If F=(f,g):X→C2 is an arbitrary polynomial mapping
with degf≤d1, degg≤d2 and generalized cusps at points a1,…,ar then ∑i=1rμai≤cX(d1,d2), where cX(d1,d2) is the number of cusps of a generic mapping from ΩX(d1,d2).
For example we have:
Corollary 7.8**.**
Let F∈Ω2(d1,d2). Assume that F has generalized
cusps at points a1,…,ar. Then ∑i=1rμai≤d12+d22+3d1d2−6d1−6d2+7.
In particular the numbers of singular germs {Fa,a∈C2} which are finitely determined and are not folds, is bounded
by the number d12+d22+3d1d2−6d1−6d2+7.
Proof.
Let Fa be a singular germ which is finitely determined. Then the curve J(Fa) is reduced. There are two possibilities:
the point F(a) is a non-singular point of Δ(F),
the point F(a) is a singular point of Δ(F).
In the case 1) we have by [10] that Fa is equivalent to the germ (x,y)→(xk,y) and since J(Fa) is reduced we have k=2, i.e. Fa is a fold.
In the case 2) Fa is a generalized cusp. Hence the number of germs Fa which are finitely determined and are not folds is bounded by the number of generalized cusps. It follows directly from Theorem 7.6 that the latter number is bounded by the number of cusps of a generic mapping from Ω2(d1,d2).
∎
Remark 7.9**.**
In the same way we can show that for the mapping F∈ΩS(d1,d2) the numbers of singular germs {Fa,a∈S} which are finitely determined and are not folds, is bounded
by the number 2(d12+d22+3d1d2−3d1−3d2+1).
8. Proper deformations
In previous sections we considered the family ΩX(d1,…,dm), of course we can consider also other families of polynomial mappings and try to investigate their properties. Let F be any algebraic family of generically-finite polynomial mappings fp:X→Cm;p∈F, where X is a smooth irreducible affine variety. We would like to know the behavior of proper mappings in such family. In general proper mappings do not form an algebraic subset of F but only constructible one. However we show that there is some regular behavior in such family. We have:
Theorem 8.1**.**
Let P,X,Y be smooth irreducible affine algebraic varieties and let F:P×X→P×Y be a generically finite mapping.
The mapping F induces a family F={fp(⋅)=F(p,⋅),p∈P}. Then either there exists a Zariski open dense subset
U⊂P such that for every p∈U the mapping fp is proper,
or there exists a Zariski open dense subset
V⊂P such that for every p∈V the mapping fp is not proper.
Moreover, in the first case we have:
a) for every non-proper mapping fp in the family F we have μ(fp)<μ(F), where μ(f) denotes the geometric degree of f,
b) generic mappings in F are topologically equivalent, i.e., there exists a Zariski open dense subset W⊂P such that for every p,q∈W the mappings fp and fq are topologically equivalent.
Proof.
First note that for every (p,x)∈P×X we have μ(p,x)(F)=μx(fp) (here μx(f) denotes the local multiplicity of f in x). In the sequel we use the fact that a mapping g:X→Y is proper
over a point y∈Y if and only if ∑g(x)=yμx(g)=μ(g) (see [12], [13]).
Let S be the non-properness set of F (see e.g. [12], [13]). If S=∅, then all mappings fp are proper.
Hence we can assume that S=∅ and consequently S is a hypersurface. Let π:S→P be the canonical projection. We have two possibilities:
(1) π(S) is dense in P.
(2) π(S) is not dense in P.
In the first case π(S) is dense and constructible so a generic mapping fp is not proper. In the second case S has dimension dimP+dimX−1 and a fiber of π has dimension at most dim X. This immediately implies that the set π(S) is a hypersurface in P. Moreover, fibers of π are the whole space X. This means that for all p∈π(S) we have μ(fp)<μ(F). Of course outside π(S) the mappings fp are proper. Two generic mappings are topologically equivalent by [11], Theorem 4.3.
∎
Now we state the main result of this section:
Theorem 8.2**.**
Let X⊂Cn be a smooth irreducible affine variety of dimension k and let F:X→Cm be a polynomial mapping.
If m≥k, then there exists a Zariski open dense subset U in the space of linear mappings L(Cn,Cm)
such that:
a) for every L∈U the mapping F+L is a finite mapping.
b) for all L∈U the mappings F+L are topologically equivalent.
c) for all L∈U the mappings F+L have only generic singularities,i.e., F is transversal to the Thom-Boardman strata.
Proof.
Let G:X∋x↦(x,F(x))∈X×Cm and X~=graph(G)≅X. Since m≥dimX~ a generic linear projection
π:X~→Cm is a proper mapping. Hence also the mapping π∘G is proper.
Consequently for a general matrix A∈GL(m,m) and general linear mapping L∈L(Cn,Cm) the mapping H(A,L)=A(F1,…,Fm)T+L is proper. Hence also the mapping A−1∘H(A,L) is proper. This means that the mapping F+A−1L is proper. But we can specialize the matrix A to the identity and the mapping L to a given linear mapping L0∈L(Cn,Cm). Hence we see that there is a dense subset of linear mappings L∈L(Cn,Cm) such that the mapping F+L:X→Cm is proper. Consider the algebraic family F={F+L,L∈L(Cn,Cm)}. By Theorem
8.1 there exists a Zariski dense open subset U⊂L(Cn,Cm) such that every mapping F+L;L∈U is proper and all these mappings are topologically equivalent. Statement c) follows from Corollary 2.2.
∎
In particular for a given mapping F:C2→C2 we can consider the “linear” deformation FL=F+L;L∈L(C2,C2).
A general member of this deformation is locally stable and proper. If F is not A finitely determined, then this deformation gives in general a different number of cusps and folds than a “generic” deformation considered in this paper. We give here an example of a finitely K determined germ F which has at least two non-equivalent stable deformations.
Example 8.3**.**
Take F(x,y)=(x,y3). This germ is finitely K determined, because it is K equivalent to the cusp G(x,y)=(x,y3+xy), which is stable. Consider two deformations of F: the first one linear Ft=(x,y3+ty) and the second one given by Gt(x,y)=(x,y3+txy). The members of the first family do not have a cusp at all and the members of the second family have exactly one cusp at [math].
This means that (contrary to the case of A finitely determined germs) we can not define the numbers c(F) and d(F) for F using stable deformations.
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