Rings Graded Equivalent to the Weyl Algebra

Journal of Algebra 321 (2009) 495531

Contents lists available at ScienceDirect

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ings graded equivalent to the Weyl algebra

san J. Sierra

partment of Mathematics, University of Washington, Seattle, WA 98195-4350, United States

r t i c l e i n f o a b s t r a c t

ticle history:ceived 9 November 2007ailable online 12 November 2008mmunicated by Em Zelmanov

ywords:aded Morita theoryaded module categorytegory equivalenceeyl algebra

We consider the rst Weyl algebra, A, in the Euler gradation, andcompletely classify graded rings B that are graded equivalent to A:that is, the categories gr-A and gr-B are equivalent. This includessome surprising examples: in particular, we show that A is gradedequivalent to an idealizer in a localization of A.We obtain this classication as an application of a general Morita-type characterization of equivalences of graded module categories.Given a Z-graded ring R , an autoequivalence F of gr-R , and anitely generated graded projective right R-module P , we showhow to construct a twisted endomorphism ring EndFR (P ) and prove:

Theorem. The Z-graded rings R and S are graded equivalent if and onlyif there are an autoequivalenceF of gr-R and a nitely generated gradedprojective right R-module P such that the modules {Fn P } generate gr-Rand S = EndFR (P ).

2008 Elsevier Inc. All rights reserved.

ntents

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4962. Graded equivalences and twisted endomorphism rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4983. The Picard group of a graded module category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5054. Graded modules over the Weyl algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5085. The Picard group of gr-A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5136. Classifying rings graded equivalent to the Weyl algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5197. The graded K -theory of the Weyl algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529knowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531ferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531

E-mail address: [email protected].

21-8693/$ see front matter 2008 Elsevier Inc. All rights reserved.

i:10.1016/j.jalgebra.2008.10.011

496 S.J. Sierra / Journal of Algebra 321 (2009) 495531

1.

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The rst Weyl algebra A = k{x, y}/(xy yx 1), where k is an algebraically closed eld of charac-ristic 0, is in many ways the fundamental ring of noncommutative algebra. Remarkably, although its been studied since the 1930s, it continues to inspire new developments in the eld.One of the most fruitful areas of recent activity has come from considering rings Morita equiv-

ent to A. The starting point of these results was a theorem of Stafford [17, Corollary B]: up toomorphism, the domains Morita equivalent to A are in natural bijection with the orbits of Autk(A)the set of (module) isomorphism classes of right ideals of A. Using earlier work of Cannings and

olland [6], Berest and Wilson [4] then exhibited a rich geometry related to Staffords result. Theyowed [4, Theorem 1.1] that isomorphism classes of right ideals of A correspond to points in thelogeroMoser space C =N0 CN , where each CN should be thought of as a noncommutative defor-ation of the Hilbert scheme of N points in the ane 2-plane, and that Autk(A) also acts naturally on. Further, they proved [4, Theorem 1.2] that the orbits of Autk(A) on C are precisely the CN . (The factat the orbits of Autk(A) acting on the set of isomorphism classes of right ideals are parameterizedthe nonnegative integers was independently proved by Kouakou [9].) Thus we have:

eorem 1.1 (BerestWilson, Kouakou, Stafford). The isomorphism classes of domains Morita equivalent to Ae in 11 correspondence with the nonnegative integers.

In this paper, we solve the analogous problem for graded module categories. That is, we grade Agiving x degree 1 and y degree 1. We ask: what are the Z-graded rings B such that the category-A of graded right A-modules is equivalent to the category gr-B of graded right B-modules? Weill call such an equivalence a graded equivalence and will say that A and B are graded equivalent,though we caution that this terminology is not universally agreed upon in the literature.Of course, any progenerator P for the full module category of A that happens to be graded inducesgraded equivalence between A and EndA(P ); we refer to such an equivalence of categories as aaded Morita equivalence. Since we are interested in nontrivial graded equivalences, we ask: what aree graded Morita equivalence classes of Z-graded rings B such that A and B are graded equivalent?The answer to this question is quite surprising. Given a positive integer n and a set J {0, . . . ,

1}, dene a ring S( J ,n) as follows: rst, let

f =n1i=0i / J

(z + i).

t W = W ( J ) be the ring generated over k by X , Y , and z, with relations

Xz zX = nX, Y z zY = nY ,XY = f , Y X = f (z n).

Dene

S( J ,n) = k[z] +(

i J(z + i)

)W W .

en we prove:

eorem 1.2 (Theorem 6.3). Let S be a Z-graded ring. Then S is graded equivalent to A if and only if S is

aded Morita equivalent to some S( J ,n).

S.J. Sierra / Journal of Algebra 321 (2009) 495531 497

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hoinalPiFurthermore, we give a precise enumeration of the graded Morita equivalence classes of ringsaded equivalent to A:

eorem 1.3 (Corollary 6.4). The graded Morita equivalence classes of rings graded equivalent to A are in 11rrespondence with pairs (J,n) where n is a positive integer and J is a necklace of n black and white beads.

Theorem 1.2 says in particular that A = S(,1) is graded equivalent to B = k+ xA[y1] = S({0},1).e ring B is the idealizer in A[y1] of the right ideal xA[y1]; that is, B = { A[y1] | x [y1]} is the maximal subring of A[y1] in which xA[y1] is a two-sided ideal. While it is knownat simplicity is not necessarily preserved under graded equivalence, it is still unexpected to nd anuivalence between A and an idealizer in a localization of A. For example, B has a 1-dimensionaladed representation, while all A-modules are innite-dimensional. By construction, B has a non-ivial two-sided ideal, and is not a maximal order, while A, as is well known, is simple. In fact, B is aandard example of a ring that fails the second layer condition governing relationships between primeeals, whereas A trivially satises the second layer condition. (See the discussion of Example 6.15 fornitions.) Thus we obtain:

rollary 1.4. The properties of having a nite-dimensional graded representation, being a maximal order, andtisfying the second layer condition are not invariant under graded equivalence.

Another ring occurring in Theorem 1.2 is the Veronese ring A(2) =nZ A2n = S(,2). By The-em 1.2, A and A(2) are graded equivalent. Of course one expects that Proj A (in the appropriatense) and Proj A(2) will be equivalent, but, as far as we are aware, this is the rst nontrivial examplean equivalence between the graded module categories of a ring and its Veronese.Theorem 1.2 is an application of general results on equivalences of graded module categories. Given

graded ring R , an autoequivalence F of gr-R , and a nitely generated graded right R-module P , weow that there is a natural way to construct a twisted endomorphism ring EndFR (P ). We prove:

eorem 1.5 (Theorem 2.12). Let R and S be Z-graded rings. Then R and S are graded equivalent if and only ifere are a nitely generated graded projective right R-module P and an autoequivalence F of gr-R such thatn P }nZ generates gr-R and S = EndFR (P ).

If F is the shift functor in gr-R , then EndFR (P ) is the classical endomorphism ring of P ; thuseorem 1.5 generalizes the Morita theorems.If R is a Z-graded ring, we dene (following [2]) the Picard group of gr-R to be the group of autoe-ivalences of gr-R , modulo natural isomorphism. Theorem 1.5 illustrates the fundamental techniquer all of our results: relate equivalences between graded module categories to the Picard groups ofe categories. Given Z-graded rings R and S , an equivalence of categories : gr-R gr-S inducesisomorphism of Picard groups, which we write : Pic(gr-S) Pic(gr-R); we refer to applying pulling back along . Let F be the autoequivalence of gr-R that comes from pulling back the shiftnctor in gr-S along . It turns out that properties of and S can be deduced from the propertiesF ; in particular, F is the autoequivalence F from Theorem 1.5.This paper is a companion to the work in [14]. There, using results of hn and Mrki [1], weve a Morita-type characterization of graded equivalences in terms of certain bigraded modules. Thisaracterization tells us that the rings graded equivalent to A are endomorphism rings (in an appro-iate sense) of these modules. However, the modules are quite large and correspondingly dicult toork with. One motivation for the current paper is to develop techniques to make the analysis in [14]ncrete in a specic case.This paper has seven sections. In Sections 2 and 3 we prove Theorem 1.5 and other results thatld for any Z-graded ring. In particular, we characterize graded Morita equivalences and Zhang twiststerms of the Picard group. In Sections 4 and 5 we analyze the graded module category of the Weylgebra and its Picard group. The key result in these two sections is Corollary 5.11, describing the

card group of gr-A explicitly.


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sefrriRreidIn Section 6 we use the results of the previous two sections to prove Theorem 1.2 and Theorem 1.3,d give several examples. In Section 7 we describe the graded K -theory of A, and in particular showat, in contrast to the ungraded case, if P Q = P Q where P , Q , and Q are nitely generatedaded projective modules, then Q = Q .In the remainder of the Introduction, we establish notation. We x a commutative ring k. (Be-

nning in Section 4, k will be an algebraically closed eld of characteristic 0.) For us, a graded ringeans a Z-graded k-algebra, and all categories and category equivalences are assumed to be k-linear.R is a graded ring, the category gr-R consists of all Z-graded right R-modules, and mod-R is thetegory of all right R-modules; similarly, we form the left module categories R-gr and R-mod. Mor-isms in gr-R and R-gr are homomorphisms that x degree; if M and N are graded R-modules, werite homR(M,N) to mean Homgr-R(M,N), and write HomR(M,N) to denote Hommod-R(M,N). Wemilarly write ExtiR and ext

iR for the derived functors of HomR and homR , respectively.

If R is a graded ring, then the shift functor on gr-R (or R-gr) sends a graded right or left moduleto the new module M1 = jZM1 j , dened by M1 j = M j1. We will write this functor as:

SR : M M1.

e caution that this is the opposite of the standard convention.)Since we will work primarily with right module categories, for the remainder of the paper unless

herwise specied an R-module M is a right module.

Graded equivalences and twisted endomorphism rings

Let R and S be graded equivalent graded rings; we write gr-R gr-S . Let : gr-R gr-S be anuivalence of categories. In this section, we show that, similarly to the classical Morita theorems, weay construct S and as a twisted endomorphism ring and twisted Hom functor, respectively.Let : gr-S gr-R be a quasi-inverse for and let G be an autoequivalence of gr-S . We dene

e pullback of G along to be the autoequivalence

G = G

gr-R; likewise, we may push forward an autoequivalence F of gr-R to the autoequivalence F =F of gr-S .We will show that the ring S and the equivalence of categories may be reconstructed from

SS . To do this, let P = (S). Then we show that there is a natural way to dene a twisted Hom func-r HomFR (P , ) and a twisted endomorphism ring End

FR (P ) = HomFR (P , P ) such that S = EndFR (P )

d is naturally isomorphic to HomFR (P , ). We prove Theorem 1.5, which, by characterizingaded equivalences in terms of twisted Hom functors, generalizes the Morita theorems. As a corollary,e obtain a simple proof of a result of Gordon and Green characterizing graded Morita equivalences.e also describe the twisted Hom functors that correspond to Zhang twists.This section is somewhat technical because of the need to be careful with subtleties of notation;

e Remark 2.14 for an indication of the potential pitfalls. One helpful notational technique, borrowedom the authors previous paper [14], is to work with Z-algebras, which are more general than gradedngs. Recall that a Z-algebra is a ring R without 1, with a (ZZ)-graded vector space decomposition=i, jZ Rij , such that for all i, j, l Z we have Rij R jl Ril , and if j = j , then Rij R jl = 0. Wequire that the diagonal subrings Rii have units 1i that act as a left identity on all Rij and a rightentity on all R ji . If R is a Z-graded ring, the Z-algebra associated to R is the ring

R =

Rij,

i, jZ


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where Rij = R ji . Let modu-R denote the category of unitary right R-modules; that is, modules Mch that MR = M . Then it is easy to see that the categories gr-R and modu-R are isomorphic, ande will identify them throughout.We recall some terminology and results from [14]. Recall that a Z-algebra E is called principal if

ere is a k-algebra automorphism of E that maps Eij Ei+1, j+1 for all i, j Z; is called aincipal automorphism of E . We recall:

oposition 2.1. (See [14, Proposition 3.3].) Let E be a Z-algebra. Then E is principal if and only if there is aaded ring B such that E = B via a degree-preserving map.

We review the constructions in Proposition 2.1. Suppose that E is principal, and let be a princi-l automorphism of E . Then we may construct a ring E , which we call the compression of E by .a graded vector space, E = E0 =iZ E0i . The multiplication on E is dened as follows: if (E )i = E0i and b (E ) j = E0 j , then

a b = a i(b) E0i Ei,i+ j E0,i+ j =(E)i+ j .

e proof of [14, Proposition 3.3] shows that E = E . Conversely, if S is a graded ring, then S has anonical principal automorphism , given by the identications Sij = S ji = Si+1, j+1.Suppose that M =i, jZMij is a bigraded right R-modulethat is, each Mi = jZMij is a

aded R-submodule of M and M =iZMi . Assume for simplicity that each Mi is nitely gener-ed. Then we dene the endomorphism Z-algebra of M to be

HR(M,M) =i, jZ

HR(M,M)i j,

here each HR(M,M)i j = homR(M j,Mi). Further, for any graded R-module N =

iZ Ni there is aaded object

HR(M,N) =jZ

homR(M j,N).

t H = HR(M,M). Then HR(M,N) is naturally a right H-module, since M has a left H-action. Ifere is a graded ring S with an isomorphism S = H , then HR(M, ) may be regarded as a functorom gr-R to gr-S .We recall from [14] the following characterization of graded equivalences:

oposition 2.2. (See [14, Proposition 2.1].) Let R and S be graded rings. Then gr-R gr-S if and only if therea bigraded right R-module MR =i, jZMij such that) MR is a projective generator for gr-R with each Mi = jZMij nitely generated;) the Z-algebras S and HR(M,M) are isomorphic via a degree-preserving map.

Further, if M is as above, then HR(M, ) : gr-R gr-S and S M : gr-S gr-R are quasi-inverseuivalences; and if : gr-S gr-R is an equivalence of categories, then M = (S S ) satises (1) and (2),d = S M.

Let R , S , and M satisfy (1) and (2) of Proposition 2.2; let H = HR(M,M). Then H = S is principal,

ith a principal automorphism induced from the canonical principal automorphism of S . Thus


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Thi, = S , and this isomorphism gives the equivalence between gr-R and gr-S . That is, if N =iZ Nia graded R-module, the S-action on HR(M,N) is given by

f s = f j(s) (2.3)

r f HR(M,N) j = homR(M j,N) and s Si = (H)i = H0i .We now dene twisted endomorphism rings. Let R be a graded ring, let F be an autoequivalencegr-R , and let P be a nitely generated graded projective R-module. We say that F is P-generativethe modules {F i P }iZ generate gr-R; more generally F is generative if it is P -generative for someitely generated graded projective P . It is easy to see that if : gr-R gr-S is a category equiva-nce, then the pullback SS is generative in gr-R . The next proposition shows that this process alsoorks in reverse: there is a natural way to obtain a ring graded equivalent to R from a generativetoequivalence of gr-R .We pause to discuss a subtlety of notation. If F is an autoequivalence of gr-R that is not antomorphism, we must be careful about dening the powers F i for i < 0. Let F1 be a quasi-inverseF . If i < 0, we dene F i = (F1)i ; we dene F0 = Idgr-R .In fact, we will work in even more generality. Let F = {Fi}iZ be a sequence of autoequivalencesgr-R . We will say that F is a Z-sequence if:

) F0 = Idgr-R ;) There are natural compatibility isomorphisms i j : Fi+ j FiF j for all i, j Z that satisfy:

For all j, the maps 0 j, j0 : F j F j are the identity (2.4)

and

for all i, j, l Z, the following diagram commutes:

Fi+ j+li, j+l

i+ j,l

FiF j+l

Fi( jl)

Fi+ jFli jFl

FiF jFl.

(2.5)

For example, for any autoequivalence F , the sequence {F j} jZ is a Z-sequence. For the compati-lity isomorphisms i j : F i+ j F iF j , let

1,1 : Idgr-R F1 F

any natural isomorphism. Then for any N gr-R , dene

1,1N : N F F1N

be the element of homR(N,F F1N) corresponding under F1 to

1,1F1N : F1N F1F F1N.

en (2.5) is satised for (i, j, l) = (1,1,1), and there is a unique way to dene i j for arbitrary

j so that (2.5) holds for all i, j, l.


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Otooposition 2.6. Let R be a graded ring and let F be an autoequivalence of gr-R. Then F is generative if andly if there are a graded ring S and an equivalence of categories : gr-R gr-S such that F = SS .

oof. Given an equivalence : gr-R gr-S with quasi-inverse , by Proposition 2.2 the R-module(S S) =iZ(SS )i( S) generates gr-R , so the autoequivalence SS is generative.Conversely, suppose that F is a P -generative autoequivalence of gr-R and that F is a Z-sequence

ith compatibility isomorphisms i j , such that F1 = F . Let M =iZ Fi P , and let H = HR(M,M).e autoequivalence F and the maps i j induce a principal automorphism of H as follows: ifHki = homR(Fi P ,Fk P ), dene (s) to be the composition

Fi+1P1i F Fi P

F(s) F Fk P(1k)1 Fk+1P

an element of Hk+1,i+1 = homR(Fi+1P ,Fk+1P ). The compatibility condition (2.5) ensures that weve

j(s) = ( jk)1 F j(s) ji (2.7)r all j Z.Let S be the compressed ring H . Then by Propositions 2.1 and 2.2, the functor HR(M, ) is anuivalence between gr-R and gr-S . We need to show that F is isomorphic to the pullback of theift functor in gr-S , or, equivalently, that SS HR(M, ) = HR(M, ) F .Let N gr-R . Dene a map N : SS(HR(M,N)) HR(M,FN) as follows: if n (SS HR(M,N)) j =

R(M,N) j1 = homR(F j1P ,N), let N(n) be the composition

N(n) : F j P1, j1 F F j1P

F(n)FN.

early N is an isomorphism of graded k-modules; we verify that it is an S-module map.Let s Si = homR(Fi P , P ) and let n SS(HR(M,N)) j = HR(M,N) j1. We need to prove that(ns) = N(n)s. From (2.3) we have that the S-action on HR(M,N) is dened by ns = n j1(s).erefore, N (ns) is given by the diagram

Fi+ j P

1,i+ j1

F F j1PF(n)

FN.

F Fi+ j1P

F( j1(s))

F( j1,i)F F j1Fi P

F F j1(s)

n the other hand, the S-action on HR(M,FN) is given by N (n)s = N (n) j(s), which correspondsthe diagram

Fi+ j Pji

j(s)

F jFi P F j P1, j1

N (n)

F F j1P FN.

F j(s) F(n)


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ridecawe need to prove the commutativity of

Fi+ j P ji

1,i+ j1

F jFi PF j(s) F j P

1, j1 F F j1PF(n)

FN.

F Fi+ j1PF( j1,i) F F j1Fi P

F F j1(s)

is follows from the compatibility condition (2.5) and the naturality of 1, j1.The naturality of is even easier. Let f : N N be a homomorphism of graded R-modulesd let n SS(HR(M,N)) j as above. We leave it to the reader to verify that the naturality of theomorphism reduces to establishing the commutativity of

F j P1, j1 F F j1P

F(n)

F( f n)

FNF( f )

FN .

is is immediate from functoriality of F . enition 2.8. We call the ring S = H =nZ homR(Fn P , P ) from the above proof the F -twisteddomorphism ring of P , and we write it EndFR (P ). We call the functor

HR

(nZ

Fn P ,)

: gr-R gr- EndFR (P )

e F -twisted Hom functor of P , and we write it HomFR (P , ).

Thus we have proved:

oposition 2.9. Let R be a graded ring. Let P be a nitely generated projective graded R-module and let F beP -generative autoequivalence of gr-R. Let S = EndFR (P ) and let = HomFR (P , ) : gr-R gr-S. Thenis an equivalence of categories, and F = SS .

mark 2.10. We comment that if P is a nitely generated graded projective right R-module andis a P -generative autoequivalence of gr-R , one may also use F , or more properly the principaltomorphism used in the proof of Proposition 2.6, to dene a natural ring structure on

nZhomR

(P ,Fn P).

is ring is clearly isomorphic to EndFR (P ) as we have dened it above.

It is easy to see that the SR -twisted endomorphism ring EndSRR (P ) is isomorphic to the ungradeddomorphism ring EndR(P ), and that Hom

SRR (P , )

= HomR(P , ). Thus twisted endomorphismngs generalize classical endomorphism rings. We also note that EndFR (P ) and HomFR (P , ) may bened for any graded R-module P , although by Proposition 2.2 they will not give an equivalence of

tegories unless P is nitely generated projective and F is P -generative.


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ThusA priori, the ring EndFR (P ) and the functor HomFR (P , ) may depend on the full Z-sequence F .owever, in fact they do not. To see this, let F be a P -generative Z-sequence with compatibilityomorphisms i j . Let B = F1, and let B1 = F1. Then B1 is a quasi-inverse for B; let 1,1 :gr-R B1B be induced from 1,1. Let { i j} be the (induced) compatibility isomorphisms onn} satisfying (2.5), as discussed before Proposition 2.6. Then for all j Z there are unique inducedtural isomorphisms j : B j F j , such that the diagram

Bi+ ji+ j

i j

Fi+ j

i j

BiB ji j

FiF j

mmutes for all i, j Z. This commutativity implies that the twisted endomorphism rings EndF1R (P )d EndBR (P ) are isomorphic, and that Hom

F1R (P , )

= HomBR (P , ). We leave the tedious but rou-e verications to the reader. Thus we have:

oposition 2.11. Let F be a P -generative autoequivalence of gr-R. Then the F -twisted endomorphism ringd F -twisted Hom functor of P are well-dened, and depend only on the equivalence class of F in Pic(gr-R).

We are ready to reframe Proposition 2.2 in terms of autoequivalences of gr-R .

eorem 2.12. Let R and S be graded rings. Then gr-R gr-S if and only if there are

) a nitely generated graded projective module P and a P-generative autoequivalence F of gr-R such that) S = EndFR (P ) via a degree-preserving map.

Further, if : gr-R gr-S is an equivalence of categories with quasi-inverse , then F = SS and= S satisfy (1) and (2), and = HomFR (P , ).

oof. One direction is Proposition 2.9. For the other, suppose that : gr-R gr-S is an equiv-ence with quasi-inverse . Let F = SS , and let P = S S . By Proposition 2.2, we have that= HR(M, ), where

M = (S S) = (

nZSnS ( P )

),

d S = HR(M,M) , where is the principal automorphism of HR(M,M) induced from the identi-tions

homR( S jS P , S iS P

)= S ji = homR( S j+1S P , S i+1S P).at is, is given by the autoequivalence F of gr-R and by the compatibility isomorphisms on the-sequence F = { S jS} = {S jS }. Since F1 = F , by Proposition 2.11 we have that S = EndFR (P )d that = HomFR (P , ). We comment that although Proposition 2.2 and Theorem 2.12 are closely related, in many respectseorem 2.12 is superior. Theorem 2.12 is a much more functorial way of constructing equivalences,

es only one nitely generated graded R-module, and more compactly and clearly generalizes the


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suisgraded Morita theorems. Theorem 1.2, which is the main application in this paper of Theorem 2.12,ves one indication of the power of our approach.In the remainder of this section and in Section 3, we give some general applications of The-

em 2.12. We rst note that Theorem 2.12, when combined with the seemingly purely formaloposition 2.11, has surprising power. For example, we immediately obtain the following result ofordon and Green [7, Corollary 5.2, Proposition 5.3]:

rollary 2.13. Let R and S be graded rings, and let : gr-R gr-S be an equivalence of categories. Then a graded Morita equivalence if and only if commutes with shifting in the sense that SS = SR .

oof. If is a graded Morita equivalence, then we have = HomR(P , ) = HomSRR (P , ) forme graded module P that is a progenerator in mod-R , and S = EndR(P ) = EndSRR (P ). By Proposi-on 2.9, SR = SS .Conversely, suppose that SS = SR . Let be the quasi-inverse to , let P = (S), and let= SS . Since F = SR , by Proposition 2.11 EndR(P ) = EndSRR (P ) is isomorphic to EndFR (P ), whichTheorem 2.12 is isomorphic to S . Furthermore, applying the same results we see that

= HomFR (P , ) = HomSRR (P , ) = HomR(P , )

a graded Morita equivalence. mark 2.14. Versions of many the results of this section also hold for rings graded by arbitraryoups. Suppose that G and G are groups, and that R is a G -graded ring. The idea is to workith G-sequences F = {Fg}gG of autoequivalences of gr-R , dened as sequences with appropriatempatibility isomorphisms f ,g : F f g F f Fg . Given a G-sequence F and a graded R-module P ,e may construct a G-graded ring EndFR (P ) and a functor Hom

FR (P , ), and suitably modied

rsions of Proposition 2.6 and Theorem 2.12 still hold. However, Proposition 2.11 does not hold forbitrary grading groups: it is possible to have two G-sequences B and C such that for all g G weve Bg = Cg , but the functors HomBR (P , ) and HomCR (P , ) are not isomorphic.This means that Corollary 2.13 does not hold for rings graded by arbitrary groups. In [3, Exam-

e 4.8] the authors construct a ring R graded by the symmetric group 3 with the property that thet of autoequivalences of gr-R such that S gR = S gR for all g 3 is strictly larger than the setgraded Morita equivalences from gr-R to gr-R . We note that part of the proof of this statement isntained in [12].

For Z-graded rings, Corollary 2.13 is extremely powerful, since it provides an easy way to check ifo rings in the graded equivalence class of R are Morita equivalent. The following application willparticularly helpful for us:

oposition 2.15. Let R be a graded ring, let P and P be nitely generated graded projective R-modules, andppose that F and F are autoequivalences of gr-R such that F is P -generative and F is P -generative. ThendFR (P ) and End

F R (P

) are graded Morita equivalent if and only if F and F are conjugate in Pic(gr-R).

oof. Let T = EndFR (P ) and let T = EndF

R (P). Let = HomFR (P , ) : gr-R gr-T ; likewise let

= HomF R (P , ) : gr-R gr-T . By Theorem 2.12, and are equivalences of categories; let d be their respective quasi-inverses.First assume that F and F are conjugate in Pic(gr-R), so there is an autoequivalence of gr-Rch that F = F . Let 1 be a quasi-inverse to . Dene = : gr-T gr-T ; note that 1a quasi-inverse to .We apply Corollary 2.13 and compute ST . We have:

1 ( 1 ) ST = ST = ( ) (ST ) .


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(1(2nce by Proposition 2.9 we have ST = F and ( )F = ( )( )ST = ST , we see that

ST = (1(ST )

)= (1F )= (F ) = ST .us by Corollary 2.13, is a graded Morita equivalence.Conversely, suppose that : gr-T gr-T is a graded Morita equivalence. By Corollary 2.13, weve ST = ST . Let = Pic(gr-R). Then we have that

F = ST = ST = ST = ST = ST = F .

us F and F are conjugate in Pic(gr-R). The Picard group of a graded module category

If S is an ungraded ring, there is a classical map from Aut S Pic S . Beattie and del Ro [2] haveserved that this extends, using Z-algebras, to graded categories. That is, given a graded ring R ,ere is a group homomorphism

: Aut R Pic(gr-R), ( ) . (3.1)

e review the construction of the map (3.1). Let be an automorphism of the Z-algebra R , andt M be an R-module. Then there is a twisted module M that equals M as a vector space, withultiplication dened by

m r =m1(r).

e functor ( ) is isomorphic to R R , and clearly if MR is unitary, so is M . Since unitarymodules and graded R-modules are the same, we may regard ( ) = ( ) as an element ofc(gr-R); it is a quick computation that () = () ().As an example, let be the canonical principal automorphism of R . Then ( ) = SR : if M is a

aded R-module, then (M)i = M 1i = M 1(1i) = M 1i1 = Mi1. This isomorphism is easilyen to be natural.In this section, we explore applications of the map (3.1) to the study of equivalences of graded

odule categories. We derive implications for the equivalences known as Zhang twists, and also studye kernel of (3.1).To begin, we move beyond graded Morita equivalences to investigate the other well-known exam-

es of graded equivalences: the equivalences induced by twisting systems, as dened by Zhang [20].e recall that a twisting system is a set = {n} of graded k-vector space automorphisms of Rtisfying relations

n(a)n+m(b) = n(am(b)

)r all n,m, l Z, a Rm , and b Rl . Given a ring R and a twisting system on R , we may formtwisted algebra R , with multiplication dened by a b = am(b) for a Rm and b Rl . Theseisted algebras are often referred to as Zhang twists. In [14, Corollary 4.4], the author characterizedang twists as follows:

eorem 3.2. Let R and S be graded rings. Then the following are equivalent:

) S is isomorphic to a Zhang twist of R via a degree-preserving map.

) The Z-algebras R and S are isomorphic via a degree-preserving map.


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M) There is a principal automorphism of R so that S = R via a degree-preserving map.) There is an equivalence : gr-R gr-S such that for all n Z, (Rn) = Sn.

We call a functor satisfying Theorem 3.2(4) a twist functor. We will prove a result analogous torollary 2.13, characterizing twist functors : gr-R gr-S in terms of the pullbacks SS .We rst make a technical comment: all twist functors may all be written in a standard form.ppose that we are given a graded ring R and a principal automorphism of R . Let be thenonical principal automorphism of R . Any graded R-module is also an R-module and so an R -odule. Let R denote the R-action on M , and let denote the R -action. Then the R -action on Mgiven as follows: for any m Mi and s Rj we have

m s =m R i i(s) (3.3)

ote that i i(s) R0 j = R j). This gives an equivalence gr-R gr-R , which we call a standardist functor.We leave to the reader the verication of the following:

mma 3.4. Let R and S be graded rings, and let : gr-R gr-S be a twist functor. Let be the principaltomorphism of R induced by SS and , and let R = R . Let Z : gr-R gr-R be the standard twist functor.t : S R be the induced ring isomorphism. Then the diagram

gr-RZ

gr-R

gr-S

mmutes. That is, without loss of generality any twist functor is standard.

We now give the desired version of Corollary 2.13 for twist functors.

oposition 3.5. Let R be a graded ring, and let F be an autoequivalence of gr-R. Then HomFR (R, ) is aist functor if and only if F = ( ) for some principal automorphism of R.

oof. Let denote the canonical principal automorphism of R .() Suppose that = HomFR (R, ) : gr-R gr-S is a twist functor. By Proposition 2.9, F =SS . Let be a quasi-inverse for . By Lemma 3.4, without loss of generality we may assume that

= R for some principal automorphism of R , and that and are standard twist functors. Weill denote multiplication in gr-R by R ; similarly for S ,R . If M is a graded R-module, then as aaded k-module (M) = M , with the S-module structure on (M) given as follows: if m Mi and S j = R0 j = R j , then by (3.3)

m S s =m R i i(s) (3.6)

d likewise

m R s =m S ii(s). (3.7)

We claim that SS = ( ) . Let M be a graded R-module. Note that (FM)i = ( SSM)i =1

i1 = M R (1i) = (M)i . We verify that FM and M are isomorphic as R-modules. Let m


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Cogr)i = (FM)i = Mi1, and let r R j . Let 1 denote the R-action on M , and let 2 denote the-action on FM . Then we have that

m 1 r =m R 1i(r) =m R (i1)1i(r).

n the other hand, since m (FM)i = ( SSM)i , we have by (3.6) and (3.7) that

m 2 r =m S ii(r) =m R (i1) i1ii(r)=m R (i1)1i(r) =m 1 r.

us FM = M ; we leave to the reader the verication that this isomorphism is natural in M .() Suppose that F = ( ) , where is a principal automorphism of R . Let S = R . As men-ned in the discussion after Proposition 2.1, the proof of [14, Proposition 3.3] shows that R = S .Theorem 3.2, there is an equivalence of categories : gr-R gr-S , where is a twist functor.

t G = SS . Then by Theorem 2.12, = HomGR (R, ). But by the discussion above, we see that= ( ) = F ; and by Proposition 2.11, HomFR (R, ) is isomorphic to HomGR (R, ) and so is aist functor. rollary 3.8. Let R and S be graded rings. An equivalence : gr-R gr-S is a Zhang twist followed by aaded Morita equivalence if and only if SS = ( ) for some principal automorphism of R.

oof. The proof is similar to the proof of Corollary 2.13; since we will not use the result in thequel, we omit the details. Let R be a graded ring, and let be the canonical principal automorphism of R . Since we have

en that ( ) = SR , we immediately have the following corollary:

rollary 3.9. Let R and S be graded equivalent graded rings, and let be the map dened in (3.1). If :-R gr-S is either a graded Morita equivalence or a twist functor, then SS Im .

We now change focus from the image of (3.1) to its kernel. Beattie and del Ros original interestthe map (3.1) was to extend the classical short exact sequence for an ungraded ring S

0 Inn S Aut S Pic S,

rings with local units, in particular to Z-algebras, and to characterize the kernel of (3.1). Let R begraded ring. We will say that an automorphism of R is inner if lies in the kernel of (3.1). Thellowing is a special case of one of the results of [2]:

eorem 3.10. (See [2, Theorem 1.4].) Let R be a graded ring, and let be a graded automorphism of R ofgree 0. Then is inner if and only if for all m,n Z there are fm Rmm = R0 and gn Rnn = R0 so thatr all w Rmn, (w) = fmwgn.

The image of (3.1) is almost never trivial, since SR is not (usually) the identity in gr-R . However,all degree 0 automorphisms of gr-R are inner, then we expect the image of (3.1) to be small. Inis situation, there are strong consequences for gr-R . We rst see that there are no nontrivial Zhangists of R .

rollary 3.11. Let R be a graded ring such that all automorphisms of R of degree 0 are inner. If : gr-R

-S is a twist functor, then S = R and is naturally isomorphic to Idgr-R .


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Provoof. Let : gr-R gr-S be a twist functor; let F = SS . By Proposition 3.5, we know that F =) , for some principal automorphism of R . Let be the canonical principal automorphism

R; then 1 has degree 0 and so by assumption is inner. Thus F = ( ) = ( ) = SR . Byeorem 2.12 S = EndSRR (R) = R , and = HomSRR (R, ) = Idgr-R . As a second consequence, we show that all autoequivalences of gr-R are determined by theirtion on (isomorphism classes of) R-modules. Formally, let F and F be autoequivalences of thetegory gr-R . We will say that F and F are weakly isomorphic, written F =w F , if for all nitelynerated graded R-modules M , we have F(M) = F (M). There is no assumption on the naturalitythis isomorphism.

rollary 3.12. Let R be a ring such that all automorphisms of R of degree 0 are inner, and let F and F betoequivalences of gr-R. Then F =w F if and only if F and F are naturally isomorphic.

oof. Suppose that F =w F . Let G = F(F )1. Then G is weakly isomorphic to Idgr-R , and so forl n, we have G(Rn) = Rn. That is, G : gr-R gr-R is a twist functor. But now by Corollary 3.11,= Idgr-R and so F = F . Graded modules over the Weyl algebra

From this point on, we will work with the Weyl algebra and its graded module category. Let k bealgebraically closed eld of characteristic 0, and let A = k{x, y}/(xy yx 1) be the (rst) Weyl

gebra over k. We will grade A by the Euler gradation, where we put deg x= 1 and deg y = 1.The goal of the next two sections is to understand the graded Picard group of A. In Section 6 we

ill use these results and the methods of Sections 2 and 3 to prove Theorem 1.2.We x notation, which will remain in force for the remainder of the paper. We will let z = xy, so

at A0 = k[z]. Let be the automorphism of k[z] given by (z) = z + 1. We will constantly use thect that graded right A modules are actually graded (k[z], A)-bimodules: since the grading on A isven by commuting with z, we have that k[z] acts on a right A-module M by

m =mi()

r any m Mi .Our rst step in analyzing the category gr-A is to understand simple objects and the extensions be-een them. There are very few of these: if M and M are graded simple modules, then ExtA(M,M )either 0 or k. On the other hand, McConnell and Robson [10] have shown that, while ExtA(M,N)always nite-dimensional for ungraded simple A-modules M and N , there are simple M and Nith innitely many nonisomorphic extensions of M by N . This is one indication, that, as we willpeatedly see, the graded structure of A is much more rigid than the ungraded structure.Lemmas 4.1 and 4.3 are elementary and presumably known, but we have not been able to nd

ese specic results in the literature.As A is hereditary, we will use the abbreviation ExtA for Ext1A , and extA for ext

1A .

mma 4.1. Up to isomorphism, the graded simple A-modules are precisely:

X = A/xA and its shifts Xn for n Z; Y = A/yA1 and its shifts Y n; M = A/(z + )A for k \ Z.

oof. The graded A-modules are precisely those that decompose as a sum of weighted subspaceser k[z]. By [5, Theorem 3.2], up to ungraded isomorphism the simple k[z]-weighted A-modules are A/xA;


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SiAp A/yA; One module M for each coset of (k \ Z)/Z.

e observe that if k \ Z, then M1 = M+1, and M = M in gr-A if = . Thus the gradedomorphism classes of graded simples correspond to the shifts Xn and Y n, plus one module Mr each element of k \ Z. Another indication of the comparative rigidity of gr-A is that there is only one outer automorphismA that preserves the graded structure. This is the map : A A dened by (x) = y and (y) =

x. We will also denote the induced autoequivalences of gr-A, mod-A, and A-mod by . We remarkat the action of on the simples Xn, Y n is given by (Xn) = Y n 1 and (Y n) =n 1.For notational convenience, we will also dene on graded k-vector spaces as the functor that

verses grading: if V is a graded vector space, we write V for the graded vector space given byV )n = Vn . Then gives isomorphisms of graded vector spaces

HomA(M,M) = HomA(M,M ),

ExtA(M,M) = ExtA(M,M ) (4.2)

r any M and M in gr-A. If either M or M is a graded A-bimodule, then it is an easy computationat the maps in (4.2) are A-module maps.We now compute extensions between simple objects in gr-A. We remind the reader that homAd extA = ext1A refer to homomorphisms in the category gr-A, and HomA and ExtA refer to mod-A.

mma 4.3.

) ExtA(X, A) = (A/Ax)1, and ExtA(Y , A) = A/Ay.) As a graded vector space, ExtA(X, Y ) = ExtA(Y , X) = k, concentrated in degree 0.) ExtA(X, X) = ExtA(Y , Y ) = 0.) If = , then extA(M,M) = 0, but extA(M,M) = k.) Let k \ Z and let S {X, Y }. Then ExtA(M, S) = ExtA(S,M) = 0.

oof. (1) We apply HomA( , A) to the short exact sequence

0 xA A X 0 (4.4)

obtain:

0 A Ax1 ExtA(X, A) 0.

us we have computed that ExtA(X, A) = Ax1/A = (A/Ax)1. Applying , we obtain from (4.2)at ExtA(Y 1, A) = (A/Ay)1, and so ExtA(Y , A) = ExtA(Y 1, A)1 = A/Ay.(2) Applying HomA(Y , ) to (4.4), we obtain

0 ExtA(Y , A)1 ExtA(Y , A) ExtA(Y , X) 0.

nce by (1) we know that ExtA(Y , A) j is 0 if j 1 and k if j 0, we see that ExtA(Y , X) = k.plying , we obtain that ( )k = (k) = ExtA(Y ,X) = ExtA X1, Y 1 = ExtA(X, Y ).


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le(3) By applying HomA(X, ) to (4.4), we obtain a long exact sequence:

0 HomA(X, X) ExtA(X, A)1 ExtA(X, A) ExtA(X, X) 0

d a similar computation shows that ExtA(X, X) = 0. Applying again we see that ExtA(Y , Y ) = 0.(4) We apply homA( ,M) to the exact sequence

0 Az+

A M 0

obtain

0 homA(M,M) k[z]/(z +) z+ k[z]/(z +) extA(M,M) 0.

is immediate that extA(M,M) = homA(M,M) and is 0 if = and k if = .(5) This proof is similar to the others, and we omit it. We note one consequence of Lemma 4.3: since the sequences

0 Y A/zA X 0 (4.5)

d

0 X(A/(z 1)A)1 Y 0 (4.6)

not split, they and their shifts are the only nonsplit extensions in either gr-A or mod-A involvinge shifts of X and Y . In fact, in general we have:

mma 4.7. Let Z be an indecomposable graded A-module of nite length. Then Z is determined up to iso-orphism by its JordanHlder quotients.

oof. This follows by induction from Lemmas 4.1 and 4.3. The left k[z]-action on the modules (A/zA)n and (A/(z1)A)n+1 is especially nice. We record

is as:

mma 4.8. Let n Z. Then (z + n)(Xn) = 0= (z + n)(Y n), and as graded left k[z]-modules, we have

(A/zA)n = (A/(z 1)A)n+ 1 =jZ

k[z](z + n) .

oof. This follows from the exact sequences (4.6) and (4.5) and the computation (z + n)xn = xnz +1A in the quotient ring of A. Since M = jZ k[z]/(z + ) as a left k[z]-module, we see that any graded A-module of nitength, when considered as a left k[z]-module, is supported at nitely many k-points of Speck[z].


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(1enition 4.9. If M is a graded right A-module of nite length, we dene the support of M , SuppM ,be the support of M as a left k[z]-module. We are particularly interested in the cases whenppM Z (we say that M is integrally supported) or when SuppM = {n} for some n Z (we sayat M is simply supported at n). Lemma 4.1 and Lemma 4.8 show that Xn and Y n are the uniquemples simply supported at n.

We now turn to understanding graded projective A-modules. We will say that an injection: P Q of rank 1 graded projective modules is a maximal embedding if there is no f : P Qith Im( f ) Im( f ).

mma 4.10. Let P and Q be rank 1 graded projective A-modules, and let f : P Q be a maximal embed-ng. Then the module Q / f (P ) is semisimple and integrally supported.

oof. Let N = Q / f (P ). As Q A is 1-critical, N has nite length. By Lemma 4.1 N has a nite compo-tion series whose factors are isomorphic to Xn, Y n, or M , with / Z.Suppose that some M is a subfactor of N . Then we have f (P ) Q 1 Q 2 Q , with Q 2/Q 1 = M ,d so Q 1 = (z+)Q 2. But now (z+)1 f (P ) (z+)1Q 1 = Q 2 Q , contradicting the maximalityf .Thus N is integrally supported. If N is not semisimple, then by Lemma 4.3, N has a subfactor

omorphic to either (A/zA)n or (A/(z 1)A)n + 1. In either case, arguing as above we have+ n)1 f (P ) Q , contradicting the maximality of f . We have the following easy consequence of Lemma 4.10:

mma 4.11. Let P be a rank 1 graded projective A-module. Then for all n 0, we have homA(P , Xn) =mA(P , Y n) = k.

oof. Let

0 P A M 0

a maximal embedding. Then there are exact sequences

HomA(M, X) HomA(A, X) HomA(P , X) ExtA(M, X)

d

HomA(M, Y ) HomA(A, Y ) HomA(P , Y ) ExtA(M, Y ).

Lemma 4.10, M is semisimple and integrally supported. By Lemma 4.3, the vector spacesomA(M, X), HomA(M, Y ), ExtA(M, X), and ExtA(M, Y ) are nite-dimensional; so for n 0 we havemA(P , Xn) = homA(A, Xn) = Xn = k and homA(P , Y n) = homA(A, Y n) = Yn = k. In fact, more is true: the simple factors of a module P partition the set of integrally supported

mples.

oposition 4.12. Let P be a rank 1 graded projective A-module, and let n Z. Then exactly one of thellowing is true:

) We have ( ) ( )

extA Y n, P = homA P , Xn = k


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) we have

homA(P , Xn)= extA(Y n, P)= 0

and

extA(Xn, P)= homA(P , Y n)= k.

oof. We rst prove the result for n = 0. By Lemma 4.3, we have extA(X, A) = 0= Y0 = homA(A, Y )d extA(Y , A) = k = X0 = homA(A, X). Thus the claim holds for A = P . For general P , let

0 P A M 0 (4.13)

a maximal embedding of P in A. By Lemma 4.10, M is semisimple and integrally supported. Sinceis not a factor of A, we know that Y is not a factor of M , and so homA(M, Y ) = homA(Y ,M) =tA(M, X) = extA(X,M) = 0. Further, by Lemma 4.3, there are k-vector space isomorphisms

homA(M, X) = extA(M, Y ) = extA(Y ,M) = homA(X,M). (4.14)

Via the long exact Ext sequence, (4.13) induces a diagram with exact rows

0 homA(X,M)

=

extA(X, P ) extA(X, A) = 0

0 homA(P , Y ) extA(M, Y ) 0

here the vertical isomorphism comes from (4.14). Thus extA(X, P ) = homA(P , Y ). Applying , wee that extA(Y , P ) = homA(P , X), where P = (P )1. As P ranges over all rank 1 graded projec-ves, so does P . This shows that for all P we have extA(Y , P ) = homA(P , X).To see that exactly one of (1) and (2) holds, observe that (4.13) also induces a diagram with exact

ws

0 extA(Y , P ) extA(Y , A) extA(Y ,M)

=

0

0 homA(P , Y ) extA(M, Y ) 0,

here again the vertical map comes from (4.14). Thus we see that

k = extA(Y , A) = extA(Y , P ) homA(P , Y ).

e claim follows, and by shifting degrees we obtain the result for all n. enition 4.15. For any integer j and rank 1 graded projective P , let F j(P ) be the unique gradedmple factor of P supported at j. We will say that two rank 1 projectives P and Q are j-congruent

F j(P ) = F j(Q ), and are j-opposite if they are not j-congruent.


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PriNote that F j(P ) is well-dened by Proposition 4.12.

mma 4.16. If P and Q are rank 1 graded projective A-modules that are j-congruent for all j Z, then= Q .

oof. Let

0 P Q M 0

a maximal embedding of P in Q . By Lemma 4.10, M is semisimple and integrally supported. Letbe a simple summand of M . Consider the sequence

0 homA(M,N) homA(Q ,N) homA(P ,N) extA(M,N).

ow, by Proposition 4.12, there is no j such that M contains both X j and Y j as compositionctors, and dimk homA(Q ,N) = 1. Then, by Lemma 4.3, extA(M,N) = 0, and so homA(P ,N) = 0. Thisplies that for any j SuppM , the modules P and Q are j-opposite. Since P and Q are alwayscongruent, we must have M = 0 and P = Q . This lemma again shows the comparative rigidity of the graded category: it is certainly not true

at the (ungraded) simple factors of a projective module determine it up to isomorphism, since everyojective module is a generator for mod-A.We record for future reference the following routine consequence of the fact that A is hereditary:

mma 4.17. Let P be a nitely generated graded projective A-module. Then P splits completely as a directm of rank 1 graded projective modules.

The Picard group of gr-A

In this section we will calculate the Picard group of gr-A. Let D denote the innite dihedraloup, and let (Z/2Z)(Z) be the direct sum of countably many copies of Z/2Z. We will show thatere is an exact sequence of groups

1 (Z/2Z)(Z) Pic(gr-A) D 1

d that in fact Pic(gr-A) is isomorphic to the restricted wreath product

(Z/2Z)wrZ D = (Z/2Z)(Z) D.

We rst investigate the automorphism group of the Z-algebra A and the map (3.1) for A. Wetablish notation: dene mij Aij = A ji to be the canonical k[z]-module generator of A ji ; that is,i j is x ji if j i and yi j if i > j.Recall that an automorphism of A is inner if it is in the kernel of (3.1).

mma 5.1. Let be an automorphism of A of degree 0. Then is inner.

oof. For all i, j Z there is a unit i j k[z] such that (mij) = i jmij . Thus i j k , and in particularj is central in A. Further, for all n Z we have nn = 1. Applying to the identitymn,n+1mn+1,n mn,n1mn1,n = 1n,


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CoPie obtain that n,n1n1,n = 1 for all n, and so (z 1n) = z 1n . This implies that for any f k[z]d i, j Z, we have

( f mij) = i j f mij,

d in particular that i j j = i for all i, j, Z. Thus if v Aij , we have (v) = i j v = i00 j v =0v0 j , and so is inner by Theorem 3.10. rollary 5.2. If : gr-A gr-S is a twist functor, then S = A and = Idgr-A .

oof. This follows from Lemma 5.1 and Corollary 3.11. The next result is the most important application of the rigidity of the category gr-A: the innite

hedral group D is a factor as well as a subgroup of Pic(gr-A). This gives us two integer invariantssociated to each element of Pic(gr-A).

eorem 5.3. Let F be an autoequivalence of gr-A. Then there are unique integers a = 1 and b such that forl n Z, we have {F(Xn),F(Y n)}= {Xan+ b, Y an+ b}d for all k \ Z,

F(M) = Ma+b.

oof. If a and b exist, they are clearly unique.We observe from Lemmas 4.1 and 4.3 that F must map the simple module M to some other

mple M , since these are the only simples with nonsplit self-extensions. Therefore, F must maptegrally supported simples to integrally supported simples. Further, by Lemma 4.3, for all n Z their {Xn, Y n} must map to some other pair {Xn, Y n}, as these pairs form the only nonsplittensions of two nonisomorphic simples. In other words, there is a bijection g : k k such that:

) If Z, then g() Z and F({X, Y }) = {Xg(), Y g()}.) If k \ Z, then g() / Z and F(M) = Mg() .

Now consider the functor F0 = F( k[z] A)0 :mod-k[z] mod-k[z]. We claim that for all k,e have F0(k[z]/(z + )) = k[z]/(z + g()). This follows from Lemma 4.8 if Z, and from thenition of M and Mg() if / Z. In particular, as g is a bijection, F0 is invertible and so antoequivalence of mod-k[z]. But the only such functors act via automorphisms of k[z], so there arenstants a,b k such that g(n) = an+ b for all n. As g maps Z bijectively to Z, a must be 1 and bust be an integer. enition 5.4. If F is an autoequivalence of gr-A, the integer b above is the rank of F , and theteger a is the sign of F . We say that F is odd if it has sign 1, and is even if it has sign 1. We sayat F is numerically trivial if it is even and has rank 0.

For example, SnA is even of rank n, and is odd of rank 1.Since the set of maps g : Z Z of the form n n + b is isomorphic to D , and rank and

gn clearly behave well with respect to composition of functors, from Theorem 5.3 we immediatelytain:

rollary 5.5. Let Pic0(gr-A) be the subgroup of Pic(gr-A) of numerically trivial autoequivalences. Then

c(gr-A) = Pic0(gr-A) D .


Pr

is

su

Coin

PrFgeisfo

PFFLe

la

F

PrY

an

Prpr

Prth

is

mthoof. This follows from Theorem 5.3 and the fact that the subgroup

,SA Pic(gr-A)

isomorphic to D . As a second corollary, we see that to check if two autoequivalences of gr-A are isomorphic, itces to see if they agree on integrally supported simples.

rollary 5.6. Let F and F be autoequivalences of gr-A. Then F = F if and only if F(S) = F (S) for alltegrally supported simples S.

oof. Suppose that F(S) = F (S) for all integrally supported simples S . Then by Theorem 5.3, F and have the same rank and sign, and so F(S) = F (S) for any simple module S . Let N be a nitelynerated graded A-module; we write N = P Z where P is torsion-free (and hence projective) and Ztorsion. Let Z = Z1 Z2 Zn , where the Zi are indecomposable. By Lemma 4.7, F(Zi) = F (Zi)r all i, so F(Z) = F (Z).By Lemma 4.17, P splits completely as a direct sum of rank 1 graded projectives; we write P =

1 Pn , where the Pi have rank 1. Since F and F agree on simples, for each i the modulesPi and F Pi have isomorphic simple factors and so by Lemma 4.16, F Pi = F Pi for all i. ThusP = F P , and so FN = F P F Z = F P F Z = F N . Thus F and F are weakly isomorphic; bymma 5.1 and Corollary 3.12, F = F .The other direction is trivial. A priori, it is not clear that Pic0(gr-A) is nontrivial. It turns out, however, that this group is quite

rge. We exhibit generators for it in the next proposition.We recall the notation from Denition 4.15 that if P is a rank 1 graded projective module, then

j(P ) is the unique simple factor of P supported at j.

oposition 5.7. For any j Z there is a numerically trivial automorphism j of gr-A such that j(X j) = j, j(Y j) = X j, and if j = j, then j(X j) = X j and j(Y j) = Y j. Further, for any i, j Z,

S iA j = i+ jS iA, (5.8)

d 2j= Idgr-A .

oof. Suppose that 0 exists as described. Then the automorphism j = S jA0S jA has the requiredoperties, and (5.8) is satised by construction. Thus it suces to construct 0.We rst dene the action of 0 on rank 1 graded projective modules. Let P be such a module. By

oposition 4.12, we know that homA(P , X Y ) = k; thus there is a unique submodule N of P suchat the sequence

0 N P X Y

exact. Formally, N is the reject of X Y in P ; we remark that N is 0-opposite to P . Let 0(P ) = N .Suppose that Q is a rank 1 projective that is 0-opposite to P . Let f : Q P be a nonzero

ap. We claim that f (Q ) N . To see this, let S = F0(P ). Since Q is 0-opposite to P , we haveat homA(Q , S) = 0. Thus if we let M = Coker f , from the exact sequence0 homA(M, S) homA(P , S) homA(Q , S) = 0


wbu

PthPar

tia

Le

gr

thclCo

0

w

an

inTh

gi

Thon

Fha

tooffo

The see that homA(M, S) = homA(P , S) = k. Thus there is some N with f (Q ) N P and P/N = S;t now since S = P/N and homA(P , S) = k, we have N = N and so f (Q ) N , as claimed.We next dene the action of 0 on morphisms between rank 1 projectives. Let f : P P , whereand P are rank 1 graded projective modules. Dene 0( f ) : 0(P ) 0(P ) to be f |0(P ) . We verifyat this is well-denedthat is, that f (0(P )) 0(P ). If P and P are 0-congruent, then 0(P ) and are 0-opposite, so by the claim above we have f (0(P )) 0(P ). On the other hand, if P and Pe 0-opposite, then f (P ) 0(P ) so certainly f (0(P )) 0(P ).It is clear that 0 behaves functorially on maps between rank 1 projectives; that is, it sends iden-

ties to identities and preserves commuting triangles. By standard arguments, 0 extends uniquely tofunctor dened on all modules and morphisms in gr-A.We show that 0 has the properties claimed. Let P be a rank 1 graded projective module. Bymma 4.8 we have 20P = zP . Without loss of generality, we may regard P as a submodule of theaded quotient ring of A; thus if we dene 10 (P ) = z10(P ) and extend to all of gr-A it is clearat 0

10 = 10 0 = Idgr-A . Thus 0 is an automorphism of gr-A. We also note that if 0 behaves as

aimed on simple modules, then for any integrally supported simple S , we have 20(S)= S , and so by

rollary 5.6, 20= Idgr-A .

Thus it remains to establish that 0 acts as claimed on simples. First, since 0A = xA, by applyingto the exact sequence

0 xA A X 0

e obtain

0 zA xA 0(X) 0

d so 0(X) = xA/zA = Y .By Theorem 5.3, we see that 0 has rank 0 and sends Y to X . We show that 0 is even by comput-

g 0(X1). Let P = (z+1)A+x2A. It is straightforward to see that A/P = X1, and that P/x2A = X .us x2A = 0(P ), and applying 0 to the exact sequence

0 P A X1 0

ves

0 x2A xA 0(X1) 0.

us 0(X1) = xA/x2A = X1. Therefore the image of 0 in the factor group D of Pic(gr-A) actsZ by sending 0 to 0 and 1 to 1, and so must be the identity.The simple factors of 0(A) = A1 are {X j} j1 and {Y j} j0. Thus for all j = 0 we have F j(A) =

j(0(A)), which is isomorphic to 0(F j(A)) since 0 is numerically trivial. Theorem 5.3 shows that 0s the desired behavior on integrally supported simples. It is clear that i j = ji and that the subgroup of Pic(gr-A) generated by the { j} is isomorphicthe countable direct sum (Z/2Z)(Z) . It is convenient to identify this with the set of nite subsetsZ, which we denote Zn. The group operation on Zn is exclusive or, which we write as ; thusr K , J Zn, we have

K J = (K J ) \ (K J ).

e identity element of Zn is .There are multiplicative and additive actions of Z on Zn, given byn J = {nj | j J }


an

fo

is

w

FoWthposuno

Co

Th

Prph

fo

isth

Coauj

Praud

J + n = { j + n | j J }

r any n Z and J Zn. In particular, D is naturally a subgroup of Aut(Zn).To each J Zn, we may associate an automorphism of gr-A, which we write either J or [ J ]. Itdened by

[ J ] = J =j J

j,

ith inverse

1J =(

j J(z + j)1

) J .

r completeness, we dene = Idgr-A . Because 2J = Idgr-A , we refer to the functors J as involutions.e note that if P is a rank 1 graded projective A-module, then J P is the unique submodule of Pat is maximal with respect to the property that it is j-opposite to P for all j J . (In fact, it isssible to give a categorical denition of the functors J as the autoequivalences of gr-A that arebfunctors of the identity on projectives and whose square is naturally isomorphic to Idgr-A . We willt prove this assertion.)We now show that the subgroup of involutions of gr-A is in fact equal to Pic0(gr-A).

rollary 5.9. Dene

: Zn Pic0(gr-A),J J .

en is a group isomorphism.

oof. Since ( j)2 = Idgr-A for all j, we have that K J = [K J ]. Therefore is a group homomor-ism, and it is clearly injective. We prove surjectivity.Suppose that [F ] Pic0(gr-A), and let P = F(A). Since F is numerically trivial, F(F j(A)) = F j(P )

r all j; thus by Lemma 4.11, the set

J = {n F(Xn)= Y n}nite. Since for all integrally supported simples S we have J (S) = F(S), by Corollary 5.6 it followsat F = J . rollary 5.10. The automorphism group of A is generated by inner automorphisms, by the canonical principaltomorphism , and by the map : A A given by (mij) = mi, j if j i and (mij) = mi, j if< i. In particular, if is the map dened in (3.1), then

Im() = ,SA Pic(gr-A).

oof. As is clearly induced from the ring automorphism : A A, it is well-dened and is an

tomorphism of A. We note that () = and () = SA .


su

thsh

fo

Le

Th

Pim

an

Copr

Pr

Cosu

fo

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(1

(2We consider the map (3.1). Let Aut(A). By Theorem 5.3, there is some , so that(1) is a numerically trivial autoequivalence of gr-A. Thus without loss of generality we mayppose that () Pic0(gr-A). We will show that this implies that is inner.Note that if M is a cyclic graded right A-module, then M is also cyclic as an A-module and

erefore as an A-module. Now, the rank 1 cyclic graded projective right A-modules are precisely theifts An; these are the rank 1 graded projectives P so that{

i Fi(P ) = Xi}= [n,)

r some n Z.By Corollary 5.9, there is some J Zn so that () = J . Suppose that J = , and let

j =min J 1.

t P = J (A j) = (A j) . Then{i Fi(P ) = Xi}= { j} unionsq ([ j + 2,) \ J).

erefore, P is not cyclic, a contradiction.Thus J = and so () = Idgr-A ; thus by denition is inner. We note the contrast between the graded and ungraded behavior of A. Let us write the ungraded

card group of A as Pic(mod-A). Then a theorem of Stafford [17, Corollary 4.5(i)] says that the naturalap from Aut(A) Pic(mod-A) is surjective. On the other hand, Corollary 5.10 shows that

Pic(gr-A) = Pic0(gr-A) Im,

d we have seen that Pic0(gr-A) is largein particular, it is not nitely generated.

rollary 5.11. The group Pic(gr-A) is generated by SA , , and 0 , and is isomorphic to the restricted wreathoduct (Z/2Z)wrZ D = Zn D .

oof. This follows from Corollary 5.5, Corollary 5.9, (5.8), and the computation j = 1 j . rollary 5.12. Let F be an autoequivalence of gr-A. Then there are a unique integer b and a unique J Znch that if F is even, then F = SbA J , and if F is odd, then F = SbA J .

We end the section with several technical lemmas. First, it is useful to have some explicit formulaer computing involutions.

mma 5.13.

) For any J Zn ,

J A =i J

i A.

) For any i Z, consider yi A A[y1]. Then for any j Z, we have

yi A = yi A.j j+i


(3

Pri ha

m

ge

Leal

PrgrarQ

jeSi

Le

PrCo

Th

6.

anarre

eianAisrith) For any i Z,

i A =

(z + i)A + xi+1A if i 1,xA if i = 0,y A if i = 1,(z + i)A + yi A if i 2.

oof. (1) We note that J A is i-opposite to A for all i J ; therefore, J A i A for any i J . For anyZ and rank 1 graded projective P , we know that i P is a maximal submodule of P . Therefore, J As height # J in A; as the length of A/(

i J i A) is # J , we conclude that J A =

i J i A.

(2) As graded modules, j yi A = jSiA A = SiA j+i A = yi j+i A. Furthermore, both are maximal sub-odules of yi A. Therefore they must be equal.We leave the computation of (3) to the reader. We also give two lemmas that we will use to understand when an autoequivalence of gr-A isnerative.

mma 5.14. If F is an autoequivalence of gr-A, then F is P -generative if and only if the set {Fn P } generatesl integrally supported graded simple A-modules.

oof. One direction is clear. For the other, suppose that {Fn P } generates all integrally supportedaded simples. It suces to prove that {Fn P } generates all rank 1 graded projectives. Let Q be anbitrary rank 1 graded projective, and let : P Q be a maximal embedding. By Lemma 4.10,/(P ) is semisimple and integrally supported, and thus generated by {Fn P }. Thus there is a sur-ction : ji=1 Fni P Q /(P ) that by projectivity of the Fni P lifts to a map : ji=1 Fni P Q .nce Im + Im = Q , therefore {Fn P } generates Q . mma 5.15. If F is a generative autoequivalence of gr-A, then F is even and has nonzero rank.

oof. We note that SA = S1A . Let F Pic(gr-R) be generative. Suppose that F is odd, so byrollary 5.12, F = SnA J for some n Z and J Zn. Then for K = ( J + n) ( J 1), we have,

F2 = SnA JSnA J = SnA J SnA [ J 1] = K .

us, F4 = Idgr-A , contradicting the generativeness of F .Thus F is even; since the involutions J are not generative, the rank of F must be nonzero. Classifying rings graded equivalent to the Weyl algebra

We are now ready to classify rings graded equivalent to A and prove Theorem 1.2. Theorem 2.12d the analysis of Pic(gr-A) completed in the previous section are our basic tools; using them, wee able to compute twisted endomorphism rings extremely explicitly, and we obtain canonicalpresentatives for each graded Morita equivalence class.As a rst illustration of our approach, note that for any n 1, SnA is A-generative. This can be seen

ther from Lemma 5.14, or directly: since xn1A + yA = A, therefore An 1 and A1 generate Ad (by induction) all A1, . . . , An1. It is straightforward to see that if F = SnA , then EndFA (A) =

(n) =iZ Ani , the nth Veronese of A. Thus by Proposition 2.9, A is graded equivalent to A(n) , whichof course the Z/nZ-invariant ring of A. This is in marked contrast to most commutative gradedngs, where if n 2, we expect that a ring and its nth Veronese will have the same Proj, but not

e same graded module category. Note that if we grade R = k[x, y] by deg x = 1 and deg y = 1,


thto

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ThBa

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D

N[1

Ex

(1

(2

eqcaadjblneen SnR is generative exactly when n = 1; this tells us that no Veronese of R is graded equivalentR .Our goal is to compute (up to graded Morita equivalence) all rings that are graded equivalent to

-A. In general we will see a combination of idealizers, as in the ring B from the Introduction, anderonese rings, as discussed above. More specically, let f (z) be a monic polynomial in z, and let na positive integer. We dene the generalized Weyl algebra dened by f and n, denoted W ( f ,n), to bee k-algebra generated by X , Y , and z subject to the relations:

Xz zX = nX, Y z zY = nY ,XY = f , Y X = f (z n).

ese rings were dened by Bavula [5], and have been extensively studied by him. In particular,vula proves the following:

eorem 6.1. (See [5, Proposition 1.3, Corollary 3.2, Theorem 5].) If f does not have multiple roots or twostinct roots differing by a multiple of n, then W ( f ,n) is a simple hereditary Noetherian domain.

We caution the reader that Bavula uses slightly different notation; in particular, for Bavula therameter n is always equal to 1. Since W ( f ,n) = W (g,1) for an appropriate choice of g , this isrely notational.For any J Zn, dene f J = j J (z+ j). Now let n be a positive integer, and let J {0, . . . ,n1}.

ene J = {0, . . . ,n 1} \ J , and let W = W ( f J ,n). We dene a ring S( J ,n) by

S( J ,n) = k[z] + f J W W .

ote that S( J ,n) = IW ( f J W ), the idealizer of f J W in W . As a subring of W , S( J ,n) is a domain; by3, Theorem 4.3], it is hereditary and Noetherian.

ample 6.2.

) Let n = 1. There are two subsets of {0}. If J = {0}, then f J = 1 and so W = W (1,1) = A[y1].Let T = A[y1]. Then S({0},1) = IT (zT ) = IT (xT ). On the other hand, if J = , then S(,1) =W (z,1) = A.

) Now let n > 1 and let J = . Then f J = 1, and S(,n) = W (n1i=0 (z + i),n), which is isomorphicto the nth Veronese A(n) via

X xn,Y yn,z z.

We will see that the rings S( J ,n) give all the graded Morita equivalence classes of rings gradeduivalent to A. However, they are not quite unique representatives for these equivalence classes. Well a pair ( J ,n) Zn Z an admissible pair if n 1 and J {0, . . . ,n 1}. We will say that twomissible pairs ( J ,n) and ( J ,n) have the same necklace type if n = n and there is some integersuch that J + j J (mod n). The idea is that J and J both encode a string of n beads, someack and some white, and that two pairs have the same necklace type if the strings make identical

cklaces once we join the ends; we are allowed to rotate necklaces but not to turn them over. These


neis

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Cow

Pico

ta

N

Le

Pram

Th1

1)

Th2cklaces are well-known combinatorial objects. The number of necklaces of n black and white beads

1

n

d|n

(d)2n/d,

here is the Euler totient function [18, Exercise 7.112].We will prove:

eorem 6.3. If S is a Z-graded ring, then gr-A gr-S if and only if there is an admissible pair ( J ,n) suchat S is graded Morita equivalent to S( J ,n). Furthermore, S( J ,n) and S(L, r) are graded Morita equivalentand only if ( J ,n) and (L, r) have the same necklace type.

This proves Theorem 1.2 from the Introduction. We also obtain Theorem 1.3:

rollary 6.4. The gradedMorita equivalence classes of rings graded equivalent to A are in 11 correspondenceith pairs (J,n) where n is a positive integer and J is a necklace of n black and white beads.

Our program for proving Theorem 6.3 has two parts. First we will understand conjugacy classes inc(gr-A) and thus graded Morita equivalence classes of rings graded equivalent to A. Then we willmpute representative rings for each equivalence class.For the rst part, we will need several easy combinatorial lemmas. We begin by establishing no-

tion. For any positive integer n, dene an operator n on Zn by setting

n J = J ( J n).

ote that n(I J ) = n I n J .Given i,n Z with 0 i n 1 and J Zn, dene

Jni = { j Z | nj + i J }.

mma 6.5. For all n, the operator n is a one-to-one map from Zn onto{J Zn

#( Jni ) is even for 0 i n 1}.oof. We rst prove the lemma for n = 1. Let J Zn with # J = 2m and write J = {a1 < b1 < 1, and let J Zn. Then

n J =n1i=0

n(1 J

ni

)+ i.us the result for general n follows from the result for n = 1. mma 6.6.

) Let F be a generative autoequivalence of gr-A. Then F is conjugate in Pic(gr-A) to some SnA J , where( J ,n) is an admissible pair.

) If ( J ,n) and (L,m) are admissible pairs, then SmA L and SnA J are conjugate in Pic(gr-A) if and only if( J ,n) and (L,m) have the same necklace type.

oof. (1) By Lemma 5.15 F is even and has nonzero rank; thus by Corollary 5.12 F = SnAK for some= 0 and some K Zn. Since F has rank n, without loss of generality we may assume that> 0.Dene the set J {0, . . . ,n 1} to be:

J = {i #Kni is odd}.en for all 0 i n 1 we have that #( J K )ni is even, so by Lemma 6.5 there is some I Znch that n I = J K . That is,

ISnAK I = SnA[(I n) K I]= SnA[n I K ] = SnA J .

(2) Let ( J ,n) and (L,m) be admissible pairs. Note that conjugating by sends a rank n autoequiv-ence to a rank n autoequivalence, while conjugating by an even autoequivalence preserves rank.nce m,n 1, if SmA L and SnA J are conjugate in Pic(gr-A) then they are conjugate in the subgroupeven autoequivalences, and we have m = n.Suppose that m = n. Then by the above and by Corollary 5.12, SnAL and Sn J are conjugate if andly if there are some j Z and I Zn such that S jAISn J IS jA = SnL ; this is equivalent to theistence of I and j such that

J = n I (L j).

Lemma 6.5, such an I exists if and only if (L j)ni and Jni have the same parity for all 0 i n1.t since J and L are both contained in {0, . . . ,n 1} this is true if and only if (L j) J (mod n).is is equivalent to (L,n) and ( J ,n) having the same necklace type. mma 6.7. Let F = SnA I , where n > 0 and I {0, . . . ,n 1}. Then F is A-generative.

oof. We introduce notation: for j = 0 dene sets j and j by

j ={(I + n) (I + 2n) (I + jn) if j > 0,I (I n) (I + ( j + 1)n) if j < 0,

={ {0, . . . ,nj 1} if j > 0,j {1, . . . ,nj} if j < 0.


(T

alAge

If

Ag

pi

Pr

PrmS(

inge

( J

gr

S(Pity

thcarithemhus [ j]A = Anj.) Then we see that if j = 0,

F j A = [ j]SnjA A = [ j][ j]A = [ j j]A.

Let J j = j j for j = 0. By Lemma 5.14, {F j A} generates gr-A if and only if {F j A} generatesl integrally supported simple modules. Since A generates Xll0 and Y ll1 we see that F is-generative if and only if for all l 0, some F j A generates Y l, and for all l 1, some F j Anerates Xl. That is, F is A-generative if and only if

j =0

J j = Z. (6.8)

Now, if i / I , then j (nZ + i) = , and so if j > 0, then

J j (nZ + i) = j (nZ + i) ={i,n+ i, . . . ,n( j 1) + i}.

j < 0, then J j (nZ + i) = {i n, . . . , i + nj}. Thus (nZ + i) j =0 J j .Now suppose that i I . Then if j > 0, then j (nZ + i) = {n + i, . . . ,nj + i} and if j < 0, then

j (nZ+ i) = {i, i n, . . . , i +n( j + 1)}. We see that J j (nZ+ i) = ( j j) (nZ+ i) = {i, i +nj}.ain we have that (nZ + i) j =0 J j . Thus (6.8) is satised, and F is A-generative. We have gathered the ingredients for one-half of the proof of Theorem 6.3. The other necessary

ece is:

oposition 6.9. Let F = SnA J , with ( J ,n) an admissible pair. Then EndFA (A) is isomorphic to S( J ,n).

Assuming this proposition for the moment, we may now prove Theorem 6.3.

oof of Theorem 6.3. Proposition 6.9 tells us that if ( J ,n) is an admissible pair, then S( J ,n) is iso-orphic to EndFA (A) for F = SnA J . By Lemma 6.7, such an F is A-generative; thus by Theorem 2.12,J ,n) is graded equivalent to A.Now let S be a graded ring and let : gr-A gr-S be a category equivalence, with quasi-

verse . Let F = SS , let m be the rank of F , and let P = S . By Theorem 2.12, F is P -nerative. By Lemma 6.6(1), there is some F in the conjugacy class of F of the form F = SnA J with,n) an admissible pair. By Lemma 6.7, F is A-generative; thus by Proposition 2.15, S = EndFR (P ) isaded Morita equivalent to EndF

A (A). By Proposition 6.9, End

F A (A) is isomorphic to S( J ,n).

Now let (L, r) be another admissible pair, and let G = SrAL . By Proposition 6.9 and Proposition 2.15,L, r) = EndGA (A) and S( J ,n) are graded Morita equivalent if and only if G and F are conjugate inc(gr-A); but by Lemma 6.6(2) this is true if and only if (L, r) and ( J ,n) have the same necklacepe. To complete the proof of Theorem 6.3, all that remains is to prove Proposition 6.9. Before doing

is, we give a general lemma allowing us to calculate twisted endomorphism rings explicitly. Torry out our computations, we will work in two localizations of A. Let D be the graded quotientng of A and let T = A[y1]. If is the automorphism of k[z] or k(z) that sends z z+ 1, we haveat D = k(z)[x, x1; ] = k(z)[y1, y; ] and that T = k[z][y1, y; ], which we write in this way to

phasize the grading.


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bymma 6.10. Let F be an A-generative (and therefore even) rank n autoequivalence of gr-A, and let C be the-twisted endomorphism ring EndFA (A). Write F = SnA J , and dene A-submodules M( j) of D by:

M( j) =

( j

i=1 [ J + ni]1)A if j 1,A if j = 0,( j1

i=0 [ J ni])A if j 1

d a graded vector subspace C = jZ C j of the Veronese ring D(n) byC j = M( j)nj .

en C is a subring of D(n) and C = C.

oof. Since J (D) D is the graded injective hull of J A, then J (D) = D and so for any j Z, weve F j D = SnjA D = Dnj; further, if g : D D is a graded A-module map, then J (g) = g and soj(g) = SnjA (g) as maps from Dnj Dnj.Now, for each j Z there is a natural map j : F j A Dnj given by applying F j to the inclusion

0 : A D . For j 1, j is the composition

F j A SnjA [J n( j 1)] J A Dnj,

d for j 1, j is the composition

F j A SnjA [J + n( j)]1 [ J + n]1A Dnj.

otice that the image of j is precisely SnjA M( j). That is, for any j Z the module M( j) may beentied with SnjA F j A.Let H be the Z-algebra H = HA(

jZ F j A,

jZ F j A); that is,

Hij = homA(F j A,F i A).

ppose that f : F j A F i A is an element of Hij . Then by graded injectivity of Dni, there is aique map f : Dnj Dni such that the diagram

F j Af

j

F i Ai

Dnjf

Dni(6.11)

mmutes. Let E = D(n) and dene a map

: H E = HA(

jZDnj,

iZ

Dni)dening ( f ) = f .


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PrmWe claim that is a homomorphism of principal Z-algebras; that is, is a (graded) map of-algebras that commutes with the principal automorphisms of H and E . Let be the principaltomorphism of H induced from F and let be the canonical principal automorphism of E , inducedom SnA . Since F and SA are automorphisms of gr-A, in this case the compatibility isomorphismse trivial, and we have ( f ) = F( f ) and (( f )) = SnA(( f )).Given f Hij and g H jk , there is a commutative diagram

Fk Ag

k

F j Af

j

F i Ai

Dnk (g)

( f g)

Dnj ( f ) Dni.

iqueness of the lifting ( f g) gives us that ( f g) = ( f ) (g), and so is a homomorphismZ-algebras.We verify that is a morphism of principal Z-algebras. Fix f Hij . Then ( f ) is given by the

agram (6.11), where ( f ) = f . Applying F to (6.11), we obtain:

F j+1AF( f )=( f )

F( j)

F i+1AF(i)

Dn( j + 1)F(( f ))=SnA (( f ))=(( f ))

Dn(i + 1).

(6.12)

t by construction, F( j) = j+1 and F(i) = i+1, so (6.12) is precisely the diagram giving(( f )). That is, (( f )) = (( f )), and is a map of principal Z-algebras, as claimed.In particular, the canonical principal map on E restricts to a principal map on the image of .e saw that induces a natural ring structure on

jZ E0 j that makes it isomorphic to D(n) . By

mark 2.10, it likewise induces a ring structure isomorphic to D(n) on

jZ E j,0. Thus we maystrict the identications E0, j = Dnj = E j,0 to the image of to obtain a subring of D(n) . That is,e natural map

C j = homA(A,F j A) homA(A,SnjA (M( j)))= M( j)nj E j,0 = Dnj

ves a ring monomorphism

: C D(n).

erefore, C = Im = jZM( j)nj is a subring of D(n) and C = C , as claimed. oof of Proposition 6.9. The proof is by direct computation. Put S = EndFA (A). By Lemma 6.10, weay compute S via:

S =

M( j)nj D(n)

jZ


w

an

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Le(th

Puan

Th

Ithere

M( j) = SnjA F j A =

[( J + n) ( J + jn)]1A for j 1,A for j = 0,[ J ( J n) ( J + ( j + 1)n)]A for j 1.

Therefore, if j > 0 we have

S j = M( j)nj =([( J + n) ( J + jn)]1A)nj

=( j

l=1f J+nl

)1 ([( J + n) ( J + jn)]A)nj,

d if j < 0 we have

S j =([J ( J n) ( J + ( j + 1)n)]A)nj .

j = 0 then S0 = A0 = k[z].We compute the terms S j , using Lemma 5.13. We rst let j < 0. We want the ynj term of

[J ( J n) ( J + ( j + 1)n)]A = [ J ( J n) ( J + ( j + 1)n)]A.

t L = ( J n) ( J + ( j + 1)n). If i L, then since 0 > i > nj, by Lemma 5.13(3) we havei A)nj = ynjk[z]. If i J , then (i A)nj = (z + i)ynjk[z]. That is, if j < 0 we have by Lemma 5.13(1)at

S j =

i JLi A =

(i J

(z + i))ynjk[z] ynjk[z] = f J ynjk[z]. (6.13)

Now consider the terms for j > 0. Let

K = ( J + n) ( J + nj) = ( J + n) ( J + nj).

t K = K Znj1 and K = K Znj = J +nj. A similar computation shows that (K A)nj = xnjk[z],d (K A)nj = f K xnjk[z] = f J+njxnjk[z]. Now, K = K K and we see that

(K A)nj = (K A)nj (K A)nj = f J+njxnjk[z].

us,

S j =( j

l=1f 1J+nl

) (K A)nj =

( jl=1

f 1J+nl

) f J+njxnjk[z]

= f J ( j1

l=0f 1J+nl

)xnjk[z].

is straightforward to verify that this is equal to

( 1 n) jf J f J x k[z].


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thPibytoRecall the notation that J = {0, . . . ,n 1} \ J . Since xn =n1i=0 (z+ i)yn = f J f J yn , we see thatr j > 0, we have

S j = f J (f J y

n) jk[z]. (6.14)nce S0 = k[z], combining (6.13) and (6.14) we have that S = IW ( f J W ), where W = k f J yn, yn, z.t it is easy to see that W = W ( f J ,n)that is, S is precisely S( J ,n). ample 6.15. Let F = SA 0. By Theorem 6.3, A and B = EndFA (A) are graded equivalent; by Propo-tion 6.9, B is isomorphic to

S({0},1)= IT (zT ) =

zynk[z] if n 1,k[z] if n = 0,zynk[z] if n1.

It is certainly surprising that these two rings are graded equivalent, since their behavior is sofferent. It is well known that A is simple, and is therefore a maximal order: there is no ring S with S Q (A) such that aSb A for some nonzero a,b A. On the other hand, B is neither simple ormaximal order. Further, B fails the second layer condition governing relationships among prime idealsd enabling localization (see [8, Chapter 11]), while A (trivially) satises the second layer condition.so, while A has no nite-dimensional modules, B has a 1-dimensional graded representation.We explore the equivalence = HomFA (A, ) between gr-A and gr-B . We rst describe how

ealizing affects graded module categories. Let R be a graded ring with a graded right ideal I thatmaximal as a right ideal of R; let S = IR(I) be the idealizer of I in R . By [13, Theorem 1.3], thegraded category mod-S has one more simple than the category mod-R: the simple R-module R/Icomes a length 2 module over S . Thus in the graded category, idealizing corresponds to replacinge simples (R/I)n with pairs of simple modules.Now consider the category gr-B . Since T is strongly graded, by [11, Theorem I.3.4], the categories

-T and mod-k[z] are equivalent, and the simples in gr-T are naturally parameterized by the anee. The discussion above shows that gr-B may be represented by an ane line with double pointsevery integer:

: : : : .1 0 1 2

e saw in Lemma 4.1 that this is also a representation of the simple objects in gr-A.A quick computation shows that Fn A = 0n A, and since (Fn A) = Bn, we see that B is theique shift of B that generates (X). Thus (X) is the 1-dimensional module k, and maps theact sequence

0 xA A X 0

0 zT B k 0.

us the right ideal xA maps to a two-sided ideal in B .Since the existence of is rather counterintuitive, it is not surprising that is quite different from

e standard examples of graded Morita equivalences and Zhang twists. This can also be seen in thecard group of gr-A: the autoequivalence SA0 = SB is not induced from an automorphism of A,Corollary 5.10. On the other hand, by Corollary 3.9, any autoequivalence of gr-A that corresponds

a Zhang twist or a graded Morita equivalence would be in the image of (3.1).


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Pr

(1The discussion in Example 6.15 proves Corollary 1.4.

ample 6.16. For another example, we note that A is graded equivalent to an idealizer in A: put= 2 and J = {0}. Then by Proposition 6.9 the corresponding ring is C = S({0},2) = IW (zW ), where= W (z + 1,2) is isomorphic to A via

X x,Y 2y,z 2z 1.

us C is isomorphic to IA((z 1/2)A).This example is slightly less counterintuitive if we note that passing from gr-A to gr-C correspondsadding a simple module for each half-integer point; thus pictorially the equivalence between gr-Ad gr-C corresponds to scaling by a factor of 1/2. It is still surprising that this can be made to worknctorially.

mark 6.17. Paul Smith has recently shown [15] that the category gr-A is also equivalent to a gradedodule category over a commutative ring R . The ring R is graded by Zn, and gr-R naturally corre-onds to the ane line with a stacky Z/2Z-point at every integer. (See [15] for precise denitions.)is is plausible from the pictorial representation of gr-A, but it is certainly quite counterintuitiveat any module category associated to the Weyl algebra is commutative in any sense!

We remark that a similar classication to Theorem 6.3 can be carried out for other generalizedeyl algebras, in particular for primitive factors of U = U (sl2(C)). In terms of generators and rela-ons, U is generated over C by E , F , and H , subject to the relations

[H, E] = 2E, [H, F ] = 2F , [E, F ] = H .

t U be the Casimir element 4E F + H2 2H . Then the primitive factors of U are given byctoring out + for some C; we write them as U = U/( 2 2) for C. If we let ed f be the images of E and F , respectively, in U , and let h be 1/2 times the image of H , then Ugiven in terms of generators and relations by

[h, e] = e, f e = (h +

2+ 1)(

h 2

),

[h, f ] = f , ef = (h +

2

)(h

2 1).

e U are clearly (isomorphic to) generalized Weyl algebras.Stafford [16] has shown that in this parameterization, U is hereditary when / Z; that is, when

e roots of ef do not differ by an integer. If Z but = 1, then U has global dimension 2. If= 1, so the roots of ef coincide, then U has innite global dimension. (See [5] for a generalizationarbitrary generalized Weyl algebras.)For all values of , there are involutions of the category gr-U similar to the involutions j on

-A. We will not give the classication of rings graded equivalent to U , but we will note that if wet S = U[ f 1] = C[h][ f 1, f ; ], where (h) = h 1, then similar calculations to Proposition 6.9oduce the following result,

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Rings Graded Equivalent to the Weyl Algebra