The Weyl Algebras

Home , Weyl algebra

THE WEYL ALGEBRAS

David Cock

Supervisor: Dr. Daniel Chan

School of Mathematics, The University of New South Wales.

November 2004

Submitted in partial fulfillment of the requirements of the degree of Bachelor of Science with Honours Contents

Chapter 1 Introduction 1

Chapter 2 Basic Results 3

Chapter 3 Gradings and Filtrations 16

Chapter 4 Gelfand-Kirillov Dimension 21

Chapter 5 Automorphisms of A1 29

References 53

i Chapter 1

Introduction

An important result in single-variable calculus is the so-called product rule. That is, for two polynomials (or more generally, functions) f(x), g(x): R → R:

δ δ δ (fg) = ( f)g + f( g) δx δx δx

It turns out that this formula, which is firmly rooted in calculus has very interesting algebraic properties. If k[x] denotes the ring of polynomials in one variable over a characteristic 0 field k, differentiation (in the variable x) can be considered as a map δ : k[x] → k[x]. It is relatively straighforward to verify that the map δ is in fact a k-linear vector space endomorphism of k[x]. Similarly, we can define another k-linear endomorphism X by left multiplication by x ie. X(f) = xf. Consider the expression (δ · X)f(x). Expanding this gives:

(δ · X)f(x) = δ(xf(x)) applying the product rule gives

(δ · X)f(x) = δ(x)f(x) + xδf(x)

= f(x) + (X · δ)f(x)

1 noting the common factor of f(x) gives us the relation (this time in the ring of k-linear endomorphisms of k[x]):

δ · X = X · δ + 1 where 1 is the identity map. This is the defining relation of the first Weyl algebra which can be viewed as the ring of differential operators on k[x] with polynomial coefficients. There also exist higher order Weyl algebras related to the polynomial ring in n variables. The Weyl algebras arise in a number of contexts, notably as a quotient of the universal enveloping algebra of certain finite-dimensional Lie algebras (arising from the Heisenberg group) which have links to quantum mechanics. The second chapter of this paper covers some basic results on the Weyl algebras, culminating in the proof that they are simple domains. The third chapter covers gradings, filtrations and the concept of an associated graded algebra. The fourth chapter introduces the concept of the Gelfand-Kirillov dimension which is a useful invariant of finitely-generated associative algebras. The final chapter is an exposition of a proof published in [1] that characterises the automorphisms of the first Weyl algebra.

2 Chapter 2

Basic Results

In the following, k will always be a ﬁeld of characteristic 0 and all ideals are two- sided unless speciﬁcally stated otherwise.

Definition 2.1. Let D be a (not neccessarily commutative) domain. Define A(D) as the non-commutative algebra over D on the two generators p, q with defining relation qp − pq = 1 (2.1) ie. D < p, q > A(D) = (qp − pq − 1)

For a field k of characteristic 0, define the first Weyl algebra over k, denoted

th by A1 to be A(k). Deﬁne the n Weyl algebra for n > 1 by An = A(An−1)

(note that this deﬁnition assumes that An−1 is a domain, this is proved later). For convenience assume A0 = k. Note that for n > 1 there are extra (implicit) relations: qipj − pjqi = 0 for i 6= j ie. the generators of diﬀerent index commute.

δf Definition 2.2. Define linear maps X, δ : k[x] → k[x] by X(f) = xf and δ(f) = δx ie. formal differentiation. X and δ generate a sub-algebra of the ring of k-linear endomorphisms of k[x]. Applying Leibniz’ rule for the differentiation of a product

0 gives δ · X = X · δ + 1. Call this algebra A1.

For n > 1 and 1 ≤ i ≤ n deﬁne linear maps Xi, δi : k[x1, . . . , xn] → k[x1, . . . , xn]

δf by Xi(f) = xif and δi(f) = ie. formal partial diﬀerentiation with respect to xi. δxi

3 Once again, diﬀerenting the product yields the relations δiXj = Xjδi + 1 if i = j or

0 δiXj = Xjδi if i 6= j. Call this algebra An. Expressed as a quotient:

0 k < X1,...,Xn, δ1, . . . , δn > An = (δiXj − Xiδj − ∆ij) where   1 i = j ∆ij =  0 otherwise

P i j Lemma 2.3. For any domain D, every x ∈ A(D) can be expressed as aijp q for some ﬁnite set {(i, j) ∈ N × N} and aij ∈ D.

Proof. Since p, q generate An over D, every x ∈ An can be expressed as some ﬁnite sum

X r(i,1) s(i,1) r(i,n ) s(i,n ) bip q . . . p i q i i

+ where bi ∈ D, ni ∈ Z and the leading or trailing coeﬃcent (r(i,1) and s(i,ni) respectively) may be 0. Note that p and q both commute with elements of the base domain D.

For a monomial product term M, deﬁne #p(M) to be the number of p terms appearing in M. Deﬁne #q(M) similarly. Let I(M) be the number of ‘inversions’ in the term M. That is, the sum over every q term in M of the number of p terms which occur to the right. For example:

I(pmqn) = I(λ ∈ k) = 0

I(qp) = 1

I(q2p) = I(qp2) = 2

I(q2p2) = I(qp4) = 4

P Deﬁne I ( i Mi) to be maxi (I(Mi)). P Let R = i Mi be a represention of x in the form described above. If I(R) > 0, then for at least one monomial term Mi we must have I(Mi) > 0. Thus the

4 monomial Mi must contain at least one factor of the form qp ie. Mi = AqpB where A may be in k and B may be 1. Pick one such term and apply the identity qp = pq + 1 to give:

0 Mi = biApqB + biAB calculating gives:

0 I(Mi ) = max(I(ApqB),I(AB))

= max(I(mi) − 1,I(mi) − (#q(A) + #p(B)) − 1)

0 clearly therefore, I(Mi ) = I(Mi) − 1. 0 Inductively therefore, the sequence of manipulations Mi → Mi must terminate ∗ ∗ in some Mi with I(Mi ) = 0. Applying this to each term of R gives a representation in the required form.

Corollary 2.3.1. Any x ∈ An can be expressed as

X i1 in j1 jn ai1...inj1...jn p1 . . . pn q1 . . . qn

Proof. Since k is a domain, the result is true for n = 1. An is deﬁned recursively as

A(An−1). Assuming that An−1 is a domain (again, this is proved shortly) and that the result holds for n − 1, the result follows by induction on n since the generators of diﬀerent index commute.

0 P i1 in i1 in Lemma 2.4. Every x ∈ An can be expressed as aX1 ...Xn δ1 . . . δn for some ﬁnite set {(i, j) ∈ N × N}.

0 0 Proof. By writing An recursively as An−1 < Xn, δn > and using the relation δnXn =

Xnδn + 1, the result follows as for (2.3.1).

0 Lemma 2.5. The k-linear map φ : An → An deﬁned by φ(pi) = Xi and φ(qi) = δi is an algebra homomorphism.

5 Proof. By the universal property, it suﬃces to check the images of the deﬁning relations qipi − piqi − 1 and qipj − pjqi for i 6= j:

φ(qipi − piqi − 1) = δiXi − Xiδi − 1 = 0

φ(qipj − pjqi) = δiXj − Xjδi = 0

0 Lemma 2.6. For any element of An, the representation given in lemma 2.4 is unique.

0 Proof. Suppose that an element x ∈ An has two distinct representations

X i1 in j1 jn ai1...inj1...jn X1 ...Xn δ1 . . . δn

X i1 in j1 jn = bi1...inj1...jn X1 ...Xn δ1 . . . δn

Cancel all equal terms in the above sums to give two diﬀerential operators (A, B)

with coeﬃcients ai1...inj1...jn and bi1...inj1...jn such that ai1...inj1...jn 6= bi1...inj1...jn for all

∗ i1 . . . inj1 . . . jn . For each 1 ≤ k ≤ n, pick jk to be minimal with respect to the

∗ ∗ ∗ ∗ property that ai1...inj1 ...jk jk+1...jn and bi1...inj1 ...jk jk+1...jn appear as coeﬃcients of A ∗ ∗ j1 jn and B respectively, for some i1 . . . in, jk+1 . . . jn. Let p = x1 . . . xn ∈ k[x1, . . . , xn]. Apply the operators A and B to p.

j∗ ∗ X i1 in j1 jn 1 jn Ap = ai1...inj1...jn X1 ...Xn δ1 . . . δn x1 . . . xn

j∗ ∗ X i1 in j1 jn 1 jn Bp = bi1...inj1...jn X1 ...Xn δ1 . . . δn x1 . . . xn

Consider a single term of the above sums:

j∗ ∗ i1 in j1 jn 1 jn ta = ai1...inj1...jn X1 ...Xn δ1 . . . δn x1 . . . xn

j∗ ∗ i1 in j1 jn 1 jn tb = bi1...inj1...jn X1 ...Xn δ1 . . . δn x1 . . . xn

6 ∗ If for all 0 ≤ k ≤ n, jk = jk , then

∗ ∗ i1 in ta = ai1...inj1...jn (j1 ! . . . jn !)(x1 . . . xn )

∗ ∗ i1 in tb = bi1...inj1...jn (j1 ! . . . jn !)(x1 . . . xn )

∗ Suppose that the jk and jk diﬀer for some set of indices. Let l be the smallest such

∗ ∗ index. By the choice of the jk , we must have jl > jl . Since t contains a factor

∗ jl jl δl xl , t = 0. Therefore,

X ∗ ∗ i i ∗ ∗ 1 n Ap = ai1...inj1 ...jn (j1 ! . . . jn !)(x1 . . . xn )

X ∗ ∗ i i ∗ ∗ 1 n Bp = bi1...inj1 ...jn (j1 ! . . . jn !)(x1 . . . xn )

Since the above are simply polynomials in x1, . . . , xn and Ap = Bp, we can equate coeﬃcients which implies that

∗ ∗ ∗ ∗ ai1...inj1 ...jn = bi1...inj1 ...jn which is a contradiction.

Corollary 2.6.1. For any element of An, the representation given in lemma 2.3 is unique.

Proof. This follows from considering the homomorphism φ deﬁned above. If x ∈ An has distinct representations

X i1 in i1 in ai1...inj1...jn p1 . . . pn q1 . . . qn and

X i1 in i1 in bi1...inj1...jn p1 . . . pn q1 . . . qn

7 then

X i1 in i1 in φ ai1...inj1...jn p1 . . . pn q1 . . . qn

X i1 in i1 in = ai1...inj1...jn X1 ...Xn δ1 . . . δn and

X i1 in i1 in φ bi1...inj1...jn p1 . . . pn q1 . . . qn

X i1 in i1 in = bi1...inj1...jn X1 ...Xn δ1 . . . δn

0 are distinct representations of φ(x) ∈ An, a contradiction.

0 Lemma 2.7. An ' An

0 0 Proof. Take the homomorphism φ as above. By Lemma 2.4, any x ∈ An can be expressed in the form

X i1 in j1 jn ai1...inj1...jn X1 ...Xn δ1 . . . δn

Let

X i1 in j1 jn x = ai1...inj1...jn p1 . . . pn q1 . . . qn

Clearly, φ(x) = x0. Therefore φ is surjective.

Take y ∈ An and suppose φ(y) = 0. By lemma 2.3, write y as

X i1 in j1 jn ai1...inj1...jn p1 . . . pn q1 . . . qn

The image, φ(y), is therefore

X i1 in j1 jn ai1...inj1...jn X1 ...Xn δ1 . . . δn

Since the representation of φ(y) is unique, we can equate coeﬃcients which implies

that all of the ai1...inj1...jn are zero. Thus φ is injective.

8 Lemma 2.8. For i, j ∈ Z+ and p, q the generators for A(D):

qpiqj = piqj+1 + ipi−1qj

Proof. Reduce the expression step by step as in Lemma (2.3):

qpiqj = (qp)pi−1qj

= pqpi−1qj + pi−1qj

= p(qp)pi−2qj + pi−1qj

= p2qpi−2q2 + 2pi−1qj

= ...

= piqj+1 + ipi−1qj

At each step, the left-hand q is moved past one of the p factors to the right, adding the term pi−1qj. Recursively applying this step i times gives the required result.

Lemma 2.9. For i, j, l, m ∈ Z+ and p, q as above:

    j j l i j l m X i+l−r j+m−r (p q )(p q ) = r!     p q (2.2) r=0 r r

Proof. Consider the product qjplqm. By the above lemma,

qjplqm = qj−1(plqm+1 + lpl−1qm)

= qj−1plqm+1 + lqj−1pl−1qm

Let

α(qjplqm) = qj−1plqm+1 and

β(qjplqm) = lqj−1pl−1qm

9 and note that both α and β both reduce the degree of qj by one, and that if l = 0 then β(qjplqm) = 0. The above relation then becomes

qjplqm = α(qjplqm) + β(qjplqm) inductively, this must terminate after j steps giving 2j (not neccessarily distinct) terms of a form similar to αβ . . . α(qjplqm). | {z } j Note that αβ(qjplqm) = βα(qjplqm). By commuting these operations, we can write the original expression in the form

  j j j l m X j−r r j l m q p q =   α β (q p q ) r=0 r

j   X j l! = pl−rqm+(j−r)   (l − r)! r=0 r     j j l X l−r j+m−r = r!     p q r=0 r r and thus,

(piqj)(plqm) = pi(qjplqm)       j j l i X l−r j+m−r = p  r!     p q  r=0 r r     j j l X i+l−r j+m−r = r!     p q r=0 r r

P i j Deﬁnition 2.10. Take x ∈ A(D) such that x = ai,jp q . Deﬁne the degree of x, denoted deg(x) to be max (i + j) i,j|ai,j 6=0

10 i j Deﬁne deg(0) = −∞. Deﬁne the leading terms of x to be the terms ai,jp q such that i + j = deg(x) ie. the terms of maximal degree.

Lemma 2.11. Take x, y ∈ A(D). Then deg(xy) = deg(x) + deg(y).

P i j P l m Proof. Let x = i,j ai,jp q and let y = l,m bl,mp q . Expanding the product xy gives:

X X i j l m xy = ai,jbl,mp q p q i,j l,m Applying lemma (2.9) gives:

      j j l X X X i+l−r j+m−r xy = ai,jbl,m  r!     p q  i,j l,m r=0 r r

The leading terms of this sum are those of the form:

i+l j+m ai,jbl,mp q i + j = deg(x), l + m = deg(y) deg(xy) is therefore i + j + l + m = deg(x) + deg(y).

Corollary 2.11.1. An is a domain.

Proof. If x, y ∈ A(D) are non-zero, then deg(x), deg(y) ≥ 0. Thus deg(xy) = deg(x) + deg(y) ≥ 0. Therefore xy 6= 0 and hence A(D) is a domain. Applying the above inductively with k as the base gives the result.

Deﬁnition 2.12. Let R be a ring. A map δ : R → R is called a derivation if it satisﬁes δ(ab) = δ(a)b + aδ(b).

11 Example 2.13. For a ring R, deﬁne ad c by [c, −]. That is, (ad c)(x) = cx − xc. Observe that

(ad c)(x)y + x(ad c)(y) = (cx − xc)y + x(cy − yc)

= cxy − xcy + xcy − xyc

= cxy − xyc

= (ad c)(xy)

Thus, ad c is derivation on R.

Deﬁnition 2.14. A derivation δ is called an inner derivation if there exists a c ∈ R such that for all a ∈ R, δ(a) = ca − ac. Note that if a is in the centre of R, then δ(a) = 0. An ideal I ⊂ R is called a δ-ideal if δ(I) ⊆ I. The ring R is called δ-simple if the only δ-ideals of R are (0) and R.

Deﬁnition 2.15. For a ring R and a derivation δ on R, deﬁne R[x; δ] to be the set

P i of left polynomials aix with the relation ax = xa + δ(a) for a ∈ R extending (by repeated application) to give a multiplication on R[x; δ]. The proof of the fact that this is an associative multiplication which gives a ring structure on R[x; δ] is referenced in [3].

Lemma 2.16. (Paraphrased from [3]) For a Q-algebra R, if R is δ-simple for a non-inner derivation δ, then R[x; δ] is simple.

Proof. To prove this, we prove a (logically equivalent) partial converse. That is, assume that R is δ-simple, but that R[x; δ] is not simple ie. with some ideal I 6= (0),R and show that δ must be inner. Let n be the minimum degree among non-zero (left) polynomials in I, and let U be the set of leading coeﬃcients a of polynomials f ∈ I of degree n, together with 0. Observing that xf − fx = δ(a)xn + ...

12 shows that δ(U) ⊆ U. Combined with the fact that I is an ideal (and hence ab, a + b ∈ U for any a, b ∈ U) this shows that U is a non-zero δ-ideal of R. Thus since R is δ-simple, 1 ∈ U ie. there exists an element g = xn + dxn−1 + ... in I. For any b ∈ R, it follows inductively that xnb = bxn + nδ(b)xn−1 + ... thus

bg − gb = (bd − db − nδ(b))xn−1 + (terms of lower degree)

Since bg − gb ∈ I (and Q ⊆ R),

d d δ(b) = b − b n n for every b ∈ R. Thus δ is an inner derivation.

δ Lemma 2.17. For δ deﬁned as δx , An = A(An−1) ' An−1[x][y; δ] where x, y commute with all elements of An−1.

Proof. Since every element of An−1 commutes with x, the deﬁning multiplicative relation on An−1[x][y; δ] is simply:

yx = xy + δ(y)

yx = xy + 1

Deﬁne a k-linear map φ : An → An−1[x][y; δ] by ψ(p) = x and ψ(q) = y. The image ψ(qp − pq − 1) is simply yx − xy − 1 = 0. Thus ψ is an homomorphism.

Elements of An−1[x][y; δ] are (by deﬁnition) of the form:

X j pj(x)y

where pj ∈ k[x]. Splitting each term of the sum into monomial components gives an expression of the form:

X i j aijx y

13 Whence it is clear that

X i j X i j ψ aijp q = aijx y

and thus ψ is surjective. Since (as An−1 is a domain) the expressions of the form

X j pj(x)y are unique, ψ(α) = 0 implies α = 0, whence ψ is injective.

δ Lemma 2.18. Deﬁne δ as in lemma 2.17 ie. δ = δx . For a simple ring R, R[x] is δ-simple, and δ is a non-inner derivation on R[x].

Proof. The map δ is defined as formal differentiation on R[x]. Leibniz’s rule for the differentiation of a product says that δ(ab) = δ(a)b + aδ(b) and therefore δ is a derivation. The element x is (by definition) central in R[x], but δ(x) = 1 and therefore δ is not inner. Take any δ-ideal I /R[x] (other than (0)). Take i ∈ I and let d = deg(i). Thus

d d−1 i = adx + ... + a0, with ad 6= 0 and hence δ(i) = dadx + ... + a1. Since I is a δ-ideal, δ(i) ∈ I and deg(δ(i)) = d − 1. Hence by reduction, I contains an element i0 = d!ad 6= 0 with deg(i0) = 0 ie. i0 ∈ R. Thus, since R is simple, 1 ∈ I ie. I = R[x]. Therefore, R[x] is δ-simple

Theorem 2.19. An is simple.

Proof. Deﬁne X and δ as in deﬁnition 2.2. Lemma 2.18 implies that A0[x] = k[x] is δ-simple and δ is a non-inner derivation on A0[x]. Applying lemma (2.16) shows that A1 ' A0[x][y; δ] is simple.

Assume that An−1 is simple for n ≥ 2. Then by 2.18, An−1[x] is δ-simple and δ is an inner derivation on An−1[x]. Thus An ' An−1[x][y; δ] is simple by 2.16.

Remark. As the kernel of a homomorphism is a two-sided ideal, this implies that all endomorphisms of An are either injective, or zero. Whether all endomorphisms

14 are also surjective is an important open question, which if proved, would imply an important result in multi-variable calculus known as the Jacobian conjecture.

15 Chapter 3

Gradings and Filtrations

Deﬁnition 3.1. A k-algebra A is N-graded or simply graded, if there exists a set of subspaces gr(A, i) such that:

gr(A, i).gr(A, j) ⊆ gr(A, i + j) and M A = gr(A, i) i∈N Elements of the nth graded piece gr(A, n) are called homogeneous of degree n.

Example 3.2. The above statement about homogeneous elements suggests a relation to the idea of an homogeneous polynomial. Take the polynomial algebra

P = k[x1, . . . , xn] and let gr(P, i) be the vector space generated over k by the monomials of degree i in the variables x1, . . . , xn for i ≥ 0 ie. homogeneous polynomials L of degree i. Clearly gr(P, i).gr(P, j) = gr(P, i + j) for 0 ≤ i, j and P = i gr(P, i). This is called the grading by degree and the ith graded component is commonly denoted k[x1, . . . , xn]i. Take k[x, y] graded by degree. Take two elements a, b and consider their de- P P composition into homogeneous components a = i≥0 ai and b = j≥0 bj with ai ∈ k[x, y]i and bj ∈ k[x, y]j. The product ab can therefore be decomposed as a sum of the products of homogeneous terms:

X ab = aibj i,j≥0

16 Note that the (i, j)th term of the above sum has degree i + j. This gives a homogeneous decomposition of the product:

X X ab = aibj n≥0 i+j=n

Definition 3.3. A k-algebra A is N-filtered or simply filtered, if there exists a set of subspaces fp(A, i) such that:

fp(A, i) ⊆ fp(A, i + 1), ∀i ∈ N 1 ∈ fp(A, 0)

fp(A, i).fp(A, j) ⊆ fp(A, i + j) [ A = fp(A, i) i∈N

Example 3.4. Drawing on Lemma (2.3), any element of the algebra A1 can be

P mi ni expressed as x = i aip q . Defining deg(x) = maxi{mi +ni} suggests a filtration of A1 by degree. Define the nth filtered part fp(A1, n) by {x ∈ A1 : deg(x) ≤ n} S ie. not neccessarily homogeneous elements of degree ≤ n. Clearly A1 = i fp(A1, i) and 1 ∈ fp(A1, 0). Lemma (2.11) implies that fp(A1, i).fp(A1, j) ⊆ fp(A1, i + j).

Definition 3.5. Given a filtered algebra A with filtered pieces fp(A, i), define the associated graded algebra gr(A) corresponding to this filtration to be:

M fp(A, i) gr(A) = fp(A, i − 1) i deﬁning fp(A, −1) = 0 and with multiplication deﬁned on (left)-cosets by:

[x + fp(A, i − 1)] · [y + fp(A, j − 1)] = [xy + fp(A, i + j − 1)] and extended component-wise to multiplication on gr(A).

Lemma 3.6. For a ﬁltered algebra A, gr(A) is an algebra.

17 Proof. • The coset [1+fp(A, −1)] contains only the element 1, which acts as the multiplicative identity since [x + fp(A, i − 1)] · [1 + fp(A, −1)] = [x + fp(A, i + 0 − 1)]. • Likewise, the coset [0 + fp(A, −1)] contains only the element 0, which acts as the additive identity. • We need to check that the multiplication is well deﬁned. Take the cosets X = [x+fp(A, m−1)] and Y = [y +fp(A, n−1)] and take the elements a ∈ X and b ∈ Y. Note that x ∈ fp(A, m) and y ∈ fp(A, n) and write the elements as:

a = x + a0 a0 ∈ fp(A, m − 1)

b = y + b0 b0 ∈ fp(A, n − 1)

and form the product:

ab = xy + xb0 + a0y + a0b0

Applying the rule for multiplication of ﬁltered pieces gives:

xy ∈ fp(A, m + n)

xb0 ∈ fp(A, m + n − 1)

a0y ∈ fp(A, m + n − 1)

a0b0 ∈ fp(A, m + n − 2)

Thus, ab ∈ [xy+fp(A, m+n−1)]. Thus multiplication in gr(A) is well-deﬁned. • The other ring axioms follow easily from the deﬁnition.

18 Lemma 3.7. The associated graded algebra of A1 is k[p, q], the commutative polynomial ring on the variables p and q.

Proof. Take the filtration by degree defined in Example (3.4). Let gr(A) be the associated graded algebra corresponding to this filtration. Take two homogeneous elements of gr(A),

0 i i 0 x = [a0p q + ··· + aip q + fp(A, i − 1)] and

0 j j 0 y = [b0p q + ··· + bip q + fp(A, j − 1)]

Where as usual, some of the coeﬃcients a or b may be 0. Take the product xy:

0 i i 0 0 j j 0 xy = [(a0p q + ··· + aip q )(b0p q + ··· + bip q ) + fp(A, i + j − 1)]

Consider a single term of the product:

m i−m n j−n m i−m n j−n (amp q )(bnp q ) = ambnp q p q

Reducing this expression as in 2.3 gives a representation of this term as

m+n i+j−m−n ambnp q + (terms of lower degree)

Thus the product can be expressed as

X m+n i+j−m−n ambnp q + (terms of lower degree) m,n

Note that all terms in the left hand sum have degree i + j. Therefore as a member of the coset xy + fp(A, i + j − 1) this is simply

" # X m+n i+j−m−n ambnp q + fp(A, i + j − 1) m,n

19 Thus multiplication of homogeneous elements behaves exactly as in k[p, q] graded by degree. Since gr(A) is the direct sum of these homogeneous components, the multiplication extended to the whole ring is also identical to that in k[p, q]. This allows us to deﬁne a homomorphism φ : gr(A) → k[p, q] deﬁned on the graded components of gr(A) by mapping [x + fp(A, i − 1)] to the leading terms of the unique representation of x considered as polynomials in k[p, q]i and extended component-wise to gr(A). This is clearly an homomorphism as shown above. It is also easy to show that it is bijective.

Deﬁnition 3.8. A ﬁltered algebra A for which the associated graded algebra gr(A) is commutative (eg. An) is called an almost-commutative algebra.

Proposition 3.9. Any ﬁnitely-generated almost-commutative algebra A is both left and right noetherian.

Proof. See [2] proposition 7.1.

Corollary 3.9.1. An is both left and right noetherian.

Remark. It has been proved that an algebra (over k) is almost commutative if and only if it is a homomorphic image of the universal enveloping algebra of some ﬁnite dimensional Lie algebra over k.

20 Chapter 4

Gelfand-Kirillov Dimension

Let A be a ﬁnitely generated k-algebra. Let V be a ﬁnite-dimensional generating subspace for A. That is, if

2 fp(A, n) = k + V + V + ... + Vn then: [ A = fp(A, n) n≥0 Note that this is a ﬁltration of A with ﬁltered pieces fp(A, n).

Deﬁne the function dV (n): N → R by

dV (n) = dimk(fp(A, n))

Note that if the algebra A is ﬁnite dimensional, then there exists some N such that for all n > N, fp(A, n) = fp(A, N) ie. the function dV (n) becomes stationary.

Note that the function dV (n) depends on A and the choice of generating subspace V . It would be nice to remove this dependence to get an invariant of the algebra A. This is done with the following construction.

The idea here is to form an equivalence relation on the functions dV (n) by comparing their asymptotic growth rate. This turns out to be exactly the right deﬁnition to avoid the dependence on the choice of generating subspace.

21 Deﬁnition 4.1. Let Φ denote the set of eventually non-decreasing positive valued functions f : N → R ie. those for which there exists an n0 ∈ N such that for all n, f(n) ≥ 0 and for all n ≥ n0 f(n + 1) ≥ f(n)

Define a relation ≤∗ on Φ by setting f ≤∗ g iff there exist c ∈ R, m ∈ N such that for all n sufficiently large,

f(n) ≤ cg(mn)

Lemma 4.2. The relation ≤∗ is a preorder relation on the set Φ.

Proof. • (reﬂexive) Taking c = m = 1 gives f ≤∗ f. • (transitive) Take f, g, h ∈ Φ and assume that f ≤∗ g and g ≤∗ h ie.

f(n) ≤ c0g(m0n) ∀n > n0

g(n) ≤ c1h(m1n) ∀n > n1

Let n2 = max(n0, n1). Then since f and g are non-decreasing,

f(n) ≤ c0c1h(m0m1n) ∀n > n2

Therefore f ≤∗ h.

Deﬁne an equivalence relation on Φ by f ∼ g iﬀ f ≤∗ g and g ≤∗ f. Denote the partial order induced on the quotient Φ/ ∼ by ≤. For an f ∈ Φ, the equivalence class G(f) ∈ Φ/ ∼ is called the growth of f.

Lemma 4.3. [2] Let A be a ﬁnitely generated k-algebra with ﬁnite dimensional generating subspaces V and W . If dV (n) and dW (n) denote the dimensions of

Pn i Pn i i=0 V and i=0 W , respectively, then G(dV ) = G(dW ).

22 Proof. Since ∞ ∞ [ [ A = (V 0 + ... + V n) = (W 0 + ... + W n) n=0 n=0 there exist positive integers s and t such that

s t X X W ⊆ V i and V ⊆ W i i=0 i=0

Thus dW (n) ≤ dV (sn) and dV (n) ≤ dW (tn), whence dV ∼ dW .

Thus the growth of an algebra A, deﬁned to be G(dV ), is independent of the choice of generating subspace V .

Example 4.4. Let A = k[p, q], the commutative polynomial ring in two variables.

Take the generating subspace V = kp + kq. Take the basis B1 = {p, q} for V . Let

n Bn be the corresponding basis for V (formed as the product B1 · Bn−1). Assume that the basis Bn−1 consists of all monomials of degree n − 1 ie.

n−1 0 0 n−1 Bn−1 = p q , . . . , p q

Then calculating Bn simply gives

n 0 1 n−1 n−1 1 0 n n 0 0 n Bn = p q , . . . , p q , p q , . . . , p q = p q , . . . , p q

Thus, inductively, V n is the space spanned by all (commutative) monomials of S degree n. The set Bn is linearly independent and hence

n X n dV (n) = dimk V i=0 n X n = dimk V i=0 1 = (n + 1)(n + 2) 2

23 2 Thus G(dV ) = G(n ) ie. the polynomial algebra in 2 variables has quadratic growth. This example extends simply to show that the polynomial algebra in m variables has degree m polynomial growth. It will be interesting at this point to try to calculate the growth of the ﬁrst Weyl algebra A1. To do so we will need the following lemma:

i j Lemma 4.5. Let Mn ⊂ A1 be the subspace spanned by the monomials p q with i + j = n. Let V = M1 ie. kp + kq, then for n ≥ 2,

n−1 b 2 c n M V = Mn−2i i=0

Proof. As above, let Bn be a basis for Vn with B1 = {p, q}. Let Sn = B1 · Bn−1. Calculating for n = 2 gives:

S2 = B1 · B1

= {p2, pq, qp, q2}

= {p2, pq, pq + 1, q2}

This reduces to the basis:

2 2 B2 = {p , pq, q , 1}

Which proves the result for n = 2. Assume that the result holds for n − 1. That is,

n−1 b 2 c [ j ((n−1)−2i)−j Bn−1 = {p q : 0 ≤ j ≤ (n − 1) − 2i} i=0

24 Calculating the product gives

n−1   b 2 c j+1 ((n−1)−2i)−j [ {p q : 0 ≤ j ≤ (n − 1) − 2i} ∪ Sn =   j ((n−1)−2i)−j i=0 {qp q : 0 ≤ j ≤ (n − 1) − 2i}

Applying Lemma (2.8) to the appropriate terms in the above expression (those qplqm with l > 0) gives

 j+1 ((n−1)−2i)−j  b n−1 c {p q : 0 ≤ j ≤ (n − 1) − 2i} ∪ 2   [  j (n−2i)−j j−1 ((n−1)−2i)−j  Sn =  {p q + jp q : 1 ≤ j ≤ (n − 1) − 2i} ∪  i=0   {qn−2i}

For a given i, the ﬁrst and third terms in the above expression contain all monomials of degree n−2i. The elements of the second term are simply all monomials of degree n − 2i − 2 added to some monomial of degree n − 2i. They are therefore linearly dependent on the other elements of Sn and can be ignored except in the case where

n−1 n is even and i = 2 . In this case, it is easily veriﬁed that the only element of n−1 the second term in pq + 1 which, since pq is in the span when i = 2 − 1, adds

1 to the span of Sn. Thus inductively, n−1 b 2 c n M V = span(Sn) = Mn−2i i=0 Note that for all n >= 0, V n ⊂ V n+1.

Example 4.6. The subspace V = kp + kq is clearly a generating subspace for A1. Deﬁne

n [ i fp(A1, n) = V i=0

25 n Since V contains all monomials of degree n, fp(A1, n) is just the span of all monomials of degree ≤ n.

n M fp(A1, n) = Mi i=0

The dimension dim(Mi) is simply i + 1. Therefore dV (n) is simply

dV (n) = dim(fp(A1, n)) n ! M = dim Mi i=0 n X = (i + 1) i=0 1 = n(n + 1) + n 2

2 Thus G(dV (n)) = G(n ) ie. the growth of A1 is quadratic just as for k[p, q], its associated graded algebra. This is not a coincidence, and is an example of a more general result.

Example 4.7. Let Fn = k < x, y > be the free algebra on 2 variables. Let

V = kx + ky be a generating subspace for Fn. Deﬁne a ﬁltration of Fn by

[ Fn = fp(Fn, i) i≥0 where

i X j fp(Fn, i) = V j=0

26 It is not hard to see that V j has dimension 2j and that the above sum is direct. Thus

n X i dV (n) = 2 i=0 = 2n+1 − 1

n Therefore G(dV ) = G(2 ) ie. exponential growth. The growth is an important invariant of an algebra. It can, however, be some- what unwieldy to calculate in practice. Looking at the deﬁnition of the equivalence relation on growth functions, what we are really interested in is the asymptotic growth of the algebra. This is formalised in the deﬁnition of the Gelfand-Kirillov dimension.

Deﬁnition 4.8. The Gelfand-Kirillov dimension of a k-algebra A is

GKdim(A) = sup lim logn dV (n) V

Where lim denotes the limit superior and the supremum supV is taken over all ﬁnitely-generated subspaces of A.

Lemma 4.9. [2] Take f, g ∈ Φ (two eventually non-decreasing functions N → R). The following hold:

ρ • lim logn f(n) = inf{ρ ∈ R : G(f) ≤ G(n )}

• If G(f) = G(g) then lim logn f(n) = lim logn g(n)

ρ Proof. Let r denote lim logn f(n) and s denote inf{ρ ∈ R : G(f) ≤ G(n )}.

The ﬁrst part of the preceeding lemma shows that for an algebra with polynomial growth, say G(na), then the Gelfand-Kirillov dimension of the algebra is the polynomial degree, a. It also shows that for any algebra with super-polynomial growth eg. the free algebra with exponential growth, the Gelfand-Kirillov dimension is inﬁnite.

27 Note that in a previous lemma, we showed that for two generating subspaces V and W , G(dV (n)) = G(dW (n)). Thus for any generating subspace V , we can drop the supremum and write

GKdim(A) = lim logn dV (n)

The previous lemma also shows that the Gelfand-Kirillov dimension gives an equivalence which is no ﬁner than that given by the growth.

Remark. It has been proven (see [2] Chapter 2) that the range of possible values for the Gelfand-Kirillov dimension of an algebra is {0} ∪ {1} ∪ [2, ∞).

Remark. The Gelfand-Kirillov dimension can also be deﬁned in a natural way for modules over an algebra. The study of modules over the Weyl algebras is a par- ticularly interesting application. It turns out that the minimum Gelfand-Kirillov

th dimension for a module over the n Weyl algebra An is 2n. Modules of this minimal dimension are known as holonomic modules and are linked with holonomic systems of linear diﬀerential equations.

Proposition 4.10. If A is an almost-commutative algebra with associated graded algebra gr(A) then GKdim(A) = GKdim(gr(A)).

Proof. See [2] proposition 6.6.

28 Chapter 5

Automorphisms of A1

In this chapter, we show that all automorphisms of the ﬁrst Weyl algebra, A1, are

0 generated by a set of automorphisms Φn,λ, Φn,λ which are deﬁned shortly. This proof originally appears in the paper [1] by Dixmier. At this point, we need to deﬁne a number of concepts which are used in the argument that follows.

P i j Deﬁnition 5.1. If f = aijx y ∈ k[x, y], denote by E(f) the set of pairs (i, j) such that aij 6= 0. If ρ, σ are real numbers, deﬁne

vρ,σ(f) = sup (ρi + σj) (i,j)∈E(f)

(for convenience deﬁne vρ,σ(0) = −∞). Denote by E(f, ρ, σ) the set of pairs (i, j) ∈

E(f) such that ρi + σj = vρ,σ(f). If f 6= 0, we have E(f, ρ, σ) 6= ∅. If E(f) =

E(f, ρ, σ), we say that f is (ρ, σ)-homogeneous of (ρ, σ)-degree vρ,σ(f).

Remark. The above gives a grading of k[x, y] by (ρ, σ)-degree. This fact is not required and the proof is omitted, but it does demonstrate the fact that there exist more general gradings than those considered earlier.

P i j Definition 5.2. Suppose a = aijp q ∈ A1, and σ, ρ are real numbers. By analogy with the previous definition (5.1), define E(a), vρ,σ(a) and E(a, ρ, σ). The polynomial

X i j aijx y ∈ k[x, y] (i,j)∈E(a,ρ,σ)

29 is called the (ρ, σ)-associated polynomial of a.

Remark. The previous deﬁnition is an example of an associated graded algebra of

A1 corresponding to a more general ﬁltration than that previously considered.

Lemma 5.3. Let f ∈ k[x, y] be a (ρ, σ)-homogeneous polynomial of (ρ, σ)-degree v. Then,

δf δf • ρx δx + σy δy = vf. • If ρ and σ are linearly independent over Q, f is a monomial.

Proof. Let g = xiyj with ρi + σj = v. We have

δg δg ρx + σy = ρxixi−1yj + σyjxiyj−1 δx δy = (ρi + σj)g and hence the ﬁrst result. If (i, j) ∈ E(f) and (i0, j0) ∈ E(f), we have ρi + σj = ρi0 + σj0 = v, thus ρ(i − i0) = σ(j0 − j). Since ρ and σ are linearly independent, we must have i = i0 and j = j0, whence f is a monomial.

Lemma 5.4. Let f, g ∈ k[x, y] be (ρ, σ)-homogeneous polynomials of (ρ, σ) degrees v, w. Then the following hold,

δf δg δf δg δf δg σy − = wg − vf δx δy δy δx δx δx

Moreover, if both v and w are integers,

δf δg δf δg δ σy − = f −w+1gv+1 (g−vf w) δx δy δy δx δx

30 2.

δf δg δf δg δf δg −ρx − = wg − vf δx δy δy δx δy δy

Moreover, if both v and w are integers,

δf δg δf δg δ −ρx − = f −w+1gv+1 (g−vf w) δx δy δy δx δy

Proof. Applying lemma 5.3 part 1,

δf δg δf δg δf δg δf δg σy − = wg − ρx − vf − ρx δx δy δy δx δx δx δx δx δf δg = wg − vf δx δx

Suppose that v and w are integers. Then

δ δg δf (g−vf w) = g−v−1f w−1 −v f + gw δx δx δx thus, taking into account the preceeding,

δf δg δf δg δ σy − = f −w+1gv+1 (g−vf w) δx δy δy δx δx

Lemma 5.5. Take i, j, l, m integers ≥ 0. Then

1 (piqj)(plqm) = pi+lqj+m + jlpi+l−1qj+m−1 + j(j − 1)l(l − 1)pi+l−2qj+m−2 2! 1 + j(j − 1)(j − 2)l(l − 1)(l − 2)pi+l−3qj+m−3 + ... 3!

Proof. This follows by expanding the expression derived in 2.9.

31 Lemma 5.6. Take a, b, c ∈ A1 with c = ab. Suppose

X i j X i j X i j a = aijp q b = bijp q c = cijp q

Let

X i j X i j X i j f = aijx y g = bijx y h = cijx y

Then

δf δg 1 δ2f δ2g 1 δ3f δ3g h = fg + + + + ... δy δx 2! δy2 δx2 3! δy3 δx3

Proof. It suﬃces to prove for a = piqj, b = plqm, which follows from lemma 5.5.

Lemma 5.7. Take a, b, c ∈ A1 such that c = [a, b]. Suppose

X i j X i j X i j a = aijp q b = bijp q c = cijp q and let

X i j X i j X i j f = aijx y g = bijx y h = cijx y then

δf δg δf δg 1 δ2f δ2g δ2f δ2g 1 δ3f δ3g δ3f δ3g h = − + − + − + ... δx δy δy δx 2! δx2 δy2 δy2 δx2 3! δx3 δy3 δy3 δx3

Proof. This follows directly from lemma 5.6

Lemma 5.8. Take a and b non-zero elements of A1, and ρ, σ real numbers such that

ρ + σ > 0. Let v = vρ,σ(a) and w = vρ,σ(b). Let f1 and g1 be the (ρ, σ)-associated polynomials of a and b.

(i) There exists a pair (t, u) of elements of A1, possessing the following properties: (a) [x, y] = t + u

32 (b) E(t) = E(t, ρ, σ) and vρ,σ(t) = v + w − (ρ + σ)

δf1 δg1 δf1 δg1 (ii 2) δx δy − δy δx = 0 v w (ii 3) If v and w are integers, g1 is a multiple of f1

δf1 δg1 δf1 δg1 (iii) If t 6= 0, the (ρ, σ)-associated polynomial of [x, y] is δx δy − δy δx .

δf1 δg1 δf1 δg1 Proof. Introduce the notation of lemma 5.7. Then h is the sum of δx δy − δy δx , which is (ρ, σ) homogeneous of (ρ, σ) degree v + w − (ρ + σ), and a polynomial h∗ such that

∗ vρ,σ(h ) < v + w − (ρ + σ)

This proves (i), (iii) and the equivalence (ii 1)⇔(ii 2). If v and w are integers, the equivalence (ii 2)⇔(ii 3) follows from lemma 5.4.

Deﬁnition 5.9. For a ﬁeld k, denote the algebraic closure of k by k. For an algebra

A over k, denote the algebra A ⊗k k by A.

Let A be an algebra over k, and take a ∈ A. For all y ∈ A, deﬁne Vy =

P n n≥0 k(ad a) y. Denote by F (a; A), or F (a), the set of y ∈ A such that dim Vy < ∞ ie. the set of elements for which the subspace Vy is ﬁnite dimensional. We have

F (x; A) = F (x; A) ⊗k k. If λ ∈ k, denote by F (a, λ; A) the set of y ∈ F (a; A) such that (ad a − λ)ny is zero for n suﬃciently large. If λ ∈ k, deﬁne F (x, λ; A) = F (x, λ; A) ∩ A, such that

F (x, λ; A) = F (x, λ; A) ⊗k k. Denote by N(a; A) or N(a) the set F (a, 0; A). This is the set of y ∈ A such that

(ad a)|V is nilpotent. For n = 0, 1, 2,..., denote by N(a, n; A) or N(a, n) the kernel of (ad a)n+1. Note that N(a, 0) = C(a), the set of elements which commute with a.

33 For λ ∈ k, denote by D(a, λ; A) the set of y ∈ A such that (ad a)y = λy. Deﬁne

M D(x; A) = D(x, λ; A) λ∈k and deﬁne D(x) = D(x; A) = D(x; A) ∩ A.

Lemma 5.10. Take any a ∈ A1, and Φ an automorphism of A1. Then C(Φ(a)) = Φ(C(a)), N(Φ(a)) = Φ(N(a)) and D(Φ(a)) = Φ(D(a)).

Proof. Suppose that b ∈ C(a). That is,

ab − ba = 0 consider Φ(a)Φ(b) − Φ(b)Φ(a).

Φ(a)Φ(b) − Φ(b)Φ(a) = Φ(ab) − Φ(ba)

= Φ(ab − ba)

= Φ(0) = 0

Thus Φ(C(a)) ⊆ C(Φ(a)). Applying the above calculation in the other direction shows that C(Φ(a)) ⊆ Φ(C(a)). Thus Φ(C(a)) = C(Φ(a)). Suppose that b ∈ D(a, λ) that is,

ab − ba = λb consider Φ(a)Φ(b) − Φ(b)Φ(a).

Φ(a)Φ(b) − Φ(b)Φ(a) = Φ(ab − ba)

= Φ(λb)

= λΦ(b)

34 Thus Φ(D(a, λ)) ∈ D(Φ(a, λ)). Once again, applying the calculation in the other direction gives the reverse inclusion. Therefore D(Φ(a)) = Φ(D(a)). It is not hard to see that

Φ((ad a)nb) = (ad Φ(a))nΦ(b) and thus reasoning as for C(a), N(Φ(a)) = Φ(N(a)).

Corollary 5.10.1. If equality holds between any of the N(a), D(a) and C(a), then the same equality holds between the N(Φ(a)), D(Φ(a)) and C(Φ(a)). Likewise if any of the N(a), D(a) and C(a) are equal to any Φ-invariant set (eg. 0 or A1), then the same equality holds for the N(Φ(a)), D(Φ(a)) and C(Φ(a)).

Lemma 5.11. C(p) = k[p]

Proof. Clearly k[p] ⊆ C(p). Take any a ∈ C(p), and let b = [q, a], then

[b, p] = [qa − aq, p]

= (qa − aq)p − p(qa − aq)

= qap − aqp − pqa + paq

= qpa − aqp − pqa + apq since a commutes with p

= pqa + a − apq − a − pqa + apq

= 0 and hence [q, a] ∈ C(p). Suppose there exists some c ∈ C(p) with

X i j c = aijp q with at least one j non-zero. Note that if p commutes with both a and b, then p commutes with a+b. Using this, and the fact that k[p] ⊆ C(p), cancel all terms for

35 which j = 0 to get some c0 ∈ C(p) for which every term contains a positive power of q. Calculate [q, c0]:

[q, c0] = qc0 − c0q

X i j i j+1 = aij(qp q − p q ) since [−, −] is bilinear

X i j+1 i−1 j i j+1 = aij(p q + ip q − p q ) by lemma 2.8

X i−1 j = aijip q

Note that we have decreased the degree in p by one in each term, and every term for which i = 0 becomes zero. Pick i0 to be maximal with respect to the property that ai0,j is non-zero for some j. And apply the above operation i0 times to get

00 another element c ∈ C(p). By the choice of i0,

00 X j c = ai0jq j since terms with a lower power of p will vanish (note that the terms of this sum are not neccessarily unique). Equating coeﬃcients of [p, c00] implies that [p, qj] = 0, which is a contradiction. Therefore c ∈/ C(p) and C(p) = k[p].

Lemma 5.12. Let A be an algebra over k. Take λ ∈ k and a, b ∈ A such that (ad a − λ)2b = 0. Then

1. (ad a − nλ)nbn = n!((ad a − λ)b)n for n = 1, 2, 3,... 2. (ad a − nλ)n+1bn = 0

Proof. The assumption (ad a − λ)2b = 0 can be written as

(ad a − λ)b ∈ D(a, λ)

Since D(a, λ; A).D(a, µ; A) ⊂ D(a, λ + µ; A),

((ad a − λ)b)n ∈ D(a, nλ) for n = 1, 2, 3,... (5.1)

36 Equality 1 is clear for n = 1. Assume it holds for n. Then

(ad a − (n + 1)λ)n+1bn+1 = (ad a − (n + 1)λ)n+1(bn · b)

= ((ad a − nλ)n+1bn)b

+ (n + 1)((ad a − nλ)nbn)((ad a − λ)b) 1 + (n + 1)n((ad a − nλ)n−1bn)((ad a − λ)2b) 2 + ...

= ((ad a − nλ)n+1bn)b

+ (n + 1)((ad a − nλ)nbn)((ad a − λ)b)

Applying equation 5.1 and induction,

(ad a − nλ)n+1bn = (ad a − nλ)(n!((ad a − λ)b)n)

= 0 thus

(ad a − (n + 1)λ)n+1bn+1 = (n + 1)((ad a − nλ)nbn)((ad a − λ)b)

= (n + 1)n!((ad a − λ)b)n((ad a − λ)b)

= (n + 1)!((ad a − λ)b)n+1 and thus we have shown equality 1. Equality 2 is follows from equality 1 and equation 5.1.

Lemma 5.13. Take a ∈ A1. Consider F (a) as a right C(a) module, and suppose that it is ﬁnitely generated over C(a), then F (a) = C(a).

37 Proof. Suppose that N(a) 6= C(a). Let (b1,..., br) be a set of generators for N(a) as a C(a) module. There exists an integer n > 0 such that

n n (ad a) b1 = ... = (ad a) br = 0 and thus (ad a)n(N(a)) = 0. Or, there exists a b ∈ N(a) such that

(ad a)b 6= 0, (ad a)2b = 0

Applying lemma 5.12, (ad a)nbn 6= 0, a contradiction. Suppose that D(a) 6= C(a). Then D(aλ) 6= 0 implies that D(a, nλ) 6= 0 for all integers n > 0. Therefore D(a) is an inﬁnite direct sum of non-zero C(α) modules, which is a contradiction.

Lemma 5.14. Take ρ, σ integers > 0. Take a ∈ A1, b ∈ F (a), v = vρ,σ(a), w = vρ,σ(b), and let f and g be the (ρ, σ)-associated polynomials of a and b respectively. Suppose that v > ρ + σ and that f is not a monomial. Then one of the following is true:

(a) f w is a multiple of gv

σ (b) σ > ρ, σ is a multiple of ρ, and f(x, y) is of the form λxα(x ρ + µy)β, for λ, µ ∈ k, α, β integers ≥ 0

σ (c) ρ > σ, ρ is a multiple of σ, and f(x, y) is of the form λyα(y ρ + µx)β, for λ, µ ∈ k, α, β integers ≥ 0 (d) ρ = σ, and f(x, y) is of the form λ(µx+νy)α(µ0x+ν0y)β, for λ, µ, ν, µ0, ν0 ∈ k, α, β integers ≥ 0

n Proof. Let bn = (ad a) b, for n = 0, 1, 2,.... Then vρ,σ(b0) = w. It is impossible to have vρ,σ(bn) = w + n(v − ρ − σ) for all n (because v − ρ − σ > 0 and b ∈ F (a)).

38 Thus there exists an n ≥ 0 such that

vρ,σ(bm) = w + m(v − ρ + σ) for m ≤ n and

vρ,σ(bn+1) < w + (n + 1)(v − ρ − σ)

Let h be the (ρ, σ)-associated polynomial of bn. Let vρ,σ(bn) = t. Applying

t v lemma 5.8, f is a multiple of h . If n = 0, we have bn = b, h = g and t = w ie. case (a). Suppose from now on that n > 0. Consider bn−1. Let l be the

(ρ, σ)-associated polynomial of bn−1. We have vρ,σ(bn) − vρ,σ(bn−1) = v − ρ − σ, thus

vρ,σ(bn−1) = t − v + ρ + σ

Thus

δf δl δf δl σyh = σy − lemma 5.8 part 3 δx δy δy δx δ = f −t+v−ρ−σ+1lv+1 (l−vf l−v+ρ+σ) lemma 5.4 part 1 δx thus, since f t is a multiple of hv,

δ h v h v+1 f ρ+σ = σy f ρ+σ (5.2) δx fl fl

By utilising lemma 5.4 part 2 instead of part 1, it follows similarly that

δ h v h v+1 f ρ+σ = −ρx f ρ+σ (5.3) δy fl fl

Consider f,h and l as polynomials in x with coeﬃcients in k(y). Take µ ∈ k(y).

h 0 If µ is a zero of fl of order ν > 0 and a zero of f of order ν ≥ 0, the relation 5.2 shows that vν + (ρ + σ)ν0 − 1 = (v + 1)ν + (ρ + σ)ν0, which is impossible. Thus h fl dl is non-zero on k(y). Therefore, h ∈ k(y)[x]. Applying 5.3, we see similarly

39 fl that h ∈ k(x)[y]. Thus there exists a non-zero polynomial m ∈ k[x, y] such that fl = hm. Since f, h and l are (ρ, σ) homogeneous of (ρ, σ) degrees v, t and t − v + ρ + σ, m is (ρ, σ)-homogeneous and

vρ,σ(m) = v + (t − v + ρ + σ) − t = ρ + σ

The relations 5.2 and 5.3 can now be written as

δ f ρ+σ f ρ+σ = σy (5.4) δx mv mv+1 δ f ρ+σ f ρ+σ = ρx (5.5) δy mv mv+1

Consider f and m as elements of k(y)[x] (or k(x)[y]). Applying relations 5.4 and 5.5, all zeroes of f in k(y) (or k(x)) are zeroes of m in k(y) (or k(x), respectively). If m is a monomial, then we have shown that f is a monomial, contrary to assumption. Thus E(m) contains at least two elements. And yet, if (i, j) ∈ E(m), we have ρi + σj = ρ + σ. If i > 0 and j > 0, it follows that (i, j) = (1, 1). As E(m) is not simply {(1, 1)} by the preceeding argument, E(m) contains an element of the

ρ+σ form (i, 0), which is neccessarily of the form ρ , 0 , or an element of the form ρ+σ (0, j) which is neccessarily of the form 0, σ . We thus have one of the following cases:

ρ First case ρ < σ, σ is a multiple of ρ, E(m) = (1, 1), (1 + σ , 0) , and

1+ σ m(x, y) = µx ρ + νxy

with µ, ν ∈ k, µ, ν 6= 0. ρ Second case ρ > σ, ρ is a multiple of σ, E(m) = (1, 1), (0, 1 + σ ) , and

1+ ρ m(x, y) = µy σ + νxy

40 Third case ρ = σ, E(m) ⊂ {(2, 0), (1, 1), (0, 2)}, and

m(x, y) = µx2 + νxy + ζy2

with at least two of µ, ν, ζ non-zero.

Consider the ﬁrst case. Then m ∈ k(x)[y] has one zero in k(x). Thus f ∈ k(x)[y] has, for its only zero in k(x), the zero of m. Thus there exists an integer β ≥ 0

σ and an element τ(x) ∈ k(x) such that f = τ(x)(νy + µx ρ )β. Since v 6= 0, we have τ(x) ∈ k[x]. For the other part, all zeroes of f ∈ k(y)[x] in k(y) are zeroes of m ∈ k(y)[x]. This proves that the only zero of τ(x) is at x = 0, and thus τ(x) is a monomial. This places us in case (b). We see similarly that in the second case places us in case (c). Consider now the third case. If ζ = 0, we have µ, ν 6= 0. Reasoning as for the ﬁrst case, we have (d). Similarly if µ = 0. Suppose that µ, ζ 6= 0. If m(x, y) = µ(x + ηy)(x + θy) with η, θ ∈ k, f ∈ k(y)[x] has its sole zeros at −ηy, and −θy in k(y), thus

f = τ(y)(x + ηy)α(x + θy)β with τ(y) ∈ k(y) and α, β integers ≥ 0. Clearly, τ(y) ∈ k[y] and, exchanging x and y, we see that τ(y) ∈ k, and we have established (d). Finally, suppose that

m(x, y) = µ(x + ηy)(x + θy) with η, θ ∈ k \ k, η and θ conjugate over k. We have f = τ(y)(x + ηy)α(x + θy)β, this time with α = β and also τ(y) ∈ k. But then f is a multiple of a power of m, whence

ρ+σ v vρ,σ(f ) = (ρ + σ)v = vρ,σ(m )

f ρ,σ we have mv ∈ k, which is a contradiction following 5.4 and 5.5.

41 Proposition 5.15. Take ρ, σ integers > 0, a ∈ A1, v = vρ,σ(a), let f be the (ρ, σ)- associated polynomial of a. Suppose that

1. v > ρ + σ 2. f is not a monomial 3. None of the cases (b), (c), or (d) in lemma 5.14 hold Then F (a) = C(a).

Proof. Let Λ be the set of integers λ for which there exists a b ∈ F (a) with vρ,σ(b) = λ. Since F (a) is closed under addition, Λ+Λ ⊂ Λ, and in particular {0, v, 2v, . . .} ⊂

Λ. Let Λ0 be the image in N/vN of Λ under the map n 7→ n + vN. In every 0 (coset) element of Λ , choose the smallest element. Denote these elements by λ0 =

0, λ1, . . . , λr where r ≤ v. The elements of Λ are then of the following form:

0, v, 2v, 3v, . . .

λ1, λ1 + v, λ1 + 2v, λ1 + 3v, . . . . .

λr, λr + v, λr + 2v, λr + 3v, . . .

Let bi be an element of F (a) such that vρ,σ(bi) = λi. Take b ∈ F (a). We will show by induction that b ∈ k[a]b0 +k[a]b1 +...+k[a]br. It is obvious for vρ,σ(b) = 0.

Suppose that it holds for vρ,σ(b) < n and consider the case where vρ,σ(b) = n > 0.

s Then there exists an i ∈ {0, 1, . . . , r} and an integer s ≥ 0 such that vρ,σ(a bk) = n.

s Let g and h be the (ρ, σ)-associated polynomials of b and a bi. Applying lemma 5.14, gv and hv are scalar multiples of f n and thus of each other. Thus there exists

s s a λ ∈ k such that vρ,σ(b − λa bi) < n. We have therefore b − λa bi ∈ F (a), and the

s result follows by induction on b − λa bi. P Thus we have F (a) = i k[a]bi and the result follows from lemma 5.13.

42 0 Deﬁnition 5.16. For λ ∈ k and n ∈ N, deﬁne the k-linear maps Φn,λ, Φn,λ : A1 →

A1 by

n Φn,λ(p) = p Φn,λ(q) = q + λp

0 n 0 Φn,λ(p) = p + λq Φn,λ(q) = q

0 Lemma 5.17. For all λ ∈ k and n ∈ N, Φn,λ and Φn,λ are k-linear automorphisms of A1.

Proof. The image of qp − pq − 1 under Φn,λ is

(q + λpn)p − p(q + λpn) − 1 = qp + λpn+1 − pq − λpn+1 − 1

= pq + 1 + λpn+1 − pq − λpn+1 − 1

= 0

0 thus Φn,λ is an homomorphism. Similarly for Φn,λ. As mentioned at the end of chapter 2, the fact that A1 is simple implies that any non-zero endomorphism of

A1 is injective. This follows as the kernel is an ideal of A1 and is therefore either

0 n 0 or all of A1. Thus both Φn,λ and Φn,λ are injective. Let q0 = −λp + q and let n p0 = p − λq . We have,

n n Φn,λ(p) = p Φn,λ(q0) = −λp + q + λp = q

0 n n 0 Φn,λ(p0) = p + λq − λq = p Φn,λ(q) = q

0 Thus p and q are in the images of both Φn,λ and Φn,λ and they are thus both surjective.

Deﬁnition 5.18. Let G denote the group of automorphisms of A1 generated by

0 the Φn,λ, Φn,λ for all n, λ.

43 Let V be the vector space kp+kq. The group SL(V ) consists of maps θ : V → V of the form vp + wq 7→ (av + cw)p + (bv + dw)q with ad − bc = 1. Since the map θ is

0 deﬁned for p and q, we can extend it to a multiplicative k-linear map θ : A1 → A1 deﬁned by

θ0(p) = ap + bq θ0(q) = cp + dq ad − bc = 1

0 Note that θ |V = θ. Calculating,

θ(qp − pq − 1) = (cp + dq)(ap + bq) − (ap + bq)(cp + dq) − 1

= acp2 + bcpq + adqp + bdq2 − acp2 − adpq − bcqp − bdq2 − 1

= bcpq + adpq + ad − adpq − bcpq − bc − 1

= (ad − bc) − 1

= 0

0 Thus θ is an homomorphism. Let p0 = dp − bq and q0 = −cp + aq. It is simple to

0 0 0 0 calculate that θ (p0) = p and θ (q0) = q. Thus both p, q ∈ im(θ ) and therefore θ is

0 surjective. As in the previous lemma, the fact that A1 is simple implies that θ is injective.

Deﬁnition 5.19. As above, let V be the vector space kp + kq. As we have just shown, all elements of SL(V ) extend to give an automorphism of A1. We obtain

0 thus a group G of automorphisms of A1. It is not hard to see that the restrictions

0 0 0 Φ1,λ|V and Φ1,λ|V generate the group SL(V ). Thus G ⊂ G. Denote by G the 0 analogous group acting on A1. In particular, denote by Ψ the element of G such that Ψ(p) = q, Ψ(q) = −p, this is known as the Fourier transform.

44 2 2 Lemma 5.20. Take a = αp + βpq + γq ∈ A1 with α, β, γ ∈ k. There exist Φ, Θ ∈ G0 such that

Φ(a) = θpq + ζ   α0p2 + ζ0, if β2 − αγ = 0 Θ(a) =  α0p2 + γ0q2 + ζ0, if β2 − αγ 6= 0 for α0, γ0, ζ0 ∈ k and θ, ζ ∈ k.

Proof. Noting that if Φ(p) = ap + bq and Φ(q) = cp + dq,

Φ(a) = α(ap + bq)2 + β(ap + bq)(cp + dq) + γ(cp + dq)2

= αa2p2 + αabpq + αabqp + αb2q2

+ βacp2 + βadpq + βbcqp + βbdq2

+ γc2p2 + γcdpq + γcdqp + γd2q2

= (αa2 + βac + γc2)p2 + (2αab + βad + βbc + 2γcd)pq

+ (αb2 + βbd + γd2)q2 + αab + βbc + γcd

which is simply the action of Φ|V on a considered as a quadratic form in k[p, q] plus a scalar term. The result thus follows from the analogous results for real quadratic forms on two variables.

Lemma 5.21. For a ∈ k[p], N(a) = A1.

m n Proof. First note that for b, c ∈ A1, if (ad a) b = 0 and (ad a) c = 0 then if o = max(m, n), (ad a)o(b + c) = (ad a)ob + (ad a)oc = 0 and (ad a)o(bc) = ((ad a)ob)c + b((ad a)oc) = 0 ie. N(a) is both multiplicatively and additively closed.

We have p ∈ N(a) and [a, q] ∈ k[p], whence q ∈ N(a) and thus A1 ⊆ N(a).

Lemma 5.22. Take a = λp2 + µq2 + ν with λ, µ, ν ∈ k, λ 6= 0, µ 6= 0. Then

D(a) = A1.

45 Proof. By lemma 5.20, there exists a Φ ∈ G0 such that Φ(a) = ζpq +θ with ζ, θ ∈ k. Note that [pq, p] = −p and [pq, q] = −q, and hence [pq, piqj] = (m − l)piqj. Thus,

P i j P i j for any b ∈ A1 with b = bijp q , [Φ(a), b] = (m − l)ζbijp q . The individual terms of b are therefore in the components D(Φ(a), (m − l)ζbij; A1) and hence b ∈ D(Φ(a); A1). But, since b was chosen to lie within A1, b ∈ D(Φ(a); A1) ∩ A1 =

D(Φ(a); A1). Thus by corollary 5.10.1, D(a) = D(a; A1) = A1.

Lemma 5.23. Let a be an element of A1 of the form

2 r a00 + a10p + a20p + ... + ar0p + a01q + a11pq, aij ∈ k

Then there exists a Φ ∈ G such that Φ(a) is of the form

b00 + b10p + b01q + b11pq, bij ∈ k

Proof. The result is trivial if r ≤ 1. Suppose the result holds for r − 1. If a11 6= 0, we can scale such that a11 = 1. We have the following:

r r−1 r−1 Φr−1,−ar0 (a) = a00 + a10p + ... + ar0p + a01(q − ar0p ) + p(q − ar0p )

r−2 r−1 = a00 + a10p + ... + ar−2,0p + ar−1,0p + a01q + pq and, noting that the maximum degree in p has fallen by one, the result follows by induction. If a11 = 0 and a01 6= 0, we can scale such that a01 = 1. We have:

r r Φr,−ar0 (a) = a00 + a10p + ... + ar0p + q − ar0p

r−1 = a00 + a10p + ... + ar−1,0p + q and again the result follows by induction.

Lemma 5.24. Let a be an element of A1, of the form

αp2 + 2βpq + γq2 + δp + q + ζ (α, β, . . . , ζ ∈ k)

46 • If β2 − αγ = 0, there exists a Φ ∈ G such that Φ(a) ∈ k[p] • If β2 − αγ 6= 0, there exists a Φ ∈ G and λ, µ, ν ∈ k with λ 6= 0, µ 6= 0 such that Φ(a) = λp2 + µq2 + ν

2 0 Proof. If β − αγ = 0, then by lemma 5.20 there exists a Φ1 ∈ G such that

0 2 0 0 0 Φ1(a) = α p + δ p + q + ζ

If 0 = 0, the result is proved. If 0 6= 0, can scale such that 0 = 1. We have

0 2 0 0 2 0 Φ2,−α0 (Φ1(a)) = α p + δ p + q − α p + ζ

= δ0p + q + ζ0 and it suﬃces to apply some element of G0.

2 0 If β − αγ 6= 0, then again by lemma 5.20 there exists a Φ1 ∈ G such that

0 2 0 2 0 0 0 Φ1(a) = α p + γ q + δ p + q + ζ with α0, γ0 6= 0. We have

y = Φ 1 0 0−1 (Φ1(a)) 0,− 2 γ 1 1 = α0p2 + δ0p + ζ0 + γ0(q − 0γ0−1)2 + 0(q − 0γ0−1) 2 2 0 2 0 0 0 2 = α p + δ p + ζ + γ q + ζ1

0 0 2 0 2 Likewise, there exists a Φ2 ∈ G such that Φ2(y) = α p + γ q + ζ2.

P i j Lemma 5.25. Let a = aijp q ∈ A1. Let r be the smallest integer ≥ 0 such that ai0 = 0 for i > r. Let s be the smallest integer ≥ 0 such that a0j = 0 for j > s.

Suppose that there exist integers i1, j1 ≥ 0 such that ai1,j1 6= 0, (i1, j1) 6= (1, 1) and si1 + rj1 > rs. Then F (a) 6= A1.

47 Proof. If i1 = 0, we have rj1 > rs, or j1 > s, which contradicts the deﬁnition of s.

Thus i1 > 0 and similarly j1 > 0. Take real numbers ρ, σ > 0 (relatively irrational) such that

σi1 + ρj1 > ρs

σi1 + ρj1 > rσ

(eg. take ρ = r + δ, σ = s + for 0 < δs, r < 1). By the deﬁnition of vσ,ρ(a), there exist integers i2, j2 ≥ 0 such that ai2j2 6= 0, σi2 + ρj2 = vσ,ρ(a). Thus, by the maximality of vσ,ρ(a),

σi2 + ρj2 > σi1 + ρj1 > ρs

σi2 + ρj2 > σi1 + ρj1 > rσ

This implies that i2, j2 > 0 (otherwise if, say, i2 = 0 then j2 > j1, which contradicts the choice of j1). If i2 = j2 = 1, we have

σ + ρ ≤ σi1 + ρj1

≤ σi2 + ρj2

= σ + ρ

hence i1 = i2, j1 = j2 and (i1, j1) = (1, 1), contrary to assumption. Thus i2 > 1, or j2 > 1. By applying the second part of lemma 5.3, the (σ, ρ)-associated polynomial

i2 j2 of a is ai2j2 x y .

n Suppose that i2 ≤ j2. For n = 0, 1, 2,..., let bn = (ad a) q. We aim to show by induction on n that the (σ, ρ)-associated polynomial of bn is

n(i2−1) 1+n(j2−1) bnx y

48 with bn ∈ k, bn 6= 0. The result is obvious for n = 0. Suppose that the result holds for n. Applying lemma 5.8, the (σ, ρ)-associated polynomial of bn+1 = [a, bn] is

i2+n(i2−1)−1 j2+1+n(j2−1)−1 (i2 + nj2 − ni2)ai2j2 bnx y which proves the assertion for n + 1. We also have

vσ,ρ(bn) = σn(i2 − 1) + ρ(1 + n(j2 − 1))

Since i2 > 1 or j2 > 1, we see that vσ,ρ tends to inﬁnity with n. Thus q∈ / F (a) and F (a) 6= A1.

If i2 ≥ j2, we have p∈ / F (Ψ(a)), by the preceeding, thus F (Ψ(a)) 6= A1 and

F (a) 6= A1 by corollary 5.10.1.

Lemma 5.26. Take a ∈ A1 with F (a) = A1. Then there exists a Φ ∈ G such that Φ(a) posesses one of the following properties: either Φ(a) ∈ k[p] or Φ(a) is of the form λp2 + µq2 + ν with λ, µ, ν ∈ k, λ, µ 6= 0.

Proof. Introduce integers r, s as in the previous lemma. We argue inductively on r + s. If r ≤ 2 and s ≤ 2, the previous lemma shows that v1,1 ≤ 2, and we apply lemma 5.24. Suppose therefore that r > 2 or s > 2 and that the result holds for r + s < n. Consider the case r + s = n. Using the automorphism Ψ (to swap p and q if neccessary), we suppose that r ≥ s. If s ≤ 1, a is, by lemma (5.25), of the form

2 r a00 + a10p + a20p + ... + ar0p + a01q + a11pq and it suﬃces to apply lemmas 5.23 and 5.24. Suppose from now on, therefore, that r ≥ s ≥ 2 and r > 2, whence r + s < rs. If (i, j) ∈ E(a), lemma 5.25 shows

49 that either si + rj ≤ rs, or i = j = 1, in which case si + rj = s + r < rs. Thus vs,r(a) = rs and the (s, r)-associated polynomial of a is of the form

r s f(x, y) = ar0x + ... + a0sy with ar0 6= 0, a0s 6= 0 (5.6)

Since F (a) = A1 6= C(a), applying proposition 5.15 with ρ = s, σ = r implies that one of cases b, c or d of lemma 5.14 holds. Since r ≥ s, we are in either case b or case d. Suppose that we are in case b. Then r is a multiple of s and applying theorem

r s 5.6 , f is a scalar multiple of (x s + µy) with µ ∈ k, µ 6= 0. Suppose that

r s X i j a = (p s + µq) + aijp q (i,j)∈E with si + rj < rs for (i, j) ∈ E. Then

b = Φ r −1 (a) s , µ

s s X i −1 r j = µ q + aijp (q − µ p s ) (i,j)∈E

We have

−1 r vs,r(q − µ p s ) = r and vs,r(p) = s thus

  X i −1 r j vs,r  aijp (q − µ p s )  < rs (i,j)∈E

If we denote as r1 and s1 the analogous integers to r and s, but related to b, we see that s1 = s and r1 < r. Inductively, there exists a Φ ∈ G such that Φ(b) posesses one of the required properties. Thus Φ(b) = (Φ ◦ Φ r −1 )(a), and the lemma is s , µ proved in this case.

50 Suppose that we are in case d of lemma 5.14. Then r = s and f is a multiple of (x + µy)α(x + νy)r−α with µ, ν ∈ k and α an integer such that 0 ≤ α ≤ r. Multiplying a by a scalar if neccessary, suppose that

α r−α X i j a = (p + µq) (p + νq) + aijp q (i,j)∈E with i + j < r for (i, j) ∈ E. If neccessary, exchange µ and ν, and assume that α > 0. Then

b = Φ −1 (a) 1, µ

α α −1 r−α X i −1 j = µ q ((1 − νµ )p + νq) + aijp (q − µ p) (i,j)∈E

If we again denote as r1 and r1 the integers analogous to the r and s but relative to b, we see that s1 = s = r and r1 < r. Induction therefore will terminate in case b.

Lemma 5.27. Take a ∈ A1

• If N(a) = A1, then there exists a Φ ∈ G such that Φ(a) ∈ k[p] • If, in addition, C(a) = k[a], then there exists a Φ ∈ G such that Φ(a) = p

Proof. Since N(a) ⊆ F (a), lemma 5.26 implies that there exists a Φ ∈ G such that

2 2 either Φ(a) ∈ k[p] or Φ(a) = λp + µq + ν. But since N(a) * D(a), D(a) 6= A1 and the contrapositive of lemma 5.22 implies that Φ(a) 6= λp2 + µq2 + ν.

Now suppose that N(a) = A1 and C(a) = k[a]. Thanks to the ﬁrst result, we assume a ∈ k[p]. Thus p ∈ C(a) = k[a], and hence a ∈ k.1 + k.p, and the result follows.

Theorem 5.28. The group of k-linear automorphisms of A1 is generated by the

0 automorphisms Φn,λ and Φn,λ.

51 Proof. We take Φ to be an automorphism of A1 and prove that Φ ∈ G. We have

N(Φ(p)) = A1 (lemma 5.21), and C(Φ(p)) = k[Φ(p)] (5.11). Applying lemma 5.27, we reduce to the case where Φ(p) = p. Then

[p, Φ(q) − q] = Φ([p, q]) − [p, q] = 1 − 1 = 0

Thus Φ(q) ∈ q + k[p], and Φ is a product of automorphisms Φn,λ.

52 References

[1] Jacques Dixmier, Sur les algèbres de Weyl, Bull. Soc. Math. France 96 (1968), 209–247. MR MR0242897 (39 #4224) [2] Günter R. Krause and Thomas H. Lenagan, Growth of algebras and Gelfand- Kirillov dimension, revised ed., Graduate Studies in Mathematics, vol. 22, American Mathematical Society, Providence, RI, 2000. MR MR1721834 (2000j:16035) [3] T. Y. Lam, A first course in noncommutative rings, first ed., Graduate texts in mathematics, vol. 131, Springer-Verlag, New York, 1991. [4] Dmitriy Rumynin, Rings and modules, (2003).