On automorphisms of enveloping algebras
Akaki Tikaradze
[email protected]
University of Toledo, Department of Mathematics & Statistics,
Toledo, OH 43606, USA
Abstract.
Given an algebraic Lie algebra g over C, we canonically associate to it
a Lie algebra Lββ(g) defined over Cβββthe reduction of C
modulo the infinitely
large prime, and show that for a class of Lie algebras Lββ(g)
is an invariant of the derived category of g-modules.
We give two applications of this construction. First, we show that the bounded derived category of g-modules
determines algebra g for a class of Lie algebras. Second,
given a semi-simple Lie algebra g over C, we construct a canonical homomorphism
from the group of automorphisms of the enveloping algebra Ug
to the group of Lie algebra automorphisms of g,
such that its kernel does not contain a nontrivial semi-simple automorphism.
As a corollary we obtain that any finite subgroup of automorphisms of Ug
isomorphic to a subgroup of Lie algebra automorphisms of g.
Introduction
This paper is motivated by the question whether a Lie algebra g over C
can be recovered from its enveloping algebra Ug. One way to make this question
more precise is to state the isomorphism problem for enveloping algebras: given two finite dimensional
Lie algebras g1β,g2β over C, such that their enveloping algebras are isomorphic
Ug1ββ
Ug2β, does it follow that g1ββ
g2β?
This problem is widely open in general, it is known to have the positive answer for the cases of semi-simple Lie algebras (easily follows from[AP]) and low dimensional nilpotent Lie algebras (see [H]).
One is tempted to
upgrade this isomorphism problem to the following derived isomorphism problem.
Conjecture 1*.*
Let g1β,g2β be finite dimensional Lie algebras over C.
If the derived categories of bounded complexes of Ug1β-modules and Ug2β-modules are equivalent,
then g1ββ
g2β.
We show that the conjecture holds if Lie algebras g1β,g2β satisfy
Assumption 1 below (see Theorem 2.)
A closely related problem is to understand Aut(Ug)βthe group of automorphisms of the enveloping algebra.
Of particular interest are its finite subgroups.
In this regard,
the study of finite subgroups of automorphisms of Ug for semi-simple g
and the corresponding fixed point rings have been of great interest for some time now,
see [AP], [C], [CG], [J2]. For the special case of g=sl2β, all finite subgroups
of Aut(Usl2β) where classified by Fleury [F]. More specifically,
she proved that if Ξ is a finite subgroup of Aut(Usl2β), then Ξ
is conjugate to a subgroup of Aut(sl2β)βAut(Usl2β).
The proof in [F] relies on the explicit knowledge of generators of automorphism groups of
primitive quotients of Usl2β (a result by Dixmier), no such results
are known for higher rank Lie algebras.
Following ideas and results
of Kontsevich and Belov-Kanel on automorphisms of the Weyl algebra [BK], [K],
we approach these problems by reducing modulo large prime p.
In this context it is convenient to use the reduction modulo the infinitely large prime construction.
Recall that given a commutative ring R, its reduction modulo the infinitely large
prime Rββ is defined as follows (see [K], [BK])
[TABLE]
here the direct limit is taken over all finitely generated subrings SβR, and P denotes
the set of all prime numbers [K]. We have the canonical inclusion RβQβͺRββ.
In particular, if R is an integral domain, then Rββ can (and will) be viewed is
an R-algebra via the canonical embedding RβͺRββ.
Also, we have the Frobenius map Frββ:RβββRββ, defined as follows:
[TABLE]
All results in this paper are about Lie algebras satisfying the following assumption.
Examples of such Lie algebras besides semi-simple and Frobenius ones are
certain Z2β-contractions of reductive algebras,
for example slnβ(C)βCn (see [Pa]).
Assumption 1*.*
Let g be an algebraic Lie algebra over C corresponding to a connected algebraic group
G, with the trivial center, such that the following properties hold. The algebra of invariants
Sym(g)g=C[f1β,β―,fnβ] is a polynomial algebra with homogeneous generators
f1β,β―,fnβ; such that they form a regular sequence in Sym(g). Moreover,
the corresponding algebra of coinvariants A=Sym(g)/(f1β,β―,fnβ) is a normal domain,
such that the coadjoint action of G on SpecA has an open orbit.
Given a perfect Lie algebra g: [g,g]=g,
satisfying Assumption 1,
we construct a canonical group homomorphism (Section 4)
[TABLE]
where gCβββ=gβCβCββ.
We have a base change homomorphism
Frβββ:Aut(g)βAut(gCβββ)
induces by the Frobenius embedding Frββ:CβCββ.
The following is the main result of the paper.
Theorem 1**.**
Let g be a perfect Lie algebra over C satisfying Assumption 1.
Then there are no nontrivial semi-simple elements in ker(D).
Moreover, D restricts to Frβββ on Aut(g).
In particular, if Ξ is a finite subgroup of Aut(Ug),
then there exists a subgroup Ξβ² of Aut(g),
such that Ξ is isomorphic to Ξβ² as an abstract group.
Remark 1*.*
The most interesting application of the above result is for the case of a simple Lie algebra g.
In principle, this result provides a full classification of isomorphism classes of finite groups of automorphisms
of Ug.
However, although the construction of the subgroup Ξβ²βAut(g) is somewhat canonical,
we don not know if Ξβ² is conjugate to Ξ in Aut(Ug).
In fact, we do not know if a much stronger statement about linearizability holds: Given a finite subgroup
ΞβAut(Ug),
whether there exists a subgroup Ξβ²βAut(g), such that Ξβ² is conjugate to Ξ
in Aut(Ug).
Throughout the paper, given an abelian group M, we denote by Mpβ its reduction modulo p: Mpβ=M/pM.
1. Reduction modulo p Lemmas
In this section, given any finite dimensional Lie algebra g over C, we
define Cββ-Lie algebras gββ,gββ^β,
and compute them for Lie algebras satisfying Assumption 1, Lemma 3.
This construction plays the crucial role in proving our main results.
At first, recall that given an associative flat Z-algebra R and a prime number p,
then the center Z(Rpβ) of its reduction modulo p acquires the natural Poisson bracket, which
we refer to as the deformation Poisson bracket,
defined as follows. Given a,bβZ(Rpβ), let z,wβR be their lifts respectively.
Then the Poisson bracket {a,b} is defined to be
[TABLE]
Let S be a finitely generated subring of C, and
let g be a Lie algebra over S which is a finite rank free S-module.
Throughout the paper we denote by B the quotient of Ug by the augmentation ideal of its center
[TABLE]
Then we define Sββ-Lie algebras Lββ(g),Lββ²β(g) as follows.
Let p>>0 be a large prime number. Then the following augmentation ideals
[TABLE]
are easily seen to be Poisson ideals in Z(Ugpβ),Z(Bpβ) respectively.
Hence mpβ/mp2β,npβ/np2β are Lie algebras,
and we view them as Lie algebras over Spβ via the the Frobenius map Frpβ:SpββSpβ.
Denote mpβ/mp2β by Lpβ(g), and npβ/np2β
by Lpβ²β(g).
Similarly given a base change Sβk to a field of characteristic p we define k-Lie algebras
Lkβ(g),Lkβ²β(g).
We put
[TABLE]
Now, given a Lie algebra over C, we define
[TABLE]
Here, gSβ is a model of g over S: gSββSβC=g.
We have the natural surjective homomorphism Lββ(g)βLββ²β(g).
If gSβ is an algebraic Lie algebra over S, then
we construct below a canonical Lie algebra homomorphism
[TABLE]
It follows from Lemma 3 below that if g satisfies Assumption 1, then
Lββ(g) is isomorphic to a trivial central extension of
gCβββ=gβCβCββ, while
Lββ²β(g)β
gCβββ.
Next we recall a key computation of the Poisson bracket for restricted Lie
algebras due to Kac and Radul [KR]. First, we recall and fix some notations associated with enveloping algebras of restricted Lie algebras.
Let R be a commutative reduced ring of characteristic p>0.
Let g be a restricted Lie algebra over R (g is assumed to be a finite free R-module)
with the restricted structure map gβg[p],gβg.
Then by Zpβ(g) we denote the central R-subalgebra of the enveloping algebra Ug
generated by elements of the form gpβg[p],gβg. It is well-known that the map gβgpβg[p] induces
homomorphism of R-algebras
[TABLE]
where Zpβ(g) is viewed as an R-algebra via the Frobenius map F:RβR.
The homomorphism i is an isomorphism when R is perfect.
Also, recall that
the Lie algebra bracket on g defines the Kirillov-Kostant Poisson bracket on the symmetric algebra Sym(g).
The following is the key result from [KR].
We include a proof for the readerβs convenience.
Lemma 1**.**
Let R be a finitely generated integral domain over Z.
Let G be an affine algebraic group over R, let g be its Lie
algebra. Let Gpβ,gpβ be reductions modulo p of G,g respectively.
Thus Z(Ugpβ) is equipped with the deformation Poisson bracket.
Then Zpβ(gpβ) is a Poisson subalgebra of Z(Ugpβ), moreover the induced Poisson
bracket coincides with the negative of the Kirrilov-Kostant bracket:
[TABLE]
Proof.
The proof directly follows from a similar result about Weyl algebras in [BK].
More specifically, let X be a smooth affine variety over R, and let DXβ denote the algebra of
differential operators on X. Put X=Xmodp. Then the center of DXβmodp=DXβ can be identified with
(the Frobenius twist) of the functions on the cotangent bundle TXββ [BMR]. Then the deformation Poisson
bracket of Z(DXβ) is equal to the negative of the Poisson bracket coming from the symplectic structure
of the cotangent bundle of X.
Now let ΞΈ:gβDGβ be the realization of g as left-invariant vector fields
on G. Then we have the corresponding embedding
ΞΈ:UgβDGβ and the corresponding reduction modulo p
ΞΈΛ:UgpββDGpββ,
which induces the embedding Zpβ(gpβ)βZ(DGpββ).
In this way Zpβ(gpβ) is a Poisson subalgebra of Z(DGpββ) and the assertion follows.
β
It is clear that iβ1(mpβ)=gpβSymgpβ. So, in view of Lemma1 we have the canonical
Lie algebra homomorphism gpββmpβ/mp2β.
Hence, we obtain the desired canonical homomorphism
[TABLE]
We have the following
Lemma 2**.**
Let S be a finitely generated subring of C. Let g be a nilpotent Lie algebra over S. Let
p>>0 be a prime.
Let I be a Poisson ideal of Z(Ugpβ), such that Z(Ugpβ)/I=Spβ.
Then I/I2β
mpβ/mp2β as Spβ-Lie algebras.
Proof.
We claim that given any ideal I in Z(Ugpβ), such that
Iβ©Zpβ(gpβ)=(gp,gβgpβ) and Z(Ugpβ)/I=Spβ,
then I=mpβ.
Indeed, since Ugpβ/(gp,gβgpβ)Ugpβ is a nilpotent Spβ-algebra,
it follows that Z(Ugpβ)/(gp,gβgpβ)Z(Ugpβ) is also a nilpotent Spβ-algebra:
mpplββ(gp,gβgpβ)Z(Ugpβ) for large enough l.
If a+yβI,aβSpβ,yβmpβ, then
[TABLE]
so a=0 and Iβmpβ.
Put Iβ²=Iβ©Zpβ(g). Then Iβ² is a Poisson ideal and Zpβ(g)/Iβ²=Spβ.
Hence, Iβ²=(gpβΟ(g),gβgpβ), where Ο:gpββSpβ is a Lie algebra
homomorphism. Then Ο(g)=gβΟ(g) induces an automorphism Ο:UgpββUgpβ,
such that Ο(mpβ)=I. Moreover Ο admits a lift over S/p2S. Indeed, take any character
Ο^β:g/p2gβS/p2 that lifts Ο and put Ο~β(g)=g~ββΟ(g).
Then Ο~β:U(g/p2g)βU(g/p2g) is an automorphisms
such that Ο=Ο~βmodp2.
Hence, Ο is a Poisson automorphism of Z(Ugpβ). Thus we obtain the desired isomorphism
of Spβ-Lie algebras I/I2β
mpβ/mp2β.
β
From now on we fix a Lie algebra g satisfying Assumption 1.
It follows that we may choose a finitely generated subring SβC,
and a Lie algebra gSβ over S which is finite free S-module, such that
g=gSββSβC,fiββSymgSβ and
ASβ=SymgSβ/(f1β,β―,fnβ) is a normal integral domain. Moreover
SpecASβ has a nonempty open subset USβ which is symplectic over S under the
Kirillov-Kostant bracket.
Denote by giβ the image of fiβ under the symmetrization isomorphism
Sym(g)gβZ(Ug). Hence gr(giβ)=fiβ.
We may assume that S is large enough so that giββUgSβ,1β€iβ€n.
Just as before, we put BSβ=UgSβ/(g1β,β―,gnβ).
Hence B=BSββSβC. Given a commutative S-algebra R, we denote by
gRβ,BRβ and ARβ the base changes of gSβ,BSβ,ASβ
respectively. In particular, for a base change Rβk, where k
is an algebraically closed field, Akβ is a normal integral domain (for p>>0).
We denote images of fiβ,giβ in Symgpβ,Z(Ugpβ) by
fiβΛβ,giβΛβ. Given a commutative Spβ-algebra R, we denote by
mRβ( respectively nRβ), the augmentation ideal
Z(UgRβ)β©gRβUgRβ (respectively
Z(BRβ)β©gRβBRβ.)
Then we have the following
Lemma 3**.**
Let g be as above. Then,
Z(Ugpβ) is a free Zpβ(gpβ)-module with the basis
[TABLE]
*In particular, Lββ²β(g)=gSβββ, while Lββ(g) is a trivial central
extension of gSβββ.
Moreover, given a Poisson ideal I in Z(Ugpβ), such that Z(Ugpβ)/I=S/pS,
then I/I2 is a trivial central extension of gpβ as an Spβ-Lie algebra.
If g is perfect, then npβ is the unique Poisson ideal of
Z(Bpβ),
such that Z(Bpβ)/npβ=Spβ.
*
Proof.
It is enough to verify above statements after a base change S/pSβk, where k is an algebraically closed field
of characteristic p.
At first, since Akβ=Sym(gkβ)/(f1βΛβ,β―,fnβΛβ) is a normal domain and the Poisson bracket is symplectic on a nonempty open subset of SpecAkβ,
it follows easily that the Poisson center of Akβ is
Akpβ (see, for example, Lemma 2.4[T]).
It suffices to check that the Poisson center of Symgkβ is a free module
over (Symgkβ)p with the basis
[TABLE]
Indeed, let g be in the Poisson center of Symgkβ.
Denote the ideal (f1βΛβ,β―,fnβΛβ)βSymgkβ by I.
We claim that gβSym(gkβ)p[f1βΛβ,β―,fnβΛβ]+Im, for all m.
We proceed by induction on m.
Since gΛβ=gmodI belongs to the Poisson center of
Akβ, it follows that
gΛββAkpβ. Hence gβSym(gkβ)p+I.
Assume that gβSym(gkβ)p[f1βΛβ,β―,fnβΛβ]+Im for some mβ₯1.
So, there exists xβg+Sym(gkβ)p[f1βΛβ,β―,fnβΛβ], such that
[TABLE]
Then, for any yβgkβ, we have
[TABLE]
Since the sequence (f1βΛβ,β―,fnβΛβ) is regular, we may conclude that
xΞ±βΛβ=xΞ±βmodIβAkpβ.
Hence,
[TABLE]
Therefore,
[TABLE]
Now, suppose that
[TABLE]
Such that either xΞ±β=0 or xΞ±ββ/I.
Let m be such that xΞ±β=0 for all β£Ξ±β£<m in the above sum. Since Im/Im+1
is a free Akβ-module with basis {f1βΛβΞ±1ββ―fnβΛβΞ±nβ,βΞ±iβ=m},
we get that xΞ±pβmodI=0, hence xΞ±ββI. So xΞ±β=0.
Thus, elements {f1βΛβΞ±1ββ―fnβΛβΞ±nβ,0β€Ξ±iβ<p} are linearly independent over (Symgkβ)p
as desired. It follows that gkββ¨βi=1nβkgiβΛβ surjects onto
Lkβ(g)=mkβ/mk2β,
and since the center of gkβ is trivial, we conclude that Lkβ(g) is a direct sum
of gkβ with a central subalgebra spanned by images of gΛβiβ.
Denote by f~βiβ the image of fiβΛβp under the isomorphism Sym(gkβ)β
Zpβ(gkβ).
Then gr(fiβ~β)=fΛβipβ.
Since Akβ is a domain, we have
[TABLE]
Therefore
[TABLE]
Hence, we conclude that the map i:SymgkββZ(Bkβ) induces an isomorphism Akββ
Z(Bkβ).
In particular, Lkβ²β(g)=nkβ/nkβ2β
gkβ.
Now let IβZ(Ugpβ) be a Poisson ideal such that Z(Ugpβ)/I=Spβ.
Proceeding as in the proof of Lemma 2,
without loss of generality we may assume that (gpβg[p],gβgpβ)βI.
Let giβΛββaiββI,aiββSpβ. In the above proof, replacing gΛβiβ with gΛβiββaiβ,
we conclude that
[TABLE]
If gpβ is perfect and
I is Poisson ideal in Bpβ, such that Bpβ/I=S/pS, then if follows that gpβg[p]βΟ(g)βI,gβgpβ,
such that Ο:gpββSpβ is a character. Hence, Ο must be trivial.
Since gpβg[p],gβgpβ generate Z(Bpβ), we get that I=npβ.
Example 1*.*
Let g=sl2β with the usual generators e,f,h.
Then the center of Usl2β(Fpβ)
is generated a Fpβ-algebra by x=ep,y=fp,z=hpβh,Ξ=4fe+h2+2h subject to the relation
[TABLE]
Thus the augmentation ideal mpβ=Z(Ugpβ)β©gpβ(Ugpβ)
is generated by x,y,z,Ξ and mpβ/mp2β has an Fpβ-basis
{xΛ,yΛβ,zΛ,ΞΛ}βimages of x,y,z,Ξ respectively. Thus we see that
Lpβ(sl2β)=mpβ/mp2ββ
sl2β(Fpβ)βFpβΞΛ.
Similarly npββthe augmentation ideal of Z(Bpβ)=Z(Usl2β(Fpβ)/(Ξ)), is generated
by x,y,z, and npβ/np2β=FpβxΛβFpβyΛββFpβzΛ.
So Lpβ²β(sl2β)β
sl2β(Fpβ).
β
2. Derived isomorphisms
As usual, S is a finitely generated subring of C.
Throughout, given two algebras A,B, we say that they are derived equivalent if
the respective derived categories of bounded complexes of (left) modules are equivalent.
We use the following easy consequence of [R].
Lemma 4**.**
Let A,B be flat S-algebras that are derived equivalent. Then Z(A/pA)β
Z(B/pB)
as Poisson algebras for all p>>0.
Proof.
It follows that algebras A/p2A,B/p2B are derived equivalent.
We have the following exact sequence of A/p2A-bimodules
[TABLE]
where A/pAβA/pA is the quotient map and A/pAβA/p2A is the multiplication by p.
It follows from [R] that the connecting map of the Hochschild cohomologies
[TABLE]
corresponding to
the exact sequence above commutes with isomorphisms of Hochschild cohomologies induced by the derived
equivalence
[TABLE]
Now, since the deformation Poisson bracket on Z(A/pA) is defined as
[TABLE]
the desired result follows.
β
Now we can easily prove the following.
Theorem 2**.**
Suppose that Lie algebras g,gβ² satisfy Assumption 1.
If Ug is derived equivalent to Ugβ², then gβ
gβ².
Proof.
There exists a finitely generated subring S of C, and Lie algebras
gSβ,gβ²Sβ over S, such that
[TABLE]
and U(gSβ) is derived equivalent to U(gβ²Sβ).
Hence,
[TABLE]
as Poisson algebras by Lemma 4.
Let IβZ(U(gSpββ)) be a Poisson ideal
such that Z(U(gSpββ))/I=Spβ.
Let Iβ²βZ(U(gβ²Spββ)) be the corresponding ideal under the above isomorphism.
Hence I/I2β
Iβ²/Iβ²2 as Spβ-Lie algebras.
Now using Lemma 3 we conclude that trivial central extensions of
(g)Cβββ and (gβ²)Cβββ
are isomorphic. Since dimg=dimgβ²,
this implies that gβ
gβ².
β
3. The homomorphisms D,D~
We assume that g is a perfect Lie algebra over C satisfying Assumption 1.
Let S be a large enough finitely generated subring of C, and let gSβ
be a model of g over S (just as in the paragraph preceding Lemma 3).
Then we construct canonical homomorphisms
[TABLE]
as follows. At first, remark that since g is perfect, any automorphism of Ug must
preserve the ideal Z(Ug)β©gUg. Therefore we have
the restriction homomorphism Aut(Ug)βAut(B).
Let p>>0 be a sufficiently large prime.
Let ΟβAut(BSβ).
Reducing Οmodp, we obtain ΟΛββAut(Z(Bpβ)).
Since Bpβ is obtained by the reduction modulo p, we have the corresponding deformation Poisson bracket
on its center. Hence, ΟΛβ preserves the Poisson
bracket on Z(Bpβ).
Now it follows from Lemma 2 that ΟΛβ preserves npβ, thus it induces a Lie algebra automorphism on
npβ/np2ββ
gpβ, which we denote by (D~Spββ)(Ο).
Hence, we obtain a canonical homomorphism D~Spββ:Aut(BSβ)βAut(gpβ).
Also, given a base change Spββk, we denote the corresponding homomorphism
D~SpβββSpββk:Aut(BSβ)βAut(gkβ)
by D~kβ. Also, denote by Dkβ:Aut(UgSβ)βAut(gkβ)
the composition of D~kβ with the restriction Aut(UgSβ)βAut(BSβ).
The element βpβ(D~Spββ)(Ο) gives rise
to an element of Aut(gSβββ),
which we denote by D~Sβ(Ο). This way we obtain the desired homomorphisms
D~Sβ.
We define DSβ:Aut(UgSβ)βAut(gSβββ) as the composition
of DSβ~β with the restriction Aut(UgSβ)βAut(BSβ).
Now, taking the direct limit of the above homomorphisms D~Sβ,DSβ over finitely generated subsrings SβC,
we obtain the sought after homomorphisms
[TABLE]
It is clear from the construction and Lemma3 that D(Aut(g))=Frβββ,
where Frββ:CβCββ is the canonical inclusion
followed by the Frobenius map.
4. The proof
We start by the following.
Proposition 1**.**
Let g be a perfect Lie algebra satisfying Assumption 1.
Then the kernel of the restriction homomorphism Aut(Ug)βAut(B) contains
no nontrivial semi-simple automorphisms.
We remark that in general, the homomorphism Aut(Ug)βAut(B)
is not injective.
For example, in the case of g=sl2β, this follows from existence
of a non tame automorphism of Usl2β, proved by Joseph [J1].
For the proof of Lemma1, we need a specific set of linearly independent elements of HH2(B).
Recall that for a semi-simple g, one has dimHH2(B)=n, as follows immediately from Soergelβs result [S].
Recall that Z(Ug)=C[g1β,β―,gnβ],giββgUg and (g1β,β―,gnβ) is a regular
sequence in Ug.
Put I=(g1β,β―,gnβ). So B=Ug/I. Let us put Bβ²=Ug/I2.
Then we have a short exact sequence
[TABLE]
We have I/I2=β¨i=1nβBgiβΛβ, where giβΛβ denotes the image of giβ under the quotient map UgβBβ².
Denote by
[TABLE]
the cohomology class corresponding to the above short exact sequence.
Let us put Ο=βi=1nβΟiβgiβΛβ where ΟiββHH2(B).
Under these notations we have the following.
Lemma 5**.**
Elements Οiβ,1β€iβ€n are linearly independent .
Proof.
Let βi=1nβciβΟiβ=0,ciββC. Let cjβξ =0 for some j.
Let B1β be the quotient of Bβ² by the ideal βiξ =jβBβ²giβΛβ.
Then it follows that the quotient map B1ββB1β/B1βgjβΛβ=B admits a C-algebra splitting.
Let Ο:BβB1β be such a splitting. Let us write
[TABLE]
This implies that gjβΛββgjβΛβ(gBβ²).
Therefore
[TABLE]
Thus, there exists Ξ±βUgβgUg such that
Ξ±gjβββiξ =jβgiβUg. Now the regularity of the sequence (g1β,β―,gnβ) implies that
Ξ±βI, which is a contradiction.
β
Proof of Proposition1.
Let Ο be a semi-simple automorphism of Ug that restricts to the identity on B.
Denote by Ο~β the restriction of Ο on Bβ².
Therefore Ο~β fixes Ο=βΟiβgiβΛββH2(B,I/I2).
Let Ο~β:I/I2βI/I2 be represented by the matrix
[TABLE]
Entries Οijβ are scalars since Ο preserves C[g1β,β―,gnβ].
Thus βΟijβΟjβ=Οiβ.
Since Οjβ are linearly independent by Lemma 5, we conclude that Οβ£I/I2β=Id. Hence Οβ£In/In+1β=Id for all n.
Since Ο is semi-simple, we get that Ο=Id.
β
Remark 2*.*
It was proved by Polo[P]
that when g is semi-simple, the action of an automorphism of Ug on its center is given by a Dynkin diagram automorphism
of g.
Proof of Theorem 1..
In view of Proposition 1, it suffices to check that ker(D~) has no nontrivial semi-simple automorphisms.
Assume that ΟβAut(B) is a non-trivial semi-simple automorphism.
Therefore there exists a finitely generated subring SβC,
and a finite free Ο-invariant S-submodule VβBSβ,
such that Οβ£Vβ is semi-simple over S and V generates BSβ as an S-algebra.
(we are using notations from the paragraph preceding Lemma 3).
We show that for sufficiently large S, and for all p>>0,
given any homomorphism Spββk, where k is a field,
then D~kβ(Ο)ξ =Idgkββ.
Let yβV,1ξ =Ξ±βS, such that Ο(y)=Ξ±y.
Let us write y=βaiβeiβ,0ξ =aiββS, where eiβ are basis elements of BSβ as a free S-module.
We may assume that aiβ,1βΞ± are invertible is S. Now let p>>0 and
Ο:Sβk be a base change to a field such that
D~kβ=Idgkββ.
Thus, we have a non-trivial semi-simple automorphism ΟΛββAut(Bkβ), such that
Οβ£Λβmkβ/mk2ββ=Id.
Then ΟΛβ acts trivially on mknβ/mkn+1β for all n.
Since the action of Ο on Z(Bkβ) is semi-simple, it follows that the action of ΟΛβ on Z(Bkβ) is trivial.
Then by the Noether-Skolem theorem there exists aβBkβ, such that
[TABLE]
But, since ΟΛβ(yΛβ)=Ξ±ΛyΛβ and 1ξ =Ξ±Λβk,0ξ =yΛββBkβ, we
get that ayΛβ=Ξ±ΛyΛβa. Recall that under the PBW filtration on Bkβ,
gr(Bkβ)=Akβ is a commutative domain. Hence,
[TABLE]
which is a contradiction.
Let ΞΈ:CβββC be a homomorphism.
Then we define a (non-canonical) homomorphism
Dβ:Aut(Ug)βAut(g) as the composition of D with the base change homomorphisms
ΞΈβ:Aut(gCβββ)βAut(g).
Next we show that ker(Dβ) contains no nontrivial finite order elements.
This implies that given a finite subgroup ΞβAut(Ug),
then Ξβ²=Dβ(Ξ)βAut(g),Ξβ
Ξβ².
Let ΟβAut(Ug) such that Οm=1,Οξ =1. We may choose a finitely generated
subring SβC,m1ββS containing all m-th roots of unity,
such that ΟβAut(UgSβ).
As it was shown in the preceding paragraph, by enlarging S if necessary, for all p>>0 and
a base change Spββk to a field, we have
Dkβ(Ο)ξ =Idgkββ.
Let fpβ(t)βSpβ[t] denote the characteristic polynomial of DSpββ(Ο)βAut(gSpββ).
Put l=dimg.
We show that
ΞΈ(βpβfpβ(t))ξ =(tβ1)l in Sββ[t].
Indeed, let us write Spβ=βiβSp,iβ, where
each Sp,iβ is a domain (since p is unramified in S).
Since the image of DSpββ(Ο) in Aut(gSp,iββ) has order m,
it follows that
the image of fpβ(t) in Sp,iβ[t] is not equal to (tβ1)l and is of the form βj=1mβ(tβajβ),ajmβ=1.
Denote by Ξ¨βS[t] the finite set of all degree l monic polynomials not equal to (tβ1)l,
of the form βj=1mβ(tβajβ),ajββS,ajmβ=1.
For each gβΞ¨, denote by Igβ the set of pairs (p,i) for which fpβ(t)=g in Sp,iβ.
Then we have βpβSpβ=βgβΞ¨β(β(p,i)βIgββSp,iβ).
Now, suppose gβΞ¨ is such that β(p,i)βIgββSp,iβξ βker(ΞΈ).
Then it follows that ΞΈ(βpβfpβ(t))=ΞΈ(g)ξ =(tβ1)l.
Hence Οβ/ker(Dβ) as desired.
β
Acknowledgement: I am very grateful to the anonymous referee for the numerous useful suggestions.