-
Journal of Algebra 321 (2009) 495531
Contents lists available at ScienceDirect
R
Su
De
a
ArReAvCo
KeGrGrCaW
Co
AcRe
00doJournal of Algebra
www.elsevier.com/locate/jalgebra
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
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496 S.J. Sierra / Journal of Algebra 321 (2009) 495531
1.
teha
alisonHshCamCthby
Thar
bygrwal
agrth
n
Le
Th
ThgrIntroduction
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
gr
Thco
ThxAtheqgrtrstidde
Cosa
orseof
ash
Thth{F
Th
qufothanasfuof
gachprwco
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.
-
498 S.J. Sierra / Journal of Algebra 321 (2009) 495531
anthgr
gimIfcaphwsi
M
(W
ot
2.
eqm
th
of
toangrwW
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
-
S.J. Sierra / Journal of Algebra 321 (2009) 495531 499
wsuw
thpr
Prgr
paAsa
Thca
grat
wgr
Lethfr
Pris
(1
(2
eqan
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
-
500 S.J. Sierra / Journal of Algebra 321 (2009) 495531
His
fo
ofifnlewau
auto
of
(1(2
bi
be
to
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.
-
S.J. Sierra / Journal of Algebra 321 (2009) 495531 501
Pron
Pr
wThs
asha
fo
eqsh
H
Cl
NTh
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)
-
502 S.J. Sierra / Journal of Algebra 321 (2009) 495531
So
Th
anis
Th
Den
th
Pra
ReFau
Th
en
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.
-
S.J. Sierra / Journal of Algebra 321 (2009) 495531 503
HisId{Bna
co
antin
Pran
Th
(1(2
P
Pral
anca
ThZ
an
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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504 S.J. Sierra / Journal of Algebra 321 (2009) 495531
ungi
orPrG
Cois
Pr
soti
Fby
is
Regrwco
wvearha
plseofco
twbe
PrsuEn
Pr
an
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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 505
Si
Th
ha
Th
3.
obth
Wlem
ThR-Pi
grse
mth
plWsa
foatwZh
Th
(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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506 S.J. Sierra / Journal of Algebra 321 (2009) 495531
(3(4
Co
Sucamis
(ntw
LeauLe
co
Prtw
Pr
Swgrs
an
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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 507
(MR
O
Th
tioByLe
Gtw
Cogr
Prse
se
Cogr
in
toafo
Thdefo
ifthtw
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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508 S.J. Sierra / Journal of Algebra 321 (2009) 495531
Pr(
of
Th
accageof
Coau
PralG
4.
anal
w
thfagi
fo
twisiswre
th
Le
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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 509
Wisfo
ofthX
re(
foth
an
Le
(1(2(3(4(5
Pr
to
Thth
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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510 S.J. Sierra / Journal of Algebra 321 (2009) 495531
an
to
It
an
doth
Lem
Pr
th
Le
Prxn
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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 511
DtoSuthsi
fw
Ledi
Prsi
anof
is(z
Leho
Pr
be
an
ByHho
si
Prfo
(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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512 S.J. Sierra / Journal of Algebra 321 (2009) 495531
(2
Pran
beYex
wseti
ro
w
Th
Dsiifand
homA(P , Y n)= extA(Xn, P)= 0; or
) 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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 513
LeP
Pr
beN
Nfaimj-
thpr
Lesu
5.
grth
an
esm
Le
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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514 S.J. Sierra / Journal of Algebra 321 (2009) 495531
wan
ani
Co
Pr
dias
Thal
an
Pr
siinpaex
(1(2
wdeaucom
Dinth
siob
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 .
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S.J. Sierra / Journal of Algebra 321 (2009) 495531 515
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
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516 S.J. Sierra / Journal of Algebra 321 (2009) 495531
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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 }
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S.J. Sierra / Journal of Algebra 321 (2009) 495531 517
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 .
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518 S.J. Sierra / Journal of Algebra 321 (2009) 495531
su
thsh
fo
Le
Th
Pim
an
Copr
Pr
Cosu
fo
Le
(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
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S.J. Sierra / Journal of Algebra 321 (2009) 495531 519
(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,
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520 S.J. Sierra / Journal of Algebra 321 (2009) 495531
thto
grVbeth
ThBa
Thdi
papu
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
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S.J. Sierra / Journal of Algebra 321 (2009) 495531 521
neis
w
Ththif
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.
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S.J. Sierra / Journal of Algebra 321 (2009) 495531 523
(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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524 S.J. Sierra / Journal of Algebra 321 (2009) 495531
LeF
an
Th
Pr
ha
F
an
N
id
Suun
co
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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 525
Z
aufrar
Unof
di
Bu
W
Re
reth
gi
Th
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
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526 S.J. Sierra / Journal of Algebra 321 (2009) 495531
w
an
If
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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S.J. Sierra / Journal of Algebra 321 (2009) 495531 527
fo
SiBu
Exsi
diAaanAl
idisunbeth
grlinat
W
unex
to
Th
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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528 S.J. Sierra / Journal of Algebra 321 (2009) 495531
ExnW
Th
toanfu
RemspThth
Wti
Lefaanis
Th
th
to
grlepr
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,