Polynomials in two variables and their trees at infinity Pierrette - - PowerPoint PPT Presentation

Polynomials in two variables and their trees at infinity

Pierrette Cassou-Nogu` es, Daniel Daigle IMB (Universit´ e de Bordeaux ), Universit´ e d’Ottawa Warsaw, May 2018

Let f : C2 → C be a polynomial map. Let C2 ⊂ X be a compactification of C2 where X is a smooth rational compact sur- face and such that there exists a holomorphic map Φ : X → P1 which extends f. Put D = X \ C2; D is a curve whose irreducible com- ponents are smooth rational compact curves and all its singularities are ordinary double

• points. The dual graph is a tree.

We are interested in this tree, and we want to anal- yse its shape.

THE TREE 1: Resolution of singularities Let f(x, y) = a0(x, y) + a1(x, y) + · · · ad(x, y) where aj(x, y) is homogeneous of degree j. Let F(x, y, z) = d

0 zd−jaj(x, y). Consider the

rational map: Φ0 : P2 → P1 [x : y : z] → [F(x, y, z) : zd] It is well defined outside A(f) = {[x0 : y0 : 0]|ad(x0, y0) = 0}

The composition π : X → P2 of a suitable sequence of blowing-up maps over A(f) gives the required compactification, where Φ = Φ0 ◦ π. D = π−1(L∞) Let E be an irreducible component of D. If Φ(E) = P1, we say that E is a dicritical com- ponent of D. We denote by Ddic the set of dicritical components. We set D∞ = Φ−1(∞). We have that D∞ is connected and every component of Ddic intersects D∞.

If Φ(E) = ∞, we denote by m(E) the order

• f the pole of Φ on E. If E is a dicritical we

set m(E) = 0. The affine curve f(x, y) = t, t generic in- tersects each dicritical in a finite number of points that we call the degree of the dicriti- cal.

2: The tree. We consider the dual graph of D. This means that we represent each irreducible compo- nent of D by a vertex and we put an edge between two vertices when the correspond- ing components intersect. We represent the branches of the curve by arrows.

E3 E1 E2 E4 E4 E2 E1 E3

We call valency of a vertex, the number of edges incident to this vertex. We keep information of the multiplicity of the vertices. To simplify the tree, we delete vertices of valency 2, except the one representing L∞, (that we call the root, and denote by v0), and the dicriticals. We shall assume that the valency of the root is at least 2.

(0) Nv = avNv1 Nv Nv1 av

If we have a vertex of valency 1 we replace it by an arrow decorated by 0 so that all vertices of valency 1 are arrows. We call that a dead end.

If v is not a dicritical vertex, we have that Nv1|Nv and we decorate the edge between v and v1 near v by av = Nv/Nv1. There is at most one dead end attached to a vertex.

Then we end up with a rooted tree.

• Each vertex of valency 1 is an arrow.
• The vertex near an arrow which is not

decorated with (0), is a dicritical vertex, whose degree is the number of arrows, not decorated by (0), attached to it.

• Each vertex is decorated by its

multiplicity and each dead end is decorated near the vertex v by a positive integer a such that a|Nv.

(2) (2) (0) 2 (3) (0) 3 (21) (3) (3) (3) (21) (21) (42) (42) (3) v0

We introduce a partial order on the tree, by saying that v < v′ if and only if v is on the path between v0 and v′.

3: More properties and simplifications of the tree For each dicritical vertex u there is a unique vertex v = u, v = v0 such that the path between u and v is linear. We call v the companion of u. A node is a vertex which is not a dicritical and which is the companion of at least one dicritical vertex. The set of nodes is denoted by Nd.

If v ∈ Nd we denote by Dv the set of dicriticals attached to v. The type of a node v is the unordered sequence of the degrees of the dicriticals in Dv. From now

• n we shall not represent the dicriticals,
• nly the nodes, their multiplicities,av and

the type. (The data Nv, av, type is called the signature of v).We denote by N ∗ the set of vertices that remain when we have deleted the dicritical vertices and eventually the root. N ∗ is a tree.

(2) (2) (0) 2 (3) (0) 3 (21) (3) (3) (3) (21) (21) (42) (42) (3) (0) (0) 2 3 3,1,[3,3,3] 2,2,[2] 3,3 21,1,[21] 42,1,[21,42]

Proposition 1 ∀v ∈ Nd, ∀u ∈ Dv, Nv = kudu We introduce some notation:

• 1. If v ∈ Nd, σ(v) =

u∈Dv(ku − 1)du,

dv = gcd

u∈Dv

du. If v ∈ N ∗ \ Nd, σ(v) = 0

• 2. For v ∈ N ∗, ǫ(v) = |{x ∈ N ∗|x is adjacent to v}|.

It is the valency in N ∗.

• 3. For v ∈ N ∗

˜ ∆(v) = σ(v)+(ǫ(v)−2)(Nv−1)+Nv(1−1/av)

Remarks:

• 1. If Nv = 1, then av = 1, ku = 1∀u ∈ Dv,

and ˜ ∆(v) = 0. Moreover ǫ(v) = 1.

• 2. If Nv = 1

(a) ǫ(v) > 2 implies ˜ ∆(v) > 0 (b) ǫ(v) = 2 and ˜ ∆(v) = 0 implies Nv, 1, [Nv, · · · , Nv] (c) ǫ(v) = 1 and ˜ ∆(v) ≤ 0 implies Nv, av, [dv, Nv, · · · , dv|Nv and if dv = Nv, then av = 1.

Let ˜ ∆(N ∗) =

• v∈N ∗

˜ ∆(v) In the following picture the tree is decorated with ˜ ∆(v)

(0) (0) 2 3 ˜ ∆ = 0 ˜ ∆ = 20 ˜ ∆ = −20 ˜ ∆ = −2 ˜ ∆ = 2

We have Theorem 2 ˜ ∆(N ∗) = 2g where g is the genus of the generic fiber of f. We want to study the complexity of the tree in terms of the genus of the generic fiber. We assume that N ∗ has at least 2 elements.

N ∗ is the set of vertices that remain when we have deleted the dicritical vertices and eventually the root. ∀v ∈ Nd, ∀u ∈ Dv, Nv = kudu For v ∈ N ∗ ˜ ∆(v) =

• u∈Dv

(ku−1)du+(ǫ(v)−2)(Nv−1)+Nv(1−1/av) ˜ ∆(N ∗) =

• v∈N ∗

˜ ∆(v) ˜ ∆(N ∗) = 2g

We introduce more notation:

u e N(u, e)

and ˜ ∆(u, e) =

v∈N(u,e) ˜

∆(v)

We say that (u′, e′) < (u, e) if the path γu,u′ contains e but not e′

e u u’ e’

Finally, we introduce the characteristic sequence: If (u, e) is minimal, i.e. e = (u, u0), and ǫ(u0) = 1, we define c(u, e) = d(u0)

au0 .

If (u, e) is not minimal c(u, e) = gcd(d(u0), c(u0, e1), · · · , c(u0, en)) au0

en u e u0 e1

u e N(u, e)

and ˜ ∆(u, e) =

v∈N(u,e) ˜

∆(v) We say that (u′, e′) < (u, e) if the path γu,u′ contains e but not e′

e u u’ e’

The characteristic sequence c(u, e) is a decreasing sequence defined by induction

4: Last stage of simplification: The skeleton We say that a path (z1, · · · , zn) is ˜ ∆-trivial if for i = 2, · · · , n − 1, ǫ(zi) = 2 and ˜ ∆(zi) = 0. We consider Γ = {γ = (z1, · · · , zn), ǫ(z1) = 1, ˜ ∆(z1) ≤ 0, ˜ ∆(zn) > 0, zn−1 > zn, ˜ ∆ − trivial} If (z1, · · · , zn) ∈ Γ, we say that (zn, (zn, zn−1)) is a tooth. We can prove that c(zn, (zn, zn−1)) = d(z1)

az1 .

Remarks:

• 1. If Nv = 1

(a) ǫ(v) = 2 and ˜ ∆(v) = 0 implies Nv, 1, [Nv, · · · , Nv] (b) ǫ(v) = 1 and ˜ ∆(v) ≤ 0 implies Nv, av, [dv, · · · , Nv], dv|Nv and if dv = Nv, then av = 1.

From N ∗ we define the skeleton S: If (z1, · · · , zn) ∈ Γ, we delete z1, · · · , zn−1 and the edges between them and between zn and zn−1. We keep zn and the information dz1/az1 and ˜ ∆(z1). S is a tree.

3,3 42,1,[21,42] 3,-2 1,0 21,1,[21]

Description of the Skeleton Let Ω = {z ∈ S|ǫ(z) = 1, ˜ ∆(z) ≤ 0} Proposition 3 |Ω| ≤ 2 If |Ω| = 2, N ∗ = {z1, · · · , zn}, with ǫ(zj) = 1, ˜ ∆(zj) ≤ 0, j = 1, n, ǫ(zi) = 2, ˜ ∆(zi) = 0, i = 2, · · · , n − 1 We assume |Ω| ≤ 1.

We say that the tree is a brush if |S| = 1. We assume that the tree is not a brush. We define In(S) by In(S) = Ω when Ω = ∅, and In(S) = {z ∈ S|δ∗(z) = 1} where δ∗ is the valency in S. We shall describe the skeleton from a z ∈ In(S).

For v ∈ N ∗, we define ev as the edge incident to v on the path γv,z. We can consider the sequences c(v, ev) and ˜ ∆(v, ev). We shall use also η(v, ev) = ˜ ∆(v, ev) − (1 − c(v, ev)). One can prove that η(v, ev) is an increasing sequence.

ev2 v1 z ev1 v2

c(v1, ev1) = 21; ˜ ∆(v1, ev1) = −20; η(v1, ev1) = 0 c(v2, ev2) = 3; ˜ ∆(v2, ev2) = −2; η(v2, ev2) = 0

We say that (u, eu) is a comb over (u′, eu′) if (u′, eu′) <= (u, eu), and η(u, eu) = η(u′, eu′). Remark that (u, eu) is a comb over itself. If (u, eu) is a comb over (u′, eu′), we have for all (v, ev) such that (u′, eu′) <= (v, ev) < (u, eu)

• 1. ǫ(v) = 2 and Nv, av, [dv, Nv, · · · , Nv], dv|Nv

and dv = Nv implies av = 1

• 2. ǫ(v) = 3 and Nv, 1, [Nv, Nv, · · · , Nv] and

(v, g) is a tooth.

Let O = {(u, eu), u ∈ S}. We define an equivalence relation on O: (u, eu) ∼ (u′, eu′) if and only if one of (u, eu) and (u′, eu′) is a comb over the other. This give us a partition of O and we denote by O the set

• f equivalence classes.

Theorem 4 If the tree is not a brush and |Ω| ≤ 1, then

• 1. |O| ≤ 1 + 2g
• 2. If |O| ≥ 2g then Ω = ∅.

Rational polynomials A rational polynomial is a polynomial whose generic fiber is a rational curve. They are also called field generators because f ∈ k[x, y] is a rational polynomial if and

• nly if there exists g ∈ k(x, y) such that

k(f, g) = k(x, y). Rational polynomials such that all dicriticals have degree 1 (also called simple rational polynomials) are classified:

(Miyanishi and Sugie, Neumann and Norbury, C.N and Daigle). The case where all dicriticals have degree

• ne except one dicritical with degree 2 has

been done by Sasao. Let f be a rational polynomial. We assume that f is not a variable. We know (Russell) that we can consider trees with valency of the root equal to 2. Rational trees with |N ∗| ≥ 2

• 1. We don’t know if there exist rational brushes.

What we know is that if they do exist they have at most 3 teeth.

• 2. N ∗ = {z1, · · · , zn}, with ǫ(zj) = 1, ˜

∆(zj) ≤ 0, j = 1, n, ǫ(zi) = 2, ˜ ∆(zi) = 0, i = 2, · · · , n− 1 do exist. Russell’s polynomial of de- gree 21 is an example.

• 3. Theorem 5 Otherwise, we have N ∗ =

v z

z ∈ Ω, one comb with η = 0, and at the end we have at most 3 teeth.

Genus 1 with |N ∗| ≥ 2 We assume |Ω| ≤ 1 and that the tree is not a brush. We have that |O| = 1, 2, 3 and if |O| = 1, Ω = ∅.

z z Ω = ∅ u0 u1 u2 η = 0 |O| = 3 ˜ ∆-trivial ˜ ∆-trivial u0 u1 ˜ ∆-trivial η = 0 |O| = 2 |O| = 1 z u0 η = 0 z u0 ˜ ∆-trivial Ω = ∅

The signatures of u0, u1 and u2 are not completely determined.