The starred Dixmier conjecture for $A_1$

0.15 MB·

by  Christian Valqui

THE STARRED DIXMIER CONJECTURE FOR A
1
VEREDMOSKOWICZANDCHRISTIANVALQUI
Abstract. LetA1(K)=Khx,y|yx−xy=1ibethefirstWeylalgebraovera
characteristiczerofieldKandletαbetheexchangeinvolutiononA1(K)given
by α(x) = y and α(y) = x. The Dixmier conjecture of Dixmier (1968) asks:
4
IseveryalgebraendomorphismoftheWeylalgebraA1(K)anautomorphism?
1
The aim of this paper is to prove that each α-endomorphism of A1(K) is an
0
automorphism. Hereanα-endomorphismofA1(K)isanendomorphismwhich
2
preservestheinvolutionα. WealsoproveananalogueresultfortheJacobian

n conjectureindimension2,calledα−JC2.
a
J

1
2
1 Introduction

]
A
By definition, the n’th Weyl algebra A (K) = A is the unital associative K-
R n n
algebra generated by 2n elements x ,...,x ,y ,...,y subject to the following
1 n 1 n
.
h defining relations: [y ,x ] = δ , [x ,x ] = 0 and [y ,y ] = 0, where δ is the
i j ij i j i j ij
t Kronecker delta.
a
m HerewewillonlydealwiththefirstWeylalgebraA (K)=Khx,y|yx−xy =1i,
1
where Char(K)=0. (Only in Proposition2.6 the field K is not necessarilyof zero
[
characteristic).

1
In[1]AdjamagboandvandenEssenremarkedthatA wasfirststudiedbyDirac
1
v
in[7]. Hence,theysuggesttocallit“Diracquantumalgebra”insteadof(first)Weyl
1
4 algebra. Similarly,theysuggesttocallAn “n’thDiracquantumalgebra”insteadof
1 n’th Weyl algebra. We truely do not know which name is better. For convenience,
5 in order to maintain the same terminology used by most of the authors, we shall
.
1 continue to call A1 the first Weyl algebra (and An the n’th Weyl algebra).
0 In [8], Dixmier asked six questions about the first Weyl algebra A (K), where
1
4 K is a zero characteristic field; the first question is the following: Is every algebra
1
endomorphism of A (K) an automorphism?
: 1
v Usually,Dixmier’s firstquestionisbroughtasaconjecture;namely, the Dixmier
i conjecture says that every algebra endomorphism of A (K) is an automorphism.
X 1
In order to define a K-algebrahomomorphismf :A −→A , it is enough to fix
r 1 1
a f(x) and f(y) such that [f(y),f(x)] = f(y)f(x)−f(x)f(y) = 1, and extend it as
an algebra homomorphism, and similarly for an antihomomorphism.
Recallthatanantihomomorphismsatisfiesf(ab)=f(b)f(a)andnoticethatthe
mapping α : A −→ A defined by α(x) = y and α(y) = x, is an involution on
1 1
A . Indeed, α is an antihomomorphismof order 2, so it is an antiautomorphismof
1
order 2. This mapping α is sometimes called the exchange involution.
Of course, there are other involutions on A . For example, given any automor-
1
phism g of A , g−1αg is clearly an involution on A .
1 1
Generally,it iseasyto see thateachinvolutiononA is ofthe formhα,where h
1
isanautomorphismofA whichsatisfiesthefollowingconditionhαhα=1. Indeed,
1
2010 Mathematics Subject Classification. Primary16W20, 14R15;Secondary 16S32.
VeredMoskowiczwassupportedpartiallybyanIsrael-USBSFgrant#2010/149.
ChristianValquiwassupportedbyPUCP-DGI-2012-0011andPUCP-DGI-2013-3036.
1
2 VEREDMOSKOWICZANDCHRISTIANVALQUI
let β be any involution on A . Then βα is an automorphismof A , call it h. From
1 1
βα=h follows β =hα. Of course, since β2 =1, we get hαhα=1.
Definition 1.1. An α-endomorphism f of A is an endomorphism of A which
1 1
preservestheinvolutionα. Preservingtheinvolutionαmeansthatforeveryw ∈A ,
1
f(α(w)) = α(f(w)). So an α-endomorphism of A , f, is an endomorphism of A
1 1
which commutes with α (f ◦α=α◦f).
It is easy to see that
f(α(w))=α(f(w)), ∀w∈A ⇐⇒ f(α(x))=α(f(x)), f(α(y))=α(f(y)).
1
Therefore,f isanα-endomorphismofA ,iff isanendomorphismofA ,forwhich
1 1
f(α(x))=α(f(x)) and f(α(y))=α(f(y)).
Now, one may pose the “α-Dixmier conjecture” or the “starredDixmier conjec-
ture”: Every α-endomorphism of A (K) (Char(K)=0) is an automorphism.
1
Remark 1.2. Theexchangeinvolutionαmaybedenotedby“∗ontheright”,instead
of “α on the left” (namely, x∗ =y and y∗ =x instead of α(x) =y and α(y) =x),
hence the name the “starred Dixmier conjecture”.
In the following example we describe a family of α-endomorphisms of any given
even degree 2n (this family also includes degree 1 α-endomorphisms), which is
actually a family of α-automorphisms:
Example 1.3. Let n ∈ N and a,b,c ,...,c ∈ K, with a2−b2 = 1, and define
0 0 n
S :=c (x−y)2j. The following f defines an α-automorphism:
j j
n n
f(x):=ax+by+ S and f(y):=ay+bx+ S .
j j
Xj=0 Xj=0
In fact, clearly α(S ) = S and so f(α(x)) = α(f(x)) and f(α(y)) = α(f(y)).
j j
Moreover,
n n
[f(y),f(x)]=[ay+bx,ax+by]+ay+bx±ax, Sj+ Sj,ax+by±bx
Xj=0 Xj=0
   
n n
=a2−b2+(a+b)x, Sj+ Sj,(a+b)x=1,
Xj=0 Xj=0
   
where the second equality follows from the fact that [x−y,S ]=0. A straightfor-
j
ward computation shows that for
x
f :x7→ , y 7→(a−b)y, f :x7→x+y, y 7→y, f :x7→x−y, y 7→y
2 3 4
a−b
and
n 2j
bx x
f :x7→x, y 7→y− − c ,
1 j
a−b (cid:18)a−b(cid:19)
Xj=0
we have id=f ◦f ◦f ◦f ◦f and so f−1 =f ◦f ◦f ◦f , explicitly
1 2 3 4 4 1 2 3
n n
f−1(x)=ax−by+(b−a) S˜ and f−1(y)=ay−bx+(b−a) S˜ ,
j j
Xj=0 Xj=0
2j
x−y
where S˜ :=c .
j j
(cid:18)a−b(cid:19)
Surprisingly, the family of α-automorphisms of Example 1.3 describes all the
α-endomorphisms of A (K) (see Corollary 2.8). Therefore, the starred Dixmier
1
conjecture is true.
THE STARRED DIXMIER CONJECTURE FOR A1 3
2 Proof of the starred Dixmier conjecture
x+y
Consider the automorphism ϕ of A given by ϕ(x) := and ϕ(y) := y−x.
1
2
y y
Thenϕ−1(x)=x− , ϕ−1(y)=x+ andtheantihomomorphismβ :=ϕ−1◦α◦ϕ
2 2
is given by β(x)=x and β(y)=−y. Notice that β is an involution.
Fortherestofthesectionwefixanα-endomorphismf. Thentheendomorphism
f˜:=ϕ−1◦f satisfies
β(f˜(x))=ϕ−1◦α◦ϕ(ϕ−1◦f(x))=ϕ−1α(f(x))=ϕ−1(f(α(x)))=f˜(y)
and similarly β(f˜(y))=f˜(x). (Note that f˜is NOT a β-endomorphism).
We set P :=f˜(x) andQ:=f˜(y). Since f˜is anendomorphism,we have[Q,P]=1.
We decompose P into its symmetric and antisymmetric terms with respect to β:
P =P +P withP =(P+β(P))/2andP =(P−β(P))/2. Notethatβ(P )=P
0 1 0 1 0 0
and β(P )=−P . Now β(P)=Q implies Q=P −P , and so
1 1 0 1
1=[Q,P]=[P −P ,P +P ]=2[P ,P ],
0 1 0 1 0 1
hence [P ,P ]=1/2.
0 1
We recall some definitions and notations of [9]. Let L := K[X,Y] be the poly-
nomial K-algebrain twovariablesandlet Ψ: A →L be the K-linearmapdefined
1
by Ψ xiyj :=XiYj. Let
(cid:0) (cid:1) V:={(ρ,σ)∈Z2 :gcd(ρ,σ)=1 and ρ+σ ≥0}
and
V:={(ρ,σ)∈V:ρ+σ >0}.
Note that V=V∪{(1,−1),(−1,1)}.
Definition 2.1. For all (ρ,σ)∈V and (i,j)∈Z×Z we write
v (i,j):=ρi+σj.
ρ,σ
Notations 2.2. Let (ρ,σ)∈V. For P = a XiYj ∈L\{0}, we define:
ij
- The support of P as P
Supp(P):={(i,j):a 6=0}.
ij
- The (ρ,σ)-degree of P as v (P):=max{v (i,j):a 6=0}.
ρ,σ ρ,σ ij
- The (ρ,σ)-leading term of P as
ℓ (P):= a XiYj.
ρ,σ ij
{ρi+σjX=vρ,σ(P)}
- w(P):=(i ,i −v (P)) such that
0 0 1,−1
i =max{i:(i,i−v (P))∈Supp(ℓ (P))},
0 1,−1 1,−1
- w(P):=(i −v (P),i ) such that
0 −1,1 0
i =max{i:(i−v (P),i)∈Supp(ℓ (P))},
0 −1,1 −1,1
Notations 2.3. Let (ρ,σ)∈V. For P ∈A \{0}, we define:
1
- The support of P as Supp(P):=Supp Ψ(P) .
- The (ρ,σ)-degree of P as v (P):=v(cid:0) Ψ(P(cid:1)) .
ρ,σ ρ,σ
- The (ρ,σ)-leading term of P as ℓ (P):=(cid:0) ℓ (cid:1)Ψ(P) .
ρ,σ ρ,σ
- w(P):=w Ψ(P) and w(P):=w Ψ(P) . (cid:0) (cid:1)
(cid:0) (cid:1) (cid:0) (cid:1)
4 VEREDMOSKOWICZANDCHRISTIANVALQUI
Definition 2.4. Let (ρ,σ)∈V and let P ∈W \{0}.
- If (ρ,σ)6=(1,−1), then the starting point of P with respect to (ρ,σ) is
st (P)=w(ℓ (P)).
ρ,σ ρ,σ
- If (ρ,σ)6=(−1,1), then the end point of P with respect to (ρ,σ) is
en (P)=w(ℓ (P)).
ρ,σ ρ,σ
Lemma 2.5. Let P,Q∈A with [Q,P]=1. If ℓ (Q)=λXsY for some λ∈K∗,
1 1,0
then s=0.
Proof. Assume by contradiction that s>0 and take T such that [Q,T ]=1 and
0 0
v (T )=min{v (T)| [Q,T]=1, T ∈A }.
1,0 0 1,0 1
Notice that v (T ) > 0, since otherwise T ∈ K[y] and then T = p(y) for some
1,0 0 0 0
polynomial p(y) and [Q,T ] =−λp′(y)y 6=ℓ (1), a contradiction.
0 1,0 1,0
If [Q,T ] 6=0 (See [9, Definition 2.2]), then we obtain the contradiction
0 1,0
0=v ([Q,T ])=v (Q)+v (T )−1>0.
1,0 0 1,0 1,0 0
Ontheotherhand,if[Q,T ] =0thenby[9,Theorem2.11]thereexistR∈L,
0 1,0
λ ,λ ∈K∗ and n,m∈N such that
Q P
ℓ (Q)=λ Rn and ℓ (T )=λ Rm.
1,0 Q 1,0 0 t
Clearly n = 1 and so ℓ1,0(T0) = λλmt (ℓ1,0(Q))m, hence T1 := T0− λλmt Qm satisfies
Q Q
[Q,T ] = 1 and v (T ) < v (T ), which contradicts the minimality of v (T )
1 1,0 1 1,0 0 1,0 0
and concludes the proof. (cid:3)
In the following proposition K is any field with Char(K)6=2, not necessarily a
zero characteristic field.
Proposition 2.6. Let P ,P ∈A (K). Assume 2r6=0 in K, where r =deg (P ).
0 1 1 x 0
If β(P )=P , β(P )=−P and [P ,P ]= 1, then
0 0 1 1 0 1 2
n
1
P =λy and P =− x+ α y2j, for some λ6=0 and α ∈K.
1 0 j j
2λ
Xj=0
Proof. Write ℓ (P )=Xrg(Y) and ℓ (P )=Xsh(Y). Then
1,0 0 1,0 1
ℓ (β(P ))=ℓ (g(−y)xr)=Xrg(−Y),
1,0 0 1,0
and so g(−y)=g(y), i.e., g is even, which means g(y)∈K[y2]. Similarly
ℓ (β(P ))=ℓ (h(−y)xs)=Xsh(−Y),
1,0 1 1,0
and so h(−y)=−h(y), i.e., h is odd, which means h(y)∈yK[y2].
Assume s = 0, then en (P ) = (0,2k+1) for some k ∈ N . If en (P ) ∼
1,0 1 0 1,0 1
en (P ) then P ∈ K[y] which leads to the contradiction [P ,P ] = 0. Hence
1,0 0 0 0 1
en (P ) ≁ en (P ) and so, by [9, Corollary 2.7] we have [P ,P ] 6= 0 and
1,0 1 1,0 0 0 1 1,0
by [9, Proposition2.4]
en (P )+en (P )−(1,1)=en ([P ,P ])=(0,0).
1,0 0 1,0 1 1,0 0 1
It follows that k = 0, en (P ) = (1,0) and P = λy for some λ ∈ K∗. But
1,0 0 1
then P = − 1 x +t(y) for some t(y) ∈ K[y] and from β(P ) = P we deduce
0 2λ 0 0
t(y)= n α y2j for some α ∈K.
j=0 j j
InorPderto finish the proofitsuffices to discardthe cases>0. So assume s>0
and note that the assumptions require r > 0. We will consider the two highest
order terms in deg . By [9, Proposition 1.6] for all t(y)∈K[y] and i∈N we have
x
i
[t(y),xi]=ixi−1t′(y)+ xi−2t′′(y)+..., (2.1)
(cid:18)2(cid:19)
THE STARRED DIXMIER CONJECTURE FOR A1 5
where from now on “...” denotes terms of lower x-degree. Write now
P =xrg(y)+xr−1g (y)+... and P =xsh(y)+xs−1h (y)+....
0 1 1 1
Then
β(P )=g(y)xr+g (−y)xr−1+...=xrg(y)+rxr−1g′(y)+xr−1g (−y)+...
0 1 1
Fromβ(P )=P wededuceg (y)=rg′(y)+g (−y). Decomposingg intoitseven
0 0 1 1 1
andoddpartsweobtaing (y)=g (y)+g (y)withg ∈K[y2]andg ∈yK[y2]
1 10 11 10 11
and so
r
g (y)=g (y)+ g′(y). (2.2)
1 10
2
Similarly we obtain
s
h (y)= h′(y)+h (y) for some h ∈yK[y2]. (2.3)
1 11 11
2
From (2.1) we deduce
[P ,P ]=xr+s−1(sg′h−rh′g)
0 1
s r
+xr+s−2 g′′h− h′′g+(s−1)g′h −rh′g+sg′h−(r−1)h′g +...
(cid:18)(cid:18)2(cid:19) (cid:18)2(cid:19) 1 1 1 1(cid:19)
Weinsertinthecoefficientcorrespondingtoxr+s−2 thevaluesofg andh accord-
1 1
ing to (2.2) and (2.3):
s r
g′′h− h′′g+(s−1)g′h −rh′g+sg′h−(r−1)h′g =
(cid:18)2(cid:19) (cid:18)2(cid:19) 1 1 1 1
=(r+s−1)(sg′′h+sg′h′−rh′g′−rh′′g)+2rgh′′+
+(s−1)g′h −rh′ g+sg′ h−(r−1)h′g .
11 11 10 10
Now sg′h−rh′g =0 implies sg′′h+sg′h′−rh′g′−rh′′g =0 and moreover
(s−1)g′h −rh′ g+sg′ h−(r−1)h′g
11 11 10 10
is even, hence the odd part of the coefficient is 2rgh′′ = 0. By assumption 2r 6= 0
andg 6=0,whichleadstoh′′ =0. Butthenh=λyforsomeλ∈K∗. ByLemma2.5
this implies s=0, a contradiction which concludes the proof. (cid:3)
Corollary 2.7. AssumeChar(K)=0. For anyα-endomorphism f ofA (K)there
1
exist n∈N , λ∈K∗ and α ,...,α ∈K such that
0 0 n
n n
y+x y+x
f(x)=λ(y−x)− + α (y−x)2j , f(y)=−λ(y−x)− + α (y−x)2j.
j j
4λ 4λ
Xj=0 Xj=0
Proof. From the discussion above Lemma 2.5 we know that
f(x)=ϕf˜(x)=ϕ(P +P ) and f(y)=ϕf˜(y)=ϕ(P −P ),
0 1 0 1
for some P ,P satisfying β(P ) = P , β(P ) = −P and [P ,P ] = 1. Moreover
0 1 0 0 1 1 0 1 2
r := deg (P ) = 0 is impossible since it would lead to P ∈ K[y2] which implies
x 0 0
[P ,P ] ∈/ K∗ (compute ℓ [P ,P ]). Hence 2r 6= 0 and by Proposition 2.6 we
0 1 1,0 0 1
know that there exist n∈N , λ∈K∗ and α ,...,α ∈K such that P =λy and
0 0 n 1
P =− 1 x+ n α y2j. Astraightforwardcomputationconcludesthe proof. (cid:3)
0 2λ j=0 j
P
Corollary 2.8. Assume Char(K) = 0. Any α-endomorphism f of A (K) is an
1
α-automorphism of the same form as in Example 1.3.
6 VEREDMOSKOWICZANDCHRISTIANVALQUI
Proof. Let f be an α-endomorphism. Apply Corollary 2.7 to f to get n ∈ N ,
0
λ∈K∗ and α ,...,α ∈K such that
0 n
n n
y+x y+x
f(x)=λ(y−x)− + α (y−x)2j , f(y)=−λ(y−x)− + α (y−x)2j.
j j
4λ 4λ
Xj=0 Xj=0
Hence,
n n
f(x)=ax+by+ α (y−x)2j and f(y)=ay+bx+ α (y−x)2j,
j j
Xj=0 Xj=0
where a = (−4λ2 −1)/4λ and b = (4λ2 −1)/4λ satisfy a2 −b2 = 1, and so the
endomorphism is of the desired form. (cid:3)
Theorem 2.9 (The starred Dixmier conjecture is true). Assume Char(K) = 0.
Any α-endomorphism f of A (K) is an α-automorphism.
1
Proof. Clear from Corollary 2.8 and Example 1.3. (cid:3)
3 Related topics
We suggest to consider the following topics. In those topics we usually take the
exchange involution α, although one may take other involutions as well.
3.1 Prime characteristic case
When K is of prime characteristic, A (K) is not simple. Bavula asked the fol-
1
lowing question: Is every algebra endomorphism of the first Weyl algebra A (K),
1
Char(K) = p > 0, a monomorphism? Makar-Limanov [11] gave a positive an-
swer to that question. However, the following endomorphism f (f is necessarily a
monomorphism) is not onto, since x is not in the image of f:
f :A −→A such that f(x)=x+xp and f(y)=y.
1 1
Note that Proposition 2.6 is valid for Char(K) 6= 2. This suggest that the
followingis true (the starredDixmier conjecturein the positivecharacteristiccase,
compare with [1]):
“Let Char(K) = p > 2. Any α-endomorphism of A (K) is an automorphism if
1
andonlyif its restrictionto the centerofA induces afield extensionofdegreenot
1
a multiple of p and the jacobian of this restriction is a nonzero element of K.”
3.2 Higher Weyl algebras
Assume Char(K) = 0 and let A (K) be the n’th Weyl algebra and α : A → A
n n n
an involution, for example the anti-homomorphism given by x 7→ y , y 7→ x . If
i i i i
one generalizesthe geometricmethods of[9]to higherdimensions, onecouldtry to
prove the n’th starredDixmier conjecture (α−D ): “Any α-endomorphism of A
n n
is an automorphism.”
3.3 Same questions for other algebras
One may wish to ask similar questions for other algebras, see, for example, [4]. In
algebraswhere an involutioncan be defined, one may wishto see if the presence of
an involution may be of any help in solving such questions.
THE STARRED DIXMIER CONJECTURE FOR A1 7
3.4 Connection to the Jacobian conjecture
The connection between Dixmier’s problem 1 and the Jacobian conjecture is as
follows:
(1) Then’thDixmierconjecture,D (EveryendomorphismofA (K)isanau-
n n
tomorphism),implies JC ,the n-dimensionalJacobianconjecture,see [13,
n
Theorem 4.2.8].
(2) JC ⇒ D .
2n n
Interestingly,theJacobianconjecture-2nimpliesthen’thDixmierconjecture. This
was proved independently by Tsuchimoto [12] and by Belov and Kontsevich [5].
A shorter proof can be found in [3]. For a detailed background on the Jacobian
conjecture, see, for example, [2] or [13]. One may pose the α-Jacobian conjecture
(α-JC ) in dimension 2n: “Let α : X ↔ X be the exchange involution on
2n 2n−1 2n
K[X ,...,X ]. If f : K[X ,...,X ]→K[X ,...,X ] is an α-morphism which
1 2n 1 2n 1 2n
satisfies Jac(f(X ),...,f(X ))=1, then f is invertible.”
1 2n
One may ask the following question: α−JC ⇒ α−D ?
2n n
In the previous section we proved α−D and in the last section we will prove
1
α−JC ; however we don’t know if one can deduce α−D directly from α−JC .
2 1 2
4 Analogue results for the Jacobian conjecture
There is a closedconnection between the shape of possible counterexamples to the
Jacobianconjectureindimension2andtheshapeofpossiblecounterexamplestothe
Dixmier conjecture (in dimension 1). Let α:K[X,Y]→K[X,Y] be the exchange
involution X ↔ Y. We say that f : K[X,Y] → K[X,Y] is an α-morphism, if
f◦α=α◦f. Notethatthesymmetrydeterminedbyαisnotanyofthesymmetries
analyzed in [6] or [14].
One has the following result:
Proposition 4.1. (α−JC is true.) If K is a field of characteristic zero and
2
f :K[X,Y]→K[X,Y] is an α-morphism that satisfies Jac(f(X),f(Y))=1, then
f is an automorphism.
X +Y
Proof. Let ϕ : K[X,Y] → K[X,Y] be given by ϕ(X) := and ϕ(Y) :=
2
Y Y
Y − X. Then ϕ−1(X) = X − , ϕ−1(Y) = X + and the homomorphism
2 2
β := ϕ−1 ◦α◦ϕ is given by β(X) = X and β(Y) = −Y. Notice that β is an
involution which commutes with f˜:=ϕ−1◦f ◦ϕ, since f is an α-morphism. This
means that P :=f˜(X) and Q:=f˜(Y) satisfy
β(P)=β(f˜(X))=f˜(β(X))=P, β(Q)=β(f˜(Y))=f˜(β(Y))=−Q
and Jac(P,Q) = 1. We will prove that for P,Q ∈ K[X,Y] with β(P) = P,
β(Q) = −Q and Jac(P,Q) = 1 there exist λ ∈ K× and g ∈ K[Y2] such that
Q=λY and P = X +g(Y).
λ
Thisimplies thatf˜isinvertible,withf˜−1(X)=λX−λg Y andf˜−1(Y)= Y ;
λ λ
hence f is also invertible. (cid:0) (cid:1)
Weadoptthenotationsandresultsof[10],inparticular[P,Q]:=Jac(P,Q). Let
P and Q be as above. It is clear that P ∈ K[X,Y2] and Q ∈ YK[X,Y2]. Then
w (P)≥0 and w (Q)≥1 and so, since by [10, Proposition 1.11] we have
0,1 0,1
0=w ([P,Q])≥w (P)+w (Q)−(0+1),
0,1 0,1 0,1
8 VEREDMOSKOWICZANDCHRISTIANVALQUI
itfollowsthatw (P)=0,w (Q)=1and06=[ℓℓ (P),ℓℓ (Q)]=ℓℓ ([P,Q]),
0,1 0,1 0,1 0,1 0,1
by the same proposition. Write ℓℓ (P)=h (X) and ℓℓ (Q)=Yh (X), then
0,1 p 0,1 q
1=[ℓℓ (P),ℓℓ (Q)]=h′(X)h (X),
0,1 0,1 p q
hence
1
ℓℓ (Q)=λY and ℓℓ (P)= X,
0,1 0,1
λ
for some λ∈K×. We prove now that
1
ℓ (Q)=λY and ℓ (P)= X. (4.4)
1,0 1,0
λ
Let (ρ,σ):=min{Succ (0,1),Succ (0,1)}. We first prove that
P Q
1
(ρ,σ)=(−1,1), ℓ (P)= X and ℓ (Q)=λY. (4.5)
1,−1 1,−1
λ
Note that by [10, Proposition3.13]we have st (P)=(1,0) and st (Q)=(0,1).
ρ,σ ρ,σ
Moreover, using the definition of the order on directions of [10, Section 3], the
inequalities (−1,1)≥(ρ,σ)>(0,1), imply σ ≥−ρ>0. Hence
v (en (P))=v (st (P))=v (1,0)=ρ<0
ρ,σ ρ,σ ρ,σ ρ,σ ρ,σ
and
v (en (Q))=v (st (Q))=v (0,1)=σ >0,
ρ,σ ρ,σ ρ,σ ρ,σ ρ,σ
and so en (P) ≁ en (Q), since two non-zero points A,B in the first quadrant
ρ,σ ρ,σ
are aligned if and only if A = γB for some γ > 0. But then, by [10, Proposition
2.3(2)] we have [ℓℓ (P),ℓℓ (Q)]6=0 and so, by [10, Proposition 2.5(2)],
ρ,σ ρ,σ
en (P)+en (Q)−(1,1)=en ([P,Q])=(0,0).
ρ,σ ρ,σ ρ,σ
But the only non-aligned points in N ×N which sum (1,1), are (1,0) and (0,1),
0 0
hence the only possibility is
en (P)=st (P)=(1,0)
ρ,σ ρ,σ
and
en (Q)=st (Q)=(0,1).
ρ,σ ρ,σ
Since (ρ,σ)∈Dir(P)∪Dir(Q) it follows that (ρ,σ)=(−1,1), ℓ (P)= 1X and
1,−1 λ
ℓ (Q)=λY, which is (4.5).
1,−1
We claim that
(ρ ,σ ):=min{Succ (1,−1),Succ (1,−1)}≥(1,0). (4.6)
1 1 P Q
Infactthegeometricargumentisthesameasbefore: Assume(ρ ,σ )<(1,0). Note
1 1
that by [10, Proposition 3.12] we have st (P) = (1,0) and st (Q) = (0,1).
ρ1,σ1 ρ1,σ1
Moreover, using the definition of the order on directions of [10, Section 3], the
inequalities (1,−1)≤(ρ ,σ )<(1,0), imply ρ ≥−σ >0. Hence
1 1 1 1
v (en (P))=v (st (P))=v (1,0)=ρ >0
ρ1,σ1 ρ1,σ1 ρ1,σ1 ρ1,σ1 ρ1,σ1 1
and
v (en (Q))=v (st (Q))=v (0,1)=σ <0,
ρ1,σ1 ρ1,σ1 ρ1,σ1 ρ1,σ1 ρ1,σ1 1
andsoen (P)≁en (Q),sincetwonon-zeropointsA,B inthefirstquadrant
ρ1,σ1 ρ1,σ1
are aligned if and only if A = γB for some γ > 0. But then, by [10, Proposition
2.3(1)] we have [ℓ (P),ℓ (Q)]6=0 and so, by [10, Proposition 2.4(2)],
ρ1,σ1 ρ1,σ1
en (P)+en (Q)−(1,1)=en ([P,Q])=(0,0).
ρ1,σ1 ρ1,σ1 ρ1,σ1
But the only non-aligned points in N ×N which sum (1,1), are (1,0) and (0,1),
0 0
hence the only possibility is
en (P)=st (P)=(1,0)
ρ1,σ1 ρ1,σ1
THE STARRED DIXMIER CONJECTURE FOR A1 9
and
en (Q)=st (Q)=(0,1).
ρ1,σ1 ρ1,σ1
But this contradicts the fact that (ρ ,σ )∈Dir(P)∪Dir(Q) and proves (4.6).
1 1
Now from (4.6) and [10, Proposition 3.12] it follows that st (Q) = (0,1) and
1,0
st (P)=(1,0), hence
1,0
v (Q)=v (st (Q))=0 and v (P)=v (st (P))=1,
1,0 1,0 1,0 1,0 1,0 1,0
which implies
Q=g (Y) and P =g(Y)+Xg (Y), for some g (Y),g(Y),g (Y)∈K[Y],
Q P Q P
with g(Y)∈K[Y2]. But then
1=[P,Q]=g (Y)g′ (Y),
P Q
henceg (Y)=λY andg (Y)= 1,whichproves(4.4)andconcludestheproof. (cid:3)
Q P λ
Remark 4.2. From Proposition 4.1, it can easily be seen that each α-morphism f
which satisfies Jac(f(X),f(Y))=1 is of the following form:
n n
f(X):=aX +bY + T and f(Y):=aY +bX+ T
j j
Xj=0 Xj=0
where n ∈ N , a,b ∈ K, with a2 − b2 = 1, and T := c (X − Y)2j for some
0 j j
c ,...,c ∈K.
0 n
One can structure the proof of α−JC similar to the proof of α−D , however
2 1
the keydifferenceisthe following: Inthe proofofα−D weobtainacontradiction
1
using the second highest order term via (2.1), which is not present in α−JC , if
2
we identify the commutator with the Jacobian bracket. On the other hand, in the
proof of α−JC we make use of ℓℓ , which is not well-behaved in A (K).
2 ρ,σ 1
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10 VEREDMOSKOWICZANDCHRISTIANVALQUI
DepartmentofMathematics,Bar-IlanUniversity, Ramat-Gan52900,Israel.
E-mail address: [email protected]
PontificiaUniversidadCato´licadelPeru´PUCP,Av. Universitaria1801,SanMiguel,
Lima 32,Peru´.
InstitutodeMatema´ticayCienciasAfines(IMCA)CalleLosBio´logos245. UrbSan
C´esar. La Molina,Lima 12,Peru´.
E-mail address: [email protected]