On automorphisms of enveloping algebras

TL;DR

This paper explores the relationship between automorphisms of universal enveloping algebras and Lie algebra automorphisms, introducing a new invariant derived from reduction mod infinitely large primes and establishing connections with derived categories.

Contribution

It introduces a canonical construction of a Lie algebra invariant over an extended field and links automorphisms of enveloping algebras to Lie algebra automorphisms, providing new structural insights.

Findings

The invariant $rak{g}_ ext{infty}$ is determined by the derived category of modules.

A canonical homomorphism from automorphisms of $rak{U}rak{g}$ to automorphisms of $rak{g}$ is constructed.

Finite automorphism groups of $rak{U}rak{g}$ are subgroups of Lie algebra automorphisms.

Abstract

Given an algebraic Lie algebra $\mathfrak{g}$ over $\mathbb{C}$, we canonically associate to it a Lie algebra $\mathfrak{g}_{\infty}$ defined over $\mathbb{C}_{\infty}$-the reduction of $\mathbb{C}$ mod infinitely large prime, and show that for a class of Lie algebras $\mathfrak{g}_{\infty}$ is an invariant of the derived category of $\mathfrak{g}$-modules. We give two applications of this construction. First, we show that the bounded derived category of $\mathfrak{g}$-modules determines algebra $\mathfrak{g}$ for a class of Lie algebras. Second, given a semi-simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$, we construct a canonical homomorphism from the group of automorphisms of the enveloping algebra $\mathfrak{U}\mathfrak{g}$ to the group of Lie algebra automorphisms of $\mathfrak{g}$, such that its kernel does not contain a nontrivial semi-simple automorphism. As a corollary we obtain that any finite subgroup of automorphisms of $\mathfrak{U}\mathfrak{g}$ isomorphic to a subgroup of Lie algebra automorphisms of $\mathfrak{g}.$

Equations

$$R_{\infty }=\underset{f.g.S\subset R}{\underrightarrow{\lim }}(\prod _{p\in P}S/pS)/(\underset{p\in P}{⨁}S/pS),$$

$${\text{Fr}}_{\infty }(\prod _p{x}_p)=\prod _p{x}_p^p.$$

$$D:\text{Aut}(\mathfrak{Ug})\to \text{Aut}({\mathfrak{g}}_{{\mathbb{C}}_{\infty }}),$$

$$\frac{1}{p}[z,w]\mathrm{mod}p\in Z(R_p).$$

$$B=\mathfrak{Ug}/(Z(\mathfrak{Ug})\cap \mathfrak{gUg})\mathfrak{Ug}.$$

$${\mathfrak{m}}_p=Z(\mathfrak{U}{\mathfrak{g}}_p)\cap {\mathfrak{g}}_p(\mathfrak{U}{\mathfrak{g}}_p),{\mathfrak{n}}_p=Z(B_p)\cap {\mathfrak{g}}_p{B}_p$$

$${\mathcal{L}}_{\infty }(\mathfrak{g})=\prod _{p\in P}{\mathcal{L}}_p(\mathfrak{g})/\underset{p\in P}{⨁}{\mathcal{L}}_p(\mathfrak{g}),{\mathcal{L}}_{\infty }^{\prime }(\mathfrak{g})=\prod _{p\in P}{\mathcal{L}}_p^{\prime }(\mathfrak{g})/\underset{p\in P}{⨁}{\mathcal{L}}_p^{\prime }(\mathfrak{g}),$$

$${\mathcal{L}}_{\infty }(\mathfrak{g})=\underset{f.g.S\subset \mathbb{C}}{\underrightarrow{\lim }}{\mathcal{L}}_{\infty }({\mathfrak{g}}_S),{\mathcal{L}}_{\infty }^{\prime }(\mathfrak{g})=\underset{f.g.S\subset \mathbb{C}}{\underrightarrow{\lim }}{\mathcal{L}}_{\infty }^{\prime }({\mathfrak{g}}_S).$$

$${\mathfrak{g}}_{S_{\infty }}={\mathfrak{g}}_S{\otimes }_S{S}_{\infty }\to {\mathcal{L}}_{\infty }(\mathfrak{g}).$$

$$i:\text{Sym}(\mathfrak{g})\to Z_p(\mathfrak{g}),$$

$$\{a^p-a^{[p]},b^p-b^{[p]}\}=-([a,b{]}^p-[a,b{]}^{[p]}),a\in {\mathfrak{g}}_p,b\in {\mathfrak{g}}_p.$$

$${\mathfrak{g}}_{S_{\infty }}\to {\mathcal{L}}_{\infty }(\mathfrak{g}).$$

$$(a+y{)}^{p^l}\in a^{p^l}+(g^p,g\in {\mathfrak{g}}_p)Z(\mathfrak{U}{\mathfrak{g}}_p),$$

$$\{{\overline{g_1}}^{{\alpha }_1}\cdots {\overline{g_n}}^{{\alpha }_n},0\le {\alpha }_i<p,1\le i\le n\}.$$

$$\{{\overline{f_1}}^{{\alpha }_1}\cdots {\overline{f_n}}^{{\alpha }_n},0\le {\alpha }_i<p\}.$$

$$x=\sum _{{\alpha }_1+\cdots +{\alpha }_n=m}{\overline{f_1}}^{{\alpha }_1}\cdots {\overline{f_n}}^{{\alpha }_n}{x}_{\alpha },x_{\alpha }\in \text{Sym}({\mathfrak{g}}_{\mathbb{k}}),.$$

$$0=\{y,x\}=\sum _i{\overline{f_1}}^{{\alpha }_1}\cdots {\overline{f_n}}^{{\alpha }_n}\{y,x_{\alpha }\}.$$

$$g\in \text{Sym}({\mathfrak{g}}_{\mathbb{k}}{)}^p[\overline{f_1},\cdots ,\overline{f_n}]+I^{m+1}.$$

$$g\in \bigcap _{m\ge 1}(\text{Sym}({\mathfrak{g}}_{\mathbb{k}}{)}^p[\overline{f_1},\cdots ,\overline{f_n}]+I^m)=\text{Sym}({\mathfrak{g}}_{\mathbb{k}}{)}^p[\overline{f_1},\cdots ,\overline{f_n}].$$

$$0=\sum _{{\alpha }_1,\cdots ,{\alpha }_n}{\overline{f_1}}^{{\alpha }_1}\cdots {\overline{f_n}}^{{\alpha }_n}{x}_{\alpha }^p,x_{\alpha }\in \text{Sym}({\mathfrak{g}}_{\mathbb{k}}),$$

$$(\text{Sym}{\mathfrak{g}}_{\mathbb{k}}{)}^p\cap I=\sum _i{\overline{f_i}}^p\text{Sym}{\mathfrak{g}}_{\mathbb{k}}.$$

$$Z_p({\mathfrak{g}}_{\mathbb{k}})\cap (\overline{g_1},\cdots ,\overline{g_n})=\sum \widetilde{f_i}{Z}_p({\mathfrak{g}}_{\mathbb{k}}).$$

$$I/I^2={\mathfrak{g}}_p\oplus \sum _{i=1}^n{S}_p(\overline{g_i}-a_i).$$

$$(\Delta +1{)}^p-2(\Delta +1{)}^{\frac{p+1}{2}}+(\Delta +1)=4xy+z^2.$$

$$0\to A/pA\to A/p^{2}A\to A/pA\to 0,$$

$$i_m:{\text{HH}}^m(A/pA)\to {\text{HH}}^{m+1}(A/pA)$$

$${\text{HH}}^{\ast }(A/p^{2}A)\cong {\text{HH}}^{\ast }(B/p^{2}B),{\text{HH}}^{\ast }(A/pA)\cong {\text{HH}}^{\ast }(B/p^{2}B).$$

$$\{a,-\}=i_0(a){|}_{Z(A/pA)},a\in Z(A/pA),$$

$$\mathfrak{g}={\mathfrak{g}}_S{\otimes }_S\mathbb{C},{\mathfrak{g}}^{\prime }={{\mathfrak{g}}^{\prime }}_S{\otimes }_S\mathbb{C},$$

$$Z(\mathfrak{U}({\mathfrak{g}}_{S_p}))\cong Z(\mathfrak{U}({{\mathfrak{g}}^{\prime }}_{S_p})).$$

Full text

Introduction

This paper is motivated by the question whether a Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ can be recovered from its enveloping algebra $\mathfrak{U}\mathfrak{g}.$ One way to make this question more precise is to state the isomorphism problem for enveloping algebras: given two finite dimensional Lie algebras $\mathfrak{g}_{1},\mathfrak{g}_{2}$ over $\mathbb{C},$ such that their enveloping algebras are isomorphic $\mathfrak{U}\mathfrak{g}_{1}\cong\mathfrak{U}\mathfrak{g}_{2}$, does it follow that $\mathfrak{g}_{1}\cong\mathfrak{g}_{2}?$ 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 $\mathfrak{g}_{1},\mathfrak{g}_{2}$ be finite dimensional Lie algebras over $\mathbb{C}.$ If the derived categories of bounded complexes of $\mathfrak{U}\mathfrak{g_{1}}$-modules and $\mathfrak{U}\mathfrak{g_{2}}$-modules are equivalent, then $\mathfrak{g_{1}}\cong\mathfrak{g}_{2}.$

We show that the conjecture holds if Lie algebras $\mathfrak{g}_{1},\mathfrak{g}_{2}$ satisfy Assumption 1 below (see Theorem 2.)

A closely related problem is to understand $\text{Aut}(\mathfrak{U}\mathfrak{g})$–the 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 $\mathfrak{U}\mathfrak{g}$ for semi-simple $\mathfrak{g}$ 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 $\mathfrak{g}=\mathfrak{sl}_{2}$, all finite subgroups of $\text{Aut}(\mathfrak{U}\mathfrak{sl}_{2})$ where classified by Fleury [F]. More specifically, she proved that if $\Gamma$ is a finite subgroup of $\text{Aut}(\mathfrak{U}\mathfrak{sl}_{2})$, then $\Gamma$ is conjugate to a subgroup of $\text{Aut}(\mathfrak{sl}_{2})\subset\text{Aut}(\mathfrak{U}\mathfrak{sl}_{2}).$ The proof in [F] relies on the explicit knowledge of generators of automorphism groups of primitive quotients of $\mathfrak{U}\mathfrak{sl}_{2}$ (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 $R_{\infty}$ is defined as follows (see [K], [BK])

[TABLE]

here the direct limit is taken over all finitely generated subrings $S\subset R$, and $P$ denotes the set of all prime numbers [K]. We have the canonical inclusion $R\otimes\mathbb{Q}\hookrightarrow R_{\infty}.$ In particular, if $R$ is an integral domain, then $R_{\infty}$ can (and will) be viewed is an $R$-algebra via the canonical embedding $R\hookrightarrow R_{\infty}.$ Also, we have the Frobenius map $\text{Fr}_{\infty}:R_{\infty}\to R_{\infty},$ 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 $\mathbb{Z}_{2}$-contractions of reductive algebras, for example $\mathfrak{sl}_{n}(\mathbb{C})\ltimes\mathbb{C}^{n}$ (see [Pa]).

Assumption 1*.*

Let $\mathfrak{g}$ be an algebraic Lie algebra over $\mathbb{C}$ corresponding to a connected algebraic group $G,$ with the trivial center, such that the following properties hold. The algebra of invariants $\text{Sym}(\mathfrak{g})^{\mathfrak{g}}=\mathbb{C}[f_{1},\cdots,f_{n}]$ is a polynomial algebra with homogeneous generators $f_{1},\cdots,f_{n};$ such that they form a regular sequence in $\text{Sym}(\mathfrak{g}).$ Moreover, the corresponding algebra of coinvariants $A=\text{Sym}(\mathfrak{g})/(f_{1},\cdots,f_{n})$ is a normal domain, such that the coadjoint action of $G$ on $\text{Spec}A$ has an open orbit.

Given a perfect Lie algebra $\mathfrak{g}$: $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g},$ satisfying Assumption 1, we construct a canonical group homomorphism (Section 4)

[TABLE]

where $\mathfrak{g}_{\mathbb{C}_{\infty}}=\mathfrak{g}\otimes_{\mathbb{C}}\mathbb{C}_{\infty}.$ We have a base change homomorphism $\text{Fr}_{\infty}^{*}:\text{Aut}(\mathfrak{g})\to\text{Aut}(\mathfrak{g}_{\mathbb{C}_{\infty}})$ induces by the Frobenius embedding $\text{Fr}_{\infty}:\mathbb{C}\to\mathbb{C}_{\infty}.$ The following is the main result of the paper.

Theorem 1**.**

Let $\mathfrak{g}$ be a perfect Lie algebra over $\mathbb{C}$ satisfying Assumption 1. Then there are no nontrivial semi-simple elements in $\ker(D).$ Moreover, $D$ restricts to $\mathrm{Fr}_{\infty}^{*}$ on $\mathrm{Aut}(\mathfrak{g}).$ In particular, if $\Gamma$ is a finite subgroup of $\mathrm{Aut}(\mathfrak{U}\mathfrak{g}),$ then there exists a subgroup $\Gamma^{\prime}$ of $\mathrm{Aut}(\mathfrak{g}),$ such that $\Gamma$ is isomorphic to $\Gamma^{\prime}$ as an abstract group.

Remark 1*.*

The most interesting application of the above result is for the case of a simple Lie algebra $\mathfrak{g}.$ In principle, this result provides a full classification of isomorphism classes of finite groups of automorphisms of $\mathfrak{U}\mathfrak{g}.$ However, although the construction of the subgroup $\Gamma^{\prime}\subset\text{Aut}(\mathfrak{g})$ is somewhat canonical, we don not know if $\Gamma^{\prime}$ is conjugate to $\Gamma$ in $\text{Aut}(\mathfrak{U}\mathfrak{g}).$ In fact, we do not know if a much stronger statement about linearizability holds: Given a finite subgroup $\Gamma\subset\text{Aut}(\mathfrak{U}\mathfrak{g}),$ whether there exists a subgroup $\Gamma^{\prime}\subset\text{Aut}(\mathfrak{g}),$ such that $\Gamma^{\prime}$ is conjugate to $\Gamma$ in $\text{Aut}(\mathfrak{U}\mathfrak{g}).$

Throughout the paper, given an abelian group $M$, we denote by $M_{p}$ its reduction modulo $p$: $M_{p}=M/pM.$

1. Reduction modulo $p$ Lemmas

In this section, given any finite dimensional Lie algebra $\mathfrak{g}$ over $\mathbb{C},$ we define $\mathbb{C}_{\infty}$-Lie algebras $\mathfrak{g}_{\infty},\hat{\mathfrak{g}_{\infty}},$ 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 $\mathbb{Z}$-algebra $R$ and a prime number $p,$ then the center $Z(R_{p})$ of its reduction modulo $p$ acquires the natural Poisson bracket, which we refer to as the deformation Poisson bracket, defined as follows. Given $a,b\in Z(R_{p})$, let $z,w\in R$ be their lifts respectively. Then the Poisson bracket $\{a,b\}$ is defined to be

[TABLE]

Let $S$ be a finitely generated subring of $\mathbb{C},$ and let $\mathfrak{g}$ be a Lie algebra over $S$ which is a finite rank free $S$-module. Throughout the paper we denote by $B$ the quotient of $\mathfrak{U}\mathfrak{g}$ by the augmentation ideal of its center

[TABLE]

Then we define $S_{\infty}$-Lie algebras $\mathcal{L}_{\infty}(\mathfrak{g}),\mathcal{L}_{\infty}^{\prime}(\mathfrak{g})$ as follows. Let $p>>0$ be a large prime number. Then the following augmentation ideals

[TABLE]

are easily seen to be Poisson ideals in $Z(\mathfrak{U}\mathfrak{g}_{p}),Z(B_{p})$ respectively. Hence $\mathfrak{m}_{p}/\mathfrak{m}_{p}^{2},\mathfrak{n}_{p}/\mathfrak{n}_{p}^{2}$ are Lie algebras, and we view them as Lie algebras over $S_{p}$ via the the Frobenius map $\text{Fr}_{p}:S_{p}\to S_{p}.$ Denote $\mathfrak{m}_{p}/\mathfrak{m}_{p}^{2}$ by $\mathcal{L}_{p}(\mathfrak{g})$, and $\mathfrak{n}_{p}/\mathfrak{n}_{p}^{2}$ by $\mathcal{L}_{p}^{\prime}(\mathfrak{g}).$ Similarly given a base change $S\to\mathbb{k}$ to a field of characteristic $p$ we define $\mathbb{k}$-Lie algebras $\mathcal{L}_{\mathbb{k}}(\mathfrak{g}),\mathcal{L}_{\mathbb{k}}^{\prime}(\mathfrak{g}).$ We put

[TABLE]

Now, given a Lie algebra over $\mathbb{C},$ we define

[TABLE]

Here, $\mathfrak{g}_{S}$ is a model of $\mathfrak{g}$ over $S:$ $\mathfrak{g}_{S}\otimes_{S}\mathbb{C}=\mathfrak{g}.$ We have the natural surjective homomorphism $\mathcal{L}_{\infty}(\mathfrak{g})\to\mathcal{L}_{\infty}^{\prime}(\mathfrak{g}).$ If $\mathfrak{g}_{S}$ 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 $\mathfrak{g}$ satisfies Assumption 1, then $\mathcal{L}_{\infty}(\mathfrak{g})$ is isomorphic to a trivial central extension of $\mathfrak{g}_{\mathbb{C}_{\infty}}=\mathfrak{g}\otimes_{\mathbb{C}}\mathbb{C}_{\infty},$ while $\mathcal{L}_{\infty}^{\prime}(\mathfrak{g})\cong\mathfrak{g}_{\mathbb{C}_{\infty}}.$

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 $\mathfrak{g}$ be a restricted Lie algebra over $R$ ($\mathfrak{g}$ is assumed to be a finite free $R$-module) with the restricted structure map $g\to g^{[p]},g\in\mathfrak{g}.$ Then by $Z_{p}(\mathfrak{g})$ we denote the central $R$-subalgebra of the enveloping algebra $\mathfrak{U}\mathfrak{g}$ generated by elements of the form $g^{p}-g^{[p]},g\in\mathfrak{g}.$ It is well-known that the map $g\to g^{p}-g^{[p]}$ induces homomorphism of $R$-algebras

[TABLE]

where $Z_{p}(\mathfrak{g})$ is viewed as an $R$-algebra via the Frobenius map $F:R\to R.$ The homomorphism $i$ is an isomorphism when $R$ is perfect. Also, recall that the Lie algebra bracket on $\mathfrak{g}$ defines the Kirillov-Kostant Poisson bracket on the symmetric algebra $\text{Sym}(\mathfrak{g}).$

The following is the key result from [KR]. We include a proof for the reader’s convenience.

Lemma 1**.**

Let $R$ be a finitely generated integral domain over $\mathbb{Z}.$ Let $G$ be an affine algebraic group over $R$, let $\mathfrak{g}$ be its Lie algebra. Let $G_{p},\mathfrak{g}_{p}$ be reductions modulo $p$ of $G,\mathfrak{g}$ respectively. Thus $Z(\mathfrak{U}\mathfrak{g}_{p})$ is equipped with the deformation Poisson bracket. Then $Z_{p}(\mathfrak{g}_{p})$ is a Poisson subalgebra of $Z(\mathfrak{U}\mathfrak{g}_{p}),$ 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 $D_{X}$ denote the algebra of differential operators on $X.$ Put $\overline{X}=X\mod p$. Then the center of $D_{X}\mod p=D_{\overline{X}}$ can be identified with (the Frobenius twist) of the functions on the cotangent bundle $T^{*}_{\overline{X}}$ [BMR]. Then the deformation Poisson bracket of $Z(D_{\overline{X}})$ is equal to the negative of the Poisson bracket coming from the symplectic structure of the cotangent bundle of $\overline{X}.$

Now let $\theta:\mathfrak{g}\to D_{G}$ be the realization of $\mathfrak{g}$ as left-invariant vector fields on $G.$ Then we have the corresponding embedding $\theta:\mathfrak{U}\mathfrak{g}\to D_{G}$ and the corresponding reduction modulo $p$ $\bar{\theta}:\mathfrak{U}\mathfrak{g}_{p}\to D_{G_{p}}$, which induces the embedding $Z_{p}(\mathfrak{g}_{p})\to Z(D_{G_{p}}).$ In this way $Z_{p}(\mathfrak{g}_{p})$ is a Poisson subalgebra of $Z(D_{G_{p}})$ and the assertion follows. ∎

It is clear that $i^{-1}(\mathfrak{m}_{p})=\mathfrak{g}_{p}\text{Sym}\mathfrak{g}_{p}.$ So, in view of Lemma1 we have the canonical Lie algebra homomorphism $\mathfrak{g}_{p}\to\mathfrak{m}_{p}/\mathfrak{m}_{p}^{2}.$ Hence, we obtain the desired canonical homomorphism

[TABLE]

We have the following

Lemma 2**.**

Let $S$ be a finitely generated subring of $\mathbb{C}.$ Let $\mathfrak{g}$ be a nilpotent Lie algebra over $S.$ Let $p>>0$ be a prime. Let $I$ be a Poisson ideal of $Z(\mathfrak{U}\mathfrak{g}_{p}),$ such that $Z(\mathfrak{U}\mathfrak{g}_{p})/I=S_{p}.$ Then $I/I^{2}\cong\mathfrak{m}_{p}/\mathfrak{m}_{p}^{2}$ as $S_{p}$-Lie algebras.

Proof.

We claim that given any ideal $I$ in $Z(\mathfrak{U}\mathfrak{g}_{p}),$ such that $I\cap Z_{p}(\mathfrak{g}_{p})=(g^{p},g\in\mathfrak{g}_{p})$ and $Z(\mathfrak{U}\mathfrak{g}_{p})/I=S_{p},$ then $I=\mathfrak{m}_{p}.$ Indeed, since $\mathfrak{U}\mathfrak{g}_{p}/(g^{p},g\in\mathfrak{g}_{p})\mathfrak{U}\mathfrak{g}_{p}$ is a nilpotent $S_{p}$-algebra, it follows that $Z(\mathfrak{U}\mathfrak{g}_{p})/(g^{p},g\in\mathfrak{g}_{p})Z(\mathfrak{U}\mathfrak{g}_{p})$ is also a nilpotent $S_{p}$-algebra: $\mathfrak{m}_{p}^{p^{l}}\subset(g^{p},g\in\mathfrak{g}_{p})Z(\mathfrak{U}\mathfrak{g}_{p})$ for large enough $l.$ If $a+y\in I,a\in S_{p},y\in\mathfrak{m}_{p}$, then

[TABLE]

so $a=0$ and $I\subset\mathfrak{m}_{p}.$ Put $I^{\prime}=I\cap Z_{p}(\mathfrak{g}).$ Then $I^{\prime}$ is a Poisson ideal and $Z_{p}(\mathfrak{g})/I^{\prime}=S_{p}.$ Hence, $I^{\prime}=(g^{p}-\chi(g),g\in\mathfrak{g}_{p})$, where $\chi:\mathfrak{g}_{p}\to S_{p}$ is a Lie algebra homomorphism. Then $\phi(g)=g-\chi(g)$ induces an automorphism $\phi:\mathfrak{U}\mathfrak{g_{p}}\to\mathfrak{U}\mathfrak{g_{p}},$ such that $\phi(m_{p})=I.$ Moreover $\phi$ admits a lift over ${S/p^{2}S}.$ Indeed, take any character $\hat{\chi}:\mathfrak{g}/p^{2}\mathfrak{g}\to S/p^{2}$ that lifts $\chi$ and put $\tilde{\phi}(g)=\tilde{g}-\chi(g).$ Then $\tilde{\phi}:\mathfrak{U}(\mathfrak{g}/p^{2}\mathfrak{g})\to\mathfrak{U}(\mathfrak{g}/p^{2}\mathfrak{g}$) is an automorphisms such that $\phi=\tilde{\phi}\mod p^{2}.$ Hence, $\phi$ is a Poisson automorphism of $Z(\mathfrak{U}\mathfrak{g}_{p}).$ Thus we obtain the desired isomorphism of $S_{p}$-Lie algebras $I/I^{2}\cong\mathfrak{m}_{p}/\mathfrak{m}_{p}^{2}.$

∎

From now on we fix a Lie algebra $\mathfrak{g}$ satisfying Assumption 1. It follows that we may choose a finitely generated subring $S\subset\mathbb{C},$ and a Lie algebra $\mathfrak{g}_{S}$ over $S$ which is finite free $S$-module, such that $\mathfrak{g}=\mathfrak{g}_{S}\otimes_{S}\mathbb{C},f_{i}\in\text{Sym}\mathfrak{g}_{S}$ and $A_{S}=\text{Sym}\mathfrak{g}_{S}/(f_{1},\cdots,f_{n})$ is a normal integral domain. Moreover $\text{Spec}A_{S}$ has a nonempty open subset $U_{S}$ which is symplectic over $S$ under the Kirillov-Kostant bracket. Denote by $g_{i}$ the image of $f_{i}$ under the symmetrization isomorphism $\text{Sym}(\mathfrak{g})^{\mathfrak{g}}\to Z(\mathfrak{U}\mathfrak{g}).$ Hence $\text{gr}(g_{i})=f_{i}.$ We may assume that $S$ is large enough so that $g_{i}\in\mathfrak{U}\mathfrak{g}_{S},1\leq i\leq n.$ Just as before, we put $B_{S}=\mathfrak{U}\mathfrak{g}_{S}/(g_{1},\cdots,g_{n}).$ Hence $B=B_{S}\otimes_{S}\mathbb{C}.$ Given a commutative $S$-algebra $R$, we denote by $\mathfrak{g}_{R},B_{R}$ and $A_{R}$ the base changes of $\mathfrak{g}_{S},B_{S},A_{S}$ respectively. In particular, for a base change $R\to\mathbb{k}$, where $\mathbb{k}$ is an algebraically closed field, $A_{\mathbb{k}}$ is a normal integral domain (for $p>>0$). We denote images of $f_{i},g_{i}$ in $\text{Sym}\mathfrak{g}_{p},Z(\mathfrak{U}\mathfrak{g}_{p})$ by $\bar{f_{i}},\bar{g_{i}}.$ Given a commutative $S_{p}$-algebra $R$, we denote by $\mathfrak{m}_{R}$( respectively $\mathfrak{n}_{R}),$ the augmentation ideal $Z(\mathfrak{U}\mathfrak{g}_{R})\cap\mathfrak{g}_{R}\mathfrak{U}\mathfrak{g}_{R}$ (respectively $Z(B_{R})\cap\mathfrak{g}_{R}B_{R}.$)

Then we have the following

Lemma 3**.**

Let $\mathfrak{g}$ be as above. Then, $Z(\mathfrak{U}\mathfrak{g}_{p})$ is a free $Z_{p}(\mathfrak{g}_{p})$-module with the basis

[TABLE]

*In particular, $\mathcal{L}_{\infty}^{\prime}(\mathfrak{g})=\mathfrak{g}_{S_{\infty}},$ while $\mathcal{L}_{\infty}(\mathfrak{g})$ is a trivial central extension of $\mathfrak{g}_{S_{\infty}}.$ Moreover, given a Poisson ideal $I$ in $Z(\mathfrak{U}\mathfrak{g}_{p}),$ such that $Z(\mathfrak{U}\mathfrak{g}_{p})/I=S/pS,$ then $I/I^{2}$ is a trivial central extension of $\mathfrak{g}_{p}$ as an $S_{p}$-Lie algebra. If $\mathfrak{g}$ is perfect, then $\mathfrak{n}_{p}$ is the unique Poisson ideal of $Z(B_{p}),$ such that $Z(B_{p})/\mathfrak{n}_{p}=S_{p}.$ *

Proof.

It is enough to verify above statements after a base change $S/pS\to\mathbb{k}$, where $\mathbb{k}$ is an algebraically closed field of characteristic $p.$ At first, since $A_{\mathbb{k}}=\text{Sym}(\mathfrak{g}_{\mathbb{k}})/(\bar{f_{1}},\cdots,\bar{f_{n}})$ is a normal domain and the Poisson bracket is symplectic on a nonempty open subset of $\text{Spec}A_{\mathbb{k}},$ it follows easily that the Poisson center of $A_{\mathbb{k}}$ is $A_{\mathbb{k}}^{p}$ (see, for example, Lemma 2.4[T]). It suffices to check that the Poisson center of $\text{Sym}\mathfrak{g}_{\mathbb{k}}$ is a free module over $(\text{Sym}\mathfrak{g}_{\mathbb{k}})^{p}$ with the basis

[TABLE]

Indeed, let $g$ be in the Poisson center of $\text{Sym}\mathfrak{g}_{\mathbb{k}}.$ Denote the ideal $(\bar{f_{1}},\cdots,\bar{f_{n}})\subset\text{Sym}\mathfrak{g}_{\mathbb{k}}$ by $I.$ We claim that $g\in\text{Sym}(\mathfrak{g}_{\mathbb{k}})^{p}[\bar{f_{1}},\cdots,\bar{f_{n}}]+I^{m},$ for all $m.$ We proceed by induction on $m.$ Since $\bar{g}=g\mod I$ belongs to the Poisson center of $A_{\mathbb{k}},$ it follows that $\bar{g}\in A_{\mathbb{k}}^{p}.$ Hence $g\in\text{Sym}(\mathfrak{g}_{\mathbb{k}})^{p}+I.$ Assume that $g\in\text{Sym}(\mathfrak{g}_{\mathbb{k}})^{p}[\bar{f_{1}},\cdots,\bar{f_{n}}]+I^{m}$ for some $m\geq 1.$ So, there exists $x\in g+\text{Sym}(\mathfrak{g}_{\mathbb{k}})^{p}[\bar{f_{1}},\cdots,\bar{f_{n}}],$ such that

[TABLE]

Then, for any $y\in\mathfrak{g}_{\mathbb{k}}$, we have

[TABLE]

Since the sequence $(\bar{f_{1}},\cdots,\bar{f_{n}})$ is regular, we may conclude that $\bar{x_{\alpha}}=x_{\alpha}\mod I\in A_{\mathbb{k}}^{p}.$ Hence,

[TABLE]

Therefore,

[TABLE]

Now, suppose that

[TABLE]

Such that either $x_{\alpha}=0$ or $x_{\alpha}\notin I.$ Let $m$ be such that $x_{\alpha}=0$ for all $|\alpha|<m$ in the above sum. Since $I^{m}/I^{m+1}$ is a free $A_{\mathbb{k}}$-module with basis $\{\bar{f_{1}}^{\alpha_{1}}\cdots\bar{f_{n}}^{\alpha_{n}},\sum\alpha_{i}=m\},$ we get that $x_{\alpha}^{p}\mod I=0$, hence $x_{\alpha}\in I.$ So $x_{\alpha}=0.$ Thus, elements $\{\bar{f_{1}}^{\alpha_{1}}\cdots\bar{f_{n}}^{\alpha_{n}},0\leq\alpha_{i}<p\}$ are linearly independent over $(\text{Sym}\mathfrak{g}_{\mathbb{k}})^{p}$ as desired. It follows that $\mathfrak{g}_{\mathbb{k}}\bigoplus\oplus_{i=1}^{n}\mathbb{k}\bar{g_{i}}$ surjects onto $\mathcal{L}_{\mathbb{k}}(\mathfrak{g})=\mathfrak{m}_{\mathbb{k}}/\mathfrak{m}_{\mathbb{k}}^{2},$ and since the center of $\mathfrak{g}_{\mathbb{k}}$ is trivial, we conclude that $\mathcal{L}_{\mathbb{k}}(\mathfrak{g})$ is a direct sum of $\mathfrak{g}_{\mathbb{k}}$ with a central subalgebra spanned by images of $\bar{g}_{i}.$

Denote by $\tilde{f}_{i}$ the image of $\bar{f_{i}}^{p}$ under the isomorphism $\text{Sym}(\mathfrak{g}_{\mathbb{k}})\cong Z_{p}(\mathfrak{g}_{\mathbb{k}}).$ Then $\text{gr}(\tilde{f_{i}})=\bar{f}_{i}^{p}.$ Since $A_{\mathbb{k}}$ is a domain, we have

[TABLE]

Therefore

[TABLE]

Hence, we conclude that the map $i:\text{Sym}\mathfrak{g_{\mathbb{k}}}\to Z(B_{\mathbb{k}})$ induces an isomorphism $A_{\mathbb{k}}\cong Z(B_{\mathbb{k}}).$ In particular, $\mathcal{L}_{\mathbb{k}}^{\prime}(\mathfrak{g})=\mathfrak{n}_{\mathbb{k}}{/\mathfrak{n}_{\mathbb{k}}}^{2}\cong\mathfrak{g}_{\mathbb{k}}$.

Now let $I\subset Z(\mathfrak{U}\mathfrak{g}_{p})$ be a Poisson ideal such that $Z(\mathfrak{U}\mathfrak{g}_{p})/I=S_{p}.$ Proceeding as in the proof of Lemma 2, without loss of generality we may assume that $(g^{p}-g^{[p]},g\in\mathfrak{g}_{p})\subset I.$ Let $\bar{g_{i}}-a_{i}\in I,a_{i}\in S_{p}.$ In the above proof, replacing $\bar{g}_{i}$ with $\bar{g}_{i}-a_{i},$ we conclude that

[TABLE]

If $\mathfrak{g}_{p}$ is perfect and $I$ is Poisson ideal in $B_{p},$ such that $B_{p}/I=S/pS,$ then if follows that $g^{p}-g^{[p]}-\chi(g)\in I,g\in\mathfrak{g}_{p},$ such that $\chi:\mathfrak{g}_{p}\to S_{p}$ is a character. Hence, $\chi$ must be trivial. Since $g^{p}-g^{[p]},g\in\mathfrak{g}_{p}$ generate $Z(B_{p}),$ we get that $I=\mathfrak{n}_{p}.$

Example 1*.*

Let $\mathfrak{g}=\mathfrak{sl}_{2}$ with the usual generators $e,f,h.$ Then the center of $\mathfrak{U}\mathfrak{sl}_{2}(\mathbb{F}_{p})$ is generated a $\mathbb{F}_{p}$-algebra by $x=e^{p},y=f^{p},z=h^{p}-h,\Delta=4fe+h^{2}+2h$ subject to the relation

[TABLE]

Thus the augmentation ideal $m_{p}=Z(\mathfrak{U}\mathfrak{g}_{p})\cap\mathfrak{g}_{p}(\mathfrak{U}\mathfrak{g}_{p})$ is generated by $x,y,z,\Delta$ and $m_{p}/m_{p}^{2}$ has an $\mathbb{F}_{p}$-basis $\{\bar{x},\bar{y},\bar{z},\bar{\Delta}\}$–images of $x,y,z,\Delta$ respectively. Thus we see that $\mathcal{L}_{p}(\mathfrak{sl}_{2})=m_{p}/m_{p}^{2}\cong\mathfrak{sl}_{2}(\mathbb{F}_{p})\oplus\mathbb{F}_{p}\bar{\Delta}.$ Similarly $n_{p}$–the augmentation ideal of $Z(B_{p})=Z(\mathfrak{U}\mathfrak{sl}_{2}(\mathbb{F}_{p})/(\Delta)),$ is generated by $x,y,z$, and $n_{p}/n_{p}^{2}=\mathbb{F}_{p}\bar{x}\oplus\mathbb{F}_{p}\bar{y}\oplus\mathbb{F}_{p}\bar{z}.$ So $\mathcal{L}_{p}^{\prime}(\mathfrak{sl}_{2})\cong\mathfrak{sl}_{2}(\mathbb{F}_{p}).$

∎

2. Derived isomorphisms

As usual, $S$ is a finitely generated subring of $\mathbb{C}.$ 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 $Z(A/pA)\cong Z(B/pB)$ as Poisson algebras for all $p>>0.$

Proof.

It follows that algebras $A/p^{2}A,B/p^{2}B$ are derived equivalent. We have the following exact sequence of $A/p^{2}A$-bimodules

[TABLE]

where $A/pA\to A/pA$ is the quotient map and $A/pA\to A/p^{2}A$ 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.

∎

Now we can easily prove the following.

Theorem 2**.**

Suppose that Lie algebras $\mathfrak{g},\mathfrak{g^{\prime}}$ satisfy Assumption 1. If $\mathfrak{U}\mathfrak{g}$ is derived equivalent to $\mathfrak{U}\mathfrak{g^{\prime}}$, then $\mathfrak{g}\cong\mathfrak{g^{\prime}}.$

Proof.

There exists a finitely generated subring $S$ of $\mathbb{C}$, and Lie algebras $\mathfrak{g}_{S},\mathfrak{g^{\prime}}_{S}$ over $S$, such that

[TABLE]

and $\mathfrak{U}(\mathfrak{g}_{S})$ is derived equivalent to $\mathfrak{U}(\mathfrak{g^{\prime}}_{S}).$ Hence,

[TABLE]

as Poisson algebras by Lemma 4. Let $I\subset Z(\mathfrak{U}(\mathfrak{g}_{S_{p}}))$ be a Poisson ideal such that $Z(\mathfrak{U}(\mathfrak{g}_{S_{p}}))/I=S_{p}.$ Let $I^{\prime}\subset Z(\mathfrak{U}(\mathfrak{g^{\prime}}_{S_{p}}))$ be the corresponding ideal under the above isomorphism. Hence $I/I^{2}\cong I^{\prime}/I^{\prime 2}$ as $S_{p}$-Lie algebras. Now using Lemma 3 we conclude that trivial central extensions of $(\mathfrak{g})_{\mathbb{C}_{\infty}}$ and $(\mathfrak{g^{\prime}})_{\mathbb{C}_{\infty}}$ are isomorphic. Since $\dim\mathfrak{g}=\dim\mathfrak{g^{\prime}}$, this implies that $\mathfrak{g}\cong\mathfrak{g^{\prime}}.$ ∎

3. The homomorphisms $D,\tilde{D}$

We assume that $\mathfrak{g}$ is a perfect Lie algebra over $\mathbb{C}$ satisfying Assumption 1. Let $S$ be a large enough finitely generated subring of $\mathbb{C}$, and let $\mathfrak{g}_{S}$ be a model of $\mathfrak{g}$ over $S$ (just as in the paragraph preceding Lemma 3). Then we construct canonical homomorphisms

[TABLE]

as follows. At first, remark that since $\mathfrak{g}$ is perfect, any automorphism of $\mathfrak{U}\mathfrak{g}$ must preserve the ideal $Z(\mathfrak{U}\mathfrak{g})\cap\mathfrak{g}\mathfrak{U}\mathfrak{g}.$ Therefore we have the restriction homomorphism $\text{Aut}(\mathfrak{U}\mathfrak{g})\to\text{Aut}(B).$ Let $p>>0$ be a sufficiently large prime. Let $\phi\in\text{Aut}(B_{S}).$ Reducing $\phi\mod p,$ we obtain $\bar{\phi}\in\text{Aut}(Z(B_{p})).$ Since $B_{p}$ is obtained by the reduction modulo $p$, we have the corresponding deformation Poisson bracket on its center. Hence, $\bar{\phi}$ preserves the Poisson bracket on $Z(B_{p}).$ Now it follows from Lemma 2 that $\bar{\phi}$ preserves $\mathfrak{n}_{p},$ thus it induces a Lie algebra automorphism on $\mathfrak{n}_{p}/\mathfrak{n}_{p}^{2}\cong\mathfrak{g}_{p},$ which we denote by $(\tilde{D}_{S_{p}})(\phi).$ Hence, we obtain a canonical homomorphism $\tilde{D}_{S_{p}}:\text{Aut}(B_{S})\to\text{Aut}(\mathfrak{g}_{p}).$ Also, given a base change $S_{p}\to\mathbb{k}$, we denote the corresponding homomorphism $\tilde{D}_{S_{p}}\otimes_{S_{p}}\mathbb{k}:\text{Aut}(B_{S})\to\text{Aut}(\mathfrak{g}_{\mathbb{k}})$ by $\tilde{D}_{\mathbb{k}}.$ Also, denote by $D_{\mathbb{k}}:\text{Aut}(\mathfrak{U}\mathfrak{g}_{S})\to\text{Aut}(\mathfrak{g}_{\mathbb{k}})$ the composition of $\tilde{D}_{\mathbb{k}}$ with the restriction $\text{Aut}(\mathfrak{U}\mathfrak{g}_{S})\to\text{Aut}(B_{S}).$ The element $\prod_{p}(\tilde{D}_{S_{p}})(\phi)$ gives rise to an element of $\text{Aut}(\mathfrak{g}_{S_{\infty}})$, which we denote by $\tilde{D}_{S}(\phi).$ This way we obtain the desired homomorphisms $\tilde{D}_{S}.$ We define $D_{S}:\text{Aut}(\mathfrak{U}\mathfrak{g}_{S})\to\text{Aut}(\mathfrak{g}_{S_{\infty}})$ as the composition of $\tilde{D_{S}}$ with the restriction $\text{Aut}(\mathfrak{U}\mathfrak{g}_{S})\to\text{Aut}(B_{S}).$

Now, taking the direct limit of the above homomorphisms $\tilde{D}_{S},D_{S}$ over finitely generated subsrings $S\subset\mathbb{C},$ we obtain the sought after homomorphisms

[TABLE]

It is clear from the construction and Lemma3 that $D(\text{Aut}(\mathfrak{g}))=\text{Fr}_{\infty}^{*},$ where $\text{Fr}_{\infty}:\mathbb{C}\to\mathbb{C}_{\infty}$ is the canonical inclusion followed by the Frobenius map.

4. The proof

We start by the following.

Proposition 1**.**

Let $\mathfrak{g}$ be a perfect Lie algebra satisfying Assumption 1. Then the kernel of the restriction homomorphism $\mathrm{Aut}(\mathfrak{U}\mathfrak{g})\to\mathrm{Aut}(B)$ contains no nontrivial semi-simple automorphisms.

We remark that in general, the homomorphism $\text{Aut}(\mathfrak{U}\mathfrak{g})\to\text{Aut}(B)$ is not injective. For example, in the case of $\mathfrak{g}=\mathfrak{sl}_{2},$ this follows from existence of a non tame automorphism of $\mathfrak{U}\mathfrak{sl}_{2},$ proved by Joseph [J1].

For the proof of Lemma1, we need a specific set of linearly independent elements of $\text{HH}^{2}(B)$. Recall that for a semi-simple $\mathfrak{g}$, one has $\dim\text{HH}^{2}(B)=n,$ as follows immediately from Soergel’s result [S].

Recall that $Z(\mathfrak{U}\mathfrak{g})=\mathbb{C}[g_{1},\cdots,g_{n}],g_{i}\in\mathfrak{g}\mathfrak{U}\mathfrak{g}$ and $(g_{1},\cdots,g_{n})$ is a regular sequence in $\mathfrak{U}\mathfrak{g}.$ Put $I=(g_{1},\cdots,g_{n}).$ So $B=\mathfrak{U}\mathfrak{g}/I$. Let us put $B^{\prime}=\mathfrak{U}\mathfrak{g}/I^{2}$. Then we have a short exact sequence

[TABLE]

We have $I/I^{2}=\bigoplus_{i=1}^{n}B\bar{g_{i}},$ where $\bar{g_{i}}$ denotes the image of $g_{i}$ under the quotient map $\mathfrak{U}\mathfrak{g}\to B^{\prime}.$ Denote by

[TABLE]

the cohomology class corresponding to the above short exact sequence. Let us put $\omega=\sum_{i=1}^{n}\omega_{i}\bar{g_{i}}$ where $\omega_{i}\in\text{HH}^{2}(B).$ Under these notations we have the following.

Lemma 5**.**

Elements $\omega_{i},1\leq i\leq n$ are linearly independent .

Proof.

Let $\sum_{i=1}^{n}c_{i}\omega_{i}=0,c_{i}\in\mathbb{C}.$ Let $c_{j}\neq 0$ for some $j.$ Let $B_{1}$ be the quotient of $B^{\prime}$ by the ideal $\sum_{i\neq j}B^{\prime}\bar{g_{i}}.$ Then it follows that the quotient map $B_{1}\to B_{1}/B_{1}\bar{g_{j}}=B$ admits a $\mathbb{C}$-algebra splitting. Let $\psi:B\to B_{1}$ be such a splitting. Let us write

[TABLE]

This implies that $\bar{g_{j}}\in\bar{g_{j}}(\mathfrak{g}B^{\prime}).$ Therefore

[TABLE]

Thus, there exists $\alpha\in\mathfrak{U}\mathfrak{g}\setminus\mathfrak{g}\mathfrak{U}\mathfrak{g}$ such that $\alpha g_{j}\in\sum_{i\neq j}g_{i}\mathfrak{U}\mathfrak{g}.$ Now the regularity of the sequence $(g_{1},\cdots,g_{n})$ implies that $\alpha\in I$, which is a contradiction. ∎

Proof of Proposition1.

Let $\phi$ be a semi-simple automorphism of $\mathfrak{U}\mathfrak{g}$ that restricts to the identity on $B.$ Denote by $\tilde{\phi}$ the restriction of $\phi$ on $B^{\prime}.$ Therefore $\tilde{\phi}$ fixes $\omega=\sum\omega_{i}\bar{g_{i}}\in H^{2}(B,I/I^{2}).$ Let $\tilde{\phi}:I/I^{2}\to I/I^{2}$ be represented by the matrix

[TABLE]

Entries $\phi_{ij}$ are scalars since $\phi$ preserves $\mathbb{C}[g_{1},\cdots,g_{n}].$ Thus $\sum\phi_{ij}\omega_{j}=\omega_{i}.$ Since $\omega_{j}$ are linearly independent by Lemma 5, we conclude that $\phi|_{I/I^{2}}=\text{Id}.$ Hence $\phi|_{I^{n}/I^{n+1}}=\text{Id}$ for all $n.$ Since $\phi$ is semi-simple, we get that $\phi=\text{Id}.$ ∎

Remark 2*.*

It was proved by Polo[P] that when $\mathfrak{g}$ is semi-simple, the action of an automorphism of $\mathfrak{U}\mathfrak{g}$ on its center is given by a Dynkin diagram automorphism of $\mathfrak{g}.$

Proof of Theorem 1..

In view of Proposition 1, it suffices to check that $\ker(\tilde{D})$ has no nontrivial semi-simple automorphisms. Assume that $\phi\in\text{Aut}(B)$ is a non-trivial semi-simple automorphism. Therefore there exists a finitely generated subring $S\subset\mathbb{C},$ and a finite free $\phi$-invariant $S$-submodule $V\subset B_{S},$ such that $\phi|_{V}$ is semi-simple over $S$ and $V$ generates $B_{S}$ 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 $S_{p}\to\mathbb{k}$, where $\mathbb{k}$ is a field, then $\tilde{D}_{\mathbb{k}}(\phi)\neq\text{Id}_{\mathfrak{g}_{\mathbb{k}}}.$ Let $y\in V,1\neq\alpha\in S,$ such that $\phi(y)=\alpha y.$ Let us write $y=\sum a_{i}e_{i},0\neq a_{i}\in S,$ where $e_{i}$ are basis elements of $B_{S}$ as a free $S$-module. We may assume that $a_{i},1-\alpha$ are invertible is $S.$ Now let $p>>0$ and $\rho:S\to\mathbb{k}$ be a base change to a field such that $\tilde{D}_{\mathbb{k}}=\text{Id}_{\mathfrak{g}_{\mathbb{k}}}.$ Thus, we have a non-trivial semi-simple automorphism $\bar{\phi}\in\text{Aut}(B_{\mathbb{k}}),$ such that $\bar{\phi|}_{m_{\mathbb{k}}/m_{\mathbb{k}}^{2}}=\text{Id}.$ Then $\bar{\phi}$ acts trivially on $m_{\mathbb{k}}^{n}/m_{\mathbb{k}}^{n+1}$ for all $n.$ Since the action of $\phi$ on $Z(B_{\bf{k}})$ is semi-simple, it follows that the action of $\bar{\phi}$ on $Z(B_{\bf{k}})$ is trivial. Then by the Noether-Skolem theorem there exists $a\in B_{\bf{k}},$ such that

[TABLE]

But, since $\bar{\phi}(\bar{y})=\bar{\alpha}\bar{y}$ and $1\neq\bar{\alpha}\in\mathbb{k},0\neq\bar{y}\in B_{\mathbb{k}}$, we get that $a\bar{y}=\bar{\alpha}\bar{y}a.$ Recall that under the PBW filtration on $B_{\bf{k}}$, $\text{gr}(B_{\mathbb{k}})=A_{\mathbb{k}}$ is a commutative domain. Hence,

[TABLE]

which is a contradiction.

Let $\theta:\mathbb{C}_{\infty}\to\mathbb{C}$ be a homomorphism. Then we define a (non-canonical) homomorphism $D^{*}:\text{Aut}(\mathfrak{U}\mathfrak{g})\to\text{Aut}(\mathfrak{g})$ as the composition of $D$ with the base change homomorphisms $\theta^{*}:\text{Aut}(\mathfrak{g}_{\mathbb{C}_{\infty}})\to\text{Aut}(\mathfrak{g}).$ Next we show that $\ker(D^{*})$ contains no nontrivial finite order elements. This implies that given a finite subgroup $\Gamma\subset\text{Aut}(\mathfrak{U}\mathfrak{g}),$ then $\Gamma^{\prime}=D^{*}(\Gamma)\subset\text{Aut}(\mathfrak{g}),\Gamma\cong\Gamma^{\prime}.$ Let $\phi\in\text{Aut}(\mathfrak{U}\mathfrak{g})$ such that $\phi^{m}=1,\phi\neq 1.$ We may choose a finitely generated subring $S\subset\mathbb{C},\frac{1}{m}\in S$ containing all $m$-th roots of unity, such that $\phi\subset\text{Aut}(\mathfrak{U}\mathfrak{g}_{S}).$ As it was shown in the preceding paragraph, by enlarging $S$ if necessary, for all $p>>0$ and a base change $S_{p}\to\mathbb{k}$ to a field, we have $D_{\mathbb{k}}(\phi)\neq\text{Id}_{\mathfrak{g}_{\mathbb{k}}}.$ Let $f_{p}(t)\in S_{p}[t]$ denote the characteristic polynomial of $D_{S_{p}}(\phi)\in\text{Aut}(\mathfrak{g}_{S_{p}}).$ Put $l=\dim\mathfrak{g}.$ We show that $\theta(\prod_{p}f_{p}(t))\neq(t-1)^{l}$ in $S_{\infty}[t].$ Indeed, let us write $S_{p}=\prod_{i}S_{p,i}$, where each $S_{p,i}$ is a domain (since $p$ is unramified in $S).$ Since the image of $D_{S_{p}}(\phi)$ in $\text{Aut}(\mathfrak{g}_{S_{p,i}})$ has order $m$, it follows that the image of $f_{p}(t)$ in $S_{p,i}[t]$ is not equal to $(t-1)^{l}$ and is of the form $\prod_{j=1}^{m}(t-a_{j}),a_{j}^{m}=1.$ Denote by $\Psi\subset S[t]$ the finite set of all degree $l$ monic polynomials not equal to $(t-1)^{l},$ of the form $\prod_{j=1}^{m}(t-a_{j}),a_{j}\in S,a_{j}^{m}=1.$ For each $g\in\Psi,$ denote by $I_{g}$ the set of pairs $(p,i)$ for which $f_{p}(t)=g$ in $S_{p,i}.$ Then we have $\prod_{p}S_{p}=\prod_{g\in\Psi}(\prod_{(p,i)\in I_{g}}S_{p,i}).$ Now, suppose $g\in\Psi$ is such that $\prod_{(p,i)\in I_{g}}S_{p,i}\not\subset\ker(\theta).$ Then it follows that $\theta(\prod_{p}f_{p}(t))=\theta(g)\neq(t-1)^{l}.$ Hence $\phi\notin\ker(D^{*})$ as desired.

∎

Acknowledgement: I am very grateful to the anonymous referee for the numerous useful suggestions.