PRIMITIVE AND POISSON SPECTRA OF NON-SEMISIMPLE TWISTS OF POLYNOMIAL ALGEBRAS by MARY-KATHERINE BRANDL A DISSERTATION Presented to the Department of Mathematics and the Graduate School of the University of Oregon in partial ful llment of the requirements for the degree of Doctor of Philosophy July 2001 ii \Primitive and Poisson Spectra of Non-semisimple Twists of Polynomial Algebras," a dissertation prepared by Mary-Katherine Brandl in partial ful llment of the require- ments for the Doctor of Philosophy degree in the Department of Mathematics. This dissertation has been approved and accepted by: Professor Brad Shelton, Chair of the Examining Committee Date Committee in charge: Professor Brad Shelton, Chair Professor Frank Anderson Professor Bruce Barnes Professor Jon Brundan Professor Lynn Kahle Accepted by: Vice Provost and Dean of the Graduate School iii An Abstract of the Dissertation of Mary-Katherine Brandl for the degree of Doctor of Philosophy in the Department of Mathematics to be taken July 2001 Title: PRIMITIVE AND POISSON SPECTRA OF NON-SEMISIMPLE TWISTS OF POLYNOMIAL ALGEBRAS Approved: Professor Brad Shelton We examine a family of twists of the complex polynomial ring on n generators by a non-semisimple automorphism. In particular, we consider the case where the automorphism is represented by a single Jordan block. The multiplication in the twist determines a Poisson structure on a ne n-space. We demonstrate that the primitive ideals in the twist are parameterized by the symplectic leaves associated to this Poisson structure. Moreover, the symplectic leaves are determined by the orbits of a regular unipotent subgroup of the complex general linear group. iv ACKNOWLEDGEMENTS \For everything I learn there are two I don't understand..." Emily Saliers I am fortunate to be surrounded by so many exceptional people. I would like to thank my parents John and Rochelle Brandl for all of their love and support, and for believing that what I do is important. I would also like to thank Frank Anderson for his encouragement, especially during my rst year at the University of Oregon, and Michaela Vancli for her advice and instruction during the last two years. I can't imagine what graduate school would have been like without Inga Johnson and Jessica Sklar, and I know that I would not have made it without them. Finally, I want to thank Brad Shelton who has been an excellent advisor and a great friend. v CURRICULUM VITA NAME OF AUTHOR: Mary-Katherine Brandl PLACE OF BIRTH: Fort Belvoir, Virginia DATE OF BIRTH: July 11, 1963 GRADUATE AND UNDERGRADUATE SCHOOLS ATTENDED: University of Oregon University of California, Santa Cruz DEGREES AWARDED: Master of Science in Mathematics, University of Oregon, 1997 Bachelor of Arts, University of California, Santa Cruz, 1995 AREAS OF SPECIAL INTEREST: Noncommutative Ring Theory Algebraic Geometry PROFESSIONAL EXPERIENCE: Graduate Teaching Assistant, Department of Mathematics, University of Oregon, Eugene, 1995-2001 vi AWARDS AND HONORS: Johnson Fellowship, Summer 1997 vii TABLE OF CONTENTS Chapter Page I.INTRODUCTION.............................. 1 I.1.BackgroundofProblem ....................... 1 I.2.StatementofTheorems ....................... 3 II.PRELIMINARIES.............................. 6 II.1.Non-CommutativeAlgebra ..................... 6 II.2.PoissonGeometry .......................... 10 III.THETWISTEDALGEBRA ........................ 13 III.1. Non-Semisimple Twists . . . . . . . . . . . . . . . . . . . . . . . 13 III.2.Formulas ............................... 22 III.3.PrimitiveIdeals ........................... 24 IV. THE POISSON MANIFOLD . . . . . . . . . . . . . . . . . . . . . . . . 27 IV.1.ThePoissonBracket......................... 27 IV.2.SymplecticLeaves .......................... 29 V.ALGEBRAICGROUP ........................... 34 V.1. Orbits in An ............................. 34 V.2.MomentumMap........................... 37 VI. EXAMPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 VI.1.StandardExamples ......................... 41 VI.2.MoreExamples............................ 44 viii Page BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 1 CHAPTER I INTRODUCTION I.1. Background of Problem One current strategy in noncommutative ring theory is to associate geometric objects to noncommutative algebras. Algebraists have been very successful analyzing primitive ideals by considering them as geometric objects. For example, a geometric focus was used in [1] to classify algebras with nice homological properties (similar to polynomial rings) in terms of the geometric structure of a collection of graded indecomposable modules. We refer to this geometric philosophy as noncommutative algebraic geometry. In this dissertation, we focus on a family of twists B of the polynomial algebra S = C[x1;:::;xn]. Our goal is to give a geometric description of the primitive spectrum of B. We o er the following examples as motivation. The primitive ideals of the uni- versal enveloping algebra of an algebraic solvable Lie algebra, g, are parametrized by the symplectic leaves in the Poisson manifold g , [2]. Furthermore, these leaves are the orbits of the adjoint algebraic group of g. Hodges and Levasseur use the quan- tum group Oq(SLn) to de ne a Poisson structure on the manifold SLn. They then 2 demonstrate that the primitive ideals in Oq(SLn) are parametrized by the symplectic leaves, [7]. In [10] M. Vancli describes the primitive spectra of a family of twists, B(m), of S, parametrized by the maximal ideals of an algebra R. Each twist B(m)of S is determined by a semisimple automorphism m of Pn-1. The multiplication in the twist induces a Poisson structure on Cn. Vancli restricts to the setting where the symplectic leaves for this Poisson structure are algebraic. She de nes an associated algebraic group, G, whose orbits are the symplectic leaves. She then proves that the primitive ideals in B(m) are parametrized by the symplectic leaves for the Poisson structure if and only if m has a representative in G. In this dissertation we extend Vancli 's results to a family of twists S of S in which the automorphism is not semisimple. In particular, we consider the twist of S by an automorphism that is represented by a single Jordan block. In this setting we nd that the symplectic leaves of the associated Poisson structure are always algebraic. Furthermore, we nd that as in Vancli 's case, the symplectic leaves are the orbits of an algebraic group, and that the primitive ideals are parametrized by these leaves. Much of the work in Vancli 's analysis, is due to having more than one eigen- value. In her setting, the commutator of xi and xj is a di erence of eigenvalues times xixj, and this fact makes analyzing prime and primitive ideals straightforward. Problems only arise for certain bad combinations of eigenvalues (i.e. when ratios of di erences of eigenvalues are roots of unity). We avoid these eigenvalue complications 3 in our setting, because 1 is the only eigenvalue. On the other hand, commutators of monomials are no longer monomials, and thus it is much more di cult to analyze the prime and primitive ideals. We are required to take a di erent approach to the problem, and are a orded a more intricate primitive spectrum. In Vancli 's work, the Poisson geometry is relatively straightforward to analyze because the symplectic structure is evident. In our setting, some of the symplectic leaves are evident, but we must make a careful analysis of certain di erential operators to nd the others. I.2. Statement of Theorems Let be the automorphism of Pn-1 which is represented by the matrix with ones on the diagonal and superdiagonal, and zeros everywhere else, and let B be the twist of S by . In section II.1.1 we will see that B is isomorphic to a quotient of the free algebra Chy1;:::;yni by a homogeneous quadratic ideal. We identify B with this quotient, and retain the notation yi for the image of yi in B. The algebra B is well understood as a projective object, [11], however, we are interested in understanding B as an a ne object. This thesis is organized as follows. Chapter II gives background information per- taining to the problem. In Chapter III we investigate the primitive spectrum of the twisted algebra. Our main result is the description of the primitive ideals of B. 4 Theorem III.1.6. The maximal ideals in B are the ideals hy1;:::;yn-1;yn - i, 2 C. The remaining primitive ideals are hy1;:::;yn-2i, together with a family of homogeneous ideals. These homogeneous ideals are of the form hy1;:::;yk;f1;:::fji; where k =0;:::;n- 3, j = n - k - 2 2 ,eachfi is degree 2, and each collection ff1;:::fjg is determined by a unique element of Cn-k. For notation necessary for the precise statement of Theorem III.1.6, please see Con- struction III.1.5. From Theorem III.1.6 we see that the non-maximal primitive ideals are parametrized by the set C8 = f 2 Cn-kjk =2;:::;ng; where C0 =1. In Chapter IV, we construct the Poisson structure associated to the twist. Here we de ne a di erential operator !, which is the key to the symplectic structure. In fact, this operator represents the crucial di erence between this case and the diagonal case. Each leaf is obtained by constructing a sequence of elements f1;:::;fj, such that !f1 =0,and!fi = fi-1. That is, we determine the symplectic leaves by integrating with respect to !. After a change in variables, we recognize the two dimensional symplectic leaves as open a ne subsets of classical surfaces. 5 Proposition IV.2.1. The 0-dimensional symplectic leaves associated to the Poisson structure are the points (0;:::;0;gamma) 2 An. The remaining leaves are two dimensional, and each of these leaves is an open subset of the image in An of a Veronese surface. For a precise statement of Propositions IV.2.1, the reader is refered to section IV.2. After describing the Poisson structure, we note that the primitive ideals are also parametrized by C8 . Corrolary IV.2.2. There is a natural one to one correspondence between the prim- itive ideals in B = S and the symplectic leaves for the symplectic structure induced by . In Chapter V we realize the two-dimensional leaves as orbits of a unipotent sub- group of the general linear group. Proposition V.1.1. The 2-dimensional symplectic leaves for S are the orbits in An of a regular unipotent algebraic subgroup G of GLn(C). Furthermore, G acts transitively on the 0-dimensional leaves. Finally, in Chapter VI, we give examples of our result, and a three dimensional twist example where the automorphism has two Jordan blocks. 6 CHAPTER II PRELIMINARIES II.1. Non-Commutative Algebra Our primary goal is to describe the ideal structure of the twist of a polynomial al- gebra by a degree zero automorphism. Such an algebra is a noncommutative analogue of a homogeneous coordinate ring [1]. It is de ned more simply below. II.1.1. Twisted Algebras. Given a commutative graded k-algebra A = Ad,anda degree 0 automorphism of A, we form the twisted algebra A , with multiplication de ned on homogeneous elements by a b = a r(b), where r =dega,and denotes usual multiplication in A. This new algebra A retains many of the properties of the original algebra. For example, the properties of being a domain and of being Noetherian are invariant under twisting [11]. In fact, J. Zhang has shown that twisting de nes an equivalence relation on the category of graded k-algebras that is analogous to Morita equivalence, in the following sense. Let Gr-A be the category of graded A-modules, with morphisms being graded degree 0 homomorphisms. Then a graded k-algebra B is a twisted algebra of A if and only if the categories Gr -A and Gr -B are equivalent if and only if the categories Gr -A and Gr -B are isomorphic. 7 Let B = A .SinceB = A as sets, each element of f 2 B is also an element of A. We will write f0 2 A when f is viewed as an element of A. For an ideal I in B,let I0 = ff0jf 2 Ig. For homogeneous F 2 I0i and G 2 A, FG = F -i(G) 2 I0.It follows that if I is a homogeneous ideal in B,thenI0 is a homogeneous ideal in A.Ifin addition, (I0) = I0,thenB=I is A=I0 as a graded vector space, with multiplication inherited from B, so in fact, B=I = parenleftbigA=I0 ,where is the automorphism induced by . Let f;g 2 B be homogeneous of degrees i and j respectively. Write F = f0 2 Ai, and G = g0 2 Aj, and assume that F = F. De ne f(g)by[ f(g)]0 = G i .Then [ f(g) f]0 = G i F j = FG i =(f g)0,sothat f(g) f = f g.Fromthiswesee that if f is homogeneous with (f0) = f0,thenf is normal in B. We will say that f 2 B is -invariant if (f0) = f0. II.1.2. Note. For homogeneous -invariant element f 2 B, f is an automorphism of B. Furthermore, homogeneous -invariant elements of the same degree are associated to the same automorphism. We write (F1;:::;Fd) for the ideal generated by the elements F1;:::;Fd in the commutative algebra A,andwritehf1;:::;fdi for the ideal generated by f1;:::;fd in the noncommutative algebra B.Letf 2 Bi be homogeneous and -invariant, and let 8 I = hfi.Sincef is normal, I = f B,so I0 = f(f g)0jg 2 Bg = ff0(g0) i jg 2 Bg = f0A: For G 2 A,(f0G) = f0G 2 I0,soI0 is -invariant. It follows that B=hfi = [A=(f0)] ,wherewhere is the automorphism induced by . II.1.3. Notes. 1. The preceding paragraph shows that if f is -invariant and irreducible, then the ideal hfi is prime. 2. Let f be -invariant, and let P be a prime ideal in B with f 62 P.Sincef is normal, hfi = B f.Ifg 2 B,withf g 2 P,thenhfi hgi = B f B g B = B f g B = hg fi P.Buttheng 2 P. It follows that f is regular modulo P. Now, let S = Sn = C[x1;:::xn] be the polynomial algebra in n variables over the complex numbers, with grading given by deg(xi) = 1. A graded automorphism of Sn is determined by its restriction to the vector space Sn1 of degree one elements, so is represented by an upper-triangular (n n)-matrix. Furthermore, scalar multiples of this matrix give rise to isomorphic twisted algebras, so we can take to be an automorphism of Pn-1. II.1.4. Primitive Ideals. Let R be a ring. A module MR is faithful if AnnR(M)= 0, that is, if r 2 R with Mr =0,thenr =0.WesaythatR is (left) right primitive 9 if R has a simple faithful (left) right module. Although a right primitive ring need not be left primitive, we will usually omit the word `right'. An ideal P in R is primitive if R=P is a primitive ring. A primitive ideal is prime, and each maximal ideal is primitive [5]. Furthermore, in a commutative ring an ideal is primitive if and only if it is maximal. It is not surprising then that the primitive ideals in a non-commutative ring play a role analogous to that of maximal ideals in a commutative ring. AringR has the endomorphism property if for every simple R[z] module, M,End(M) is algebraic over k.Ifk is an uncountable eld, and R is a countably generated k algebra, then R has the endomorphism property [9]. Proposition II.1.5. Let k be an uncountable algebraically closed eld, and let R be a primitive k-algebra. Then the center of the quotient ring Q(R) is k. Proof. Let Z be the center of Q(R), and z 2 Z. Write z = rs-1,withr; s 2 R,and s regular. Since z is central, it follows that for each p = p(z) 2 R[z], psn 2 R,where n is the z degree of p.LetL be a simple faithful R-module, and let L = L circlemultiplyR R[z]. We claim that L is a simple faithful R[z]-module. As an R-module, L = X i2Z L,is a sum of faithful modules, so AnnR[z]( L) \ R =0. ButR is essential in R[z], so AnnR[z]( L) = 0. Now, suppose that A is a nonzero R[z]-submodule of L,andletu be nonzero in A. Write u = xcirclemultiply p,wherep = nX i=0 izi.TheR-module uR is contained is the module nX i=0 Lzi, whose simple factors are all isomorphic to L.SinceL is faithful, AnnL(sn) 6= L,sothereisanonzeroelementv 2 uR such that vsn 6= 0. Write 10 v = y circlemultiplyq where q 2 R[z]hasz degree less than or equal to n.Thenvsn = yqsn circlemultiply1is in uR A and generates A,so L is in fact simple. Now, R[z] has the endomorphism property, so End( L) is algebraic over k, hence equal to k. But multiplication by z is an endomorphism on L,soz acts as for some 2 k.But L is faithful, so z = 2 k, and we are done. II.1.6. Note. In the proof of Proposition II.1.5, we actually showed that if R and R[z] are primitive algebras over an uncountable algebraically closed eld, then R[z]= R. II.1.7. Remark. A regular normal element r in a ring R determines an automor- phism 'r of R by xr = r'r(x). Suppose R is a primitive k-algebra. If r and s are elements of R that determine the same automorphism, then the element rs-1 is central in the quotient ring Q(R). Then by Proposition II.1.5, r = cs for some c 2 k. II.2. Poisson Geometry II.2.1. Poisson Manifolds. Let A be a C-algebra. A Poisson bracket on A is a Lie bracket f;g on A that is a derivation in each variable. So f;g is a skew-symmetric bilinear form that satis es (i) fx;fy;zgg + fy;fz;xgg + fz;fx;ygg =0; and (ii) fx;yzg = yfx;zg + fx;ygz: 11 The pair (A;f;g) is called a Poisson algebra.AnidealI in A is a Poisson ideal if fI;Ag I, and an element f 2 A is a Poisson element if (f) is a Poisson ideal in A.LetM be a di erentiable complex manifold. A Poisson structure on M is determined by choosing a Poisson bracket f;g from C1(M;C) C1(M;C) ! C1(M;C). The pair (M;f;g) is called a Poisson manifold. For any Poisson manifold (M;f;g), there is a unique di erentiable eld of twice contravariant, skew-symmetric tensors such that for any pair f;g 2 C1(M;C), ff;gg = (df ; dg ): For a point x 2 M, the rank of the 2-tensor (x) is called the rank of the Poisson structure at x.Asymplectic leaf is a maximal connected Poisson submanifold N of M such that the rank of the Poisson structure at each point of N isequaltothe dimension of N. By standard theory, the symplectic leaves have even dimension, and M is a disjoint union of symplectic leaves [8]. The collection of symplectic leaves is called a foliation of M, and we say that M is foliated by its symplectic leaves. Suppose that (M;f;g) is a Poisson manifold, with M = Cn. Then the bracket f;g is determined by its restriction to S = C[x1;:::;xn], [8]. If in fact the bracket maps C[x1;:::;xn] C[x1;:::;xn]intoC[x1;:::;xn], then we may determine the Poisson structure by studying the Poisson algebra (S;f;g). II.2.2. Drinfel'd. Let S be the polynomial algebra on n generators over C.The Poisson bracket due to Drinfel'd is de ned as follows. Let R be a commutative k- algebra which is a PID but not a eld, and let A be an R-algebra. Further assume 12 that A is flat as an R-module, and that there exists a maximal ideal m0 =(H)ofR which is unique with the property that A=hm0i = S.ForF;G 2 S choose preimages eF; eG 2 A, and de ne the bracket of F and G to be fF;Gg = eF eG - eG eF H modhHi: Then f;g is a Poisson bracket on S [3]. 13 CHAPTER III THE TWISTED ALGEBRA III.1. Non-Semisimple Twists III.1.1. The Twisted Algebra. In [10], Vancli describes geometrically the primi- tive spectrum of the twist of a polynomial algebra by a diagonalizable automorphism. We are interested in the case where the automorphism is not diagonalizable. In partic- ular, we present the case where the automorphism is represented by the Jordan block with ones on the diagonal and on the superdiagonal. Let S = Sn be the polynomial algebra with n variables over the complex numbers, and let = 0 BB BB B@ 11 11 ... ... 11 1 1 CC CC CA: Then using the convention that x0 = 0, and writing F for (F), we have x i = xi + xi-1. The twisted algebra Bn = S has multiplication xi xj = xixj + xixj-1: Notice that for each i n, we have an embedding Bi , Bn given by xi 7! xi. 14 To avoid the notation, we write yi1yi2 yit for the element xi1 xi2 xit . Then Bn is a quotient of the free algebra Chy1;:::;yni by a homogeneous quadratic ideal. Recall that each f 2 Bn corresponds to a unique polynomial f0 2 S.For example, y0i = xi,and(yiyj)0 = xix j = xixj + xixj-1. III.1.2. Remark: The goal is to describe the primitive ideal structure of Bn.The element y1 is homogeneous and -invariant, so by III.1.1, Bn=hy1i = Bn-1.By induction, we will understand the primitive ideal structure of Bn once we describe the primitive ideals in each Bi, i n, that do not contain y1. III.1.3. Example. Let n =2,soB2 = Chy1;y2i=hy1y2 -y2y1 -y21 i. We will show in Lemma III.2.3 that every primitive ideal in Bn contains a homogeneous, -invariant element. Suppose F = dX j=0 jxd-j1 xj2 2 S2 is -invariant. F - F = dX j=1 jxd-j1 [(xj2) - xj2] = dX j=1 jxd-j1 j-1X i=0 j i xj-i1 xi2 = dX j=1 j-1X i=0 j j i xd-i1 xi2 = d-1X i=0 dX j=i+1 j j i xd-i1 xi2: Then for each i, dX j=i+1 j j i = 0, and it follows that j = 0 for j =1;:::;d.This means that the only homogeneous -invariant elements of B2 are powers of y1.But y1 is normal, so every non-zero primitive ideal contains y1. The primitive ideals in 15 the commutative algebra B2=hy1i = C[y2] are the maximal ideals hy2 - gammai, gamma 2 C,so the non-zero primitive ideals in B2 are the ideals hy1;y2 - gammai, gamma 2 C. Finally, 0 is a prime ideal which is not the intersection of strictly larger primitive ideals, so 0 itself must be primitive [9]. Thus the primitive ideals in B2 are h0i,andhy1;y2 -gammai gamma 2 C: III.1.4. Note. From Example III.1.3, we see that for each n, the primitive ideals in Bn that contain y1;:::yn-2 are hy1;:::;yn-2i,andhy1;:::;yn-1;yn - gammai, gamma 2 C. Moreover, the ideal hy1;:::;yn-1i is prime but not primitive. III.1.5. Construction. Let n>2, and =( 1;:::; n-2) 2 Cn-2.Foreachj =1;:::;n- 2, let fj = " jX i=1 j-i+1y1yi # +(j - 1)y1yj+1 +(j +1)y1yj+2 - y2yj+1; and let I = hf1 ;:::;fn-2 i. We want to show that every primitive ideal in Bn that does not contain y1,containsI , for some 2 Cn-2. In fact, we will show that if P is primitive with y1 62 P then there is a unique 2 Cn-2 so that I P.Let g1 = 1y21 +2y1y3 - y22 , and for j 2, let gj = " jX k=1 kX i=1 (-1)k-i iy1yj-k+1 # + "j+1X i=3 (-1)j-iy1yi # +(j +1)y1yj+2 + "j+1X i=2 (-1)j-iy2yi # : 16 Then g1 = f1 ,and gj + gj-1 = jX k=1 kX i=1 (-1)k-i iy1yj-k+1 + j-1X k=1 kX i=1 (-1)k-i iy1yj-k -y1yj+1 +(j +1)y1yj+2 + jy1yj+1 - y2yj+1 = jX k=1 kX i=1 (-1)k-i iy1yj-k+1 + jX k=2 k-1X i=1 (-1)k-i-1 iy1yj-k+1 (j - 1)y1yj+1 +(j +1)y1yj+2 - y2yj+1 = 1y1yj + jX k=2 ky1yj-k+1 +(j - 1)y1yj+1 +(j +1)y1yj+2 - y2yj+1 = jX k=1 ky1yj-k+1 +(j - 1)y1yj+1 +(j +1)y1yj+2 - y2yj+1 = fj : Then hf1 ;:::;fn-2 i = hg1 ;:::;gn-2 i.LetG0 = 0, and for j =1;:::n- 2, let Gj =(gj )0 2 S.ThenG1 = 1x21 + x1x2 +2x1x3 - x22, and for j 2, Gj = " jX k=1 kX i=1 (-1)k-i ix1x j-k+1 # + "j+1X i=3 (-1)j-ix1x i # +(j +1)x1x j+2 "j+1X i=2 (-1)j-ix2x i # : We have 17 Gj = " jX k=1 kX i=1 (-1)k-i ix1(xj-k+1 + xj-k) # + "j+1X i=3 (-1)j-ix1(xi + xi-1) # +(j +1)x1(xj+2 + xj+1) + "j+1X i=2 (-1)j-ix2(xi + xi-1) # = " jX k=1 kX i=1 (-1)k-i ix1xj-k+1 # + "j-1X k=1 kX i=1 (-1)k-i ix1xj-k # + "j+1X i=3 (-1)j-ix1xi # + "j+1X i=3 (-1)j-ix1xi-1) # +(j +1)x1xj+2 +(j +1)x1xj+1 + "j+1X i=2 (-1)j-ix2xi # + "j+1X i=2 (-1)j-ix2xi-1 # = "j-1X k=0 k+1X i=1 (-1)k-i+1 ix1xj-k # + "j-1X k=1 kX i=1 (-1)k-i ix1xj-k # + "j+1X i=3 (-1)j-ix1xi # + " jX i=2 (-1)j-i-1x1xi) # +(j +1)x1xj+2 +(j +1)x1xj+1 + "j+1X i=2 (-1)j-ix2xi # + " jX i=1 (-1)j-i-1x2xi # = j-1X k=1 "k+1X i=1 (-1)k-i+1 ix1xj-k + kX i=1 (-1)k-i ix1xj-k # + 1x1xj -x1xj+1 +(-1)j-1x1x2 +(j +1)x1xj+2 +(j +1)x1xj+1 -x2xj+1 +(-1)j-2x2x1 = "j-1X k=1 k+1x1xj-k # + 1x1xj + jx1xj+1 +(j +1)x1xj+2 - x2xj+1 = " jX i=1 j-i+1x1xi # + jx1xj+1 +(j +1)x1xj+2 - x2xj+1: 18 Now, (Gj ) - Gj = jX i=1 j-i+1x1(x i - xi)+jx1(x j+1 - xj+1) +(j +1)x1(x j+2 - xj+2) - (x 2 x j+1 - x2xj+1) = jX i=1 j-i+1x1xi-1 + jx1xj +(j +1)x1xj+1 -(x1 + x2)(xj + xj+1)+x2xj+1 = jX i=1 j-i+1x1xi-1 + jx1xj +(j +1)x1xj+1 - x1xj - x1xj+1 -x2xj - x2xj+1 + x2xj+1 = jX i=2 j-i+1x1xi-1 +(j - 1)x1xj + jx1xj+1 - x2xj = j-1X i=1 j-ix1xi +(j - 1)x1xj + jx1xj+1 - x2xj = Gj-1 : It follows that the ideal (G1 ;:::;Gn-2 )is -invariant, and we claim that I0 = (G1 ;:::;Gn-2 ). Since I is homogeneous, I0 is homogeneous, so it su ces to show that every homogeneous element in I0 lies in (G1 ;:::;Gn-2 ). Suppose F 2 I0 is homogeneous of degree t.ThenF = f0,withf0 2 I , and deg(f)=t. Since each gi is normal modulo hg1 ;:::gi-1 i, we can write f = g1 f1 + + gn-2 fn-2,with fi 2 Bt-2.ThenF = G1 (f01 ) 2 + + Gn-2 (f0n-2) 2 2 (G1 ;:::;Gn-2 ), so in fact (G1 ;:::;Gn-2 )=I0 . This means that B=I = [S=(G1 ;:::;Gn-2 )] . Let v1 = x1;v2 = x2,andv3 = 1x1 + x2 + x3,soG1 = v1v3 - v22 . Assume that we have de ned vk = kX i=1 akixi, akk 6= 0 for each k =1;:::;j+ 2, and that the ideal (G1 ;:::;Gj ) is equal to the ideal (v1vi - v2vi-1;i =3;:::;j + 2). Note that this 19 means that there are bik, bkk 6=0sothatxi = iX k=1 bikvk.Wehave Gj+1 = j+1X i=1 j-i+2x1xi +(j +1)x1xj+2 +(j +2)x1xj+3 - x2xj+2 = v1 " j+1X i=1 j-i+2xi ! +(j +1)xj+2 +(j +2)xj+3 # - v2xj+2 = v1 " j+1X i=1 j-i+2xi ! +(j +1)xj+2 +(j +2)xj+3 # - v2 j+2X k=1 bj+2;kvk ! ; so Gj+1 - j+1X k=1 bj+2;k(v1vk+1 - v2vk) = v1 " j+1X i=1 j-i+2xi ! +(j +1)xj+2 +(j +2)xj+3 # - j+1X k=1 bj+2;kv1vk+1 ! - bj+2;j+2v2vj+2 = v1 " j+1X i=1 j-i+2xi ! +(j +1)xj+2 +(j +2)xj+3 - j+1X k=1 bj+2;kvk+1 # -bj+2;j+2v2vj+2: = v1 " j+1X i=1 j-i+2xi ! +(j +1)xj+2 +(j +2)xj+3 - j+1X k=1 bj+2;k k+1X i=1 ak+1;ixi # -bj+2;j+2v2vj+2: Set vj+3 = 1b j+2;j+2 "j+1X i=1 ixj-i+1 +(j +1)xj+2 +(j +2)xj+3 - j+1X k=1 k+1X i=1 bj+2;kak+1;ixi # : Then Gj+1 = "j+1X k=1 bj+2;k(v1vk+1 - v2vk) # + bj+2;j+2(v1vj+3 - v2vj+2). By induction, we have a change of variables so that I0 =(G1 ;:::;Gn-2 )=(v1vj - v2vj-1;j=3;:::n): 20 We claim that (v1vj - v2vj-1;j=3;:::n)=(v1;v2) \ (vivj+1 - vi+1vj;i;j=1;:::;n- 1): Set P1 =(v1;v2), and P2 =(vivj+1 - vi+1vj;i;j =1;:::;n- 1). It is clear that (v1vj-v2vj-1;j=3;:::n) P1 \P2, so it su ces to show that \n-1j=3 V(v1vj-v2vj+1) V(P1 \P2)=V(P1)[V(P2). Take p =(p1;:::;pn) 2\n-1j=3 V(v1vj-v2vj+1), and assume p 62 V(P1). If p1 =0,thensincep1p3 - p22 =0,wemusthavep2 = 0, contradicting that p 62 V(P1). Then p1 6= 0. Write p1 = tn-1,andp2 = tn-2u. p1p3 = p22,so tn-1p3 = t2n-4u2,andp3 = tn-3u2. Assume pj = tn-juj-1.Thenp1pj+1 = p2pj,so tn-1pj+1 = tn-2utn-juj-1 = t2n-j-2uj. It follows that pj+1 = tn-j-1uj. By induction, p =(tn-1;tn-2u;:::;tun-2;un-1) 2V(vivj+1 - vi+1vj) for all i and j. Now, P2 is the kernel of the map C[v1;:::;vn] ! C[t;u] that sends vi to tn-iui-1, so P2 is prime. Then I0 has primary decomposition I0 = P1 \ P2,withPi prime, P 1 = P1, and we claim that P2 is also -invariant. First we note that P 2 is an ideal in S. In fact, since P2 is a prime ideal, P 2 is also prime. Then we have I0 =(I0 ) =(P1 \ P2) = P 1 \ P 2 = P1 \ P 2 : But I0 = P1 \ P2, so by the uniqueness part of primary decomposition, [4], we must have P 2 = P2. 21 For i =2;:::n- 2, and j = i +1;:::;n- 1, let Hij = vivj+1 - vi+1vj, and de ne hij 2 B by (hij )0 = Hij .Set P = hf1 ;:::;fn-2 ;hij ;i=2;:::;n;j= i +2;:::;ni; so P0 = P2 =(G1 ;:::;Gn-2 ;Hij ;i=2;:::;n;j= i +2;:::;n): Then B=P = S=(P0 ) ; is a twist of the algebra S=(P0 ) = C[tn-1;utn-2;:::;un-2t;un-1]; so P is prime. For f = f(y1;:::;yn-k) 2 Bn-k Bn, de ne CB k f to be f(yk+1;:::;yn). We want to show: Theorem III.1.6. If P is a primitive ideal in Bn that does not contain y1,then P = P for some 2 Cn-2. By induction, the primitive ideals in Bn are hy1;:::;yn-1;yn - i; 2 C; hy1;:::;yn-2i;and hy1;:::;yk; CB k f1 ;:::; CB k fn-k-2 ; CB k hij j i =2;:::;n- k - 2;j= i +2;:::;n- ki k =0;1;:::;n- 3; 2 Cn-k-2: In the next section we establish some preliminary results, and then complete the proof of Theorem III.1.6. 22 III.2. Formulas Let ' =ad(y1) 2 Aut(B). Let f 2 Bd and F = f0 2 S.Then['(f)]0 = x1F - Fx d1 = x1(F - F), so '(f)=0ifandonlyiff is -invariant. We have the following formulas. Lemma III.2.1. (i) '(yd)=yd-1y1. (ii) 'd-1(yd)=yd1 . (iii) If f 2 Bd is -invariant and F = f0,then[f;y2]=dy1f. In particular, [yd1 ;y2]= dyd+11 . (iv) 'N(fg)= NX j=0 N j 'j(f)'N-j(g). Proof. (i) ('(yd))0 =(y1yd -ydy1)0 = x1x d -xdx 1 = x1(xd +xd-1)-xdx1 = x1xd-1 = (yd-1y1)0 (ii) Induct. If 'd-1(yd)=yd1 ,then'd(yd+1)='d-1('(yd+1)) = 'd-1(ydy1)= 'd-1(yd)y1 = yd+11 . (iii) [f;y2]0 = F(x d2 ) - x2F = F(x2 + dx1) - x2F = dx1F = (dy1f)0. (iv) This is the product rule for derivations. III.2.2. Notes. 1. Lemma III.2.1(ii) implies that 'd(yd) = 0, so (iv) implies that for each f 2 B there exists N such that 'N(f)=0. 23 2. If P is an ideal containing a = 0 + tX i=1 iyi, t 6= 0, then Lemma III.2.1(ii) implies that P contains 't-1(a)=yt1.ThusifP is a prime ideal, and P contains a linear element, then P contains y1. 3. If I is a -invariant ideal in B and g 2 B, we will say that g is -invariant modulo I if (g0) - g0 2 I0. We can use the argument used in the proof of Lemma III.2.1(iii) to show that if f 2 Bj is -invariant modulo I,withF = f0,then[f;y2] jy1f modulo I. Lemma III.2.3. Let I and P be prime ideals in B, such that I & P, I0 is - invariant, and y1 62 P.ThenP contains an element that is nonzero, homogeneous, irreducible, and -invariant modulo I. Proof. Let g 2 P n I.ChooseN minimal with 'N(g) 2 I, and set f = 'N-1(g). Then '(f)0 = x1[(f0) - f0] 2 I0,withx1 62 I0 so f is -invariant modulo I. Write f = dX i=0 fi with fi homogeneous of degree i.Eachfi is -invariant modulo I,soby III.2.2.3, [f;y2] - dX i=0 ifiy1 2 I.Thendf y1 - [f;y2] 2 P with df y1 - [f;y2] df y1 - dX i=0 ifiy1 modulo I d-1X i=0 (d - i)fiy1 modulo I: By induction on d, we may assume that f is homogeneous. Finally, suppose f0 = F1F2 Ft,withFi irreducible. Since (f0) = f0, permutes fF1;:::;Ftg,sothere 24 exists s so that F si = Fi for each i.But is a unipotent automorphism of each of the vector spaces Sj,sowemusthaveF i = Fi. Theorem III.2.4. Every prime ideal in B that does not contain y1 is of the form hg1;g2;:::;gti,whereg1;g2;:::;gt is a regular sequence with gi homogeneous irre- ducible and -invariant modulo hg1;:::gi-1i. Proof. Let P be a primitive ideal. The ring B is prime Noetherian, so by Lemma III.2.3 it su ces to show that if I is a prime ideal in P,andg is nonzero, homogeneous, irreducible, and -invariant element modulo I,theng is regular, and the ideal I +hgi is prime. These follow from Notes II.1.3. III.3. Primitive Ideals Lemma III.3.1. Every primitive ideal that does not contain y1 contains I for some 2 Cn-2. Proof. Let P be primitive, with y1 62 P and let ; 2 Cn-2 with 1 6= 1.We want to nd gamma so that f1gamma 2 P, so assume f1 ;f1 62 P. The elements f1 and f1 are regular and normal in B, hence regular and normal modulo the prime ideal P,(Note II.1.3.2). By Remark II.1.2, f1 and f1 determine the same automorphism of B=P. Then by Remark II.1.7 there exists c 2 C so that f1 - cf1 2 P.Butf1 - cf1 = 1y21 +2y1y3 - y22 - c 1y21 - 2cy1y3 - cy22 =( 1 - c 1)y21 +2(1- c)y1y3 - (1 - c)y22 .If 25 c =1,thensince 1 6= 1, P contains y21 which would imply that y1 2 P.Thusc 6=1, and P contains f1gamma for every gamma 2 Cn-2 with gamma1 = 1 - c 11 - c . Assume we have gamma1;:::;gammaj, such that if 2 Cn-2 with i = gammai, for i =1;:::j,thenfi 2 P for each i =1;:::;j. Let =(gamma1;:::;gammaj; j+1;:::; n-2), and =(gamma1;:::;gammaj; j+1;:::; n-2), with j+1 6= j+1, and assume fj+1 ;fj+1 62 P. fj+1 and fj+1 are -invariant modulo P, hence regular and normal modulo P. As above, there exists b 6=1sothatfj+1 -bfj+1 2 P. Set gammaj+1 = j+1 - b j+11 - b ,sofj+1gamma 2 P. By induction, we can thus construct gamma so that Igamma P. We are now ready to prove Theorem III.1.6 Proof of Theorem III.1.6. Let P be primitive in B, and assume y1 62 P. By Lemma III.3.1, P contains I for some , so by Lemma III.2.3, P = hI ;q1;:::;qsi,withqi ho- mogeneous, -invariant and irreducible modulo hI ;q1;:::;qi-1i.ThusP corresponds to a prime ideal P0 in S containing (G1 ;:::;Gn-2 )=(v1;v2) \ (vivj+1 - vi+1vj;i;j=1;:::;n- 1): Then P0 contains either (v1;v2)orP0 =(vivj+1 -vi+1vj;i;j=1;:::;n-1) but since y1 62 P,wemusthaveP0 P0.NowP0 is coheight two, so if P0 6= P0 ,thenP0 is coheight one or zero. In either case, P0 contains a linear polynomial. But then by Note III.2.2.2, P0 contains x1, contradicting that y1 62 P.Wehavethenshownthat the primitive ideals in B that do not contain y1 are of the form P = hf1 ;:::;fn-2 ;hij ;i=2;:::;n;j= i +2;:::;ni: 26 Now suppose P is primitive, and P contains y1;:::;yk, but yk+1 62 P.ThenP corresponds to a primitive ideal P in B=hy1;:::;yki = Bn-k. Under this isomorphism, image of yk+1 is y1,soP corresponds to a primitive ideal in Bn-k that does not contain y1. By the above, the image of P in Bn-k is P for some . Since the preimage of f 2 Bn-k is CB k f,wehave P = hy1;:::;yk; CB k f1 ;:::; CB k fn-k-2 ; CB k hij ;i=2;:::;n- k - 2;j= i+2;:::;n- ki: 27 CHAPTER IV THE POISSON MANIFOLD Here we describe the Poisson manifold associated to the twisted algebra Bn. IV.1. The Poisson Bracket Let R = C[h]. Grade the polynomial ring R[x1;:::;xn]bydeg(xi) = 1, and deg(h)=0. LetA = R[x1;:::;xn] h ,where h is given on degree one elements in coordinates x1;:::;xn, by right multiplication by h = 0 BB BB B@ 1 h 1 h ... ... 1 h 1 1 CC CC CA: Each element f 2 A corresponds to a unique polynomial f+ 2 C[h;x1;:::;xn]. Eval- uating f+ at h = 0 gives a polynomial in S. The map from A to S that takes f to f+(0;x1;:::;xn) is a ring epimorphism, whose kernel is hhi,soA=hhi = S. Similarly, the map from A to B that takes f to the unique element e 2 B with f+(1;x1;:::;xn)=(e)0, is an epimorphism with kernel hh - 1i,soA=hh - 1i = Bn. The Drinfel'd Poisson bracket (II.2.2) on S is given by fxi;xjg = xi xj - xj xih mod hhi; 28 where is multiplication in A. Again, using the convention that x0 =0,wehave fxi;xjg = xixj + hxixj-1 - xjxi - hxjxi-1h mod hhi = xixj-1 - xjxi-1 mod hhi: This yields the following formulas: fx1;xjg = x1xj-1;j>1; fxi;xjg = xixj-1 - xi-1xj;i;j>1: We de ne di erential operators ! = nX j=2 xj-1 @@x j ,and = nX j=1 xj @@x j , and observe that fxi;-g = xi! - xi-1 : Note that for homogeneous f 2 Sj, f = jf. It follows that if I is a homogeneous ideal of S,with!I I,thenI is Poisson. In particular, the ideal (x1)inS is Poisson, and the variety of x1, V(x1), is a Poisson submanifold of (An;S;f;g)which is isomorphic to (An-1;Sn-1;f;g). This means that as in the analysis of the primitive ideal structure, we can concentrate on describing the symplectic leaves that are not contained in V(x1). IV.1.1. Example. Let n =2,soS = C[x1;x2]. The Poisson bracket is given by fx1;x2g = x21. A 0-dimensional symplectic leaf is the variety of a maximal ideal m, with fm;Sg m. These are the ideals (x1;x2 - gamma), gamma 2 C. The form determined by f;g has rank 2 at each p 2 A2 nV(x1). It follows that the symplectic leaves are the points f(0;gamma)g, gamma 2 C and the 2-dimensional leaf A2 nV(x1). 29 IV.1.2. Construction. Let n>2. For each =( 1;:::; n-2) 2 Cn-2,let Fj = jX i=1 j-i+1x1xi +(j +1)x1xj+2 - x2xj+1; and let Q be the ideal (F1 ;:::;Fn-2 ). We have !F1 = !( 1x21 +2x1x3 - x22) =2x1x2 - 2x1x2 =0: For j>1, !(Fj )= " jX i=1 j-i+1x1!(xi) # +(j +1)x1!(xj+2) - !(x2)xj+1 - x2!(xj+1) = " jX i=2 j-i+1x1xi-1 # +(j +1)x1xj+1 - x1xj+1 - x2xj = "j-1X i=1 j-ix1xi # + jx1xj+1 - x2xj = Fj-1 : Thus each Q is a Poisson ideal. Furthermore, for p =(p1;:::;pn) 2 An nV(x1)there is a unique so that p 2V(Q ): indeed, set 1 = -2p1p3 + p 2 2 p21 ; and j = - Pj-1 i=1 j-ip1pi+1 - (j +1)p1pj+2 + p2pj+1 p21 ;j>1: Since Fj = jx21 + jX i=2 j-i+1x1xi+1 +(j +1)x1xj+2 - x2xj+1, =( 1;:::; n-2)is the unique element of Cn-2 with p 2V(Q ). We will show that each symplectic leaf for f;g, that is not contained in V(x1), is an open subvariety of V(Q ) for some . 30 IV.2. Symplectic Leaves Let V = V(Q ) nV(x1). Then An nV(x1) is the disjoint union of V , 2 Cn-2. For f = f(x1;:::;xn-k) 2 Sn-k Sn,let CB k f = f(xk+1;xk+2;:::;xn). Then for each 2 Cn-k-2,let CB k Q =(x1;:::;xk; CB k F1 ;:::; CB k Fn-2 ), and let CB k V = V( CB k Q ) nV(x1;:::xn-1). We want to show: Proposition IV.2.1. The symplectic foliation of An associated to f;g consists of the 0-dimensional leaves f(0;:::;0;gamma)g, gamma 2 C, and the two dimensional leaves V(x1;:::;xn-2) and CB k V = V(x1;:::;xk; CB k F1 ;:::; CB k Fn-k-2 ) nV(x1;:::;xn-1), k =0;1;:::;n- 3, 2 Cn-k-2. As an immediate corollary, we have Corollary IV.2.2. There is a one to one correspondence between primitive ideals in the twisted algebra B = S and the symplectic leaves for the Poisson structure induced by . To prove Proposition IV.2.1 inductively, it su ces to show that the symplectic leaves for f;g that are not contained in V(x1) are the varieties V = V(F1 ;:::;Fn-2 )n V(x1;:::;xn-1), 2 Cn-2. We start by showing that each V is an irreducible 2- dimensional variety. Lemma IV.2.3. For each , there is a change of coordinates so that V = f(tn-1;tn-2u;:::;tun-2;un-1)jt;u 2 C;t6=0g: 31 Proof. Let v1 = x1;v2 = x2,andv3 = 1x1 +2x3.ThenF1 = v1v3 - v22 . Assume that for for k =3;:::;d+ 2, we have de ned vj = jX i=1 cjixi, cji 2 C, cjj 6=0sothat the ideal (F1 ;:::;Fd ) is equal to the ideal (v1vj - v2vj-1;j=3;:::;d+2). Note that this means that for each i,thereexisteij so that xi = iX j=1 eijvj.Wehave Fd+1 = d+1X i=1 d-i+2x1xi ! +(d +2)x1xd+3 - x2xd+2 = v1 " d+1X i=1 d-i+2xi ! +(d +2)xd+3 # - v2 d+2X j=1 ed+2;jvj = v1 " d+1X i=1 d-i+2xi ! +(d +2)xd+3 # - d+2X j=1 ed+2;jv2vj: Then Fd+1 - d+1X j=1 ed+2;j(v1vj-1 - v2vj) = v1 " d+1X i=1 d-i+2xi ! +(d +2)xd+3 # - "d+1X j=1 ed+2;jv1vj-1 # - ed+2;d+2v2vd+2 = v1 " d+1X i=1 d-i+2xi ! +(d +2)xd+3 # - d+2X j=1 ed+2;jv1 j-1X i=1 cj-1;ixi ! - ed+2;d+2v2vd+2 = v1 d+1X i=1 d-i+2xi ! +(d +2)xd+3 - d+2X j=1 j-1X i=1 ed+2;jcj-1;ixi - ed+2;d+2v2vd+2: Set vd+3 = 1e d+2;d+2 " d+1X i=1 d-i+2xi ! +(d +2)xd+3 - d+2X j=1 j-1X i=1 ed+2;jcj-1;ixi # ,so 32 Fd+1 = ed+2;d+2(v1vd+3 - v2vd+2)+ d+1X j=1 ed+2;j(v1vj-1 - v2vj). By induction, Q =(v1vi - v2vi-1;i=3;:::;n) =(v1;v2) \ (vivj - vi+1vj-1ji +1