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 : C 2 → C be a polynomial map. Let C 2 ⊂ X be a compactification of C 2 where X is a smooth rational compact sur- face and such that there exists a holomorphic map Φ : X → P 1 which extends f . Put D = X \ C 2 ; 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 ) = a 0 ( x, y ) + a 1 ( x, y ) + · · · a d ( x, y ) where a j ( x, y ) is homogeneous of degree j . Let F ( x, y, z ) = � d 0 z d − j a j ( x, y ). Consider the rational map: P 2 P 1 Φ 0 : → [ F ( x, y, z ) : z d ] [ x : y : z ] �→ It is well defined outside A ( f ) = { [ x 0 : y 0 : 0] | a d ( x 0 , y 0 ) = 0 }

The composition π : X → P 2 of a suitable sequence of blowing-up maps over A ( f ) gives the required compactification, where Φ = Φ 0 ◦ π . D = π − 1 ( L ∞ ) Let E be an irreducible component of D . If Φ( E ) = P 1 , we say that E is a dicritical com- ponent of D . We denote by D dic the set of dicritical components. We set D ∞ = Φ − 1 ( ∞ ). We have that D ∞ is connected and every component of D dic intersects D ∞ .

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

E 2 E 1 E 2 E 1 E 4 E 3 E 3 E 4

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 v 0 ), and the dicriticals. We shall assume that the valency of the root is at least 2.

N v a v (0) N v = a v N v 1 N v 1 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 N v 1 | N v and we decorate the edge between v and v 1 near v by a v = N v /N v 1 . 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 | N v .

v 0 (2) (3) (21) (42) (2) (42) 2 3 (0) (0) (3) (21) (21) (3) (3) (3) We introduce a partial order on the tree, by saying that v < v ′ if and only if v is on the path between v 0 and v ′ .

3: More properties and simplifications of the tree For each dicritical vertex u there is a unique vertex v � = u, v � = v 0 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 D v the set of dicriticals attached to v . The type of a node v is the unordered sequence of the degrees of the dicriticals in D v . From now on we shall not represent the dicriticals, only the nodes, their multiplicities, a v and the type. (The data N v , a v , 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) (3) (21) (42) (2) (42) 2 3 (0) (0) (3) (21) (21) (3) (3) (3) 2,2,[2] 3,3 21,1,[21] 42,1,[21,42] 2 3 3,1,[3,3,3] (0) (0)

Proposition 1 ∀ v ∈ Nd, ∀ u ∈ D v , N v = k u d u We introduce some notation: 1. If v ∈ Nd, σ ( v ) = � u ∈D v ( k u − 1) d u , d v = gcd d u . u ∈D v 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)( N v − 1)+ N v (1 − 1 /a v )

Remarks: 1. If N v = 1, then a v = 1 , k u = 1 ∀ u ∈ D v , and ˜ ∆( v ) = 0. Moreover ǫ ( v ) = 1. 2. If N v � = 1 (a) ǫ ( v ) > 2 implies ˜ ∆( v ) > 0 (b) ǫ ( v ) = 2 and ˜ ∆( v ) = 0 implies N v , 1 , [ N v , · · · , N v ] (c) ǫ ( v ) = 1 and ˜ ∆( v ) ≤ 0 implies N v , a v , [ d v , N v , · · · , d v | N v and if d v � = N v , then a v = 1.

Let ∆( N ∗ ) = ˜ ˜ � ∆( v ) v ∈N ∗ In the following picture the tree is decorated with ˜ ∆( v ) ˜ ˜ ˜ ∆ = 0 ∆ = 2 ∆ = 20 ˜ ∆ = − 20 2 3 (0) (0) ˜ ∆ = − 2

We have Theorem 2 ∆( N ∗ ) = 2 g ˜ 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 ∈ D v , N v = k u d u For v ∈ N ∗ ˜ � ∆( v ) = ( k u − 1) d u +( ǫ ( v ) − 2)( N v − 1)+ N v (1 − 1 /a v ) u ∈D v ∆( N ∗ ) = ˜ ˜ � ∆( v ) v ∈N ∗ ∆( N ∗ ) = 2 g ˜

We introduce more notation: u e N ( u, e ) and ˜ v ∈N ( u,e ) ˜ ∆( u, e ) = � ∆( v )

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

Finally, we introduce the characteristic sequence: If ( u, e ) is minimal, i.e. e = ( u, u 0 ), and ǫ ( u 0 ) = 1, we define c ( u, e ) = d ( u 0 ) a u 0 . If ( u, e ) is not minimal c ( u, e ) = gcd( d ( u 0 ) , c ( u 0 , e 1 ) , · · · , c ( u 0 , e n )) a u 0 u u 0 e 1 e e n

u e N ( u, e ) and ˜ v ∈N ( u,e ) ˜ ∆( u, e ) = � ∆( v ) We say that ( u ′ , e ′ ) < ( u, e ) if the path γ u,u ′ contains e but not e ′ u u’ e 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 ( z 1 , · · · , z n ) is ˜ ∆-trivial if for i = 2 , · · · , n − 1, ǫ ( z i ) = 2 and ˜ ∆( z i ) = 0. We consider Γ = { γ = ( z 1 , · · · , z n ) , ǫ ( z 1 ) = 1 , ˜ ∆( z 1 ) ≤ 0 , ∆( z n ) > 0 , z n − 1 > z n , ˜ ˜ ∆ − trivial } If ( z 1 , · · · , z n ) ∈ Γ, we say that ( z n , ( z n , z n − 1 )) is a tooth. We can prove that c ( z n , ( z n , z n − 1 )) = d ( z 1 ) a z 1 .

Remarks: 1. If N v � = 1 (a) ǫ ( v ) = 2 and ˜ ∆( v ) = 0 implies N v , 1 , [ N v , · · · , N v ] (b) ǫ ( v ) = 1 and ˜ ∆( v ) ≤ 0 implies N v , a v , [ d v , · · · , N v ], d v | N v and if d v � = N v , then a v = 1.

From N ∗ we define the skeleton S : If ( z 1 , · · · , z n ) ∈ Γ, we delete z 1 , · · · , z n − 1 and the edges between them and between z n and z n − 1 . We keep z n and the information d z 1 /a z 1 and ˜ ∆( z 1 ). S is a tree. 3,3 21,1,[21] 42,1,[21,42] 1,0 3,-2

Description of the Skeleton Let Ω = { z ∈ S| ǫ ( z ) = 1 , ˜ ∆( z ) ≤ 0 } Proposition 3 | Ω | ≤ 2 If | Ω | = 2, N ∗ = { z 1 , · · · , z n } , with ǫ ( z j ) = 1 , ˜ ∆( z j ) ≤ 0 , j = 1 , n, ǫ ( z i ) = 2 , ˜ ∆( z i ) = 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 e v as the edge incident to v on the path γ v,z . We can consider the sequences c ( v, e v ) and ˜ ∆( v, e v ). We shall use also η ( v, e v ) = ˜ ∆( v, e v ) − (1 − c ( v, e v )). One can prove that η ( v, e v ) is an increasing sequence. v 2 v 1 z e v 2 e v 1 c ( v 1 , e v 1 ) = 21; ˜ ∆( v 1 , e v 1 ) = − 20; η ( v 1 , e v 1 ) = 0 c ( v 2 , e v 2 ) = 3; ˜ ∆( v 2 , e v 2 ) = − 2; η ( v 2 , e v 2 ) = 0

We say that ( u, e u ) is a comb over ( u ′ , e u ′ ) if ( u ′ , e u ′ ) < = ( u, e u ), and η ( u, e u ) = η ( u ′ , e u ′ ). Remark that ( u, e u ) is a comb over itself. If ( u, e u ) is a comb over ( u ′ , e u ′ ), we have for all ( v, e v ) such that ( u ′ , e u ′ ) < = ( v, e v ) < ( u, e u ) 1. ǫ ( v ) = 2 and N v , a v , [ d v , N v , · · · , N v ] , d v | N v and d v � = N v implies a v = 1 2. ǫ ( v ) = 3 and N v , 1 , [ N v , N v , · · · , N v ] and ( v, g ) is a tooth.

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