Embeddings of Affine Spaces into Quadrics
Jérémy Blanc
Jérémy Blanc, Universität Basel, Departement Mathematik und Informatik, Spiegelgasse 1, CH-4051 Basel, Switzerland
[email protected]
and
Immanuel van Santen (né Stampfli)
Immanuel van Santen,
Fachbereich Mathematik der Universität Hamburg,
Bundesstraße 55, DE-20146 Hamburg, Germany
[email protected]
Abstract.
This article provides, over any field, infinitely many algebraic embeddings of the
affine spaces A1 and A2
into smooth quadrics of dimension two and three respectively, which are pairwise non-equivalent under automorphisms of the smooth quadric.
Our main tools are the study
of the birational morphism SL2→A3
and the fibration
SL2→A3→A1 obtained by projections,
as well as degenerations of variables of polynomial rings,
and families of A1-fibrations.
2010 Mathematics Subject Classification:
14R10, 14R25, 14J70, 14J50, 14E05
Contents
-
1 Introduction
-
2 The smooth quadric of dimension 2 and the proof of Theorem 1
-
2.1 The isomorphism with the complement of the diagonal in Pk1×Pk1
-
2.2 Families of embeddings
-
3 Variables of polynomial rings
-
3.1 Variables of polynomial rings in two variables
-
3.2 Non-trivial embeddings in positive characteristic
-
4 Liftings of automorphisms and the proof of Theorem 2
-
4.1 Lifting of automorphisms of Ak3 to affine modifications
-
4.2 Application of liftings to the case of SL2
-
5 Fibered embeddings of Ak2 into SL2
and the proof of Theorem 3
-
5.1 Polynomials associated to fibred embeddings
-
5.2 Determining when two fibred embeddings are equivalent
-
5.3 Examples of non-equivalent embeddings
-
5.4 Embeddings of Ak2 into SL2 of small degree
-
6 A non-trivial embedding of A1 into SL2, over the reals
1. Introduction
In the sequel we denote by k the ground field of our algebraic varieties.
Given two affine algebraic varieties X,Y, we say that two closed embeddings ρ,ρ′:X↪Y are equivalent if there exists an automorphism φ∈Aut(Y) such that ρ′=φ∘ρ. Similarly, we say that two closed subvarieties X,X′⊂Y are equivalent if there exists an automorphism φ∈Aut(Y) such that X′=φ(X). If two closed embeddings are equivalent,
then their images are equivalent, but the converse is not always true and is related to the extension of automorphisms.
In the Bourbaki Seminar Challenging problems on affine n-space
[Kra96], Hanspeter Kraft gives a list of eight
fundamental problems related to the affine n-spaces. The third one is the following:
Embedding Problem. Is every closed embedding Akm↪Akn equivalent to the standard embedding (x1,…,xm)↦(x1,…,xm,0,…,0)?
This question, asked over the ground field k=C in [Kra96], has until now no negative answer. For k=R, it is easy to find counterexamples for m=1 and n=3, by taking embeddings which are not topologically trivial (non-trivial knots),
see for instance the example of [Sha92], reproduced below in
Example 6.1.
In positive characteristic, there are counterexamples when m=n−1
(see Proposition 3.16). The embedding problem has however a positive answer in the following cases:
- (1)
m=1, n=2, char(k)=0 (Abhyankar-Moh-Suzuki Theorem) [AM75, Suz74], [vdE00, Theorem 2.3.5];
2. (2)
n≥2m+2, k infinite (Theorem of Kaliman, Nori and
Srinivas [Kal91, Sri91]).
The case of hypersurfaces (m=n−1) is of particular interest.
In this case, the image is given by the zero set of an irreducible polynomial equation f∈k[An]. One necessary condition
for an embedding to be equivalent to the standard embedding
consists of asking that the other fibres of f:Akn→Ak1
are affine spaces. In fact, for any field k and any n≥1, there is no known example of a hypersurface X⊂Akn isomorphic to Akn−1 and given by f=0, f∈k[An] irreducible such that another fibre f=λ is not isomorphic to Akn−1. It is conjectured by Abhyankar and
Sathaye that no such examples exist, at least when char(k)=0,
see [vdE00, §3, page 103], even if this is quite strong and seems “unlikely” (as Arno van den Essen says in [vdE00, §3, page 103]). Moreover, for n=3 and char(k)=0, the fact that infinitely many
fibres f=λ are isomorphic to Ak2 implies that the fibration is equivalent to the standard one, and in particular that all fibres are isomorphic to Ak2 [KZ01, Kal02, DK09].
For char(k)>0, there is until now no known counterexample to the above conjecture, which is open even in dimension n=2 (and corresponds to a question of Abhyankar, see [Gan11, Question 1.1]).
In this paper, we replace the affine space at the target by some analogue varieties, namely affine smooth quadrics. This simplifies the question in such a way that one can actually give an answer. Moreover, it also gives some idea on what kind of behaviour one could expect in a general situation.
In dimension n=2, the most natural quadric is
[TABLE]
In fact, if k is an algebraically closed field, then every smooth quadric hypersurface Q⊂Ak3 is isomorphic to Ak2,
(Ak1∖{0})×Ak1 or Q2, as one can see
using the classification
of quadratic forms. As all embeddings of
Ak1 into (Ak1∖{0})×Ak1 are constant on the first factor,
they are all equivalent.
Over any field, the group of automorphisms of Q2 is similar to the one of Ak2, as it is an amalgamated product of two factors, corresponding to affine maps and triangular maps [BD11, Theorem 5.4.5(7)(a)]. This is also the case for the affine surface Pk2∖Γ, where Γ⊂Pk2
is any smooth conic having a k-point (see for instance [DD16, Theorem 2]). If char(k)=0, there is exactly one (respectively two) closed
curve C⊂Ak2 (respectively C⊂Pk2∖Γ) isomorphic to Ak1, up to automorphism of the surface. This follows from
the Abhyankar-Moh-Suzuki Theorem for Ak2 and from [DD16] for
Pk2∖Γ. In particular, all automorphisms of the corresponding curves extend to automorphisms of Ak2 or Pk2∖Γ.
Similarly, a complex toric affine surface admits only finitely many embeddings of
AC1, up to equivalence [AZ13].
By contrast, we prove the following result:
Theorem 1**.**
For each field k, there is an infinite set of closed curves
[TABLE]
which are pairwise non-equivalent up to automorphism of Q2, such that each Ci is isomorphic to Ak1 and such that the identity is the only automorphism of C that extends to an automorphism of Q2.
Moreover, if k is uncountable, then one can choose the same for I.
In dimension n=3, the most natural quadric is
[TABLE]
Similarly as in dimension two, over an algebraically closed field k,
every quadric hypersurface in Ak4
is isomorphic to Ak3, (Ak1∖{0})×Ak2,
Q2×Ak1 or SL2. Moreover, the quotient of SL2 by its maximal torus yields a morphism SL2→SL2/T≃Q2, which is the “universal torsor” (also called the Cox quotient presentation or the characteristic space), see [ADHL15, Examples 4.5.13–4.5.14].
We consider the quadric hypersurface SL2 more closely.
Its automorphism group shares similar properties with the one of Aut(Ak3) (see [LV13, BFL14, Mar15]). Both are known to be complicate, as they contain “wild” automorphisms [LV13], and do not preserve any fibration, as it is the case for other varieties being topologically closer to Ak3, like the Koras-Russell threefold. However, in contrast to the quadric
Q2, the quadric SL2 is closer to a contractible affine variety
in the sense that the ring of regular functions on SL2
is a unique factorisation domain (see Lemma 4.4).
The first difference concerning embeddings
of affine spaces with the surfaces Q2,Ak2,Pk2∖Γ and with Ak3 is that the “simplest embedding” Ak2↪SL2 is more rigid in the following sense:
Theorem 2**.**
Let k be any field and let
[TABLE]
be the “standard” embedding. Then, an automorphism (s,t)↦(f(s,t),g(s,t)) of Ak2 extends to an automorphism of SL2, via ρ1, if and only if it has Jacobian determinant ∂s∂f∂t∂g−∂t∂f∂s∂g∈k∗
equal to ±1. In particular, the following holds:
- (1)
every embedding
Ak2↪SL2 with image
ρ1(Ak2) is equivalent to an embedding
[TABLE]
for a certain λ∈k∗. Moreover,
ρλ and ρλ′ are equivalent if
and only if λ′=±λ;
2. (2)
if k has at least 4 elements,
then not all automorphisms of Ak2 extend to SL2 via ρ1.
Remark 1.1*.*
Let us make some comments on
Theorem 2:
- (1)
Over the field of complex numbers k=C,
we show that all algebraic automorphisms of Ak2 extend
via the standard embedding ρ1 to holomorphic automorphisms of SL2,
see Remark 4.7.
2. (2)
If all component functions of a closed
embedding f:Ak2↪SL2
are polynomials of degree ≤2, then f is equivalent to ρλ
for a certain λ∈k∗ (Proposition 5.19).
Next, we focus on the closed embeddings Ak2↪SL2
that are compatible with the simplest A2-fibration of SL2. More precisely:
Definition 1.2**.**
A closed embedding ρ:Ak2↪SL2 is said to be a fibred embedding if it is of the form
[TABLE]
for some p,q,r∈k[s,t]. This corresponds to the commutativity of the diagram
[TABLE]
where π1:Ak2→Ak1, π2:SL2→Ak1 are respectively given by (s,t)↦t and (xuty)↦t.
As we will show, there are a lot of fibred embeddings (i.e. embeddings of the
form (1.2)):
Theorem 3**.**
Let k be any field, let P∈k[t,x,y] be a polynomial that is a variable of the k(t)-algebra k(t)[x,y] (which means that P is the image of x by some automorphism of the
k(t)-algebra \mathbf{\textbf{k}}(t)[x,y]$$), and let HP⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1)) and ZP⊂Ak3=Spec(k[t,x,y]) be the hypersurfaces given by P=0.
- (1)
The following conditions are equivalent:
- (a)
The hypersurface HP⊂SL2 is isomorphic to Ak2.
2. (b)
The hypersurface HP⊂SL2 is the image of a fibred embedding Ak2↪SL2.
3. (c)
The fibre
of ZP→Ak1, (t,x,y)↦t over every closed point
of Ak1∖{0}
is isomorphic to A1 and the
polynomial P(0,x,y)∈k[x,y] is of the form μxm(x−λ) or
μym(y−λ)
for some μ,λ∈k∗
and some m≥0.
2. (2)
If P,Q∈k[t,x,y] are two polynomials of the above form satisfying the conditions \eqrefThmEqa−\eqrefThmEqb−\eqrefThmEqc, such that HP,HQ⊂SL2 are equivalent under an automorphism of SL2, then ZP,ZQ⊂Ak3 are equivalent under an automorphism of Ak3.
3. (3)
There are infinitely many fibred embeddings Ak2↪SL2
having pairwise non-equivalent images in SL2. If k is uncountable, we can moreover choose uncountably many such embeddings.
Remark 1.3*.*
Let us make some comments on
Theorem 3:
- (1)
It is possible that HP,HQ are non-equivalent,
even if ZP,ZQ are equivalent (Lemma 5.11).
2. (2)
If char(k)=0, then every image of a fibred embedding Ak2↪SL2 is of the form HP as above (Lemma 5.2(2)).
This is false if char(k)>0 (Lemma 5.3).
Let us make the following comment concerning
embeddings of Ak1 into the smooth quadric SL2
over the field k=C.
Although there are infinitely many
non-equivalent embeddings of AC2 into SL2,
it is not known, whether all embeddings of AC1 into SL2
are equivalent under an algebraic automorphism. It seems
that this questions is as difficult, as the question of
non-equivalent embeddings
AC1↪AC3. However, up to holomorphic
automorphisms, all embeddings of AC1 into AC3
and into SL2 are equivalent, see [Kal92, Sta15].
In the last section (Lemma 6.2), we give an example of
an embedding AR1↪SL2 which is
non-equivalent to the standard embedding.
Acknowledgement
We would like to thank Peter Feller for many fruitful discussions, and Ivan Arzhantsev for indicating us some references.
2. The smooth quadric of dimension 2 and the proof of Theorem 1
2.1. The isomorphism with the complement of the diagonal in Pk1×Pk1
In this section, we study the smooth quadric Q2⊂Ak3 given by
[TABLE]
and more particularly closed embeddings Ak1↪Q2. Since the closure of Q2 in Pk3 is a smooth quadric, isomorphic to Pk1×Pk1, we get the following classical isomorphism:
Lemma 2.1**.**
The morphism
[TABLE]
yields an isomorphism Q2⟶≃(Pk1×Pk1)∖Δ, where Δ⊂Pk1×Pk1 is the diagonal, with an inverse given by
[TABLE]
Proof.
We first check that ρ((x,y,z))∈(Pk1×Pk1)∖Δ for each (x,y,z)∈Q2. If z=0 then [y:z]=[z:x], since xy−z2=z=0. If z=0, then xy=0, whence [z+1:x]=[1:x]=[y:1]=[y:z+1].
It remains then to check that ρ∘ψ=id(Pk1×Pk1)∖Δ and ψ∘ρ=idQ2, which follows from a straight-forward calculation.
∎
2.2. Families of embeddings
The following result is the key step in the proof of Theorem 1.
Lemma 2.2**.**
- (1)
For each polynomial p∈k[t], the morphism νp:Ak1↪Q2 given by
[TABLE]
is a closed embedding.
2. (2)
If p,q∈k[t] are polynomials of degree ≥3
such that ανp=νqβ for some β∈Aut(Ak1) and α∈Aut(Q2), then there exist μ∈k and λ∈k∗
such that
[TABLE]
Proof.
Using the isomorphism ρ:Q2⟶≃(Pk1×Pk1)∖Δ of Lemma 2.1, we obtain that ρ∘νp:Ak1→Pk1×Pk1 is given by t↦([1:t],[p(t):1+tp(t)]), which is the restriction of the closed embedding
[TABLE]
where d=deg(p) and P(u,v)=p(uv)ud is the homogenisation of p. This implies that Γp=ν^p(Pk1)⊂Pk1×Pk1 is a smooth closed curve (isomorphic to Pk1), and since Γp∩Δ is given by
u(ud+1+vP(u,v))−vuP(u,v)=0, i.e. ud+2=0, this shows that νp is a closed embedding, and thus yields (1).
It remains to prove Assertion (2).
We fix two polynomials p,q∈k[t] of degree ≥3 such that ανp=νqβ for some β∈Aut(Ak1) and α∈Aut(Q2). This implies in particular that the automorphism α′=ρ−1αρ∈Aut((Pk1×Pk1)∖Δ) sends Γp∖Δ onto Γq∖Δ.
We first prove that α′∈Aut((Pk1×Pk1)∖Δ) extends to an automorphism α^∈Aut(Pk1×Pk1). Assume for contradiction that this is not the case. The map α′ would then extend to a birational map α^:Pk1×Pk1\dasharrowPk1×Pk1, which is not an automorphism. We consider the minimal resolution of α^, which is
[TABLE]
where χ1, χ2 are birational morphisms. The resolution being minimal, every (−1)-curve E⊂Z contracted by χ2 is not contracted by χ1, so χ1(E)⊂Pk1×Pk1 is contracted by α^. There is thus a unique (−1)-curve contracted by χ2, which is the strict transform Δ~ of Δ, and satisfies χ1(Δ~)=Δ. As Δ2=2 and (Δ~)2=−1, there are exactly three base-points of χ1−1 that lie on the curve Δ (as proper point or infinitely near points). Since Γp is smooth of bidegree (1,1+degp), we get
Γp⋅Δ=2+degp≥5, which implies that the strict transforms of Γp and Δ on Z satisfy Γ~p⋅Δ~≥2 (as only three points belonging to Δ have been blown-up). As the curve Δ~ is contracted by χ2, the curve χ2(Γ~p) is singular. This contradicts the equality χ2(Γ~p)=Γq, which follows from the fact that α^(Γp∖Δ)=Γq∖Δ.
We have shown that the extension of α′=ρ−1αρ∈Aut((Pk1×Pk1)∖Δ) is an automorphism α^∈Aut(Pk1×Pk1), which satisfies therefore α^(Δ)=Δ and α^(Γp)=Γq. The curves Γp and Γq being of bidegree (1,1+degp) and (1,1+degq), we get degp=degq and we obtain that α^ does not exchange the two factors of Pk1×Pk1. Moreover, as the point ([0:1],[0:1])=Δ∩Γp=Δ∩Γq is fixed, and as
the diagonal Δ is invariant, we can write α^ as
[TABLE]
for some λ∈k∗, μ∈k.
The equality ανp=νqβ implies that α^ν^p=ν^qβ^, for some automorphism β^∈Aut(Pk1), which is the extension of β and therefore it is
of the form [u:v]↦[u:λv+μu]. We then compute
[TABLE]
and obtain that P(u,v)=λQ(u,λv+μu). Remembering that P(u,v)=p(uv)ud and Q(u,v)=q(uv)ud we obtain that p(t)=λq(λt+μ). We then compute the explicit form of α by conjugating α^ with ρ−1.
∎
Example 2.3*.*
For each n≥1, let pn(t)=tn(t+1)n+1.
The closed curve Cn=νpn(Ak1)⊂Q2 is
isomorphic to Ak1, via
[TABLE]
Then Lemma 2.2(2) shows that all curves Cn are non-equivalent for different n≥1,
and that the identity is the only automorphism of Cn that extends to Q2.
The proof of Theorem 1 is now a consequence of Lemma 2.2.
Proof of Theorem 1.
If k is the field with two elements, then we conclude by
Example 2.3.
Hence we can assume that k contains more than two elements.
For each n≥1 and each λ∈k,
ε∈k∖{0,1},
one defines pn,ε(t)=tn(t+1)n+1(t+ε)n+2∈k[t], and let Cn,ε⊂Q2 be the closed curve given by νpn,ε(Ak1), which is isomorphic to Ak1 (Lemma 2.2(1)).
Lemma 2.2(2) implies that the identity is the only automorphism of Cn,ε that extends to an automorphism of Q2, since λpn,ε(λt+μ)=pn,ε(t), for
(λ,μ)∈(k∗×k)∖{(1,0)}.
Similarly, Lemma 2.2(2) shows that Cn,ε is equivalent to Cn′,ε′ if and only if n=n′ and ε=ε′.
∎
3. Variables of polynomial rings
In this section, we give some results on variables of polynomial rings. Most of them are classical or belong to the folklore. We include them for self-containedness and for lack of precise references.
Definition 3.1**.**
Let S be a ring and let R⊂S be a subring. We denote by AutR(S) the group of automorphisms of the R-Algebra S. More precisely,
[TABLE]
Definition 3.2**.**
Let R be a domain and S be a polynomial ring in n≥1 variables over
R, i.e. R⊂S and there exist x1,…,xn∈S such that each element of S can be written in a unique way as f(x1,…,xn), where f is
a polynomial in the xi with coefficients in R. An element v∈S is called variable of the R-algebra S
if there exists f∈AutR(S) such that f(v)=x1.
In the sequel, we often denote by R[t] or R[x] the polynomial ring in one variable over R, by R[x,y] the polynomial ring in two variables over R
and by R[x1,…,xn] the polynomial ring in n variables over R.
Lemma 3.3**.**
Let R be a domain, let S=R[x1,…,xn] be the polynomial ring in n variables over R and let v∈S. The following conditions are equivalent:
- (1)
v* is a variable of the R-algebra S;*
2. (2)
The R[t]-algebra S[t]/(v−t) is isomorphic to a polynomial ring in n−1 variables over R[t].
Proof.
If v is a variable, then
there exists f∈AutR(S) such that f(v)=x1. Using the natural inclusion AutR(S)↪AutR[t](S[t]), we get isomorphisms of R[t]-algebras
[TABLE]
Conversely, suppose that the R[t]-algebra S[t]/(v−t) is isomorphic to a polynomial ring in n−1 variables over R[t]. This yields an R[t]-isomorphism ψ:S[t]/(t−v)⟶≃R[t][x2,…,xn]. We then compose the isomorphisms of R-algebras
[TABLE]
and
[TABLE]
and obtain an element of AutR(S) that sends v onto x1.
∎
Lemma 3.4**.**
Let k be a field, let k[x1,…,xn] be the polynomial ring in n≥1 variables over k and let w∈k[x1,…,xn] be a variable of this k-algebra.
Then k[w] is factorially closed in k[x1,…,xn], i.e. for all f,g∈k[x1,…,xn]∖{0}, we have
fg∈k[w]⇔f∈k[w] and g∈k[w].
Proof.
If f∈k[w] and g∈k[w], then fg∈k[w], since k[w] is a subring of
k[x1,…,xn].
Conversely, suppose that fg∈k[w]. Choose ψ∈Autk(k[x1,…,xn]) such that ψ(w)=x1. Then, ψ(f),ψ(g)∈k[x1,…,xn] are two polynomials such that ψ(f)⋅ψ(g)∈k[x1]. For each i≥2, the degree in xi satisfies degxi(ψ(f))+degxi(ψ(f))=degxi(ψ(f)⋅ψ(g))=0, so degxi(ψ(f))=degxi(ψ(f))=0 since both elements are non-zero. Hence, ψ(f),ψ(g)∈k[x1].
Applying ψ−1, we get f∈k[w] and g∈k[w].
∎
3.1. Variables of polynomial rings in two variables
We will need the following two technical lemmas:
Lemma 3.5**.**
Let k be a field, let w be a variable of the k-algebra k[x,y] and let v∈k[x,y] be a polynomial. The following conditions are equivalent:
- (1)
v∈k[w];
2. (2)
For each u∈k[w], the elements u and v are algebraically dependent over k.
3. (3)
There exists u∈k[w]∖k such that u and v are algebraically dependent over k.
Proof.
The
implications (\refvkw1)⇒(\refvkw2)⇒(\refvkw3)
being clear, we only need to prove (\refvkw3)⇒(\refvkw1).
Replacing v and w with f(v) and f(w), for some f∈Autk(k[x,y]),
we can assume that w=x. Denoting by k the algebraic closure of
k, we have k[x]∩k[x,y]=k[x], so we can assume that
k=k.
We then consider the morphism τ:Ak2→Ak2 given by (x,y)↦(u(x),v(x,y)), which is dominant if and only if u,v are algebraically independent over k. It remains then to see that τ is dominant if u∈k[x]∖k and v∈k[x].
Let v(x,y)=∑i=0dvi(x)yi, where vd=0 and d>0.
For a general a∈k, u(x)=a
has a solution x0 such that vd(x0)=0,
since k is algebraically closed.
Hence v(x0,y)=b has a solution for all b∈k. This proves that
τ is dominant.
∎
Lemma 3.6**.**
Let k be a field, let p∈k[t] be an irreducible element and let kp=k[t]/(p) be the corresponding residue field.
Let u,v∈k[t][x,y] be elements such that k(t)[u,v]=k(t)[x,y]. Then, the classes u0,v0∈kp[x,y] of u,v satisfy one of the following properties, depending on the Jacobian determinant ν=∂x∂u⋅∂y∂v−∂y∂u⋅∂x∂v∈k[t]∖{0}:
- (1)
If p divides ν, then u0,v0 are algebraically dependent over kp.
2. (2)
If p does not divide ν, then kp[u0,v0]=kp[x,y]. In particular, both u0 and v0 are variables of the kp-algebra kp[x,y].
Proof.
Since k(t)[u,v]=k(t)[x,y], there are polynomials P,Q∈k(t)[X,Y] such that P(u,v)=x, Q(u,v)=y. Moreover, the polynomial ν=∂x∂u⋅∂y∂v−∂y∂u⋅∂x∂v∈k[t,x,y] belongs to k(t)∗ and thus to k[t]∖{0}. The element ν0=∂x∂u0⋅∂y∂v0−∂y∂u0⋅∂x∂v0 is then the class of ν in kp.
We write P=αP~, Q=βQ~, where P~,Q~∈k[t][X,Y], α,β∈k[t]∖{0} and such that p does not divide both α and P~ (and the same for β and Q~). We then get
[TABLE]
where P~0,Q~0∈kp[X,Y] are the classes of P~,Q~ and α0,β0∈kp are the classes of α,β.
If α0 and β0 are not equal to zero, then kp[u0,v0]=kp[x,y]. In particular, u0 and v0 are variables of the kp-algebra kp[x,y] and ν0∈kp∗, so p does not divide ν.
If α0=0, then P~0=0 and P~0(u0,v0)=0 implies that u0 and v0 are algebraically dependent over kp. The same conclusion holds when β0=0. In both cases, the Jacobian determinant ν0 is equal to zero, so p divides ν.
This yields (1) and (2).
∎
We recall the following classical result, essentially equivalent to the Jung-van der Kulk theorem:
Lemma 3.7**.**
Let k be a field, let k[x,y] be the polynomial ring in two variables over
k, let f∈Autk(k[x,y]) and u=f(x),v=f(y)∈k[x,y].
If deg(u)≥deg(v)>1, then there exists a polynomial P with coefficients in k such that deg(u−P(v))<deg(u).
Proof.
By van der Kulk’s Theorem all automorphisms
of k[x,y] are tame [Jun42, vdK53].
The statement is then a direct consequence of
[vdE00, Corollary 5.1.6].
∎
The following result is needed in the sequel. When the characteristic of k is zero, and p=t, it follows from [Fur02, Theorem 4]. We adapt here the proof of [Fur02] for our purpose.
Lemma 3.8**.**
Let k be a field, let p∈k[t] be an irreducible element and let kp=k[t]/(p) be the corresponding residue field.
If v∈k[t,x,y] is a variable of the k(t)-algebra k(t)[x,y], then its class in kp[x,y] is an element which belongs to kp[w]⊂kp[x,y] for some variable w of the kp-algebra kp[x,y].
Proof.
Let f∈Autk(t)(k(t)[x,y]) such that f(x)=v, and let us define u=f(y). We denote by v0∈kp[x,y] the class of v and will use the degree of polynomials in x,y with coefficients in k(t) or kp.
If deg(v)=1, then deg(v0)≤1. If v0∈kp the result follows by taking any variable for w, for instance w=x. Otherwise, v0=αx+βy+γ for some α,β,γ∈kp with (α,β)=(0,0). This implies that w=αx+βy is a variable, as it is the component of an element of GL2(kp), and the result follows.
We can thus assume that deg(v)>1 and prove the result by induction on the pair (deg(v),deg(u)), ordered lexicographically.
(i) If deg(u)≥deg(v), then there exists a polynomial P∈k(t)[X] such that deg(u−P(v))<deg(u) (Lemma 3.7). We can thus apply induction hypothesis to (u−P(v),v), since k(t)[u,v]=k(t)[u−P(v),v],
and obtain the result.
(ii) If deg(u)<deg(v), we first replace u with u−λ for some λ∈k(t) and assume that u∈k(t)[x,y] is a polynomial in x,y with no constant term. We then replace u with qu for some q∈k(t)∗ and assume that u∈k[t][x,y] and the greatest common divisor in k[t] of the coefficients of u (as a polynomial in x,y) is equal to 1. One can then define the class u0∈kp[x,y] of u, which is not equal to zero. Moreover, u0 does not belong to kp, since u0 has no constant term.
If v0 is a variable of the kp-algebra kp[x,y], then we are done. Otherwise, u0,v0 are algebraically dependent over kp (Lemma 3.6).
Since the pair (deg(v),deg(u)) is smaller than (deg(u),deg(v)), we can apply induction hypothesis and get a variable w∈kp[x,y] such that u0∈kp[w]. The fact that u0 and v0 are algebraically dependent over kp and that u0∈kp imply that v0∈kp[w] (Lemma 3.5).
∎
We finish this section with several results relating variables and A1-bundles.
Lemma 3.9**.**
Let k be a field and let P∈k[x,y]. Then, the following conditions are equivalent:
- (1)
The polynomial P is a variable of the k-algebra k[x,y].
2. (2)
The k[t]-algebra k[t,x,y]/(P−t) is a polynomial ring in one variable over k[t].
3. (3)
The k(t)-algebra k(t)[x,y]/(P−t) is a polynomial ring in one variable over k(t).
4. (4)
The morphism Ak2→Ak1 given by (x,y)↦P(x,y) is a trivial A1-bundle.
5. (5)
The morphism Ak2→Ak1 given by (x,y)↦P(x,y) is a trivial A1-bundle over some dense open subset U⊂Ak1.
Proof.
(\refPvkxy)⇔(\refPktxy) follows from Lemma 3.3.
(\refPvkxy)⇔(\refPtrivial): By definition, (\refPvkxy) is equivalent to the existence of f∈Autk(k[x,y]) such that f(x)=P. As f=φ∗ for some φ∈Aut(Ak2), this is equivalent to ask for φ∈Aut(Ak2) that prx∘φ is the map (x,y)↦P(x,y), where prx:Ak2→Ak1 is given by (x,y)↦x. This yields the equivalence (\refPvkxy)⇔(\refPtrivial).
(\refPktxy)⇒(\refPktxyfield) is trivially true.
(\refPktxyfield)⇒(\refPloctrivial): Assertion (\refPktxyfield) corresponds to say that the generic fibre of (x,y)↦P(x,y) is isomorphic to Ak(t)1. This yields (\refPloctrivial).
(\refPloctrivial)⇒(\refPvkxy):
Assume that the subset U given in (\refPloctrivial)
contains a k-rational point.
Replacing P with P+λ, λ∈k,
one can assume that [math] belongs to the open subset
U. One then observes that the curve
Γ⊂Ak2 given by P=0 is isomorphic to Ak1
and equivalent to a line by a birational map of Ak2, hence
can be contracted by a birational map of Ak2.
By [BFH16, Proposition 2.29], there exists
an automorphism φ∈Aut(Ak2) which sends Γ
onto the line given by x=0. This implies that P is a variable of
the k-algebra k[x,y].
If U contains no k-rational point, then k
is a finite field and thus it is perfect. For a finite Galois extension
k′⊃k the subset U contains a k′-rational point. By
the argument above, P is a variable of the k′-algebra k′[x,y]
and hence k′[x,y]=k′[P,Q] for some Q∈k′[x,y].
Since P is a polynomial with coefficients in k,
it is fixed under
the action of the Galois group G=Gal(k′/k) on k[x,y]=k′[P,Q].
For each σ∈G, there exists (aσ,bσ)∈(k′)∗⋉k′[T] with
[TABLE]
We can then find d≥0 such that {bσ∣σ∈G} is contained in the finite dimensional k′-vector subspace Vd={f∈k′[T]∣deg(P)≤d}⊂k′[T].
Thus σ↦(aσ,bσ) defines
an element of H1(G,(k′)∗⋉Vd).
As H1(G,(k′)∗)={1} and H1(G,Vd)={1} [Ser68, Proposition 1, 2, Chp. X], we have H1(G,(k′)∗⋉Vd)={1}. The fact that (aσ,bσ) is a trivial cocycle corresponds to the existence of
a polynomial Q0∈k[x,y] such that k[P,Q0]=k[x,y].
This implies that P is a variable of the k-algebra k[x,y].
∎
We recall the following classical result:
Lemma 3.10**.**
Let k be a field, let Z be an affine variety over k,
all of its irreducible components being surfaces,
let U⊆Ak1 be a dense open subset and let
π:Z→U be a dominant morphism.
Then, the following conditions are equivalent:
- (1)
The morphism π:Z→U is a trivial A1-bundle.
2. (2)
The morphism π:Z→U is a locally
trivial A1-bundle.
3. (3)
For each maximal ideal m⊂k[U],
the fibre π−1(m)⊂Z is isomorphic
to Aκ(m)1 and the generic fibre
of π is isomorphic to Ak(t)1.
Proof.
The implications \eqrefPPPtrivialA1bundle⇒\eqrefPPPloctrivialA1bundle⇒\eqrefPPPeachfibA1
are obvious.
Assume \eqrefPPPeachfibA1 holds.
Since each irreducible component of Z has
dimension two, it follows that each of these irreducible components
is mapped dominantly onto U via π.
Thus π is flat. By [Asa87, Corollary 3.2]
it follows now from \eqrefPPPeachfibA1
that k[Z]m is a polynomial ring
in one variable over k[U]m for all
maximal ideals m⊂k[U]. Hence, by
[BCW77], the morphism π is a vector bundle
with respect to the Zariski topology and since
k[U] is a principal ideal domain, π
is a trivial A1-bundle.
∎
Lemma 3.11**.**
Let k be a field, P∈k[t,x,y] be a polynomial which is a variable of the
k(t)-algebra k(t)[x,y], let U⊂Ak1=Spec(k[t]) be a
dense open subset, let Z⊂U×Ak2=Spec(k[U][x,y]) be the
hypersurface given by P=0
and let π:Z→U be the morphism (t,x,y)↦t. Then,
the following conditions are equivalent:
- (i)
P* is a variable of the k[U]-algebra k[U][x,y];*
2. (ii)
There is an isomorphism φ:U×Ak1⟶≃Z such that πφ is the projection (t,x)↦t;
3. (iii)
The morphism π:Z→U is a trivial A1-bundle;
4. (iv)
The morphism π:Z→U is a locally trivial A1-bundle;
5. (v)
For each maximal ideal m⊂k[U], the fibre π−1(m)⊂Z is isomorphic to Aκ(m)1.
Proof.
(\refPPVarktU)⇒(\refPPisoA2fibrU):
If P is a variable of the k[U]-algebra k[U][x,y], there exists
f∈Autk[U](k[U][x,y]) such that f(x)=P. The element f is then equal to ψ∗ for some ψ∈Aut(U×Ak2) such that πψ=π. Hence, ψ(Z) is the closed subset of U×Ak2 given by x=0.
Let θ:U×Ak2→U×Ak1
be the projection given by (t,x,y)↦(t,y).
The composition θ∘ψ restricts to an
isomorphism Z⟶≃U×Ak1
that we denote by φ−1. Thus pr1∘φ−1=π, where pr1:U×Ak1→U is the projection on the first factor.
This yields (\refPPisoA2fibrU).
(\refPPisoA2fibrU)⇔(\refPPtrivialA1bundleU): is the definition of a trivial A1-bundle.
(\refPPtrivialA1bundleU)⇒(\refPPloctrivialA1bundleU)⇒(\refPPeachfibA1U) are obvious.
(\refPPeachfibA1U)⇒(\refPPVarktU):
This follows
from the implication \eqrefPPPeachfibA1⇒\eqrefPPPtrivialA1bundle
of Lemma 3.10.
∎
Corollary 3.12**.**
Let k be a field, let P∈k[t,x,y] be a polynomial which is
a variable of the k(t)-algebra k(t)[x,y], let
Z⊂Ak3=Spec(k[t,x,y]) be the hypersurface given by P=0
and let π:Z→Ak1 be the morphism (t,x,y)↦t. Then, the following conditions are equivalent:
- (i)
P* is a variable of the k[t]-algebra k[t][x,y];*
2. (ii)
There is an isomorphism φ:Ak2⟶≃Z such that πφ is the projection (t,x)↦t;
3. (iii)
There is an isomorphism φ:Ak2⟶≃Z;
4. (iv)
The morphism π:Z→Ak1 is a trivial A1-bundle;
5. (v)
The morphism π:Z→Ak1 is a locally trivial A1-bundle;
6. (vi)
For each maximal ideal m⊂k[t], the fibre π−1(m)⊂Z is isomorphic to Aκ(m)1.
Proof.
Applying Lemma 3.11 with U=Ak1, we obtain the equivalence between (\refPPVarkt)-(\refPPisoA2fibr)-(\refPPtrivialA1bundle)-(\refPPloctrivialA1bundle)-(\refPPeachfibA1).
We then observe that (\refPPisoA2fibr) implies (\refPPisoA2). It remains then to prove (\refPPisoA2)⇒(\refPPtrivialA1bundle). As P is a variable of the k(t)-algebra k(t)[x,y], the generic fibre of π:Z→Ak1 is isomorphic to Ak(t)1, so π is a trivial A1-bundle over some dense open subset U⊂Ak1. The fact that Z is isomorphic to Ak2 implies then that π is a trivial Ak1-bundle. (Implication (\refPloctrivial)⇒(\refPtrivial) of Lemma 3.9).
∎
3.2. Non-trivial embeddings in positive characteristic
In this paragraph, we recall the existence of non-trivial embeddings in positive characteristic. The family of examples that we give below seems classical (the case Ak1↪Ak2 with parameters equal to 1 corresponds in particular to [vdE00, Exercise 5(iii) in §5]). We give the (simple) proof here for a lack of a precise reference and for self-containedness.
Lemma 3.13**.**
For each field k of characteristic p>0, each a∈k, b∈k∗
and each integer q≥0, the morphism
[TABLE]
is a closed embedding, with image being the closed curve of
Ak2=Spec(k[x,y]) given by
[TABLE]
Proof.
We first compute the equality
[TABLE]
which shows that ρ(Ak1) is contained in the closed curve Γ⊂Ak2 given by P=0, where
P=x−(bpyp−apxq)p=x−bp2yp2+ap2xqp∈k[x,y]. The equality also yields up∈k[up2,b1(aupq+u)]
and thus yields
k[up2,b1(aupq+u)]=k[u], which implies that ρ is a closed embedding.
It remains to see that the degree of ρ (maximum of the degree of both components) is equal to the degree of P, to obtain that P is irreducible and that it defines the irreducible curve ρ(Ak1). For a=0, this follows, since deg(ρ)=p2=deg(P). For a=0, we have deg(ρ)=max(p2,pq)=deg(P).
∎
To show that the above embeddings are not equivalent to the standard one, when q≥2 is not a multiple of p and a,b=0,
one could make argument on the degree of the components
(no one divides the other) or can use the characterisation of
variables given in Lemma 3.3
to show that
P=x+ap2xpq−bp2yp2∈k[x,y] is not a variable, by proving that k[x,y,t]/(P−t) is not a polynomial ring in one variable over k[t], as we do in
Lemma 3.14 below.
The second way has the advantage to give examples in any dimension (see Proposition 3.16).
This is related to the forms of the affine line over non-perfect fields (for more details on this subject, see [Rus70]).
Lemma 3.14**.**
For each field k of characteristic p>0, each b∈k∗ and
each integer q≥2, not a multiple of p, the curve
[TABLE]
is not isomorphic to Ak(t)1, but after extension of scalars to k(t1/p) we have an isomorphism
[TABLE]
Proof.
After extending the scalars to k(t1/p2), the curve Γ becomes
[TABLE]
Replacing y with y+bt1/p2 and applying
Lemma 3.13 we obtain an isomorphism
[TABLE]
Replacing then u with u+t1/p2 we get an isomorphism defined over
k(t1/p):
[TABLE]
It remains that no isomorphism ν^:Ak(t)1⟶≃Γk(t) exists. If ν^ exists, then ν−1ν^∈Aut(Ak(t1/p)1) would be given by u↦αu+β, with α,β∈k(t1/p), α=0. The second coordinate of ν^(u) would then be
[TABLE]
The coefficient of u being bα, we get α∈k(t). Remembering that q≥2, the coefficient of up(q−1) is equal to
baqαp(q−1)(βp+t1/p). As βp∈k(t) we have βp+t1/p∈/k(t). Impossible, since q is not a multiple of p and α=0.
∎
Corollary 3.15**.**
For each field k of characteristic p>0, each integer q≥2 which is not a multiple of p, each λ,μ∈k∗ and each integer n≥2, the polynomial
[TABLE]
is not a variable of the k-algebra k[x1,…,xn].
Proof.
Showing that f is not a variable of k[x1,…,xn] is equivalent to ask that the k[t]-algebra k[x1,x2,…,xn,t]/(f−t) is not a polynomial ring in n−1 variables over k[t] (Lemma 3.3). It suffices then to show that An=k(t)[x1,…,xn]/(f−t) is not a polynomial ring in n−1 variables over k(t).
We first prove the result for n=2. By extending the scalars, we can assume that λ=ap2 and μ=−bp2 for some a,b∈k∗. Lemma 3.14 then shows that A2=k(t)[x1,x2]/(f−t) is not a polynomial ring in one variable.
As An=A2[x3,…,xn], the positive answer to the cancellation problem for curves [AHE72] implies that An is not a polynomial ring in n−1 variables over k(t) for each n≥2.
∎
Proposition 3.16**.**
For each field k of characteristic p>0, each integer q≥2 which is not a multiple of p, each a∈k∗ and each n≥1, the morphism
[TABLE]
is a closed embedding, which is not equivalent to the standard one.
Proof.
It follows from Lemma 3.13 that ρ is a closed embedding and that its image is given by the hypersurface with equation
f=0, where
[TABLE]
It remains to show that f is not a variable of k[x1,…,xn],
which follows from Corollary 3.15.
∎
4. Liftings of automorphisms and the proof of Theorem 2
4.1. Lifting of automorphisms of Ak3 to affine modifications
Lemma 4.1**.**
Let k be a field and let
[TABLE]
where n≥1 and h∈k[t,x,y] is a polynomial such that
h0=h(0,x,y)∈k[x,y] does not belong to k[w] for each variable w∈k[x,y].
- (1)
Every element of R∖k[t,x,y] can be written as
[TABLE]
where s∈k[t,x,y], m≥1, f1,…,fm∈k[t,x,y] are polynomials of degree <n in t, and fm=0.
2. (2)
If f∈R∖k[t,x,y] is written as in (\refDescriptionR) and d=ν(fm) is the valuation of fm in t, then 0≤d<n and tmn−df=g(t,x,y)∈k[t,x,y] satisfies g(0,x,y)∈h0⋅k[x,y]∖{0}.
3. (3)
Writing I⊂k[t,x,y] for the ideal (tn,h), we have tnR∩k[t,x,y]=I.
4. (4)
Every element of Autk[t](R) preserves the sets k[t,x,y] and I.
Proof.
(1): We first prove that every element of R∖k[t,x,y] has the desired form. Every element of R can be written as ∑i=0mfiui for some polynomials fi∈k[t,x,y]. We denote by r the biggest integer such that degt(fr)≥n. If r=0 or if no such integer exists, we are done. Otherwise, we write fr=tnA+B for some A,B∈k[t,x,y] with degt(B)<n. Then, replacing fr−1ur−1+frur=fr−1ur−1+(tnA+B)ur in ∑i=0mfiui
with (fr−1+h(t,x,y)A)ur−1+Bur decreases the integer r. After finitely many such steps, we obtain the desired form.
(2): We write f=s+∑i=1mfiui=s+∑i=1mfitnihi as in (1), we write d=ν(fm), which satisfies 0≤d<n (since fm=0 and degt(fm)<n), and obtain g=tmn−df=stmn−d+∑i=1mfihitmn−d−ni. In particular, g∈k[t,x,y] and it
satisfies g(0,x,y)=(h0)m⋅r, where r∈k[x,y] is obtained by replacing t=0 in tdfm∈k[t,x,y]. From {r,h0}⊂k[x,y]∖{0}, we deduce g(0,x,y)=0.
(3): The inclusion I⊂tnR follows from {tn,h}={tn⋅1,tn⋅u}⊂tnR. To show that tnR∩k[t,x,y]⊂I, we take f∈R such that tnf∈k[t,x,y] and show that tnf∈I. If f∈k[t,x,y], then
tnf∈tnk[t,x,y]⊂I. Otherwise, we write
f=s+∑i=1mfiui as in (1) and use (2) to obtain that g=tmn−df∈k[t,x,y] with 0≤d=ν(fm)<n, and we
get g(0,x,y)=0. The fact that tnf∈k[t,x,y] implies then that n>mn−d, whence n>d>(m−1)n, so m=1. Hence
tnf=tn(s+f1u)=tns+hf1∈I.
(4): Using (3) it suffices to show that every ψ∈Autk[t](R) preserves k[t,x,y]. The algebra R is canonically isomorphic to
k[t,x,y][tnh]⊂k(t)[x,y].
Since k(t)[x,y] is the localisation of k[t,x,y][tnh] in the multiplicative
system k[t]∖{0}, we get a natural inclusion
Autk[t](R)⊂AutKK[x,y], with K=k(t).
Suppose for contradiction that some ψ∈Autk[t](R) satisfies ψ(k[t,x,y])⊂k[t,x,y]. This implies that ψ(x)∈/k[t,x,y] or ψ(y)∈/k[t,x,y].
We assume that ψ(x)∈/k[t,x,y] (the case ψ(y)∈/k[t,x,y] being similar) and use (2) to obtain an integer l>0 such that g=tlψ(x)∈k[t,x,y] satisfies g(0,x,y)∈h0⋅k[x,y]∖{0}. Since ψ∈Autk[t](R)⊂AutKK[x,y], the element
ψ(x) is a variable of K[x,y] and the same holds for g(t,x,y)=tlψ(x). By Lemma 3.8, g(0,x,y) belongs to k[w] for some variable w∈k[x,y]. The fact that g(0,x,y)∈h0⋅k[x,y]∖{0} implies then that h0∈k[w] (Lemma 3.4), contradicting the hypothesis.
∎
Corollary 4.2**.**
Let k be a field and
[TABLE]
where n≥1 and h∈k[t,x,y] is a polynomial such that
h0=h(0,x,y)∈k[x,y] does not belong to k[w] for each variable w∈k[x,y].
Writing I the ideal (tn,h)⊂k[t,x,y], we obtain a group isomorphism
[TABLE]
Proof.
According to Lemma 4.1(4), every element φ∈Autk[t](R) preserves k[t,x,y] and I, and thus restricts to an element ψ∈Autk[t](k[t,x,y]) that preserves I.
Conversely, each automorphism ψ∈Autk[t](k[t,x,y]) that preserves I induces an automorphism of R=k[t,x,y][tnI]=k[t,x,y][tnh]. This latter is uniquely determined by ψ, since the morphism Spec(R)→Spec(k[t,x,y]) given by the inclusion k[t,x,y]↪R is birational.
∎
Remark 4.3*.*
According to [KZ99, Definition 1.1 and Proposition 1.1], Spec(R) is the affine modification of Ak3=Spec(k[t,x,y]) with locus (I,tn). It is thus natural that every automorphism of Ak3 fixing the ideal lift to an automorphism of Spec(R). The interesting part of Corollary 4.2 consists then in saying that all automorphisms
of the k[t]-algebra R are of this form.
4.2. Application of liftings to the case of SL2
We will apply Corollary 4.2 to the variety SL2⊂A4 given by
[TABLE]
and obtain Proposition 4.5 below.
Before we give a proof, let us recall the following basic facts on
the coordinate ring of the variety SL2.
Lemma 4.4**.**
Let R be the coordinate ring of SL2, i.e. R=k[t,u,x,y]/(xy−tu−1).
Then R is a unique factorisation domain and the units of R
satisfy R∗=k∗.
Proof.
Since the localisation Rt=k[t,t1][x,y] is a unique
factorisation domain, we only have to see that
tR is a prime ideal of R, by [Mat89, Theorem 20.2].
This is the case, since R/tR≃k[u,x,y]/(xy−1)
is an integral domain. Moreover, we have
R∗⊆(Rt)∗={μtn∣μ∈k∗,n∈Z}.
Since tn is invertible in R if and only if n=0, it follows
that R∗=k∗.
∎
Proposition 4.5**.**
We consider the morphisms
[TABLE]
and denote by X⊂SL2 the hypersurface given by t=1 and by Γ⊂Ak3 the closed curve given by t=xy−1=0.
Then, the birational morphism η:SL2→Ak3
yields a group isomorphism
[TABLE]
We moreover have
[TABLE]
Proof.
Every automorphism g of Ak3=Spec(k[t,x,y]) yields an automorphism
g∗∈Autk(k[t,x,y]). Moreover, the condition πg=g corresponds to
g∗(t)=t, and the condition g(Γ)=Γ to g∗(I)=I, where I⊂k[t,x,y] is the ideal of Γ, generated by t and xy−1. The isomorphism Aut(Ak3)→Autk(k[t,x,y]) then yields a bijection
[TABLE]
We then want to apply Corollary 4.2 with n=1 and h=xy−1. To check that it is possible, we need to see that h does not belong to k[w] for each variable w∈k[x,y]. Indeed, xy−1∈k[w] would imply that xy∈k[w], and thus that x,y∈k[w], since k[w] is factorially closed (Lemma 3.4). This would yield k[w]=k[x,y], a contradiction.
We then apply Corollary 4.2 and obtain a group isomorphism
[TABLE]
where R=k[t,u,x,y]/(tu−xy−1). This yields then a group isomorphism
[TABLE]
It remains then to show that
[TABLE]
The inclusion “⊂” follows from the equality X=(πη)−1(1). It remains then to show the inclusion “⊃”.
To do this, we take g∈Aut(SL2) such that g(X)=X and prove that πηg=πη. The element g corresponds to an element g∗∈Autk(R).
The fact that g(X)=X is then equivalent to ask that g∗ sends the ideal generated by t−1 onto itself. Since R∗=k∗ by Lemma 4.4, so t−1 is sent onto μ(t−1) for some μ∈k∗. This implies that the restriction of g∗ yields an automorphism of k[t], corresponding to an automorphism g^∈Aut(Ak1) such that g^πη=πηg. As (πη)−1(0)
is the only fibre of πη that is not isomorphic to Ak2,
it has to be preserved under g. As the fibre π−1(1)=X is
also preserved under g, we find that g^ is the identity,
so πηg=πη as desired.
∎
Corollary 4.6**.**
The closed embedding
[TABLE]
has the following property: an automorphism of Ak2 extends to an automorphism of SL2, via ν, if and only if it has Jacobian determinant equal to ±1.
Proof.
We denote by X=ν(Ak2)⊂SL2 the closed hypersurface given by
[TABLE]
write G={g∈Aut(SL2)∣g(X)=X} and denote by τ:G→Aut(Ak2) the group homomorphism such that
g∘ν=ν∘τ(g) for each g∈G.
We first prove that the subgroup H={h∈Aut(Ak2)∣Jac(h)±1} is contained in τ(G). The group H is generated by (x,y)↦(y,x), which is induced by
[TABLE]
and by automorphisms of the form (x,y)↦(x,y+p(x)), p∈k[x], induced by
[TABLE]
It remains to take g∈G and to prove that
τ(g)∈H. Proposition 4.5 implies that g can be written as
[TABLE]
where a,b∈k[t,x,y], s∈k[t,u,x,y] and such that
g~:Ak2→Ak2
given by g~(t,x,y)=(t,a(t,x,y),b(t,x,y))
is an automorphism of Ak3 that preserves the
curve Γ given by t=xy−1=0.
The Jacobian determinant of g~ is μ=∂x∂a⋅∂y∂b−∂y∂a⋅∂x∂b∈k∗. Replacing with t=0, we obtain the automorphism of Ak2 given by (x,y)↦(a(0,x,y),b(0,x,y)), which preserves the curve with equation xy=1 and is thus of Jacobian ±1. Indeed, it is of the form (x,y)↦(ξx,ξ−1y) or (x,y)↦(ξy,ξ−1x), for some ξ∈k∗ (see [BS15, Theorem 2 (iii)]). This shows that μ=±1. Replacing then t=1 we get that the automorphism τ(g) which is given
by τ(g)(x,y)=(a(1,x,y),b(1,x,y)) has
Jacobian ±1.
∎
Proof of Theorem 2.
We first observe that the embeddings ρ1,νˉ:Ak2→SL2 given by
[TABLE]
are equivalent, under the map
(xuty)↦(u−yx−t).
The embedding
[TABLE]
satisfies νˉ=ντ, where τ∈Aut(Ak2) is the automorphism of Jacobian −1 given by τ:(x,y)↦(x,−y). Corollary 4.6 then
implies that there exists τ^∈Aut(SL2) such that τ^ν=ντ=ν, i.e. that the embeddings ν and ν are equivalent, so ν and ρ1 are equivalent.
Corollary 4.6 implies then that an automorphism of Ak2 extends to an automorphism of SL2, via ρ1, if and only if it has Jacobian determinant equal to ±1. It remains to prove Assertions (1) and (2) of Theorem 2.
Assertion (2) follows from the fact that the
group homomorphism
[TABLE]
is surjective (taking for instance diagonal automorphisms), so there are automorphisms of Jacobian determinant in k∗∖{±1} if and only if k contains at least 4 elements.
To obtain Assertion (1), we observe that every closed embedding Ak2→SL2 having image in ρ1(A2) is of the form ρ1ν for some ν∈Aut(Ak2). Writing dλ∈Aut(Ak2) the automorphism given by dλ:(s,t)↦(λs,t), λ∈k∗, we can write ν=dλν1 for some λ∈k∗ and some ν1∈Aut(Ak2) of Jacobian determinant equal to 1. The result above implies that ρ1ν is equivalent to ρ1dλ=ρλ.
It remains to observe that ρλ′=ρλdλ′λ−1,
so ρλ and ρλ′ are equivalent if
and only if λ′λ−1∈{±1}, which corresponds to λ′=±λ.
∎
Remark 4.7*.*
Over the field k=C of complex numbers, all algebraic embeddings
of C2 into SL2(C) with image equal to ρ1(C2)
are equivalent under holomorphic automorphisms of SL2(C).
Indeed, according to
Theorem 2\eqrefThm:StdEmbSL2ass1
it is enough to show that the embeddings
[TABLE]
are equivalent under a holomorphic automorphism
for all λ∈C∗. Such a holomorphic
automorphism of SL2(C) is given by
[TABLE]
where μ:C→C∗
is a holomorphic function with μ(1)=λ and μ(0)=1.
5. Fibered embeddings of Ak2 into SL2
and the proof of Theorem 3
In this section, we study fibred embeddings (as in 1.2).
We will need the following simple description of the morphism η:SL2→Ak3 already studied in Proposition 4.5:
Lemma 5.1**.**
Let Γ⊂Ak3=Spec(k[t,x,y]) be the curve given by t=xy−1=0 and let η:BℓΓ(Ak3)→Ak3 the blow-up of Γ. We then have a natural open embedding SL2↪BℓΓ(Ak3) such that the restriction of η corresponds to (t,u,x,y)↦(t,x,y).
Proof.
The blow-up of Γ can be seen as
[TABLE]
The open subset of BℓΓ(Ak3) given by v=0 is then naturally isomorphic to SL2, by identifying ((t,x,y),[u:1]) with
(t,u,x,y)∈SL2, and the birational morphism η:SL2→Ak3 sends ((t,x,y),[u:1]) onto (t,x,y).
∎
5.1. Polynomials associated to fibred embeddings
The following result associates to every fibred embedding Ak2↪SL2 a polynomial in k[t,x,y], and gives some basic properties of this polynomial (which will be studied in more details after).
Lemma 5.2**.**
Let ρ:Ak2↪SL2 be a fibred embedding, and let Z⊂Ak3 be the closure of η(ρ(Ak2)), where
[TABLE]
Then, Z is given by P(t,x,y)=0,
where P∈k[t,x,y] is a polynomial having the following properties:
- (1)
*The ring k[t,t1][x,y]/(P) is a polynomial ring in one variable over k[t,t1] *(equivalently the morphism π:Z→Ak1 given by (t,x,y)↦t is a trivial A1-bundle over
Ak1∖{0}).
2. (2)
If P is a variable of the k(t)-algebra k(t)[x,y]
(which is always true if char(k)=0
by (1) and the
Abhyankar-Moh-Suzuki Theorem), then the polynomial
P(0,x,y)∈k[x,y] is given by μxm(x−λ) or
μym(y−λ)
for some μ,λ∈k∗
and some m≥0,
and ρ(Ak2)⊂SL2 is the
hypersurface given by P=0.
Proof.
We consider the morphisms
[TABLE]
and observe that η yields an isomorphism between the two open subset (SL2)t⊂SL2 and (Ak3)t⊂Ak3 given by t=0. The morphism ηρ:Ak2→Ak3 restricts thus to a closed embedding (Ak2)t↪(Ak3)t, where
(Ak2)t⊂Ak2 is the open subset where t=0. This yields (1).
We now assume that P∈k[t,x,y] is a variable of the k(t)-algebra k(t)[x,y]. Applying Lemma 3.8 we obtain that P0=P(0,x,y)∈k[w] for some variable w∈k[x,y]. In particular, xy−1 does not divide P0 (otherwise, by Lemma 3.4 we would have xy−1∈k[w] and then x,y∈k[w], impossible). This implies that Z∩Γ is a [math]-dimensional scheme
(which is a priori not reduced), where Γ⊂Ak3 is the closed curve given by t=xy−1=0.
Recall that SL2 is an open subset of BℓΓ(Ak3) (Lemma 5.1) and that the exceptional divisor E⊂SL2 is simply given by t=xy−1=0 and is a trivial A1-bundle over Γ. Since
the pull-back H⊂SL2 of Z on SL2, given by the equation P=0
has all its irreducible components of pure codimension 1 we get
H=ρ(Ak2)≃Ak2. As ρ is a fibred embedding, the morphism H→Ak1 given by the projection on t is a trivial A1-bundle.
This implies that Z∩Γ consists of a single reduced point, which
is defined over k and thus of the form q=(0,λ,λ1)∈Γ for some λ∈k∗.
We can thus write P0∈k[w] as P0=ab where a,b∈k[w] are such that b(q)=0, a is irreducible and a(q)=0. This implies that a is a polynomial of degree 1 in w, so we can assume that a=w (by replacing w with a).
We now show that w=x−λ or w=y−λ1 (after replacing w with μw, μ∈k∗). As w is a variable in k[x,y], the curve C⊂Ak2 defined by w=0 is isomorphic to Ak1, and its closure in Pk2 is a curve C passing through exactly one point q0 of the line at infinity L∞=Pk2∖Ak2. The closure of Γ is then Γ⊂Pk2 given by xy−z2=0, and Γ∖Γ={[1:0:0],[0:1:0]}. To get w=x−λ or w=y−λ1 we then only need to show that C is a line through [1:0:0] or [0:1:0]. We extend the scalars to an algebraically closed field and apply Bézout Theorem to get 2deg(C)=1+∑mq0′ where the sum is taken over all points
q0′ of C infinitely near to q0 and lying on Γ
and mq0′ denotes the multiplicity of C at q0′
(follows from the fact that C∩Γ is the reduced point q). If C is a line, we get the result, since q∈C. If C has degree at least 2, then it is tangent to L∞ at the point q0,
because, otherwise
mq0=(C,L∞)q0=deg(C) by Bézout’s Theorem and if L denotes the tangent line of C at q0,
then deg(C)≥(C,L)q0≥1+mq0,
a contradiction.
As L∞ and Γ have transversal intersections, this implies that only q0 belongs to Γ and yields
2deg(C)−1=mq0≤deg(C), yielding deg(C)≤1 as desired.
Now that w=x−λ is proven (respectively w=y−λ1), we obtain P0=wb for some b∈k[x] (respectively b∈k[y]) which does not vanish on any point of Γ. Hence, P0 is equal to xm(x−λ) or ym(y−λ1) for some m≥0, after replacing P with μP, μ∈k∗.
∎
We now give an example which shows that the polynomial P given in Lemma 5.2 is not always a variable of the k(t)-algebra k(t)[x,y] (even if P is always such a variable when char(k)=0).
Lemma 5.3**.**
Let k be a field of characteristic p>0 and let q≥2 be an integer that does not divide p. Then, the polynomial
[TABLE]
has the following properties:
- (1)
P* is not a variable of the k(t)-algebra k(t)[x,y].*
2. (2)
*The hypersurface ZP⊂Ak3=Spec(k[t,x,y])
given by P=0 satisfies that ZP→Ak1, (t,x,y)↦t
is a trivial A1-bundle *(in particular, ZP is isomorphic to Ak2).
3. (3)
*The hypersurface HP⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1))
given by P=0 is the image of a fibred embedding
Ak2↪SL2 *(in particular, HP
is isomorphic to Ak2).
Proof.
(1):
Replacing x with x+1, it suffices to show that x−tp(yp−xq)p is not a variable of the k(t)-algebra k(t)[x,y]. This follows from Corollary 3.15.
(2) We consider the morphisms
[TABLE]
and check that τ∘χ=idZP,
χ∘τ=idAk2.
(3): The morphism η:HP→ZP, (t,u,x,y)↦(t,x,y) being an isomorphism on the subsets given by t=0, the morphism
π∘η:HP→Ak1, (t,u,x,y)↦t is a trivial A1-bundle over Ak1∖{0}. The zero fibre is moreover isomorphic to Ak1 since P(0,x,y)=x−1 and the line {x=1} intersects the conic
{xy=1} transversally in one point
(follows from Lemma 5.1).
By Lemma 3.10 it follows that
π∘η:HP→Ak1 is a trivial A1-bundle.
Hence HP is isomorphic to Ak2
and is the image of a fibred embedding
Ak2↪SL2.
∎
We now start from a polynomial P∈k[t,x,y] that is a variable of the k(t)-algebra k(t)[x,y] and determine when this one comes from a fibred embedding Ak2↪SL2, by the process determined in Lemma 5.2. This yields the following result, which corresponds to Part (1) of Theorem 3.
Proposition 5.4**.**
Let k be any field, let P∈k[t,x,y] be a polynomial that is a variable of the k(t)-algebra k(t)[x,y], and let HP⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1)) and ZP⊂Ak3=Spec(k[t,x,y]) be the hypersurfaces
given by P=0.
The following conditions are equivalent:
- (a)
The hypersurface HP⊂SL2 is isomorphic to Ak2.
2. (b)
The hypersurface HP⊂SL2 is the image of a fibred embedding Ak2↪SL2.
3. (c)
The fibre
of ZP→Ak1, (t,x,y)↦t over every closed point
of Ak1∖{0}
is isomorphic to A1 and the
polynomial P(0,x,y)∈k[x,y] is of the form μxm(x−λ) or
μym(y−λ)
for some μ,λ∈k∗
and some m≥0.
Proof.
We will use the morphisms
[TABLE]
\eqrefThmEqa⇒\eqrefThmEqb:
Proving that HP is the image of a fibred embedding
Ak2↪SL2 is equivalent to ask that
π∘η:HP→Ak1 is a trivial A1-bundle.
Since P is a variable of the k(t)-algebra
k(t)[x,y], it follows that the generic fibre of π:ZP→Ak1
is isomorphic to Ak(t)1.
Moreover, η:SL2→Ak3 is an isomorphism
over {t=0}, so the generic fibre of π∘η
is also isomorphic to Ak(t)1.
The fact that HP is isomorphic to Ak2
(which is the hypothesis (a)) implies that
π∘η:HP→Ak1
is a trivial A1-bundle, by
Lemma 3.9
(\eqrefPktxyfield⇒\eqrefPtrivial).
\eqrefThmEqb⇒\eqrefThmEqc: Follows from
Lemma 5.2(1) and
(2).
\eqrefThmEqc⇒\eqrefThmEqa: Since
η:SL2→Ak3 is an isomorphism over the open subset
{t=0}, it follows that all fibres of
π∘η:HP→Ak1 over closed points
of Ak1∖{0}
are isomorphic to A1. Moreover, the fibre
of π∘η over [math] is isomorphic to Ak1,
since the restriction
η∣{t=0}:{t=0}→{t=xy−1=0}⊂{0}×Ak2
is a trivial A1-bundle over the curve {t=xy−1=0}
and since {P(0,x,y)=0} intersects {xy=1} in exactly one
point, transversally. The generic fibre of π∘η:HP→Ak1
being isomorphic to Ak(t)1, it follows from Lemma 3.10
that π∘η:HP→Ak1 is a trivial A1-bundle
and thus HP is isomorphic to the affine plane Ak2,
which proves (a).
∎
Example 5.5*.*
For each n≥1,m≥0, μ∈k∗ and q∈k[t,x], the polynomial
[TABLE]
defines an hypersurface HP⊂SL2 which is the image of a fibred embedding. Indeed, since P has degree 1 in y with coefficent tn, it is a variable of k[t,t−1][x,y]. We can thus apply Proposition 5.4 and only need to check that P(0,x,y)=μxm(x−1)
is of the desired form (as in Assertion (c)).
5.2. Determining when two fibred embeddings are equivalent
In this section, we consider embeddings satisfying the conditions of
Proposition 5.4 (or equivalently of Theorem 3(1)) and determine when two of these are equivalent, by proving Theorem 3(2). We first characterise the case where the integer m of Proposition 5.4 (or equivalently of Lemma 5.2 or Theorem 3(1)) is equal to zero.
Lemma 5.6**.**
Let k be any field and P∈k[t,x,y] be a polynomial that is a variable of the k(t)-algebra k(t)[x,y], and let HP⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1)) and ZP⊂Ak3=Spec(k[t,x,y])
be the hypersurfaces given by P=0.
Assume that HP is isomorphic to Ak2, which implies that P(0,x,y)∈k[x,y] is of the form μxm(x−λ) or μym(y−λ) for some μ,λ∈k∗ and some m≥0. Then, the following conditions are equivalent:
- (a)
m=0;
2. (b)
P* is a variable of the k[t]-algebra k[t][x,y];*
3. (c)
There is an isomorphism φ:Ak2⟶≃ZP
such that πφ is the projection (t,x)↦t.
4. (d)
There is an isomorphism φ:Ak2⟶≃ZP.
5. (e)
There exist φ∈Aut(SL2) such that
φ(HP)=ρ1(Ak2), where ρ1
is the standard embedding;
6. (f)
There exist φ∈Aut(SL2) such that
φ(HP)=ρ1(Ak2) and
φ∗(t)=t.
Proof.
As before, we use the morphisms
[TABLE]
Proposition 5.4 says that HP⊂SL2 is the image of a fibred embedding Ak2↪SL2, which corresponds to say that πη:HP→Ak1 is a trivial A1-bundle. Since
η:SL2→Ak3 is an isomorphism over the open subset
{t=0}, we obtain that π:ZP→Ak1 is a trivial
A1-bundle over Ak1∖{0}.
We first prove (\refm0)⇔(\refPvarkt)⇔(\refZA2fibr)⇔(\refZA2), using Corollary 3.12. We observe that (\refPvarkt), (\refZA2fibr) and (\refZA2) correspond respectively to the equivalent assertions (\refPPVarkt), (\refPPisoA2fibr) and (\refPPisoA2) of Corollary 3.12. Moreover, the condition m=0 (which is (\refm0)) corresponds to say that the [math]-fibre of π:ZP→Ak1 is isomorphic to Ak1. Since π:ZP→Ak1 is a trivial
A1-bundle over Ak1∖{0}, assertion (\refm0) corresponds to assertion (\refPPeachfibA1) of Corollary 3.12. Thus Corollary 3.12 yields
[TABLE]
It remains to show that these are also equivalent to (\refAutoStd) and (\refAutofibStd).
(\refPvarkt)⇒(\refAutofibStd): Applying an automorphism of the form
[TABLE]
for some μ∈k∗, we can assume that P(0,x,y)=x−1. Since P is a variable of the k[t]-algebra k[t][x,y], there exists f∈Autk[t](k[t,x,y]) such that f(x−1)=P. The element ψ∈Aut(Ak3) satisfying ψ∗=f is then such that πψ=π and sends ZP onto the hypersurface of Ak3 given by x=1. The restriction of ψ to the hypersurface given by t=0 is an automorphism of the form (0,x,y)↦(0,ν(x,y),ρ(x,y)) which preserves the curve given by x−1=0. Replacing ψ with its composition with
the inverse of (t,x,y)↦(t,ν(x,y),ρ(x,y)), we can assume that the restriction of ψ to the hypersurface t=0 is the identity, so ψ(Γ)=Γ, where Γ is the curve given by t=xy−1=0. Proposition 4.5 implies then that ψ lifts to an automorphism φ of SL2 sending HP onto ρ1(Ak2).
We moreover have φ∗(t)=t, since ψ∗(t)=t.
(\refAutofibStd)⇒(\refAutoStd) being clear, it
remains to show (\refAutoStd)⇒(\refm0).
For this implication, one can assume that k is algebraically closed.
Assertion (\refAutoStd) yields an automorphism φ∈Aut(SL2)
such that φ(HP)=ρ1(Ak2). Hence the automorphism
φ∗∈Autk(R), where R=Spec(k[t,u,x,y]/(xy−tu−1)), sends the
ideal (x−1)⊂R onto the ideal (P)⊂R. It follows from
Lemma 4.4 that x−1 is sent onto μP, for some
μ∈k∗. In particular, for a general a∈k, the variety
HP−a⊂SL2 given by P−a=0 is isomorphic to Ak2.
It remains to show that this implies that m=0.
Since P is a variable of the k(t)-algebra k(t)[x,y], so is P−a.
There exists then an open dense subset U⊂Ak1
such that U×Ak2→U×Ak1, (t,x,y)↦(t,P(t,x,y)−a)
is a trivial A1-bundle. This implies that
qa:HP−a→Ak1 (t,u,x,y)↦t
is a trivial A1-bundle
over U. By Lemma 3.9,
qa is a trivial A1-bundle (since HP−a≃Ak2),
so the fibre (qa)−1({0}) needs to be isomorphic to an affine line.
Since (qa)−1({0})
is given by the equations xy−1=P(0,x,y)−a=t=0
in the affine 4-space Ak4=Spec(k[t,u,x,y]) and since
P(0,x,y)−a is equal to μxm(x−λ)−a or
μym(y−λ)−a and a∈k is general
(k is algebraically closed),
this implies that m=0 and yields (\refm0) as desired.
∎
Remark 5.7*.*
Lemma 5.6 shows in particular that if HP,HQ⊂SL2 are two hypersurfaces given by two polynomials P,Q∈k[t,x,y] as in Theorem 3 (or as in the previous results), and if one of the two integers m,m′∈N associated to P,Q is equal to zero, then HP,HQ are equivalent if and only if m=m′=0.
Proposition 5.8**.**
Let k be any field, let P,Q∈k[t,x,y] be polynomials that are variables of the k(t)-algebra k(t)[x,y], and let HP,HQ⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1)) and ZP,ZQ⊂Ak3=Spec(k[t,u,x]) be the hypersurfaces given by P=0 and Q=0 respectively.
Suppose that HP is isomorphic to Ak2 but that ZP is not isomorphic to Ak2, and that there exists φ∈Aut(SL2) that sends HP onto HQ. Then, the following hold:
- (1)
There exists μ∈k∗ such that φ∗(t)=μt.
2. (2)
The birational map ψ=ηφη−1 is an automorphism of Ak3 which sends ZP onto ZQ, where η:SL2→Ak3 is as before given by (t,u,x,y)↦(t,x,y).
3. (3)
*There exists m≥1 such that P(0,x,y) and Q(0,x,y) are of the form μxm(x−λ) or
μym(y−λ)
for some μ,λ∈k∗
*(the integer m is the same for P,Q but
μ,λ and the choice between x and y
depend on P,Q).
Proof.
Since HP is isomorphic to Ak2, the same holds for HQ. The hypersurfaces HP,HQ⊂SL2 are thus image of fibred embeddings Ak2↪SL2 and there are thus integers m,m′≥0 and λ,λ′,μ′,μ∈k∗ such that P(0,x,y)∈{μxm(x−λ),μym(y−λ)} and Q(0,x,y)∈{μ′xm′(x−λ′),μ′ym′(y−λ′)} (Proposition 5.4). Moreover, the fact that ZP is not isomorphic to Ak2 is equivalent to m>0 and to the fact that HP is not equivalent to the image ρ1(A2) of the standard embedding (Lemma 5.6). As HP and HQ are equivalent, the same hold for HQ, so m′>0.
The main part of the proof consists in proving (1). To do this, one
can extend the scalars and assume k to be algebraically closed.
We moreover have φ∗(Q)=ξP for some ξ∈k∗
(follows from Lemma 4.4). Replacing P with ξP,
we can assume that φ∗(Q)=P. For each a∈k∗, the element
φ then sends HP−a onto HQ−a, where
HP−a,HQ−a⊂SL2 are given by the polynomials
P−a,Q−a∈k[t,x,y]. If a is chosen general, then HP−a,HQ−a
are smooth hypersurfaces of SL2 (since this is true for a=0), and the
same holds for the hypersurfaces ZP−a,ZQ−a⊂Ak3 given by
P−a and Q−a respectively. Since P,Q are variables of the k(t)-algebra
k(t)[x,y] and because the t-projections ZP→A1 and
ZQ→A1 are trivial A1-bundles over Ak1∖{0}
(Lemma 5.2(1)),
the polynomials P,Q are also
variables of the k[t,t1]-algebra k[t,t1][x,y]
(follows from Lemma 3.11 with U=A1∖{0}).
Hence, the same holds for P−a and Q−a. The morphisms
HP−a,HQ−a,ZP−a,ZQ−a→Ak1 given by the projection on
t are therefore trivial
A1-bundles over Ak1∖{0}. We can thus see these varieties
as open subsets of smooth projective surfaces HP−a,HQ−a,ZP−a,ZQ−a obtained by blowing-up some Hirzebruch surfaces, so that the projection on t is the restriction of the morphism to Pk1 given by a P1-bundle of the Hirzebruch surface and having only one singular fibre. We can moreover assume that the boundary is a union of smooth rational curves of self-intersection [math] or ≤−2 (in particular the projectivisation is minimal). Indeed, if a component of the singular fibre has self-intersection −1 and is in the boundary, we can contract it, and if the section has self-intersection −1, then we blow-up a general point of the smooth fibre contained in the boundary and then contract the strict transform of this fibre to obtain a section of self-intersection [math]. The zero fibre of ZP−a→Ak1 is given by t=μxm(x−λ)−a=0 or t=μym(y−λ)−a=0 and is thus a disjoint union
C≃∐i=1m+1Ak1
of m+1 affine curves isomorphic to Ak1. Similarly the zero fibre of ZQ−a→Ak1 is a disjoint union
C′≃∐i=1m′+1Ak1
of m′+1 affine curves isomorphic to Ak1. The closure of C is contained in the singular fibre F0 of ZP−a→Pk1, which is a tree of smooth rational curves of self-intersection ≤−1, being a SNC divisor. Hence, the closure of each component of C is a smooth rational curve of self-intersection ≤−1, which intersects the boundary into a component lying in F0. A similar description holds for C′.
The curves C, C′ meet transversally the conic Γ given by xy=1 (because of the form of P(0,x,y)−a and Q(0,x,y)−a).
The surfaces HP−a,HQ−a
are then obtained by blowing-up some
points in each of the components of C, C′
and removing these components, so we can choose the a minimal
projectivisations of HP−a,HQ−a to be blowing-ups of the above points in ZP−a,ZQ−a and get a dual graph of
the boundary of these surfaces which is not a chain (or which is not “linear” or not a “zigzag”). This implies that the A1-fibration given by the t-projection is unique up to automorphisms of the target ([Ber83, Theorème 1.8]). As the zero fibre of
HP−a,HQ−a→Ak1 is the unique degenerate fibre,
there exist μa∈k∗ and qa∈k[t,u,x,y] such that φ∗(t)=μat+qa⋅(P−a). Since this holds for a general a, we get φ∗(t)=μt for some μ∈k∗. Indeed, replacing t with [math] in φ∗(t) yields an element of k[u,x,y]/(xy−1) which is divisible by P−a for infinitely many a.
This element is thus equal to zero.
We now show how Assertion (1) implies the two others. We write φ=φ1φ2 where (φ1)∗(t)=t and φ2 is given by
[TABLE]
The fact that (φ1)∗(t)=t implies that ψ1=ηφ1η−1 is an automorphism of Ak3 (Proposition 4.5). Since ψ2=ηφ2η−1 is a diagonal automorphism of Ak3, the element ψ=ψ1ψ2=ηφη−1 is an automorphism of Ak3. As φ sends HP onto HQ, the automorphism ψ sends ZP onto ZQ, which yields (2). As ψ∗(t)=μt, the hyperplane W⊂Ak3 given by t=0 is invariant, this implies that m=m′ and thus yields (3).
∎
Lemma 5.6 and Proposition 5.8 yield then the following result, which yields in particular Assertion (2) of Theorem 3:
Corollary 5.9**.**
If P,Q∈k[t,x,y] are polynomials which are variables of the
k(t)-algebra k(t)[x,y] and if the corresponding
hypersurfaces HP,HQ⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1))
are equivalent
and isomorphic to Ak2, the following hold:
- (1)
HP,HQ* are the image of fibred embeddings
Ak2↪SL2.*
2. (2)
There exists φ∈Aut(SL2) such that φ(HP)=HQ
and φ∗(t)=μt for some μ∈k∗.
In particular, the element ψ=ηφη−1∈Aut(Ak3) satisfies
ψ∗(t)=μt, ψ(ZP)=ZQ and ψ(Γ)=Γ, where
η:SL2→Ak3 is the morphism
(t,u,x,y)↦(t,x,y),
ZP,ZQ⊂Ak3 are the two hypersurfaces given by P=0, Q=0
and Γ⊂Ak3 is the conic given
by t=xy−1=0.
Proof.
Assertion (1) follows from Proposition 5.4. It remains then to show (2). We denote by φ0∈Aut(SL2) an element such that φ0(HP)=HQ.
(i) If φ0∗(t)=μt for some μ∈k∗, we choose φ=φ0 and denote by θ∈Aut(SL2) the element
[TABLE]
to obtain (φ0θ)∗(t)=t. Proposition 4.5 shows that
ψ^=η(φ0θ)η−1∈Aut(Ak3) and
ψ^(Γ)=Γ. Since θ~=ηθη−1 is the
automorphism of Ak3 given by (t,x,y)↦(μ−1t,x,y), we have
ψ=ηφ0η−1∈Aut(Ak3) and ψ(Γ)=Γ. The
fact that φ0∗(t)=μt and φ0(HP)=HQ yields then
ψ∗(t)=μt and ψ(ZP)=ZQ.
(ii) If φ0∗(t)∈{μt∣μ∈k∗}, then Proposition 5.8(1) does not hold, so ZP is isomorphic
to Ak2. Applying the same argument to φ0−1 shows that ZQ is isomorphic to Ak2. Lemma \refLem:M0(\eqrefZA2⇒\eqrefAutofibStd) then shows that there exist φ1,φ2∈Aut(SL2)
such that φ1(HP)=φ2(HQ)=ρ1(Ak2)
and (φ1)∗(t)=(φ2)∗(t)=t. We then choose
φ=(φ2)−1φ1 and apply case (i).
∎
5.3. Examples of non-equivalent embeddings
Lemma 5.10**.**
To each polynomial r∈k[t], we associate the polynomial
[TABLE]
and denote by let HPr⊂SL2=Spec(k[t,u,x,y]/(xy−tu−1)) and ZPr⊂Ak3=Spec(k[t,x,y]) the hypersurfaces given by Pr=0. Then,
- (1)
For each r∈k[t], the surface HPr is the image of a fibred embedding Ak2↪SL2.
2. (2)
For each r,s∈k[t], the following are equivalent
- (i)
There exists φ∈Aut(SL2) such that φ(HPr)=HPs.
2. (ii)
There exists φ∈Aut(Ak3) such that φ(ZPr)=ZPs.
3. (iii)
The surfaces ZPr and ZPs are isomorphic.
4. (iv)
r=s.
Proof.
For each r∈k[t], we write Sr(t,x)=(x−t)(x−1−t2r(t))∈k[t,x] and observe that Pr(t,x,y)=ty−Sr(t,x).
(1): Since Pr is of degree 1 in y, it is a variable of
the k(t)-algebra k(t)[x,y]. Moreover, Pr(0,x,y)=Sr(0,x)=x(x−1) is of the form μxm(x−λ) (with μ,λ∈k∗ and m≥0).
The coefficient of y in Pr being t, the morphism ZPr→Ak1, (t,x,y)↦t is a trivial A1-bundle over Ak1∖{0}. Proposition 5.4 (\eqrefThmEqc⇒\eqrefThmEqb) then implies that HPr⊂SL2 is the image of a fibred embedding Ak2↪SL2.
It remains to show that the assertions
(\refEquivRSHeq)−(\refEquivRSZeq)−(\refEquivRSZiso)−(\refEquivRSequal) of (\refEquivRS) are equivalent.
The implications (\refEquivRSequal)⇒(\refEquivRSHeq) and (\refEquivRSZeq)⇒(\refEquivRSZiso) are trivial.
Lemma 5.6 implies that ZPr and ZPs are not isomorphic to Ak2 (the integer m being here equal to 1). We can thus appply Proposition 5.8(2), which yields (\refEquivRSHeq)⇒(\refEquivRSZeq).
It remains then to show (\refEquivRSZiso)⇒(\refEquivRSequal). According to [DP09, Proposition 3.6], the surface ZPr and ZPs are isomorphic if and only if there exist a,μ∈k∗, τ∈k[t], such that
[TABLE]
This corresponds to
[TABLE]
and thus gives two possibilities:
(I): at=μ(t−τ(t)) and 1+a2t2r(at)=μ(1+t2s(t)−τ(t)). The first equation yields τ(t)=(1−μa)t and the second yields μτ(t)≡μ−1(modt2), which gives τ=0 and then μ=1 and a=1.
The second equation thus yields r(t)=s(t).
(II): at=μ(1+t2s(t)−τ(t)) and μ(t−τ(t))=1+a2t2r(at). This yields
[TABLE]
and thus 1−μt≡−μ+at(modt2), whence μ=−1 and a=1. Replacing in the equation above, we find r(t)=s(t).
∎
The proof of Theorem 3 is now clear:
Proof of Theorem 3.
Assertion (1) corresponds to Proposition 5.4.
Assertion (2) follows from Corollary 5.9.
Assertion (3) follows from Lemma 5.10, which yields hypersurfaces HPr⊂SL2 that are
parametrised by r∈k[t], which are all images of fibred embeddings and are pairwise non-equivalent.
∎
We finish this subsection with two explicit examples:
Lemma 5.11**.**
Let us denote by P,Q∈k[t,x,y] the polynomials
[TABLE]
Then, the following hold:
- (1)
The hypersurfaces ZP,ZQ⊂Ak3 given by P=0 and Q=0 are equivalent.
2. (2)
The hypersurfaces HP,HQ⊂SL2 given by P=0 and Q=0 are both images of fibred embeddings but are not equivalent.
Proof.
To get (1), it suffices to observe that the linear automorphism θ∈Aut(Ak3) given by (t,x,y)↦(t,x,y−x) satisfies θ∗(Q)=P, so θ(ZP)=ZQ.
Since P,Q are of degree 1 in y, both are variables of the k(t)-algebra
k(t)[x,y]. Moreover, P(0,x,y)=Q(0,x,y)=−x(x+1) is of the form μxm(x−λ) (with μ,λ∈k∗ and m=1≥0).
Since the coefficient of y in P and Q is t2, the morphisms
ZP,ZQ→Ak1, (t,x,y)↦t are trivial A1-bundles
over Ak1∖{0}. Proposition 5.4 (\eqrefThmEqc⇒\eqrefThmEqb) then implies that HP,HQ⊂SL2 are images of fibred embeddings Ak2↪SL2.
To get (2), we suppose that there is φ∈Aut(SL2) such that φ(HP)=HQ and derive a contradiction. Corollary 5.9 yields an automorphism ψ∈Aut(Ak3) such that ψ∗(t)=μt for some μ∈k∗ and such that ψ(ZP)=ZQ and ψ(Γ)=Γ, where Γ⊂Ak3 is the conic given by t=xy−1=0. The restriction of ψ to the hyperplane H⊂Ak3 given by t=0 then preserves Γ and also the curve C=H∩ZP=H∩ZQ, given by t=x(x+1)=0 (which is isomorphic to two copies of A1). The fact that C is preserved implies that ψ∣H is of the form
(x,y)↦(x,ay+p(x)) or (x,y)↦(−1−x,ay+p(x)) for some a∈k∗ and p∈k[x]. The fact that Γ is preserved implies that ψ∣H=id.
The element ξ=θ−1ψ∈Aut(Ak3) then satisfies ξ(ZP)=ZP, ξ∗(t)=μt and ξ∣H is the automorphism (x,y)↦(x,x+y). To show that this is impossible, we use [DP09, Theorem 3.11]
to see that every automorphism of
ZP preserves C and its action on C corresponds to an
element of the subgroup G=G0∪G1≃G0⋊(Z/2Z)
of Aut(C) given by
[TABLE]
We then study an explicit example of a fibred embedding Ak2↪SL2 whose image is not equivalent to the standard embedding.
Example 5.12*.*
According to the above study, the “simplest” example of a hypersurface
E⊂SL2 being the image of a fibred embedding but not being
equivalent to the image of the standard embedding is given by
[TABLE]
Indeed, using the polynomial P=ty−x(x−1), which yields P(0,x,y)=−x(x−1), the surface E is the image of a fibred embedding ρ:Ak2↪SL2 (Example 5.5) but is not equivalent to ρ1(Ak2) (Lemma 5.6).
One can construct an explicit embedding
ρ:Ak2↪SL2 having image E
in the following way. First, denoting by Et⊂E and (Ak2)t⊂Ak2=Spec(k[x,t]) the open subsets given by t=0, we get isomorphisms
[TABLE]
To obtain a fibred embedding ρ:Ak2↪SL2 having image equal to E, we need to remove the denominators of the isomorphism
(Ak2)t⟶≃Et.
We then compose with the automorphism of (Ak2)t given by
(x,t)↦(t2x+t+1,t) and get isomorphisms
[TABLE]
and
[TABLE]
We can observe that all components of ρ are indeed polynomials, and that t2x−t−1∈k[E]. To show the latter,
we compute
y=tx(x−1)∈k[E], u=txy−1=t2x2(x−1)−t∈k[E], y2−ux+u=tx−1∈k[E],
which yields t2x−t−1=u−(x+1)(tx−1)2∈k[E].
Writing
[TABLE]
we get an equivalent closed embedding ρ~:Ak2→SL2, which is an isomorphism
[TABLE]
where E~={(t,u,x,y)∈Ak4∣xy−tu=1,t(y+t)=(x+t)(x+t−1)}.
The morphism ρ~ corresponds to the
closed embedding Ak2↪Ak4
[TABLE]
that we can simplify using elementary automorphisms of Ak4
to the embedding
[TABLE]
Question 5.13**.**
Is the closed embedding
[TABLE]
equivalent to the standard one?
5.4. Embeddings of Ak2 into SL2 of small degree
This last subsection consists in showing the second part of Remark 1.1, which claims that if all component functions of a closed
embedding f:A2↪SL2
are polynomials of degree ≤2, then f is equivalent to ρλ
for a certain λ∈k∗. This will be done in Proposition 5.19 below, after a few lemmas.
We first make the following easy observation:
Lemma 5.14**.**
For each fibred embedding
[TABLE]
(with a,b,c∈k[s,t])
there is an automorphism g∈Aut(SL2) such that
gρ is a fibred embedding given by
[TABLE]
for some p,q∈k[s,t] such that p(s,0)+q(s,0)∈k∗ and such that deg(1+stp(s,t))≤deg(a), deg(1+stq(s,t))≤deg(b) (where the degree is here the degree of polynomials in s,t).
Remark 5.15*.*
The standard embedding ρ1 is of the above form with p=0 and q=1. More generally, the embeddings {ρλ}λ∈k∗ of Theorem 2 are given by p=0 and q=λ.
Proof.
Replacing t with [math] yields two elements a(s,0),b(s,0)∈k[s] such that a(s,0)⋅b(s,0)=1. This implies that a(s,0),b(s,0)∈k∗. Applying the automorphism
[TABLE]
for some μ∈k∗, we can assume that a(s,0)=b(s,0)=1. We then apply
[TABLE]
for some d∈k[t] and replace a(s,t) with a(s,t)+td(t), so can assume that a(0,t)=1. Applying similarly an automorphism of the form
[TABLE]
we can assume that b(0,t)=1. This yields p,q∈k[s,t] such that a=1+stp and b=1+stq, which yields c=s(p+q+stpq). Replacing t with [math] yields a closed embedding
[TABLE]
whence p(s,0)+q(s,0)∈k∗.
∎
Corollary 5.16**.**
Each fibred embedding
[TABLE]
where a,b,c∈k[s,t] are such that dega+degb≤4
is equivalent to the embedding
[TABLE]
for some λ∈k∗.
Proof.
Applying Lemma 5.14, one can assume that a=1+stp, b=1+stq, c=s(p+q+stpq) for some p,q∈k[s,t] with p(s,0)+q(s,0)∈k∗. If p=0, then ρ(A2) is equal to ρ1(A2), so the result follows from Theorem 2(1).
The same holds if q=0 by applying the automorphism
[TABLE]
To finish the proof, we assume that pq=0 and derive a contradiction.
The fact that dega+degb≤4 implies that p,q∈k∗. Hence,
k[s,t]=k[a,b,c,t]=k[t,st,s(p+q+stpq)]=k[t,st,s(st+ξ)]
with ξ=pqp+q=0, and thus the morphism
[TABLE]
would be a closed embedding. This is false, since
the image is properly contained in the irreducible
hypersurface given by
{(x,y,z)∈Ak3∣xz=y(y+ξ)}
(the line given by x=y+ξ=0 is missing).
∎
It remains to generalise Corollary 5.16 to the case of embeddings Ak2↪SL2 of small degree (which are fibred or not).
In the sequel we will use the following subgroups of Aut(Ak2):
Definition 5.17**.**
[TABLE]
Lemma 5.18**.**
Let k be an algebraically closed field and let ρ:Ak2→Ak2∖{0} be a morphism of the form
[TABLE]
such that f,g have degree 2 and that the homogeneous parts f2 and g2 of f,g of degree 2 are linearly independent. Then, there exist α∈Aff2 and β∈GL2 such that
[TABLE]
Proof.
We first observe that replacing ρ with βρα, where α∈Aff2 and β∈GL2, does not change the degree of f,g or the fact that f2 and g2 are linearly independent. We then observe that we can assume that f2=s2. If f2 is a square, it suffices to replace f with ρα for some α∈GL2. If f2 is not a square, we choose ξ∈k such that g2+ξf2 is a square (this is possible since the discriminant of g2+ξf2 is a polynomial of degree 2 in ξ and k is algebraically closed).
We then apply an element of GL2 at the target to replace f2,g2 with g2+ξf2,f2,and then apply as before an element of GL2 at the source, to obtain f2=s2.
For each irreducible factor P of f, we denote by CP⊂Ak2=Spec(k[s,t]) the irreducible curve given by P=0, and observe that g yields an invertible function on CP.
(a) If f is a product of factors of degree 1, all belong to k[s], since f2=s2. We can then write f=∏i=12(s−λi), for some λi∈k. If λ1=λ2, we replace s with s−λ1 and get f=s2, which yields g=s(as+bt+c)+d for some a,b,c∈k,d∈k∗. The parts of degree 2 of f and g being linearly independent, we get b=0. Replacing t with bt−as−c, we replace g with st+d. We then apply diagonal elements of the form (s,t)↦(s,μt),
μ∈k∗ at the source and target and replace d with 1, which yields the desired form. To finish case (a), it remains to see
that λ1=λ2 is impossible. To derive this contradiction, we apply an element of Aff2 at the source and get f=s(s−1) This yields g=sp(s,t)+μ, where μ∈k∗ and p∈k[s,t] is of degree 1. We moreover obtain p(1,t)∈k∖{−μ}, so p(s,t)=(s−1)ξ+ν for some ξ,ν∈k. This yields g∈k[s], which is impossible since g2 is not a multiple of f2=s2.
(b) We can now assume that f is not a product of factors of degree 1, i.e. f is irreducible, and derive a contradiction. We observe that the curve Cf⊂A2 given by f=0 is isomorphic to A1. Indeed, the closure of Cf in Pk2 is an irreducible and thus a smooth conic with one point at infinity since f2=s2 (recall that k is assumed to be algebraically closed).
This implies that the restriction g∣Cf is a non-zero constant and so g=μ+ξf for some μ∈k∗,ξ∈k. This contradicts the fact that f2 and g2 are linearly independent.
∎
Proposition 5.19**.**
Each closed embedding
[TABLE]
where f11,f12,f21,f22∈k[s,t] have at most degree 2 is
equivalent to the embedding
[TABLE]
for some λ∈k∗.
Proof.
Applying
Theorem 2,
one only needs to show the existence of an automorphism of SL2 that sends ρ(Ak2) onto ρ1(Ak2). We distinguish the following cases:
(a) Suppose first that one of the polynomials fij is constant. One can assume that it is f11 by using permutation of coordinates (with signs). The case f11=0 is impossible, since the image would then be contained in
[TABLE]
We then have f11=0 and apply a diagonal automorphism of SL2 to get f11=1, which corresponds to ρ(Ak2)=ρ1(Ak2).
(b) Suppose then that one of the fij has degree 1. Applying permutations one can assume that f12 has degree 1.
Applying an element of Aff2 at the source (see Definition 5.17), we do not change the degree of the fij′s and can assume that f12=t.
Since deg(f11f22)=deg(f12f21)≤4, the result follows from
Corollary 5.16.
(c) It remains to study the case where
deg(fij)=2 for each i,j∈{1,2}. If the homogeneous parts of f11 and f12 of degree 2 are collinear, we apply
[TABLE]
for some μ∈k and obtain deg(f12)≤1, which reduces to the cases (a), (b). To achieve the proof of (c), we now assume that the homogeneous parts of f11 and f12 of degree 2 are linearly independant and prove that this implies that k[f11,f12,f21,f22]⊊k[s,t] (which contradicts the fact that ρ is a closed embedding). To show this, one can extend the scalars and assume that k is algebraically closed. We then apply Lemma 5.18 to the morphism ν:Ak2→Ak2∖{0} given by (s,t)↦(f11(s,t),f12(s,t)), and find α∈Aff2(k), β∈GL2(k) such that βνα=(s,t)↦(s2,st+1). We write μ=det(β)∈k∗ and replace ρ with β^ρα, where β^∈Aut(SL2) is of the form
[TABLE]
This change being made, we obtain f11=s2, f12=st+1. Since 1=f11f22−f12f21=s2f22−(st+1)f21, we obtain f21=st−1+g(s,t)s2 for some g∈k[s,t]. This implies that f11,f12−1,f21+1 all belong to the maximal ideal (s,t)2⊂k[s,t], which yields the desired contradiction k[f11,f12,f21,f22]=k[f11,f12−1,f21+1,f22]⊊k[s,t].
∎
6. A non-trivial embedding of A1 into SL2, over the reals
In this section, we provide over the field k=R
an explicit example of an algebraic embedding
AR1↪SL2 which is not equivalent
to the standard embedding
[TABLE]
Example 6.1*.*
In [Sha92] the closed embedding
[TABLE]
is given. This one is not equivalent to the standard embedding
A1↪A3, t↦(t,0,0), over the field R of real numbers.
The reason is that it corresponds, as an embedding R↪R3, to the (open) trefoil knot.
The fact that γ is a closed embedding, over any field k, can be shown as follows. Writing γ1=t3−3t,γ2=t4−4t2−1,γ3=t5−10t∈k[t], we get
[TABLE]
The fact that γ:R→R3 corresponds to the open trefoil knot can be seen by looking at the three projections:
-3$$-2$$-1$$1$$2$$3$$-5$$-4$$-3$$-2$$-1$$1$$t\mapsto(t^{3}-3t,t^{4}-4t^{2}-1)
-3$$-2$$-1$$1$$2$$3$$-15$$-10$$-5$$5$$10$$15$$t\mapsto(t^{3}-3t,t^{5}-10t)
-5$$-4$$-3$$-2$$-1$$1$$-15$$-10$$-5$$5$$10$$15$$t\mapsto(t^{4}-4t^{2}-1,t^{5}-10t)
We now use Example 6.1 to provide a similar example in SL2:
Lemma 6.2**.**
- (1)
For each field k of characteristic =2, the morphism
[TABLE]
is a closed embedding.
2. (2)
If k=R, then τ is not equivalent to the standard embedding, because the fundamental group π1(SL2(R)∖τ(R)) is not isomorphic to the free group π1(SL2(R)∖τ1(R)).
Proof.
(1): The fact that τ is a closed embedding can be done explicitely by giving a formula for t, but can also be shown by using the A1-bundle
[TABLE]
Writing γ1=t3−3t,γ2=t4−4t2−1,γ3=t5−10t∈k[t] as in Example 6.1, we get
γ12(γ12−4)−γ2(γ22+9γ2+24)=16 and thus get a birational morphism
[TABLE]
from A1 to the singular affine quartic curve Γ⊂A2. We then get a morphism
[TABLE]
which satisfies p∘f=idΓ and is thus a section of
p over Γ. This implies that
[TABLE]
is a closed embedding. Since γ:A1→Γ×A1⊂A3 is a closed embedding,
the morphism
[TABLE]
is a closed embedding. Replacing γ1 and γ2 in the above formula yields the explicit form of the morphism given in the statement of the lemma.
(2): In the remaining part of the proof, we work over k=R and use the Euclidean topology. The R-bundle p:SL2(R)→R2∖{(0,0)} is trivial, as it admits a (rational) continuous section given by
[TABLE]
This yields a birational diffeomorphism
[TABLE]
In particular, SL2(R)∖τ1(R) is diffeomorphic to R2∖{(0,0),(0,1)}×R, which implies that the fundamental group π1(SL2(R)∖τ1(R)) is a free group (over two generators). It remains to show that π1(SL2(R)∖τ(R)) is not a free group. This will imply that no diffeomorphism of SL2(R) sends τ(R) onto τ1(R), and in particular no algebraic automorphism defined over R.
We extend f:Γ(R)→SL2(R) to a global continuous section
f^:R2∖{(0,0)}→SL2(R) of p
(which exists, since
p is a trivial R-bundle).
This yields a rational diffeomorphism
[TABLE]
which maps γ(R) onto τ(R).
We take an open subset U⊂R2∖{(0,0)} (for the Euclidean topology) that contains the singular curve Γ(R) and a homeomorphism
h:U⟶≃R2 which fixes Γ(R) pointwise, and which is homotopic to the inclusion U↪R2, via a homotopy that fixes Γ(R) pointwise. We can for instance take U={(x,t)∈R2∣t≤x2−21} and construct a homeomorphism and a homotopy
which preserve the fibres of the projection (x,t)↦x.
U$$-3$$-2$$-1$$1$$2$$3$$-5$$-4$$-3$$-2$$-1$$1$$2$$3
The homeomorphisms
[TABLE]
yield isomorphisms of the fundamental groups
[TABLE]
Since γ:R↪R3 is the (open) trefoil knot,
it follows that π1(R3∖γ(R)) is the braid group
with 3 strands and thus π1(p−1(U)∖τ(R)) is not a free
group. It remains then to see that the group homomorphism
[TABLE]
induced by the inclusion
p−1(U)∖τ(R)↪SL2(R)∖τ(R)
is injective (as a subgroup of a free group is free).
Every element α∈Ker(ι) lies in the kernel of the map
ι′:π1(p−1(U)∖τ(R))→π1(R3∖γ(R))
induced by the composition
[TABLE]
which corresponds simply to the composition
[TABLE]
Since h:U→R2 is homotopic to the inclusion U↪R2
via a homotopy that fixes Γ(R) pointwise, the two compositions p−1(U)∖τ(R)→R3∖γ(R) of (⋆ ‣ 6) and (⋆⋆ ‣ 6) are homotopic, so ι′ is an isomorphism. This implies that ι is injective and achieves the proof.
∎
Question 6.3**.**
Working over the field of complex numbers C, is the algebraic embedding
τ:A1→SL2 of
Lemma \refExam:RSL2R\eqrefClosedEmb
equivalent to the standard embedding
τ1:A1↪SL2?