David Andrés Dávila Díaz 200911582
Universidad de los Andes Proyecto de grado Documento nal MATE-3902
On the structure of quantum Weyl algebras
1 Introduction
This thesis is about quantized Weyl algebras over the complex numbers, This is, a C-algebra generated by variables x, ∂ modulo the relation ∂x−qx∂ = 1. When q is a root of unity it
is known thatAq is free over its centerZ(Aq) and that for a generic point maximal ideal I in Z(Aq) the ber Aq/I is a full matrix ring. The purpose of this thesis is to the describe the
bers of Aq which are not full matrix rings (so called non-Azumaya bers).
For this we will start with some basics about nite dimensional algebras, an algebra is basically a ring which has properties of vector space (a better denition is given below). Even though
the denitin for an algebra is given, some basic denitions and properties are assumend, for example, the denitions of a module, the tensor product of an algebra, morphisms of algebras and modules, etc.
The natural question that arises is: Why would we be interested in this work? The answer splits in three parts, rst, the structure of quantized Weyl algebras is important from the view point of representation theory of quantum groups, since they look locally as quantum Weyl algebras; additionally, to understand automorphisms of Weyl algebras in characteristic 0, in [7] it's used a reduction modulo large primes, ultra products and Los theorem in logic in order to relate such automorphisms to symplectoautomorphisms of ane 2n-space. It was observed in [1] that one might instead consider quantizations of automoprphisms and let the quantum parameter q converge to 1 along roots of unity. For this to work it is essential that the quantum Weyl algebra is generically Azumaya and for ner results it will be essential to understand the non-Azumaya bers. Finally, it is a good starting point to study Dixmier's conjecture, which states that for a Weyl algebra any endomorphism is also an automorphism
(i.eAut(An) =End(An)).
The document will split in 3 parts, the rst one will be a little background on nite dimensional algebras, Here we will mention some important theorems that will be very usefull later in the document, for example the Jordan Hölder Theorem, or a way to express a nite dimensional algebra as a sum of easier structures. Later we will talk about the theory of the Weyl algebras in general, describe their basis, their center, and check some important properties; Finally in the last section we will check the structure of the non-azumaya bers, which, according to [1] are the bers of the form Aq/I where I is a maximal ideal of the form(x−a, ∂−b), such that a·b = (1−q)−`(q primitive `'th root of unity), from here we will nd a projective cover for
the simple modules of this ring, and check the morphisms between them, nding this way all the possible morphisms from the ring to itself.
2 Finite dimensional algebras
First We'll make a brief introduction to nite dimensional algebras, most of the denitions are assumed to be known by the reader, for this section we will only work with nite dimensional modules over nite dimensional algebras, unless it's explicitly said otherwise.
2.1 The Jordan Hölder Theorem
Denition 2.1. Let k be a eld, a k- Algebra is a vector space A over k, where you can
multiply the elements of A, for this multiplication hold left and right distributivity and it is compatible with the scalar product
Here are some examples:
• Cis anR-algebra, with the usual multiplication and 1C= 1R
• the ring of polinomials in n variables over a eld kk[x1, x2, ..., xn]is a k-algebra
• the set of invertible n×n matrixes over C in an C-algebra (or in fact over any other eld)
Denition 2.2. A module M is called simple if the only submodules of M are 0 and M itself An example of this is any eld K seen itself as a K-Module, since any nonzero element has a multiplicative inverse
Denition 2.3. Let M a module. A composition series of M is a sequence of submodules as shown
0 =M0 (M1(...(Md=M Such that ∀i∈ {1, ..., d}, Mi/Mi−1 are simple
in this case d is called the length of the series (noted as L(M)), and theMi/Mi−1 are called
the composition factors of the series.
It is easy to see that every nite dimensional module of a nite dimensional algebra has a composition series, if the module is not simple, take M1 (M of smallest dimension possible, so it is simple, proceed to do the same toM/M1 obtainingM20 (M/M1, now, deneM2 (M such thatM20 =M2/M1, etc.
Lemma 2.4. If N is a proper submodule of M then L(N)< L(M)
Proof. Let0 =M0 (M1 (...(Md=M a composition series for M, thus
0 =N∩M0(N∩M1(...(N ∩Md=N
is a composition series for N, this is because, since the natural map φ:N∩Mi/N ∩Mi−1 →
Mi/Mi−1 is injective, and as theMi/Mi−1 are simple, thenN∩Mi/N∩Mi−1 must be simple
(and hence, equal toMi/Mi−1) or be 0.
so we haveL(N)≤L(M), let's show it can't be equal, for this it is enough to show that there
exists an i such that:
note thatdim(N) =P
idim(N ∩Mi/N ∩Mi−1), so, if N∩Mi/N ∩Mi−16= 0 for each i,
thendim(N) =dim(M) so N would not be a proper submodule. Therefore there is such an i
and we conclude that L(N)< L(M).
The next theorem shows the length of any composition series for a module is unique, and the compositions factors are unique up to isomorphism, it is known as The Jordan Hölder Theorem.
Theorem 2.5. Any composition series of a module M have the same length and the same composition factors up to isomorphism.
Proof. The proof will be done by induction on the length of the composition series for M. clearly if `(M) = 1 then M is simple, so the proof is trivial.
Suppose now that if `(M)< k the theorem holds
Let
0 =M0 (M1(...(Mk=M and 0 =N0(N1(...(Nj =M
two composition series for M, now we have two cases
• Case 1: Mk−1 = Nj−1 here we can apply the induction hypothesis thanks to lemma
2.10, sok−1 =j−1and obviouslyk=j, we also need to show that the last composition
factors are equal, but this is obvious, since both are M/Mk−1
• Case 2: Mk−1 6=Nk−1 Consider the submodule Mk−M1k+−N1j−1 (M/Mk−1, sinceMk−1 6=
Nk−1 this submodule is not 0, and since M/Mk−1 is simple they must be equal,
analo-gouslyM/Nj−1 =
Mk−1+Nj−1 Nj−1
Also, note that Mk−1+Nj−1 Nj−1
∼
= Mk−1
Mk−1∩Nj−1 and
Mk−1+Nj−1 Mk−1
∼
= M Nj−1
k−1∩Nj−1
Consider now a composition series for Mk−1∩Nj−1 as follows:
0 =L0(L1 (...(Li =Mk−1∩Nj−1
note that this composition series can be extended to a composition series forMk−1 and
Nj−1with composition factors{{Lm/Lm−1},M Mk−1
k−1∩Nj−1}and{{Lm/Lm−1},
Nj−1 Mk−1∩Nj−1}
respectively, this composition factors of Mk−1 andNj−1 are uniquely determined up to
isomorphism, so the composition factors of
0 =M0 (M1 (...(Mk =M
are{Lm/Lm−1}, Mk−1/(Mk−1∩Nj−1) andM/Mk−1 ∼=Nj−1/(Mk−1∩Nj−1)
And the composition factors of
0 =N0 (N1 (...(Nj =M
are{Lm/Lm−1}, Nj−1/(Mk−1∩Nj−1) andM/Nj−1 ∼=Mk−1/(Mk−1∩Nj−1)
so this two composition series have the same composition factors up to isomorphism.
2.2 Decomposition of nite dimensional algebras
Denition 2.6. A module P is called projective if for all modules M,N and surjections φ : M →N, if f :P →M, then there exists a mapψ:P →N such thatf =ψ◦φ.
Denition 2.7. Let M an A-module, a projective cover of M is a projective module P with a surjection ψ :P →M, such that, if Q is any other projective module and φ:Q→M is a
surjection, then there exists f :Q→P such thatφ=f◦ψ.
Theorem 2.8. Every nite dimensional A-module of a nite dimensional algebra has a pro-jective cover.
Proof. Let M an A-module, and let {m1, m2, ..., mn} a basis for M, hence, there exists a
sur-jection A⊕A⊕...⊕A(n−times)→M, sending the tuple(a1, a2, ..., an)toPni=1aimi,hence,
there exists a ,projective module ρ and a surjection π:ρ→M, now dene:
Λ ={P ⊂ρsuch thatπ |P is surjective}
Choose P in Λ such that dim(P) is minimun, so far we have π |P: P → M and π : ρ → M
both surjections, so, since ρ is projective, there exists τ :ρ→P such thatπ =τ ◦π |P.
(clearly06=Im(τ ◦i)⊂P, and since P has minimum dimension they must be equal).
Therefore, the sequence
0→P →ρ→ρ/P →0
splits with the map (τ◦i)−1◦τ :Q→P and hence Q∼=P ⊕Q/P, so P must be projective.
Now, letP0 a projective module, and letψ:P0→M a surjection, then there existsφ:P0 →P
such that π |P ◦φ =ψ, so we have 06=φ(P0) ⊂P, but since P has minimun dimension this
must be a surjection. So P is a projective cover for M.
Proposition 2.9. Let M an A-module, andP1, P2 two projective covers for M, then P1 ∼=P2.
Proof. Let P1, P2 projective covers of M an A-module, with surjections φ1, φ2 respectively.
Since P2 is projective, there exists psi:P1 →P2 such thatφ1=φ2◦ψ
Since P1 is projective, there exists η:P2→P1 such thatφ2 =φ1◦η
So η◦ψ:P1 →P1 is a surjection, so it must be an isomorphism, so η must be injective and
since it was surjective by denition, it must be an isomorphism.
Theorem 2.10. Let R a nite dimensional Algebra.Then there are nitely manyL1, L2, ..., Ld
simple R-modules (Modulo isomorphism), and if Pi→Li is projective cover
R∼=P1m1⊕P2m2⊕...⊕Pmd
d
As a left R-module
Before proving this theorem we will prove a lemma to make our life easier
Lemma 2.11. Let P a projective module, dene R=∩N∈MN where M is the set of maximal
ideals of P, then, the map q:P →P/R is a projective cover.
Proof. LetP0a projective module such that there exists a surjection ρ:Q→P/R, then there
must exist τ :Q→P such thatρ=q◦τ,(this is since q is a quotient, otherwise this woldn't
be necessary true )
Therefore we have that τ(Q) +R = P. Suppose now that τ is not a surjection, and hence (P, q)is not a projective cover of R, then there must exist a maximal submoduleM (P such
that τ(Q) ⊂M, also, by denition we know that R ⊂M, so τ(Q) +R⊂X (P, which is a contradiction.
So τ must be surjective, and (P, q) is a projective cover of R
Now we can proceed with the proof of theorem 2.10(we will prove this theorem, since it
is very important on the main objective of the document)
Proof. LetL a simpleA-Module, pickings∈S, s6= 0 there must exist a map
f :A→S a7−→as
Since S in simple, f must be surjective, so S is a composition factor of A(seeing A as a module
itself). By Jordan-Hölder we know there can be at mostL(A) =ndierent simple submodules,
let L1, L2, ..., Lnbe this simple submodules, and letP1, P2, ..., Pn their projective covers, then
thePi are indecomposable, since otherwise, ifPi=Q1⊕Q2 withQ1 6= 06=Q2by simplicity of
Li there would be a surjection either fromQ1 toLi or fromQ2toLi, but sincePi is projective
cover, this can't happen.
So far we have a set of indecomposable projecive A-Modules, we would like to check that any other indecomposable projective A-Module must be isomorphic to one of this Pi, for
this notice that if R is another indecomposable projective A-Module, then R/M, where M = ∩N∈M ax(R)N, is simple, and R → R/M is a projective cover. Since the Li are all
the simple modules (up to ismorphism) we getR/M ∼=Lk for somek∈ {1,2, ..., n}.
Take now one of the indecomposables Pi, it's easy to see that A itself is a projective module,
and since φ : A → Li is a surjection, then φˆ :A → Pi is also a surjection, then we have an
exact sequence
0−→K −→A−→Pi−→0
which splits sincePi is projective, so Pi must be a direct summand of A.
All what's left is to prove that every summand of A must be isomorphic to some Pi, for
a projective cover (Ms = ∩N∈M ax(S)N),but S/MS is simple, so it's isomorphic to some Lj,
meaning U is isomorphic to Pj.
2.3 The Center of an algebra
Denition 2.12. The center of a K-algebra A (Z(A)) consists of the elements x ∈ A such
thatxa=ax∀a∈A, ifZ(A) =A, A is called commutative, ifZ(A) =K A is called a central
simple algebra.
Proposition 2.13. the center of an Algebra Z(A) is a subalgebra.
Proof. we need to prove that Z(A) is a ring, and a vector space, let's rst check that the
restrictions of the operations from A giveZ(A) the structure of a ring, for this it is enough to
show Z(A)is closed under these operations.
Letx, y∈Z(A), we want to see thatx+y andxy are inZ(A), leta∈A a(x+y) =ax+ay
=xa+yasince x and y are central. = (x+y)a
so(x+y)inZ(A)
(xy)a=x(ya)
=x(ay)
= (xa)y
=a(xy)
soxy ∈Z(A).
So far we only haveZ(A) is a subring, however, claerlyZ(A) is compatible with the scalar
product, so it's also a vector subspace, thus, a subalgebra.
3 Basic Theory of Weyl Algebras
Denition 3.1. Let K be a Field, Dene A(K) as the non commutative algebra over D on
∂x−x∂= 1
This is
A(K) = (∂xK−hx,∂x∂−i1)
We will callA(K)the rst Weyl Algebra(as the name suggests there is a denition for the n'th
Weyl algebra, which is the n'th tensor power ofA(K), however we won't talk about them here).
For this document we will only consider the caseK=C
Denition 3.2. The quantum Weyl algebraAqis theC−algebra on two generatorsx, ∂ with
the dening relation
x∂−q∂x= 1
ie.
A(K) = (∂xK−hqx∂x,∂−i1)
whereq ∈C∗
For this document we will only be interested in the case where q is a primitive`'th root of
the unity.
Proposition 3.3. for the quantized Weyl Algebra Aq the realtions
• ∂xm = 11−−qqmxm−1+qmxm∂ and
• ∂mx= 11−−qqm∂m−1+qmx∂m
Proof. let's show ∂xm = 11−−qqmxm−1+qmxm∂ holds, for this we will proceed by induction on
m for the base case, the dening relation of the quantized Weyl algebra gives:
∂x1 = 1 +q∂x, this actually has the form we want, if we look at it as 11−−qq1x0+qx∂
Let's assume now that the relation holds for m, and let's check that:
∂xm+1= 1−1q−mq+1xm+qm+1xm+1∂
notice that
∂xm+1=∂xmx
= (11−−qqmxm−1+qmxm∂)x
= 11−−qqmxm+qmxm∂x
= 11−−qmq xm+qmxm(1 +qx∂)
= 11−−qqmxm+qmxm+qm+1xm+1∂
= (11−−qqm +qm)xm+qm+1xm+1∂
= (11−−qqm +qm1(1−−qq))xm+qm+1xm+1∂
= 1−1q−mq+1xm+qm+1xm+1∂
the proof for ∂mx= 11−−qqm∂m−1+qmx∂m is analogue
Proposition 3.4. The set {xi∂j :i, j≥0} is a basis for A(C)
Proof. notice that if you have an element of the forma∂xyou can always write it asa(1−qx∂),
so all we have to do is show they are linearly independent, in order to do so we will nd a suitable representation for the Weyl algebra.
Let t be a parameter, and dene
At:=C[t]< ∂, x > /(∂x−tx∂−1) we would like to show that
At×C[t][x]→C[t][x]
∂xn7−→(1 +t+...+tn−1)xn−1 xxn7−→xn+1
Is a representation.
For this we will prove this map satisfy the relation(∂x−tx∂−1)xn7−→0∀n∈N, this means, it satises the only relation in the Weyl algebra.
= (∂x−tx∂−1)xn=∂xn+1−tx∂xn−xn = (1 +t+...+tn)xn−tx(1 +t+...+tn−1)xn−1−xn
= (1 +t+...+tn)xn−(t+t2...+tn)xn−xn =xn+ (t+t2...+tn)xn−(t+t2...+tn)xn−xn
= 0
Let P ∈ At((C)), then, we can write P = PNi=0ai(x)∂i for some N, now we want to prove
that if P xn = 0 ∀n∈N, thenai = 0 ∀i∈ {0,1, ..., N}, since P·1 = 0, andP ·1 = a0(x) we
have a0(x) = 0.
now we continue witha1,P·x= 0, butP·x=a1(x)+a0(x)x, since we already havea0(x) = 0,
we conclude now thata1(x) = 0.
proceeding inductively on the sub index we have that forP·xn= 0 the sum gives(1 +t+...+ tn−1)an(x)+(1+t+...+tn−2)an−1(x)x+...+a0xn= 0, but everyai = 0fori∈ {0,1, ..., n−1},
so the result is an = 0. From this we can conclude that the action is faithful, so At is a free
C[t]−module.
Clearly, At=SpanC[t]{x
i∂j}, soA
t is a free C[t] module, with basisxi∂j Now note that C[t]/(t−q) =C, sinceqis a complex number. SoAq=At/(t−q), soAq is also a free module with basis xi∂j, so we can concludeAq is free with basisxi∂j, which means the elementsxi∂j
are linearly independent.
Proposition 3.5. Let Aq be a Weyl algebra over C, then, the center of Aq, when q is the
primitive L'th root of unity is isomorphic to the ring of polinomials in 2 variables over C, namely
Proof. For this we want to compute[∂, xk`]and[∂`, x]fork∈N, let's begin with the rst one, showing by induction it's 0 for every k.
[∂, x`] = 11−−qq`x`−1+q`x`∂−x`∂ = 11−−1qx`−1+x`∂−x`∂
= 0
assume now this equality holds for every naturaln≤m , let's prove it for m+1 [∂, x(k+1)`] =∂x(k+1)L−x(k+1)`∂
=∂x(k+1)`−x(k+1)`∂+xk`∂x`−xk`∂x` =∂xk`x`−xk`∂x`+xk`∂x`−xk`x`∂
= [∂, xk`]x`+xk`[∂, x`]
= 0
once again, the proof for the other identity is analogue.
It remains to prove that every element in the center has this form, suppose xk∈Z(Aq), then
we have that [∂, xk] = 0, using the denition of the commutator we get
[∂, xk] =∂xk−xk∂ = 11−−qqkxk−1+qkxk∂−xk∂
= 11−−qqkxk−1+ (qk−1)xk∂ = (1−qk)(x1k−−q1 −xk∂) = 0
since xk−1∂0 and xk∂ are linearly independent we have that 1−qk must be equal to zero,
however this only happens if k isn·`(being q an`0th root of unity).
Suppose now we have an element of the form∂k ∈Z(A
q), then[∂k, x] = 0, lets do a similar
computation
[∂k, x] =∂kx−x∂k = 11−−qqk∂k−1+qkx∂k−x∂k
= (1−qk)(∂1k−−q1 −x∂k)
So the elementsxk` and∂k` for k∈
Nare in the center.
Consider now a generic element in the center, this means, an element of the form P =
Pn
i=0αi(x, ∂), factoring the powers of∂we can write this element of the formP =Pni=0ai(x)∂i,
now we want to prove that every ai for i6=k` must be equal to zero.
Since P is in the center we know that [x, P] = 0, let's compute
[x, P] =xP −P x =xPn
i=0ai(x)∂i−
Pn
i=0ai(x)∂ix
=Pn
i=0xai(x)∂i−ai(x)∂ix
=Pn
i=0xai(x)∂i−ai(x)(1−q
i
1−q∂i
−1+qix∂i)
=Pn
i=0xai(x)∂i−ai(x)1−q
i
1−q∂ i−1−a
i(x)qix∂i
=Pn
i=0ai(x)(x∂i−qix∂i)−ai(x)1−q
i
1−q∂ i−1
since thexi∂j are linearly independent we have thatan= 0ifqi 6= 1, this is, ifi6=k`. Dening
now Q=P−an(x)∂n, we can proceed analogously, so everyai(x) = 0if i6=k`.
Notice that factoring now the powers of x we can writeP =Pn
i=0bi(∂)xi, and by a similar
computation we have that bi = 0if i6=k`. With this the proof is complete.
Denition 3.6. Let A a C-algebra free of rankN2 over its center Z, for someN ∈N, Dene the Azumaya Locus of A as
AL(A) ={m∈max(Z)such thatA/(m)∼=MN(C)} An algebra A is called Azumaya ifAL(A) =max(Z)
4 The structure of the non-Azumaya bers
in [1] it is proved that AL(Aq) = {(x` −a, ∂`−b)} such that ab 6= (1−1q)`. That is, if we
identify max(Aq)∼=C2 by sending (x`−a, ∂`−b)7→ (a, b), then the azumaya locus ofAq is the complement of the hyperbola{(a, b)s.t. ab= (1−1q)`} ⊂C2. In this section we want to
4.1 Finding the structure for the bers
Recall the denition of theq0th Weyl algebra.
Aq=C< x, ∂ > /(∂x−qx∂−1) Now dene R as follows:
R:=Aq/(x`−a, ∂`−b)
withaand b dened as above, in fact by a choice of coordinates
x7−→a1`
∂7−→b−`1
We have the next isomorphism
Aq/(m)∼=Aq/(x`−1, ∂ell− (1−1q)`)
this means, that without losing generality we can set R=Aq/(x`−1, ∂ell−(1−1q)`). Now, in
virtue of proposition 3.4, we can easily see that a basis for R would be{xi∂j}/(∂x−qx∂−1),
this is, {xi∂j : 0≤i, j < `}.
Let us start by proving a very useful relation, which we will use some times in this section. Lemma 4.1. Let ai = (∂− 1q−iq)(∂−1−qq)...(∂−q1i−−q1)(∂−q1i−+1q)...(∂− q1`−−q1). Then we have
the relation (∂− 1q−iq)ai= 0.
Proof. consider the polynomial in ∂ given by
∂`−(1−1q)` = 0
We know it has at most `roots, but we also know that, since q is primitive`0th root of unity, ∂= 1q−iq, fori∈ {0,1, ..., `−1} are roots for the polynomial, so we can factorize it as
(∂−1−1q)(∂−1−qq)(∂−1q−2q)...(∂−q1`−−q1) = 0
Or analogously
(∂− 1q−iq)ai= 0. (this is because obviously∂ commutes with ∂)
Let's write the result from the previous lemma in a more useful way, this is:
∂ai= 1q−iqai
We would like to have a decomposition of R in terms of indecomposable modules or their projective covers, in order to use theorem 2.10, for this dene Pi:=R/R(∂− q
i
1−q).
Lemma 4.2. Pi∼=R(∂−1−1q)(∂− q
1−q)...(∂− qi−1
1−q)(∂− qi+1
1−q)...(∂− q`−1
1−q) as R modules
Proof. consider the morphism φ:Pi−→Ra, mapping 1 to a.
the injectivity is a consequence of lemma 4.1, and the surjectivity is clear, since the pre image of an element of the form α·Ra would be justα.
It only remains to check that φis compatible with the module operations, letx, y∈Pi,
com-puting we have:
φ(x) +φ(y) =xa+ya
= (x+y)a
=φ(x+y)
now, letλ∈R, we have
φ(λx) = (λx)a
=λ(xa)
=λφ(x)
Clearly Pi is an R-module, it actually will turn out to be one of the direct summands of R we are looking for,
Proposition 4.3. the ascending modules chain
0⊂R(x−q−i−`+2)...(x−q−i−1)a⊂R(x−q−i−`+3)...(x−q−i−1)a⊂...⊂Pi
is a ltration for Pi
Proof. Dene b= (x−qj)ai, a as in lemma 4.1 we want to nd j such that:
∂b= q1i−+1qb
this means, j such that ∂ acts linearly. Computing we have: ∂b=∂(x−qj)ai = ([∂, x−qj] + (x−qj)∂)ai
= ([∂, x] + (x−qj)∂)a i
= (1 + (q−1)x∂+x∂−qj∂)ai = (1 +qx∂−qj∂)ai =ai+qx∂ai−qj∂ai =ai+qix ai
1−q −q j+i ai
1−q
= (1 + 1q−ixq− q1j−+qi)ai = 1−1q(1−q−qix−qj+i)ai = q1i−+1q(q−i−1−q−i+x−qj−1)ai
Therefore, what we want isq−i−1−q−i+x−qj−1=x−qj, this means:
q−i−1−q−i−qj−1=−qj
Note that settingj =−i−1 we have that the equality holds, so b= (x−q−i−1)a.
Now, let c = (x −qj)b, for this case we want j such that ∂c = q1i−+2qc. Let's do the
computation:
∂c=∂(x−qj)b = ([∂, x−qj] + (x−qj)∂)b
= (1 + (q−1)x∂+x∂−qj∂)b = (1 +qx∂−qj∂)b
=b+qx∂b−qj∂b =b+qxq1i−+1qb−qj q1i−+1qb = 1−1q(1−q+qi+2x−qj+i+1)b = q1i−+2q(q−i−2−q−i−1+x−qj−1)b
Proceeding as before, we have that the equality that must hold is:
q−i−2−q−i−1−qj−1=−qj
Since setting j=−i−2 the equality holds, so we conclude thatc= (x−q−i−2)b.
Having this, we can now proceed by induction, under the following hypothesis:
if α= (x−q−k−i)(x−q−k−i)...(x−q−i−1)asuch that∂α= qk1+−i+1q α, and we want an element β dened asβ = (x−qn)α such that∂β= qk1+−i+2q β
thenn=−k−i−1
Let's compute this
∂(β) =∂(x−qn)α = ([∂, x−qn] + (x−qn)∂)α
= ([∂, x] + (x−qn)∂)α = (1 +qx∂−qn∂)α =α+qxqk1+−i+1q α−qn qk1+−i+1q α
= (1 +xqk1+−i+2q −qn+1k−+qi+2)α = qk1+−i+2q (q−k−i−2−q−k−i−1+x−qn−1)α
This tells us we need q−k−i−2−q−k−i−1−qn−1=−qn and as expected, settingn=−k−i−1
To simplify notation name each of the modules in the composition series ofPi Fk, k ∈
{0,1, ..., L} such that the composition series is as follows:
0 =F0⊂F1 ⊂F2⊂...⊂FL
And dene the composition factors L0 =FL/FL−1,L1 =F`−1/F`−2,...,L`−1=F1/F0 =F1
Here one should notice Lj is a one dimensional module with a generatorα such that
• (∂−1q−jq)α = 0
• (x−q−j)α= 0
In particular, each Lj is a simple module
Now we would like to nd the projective covers of the composition factors, in order to use theorem 2.10, and analize the automorphisms of R, since every automorphism would be determined by the way it acts on the projective summands of R.
Recall that Pi = R/R(∂− q
i
1−q), so we have a surjection φ :Pi −→ Li, since every Pi is
indecomposable, we have that they are actually the projective covers of theLi, so in virtue of
theorem 2.10 we have that
R∼=P0⊕P1⊕...⊕P`−1
Where as dened above, Pi = Rwi where wi is the factor in lemma 4.2.So in order to
understand the automorphisms of R it is enough to understand the homomorphisms between the Pk's. If φ: Pi −→ Pj is an homomorphism, then it must keep every relation in Pi, i.e,
∀y ∈ Pi,(∂− q
i
1−q)φ(y) = 0, however, we already computed when does this happens, so, we
can conclude that
φji(1) = (x−q−i+1)(x−q−i+2)...(x−q−i+j)wj
in particular, if j = i+ 1 we get φii+1(1) = (x−q−i+1)wi+1, if j = i+ 2, the result is
we will note φii+1 just asφi.
4.2 a graphical way to describe the ber
The previous description may be complete, however, the reader could have troubles trying to understand it, so in this section we will give a graphical description of the ber, we begin with the denition of a quiver, .
Denition 4.4. a Quiver is a directed graph in which multiple edges and loops are allowed, the edges are called arrows (that's why its called quiver ). For any Quiver Q, let Q0 denote
the vertices and Q1 the arrows, and leth, t:Q1 −→Q0 be the maps assigning to each arrow
its head (the point where the arrow is pointing) and tail ( the starting point of the arrow)
respectively.
We will not dive too deep on the theory of quivers, however, this denition comes very handy to represent graphically the structure of the non azumaya bers, if we set the proyective modules Pi as the vertices, the posible morphisms between them as the arrows, and add some
where the relations:
• ei◦ej =δij
• φi◦ej =δijφi
• ej◦φi=δi+1,jφi
• φ0◦φ1◦...◦φ`−1= 0
Must hold
This last relation comes from the factors that appear applying the φ's, after one loop, we
get(x−q)(x−q2)...(x−q`) = 0, the other ones are clear form the picture.
5 Bibliography
[1] Backelin E., endomorphisms of quantized weyl algebras. Lett. Math. Phys. 97 (2011), no.3, 317-338
[2] Cock David, THE WEYL ALGEBRAS, http://web.maths.unsw.edu.au/ danielch/thesis/dcock.pdf [3] J. T. Hartwig, Locally nite simple weight modules over twisted generalized Weyl
alge-bras, Journal of Algebra 303 No. 1 (2006)42−76, http://www.math.chalmers.se/Math/Research/Preprints/Doctoral/2008/2.pdf
[4] Tsuchimoto Y., preliminaries on Diximier conjecture, http://www.math.kochi-u.ac.jp/docky/preprints/dixmier1.pdf [5] Vale R.,Topics in nite dimensional algebras, http://www.math.cornell.edu/ rvale/fdalgebras.pdf
[6] Auslander M.,Representation theory of Artin algebras
[7] A. Belov-Kanel and M. Kontsevich, Automorphisms of the Weyl algebra, Letters in Mathematical Physics, (2005) .
