Table Of ContentTHE DECOMPOSITION GROUP OF A LINE IN THE PLANE
ISAC HEDE´N AND SUSANNA ZIMMERMANN
Abstract. We show that the decomposition group of a line L in the plane, i.e. the
subgroup of plane birational transformations that send L to itself birationally, is gen-
erated by its elements of degree 1 and one element of degree 2, and that it does not
decompose as a non-trivial amalgamated product.
6
1
0 1. Introduction
2
We denote by Bir(P2) the group of birational transformations of the projective plane
r
p P2 = Proj(k[x,y,z]), where k is an algebraically closed field. Let C ⊂ P2 be a curve,
A
and let
6
2 Dec(C) = {ϕ ∈ Bir(P2), ϕ(C) ⊂ C and ϕ|C : C (cid:57)(cid:57)(cid:75) C is birational}.
This group has been studied for curves of genus ≥ 1 in [BPV2009], where it is linked to
]
G the classification of finite subgroups of Bir(P2). It has a natural subgroup Ine(C), the
A
inertia group of C, consisting of elements that fix C, and Blanc, Pan and Vust give the
h. following result: for any line L ⊂ P2, the action of Dec(L) on L induces a split exact
t sequence
a
m 0 −→ Ine(L) −→ Dec(L) −→ PGL = Aut(L) −→ 0
2
[
and Ine(L) is neither finite nor abelian and also it doesn’t leave any pencil of rational
2 curves invariant [BPV2009, Proposition 4.1]. Further they ask the question whether
v
Dec(L) is generated by its elements of degree 1 and 2 [BPV2009, Question 4.1.2].
5
2 We give an affirmative answer to their question in the form of the following result,
7 similar to the Noether-Castelnuovo theorem [Cas1901] which states that Bir(P2) is gen-
2
0 erated by σ: [x : y : z] (cid:55)(cid:57)(cid:57)(cid:75) [yz : xz : xy] and Aut(P2) = PGL3.
.
1 Theorem 1. For any line L ⊂ P2, the group Dec(L) is generated by Dec(L) ∩ PGL
0 3
6 and any of its quadratic elements having three proper base points in P2.
1
: The similarities between Dec(L) and Bir(P2) go further than this. Cornulier shows in
v
i [Cor2013] that Bir(P2) cannot be written as an amalgamated product in any nontrivial
X
way, and we modify his proof to obtain an analogous result for Dec(L).
r
a
Theorem 2. The decomposition group Dec(L) of a line L ⊂ P2 does not decompose as
a non-trivial amalgam.
Thearticleisorganisedasfollows:inSection2weshowthatforanyelementofDec(L)
we can find a decomposition in Bir(P2) into quadratic maps such that the successive
images of L are curves (Proposition 2.6), i.e. the line is not contracted to a point at any
Date: April 27, 2016.
2010 Mathematics Subject Classification. 14E07.
ThefirstnamedauthorisanInternationalResearchFellowoftheJapaneseSocietyforthePromotion
of Sciences, and this work was supported by Grant-in-Aid for JSPS Fellows Number 15F15751. Both
authors acknowledge support by the Swiss National Science Foundation Grant ”Birational Geometry”
PP00P2 153026.
1
2 ISAC HEDE´N AND SUSANNA ZIMMERMANN
time. We then show in Section 3 that we can modify this decomposition, still in Bir(P2),
into de Jonqui`eres maps where all of the successive images of L have degree 1, i.e. they
are lines. Finally we prove Theorem 1. Our main sources of inspiration for techniques
and ideas in Section 3 have been [AC2002, §8.4, §8.5] and [Bla2012]. In Section 4 we
prove Theorem 2 using ideas that are strongly inspired by [Cor2013].
Acknowledgement: The authors would like to thank J´er´emy Blanc for helpful dis-
cussions, and Yves de Cornulier for kindly answering their questions.
2. Avoiding to contract L
Given a birational map ρ: P2 (cid:57)(cid:57)(cid:75) P2, the Noether-Castelnuovo theorem states that
there is a decomposition ρ = ρ ρ ...ρ of ρ where each ρ is a quadratic map with
m m−1 1 i
three proper base points. This decomposition is far from unique, and the aim of this
section is to show that if ρ ∈ Dec(L), we can choose the ρ so that none of the successive
i
birational maps (ρ ...ρ : P2 (cid:57)(cid:57)(cid:75) P2)m contracts L to a point. This is Proposition 2.6.
i 1 i=1
Given a birational map ϕ : X (cid:57)(cid:57)(cid:75) Y between smooth projective surfaces, and a
curve C ⊂ X which is contracted by ϕ, we denote by π : Z → Y the blowup of the
1 1
point ϕ(C) ∈ Y. If C is contracted also by the birational map π−1ϕ: X (cid:57)(cid:57)(cid:75) Z , we
1 1
denote by π : Z → Z the blowup of (π−1ϕ)(C) ∈ Z and consider the birational
2 2 1 1 1
map (π π )−1ϕ: X (cid:57)(cid:57)(cid:75) Z . If this map too contracts C, we denote by π : Z → Z
1 2 2 3 3 2
the blowup of the point onto which C is contracted. Repeating this procedure a finite
numberoftimesD ∈ N,wefinallyarriveatavarietyZ := Z andabirationalmorphism
D
π := π π ···π : Z → Y such that (π−1ϕ) does not contract C. Then (π−1ϕ)| : C (cid:57)(cid:57)(cid:75)
1 2 D C
(π−1ϕ)(C) is a birational map.
Definition 2.1. In the above situation, we denote by D(C,ϕ) ∈ N the minimal number
of blowups which are needed in order to not contract the curve C and we say that C is
contracted D(C,ϕ) times by ϕ. In particular, a curve C is sent to a curve by ϕ if and
only if D(C,ϕ) = 0.
Remark 2.2. The integer D(C,ϕ) can equivalently be defined as the order of vanishing
of K −π∗(K ) along (π−1ϕ)(C).
Z Y
We recall the following well known fact, which will be used a number of times in the
sequel.
Lemma 2.3. Let ϕ ,ϕ ∈ Bir(P2) be birational maps of degree 2 with proper base points
1 2
p ,p ,p and q ,q ,q respectively. If ϕ and ϕ have (exactly) two common base points,
1 2 3 1 2 3 1 2
say p = q and p = q , then the composition τ = ϕ ϕ−1 is quadratic. Furthermore the
1 1 2 2 2 1
three base points of τ are proper points of P2 if and only if q is not on any of the lines
3
joining two of the p .
i
Proof. The lemma is proved by Figure 1, where squares and circles in P2 denote the
2
base points of ϕ and ϕ respectively. The crosses in P2 denote the base points of ϕ−1
1 2 1 1
(corresponding to the lines in P2), and the conics in P2 and P2 denote the pullback of a
2 1 2
general line (cid:96) ∈ P2.
3
If q is not on any of the three lines, the base points of τ are E ,E ,ϕ (q ). If q is on
3 1 2 1 3 3
one of the three lines, then the base points of τ are E ,E and a point infinitely close
1 2
to the E which corresponds to the line that q is on. (cid:3)
i 3
THE DECOMPOSITION GROUP OF A LINE IN THE PLANE 3
P2 C =ϕ∗((cid:96))
2
2
P2 τ∗((cid:96))
p =q
1 2 2
E
3 E
1
ϕ1 q ϕ2 P2
3 3
ϕ (q )
1 3 (cid:96)
p =q
1 1
p
3
E
2
τ
Figure 1. The composition of ϕ and ϕ in Lemma 2.3
1 2
The following lemma describes how the number of times that a line is contracted
changes when composing with a quadratic transformation of P2 with three proper base
points.
Lemma 2.4. Let ρ: P2 (cid:57)(cid:57)(cid:75) P2 be a birational map and let ϕ: P2 (cid:57)(cid:57)(cid:75) P2 be a quadratic
birational map with base points q ,q ,q ∈ P2. For 1 ≤ i < j ≤ 3 we denote by (cid:96) ⊂ P2
1 2 3 ij
the line which joins the base points q and q . If D(L,ρ) = k ≥ 1, we have
i j
k +1 if ρ(L) ∈ ((cid:96) ∪(cid:96) ∪(cid:96) )\Bp(ϕ),
12 13 23
k if ρ(L) ∈/ (cid:96) ∪(cid:96) ∪(cid:96) ,
12 13 23
D(L,ϕρ) =
k if ρ(L) = q for some i, and (ρϕ)(L) ∈ Bp(ϕ−1),
i
k −1 if ρ(L) = q for some i, and (ρϕ)(L) ∈/ Bp(ϕ−1).
i
Proof. We consider the minimal resolutions of ϕ; in Figures 2-5, the filled black dots
denote the successive images of L, i.e. ρ(L), (π−1ρ)(L) and (ηπ−1ρ)(L) respectively.
We argue by Figure 2 and 3 in the case where ρ(L) does not coincide with any of
the base points of ϕ. If ρ(L) ∈ (cid:96) for some i,j, then D(L,ϕρ) = D(L,ρ)+1, since (cid:96)
ij ij
is contracted by ϕ. Otherwise, the number of times L is contracted does not change.
Suppose that ρ(L) = q for some i. If D(L,ρ) = 1, we have (π−1ρ)(L) = E , and then
i i
(cid:96)˜ik (cid:96)˜jk (cid:96)˜ik (cid:96)˜jk
(cid:96)˜ (cid:96)˜
ij ij
π η π η
P2 P2 P2 P2 P2 P2
q q
k k
ρ ϕ ρ ϕ
q q
i i
L q L q
j j
Figure 2. D(L,ϕρ) = k+1; Figure 3. D(L,ϕρ) = k;
ρ(L) ∈ ((cid:96) ∪(cid:96) ∪(cid:96) )\Bp(ϕ). ρ(L) ∈/ (cid:96) ∪(cid:96) ∪(cid:96) .
12 13 23 12 13 23
4 ISAC HEDE´N AND SUSANNA ZIMMERMANN
clearly D(L,ϕρ) = 0 since E is not contracted by η. If D(L,ρ) ≥ 2 we argue by the
i
Figures 4 and 5.
(cid:96)˜ik (cid:96)˜jk (cid:96)˜ik (cid:96)˜jk
(cid:96)˜ (cid:96)˜
ij ij
π η π η
P2 P2 P2 P2 P2 P2
q q
k k
ρ ϕ ρ ϕ
q q
i i
L q L q
j j
Figure 4. D(L,ϕρ) = k; Figure 5. D(L,ϕρ) = k − 1;
ρ(L) = q and (ρϕ)(L) ∈ Bp(ϕ−1). ρ(L) = q and (ρϕ)(L) ∈/ Bp(ϕ−1).
i i
(cid:3)
Remark 2.5. If D(L,ρ) ≥ 2, then the point (π−1ρ)(L) in the first neighbourhood of
ρ(L) defines a tangent direction at ρ(L) ∈ P2. If we take ϕ as in Lemma 2.4 with
q ∈ Bp(ϕ) for some i, then this tangent direction coincides with the direction of one of
i
(cid:96) ,(cid:96) if and only if (ρϕ)(L) ∈ Bp(ϕ−1).
ij ik
Proposition 2.6. For any given element ρ ∈ Dec(L), there is a decomposition of ρ
into quadratic maps ρ = ρ ...ρ with three proper base points such that none of the
m 1
successive compositions (ρ ...ρ )m contract L to a point.
i 1 i=1
Proof. Let ρ = ρ ...ρ be a decomposition of ρ into quadratic maps with only proper
m 1
base points. We can assume that d := max{D(L,ρ ...ρ ) | 1 ≤ j ≤ m} > 0, otherwise
j 1
we are done. Let n := max{j | D(L,ρ ...ρ ) = d}. We denote the base points of ρ−1
j 1 n
and ρ by p ,p ,p and q ,q ,q respectively.
n+1 1 2 3 1 2 3
We first look at the case where D(L,ρ ...ρ ) = D(L,ρ ...ρ ) = d − 1. Then
n−1 1 n+1 1
composition with ρ and ρ fall under Cases 1 and 4 of Lemma 2.4, so both ρ−1
n n+1 n
and ρ have a base point at (ρ ...ρ )(L) ∈ P2. We may assume that this point is
n+1 n 1
p = q , as in Figure 6. Interchanging the roles of q and q if necessary, we may assume
1 1 2 3
that p ,p ,q are not collinear. Let r ∈ P2 be a general point, and let c and c denote
1 2 2 1 2
quadratic maps with base points [p ,p ,r] and [p ,q ,r] respectively; then the maps
1 2 1 2
τ ,τ ,τ (defined by the commutative diagram in Figure 6) are quadratic with three
1 2 3
proper base points in P2. Note that D(L,τ ...τ ρ ...ρ ) = d − 1 for i = 1,2,3.
i 1 n−1 1
Thus we obtained a new decomposition of ρ into quadratic maps with three proper base
points
ρ = ρ ...ρ τ τ τ ρ ...ρ ,
m n+2 3 2 1 n−1 1
where the number of instances where L is contracted d times has decreased by 1.
Now assume instead that D(L,ρ ...ρ ) = d and D(L,ρ ...ρ ) = d−1. Then
n−1 1 n+1 1
composition with ρ falls under Case 4 of Lemma 2.4, so (ρ ...ρ )(L) is a base point
n+1 n 1
of ρ , which we may assume to be q . Furthermore composition with ρ falls under
n+1 1 n
THE DECOMPOSITION GROUP OF A LINE IN THE PLANE 5
P2
P2 p P2
2
r
q
2
q
3
p
3
p =q
1 1
ρ−1 ρ
n n+1
c c
ρ ...ρ 1 2 ρ ...ρ
n−1 1 m n+2
L P2 P2 L
τ τ
1 τ 3
P2 2 P2
Figure 6. The decomposition of ρ ρ into quadratic maps τ ,τ ,τ
n+1 n 1 2 3
Cases 2 or 3 of Lemma 2.4, so (ρ ...ρ )(L) either does not lie on a line joining two base
n 1
points of ρ−1, or D(L,ρ ...ρ ) ≥ 2 and (ρ ...ρ )(L) is a base point of ρ−1 (which we
n n 1 n 1 n
may assume to be p , and equal to q ), at the same time as (ρ ...ρ )(L) is a base
1 1 n−1 1
point of ρ .
n
We consider the first case. If D(L,ρ ...ρ ) ≥ 2 so that L defines a tangent direction
n 1
at (ρ ...ρ )(L), then this tangent direction has to be different from at least two of the
n 1
three directions at q that are defined by the lines through q and the p , i = 1,2,3. By
1 1 i
renumbering the p , we may assume that p ,p define these two directions (no renum-
i 2 3
bering is needed if D(L,ρ ...,ρ ) = 1). Then with a quadratic map c := [q ,p ,p ]
n 1 1 1 2 3
with base points q ,p ,p , we are in Case 4 of Lemma 2.4 and obtain D(L,c ρ ...ρ ) =
1 2 3 1 n 1
D(L,ρ ...ρ ) − 1. Let r,s ∈ P2 be two general points and define c ,c ,c with three
n 1 2 3 4
properbasepointsrespectivelyas[q ,r,p ], [q ,r,s], [q ,q ,s].Notethatthecorrespond-
1 3 1 1 2
ing maps τ ,...,τ , defined in an analogous way as in Figure 6, are quadratic with three
1 5
proper base points. Note also that D(L,c ρ ...ρ ) = D(L,ρ ...ρ )−1 for i = 2,3,4.
i n 1 n 1
Only for i = 4 this is not immediately clear, so suppose that this is not the case, i.e.
D(L,c ρ ...ρ ) = D(L,ρ ...ρ ). It follows that D(L,ρ ...ρ ) ≥ 2 and that the tan-
4 n 1 n 1 n 1
gent direction corresponding to (ρ ...ρ )(L) is given by the line through q and q , but
n 1 1 2
this is not possible by the assumption that D(L,ρ ...ρ ) = d−1.
n+1 1
In the second case we have p = q and the tangent direction at p = q corresponding
1 1 1 1
to(ρ ...ρ )(L)isthedirectioneitherofthelinethroughp andp orthelinethroughp
n 1 1 2 1
andp (seeFigure4).Byinterchangingtherolesofp andp ifnecessary,wemayassume
3 2 3
that it corresponds to the direction of the line through p and p . Interchanging the roles
1 3
of q and q if necessary, we may assume that p ,q ,p are not collinear. Let r,s ∈ P2
2 3 1 2 3
be general points and define quadratic maps c ,c ,c with three proper base points
1 2 3
respectively by [p ,p ,s], [p ,r,s], [p ,r,q ]. Then the corresponding maps τ ,τ ,τ ,τ
1 2 1 1 2 1 2 3 4
are quadratic with three proper base points and D(L,c ρ ...ρ ) = D(L,ρ ...ρ )−1
i n 1 n 1
for i = 1,2,3. The latter holds for c since the direction given by p and p is different
1 1 2
from the tangent direction corresponding to (ρ ...ρ )(L), and for c it follows from
n 1 3
the assumption that the image of L is contracted d−1 times by (ρ ...ρ ) and that
n+1 1
p ,q ,p are not collinear.
1 2 3
Both in the first and second case, we again arrive at a new decomposition into qua-
dratic maps with three proper base points
ρ = ρ ...ρ τ ...τ ρ ...ρ (j ∈ {4,5}),
m n+2 j 1 n−1 1
where the number of instances where L is contracted d times has decreased by 1, and
we conclude by induction. (cid:3)
6 ISAC HEDE´N AND SUSANNA ZIMMERMANN
3. Avoiding to send L to a curve of degree higher than 1.
By Proposition 2.6, any element ρ ∈ Dec(L) can be decomposed as
ρ = ρ ...ρ
m 1
whereeachρ isquadraticwiththreeproperbasepoints,andallofthesuccessiveimages
j
((ρ ...ρ )(L))m of L are curves. The aim of this section is to show that the ρ even
i 1 i=1 j
can be chosen so that all of these curves have degree 1. That is, we find a decomposition
of ρ into quadratic maps such that all the successive images of L are lines. This means
in particular that Dec(L) is generated by its elements of degree 1 and 2.
Definition 3.1. A birational transformation of P2 is called de Jonqui`eres if it preserves
the pencil of lines passing through [1 : 0 : 0] ∈ P2. These transformations form a
subgroup of Bir(P2) which we denote by J.
Remark 3.2. In[AC2002],adeJonqui`eresmapisdefinedbytheslightlylessrestrictive
propertythatitsendsapenciloflinestoapenciloflines.Givenamapwiththisproperty,
we can always obtain an element in J by composing from left and right with elements
of PGL .
3
For a curve C ⊂ P2 and a point p in P2 or infinitely near, we denote by m (p) the
C
multiplicity of C in p. If it is clear from context which curve we are referring to, we will
use the notation m(p).
Lemma 3.3. Let ϕ ∈ J be of degree e ≥ 2, and C ⊂ P2 a curve of degree d. Suppose
that
deg(ϕ(C)) ≤ d.
Then there exist two base points q ,q of ϕ different from [1 : 0 : 0] such that
1 2
m ([1 : 0 : 0])+m (q )+m (q ) ≥ d.
C C 1 C 2
Thisinequalitycanbemadestrictincasedeg(ϕ(C)) < d,withacompletelyanalogous
proof.
Proof. Since ϕ ∈ J is of degree e, it has exactly 2e − 1 base points r := [1 : 0 :
0
0],r ,...,r of multiplicity e−1,1,...,1 respectively. Then
1 2e−2
e−1
(cid:88)
d ≥ deg(ϕ(C)) =ed−(e−1)m (r )− (m (r )+m (r ))
C 0 C 2i−1 C 2i
i=1
e−1
(cid:88)
=d+ (d−m (r )−m (r )−m (r ))
C 0 C 2i−1 C 2i
i=1
Hence there exist i such that d ≤ m (r )+m (r )+m (r ). (cid:3)
0 C 0 C 2i0−1 C 2i0
Remark 3.4. Note also that we can choose the points q ,q such that q either is a
1 2 1
proper point in P2 or in the first neighbourhood of [1 : 0 : 0], and that q either is proper
2
point of P2 or is in the first neighbourhood of [1 : 0 : 0] or q .
1
Remark 3.5. A quadratic map sends a pencil of lines through one of its base points
to a pencil of lines, and we conclude from Proposition 2.6 and Remark 3.2 that there
exists maps α ,...,α ∈ PGL and ρ ∈ J \PGL such that
1 m+1 3 i 3
ρ = α ρ α ρ α ...α ρ α
m+1 m m m−1 m−1 2 1 1
and such that all of the successive images of L with respect to this decomposition are
curves.
THE DECOMPOSITION GROUP OF A LINE IN THE PLANE 7
The following proposition is an analogue of the classical Castelnuovo’s Theorem stat-
ing that any map in Bir(P2) is a product of de Jonqui`eres maps.
Proposition 3.6. Let ρ ∈ Dec(L). Then there exists ρ ∈ J \ PGL and α ∈ PGL
i 3 i 3
such that ρ = α ρ α ρ α ...α ρ α and all of the successive images of L are
m+1 m m m−1 m−1 2 1 1
lines.
Proof. Startwithadecompositionρ = α ρ α ρ α ...α ρ α asinRemark3.5.
m+1 m m m−1 m−1 2 1 1
Denote C := (ρ α ···ρ α )(L) ⊂ P2, d := deg(C ) and let
i i i 1 1 i i
n
(cid:88)
D := max{d | i = 1,...,m}, n := max{i | D = d }, k := (degρ −1).
i i i
i=1
We use induction on the lexicographically ordered pair (D,k).
We may assume that D > 1, otherwise our goal is already achieved. We may also
assume that α ∈/ J, otherwise the pair (D,k) decreases as we replace the three
n+1
maps ρ ,α ,ρ by their composition ρ α ρ ∈ J. Indeed, either D decreases,
n+1 n+1 n n+1 n+1 n
or D stays the same while k decreases at least by degρ −1. Using Lemma 3.3, we find
n
simple base points p ,p of ρ−1 and simple base points q˜ ,q˜ of ρ , all different from
1 2 n 1 2 n+1
p := [1 : 0 : 0], such that
0
m (p )+m (p )+m (p ) ≥ D
Cn 0 Cn 1 Cn 2
and
m (p )+m (q˜ )+m (q˜ ) > D.
αn+1(Cn) 0 αn+1(Cn) 1 αn+1(Cn) 2
We choose p ,p ,q˜ ,q˜ as in Remark 3.4. By slight abuse of notation, we denote by
1 2 1 2
q = α−1 (p ), q = α−1 (q˜ ) and q = α−1 (q˜ ) respectively the (proper or infinitely
0 n+1 0 1 n+1 1 2 n+1 2
near) points in P2 that correspond to p ,q˜ , and q˜ under the isomorphism α−1 . Note
0 1 2 n+1
that p and q are two distinct points of P2 since α ∈/ J. We number the points so
0 0 n+1
that m(p ) ≥ m(p ), m(q˜ ) ≥ m(q˜ ) and so that if p (resp. q˜) is infinitely near p (resp.
1 2 1 2 i i j
q˜), then j < i.
j
We study two cases separately depending on the multiplicities of the base points.
Case (a): m(q ) ≥ m(q ) and m(p ) ≥ m(p ). Then we find two quadratic maps
0 1 0 1
τ(cid:48),τ ∈ J and β ∈ PGL so that ρ α ρ = (ρ τ−1)β(τρ ) and so that the pair
3 n+1 n+1 n n+1 n
(D,k) is reduced as we replace the sequence (ρ ,α ,ρ ) by (ρ τ−1,β,τρ ). The
n+1 n+1 n n+1 n
procedure goes as follows.
If possible we choose a point r ∈ {p ,q } \ {p ,q }. Should this set be empty, i.e.
1 1 0 0
p = q and p = q , we choose r = q instead. The ordering of the points implies
0 1 1 0 2
that the point r is either a proper point in P2 or in the first neighbourhood of p or q .
0 0
Furthermore, the assumption implies that m(p ) + m(q ) + m(r) > D, so r is not on
0 0
the line passing through p and q . In particular, there exists a quadratic map τ ∈ J
0 0
with base points p ,q ,r; then
0 0
deg(τ(C )) = 2D−m(p )−m(q )−m(r) < D.
n 0 0
Chooseβ ∈ PGL sothatthequadraticmapτ(cid:48) := βτ(α )−1 inthebelowcommutative
3 n+1
diagram is de Jonqui`eres – this is possible since τ has q as a base point. This decreases
0
the pair (D,k).
8 ISAC HEDE´N AND SUSANNA ZIMMERMANN
P2 αn+1(cid:47)(cid:47) P2
ρ−n1 ρn+1
(cid:126)(cid:126) (cid:32)(cid:32)
P2 P2
τ τ(cid:48) (cid:62)(cid:62)
(cid:32)(cid:32) (cid:15)(cid:15) (cid:15)(cid:15)
P2 (cid:47)(cid:47) P2
β
Case (b): m(p ) < m(p ). Let τ be a quadratic de Jonqui`eres map with base points
0 1
p ,p ,p .Thisispossiblesinceourassumptionimpliesthatp isaproperbasepointand
0 1 2 1
because p ,p ,p are base points of ρ−1 of multiplicity degρ −1,1,1 respectively and
0 1 2 n n
hence not collinear. Choose β ∈ PGL which exchanges p and p , let γ = α β−1 and
1 3 0 1 n+1 1
choose β ∈ PGL so that τ(cid:48) := β τβ−1 ∈ J. The latter is possible since β−1(p ) = p
2 3 2 1 1 0 1
is a base point of τ, and we have the following diagram.
P2 αn+1 (cid:47)(cid:47) P2
(cid:62)(cid:62)
ρ−n1 β1 γ ρn+1
(cid:127)(cid:127) (cid:32)(cid:32) (cid:31)(cid:31)
P2 P2 P2
τ
τ(cid:48)
(cid:31)(cid:31) (cid:15)(cid:15) (cid:15)(cid:15)
P2 β2 (cid:47)(cid:47) P2
Since deg(τρ ) = degρ − 1, the pair (D,k) stays unchanged as we replace the
n n
sequence (α ,ρ ) in the decomposition of ρ by the sequence (γ,(τ(cid:48))−1,β ,τρ ). In
n+1 n 2 n
the new decomposition of ρ the maps (τ(cid:48))−1 and γ play the roles that ρ and α
n n+1
respectively played in the previous decomposition. In the squared P2, we have
m(p ) = m(β (p )) > m(β (p )) = m(p ).
0 1 1 1 0 1
Define q(cid:48) := γ−1(p ), q(cid:48) := γ−1(q˜ ), q(cid:48) := γ−1(q˜ ), and note that q(cid:48) = β (q ), q(cid:48) = β (q )
0 0 1 1 2 2 0 1 0 1 1 1
and q(cid:48) = β (q ). In the new decomposition these points play the roles that q ,q ,q
2 1 2 0 1 2
played in the previous decomposition.
Ifm(q(cid:48)) ≥ m(q(cid:48)),wecontinueasincase(a)withthepointsp ,p ,β (p )andq(cid:48),q(cid:48),q(cid:48).
0 1 0 1 1 2 0 1 2
If m(q(cid:48)) < m(q(cid:48)), we replace the sequence (ρ ,γ) by a new sequence such that,
0 1 n+1
similar to case (a), the roles of q(cid:48) and q(cid:48) are exchanged, and we will do this without
0 1
touching p ,p ,β(b ). The replacement will not change (D,k) and we can apply case
0 1 2
(a) to the new sequence.
As m(q(cid:48)) < m(q(cid:48)), the point q(cid:48) is a proper point of P2. Analogously to the previous
0 1 1
case, there exists σ ∈ J with base points γ(q(cid:48)) = p ,γ(q(cid:48)) = q˜ ,γ(q(cid:48)) = q˜ , and there
0 0 1 1 2 2
exists δ ∈ PGL which exchanges p and q˜ . Since δ−1(p ) = q˜ is a base point of σ,
1 3 0 1 1 0 1
there furthermore exists δ ∈ PGL such that σ(cid:48) := δ σδ−1 ∈ J. Let γ := δ γ.
2 3 2 1 2 1
THE DECOMPOSITION GROUP OF A LINE IN THE PLANE 9
P2 αn+1 (cid:47)(cid:47)(cid:53)(cid:53) P2
ρ−n1 β1 γ ρn+1
(cid:127)(cid:127) (cid:32)(cid:32) (cid:127)(cid:127) δ1 (cid:31)(cid:31)
P2 P2 (cid:47)(cid:47) P2 P2
τ σ
γ2
τ(cid:48) σ(cid:48)
(cid:31)(cid:31) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:127)(cid:127)
P2 β2 (cid:47)(cid:47) P2 P2 (cid:111)(cid:111) δ2 P2
Replacing the sequence (ρ ,γ) with (ρ σ−1,δ−1,σ(cid:48),δ γ) does not change the pair
n+1 n+1 2 1
(D,k). The latest position with the highest degree is still the squared P2 but in the new
sequence we have
m(γ−1(p )) = m(β (q )) > m(β (q )) = m(γ−1(δ (q˜ )))
2 0 1 1 1 0 2 1 1
Since p ,p ,β (p ) were undisturbed, the inequality m(p ) > m(p ) still holds, and we
0 1 1 2 0 1
proceed as in case (a).
In this proof, we have used several different quadratic maps τ,τ(cid:48),σ,σ(cid:48). Note that none
ofthesecancontractC (oranimageofC),sincequadraticmapsonlycancontractcurves
of degree 1. (cid:3)
Remark 3.7. Suppose that ρ ∈ J preserves a line L. Then the Noether-equalities
imply that L passes either through [1 : 0 : 0] and no other base points of ρ, or that it
passes through exactly degρ−1 simple base points of ρ and not through [1 : 0 : 0].
Lemma 3.8. Let ρ ∈ J be of degree ≥ 2 and let L be a line passing through exactly
degρ−1 simple base points of ρ and not through [1 : 0 : 0]. Then there exist ρ ,...,ρ ∈
1 i
J of degree 2 such that ρ = ρ ···ρ and the successive images of L are lines.
m 1
Proof. Note that the curve ρ(L) is a line not passing through ρ(L). Call p := [1 : 0 :
0
0],p ,...,p the base points of ρ. Without loss of generality, we can assume that
1 2d−2
p ,...,p are the simple base points of ρ that are contained in L and that p is a
1 d−1 1
proper base point in P2. We do induction on the degree of ρ.
If there is no simple proper base point p , i ≥ d, of ρ in P2 that is not on L, choose a
i
general point r ∈ P2. There exists a quadratic transformation τ ∈ J with base points
p ,p ,r. The transformation ρτ−1 ∈ J is of degree degρ and sends the line τ(L) (which
0 1
does not contain [1 : 0 : 0]) onto the line ρ(L). The point ρ(r) ∈ P2 is a base point of
(ρτ−1)−1 not on the line ρ(L).
So, we can assume that there exists a proper base point of ρ in P2 that is not on
L, lets call it p . The points p ,p ,p are not collinear (because of their multiplicities),
d 0 1 d
hence there exists τ ∈ J of degree 2 with base points p ,p ,p . The map ρτ−1 ∈ J is of
0 1 d
degree degρ−1 and τ(L) is a line passing through exactly degρ−2 simple base points
of ρτ−1 and not through [1 : 0 : 0]. (cid:3)
Lemma 3.9. Let ρ ∈ J be of degree ≥ 2 and let L be a line passing through [1 : 0 : 0]
and no other base points of ρ. Then there exist ρ ,...,ρ ∈ J of degree 2 such that
1 m
ρ = ρ ···ρ and the successive images of L are lines.
m 1
Proof. Note that the curve ρ(L) is a line passing through [1 : 0 : 0]. We use induction
on the degree of ρ.
Assume that ρ has no simple proper base points, i.e. all simple base points are infin-
itely near p := [1 : 0 : 0]. There exists a base point p of ρ in the first neighbourhood
0 1
of p . Choose a general point q ∈ P2. There exists τ ∈ J quadratic with base points
0
10 ISAC HEDE´N AND SUSANNA ZIMMERMANN
p ,p ,q. The map ρτ−1 ∈ J is of degree degρ and τ(L) is a line passing through the
0 1
base point p of ρτ−1 of multiplicity degρ−1 and through no other base points of ρτ−1.
0
Moreover, the point ρ(q) is a (simple proper) base point of τρ−1. Therefore, τρ−1 has a
simple proper base point in P2 and sends the line ρ(L) onto the line τ(L), both of which
pass through p and no other base points.
0
So, we can assume that ρ has at least one simple proper base point p . Let p be a
1 2
base point of ρ that is a proper point of P2 or in the first neighbourhood of p or p .
0 1
Because of their multiplicities, the points p ,p ,p are not collinear. Hence there exists
0 1 2
τ ∈ J quadratic with base points p ,p ,p . The map ρτ−1 is a map of degree degρ−1
0 1 2
and τ(L) is a line passing through p and no other base points. (cid:3)
0
Lemma 3.10. Let ρ ∈ J be a map of degree 2 that sends a line L onto a line. Then
there exist quadratic maps ρ ,...,ρ ∈ J with only proper base points such that
1 n
ρ = ρ ···ρ ,
n 1
and the successive images of L are lines.
Proof. Supposefirstthatexactlytwoofthethreebasepointsofρareproper.Wenumber
the base points so that p ,p ∈ P2 and
1 2
P2 (cid:96)1 so that p3 is in the first neighbour-
hood of p , and denote by (cid:96) ⊂ P2
1 1
(cid:96)˜[−1] the line through p which has the
1 1
η r tangent direction defined by p . Choose
3
E [−1] a general point r ∈ P2, and define a
p3
quadratic map ρ with three base points
(cid:96)2 1
(cid:96)˜2[−1] p1 p2 p1,p2,r ∈ P2. A minimal resolution of
ρ is given by π and η as in Figure 7;
Ep1[−2] Ep2[−1] ρ1 it is obtained by blowing up, in order,
ρ
p ,p ,p , and then contracting in order
π 1 2 3
(cid:96)˜ := η−1((cid:96) ), (cid:96)˜ := η−1((cid:96) ) and the
2 ∗ 2 1 ∗ 1
P2 P2 exceptional divisor corresponding to p .
π ((cid:96)˜) 1
∗ 2 By looking at the pull back of a general
ρ(r)
ρ = ρ ρ−1 line in P2 with respect to ρ := ρ ρ−1,
2 1 2 1
we see that this map has three proper
˜
base points E , ρ(r), π ((cid:96) ). This
E p1 ∗ 1
E p3 gives us a decomposition of the desired
p1 π ((cid:96)˜)
∗ 1 form: ρ = ρ−1ρ . Note that since ρ
2 1
sends the line L onto a line, L has
Figure 7. Numbers in square
to pass through exactly one of the
brackets denote self-intersection.
base points of ρ, and this base point
has to be proper. Thus L is sent to a line by ρ . Using the diagram in Figure 7, we can
1
see that this line is further sent by ρ−1 to a line through E if L passes through p and
2 p1 1
˜
a line through π ((cid:96) ) if L passes through p .
∗ 1 2
If [1 : 0 : 0] is the only proper base point of ρ, we reduce to the first case as follows.
Denote by q the base point in the first neighbourhood of [1 : 0 : 0] and choose a general
point r ∈ P2. Let ρ be a quadratic map with base points [1 : 0 : 0],q,r, and let
1
ρ := ρ ρ−1. If we denote the base points of ρ−1 by q ,q ,q so that q is the proper base
2 1 1 2 3 1
point and q the base point in the first neighbourhood of q , then the base points of ρ
2 1 2
are q ,q ,ρ(r), i.e. it has exactly two proper base points.
1 2
