Hochschild cohomology of algebras of differential operators tangent to a central arrangement of lines
Francisco Kordon Mariano Suárez-Álvarez




Given a central arrangement of lines  in a 2-dimensional vector
space V over a field of characteristic zero, we study the algebra
() of differential operators on V which are logarithmic along .
Among other things we determine the Hochschild cohomology of () as a
Gerstenhaber algebra, establish a connection between that cohomology and the
de Rham cohomology of the complement M() of the arrangement, determine
the isomorphism group of () and classify the algebras of that form up
to isomorphism.


Performance Analysis of User-centric Virtual Cell Dense Networks over mmWave Channels
    Jianfeng Shi, Yinlu Wang, Hao Xu and Ming Chen
National Mobile Communications Research Laboratory

Southeast University, Nanjing, China

Email: {shijianfeng, yinluwang, 
xuhao2013, chenming}@seu.edu.cn
Benoit Champagne
Department of Electrical and Computer Engineering

	 McGill University, Montreal, Canada

Email: benoit.champagne@mcgill.ca

    December 30, 2023
===========================================================================================================================================================================================================================================================================================================================================================



Let us fix a ground field  of characteristic zero, a vector space V
and a central arrangement of hyperplanes  in V. We let S be
the algebra of polynomial functions of V, fix a defining polynomial Q∈
S for , and consider, following K. Saito <cit.>, the Lie algebra
  
    () = {δ∈(S):δ(Q)∈ QS}

of derivations of S logarithmic with respect to , which is,
geometrically speaking, the Lie algebra of vector fields on V which are
tangent to the hyperplanes of . This Lie algebra is a very interesting
invariant of the arrangement and has been the subject of a lot of work — we refer to the book of P. Orlik and H. Terao <cit.> and the one by
A. Dimca <cit.> for surveys on this subject. In particular,
using this Lie algebra we can define an important class of arrangements: we say
that an arrangement  is free if () is free as a left
S-module. For example, central arrangements of lines in the plane are
free, as are, according to a beautiful result of Terao <cit.>, the
arrangements of reflecting hyperplanes of a finite group generated by
pseudo-reflection

Now, along with () we can consider also the associative algebra ()
generated inside the algebra _(S) of linear endomorphisms of the
vector space S by () and the set of maps given by left
multiplication by elements of S: we call it the algebra of
differential operators tangent to the arrangement . When  is free,
this coincides with the algebra of differential operators on S which
preserve the ideal QS of S and all its powers, studied for example by
F. J. Calderón-Moreno <cit.> or by the second author
in <cit.>.

The purpose of this paper is to study, from the point of view of
non-commutative algebra and homological algebra, this algebra () in
the simplest case of free arrangement, that of central line arrangements.



Let us describe briefly our results. We thus assume in what follows that
 is a central arrangement of r+2 lines in a 2-dimensional vector
space V, and for simplicity we suppose that  has at least five lines,
so that r≥3. We let Q∈ S be a defining polynomial for , that is,
a square-free product of linear forms on V with the union of the
hyperplanes of  as zero locus. As S is a subalgebra of (A), we
view Q as an element of the latter.


The algebra () is a noetherian domain, it has global dimension 4
and projective dimension as a bimodule over itself also equal to 4. The
Hochschild cohomology ^∙(()) of () has Hilbert series
  
    ∑_i≥0^i(())· t^i
            = 1 + (r+2)t + (2r+3)t^2 + (r+2)t^3.

The algebra () has Hochschild homology and cyclic homology
isomorphic to those of a polynomial algebra [X], and periodic homology and
higher K-theory isomorphic to that of the ground field . It is a twisted
Calabi–Yau algebra of dimension 4, the element Q of (A) is normal, 
and the modular automorphism σ:()→() of () is the
unique one such that for all a∈() one has
  
    Qa=σ(a)Qa.



These claims are contained in Propositions prop:hh, prop:hom,
prop:CY and prop:conn. In Propositions prop:cup
and prop:bracket we describe completely the cup product and the
Gerstenhaber Lie structure on ^∙(()) — we refer to their
statements for the precise details, which are technical. The
calculations needed in order to do these computations are annoyingly involved.

We obtain a very concrete description of ^1(()) in
Proposition prop:partial:


Let Q=α_1⋯α_r+2 be a factorization of the defining
polynomial as a product of linear factors, so that
α_1, …, α_r+2 are linear polynomials on V whose zero
loci are the hyperplanes of .



  * For each i∈{1,…,r+2} there is a unique derivation
∂_i:()→() such that ∂_i(f)=0 for all f∈ S
and ∂_i(δ)=δ(α_i)/α_i for all
δ∈().


  * The set of classes of ∂_1, …, ∂_r+2
in ^1(()), which we view as the space of outer derivations of the
algebra ()), is a basis.



The elements ∂_1, …, ∂_r+2 are canonically
determined and in a natural bijection with the set of hyperplanes. We do
not have a description along the same lines of the rest of the cohomology. 
In Proposition prop:partial:sub, though,  we do obtain the following piece of
information:


The subalgebra $̋ of^∙(())generated by the
component^1(())of degree1is isomorphic to the de Rham
cohomology of the complementM()of the arrangement. It is freely
generated as a graded-commutative algebra by ther+2elements∂_1, …,∂_r+2of^1(())subject to the
relations
  
    ∂_i⌣∂_j 
            + ∂_j⌣∂_k
            + ∂_k⌣∂_i
            = 0,

one for each choice of three pairwise distinct elementsi,j,kof{1,…,r+2}. 


Using our precise description of^1(())and the techniques of
J. Alev and M. Chamarie <cit.>, we arrive in Section sect:autos
at a description of the
automorphism group of the algebra(). Since the arrangementis
central, the Lie algebra()is a gradedS-module, and that
grading turns()into a graded algebra: we will use this structure
in the following result.


Let G be the subgroup of (V) of maps which preserve the
arrangement . 

  *  There is an action of G on a vector space W of
dimension r+2 such that the semidirect product G⋉ W is isomorphic
to the group _0(()) of algebra automorphisms of () which
respect the grading.


  *  An element of () is locally -nilpotent if and only if it
belongs to S. The set ()={exp(f):f∈ S} of the
automorphisms of () obtained as exponentials of locally
-nilpotent elements is a subgroup of the full group of
automorphisms (()).


  *  There is an action of _0(()) on () such that there
is an isomorphism of groups
(()) = _0(())⋉().



This knowledge of the automorphism group of())allows us to describe
the set of normal elements of the algebra and its birational class:


The set of normal elements of () is the saturated multiplicatively
closed subset generated by Q. The maximal normal localization
of (), which is therefore ()[1Q], is isomorphic to the
localization of the Weyl algebra 𝐷𝑖𝑓𝑓(S)[1Q].


Finally, using —as it is often done— normal elements, we are able to
classify the algebras under study up to isomorphism:


Let  and ' be two central arrangements of lines in V. The
algebras () and (') are isomorphic if and only if the
arrangements  and ' themselves are linearly isomorphic.


This means, essentially, that we can reconstruct the arrangement from the
algebra()of its differential operators.



We expect most of the above results to hold in the general case of a free
arrangement of hyperplanes of arbitrary rank. As our computations here make
clear, some technology is needed in order to deal with more complicated
cases. In future work, we will show how to organize this computation using
the language of Lie–Rinehart pairs <cit.> and their cohomology theory.
On the other hand, one can interpret the second cohomology space^2(())as classifying infinitesimal deformations of
the algebra()and use^3(())and our description of the
Gerstenhaber bracket to study the deformation theory of(). This
produces a somewhat concrete interpretation of the second cohomology space
in geometrical terms. As this involves quite a bit of calculation, we defer
the exposition of these results to a later paper.

The contents of this paper are part of the doctoral thesis of the first
author.



The paper is organized as follows. We start in Section sect:algebra
by giving a concrete realization of the algebra()as an iterated Ore
extension of a polynomial ring and proving some useful lemmas. In
Section sect:res we construct a resolution for()and in
Sections sect:hh and sect:cup we present the
computation of the Hochschild cohomology^∙(())and
its Gerstenhaber algebra structure. Section sect:rest gives the much
easier determination of the Hochschild homology, cyclic homology, periodic
homology andK-theory of our algebra, followed by the proof, in
Section sect:cy, of the twisted Calabi–Yau property. Finally, in
the last section we determine the automorphism group of()and
classify the algebras of this form up to isomorphism.

Some notations We will use the symbols▹and◃to denote the left and right actions of an algebra on a
bimodule whenever this improves clarity. We will have a ground fieldof characteristic zero. All vector spaces and algebras are implicitly
defined over, and unadorned⊗andare taken with
respect to. IfMis a vector space, we will often denote by
  
    M

an element ofMabout which we do not need to be specific.

We refer to the book <cit.> for a general reference about hyperplane
arrangements and their derivations, and to C. Weibel's book <cit.>
for generalities about homological algebra and, in particular, Hochschild,
cyclic and periodic theories.




§ THE ALGEBRA OF DIFFERENTIAL OPERATORS ASSOCIATED TO A CENTRAL
ARRANGEMENT OF LINES


  
 We fix once and for all a ground fieldof characteristic
zero and putS=[x,y]. We viewSas a graded algebra as usual, with
bothxandyof degree1, and for eachp≥0we writeS_pthe
homogeneous component ofSof degreep.

We write(S)the Lie algebra of derivations ofS, which is a free left
gradedS-module, freely generated by the usual partial derivatives∂_x,∂_y:S→S, which are homogeneous elements
of(S)of degree-1. On the other hand, we write(S)the associative algebra of regular differential operators onS,
as defined, for example, in <cit.>*15.5.
As this is by definition a subalgebra of_(S), there is a tautological
structure of left(S)-module onS.

There is an injective morphism of algebrasϕ:S→(S)such thatϕ(s)(a)=asfor alls,a∈Swhich we will view as an
identification; elements in its image are the differential operators of
order zero. SinceSis a regular algebra, the algebra(S)is
generated as a subalgebra of_(S)bySand(S);
see <cit.>*Corollary 15.5.6. A consequence of this is that(S)is
generated as an algebra byx,y,∂_xand∂_y, and in
fact these elements generate it freely subject to the relations
  
    [x,y]=[∂_x,y]=[∂_y,x]=[∂_x,∂_y]=0,
             [∂_x,x]=[∂_y,y]=1.

It follows easily from this that(S)has a-grading withxandyin
degree1and∂_xand∂_yin degree-1, and that with respect
to this grading,Sis a graded(S)-module.



  
 We fix an integerr≥-1and consider a central
arrangementofr+2lines in the planeÅ^2. Up to a change of
coordinates, we may assume that the line with equationx=0is one of the
lines in, so that the defining polynomialQof the arrangement is of the
formxFfor some square-free homogeneous polynomialF∈Sof degreer+1which does not havexas a factor. Up to multiplying by a scalar,
which does not change anything substantial, we may assume that
  F=xF̅+y^r+1
for someF̅∈S_r.

We let()be the Lie
algebra of derivations ofSthat preserve the arrangement, as
in <cit.>*4.1, so that
  
    ()={d∈(S):d(Q)∈ QS}.

This a graded Lie subalgebra of(S).
The two derivations
  
    E=x∂_x+y∂_y,
             D=F∂_y

are elements of()of degrees0andr, and it follows
immediately from Saito's criterion <cit.>*Theorem 4.19 that the set{E,D}is a basis of()as a gradedS-module; this is the
content of Example 4.20 in that book.

The algebra of differential operators tangent to the
arrangement  is the subalgebra()of(S)generated bySand(). It follows immediately from the remarks above that()is generated byx,y,EandD, and a computation shows that the
following commutation relations hold in():
  
    [y,x] = 0, 
         [D,x] = 0,
             [D,y] = F, 
         [E,x] = x,
             [E,y] = y,
             [E,D] = rD.

Since these generators are homogeneous elements in(S)—withEof
degree0,xandyof degree1andDof degreer— we see that
the algebra()is a graded subalgebra of(S)and, by restricting
the structure from(S), thatSis a graded()-module.

The set of commutation relations given above is in fact a
presentation of the algebra(). More precisely, we have:

lemma:ore
The algebra () is isomorphic to the iterated Ore extension
S[D][E]. It is a noetherian domain and the set
{x^iy^jD^kE^l:i,j,k,l≥0} is a -basis for ().


Here we viewDas a derivation ofS, so that we way
construct the Ore extensionS[D], and viewEas a derivation of this
last algebra, so as to be able extend once more to obtainS[D][E].


It is clear at this point that the obvious map
π:S[D][E]→() is a surjective morphism of algebras, so we
need only prove that it is injective. To do that, let us suppose that there
exists a non-zero element L in S[D][E] whose image under the
map π is zero, and suppose that L=∑_i,j≥0f_i,jD^iE^j, with
coefficients f_i,j∈ S for all i, j≥0, almost all of which are
zero. As L is non-zero, we may consider the number
m=max{i+j:f_i,j≠0}.

Let us now fix a point p=(a,b)∈Å^2 which is not on any line of the
arrangement , so that aF(a,b)≠0, and let Ø_p be the completion
of S at the ideal (x-a,y-b) or, more concretely, the algebra of formal
series in x-a and y-b. We view Ø_p as a left module over (S) in
the tautological way and, by restriction, as a left ()-module.
There exist formal series ϕ and ψ in Ø_p such that
  
    E·ϕ = 1,
             D·ϕ = 0,
             E·ψ = 0,
             D·ψ = x^r.
    Indeed, we may choose ϕ=ln x to satisfy the first two
    conditions, and the last two ones are equivalent to the equations   ∂_xψ = -x^r-1y/F,
            ∂_yψ = x^r/F,

which can be solved for ψ, as the usual well-known sufficient
integrability condition from elementary calculus holds. If now
s, t∈_0 are such that s+t=m, a straightforward computation shows
that
  L·ϕ^sψ^t=s!t!x^rtf_s,t
in Ø_p, and this implies that f_s,t=0. This contradicts the choice
of m and this contradiction proves what we want.


  
 We will use the following two simple lemmas a few
times:

lemma:indep
Suppose that r≥2.
If α, β∈ S_1 are such that α F_x+β F_y=0,
then α=β=0.


The conclusion of this statement is false ifr<2.


Suppose that F_1, F_2 and F_3 are three distinct linear
factors of F (here is where we need the hypothesis that r is at least
2) so that F=F_1F_2F_3F' for some F'∈ S_r-2; as
F has degree at least 3, this is possible. We have
F_x≡ F_1xF_2F_3F' and F_y≡ F_1yF_2F_3F' modulo F_1, so
that (α F_1x+β F_1y)F_2F_3F'≡0 F_1. Since F is
square free, this tells us that F_1 divides α F_1x+β F_1y
and, since both polynomials have the same degree and F_1≠0, that there exists a
scalar λ such that α F_1x+β F_1y=λ F_1. Of
course, we can do the same with the other two factors F_2 and F_3. We
can state this by saying that the matrix
[ α_x β_x; α_y β_y ]
has the three vectors
[ F_1x; F_1y ],
[ F_2x; F_2y ] and
[ F_3x; F_3y ] as eigenvectors.
Since no two of these are linearly dependent, because F is square-free,
this implies that the matrix
is in fact a scalar multiple of the identity, and there is a μ∈
such that α=μ x and β=μ y. The hypothesis is then that
μ (r+1)F=μ(xF_x+yF_y)=0, so that μ=0. This proves the claim.

If α_1, …, α_r+1∈ S_1 are such that
F=∏_i=1^r+1α_i, then the set of quotients
  {Fα_1,…,Fα_r+1}
is a basis for S_r.


Suppose c_1, …, c_r+1∈ are scalars such that
∑_i=1^r+1c_iF/α_i=0. If j∈{1,…,r+1}, we
then have c_jF/α_j≡0 modulo α_j and, since
F is square-free, this implies that in fact c_j=0. The set
{Fα_1,…,Fα_r+1} is therefore linearly
independent. Since S_r=r+1, this completes the proof.



§ A PROJECTIVE RESOLUTION


  
 We keep the situation of the previous section, and write from
now onAinstead of(). Our immediate objective is to construct a
projective resolution ofAas anA-bimodule, and we do this by looking
atAas a deformation of a commutative polynomial algebra, which suggests
that it should have a resolution resembling the usual Koszul complex.



  
 IfUis a vector space andu∈U, there are
derivations∇_x^u,∇_y^u:S→S⊗U⊗SofSinto
theS-bimoduleS⊗U⊗Suniquely determined by the condition
that
  
    ∇_x^u(x) = 1⊗ u⊗ 1,
            ∇_x^u(y) = 0,
            ∇_y^u(x) = 0,
            ∇_y^u(y) = 1⊗ u⊗ 1,

and in fact we have, for everyi,j≥0, that
  
    ∇_x^u(x^iy^j) = ∑_s+t+1=ix^s⊗ u⊗ x^ty^j,
            ∇_y^u(x^iy^j) = ∑_s+t+1=jx^iy^s⊗ u⊗ y^s.


We consider the derivation∇=∇_x^x+∇_y^y:S→S⊗S_1⊗S; it is the unique derivation such that∇(α)=1⊗α⊗1for allα∈S_1.
There is, on the other hand, a unique morphism ofS-bimodulesd:S⊗S_1⊗S→S⊗Ssuch thatd(1⊗α⊗1)=α⊗1-1⊗αfor allα∈S_1, and we have
  
    d(∇(f)) = f⊗1-1⊗ f

for allf∈S. To check this last equality, it is enough to notice thatd∘∇:S→S⊗Sis a derivation and, sinceS_1generatesSas an algebra, that the equality holds whenf∈S_1.



  
 LetVbe the subspace ofAspanned byx,y,DandE.
This is a graded subspace and its grading induces on the
exterior algebraΛ^∙(V)an internal grading.
Ifωis an element of an exterior powerΛ^p(V)ofV, we write()∧ωthe map ofA-bimodules
  
    A⊗ S_1⊗ A→ A⊗Λ^p+1V⊗ A

such that(1⊗α⊗1)∧ω=1⊗α∧ω⊗1for allα∈S_1.



  
 There is a chain complexof free gradedA-bimodules of the form
  
    [column sep=1.75em]
      A|Λ^4V|A [r, "d_4"]
                A|Λ^3V|A [r, "d_3"]
                A|Λ^2V|A [r, "d_2"]
                A|V|A [r, "d_1"]
                A|A

withA^e-linear maps homogeneous of degree zero and such that

    d_1(1|v|1) = [v,1|1],     ∀ v∈ V; 
    
      d_2(1|x∧ y|1) = [x,1|y|1]-[y,1|x|1]; 
    
      d_2(1|x∧ E|1) = [x,1|E|1]-[E,1|x|1] + 1|x|1; 
    
      d_2(1|y∧ E|1) = [y,1|E|1]-[E,1|y|1] + 1|y|1; 
    
      d_2(1|x∧ D|1) = [x,1|D|1]-[D,1|x|1]; 
    
      d_2(1|y∧ D|1) = [y,1|D|1]-[D,1|y|1] + ∇(F); 
    
      d_2(1|D∧ E|1) = [D,1|E|1]-[E,1|D|1] + r|D|1; 
    
      d_3(1|x∧ y∧ D|1)
            =  [x,1|y∧ D|1] - [y,1|x∧ D|1] + [D,1|x∧ y|1] + ∇(F)∧ x;
    
      d_3(1|x∧ y∧ E|1)
            =  [x,1|y∧ E|1] - [y,1|x∧ E|1] + [E,1|x∧ y|1] - 2|x∧ y|1;  
    [0.85]
      d_3(1|x∧ D∧ E|1)
            =  [x,1|D∧ E|1] - [D,1|x∧ E|1] + [E,1|x∧ D|1] 
    
            - (r+1)|x∧ D|1;
      
    [0.9]
      d_3(1|y∧ D∧ E|1)
            =  [y,1|D∧ E|1] - [D,1|y∧ E|1] + [E,1|y∧ D|1] 
    
                    + ∇(F)∧ E - (r+1)|y∧ D|1;
      
    
      d_4(1|x∧ y∧ D∧ E|1) =
      [0.625][t]
      [x,1|y∧ D∧ E|1] - [y,1|x∧ D∧ E|1]  
    
            + [D,1|x∧ y∧ E|1] - [E,1|x∧ y∧ D|1] 
    
            + ∇(F)∧ x∧ E + (r+2)|x∧ y∧ D|1.

Thatis indeed a complex follows from a direct calculation. More
interestingly, it is exact:


The complex  is a projective resolution of A as an A-bimodule, with
augmentation d_0:A|A→ A such that d_0(1|1)=1.

For each p∈_0 we consider the subspace
  _pA = x^iy^jD^kE^l:k+l≤ p
of A. As a consequence of Lemma <ref>,
one sees that A=(_pA)_p≥0 is an exhaustive and increasing algebra
filtration on A and that the corresponding associated graded algebra (A) is
isomorphic to the usual commutative polynomial ring [x,y,D,E]. Since
V is a subspace of A, we can restrict the filtration of A to one
on V, and the latter induces as usual a filtration on each exterior
power Λ^pV. In this way we obtain a filtration on each component
of the complex , which turns out to be  compatible with its
differentials, as can be checked by inspection. The complex ()
obtained from  by passing to associated graded objects in each degree
is isomorphic to the Koszul resolution of (A) as a (A)-bimodule
and it is therefore acyclic over (A). A standard argument using the
filtration of  concludes from this that the complex  itself acyclic over A.
As its components are manifestly free A-bimodules, this proves the lemma.


  
 One almost immediate application of having a bimodule projective
resolution for our algebra is in computing its global dimension.


The global dimension of A is equal to 4.


Of course, asAis noetherian, there is no need to distinguish between the left and
the right global dimensions.


If λ∈ let M_λ be the left A-module which as a vector
space is freely spanned by an element u_λ and on which the action
of A is such that x· u_λ=y· u_λ=D· u_λ=0
and E· u_λ=λ u_λ. It is easy to see that all
1-dimensional A-modules are of this form and that M_λ≅
M_μ iff λ=μ, but we will not need this.

The complex ⊗_AM_λ is a projective resolution
of M_λ as a left A-module, and therefore the cohomology of
_A(⊗_AM_λ,M_μ) is canonically isomorphic to
_A^∙(M_λ,M_μ). Identifying as usual
_A(⊗_AM_λ,M_μ) to M_μ⊗Λ^∙
V^*⊗ M_λ^*, we compute that the complex is
  
    [ampersand replacement=&, cramped]
      M_μ⊗ M_λ^*
            [r, "δ^0"]
        & M_μ⊗ V^*⊗ M_λ^*
            [r, "δ^1"]
        & M_μ⊗Λ^2V^*⊗ M_λ^*
            [r, "δ^2"]
        &
    [ampersand replacement=&, cramped]
     [r]
        & M_μ⊗Λ^3V^*⊗ M_λ^*
            [r, "δ^3"]
        & M_μ⊗Λ^4V^*⊗ M_λ^*

with differentials given by
  
    δ^0(1)
            = (μ-λ)⊗Ê, 
    δ^1(a⊗x̂+b⊗ŷ+c⊗D̂+d⊗D̂) 
       = (λ+1-μ)a⊗x̂∧Ê
              +(λ+1-μ)b⊗ŷ∧Ê
              +(λ+r-μ)c⊗D̂∧Ê, 
      
    δ^2(a⊗x̂∧ŷ+b⊗x̂∧Ê
            +c⊗ŷ∧Ê+d⊗x̂∧D̂
            +eŷ∧D̂+fD̂∧Ê) 
       = [.8][t]
              (μ-λ-2)a⊗x̂∧ŷ∧Ê
              + (μ-λ-r-1)d⊗x̂∧D̂∧Ê
    
              + (μ-λ-r-1)e⊗ŷ∧D̂∧Ê,
              
    δ^3(a⊗x̂∧ŷ∧D̂
            +b⊗x̂∧ŷ∧Ê
            +c⊗x̂∧D̂∧Ê
            +d⊗ŷ∧D̂∧Ê) 
       = (λ+r+2-μ)a⊗x̂∧ŷ∧D̂∧Ê.

An easy computation shows that
  
    ^p_A(M_λ,M_λ+r+2) =
        
        1,     if p=3 or p=4; 
    
        0,     in any other case.

In particular, _A^4(M_λ,M_λ+r+2)≠0 and therefore
A≥4. On the other hand, we have constructed a projective
resolution of A as an A-bimodule of length 4, so that the projective
dimension of A as a bimodule is
_A^eA≤ 4. Since A≤_A^eA, the proposition
follows from this.



§ THE HOCHSCHILD COHOMOLOGY OF D(A)


  
p:hhdifferentials We want to compute the Hochschild
cohomology of the algebraA. Applying the functor_A^e(,A)to the resolutionof p:P we get, after standard
identifications, the cochain complex
  
    A [r, "d^0"]
                A⊗ V^* [r, "d^1"]
                    [l, shift left=1ex, dashed, "s^1"]
                A⊗Λ^2V^* [r, "d^1"]
                    [l, shift left=1ex, dashed, "s^2"]
                A⊗Λ^3V^* [r, "d^2"]
                    [l, shift left=1ex, dashed, "s^3"]
                A⊗Λ^4V^* [r]
                    [l, shift left=1ex, dashed, "s^4"]
                0

which we denote simply byA⊗ΛV^*, with differentials such that
  
    d^0(a)
                  = [x,a]⊗x̂
                    +[y,a]⊗ŷ
                    +[D,a]⊗D̂
                    +[E,a]⊗Ê;
                    
    [0.9]
      d^1(a⊗x̂)
                  = -[y,a]⊗x̂∧ŷ
                    +(a-[E,a])⊗x̂∧Ê
                    -[D,a]⊗x̂∧D̂
    
                    +∇_x^a(F)⊗ŷ∧D̂;
      
    
      d^1(a⊗ŷ)
                 = [x,a]⊗x̂∧ŷ
                   +(a-[E,a])⊗ŷ∧Ê
                   +(∇_y^a(F)-[D,a])⊗ŷ∧D̂;
                   
    
      d^1(a⊗D̂)
                = [x,a]⊗x̂∧D̂
                  +[y,a]⊗ŷ∧D̂
                  +(ra-[E,a])⊗D̂∧Ê;
                  
    
      d^1(a⊗Ê)
                = [x,a]⊗x̂∧Ê
                  + [y,a]⊗ŷ∧Ê
                  + [D,a]⊗D̂∧Ê;
                  
    
      d^2(a⊗x̂∧ŷ)
                   = ([D,a] - ∇_y^a(F))⊗x̂∧ŷ∧D̂
                     + ([E,a]-2a)⊗x̂∧ŷ∧Ê;
                    
    
      d^2(a⊗x̂∧Ê)
                  = -[y,a]⊗x̂∧ŷ∧Ê
                    -[D,a]⊗x̂∧D̂∧Ê
                    +∇_x^a(F)⊗ŷ∧D̂∧Ê;
                    
    
      d^2(a⊗ŷ∧Ê)
                 = [x,a]⊗x̂∧ŷ∧Ê
                   +(∇_y^a(F)-[D,a])⊗ŷ∧D̂∧Ê;
                    
    
      d^2(a⊗x̂∧D̂)
                 = -[y,a]⊗x̂∧ŷ∧D̂
                   +([E,a]-(r+1)a)⊗x̂∧D̂∧Ê;
                    
    
      d^2(a⊗ŷ∧D̂)
                 = [x,a]⊗x̂∧ŷ∧D̂
                   +([E,a]-(r+1)a)⊗ŷ∧D̂∧Ê;
                    
    
      d^2(a⊗D̂∧Ê)
                 = [x,a]⊗x̂∧D̂∧Ê
                   + [y,a]⊗ŷ∧D̂∧Ê;
                    
    
      d^3(a⊗x̂∧ŷ∧D̂)
                 = (-[E,a]+(r+2)a)⊗x̂∧ŷ∧D̂∧Ê;
                    
    
      d^3(a⊗x̂∧ŷ∧Ê)
                 = ([D,a]-∇_y^a(F))⊗x̂∧ŷ∧D̂∧Ê;
                    
    
      d^3(a⊗x̂∧D̂∧Ê)
                 = -[y,a]⊗x̂∧ŷ∧D̂∧Ê;
                    
    
      d^3(a⊗ŷ∧D̂∧Ê)
                 = [x,a]⊗x̂∧ŷ∧D̂∧Ê.

These differentials are homogeneous with respect to the natural
internal grading on the complexA⊗ΛV^*coming from the
grading ofA. We denoteγ: A⊗ΛV^*→A⊗ΛV^*the-linear map whose
restriction to each homogeneous component
of the complexA⊗ΛV^*is simply the multiplication by the
degree. There is a homotopy, drawn in
the diagram (<ref>) with dashed arrows, with
  
    s^1(a⊗x̂+b⊗ŷ+c⊗D̂+d⊗Ê)
            = d, 
    [0.9]
      s^2(a⊗x̂∧ŷ
          + b⊗x̂∧Ê
          + c⊗ŷ∧Ê
          + d⊗x̂∧D̂
          + e⊗ŷ∧D̂
          + f⊗D̂∧Ê) 
    
            = -b⊗x̂-c⊗ŷ-f⊗D̂,
      
    [0.9]
      s^3(a⊗x̂∧ŷ∧D̂
          + b⊗x̂∧ŷ∧Ê
          + c⊗x̂∧D̂∧Ê
          + d⊗ŷ∧D̂∧Ê)
            
    
            = b⊗x̂∧ŷ + c⊗x̂∧D̂
              + d⊗ŷ∧D̂,
      
    
      s^4(a⊗x̂∧ŷ∧D̂∧Ê)
            = -a⊗x̂∧ŷ∧D̂

and such thatd∘s+s∘d = γ: this tells us thatγinduces the zero map on cohomology. Since our
ground fieldhas characteristic zero, this implies that the inclusion(A⊗ΛV^*)_0→A⊗ΛV^*of the component of degree
zero of our complexA⊗ΛV^*is a quasi-isomorphism.



  
From now on and until the end of this
section, we will assume that r≥3. Let us write the complex(A⊗ΛV^*)_0simplyand let us putT=[E], which
coincides withA_0. The complexhas components
  
    ^0 = A_0, 
    ^1 = A_1⊗(x̂⊕ŷ)
            ⊕ A_r⊗D̂⊕ A_0⊗Ê, 
    [0.9]
      ^2 = A_2⊗x̂∧ŷ⊕ A_1⊗(x̂∧Ê⊕ŷ∧Ê)
            ⊕ A_r⊗D̂∧Ê
    ⊕ A_r+1⊗(x̂∧D̂⊕ŷ∧D̂),
      
    [0.9]
      ^3 = A_2⊗x̂∧ŷ∧Ê⊕ A_r+1⊗(x̂∧D̂∧Ê⊕ŷ∧D̂∧Ê) 
    ⊕ A_r+2⊗x̂∧ŷ∧D̂, 
      
    ^4 = A_r+2⊗x̂∧ŷ∧D̂∧Ê

and, sincer>2, we have
  
    A_0 = T,
             A_1 = S_1T,
             A_2 = S_2T, 
         A_r = (S_r⊕ D)T,
             A_r+1 = (S_r+1⊕ S_1D)T,
             A_r+2 = (S_r+2⊕ S_2D)T.

In fact, this is where our assumption thatr≥3intervenes: ifr≤2, then these subspaces ofAhave different descriptions.

The differentials incan be computed to be given by 
  
    δ^0(a) = xτ_1(a)⊗x̂+yτ_1(a)⊗ŷ 
                            + Dτ_r(a)⊗D̂, 
    δ^1(ϕ a⊗x̂)
            = -ϕ yτ_1(a)⊗x̂∧ŷ
              -(Fϕ_ya+ϕ Dτ_r(a))⊗x̂∧D̂
              +∇_x^ϕ a(F)⊗ŷ∧D̂, 
    δ^1(ϕ a⊗ŷ)
            = ϕ xτ_1(a)⊗x̂∧ŷ
              +(∇_y^ϕ a(F)-Fϕ_ya-ϕ Dτ_r(a))⊗ŷ∧D̂, 
    [.9]
      δ^1((ϕ +λ D)a⊗D̂)
            = (ϕ xτ_1(a)+λ xDτ_1(a))⊗x̂∧D̂
    
              + (ϕ yτ_1(a)+λ F(τ_1(a)-a)+λ yDτ_1(a))
              ⊗ŷ∧D̂,
      
    δ^1(a⊗Ê)
            = xτ_1(a)⊗x̂∧Ê
              +  yτ_1(a)⊗ŷ∧Ê
              +  Dτ_r(a)⊗D̂∧Ê, 
    δ^2(ϕ a⊗x̂∧ŷ)
            = (Fϕ_ya+ϕ Dτ_r(a)-∇_y^ϕ a(F))
                    ⊗x̂∧ŷ∧D̂, 
    [0.9]
      δ^2(ϕ a⊗x̂∧Ê)
            = -ϕ yτ_1(a)⊗x̂∧ŷ∧Ê
              -(Fϕ_ya+ϕ Dτ_r(a))⊗x̂∧D̂∧Ê
    
              +∇_x^ϕ a(F)⊗ŷ∧D̂∧Ê,
      
    [0.9]
      δ^2(ϕ a⊗ŷ∧Ê)
            = ϕ xτ_1(a)⊗x̂∧ŷ∧Ê
    
              +(∇_y^ϕ a(F)-Fϕ_ya-ϕ Dτ_r(a))
                    ⊗ŷ∧D̂∧Ê, 
      
    δ^2((ϕ+ψ D)a⊗x̂∧D̂)
            = (-ϕ yτ_1(a)-ψ F(τ_1(a)-a)-ψ yDτ_1(a))
              ⊗x̂∧ŷ∧D̂, 
    δ^2((ϕ+ψ D)a⊗ŷ∧D̂)
            = (ϕ xτ_1(a)+ψ xDτ_1(a))
                    ⊗x̂∧ŷ∧D̂, 
    [.9]
      δ^2((ϕ+λ D)a⊗D̂∧Ê)
            = (ϕ xτ_1(a)+λ xDτ_1(a))
                    ⊗x̂∧D̂∧Ê
    
              +(ϕ yτ_1(a)+λ yDτ_1(a)+λ F(τ_1(a)-a))
              ⊗ŷ∧D̂∧Ê,
      
    δ^3((ϕ+ψ D)a⊗x̂∧ŷ∧D̂)
            = 0, 
    δ^3(ϕ a⊗x̂∧ŷ∧Ê)
            = (Fϕ_ya+ϕ Dτ_r(a)-∇_y^ϕ a(F))
              ⊗x̂∧ŷ∧D̂∧Ê, 
    [.9]
      δ^3((ϕ+ψ D)a⊗x̂∧D̂∧Ê)
            
    
            = -(ϕ yτ_1(a)+ψ y Dτ_1(a) +ψ F(τ_1(a)-a))
                    ⊗x̂∧ŷ∧D̂∧Ê,
      
    δ^3((ϕ+ψ D)a⊗ŷ∧D̂∧Ê)
            = (ϕ xτ_1(a)+ψ xDτ_1(a))
                    ⊗x̂∧ŷ∧D̂∧Ê.

Here and belowτ_t:T→Tis the-linear map such thatτ_t(E^n)=E^n-(E+t)^nfor alln∈_0, andϕandϕdenote
homogeneous elements ofrof appropriate degrees andλa scalar.



  
 We proceed to compute the cohomology of the complex,
starting with degrees zero and four, for which the computation is almost
immediate. Indeed, since the kernel ofτ_1and ofτ_ris⊆T, it is clear thatH^0()=δ^0=. On the other
hand, ifψ∈S_2anda∈T, we can writeψ=ψ_1x+ψ_2yfor someψ_1,ψ_2∈S_1and there is ab∈Tsuch thatτ_1(b)=a, so that
  
    δ^3(-ψ_2Db⊗x̂∧D̂∧Ê
         + ψ_1Db⊗ŷ∧D̂∧Ê)
           = (ψ Da + S_r+2T)
                    ⊗x̂∧ŷ∧D̂∧Ê.

Similarly, we have
  δ^3(S_r+1T⊗x̂∧D̂∧Ê
           +S_r+1T⊗ŷ∧D̂∧Ê)
        = S_r+2T⊗x̂∧ŷ∧D̂∧Ê.
These two facts imply that the mapδ^3is surjective, so thatH^4()=0.



  
 Letω∈^1be a1-cocycle in. There are thena,b,c,d,e,f∈T,k∈_0andϕ_0, …,ϕ_k∈S_rsuch that eitherk=0orϕ_k≠0,
and
  
    ω
        = (xa+yb)⊗x̂
         +(xc+yd)⊗ŷ
         +(∑_i=0^kϕ_i E^i+De)⊗D̂
         +f⊗Ê.

Ife̅∈Tis such thatτ_r(e̅)=e, then by replacingωbyω-δ^0(e̅), which does not change the cohomology
class ofω, we can assume thate=0. The
formula forδ^0then shows thatωis a coboundary iff it is
equal to zero. The coefficient ofx̂∧ŷinδ^1(ω)is
  
    x^2τ_1(c) + xy(τ_1(d)-τ_1(a)) - y^2τ_1(b) = 0.

We therefore haveb,c,d-a∈. The coefficient ofD̂∧Ê,
on the other hand, isDτ_r(f)=0, so that alsof∈; exactly the
same information comes from the vanishing of the coefficients ofx̂∧Êand ofŷ∧Ê. Sinceb∈, 
the coefficient ofx̂∧D̂is
  
    -Fb-xDτ_r(a)+∑_i=0^k ϕ_ixτ_1(E^i) = 0.

We see thatτ_r(a)=0,
so thata∈, and that∑_i=0^k ϕ_ixτ_1(E^i)=Fb. This
implies thatk≤1, that-ϕ_1x=Fband therefore, sincexis not a factor ofFby hypothesis, thatϕ_1=0andb=0.

Finally, using all the information we have so far, we can see that
the vanishing of the
coefficient ofŷ∧D̂inδ^1(ω)implies thatF_xxa + F_y(xc+yd) = Fd. Together with Euler's relationF_xx+F_yy=(r+1)Fthis tells us that
  
    (cx+(d-a)y)F_y = (d-(r+1)a)F.

AsFis square-free, it follows[Suppose that u=cx+(d-a)y is not zero.
Differentiating in (<ref>) with respect to y, we find that
-raF_y=uF_yy. Since x does not divide F, we have F_yy≠0, and
then a≠0 and u divides F_y: from (<ref>) it follows then
that u^2 divides F, since the left hand side of that equality is
non-zero, and this is absurd because F is square-free.]
from this equality the polynomialcx+(d-a)yis zero
so thatc=0andd=aand, finally, thata=0.
We conclude in this way that the set of1-cocycles
  
    ϕ⊗D̂+f⊗Ê,     ϕ∈ S_r,f∈

is a complete, irredundant set of representatives for the elements ofH^1().



  
 Letω∈^3be a3-cocycle, so that
  
    ω
            = a⊗x̂∧ŷ∧D̂
            + b⊗x̂∧ŷ∧Ê
            + c⊗x̂∧D̂∧Ê
            + d⊗ŷ∧D̂∧Ê

for somea∈(S_r+2⊕S_2D)T,b∈S_2T,c,d∈(S_r+1⊕S_1D)Tandδ^3(ω)=0.
For allϕ∈S_1ande∈Twe have
  
    [.9]
      δ^2(ϕ e⊗x̂∧Ê)
            = - ϕ yτ_1(e)⊗x̂∧ŷ∧Ê
                    + A_r+1⊗x̂∧D̂∧Ê
    
                    + A_r+1⊗ŷ∧D̂∧Ê
    andδ^2(ϕ e⊗ŷ∧Ê)
            = ϕ xτ_1(e)⊗x̂∧ŷ∧Ê
                    + A_r+1⊗ŷ∧D̂∧Ê,

so that by adding toωan element ofδ^2(S_1T⊗x̂∧Ê+S_1T⊗ŷ∧Ê), 
which does not change the cohomology class ofω, we can
suppose thatb=0. Similarly, for allϕ∈S_2and alle∈Twe
have
  
    δ^2(ϕ e⊗x̂∧ŷ) = 
            (S_r+2T + ϕ Dτ_r(e))⊗x̂∧ŷ∧D̂, 
    and for all ϕ∈ S_r+1 and all e∈ T we haveδ^2(ϕ e⊗x̂∧D̂) 
            = -ϕ y τ_1(e)⊗x̂∧ŷ∧D̂andδ^2(ϕ e⊗ŷ∧D̂) 
            = ϕ xτ_1(e)⊗x̂∧ŷ∧D̂.

Using this we see that, up to changingωby adding to it a3-coboundary, we can suppose thata=0. Finally, for eachϕ∈S_rand alle∈Twe have
  
    δ^2(ϕ e⊗D̂∧Ê)
            = ϕ xτ_1(e)⊗x̂∧D̂∧Ê
              + A_r+1⊗ŷ∧D̂∧Ê,
              
    δ^2(De⊗D̂∧Ê)
            = xDτ_1(e)⊗x̂∧D̂∧Ê
              + A_r+1⊗ŷ∧D̂∧Êandδ^2(-y⊗x̂∧Ê+F̅E⊗D̂∧Ê)
            = y^r+1⊗x̂∧D̂∧Ê
              +A_r+1⊗ŷ∧D̂∧Ê,

so we can also suppose thatc∈y^r+1ET+yDT.

There arel≥0,λ_1, …,λ_l,μ_0, …,μ_l∈,ϕ_0, …,ϕ_l∈S_r+1,ψ_0, …,ψ_l∈S_1,ζ_0, …,ζ_l∈S_1such thatc=∑_i=1^lλ_i y^r+1E^i+∑_i=0^lμ_i yDE^iandd=∑_i=0^l(ϕ_i+ψ_iD)E^i. The vanishing ofδ^3(ω)means precisely that
  
    ∑_i=1^lλ_iy^r+2τ_1(E^i)
      +
      ∑_i=0^l
            (
            μ_iy^2Dτ_1(E^i)
            -μ_iyF(E+1)^i
            -ϕ_ixτ_1(E^i)
            -ψ_ixDτ_1(E^i)
            )
            = 0.

The left hand side of this equation is an element ofS_r+2T⊕S_2DT. The component inS_2DTis
  ∑_i=0^l(μ_iy^2-ψ_ix)Dτ_1(E^i) = 0
and thereforeμ_i=ψ_i=0for alli∈{1,…,l}. On the other
hand, the component inS_r+2Tis
  
    ∑_i=1^lλ_iy^r+2τ_1(E^i)
      - μ_0yF - ∑_i=0^lϕ_ixτ_1(E^i) = 0.

This implies thatλ_iy^r+2-ϕ_ix=0ifi∈{2,…,l},
so thatλ_i=ϕ_i=0for suchi, and then the equation reduces toλ_1y^r+2+μ_0yF-ϕ_1x=0.
Recalling from p:start thatF=y^r+1+xF̅,
we deduce from this thatλ_1=-μ_0andϕ_1=μ_0yF̅.
We conclude in this way that every3-cocycle is cohomologous to one of
the form
  
    (μ_0yD-μ_0y^r+1E)x̂∧D̂∧Ê
      + (ϕ_0+ψ_0D+μ_0yF̅E)ŷ∧D̂∧Ê

withμ_0∈,ϕ_0∈S_r+1andψ_0∈S_1, and a direct
computation shows that moreover every3-cochain of this form is a3-cocycle.

Let nowηbe a2-cochainηin, so that
  
    η = A_2⊗x̂∧ŷ⊕ A_r+1⊗(x̂∧D̂⊕ŷ∧D̂) 
            + u⊗x̂∧Ê 
            + v⊗ŷ∧Ê
    
            + w⊗D̂∧Ê

withu,v∈A_1andw∈A_r, and let us
suppose thatδ^2(η)is equal to the3-cocycle (<ref>).
There arel≥0,α_0, …,α_l,β_0, …,β_l∈S_1,γ_0, …,γ_l∈S_randξ_0, …,ξ_l∈such thatu=∑_i=0^lα_iE^i,v=∑_i=0^lβ_iE^iandw=∑_i=0^l(γ_i+ξ_iD)E^i. The coefficient ofx̂∧ŷ∧Êinδ^2(η)must be equal to zero, so that
  
    ∑_i=0^l(-α_iy+β_ix)τ_1(E^i) = 0,

and this implies that there are scalarsρ_1, …,ρ_l∈such thatα_i=ρ_i xandβ_i=ρ_i yfor alli∈{1,…,l}.

Looking now at the coefficient ofx̂∧D̂∧Êinδ^2(η)and comparing with (<ref>) we find that
  
    ∑_i=0^l(-Fα_iyE^i-α_iDτ_r(E^i)
            + γ_ixτ_1(E^i) + ξ_i xDτ_1(E^i))
            = μ_0yD-μ_0y^r+1 E.

This is an equality of two elements ofST⊕SDT. Considering the
components inDT, we find that
  xD∑_i=1^l(-ρ_iτ_r(E^i)+ξ_iτ_1(E^i)) = μ_0yD, 
and this tells us thatμ_0=0and that
  
    ∑_i=1^l(-ρ_iτ_r(E^i)+ξ_iτ_1(E^i)) = 0.

On the other hand, as the components inSTof the two sides
of (<ref>) are equal, we have
  
    -Fα_0y +
      ∑_i=0^lγ_ixτ_1(E^i)
            = 0,

so thatγ_i=0for alli∈{2,…,l}andFα_0y+γ_1x=0. Asxdoes not divideF, we must haveα_0y=0andγ_1=0; in particular, there is aρ_0∈such thatα_0=ρ_0 x.

Finally, considering the coefficient ofŷ∧D̂∧Êofδ^2(η)and of (<ref>) we see that
  
    ∑_i=0^l(∇_x^α_iE^i(F)
            + ∇_y^β_iE^i(F)-Fβ_iyE^i - β_iDτ_r(E^i) 
    
            + γ_iyτ_1(E^i) + ξ_iyDτ_1(E^i) - ξ_iF(E+1)^i
            )
            = ϕ_0 + ψ_0D,

which at this point we can rewrite (using in the process the
equality (<ref>) above and the fact that∇_x^xE^i(F)+∇_x^yE^i(F)=F∑_t=0^r(E+t)^i) as
  
    ρ_0xF_x+β_0F_y
            -F(β_0y+ξ_0-
            ∑_i=1^l(ρ_i∑_t=1^r(E+t)^i-ξ_i(E+1)^i)
            )
            = ϕ_0 + ψ_0D.

It follows at once thatψ_0=0and that, in fact,
  
    ρ_0xF_x+β_0F_y
            -F(β_0y+ξ_0-
            ∑_i=1^l(ρ_i∑_t=1^rt^i-ξ_i)
            )
            = ϕ_0.

The polynomialϕ_0is then in the linear span ofxF_x,xF_y,yF_yandFinsideS_r+1. Euler's relation implies that
already the first three polynomials span this subspace, and we have
  
    δ(x⊗x̂∧Ê) = xF_x⊗ŷ∧D̂∧Ê, 
    δ(x⊗ŷ∧Ê) = xF_y⊗ŷ∧D̂∧Ê, 
    δ(y⊗ŷ∧Ê-D⊗D̂Ê) 
                    = yF_y⊗ŷ∧D̂∧Ê.

We conclude in this way that the only3-coboundaries among the cocycles
of the form (<ref>) are the linear combinations of the right hand
sides of the equalities (<ref>); these three cocycles are, moreover,
linearly independent. This means that there is an isomorphism
  
    H^3() ≅ω_3
            ⊕ S_1D⊗ŷ∧D̂∧Ê⊕S_r+1/xF_x,xF_y,yF_y⊗ŷ∧D̂∧Ê,

with
  
    ω_3=
        (yD-y^r+1E)⊗x̂∧D̂∧Ê
        +yF̅E⊗ŷ∧D̂∧Ê,

and that, in particular,H^3()=r+2, since the denominator
appearing in the right hand side of this isomorphism is a3-dimensional
vector space —this follows at once from Lemma lemma:indep.



  
 We consider now a2-cocycleω∈^2anda∈S_2T,b,c∈S_1T,d,e∈S_r+1T⊕S_1DTandf∈S_rT⊕DTsuch that
  
    ω = a⊗x̂∧ŷ+b⊗x̂∧Ê
               + c⊗ŷ∧Ê
               + d⊗x̂∧D̂
               + e⊗ŷ∧D̂
               + f⊗Ê∧D̂.

Adding toωan element ofδ^1(T⊗Ê),
we can assume thatf∈S_rT; adding an element ofδ^1(S_1T⊗x̂⊕S_1T⊗ŷ), we can suppose thata=0; finally, adding
an element ofδ^1((S_rT⊕DT)⊗D̂)we can suppose thatd∈y^r+1T⊕yDT.
In this situation, there are an integerl≥0,α_0, …,α_l,β_0, …,β_l∈S_1,λ_0, …,λ_l,μ_0, …,μ_l∈,ϕ_0, …,ϕ_l∈S_r+1,ψ_0, …,ψ_l∈S_1andξ_0, …,ξ_l∈S_rsuch thatb=∑_i=0^lα_iE^i,c=∑_i=0^lβ_iE^i,d=∑_i=0^l(λ_iy^r+1+μ_iyD)E^i,e=∑_i=0^l(ϕ_i+ψ_iD)E^iandf=∑_i=0^lξ_iE^i. As
  
    δ^1(-y⊗x̂+F̅E⊗D̂)
            = y^r+1⊗x̂∧D̂
              + S_r+1⊗ŷ∧D̂,

we can assume thatλ_0=0.

The coefficient ofx̂∧ŷ∧Êinδ^2(ω)is∑_i=0^l(-α_iy+β_ix)τ_1(E^i)=0, and this implies that
there are scalarsρ_1, …,ρ_l∈such thatα_i=ρ_ixandβ_i=ρ_iyfor eachi∈{1,…,l}. The
coefficient ofx̂∧D̂∧Êinδ^2(ω)is
  
    ∑_i=0^l
            (-Fα_iyE^i-α_iDτ_r(E^i)+ξ_ixτ_1(E^i))
            = 0.

It follows that∑_i=0^lα_iDτ_r(E^i)=0, so thatα_1=⋯=α_l=0; as a consequence of this, we have thatρ_1=⋯=ρ_l=0andβ_1=⋯=β_l=0. The
equality (<ref>) also tells us that-Fα_0y+∑_i=0^lξ_ixτ_1(E^i)=0, and from this we see thatξ_2=⋯=ξ_l=0and-Fα_0y-ξ_1x=0, so thatα_0y=0andξ_1=0, sincexdoes not divideF. In
particular, there is aρ_0∈such thatα_0=ρ_0x.

The coefficient ofŷ∧D̂∧Êinδ^2(ω)is
  
    ∑_i=0^l
            (
            ∇_x^α_iE^i(F) + ∇_y^β_iE^i(F)
            - F β_iyE^i - β_iDτ_r(E^i)
            +ξ_iyτ_1(E^i)
            ) 
       = ρ_0xF_x+β_0F_y-β_0yF 
       = (ρ_0-(r+1)^-1β_0y)xF_x
           + (β_0xx+(1-(1+r)^-1)β_0yy)F_y
         = 0,

and our Lemma lemma:indep implies then thatβ_0=0andρ_0=0.
Finally, we consider the coefficient ofx̂∧ŷ∧D̂:
  
    ∑_i=0^l
            (
            -λ_iy^r+2τ_1(E^i)
            +μ_iyF(E+1)^i
            -μ_iy^2Dτ_1(E^i)
            +ϕ_ixτ_1(E^i)+ψ_ixDτ_1(E^i)
            )
            = 0.

Looking at the terms involvingDin this equation, we see that
  
    ∑_i=0^l(-μ_iy^2+ψ_ix)Dτ_1(E^i))=0,

soμ_1=⋯=μ_l=0andψ_1=⋯=ψ_l=0. The terms not
involvingDadd up to
  
    μ_0yF+∑_i=0^l(-λ_iy^r+2+ϕ_ix)τ_1(E^i)=0,

so thatλ_2=⋯=λ_l=0,ϕ_2=⋯=ϕ_l=0andμ_0yF+λ_1y^r+1-ϕ_1x=0, which implies thatλ_1=-μ_0andϕ_1=μ_0yF̅.

After all this, we see that every2-cocycle in our complex is
cohomologous to one of the form
  
    (μ_0yD-μ_0y^r+1E)x̂∧D̂
      + (ϕ_0+ψ_0D+μ_0yF̅E)ŷ∧D̂
      + ξ_0D̂∧Ê

withμ_0∈,ϕ_0∈S_r+1,ψ_0∈S_1andξ_0∈S_r. Computing we find that all elements of this form are in
fact2-cocycles.

Let us now suppose that the cocycle (<ref>), which we call againω,
is a coboundary, so that
there existk≥0,α_0, …,α_k,β_0, …,β_k∈S_1,σ_1, …,σ_k∈S_r,ζ_0, …,ζ_k∈andu∈Tsuch that if
  
    η = ∑_i=0^kα_iE^ix̂
            +∑_i=0^kβ_iE^iŷ
            +∑_i=0^k(σ_i+ζ_iD)E^iD̂
            +uÊ,

we haveδ^1(η)=ω. The coefficient ofD̂∧Êinδ^1(η)isDτ_r(u)so, comparing with (<ref>), we
see that we must haveξ_0=0andu∈; it follows from this that
the coefficients ofÊ∧Êand ofŷ∧Êinδ^1(η)vanish. On the other hand, the coefficient ofx̂∧ŷinδ^1(η)is∑_i=0^k(-α_iy+β_ix)τ_1(E^i): as this has to be zero, we
see that there existρ_1, …,ρ_k∈such thatα_i=ρ_ixandβ_i=ρ_iyfor eachi∈{1,…,k}.

The coefficient ofx̂∧D̂inδ^1(η)is
  
    ∑_i=0^k
            (
            -Fα_iyE^i-α_iDτ_r(E^i)+σ_ixτ_1(E^i)
            +ζ_ixDτ_1(E^i)
            ) = μ_0yD-μ_0y^r+1E.

This means, first, that
  ∑_i=1^k
        (
        -ρ_ixDτ_r(E^i)+ζ_ixDτ_1(E^i)
        )
        =μ_0yD
and this is only possible ifμ_0=0and
  
    ∑_i=1^k(-ρ_iτ_r(E^i)+ζ_iτ_1(E^i))=0.

Second, the equality (<ref>) implies that
  
    ∑_i=0^k
            (
            -Fα_iyE^i+σ_ixτ_1(E^i)
            )
            = -Fα_0y+∑_i=1^kσ_ixτ_1(E^i)
            = 0,

so thatσ_2=⋯=σ_k=0andFα_0y+σ_1x=0, which tells us thatσ_1=0andα_0y=0; there is then aρ_0∈such thatα_0=ρ_0x.

Finally, the coefficient ofŷ∧D̂inδ^1(η)is
  
    ∑_i=0^k
            (
            ∇_x^α_iE^i(F)+∇_y^β_iE^i(F)
            - Fβ_iyE^i
            - β_iDτ_r(E^i)
            +σ_iyτ_1(E^i)
            
    
            -ζ_iF(E+1)^i+ζ_iyDτ_1(E^i)
            )
            = ϕ_0+ψ_0D.

Looking only at the terms which are inS_1DT, we see that
  
    yD∑_i=1^k(-ρ_iτ_r(E^i)+ζ_iτ(E^i))=ψ_0D

and, in view of (<ref>), it follows from this thatψ_0=0. The
terms inS_r+1T, on the other hand, are
  
    ρ_0xF_x+β_0F_y
        +F
        (
        -β_0y-ζ_0
        +∑_i=0^k
            (
            ρ_i∑_t=1^r(E+t)^i-ζ_i(E+1)^i
            )
        ) = ϕ_0,

and proceeding as before we see thatϕ_0is in the linear span ofxF_x,xF_yandyF_y. Computing, we find that
  
    δ^1(x⊗x̂)=xF_x⊗ŷ∧D̂, 
    δ^1(x⊗ŷ)=xF_y⊗ŷ∧D̂, 
    δ^1(y⊗ŷ-DD̂)=yF_y⊗ŷ∧D̂.

We thus conclude that there is an isomorphism
  
    H^2()
            ≅ω_2
            ⊕S_r+1/xF_x,xF_y,yF_y⊗ŷ∧D̂⊕ S_1D⊗ŷ∧D̂⊕ S_r⊗D̂∧Ê,

withω_2=(yD-y^r+1E)⊗x̂∧D̂+yF̅E⊗ŷ∧D̂, and that, in particular, the dimension ofH^2()is2r+3.



  
 We can summarize our findings as follows:

prop:hh
Suppose that r≥3.
For all p≥4 we have ^p(A)=0. There are
isomorphisms 
  
    ^0(A) ≅, 
    ^1(A) ≅ S_r⊗D̂⊕⊗Ê, 
    ^2(A) ≅ω_2
            ⊕S_r+1/xF_x,xF_y,yF_y⊗ŷ∧D̂⊕ S_1D⊗ŷ∧D̂⊕ S_r⊗D̂∧Ê, 
    ^3(A) ≅ω_3
            ⊕S_r+1/xF_x,xF_y,yF_y⊗ŷ∧D̂∧Ê⊕ S_1D⊗ŷ∧D̂∧Ê,

with 
  
    ω_2=(yD-y^r+1E)⊗x̂∧D̂+yF̅E⊗ŷ∧D̂, 
    ω_3=
        (yD-y^r+1E)⊗x̂∧D̂∧Ê
        + yF̅E⊗ŷ∧D̂∧Ê.

The Hilbert series of the Hochschild cohomology of A is
  
    h_^∙(A)(t) 
              = 1 + (r+2)t + (2r+3)t^2 + (r+2)t^3 
       = (1+t)(1+(r+1)t+(r+2)t^2).


In fact, in each of the isomorphisms appearing in the statement of the
proposition we have given a set of representing cocycles. This will be
important in what follows, when we compute the Gerstenhaber algebra
structure on the cohomology ofA.

We have chosen a system of coordinates in the vector space
containing the arrangementAin such a way that one of the lines is given
by the equationx=0. This was useful in picking a basis for theS-module
of derivations()and, as a consequence, obtaining a presentation
of the algebraAamenable to the computations we wanted to carry out, but
the unnaturality of our choice is reflected in the rather unpleasant form
of the representatives that we have found for cohomology classes —a
consequence of the combination of the truth of Hermann Weyl's dictum that
the introduction of coordinates is an act of violence together with that of
the everyday observation that violence does not lead to anything good. In
the next section we will be able to obtain a more natural description.



  
 In Proposition prop:hh we considered only line
arrangements withr≥3, that is, with at least5lines. As we
explained in p:r3, without the restriction the method of
calculation that we followed has to be modified, and it turns out that this
is not only a technical difference: the actual results are different. Let
us describe what happens, starting with the factorizable cases:

  *  If there are no lines, so that r=-2, the arrangement is empty and
() is the second Weyl algebra [x,y,∂_x,∂_y].


  *  If there is one line, then () is
[x,y,x∂_x,∂_y] and this is isomorphic to (𝔰)⊗ A_1, with (𝔰) the enveloping algebra of the
non-abelian 2-dimensional Lie algebra 𝔰 and A_1 the first
Weyl algebra.


  *  If there are two lines, so that r=0, then () is
[x,y,x∂_x,y∂_y], which is isomorphic to (𝔰)⊗(𝔰). 


The Hochschild cohomology of the Weyl algebras is well-known —for
example, from <cit.>— as is that of(s). Using this and Künneth's
formula we find that when-2≤r≤0we have for alli∈_0.
  
    ^i(()) = r+2i.

Finally, we have the cases of three and four lines. Up to isomorphism of
arrangements, one can assume that the defining polynomials areQ=xy(x-y)andQ=xy(x-y)(x-λy)for someλ∈∖{0,1},
respectively. One can compute the cohomology of()in these cases
along the lines of what we did above, but the computation is surprisingly
much more involved. We have done the computation using an alternative, much
more efficient approach —using a spectral sequence that computes in
general the Hochschild cohomology of the enveloping algebra of a
Lie–Rinehart pair— on which we will report in an upcoming paper. Let us
for now simply summarize the result: whenris2or3, the Hilbert
series of^∙(A)is
  
    h_^∙(A)(t) 
            = 1 + (r+2)t + (2r+4)t^2 + (r+3)t^3.

This differs from the general case of Proposition prop:hh in the
coefficients oft^2andt^3.

For our immediate purposes, we remark that in all cases^1(())has
dimension equal to the number of lines in the arrangement, and that
its concrete description is the same in all cases.




§ THE GERSTENHABER ALGEBRA STRUCTURE ON 
HH(D(A))


  
 
LetAbe the usual bar resolution forAas anA-bimodule.
There is a morphism of complexesϕ:→Aover
the identity map ofAsuch thatϕ=ϕ_K+ϕ_Nwithϕ_K,ϕ_N:→Amaps ofA-bimodules such that
  
    ϕ_K(1|v_1∧⋯∧ v_p|1)
            = ∑_π∈ S_p(-1)^ϵ(π)
                    1|v_π(1)|⋯|v_π(p)|1,

wheneverp≥0andv_1, …,v_p∈V,
with the sum running over permutations of degreep,
and
  
    ϕ_N(1|1)
            = 0; 
    ϕ_N(1|v|1)
            = 0,   ∀ v∈ V; 
    [.9]
      ϕ_N(1|x∧ y|1)
            = ϕ_N(1|x∧ E|1)
            = ϕ_N(1|y∧ E|1)
            = ϕ_N(1|x∧ D|1) 
    
            = ϕ_N(1|D∧ E|1)
            = 0;
      
    ϕ_N(1|y∧ D|1)
            = q1|q̅2|q3|1 - F|1|1|1; 
    ϕ_N(1|x∧ y∧ E|1)
            = ϕ_N(1|x∧ D∧ E|1)
            = 0; 
    [.9]
      ϕ_N(1|x∧ y∧ D|1)
            = q1|q̅2|q3|x|1
                    - q1|q̅2|x|q3|1
                    + q1|x|q̅2|q3|1 
    
                    - F|x|1|1|1 - F|1|1|x|1;
      
    [.9]
      ϕ_N(1|y∧ D∧ E|1)
            = q1|q̅2|q3|E|1
                    - q1|q̅2|E|q3|1
                    + q1|E|q̅2|q3|1 
    
                    - F|E|1|1|1 - F|1|1|E|1.

Hereq1|q̅2|q3denotes the element∇(F)∈S⊗S_1⊗S, with an omitted sum.

On the other hand, there is a morphism of complexes ofA-bimodulesψ:A→over the identity map ofAsuch that
  
    ψ_0(1|1) = 1|1, 
    ψ_1(1|w|1) = w1|w2|w 3,
                for all standard monomials w; 
    
    
      ψ_2(1|yD|y|1) = -y|y∧ D|1 - q1|q2∧ y|q3; 
    ψ_2(1|y^r+1E|y|1) = -y^r+1|y∧ E|1; 
    ψ_2(1|E|w|1) = -w1|w2∧ E|w3
                for all standard monomials w; 
    ψ_2 (1|v|w|1) = - 1|w∧ v |1,     if v,w∈{x,y,D,E} and vw is not standard; 
    ψ_2(1|w|x|1) = -w1|x∧ w2|w3
                for all standard monomials w; 
    andψ_2(1|u|v|1) = 0

wheneveruandvare standard monomials ofAsuch that the
concatenationuvis also a standard monomial. This morphismψcan be
taken —and we will take it— to be normalized, so that it vanishes on
elementary tensors ofAwith a scalar factor.



  
 We need the comparison morphisms that we have just described in
order to compute the Gerstenhaber bracket on^∙(A), but we start
with a more immediate application: obtaining a natural basis of the first
cohomology space^1(A).

prop:partial
  *  If α is a non-zero element of S_1 that divides Q, then
there exists a unique derivation ∂_α:A→ A such that
∂_α(f)=0 for all f∈ S and
  
    ∂_α(δ) = δ(α)/α

for all δ∈(). 


  *  If Q=α_0…α_r+1 is a factorization of Q as a
product of elements of S_1, then the cohomology classes of the r+2
derivations ∂_α_0, …, ∂_α_r+1 of A
freely span the vector space ^1(A).



Here we are viewing^1(A)as the vector space of outer
derivations ofA, as usual. It should be noticed that the
derivation∂_αassociated to a linear factor ofQdoes not
change if we replaceαby one of its non-zero scalar multiples: this
means that the basis of^1(A)is really indexed by the lines of the
arrangement.

1 Let us fix a non-zero element α in S_1 dividing Q.
There is at most one derivation ∂_α:A→ A as in the
statement of the proposition simply because the algebra A is generated by
the set S∪(). In order to prove that there is such a derivation,
we need only recall from <cit.>*Proposition 4.8 that
δ(α)∈α S for all δ∈() and check that the
candidate derivation respects the relations (<ref>)
of p:start that present the algebra A.

2 We need to pass from the description of ^1(A) as the
space of outer derivations to its description in terms of the complex 
that was used to compute it: we do this with the comparison morphism 
ϕ:→ A over the identity map that we described
in p:comparison.
If δ:A→ A is a
derivation of A and δ̃:A⊗ A⊗ A→ A is the map
such that δ̃(a⊗ b⊗ c)=aδ(b)c for all
a, b, c∈ A, which is a 1-cocycle on A, the composition
δ̅∘ϕ_1:A⊗ V⊗ A→ A is a 1-cocycle in the
complex _A^e(,A) whose cohomology class corresponds to δ
in the usual description of ^1(A) as the space of outer derivations
of A. In the notation that we used in p:hhdifferentials, this
cohomology class is that of 
  
    δ(x)⊗x̂ + δ(y)⊗ŷ
        + δ(D)⊗D̂ + δ(E)⊗Ê∈ A⊗V̂.

Using this, we can now prove the second part of the proposition.
We can suppose without loss of generality that α_0=x,
and then the class of δ_α_0 in ^1(A) is that of
  
    1⊗Ê.

On the other hand, for each
i∈{1,…,r+1}, computing we find that the class
of ∂_α_i is
  
    α_iyF/α_i⊗D̂ + 1⊗Ê.

It follows easily from the second part of Lemma lemma:indep that
these r+2 classes span ^1(A) and, since the dimension of
this space is exactly r+2, do so freely.


 §.§ The cup product


  
 We describe the associative algebra
structure on^∙(A)given by the cup product.


The cup product on ^∙(A) is such that
      S_r⊗D̂⌣ S_r⊗D̂ = 0; 
   ϕD̂⌣Ê = ϕD̂∧Ê,
                  ∀ϕ∈ S_r; 
    S_r⊗D̂⌣^2(A) = 0; 
    1⊗Ê⌣ω_2 = ω_3; 
    1⊗Ê⌣κ⊗ŷ∧D̂ 
        = κ⊗ŷ∧Ê∧D̂,
              ∀κ∈ S_r+1/xF_x,xF_y,yF_y; 
    1⊗Ê⌣ψ D⊗ŷ∧D̂
        = ψ D⊗ŷ∧D̂∧Ê,
              ∀ψ∈ S_1; 
    1⊗Ê⌣ S_r⊗D̂∧Ê = 0.
  

These equalities completely describe the multiplicative
structure on^∙(A).


There is a morphism of complexes
of A-bimodules Δ:→⊗_A that lifts the canonical
isomorphism A→ A⊗_AA such that Δ=Δ_K+Δ_N, with

  * Δ_K:→⊗_A the map of A-bimodules such that
for whenever p≥0 and v_1, …, v_p∈ V we have
  
    Δ_K(1|v_1∧⋯∧  v_p|1)
        = ∑(-1)^ϵ
          1|v_i_1∧⋯∧  v_i_r|1
            ⊗1|v_j_1∧⋯∧  v_j_s|1,

with the sum taken over all decompositions r+s=p with r, s≥0, and
all permutations (i_1,…,i_r,j_1,…,j_s) of (1,…,p) such that
i_1<⋯<i_r and j_1<⋯<j_s, and where ϵ is the
signature of the permutations,


  *  and Δ_N:→⊗_A the map of A-bimodules such that
  
    Δ_N(1|1) = 0; 
    Δ_N(1|v|1) = 0,     ∀ v∈ V; 
    Δ_N(1|v∧ w|1) = 0,     if v, w∈{x,y,D,E}, v≠
      w and {v,w}≠{y,D}; 
    Δ_N(1|y∧ D|1)
            = f1|f2|f3⊗1|f4|f5; 
    Δ_N(1|x∧ y∧ D|1) = Δ_N(1|x∧ y∧ E|1)
            = Δ_N(1|x∧ D∧ E|1) = 0; 
    [.9]
      Δ_N(1|y∧ D∧ E|1) =
              - f1|f2∧ E|f3⊗1|f4|f5
    
              + f1|f2|f3⊗1|f4∧ E|f5.

Here we have written f1|f2|f3|f4|f5
the image of F under the composition
  
    S [r, "∇"]
            S ⊗ S_1⊗ S [r, "𝕀_S⊗𝕀_S_1⊗∇"]
           [4em] S ⊗ S_1⊗ S⊗ S_1⊗ S,

with an omitted sum, à la Sweedler.


We leave the verification that this does define a morphism of complexes to
the reader.

One can compute the cup product on ^∙(A) using this diagonal
morphism Δ. Indeed, we view ^∙(A) as the cohomology of
the complex _A^e(,A), and if ϕ and ψ are a p- and a
q-cocycle in that complex, the cup product of their cohomology classes is
represented by the composition
  
    P_p+q[r, "Δ_p,q"]
            P_p⊗_A P_q [r, "ϕ⊗ψ"]
            A⊗_AA=A,

with Δ_p,q the component P_p+q→ P_p⊗ P_q of the
morphism Δ. The multiplication table given in the statement of the
composition can be computed in this way, item by item.

  *  For all i, j, k∈{0,…,r+1} we have
  
    ∂_α_i⌣∂_α_j
      + ∂_α_j⌣∂_α_k
      + ∂_α_k⌣∂_α_i 
      = 0

and ^1(A)⌣^1(A)=S_r⊗D̂∧Ê. 


  *  The subalgebra $̋ of^∙(A)generated by^1(A)is the
graded-commutative algebra freely generated by its elements∂_α_0, …,∂_α_r+1of degree1subject to ther+23relations (<ref>).



This subalgebra$̋ is isomorphic to the de Rham
cohomology of the complement of the arrangement of lines . This follows
from a direct computation of this cohomology or, in fact, from the solution
of Arnold's conjecture by Brieskorn; this is discussed in detail in
<cit.>*Section 5.4.


Using Proposition prop:cup and the description given in the proof
of Proposition prop:partial for the
derivations ∂_α_i we compute immediately that
  
    ∂_α_i⌣∂_α_j
            = - α_ix   α_jx
    α_iy   α_jyQ/α_iα_j

for all i, j∈{0,…,r+1}. Using this, we see that for all
i, j, k∈{0,…,r+1} we have
  
    ∂_α_i⌣∂_α_j
      + ∂_α_j⌣∂_α_k
      + ∂_α_k⌣∂_α_i 
      =
      - α_i    α_j    α_k 
    α_ix   α_jx   α_kx
    α_iy   α_jy   α_ky
    Q/α_iα_jα_k
        = 0,

as the determinant vanishes. This proves the first claim of 1.
The second one follows immediately from the description of the cup product
of Proposition prop:cup.

2 Let ℱ=⊕_n≥0ℱ_n be the free graded-commutative
algebra generated by r+2 generators w_0, …, w_r+1 of degree 1
subject to the relations w_iw_j+w_jw_k+w_kw_i=0, one for each choice of
i, j, k∈{0,…,r+1}. We have ℱ_n=0 if n≥3: if
i, j, k∈{1,…,r+1} we have
w_iw_jw_k=(w_iw_j+w_jw_k+w_kw_i)w_k=0, because of graded-commutativity.
On the other hand, we have ℱ_2≤ r+1. To see this, we notice that
ℱ_2 is spanned by products w_iw_j with 1≤ i<j≤ r+1. If 
i+1<j then w_iw_j=-w_i+1w_j-w_i+1w_i: it follows from this that the
set of monomials {w_iw_i+1:0≤ i≤ r} already spans ℱ_2.

The first part of the proposition implies that there is a surjective
morphism of graded algebras f:ℱ→$̋ such thatf(w_i)=∂_α_ifor alli∈{0,…,r+1}, and 
this map is also injective because the
dimension of the component of degree2of$̋, which is S_r⊗D̂∧Ê, is r+1.


  
 Proposition prop:partial:sub describes
meaningfully a part of the associative algebra ^∙(A), the
subalgebra $̋ generated by^1(A), in terms of the geometry of the
arrangement. It is not clear how to make sense of the complete
algebra. We can make the following observation, though. Let us write 
  
    ^2(A)' = 
            ω_2⊕
            (S_r+1/⟨ xF_x,xF_y,yF_y⟩⊕ S_1D)⊗ŷ∧D̂,

which is a complement of^̋2in^2(A), and letQ=α_0…α_r+1be a factorization ofQas a product of
linear factors. Ifδ:A→ Ais
derivation ofA, then our description of^1(A)implies that there exist 
scalarsδ_0, …,δ_r+1∈and an elementu∈ Asuch thatδ=∑_i=0^r+1δ_i∂_α_u+(u), and
it follows easily from Proposition prop:cup that the map
  
    ζ∈^2(A)' ↦δ⌣ζ∈^3(A)

is either zero or an isomorphism, provided∑_i=0^r+1δ_iis
zero or not.



 §.§ The Gerstenhaber bracket


  
 Using the comparison morphisms of p:comparison, we can
now compute the Gerstenhaber bracket. As usual, this is very
laborious.

prop:bracket
In ^∙(A) we have
  
    [0,∙]   {  
          [^0(A),^∙(A)] 
              = 0, 
          .
          
    
      [1,1]   {  
          [^1(A),^1(A)] 
              = 0, 
          .
          
    
      [1,2]   {  
          [^1(A),S_r⊗D̂∧Ê] 
              = 0, 
              
    
          [u⊗D̂+λ⊗Ê,(v+wD)⊗ŷ∧D̂]
              = uw⊗ŷ∧D̂, 
              
    
          [u⊗D̂+λ⊗Ê,ω_2]
              = ((μ-λ)yF_x+μ yF̅-y^2u̅)⊗ŷ∧D̂, 
          .
          
    
      [1,3]   {  
          [u⊗D̂+λ⊗Ê,(v+wD)⊗ŷ∧D̂∧Ê]
              = uw⊗ŷ∧D̂∧Ê, 
    
          [u⊗D̂+λ⊗Ê,ω_3]
              = ((μ-λ)yF_x+μ yF̅-y^2u̅)
                    ⊗ŷ∧D̂∧Ê, 
          .
          
    
      [2,2]   {  
          [S_r⊗D̂∧Ê,S_r⊗D̂∧Ê] 
              = 0, 
              
    
          [u⊗D̂∧Ê,(v+wD)⊗ŷ∧D̂]
              = uw⊗ŷ∧D̂∧Ê, 
              
    
          [u⊗D̂∧Ê,ω_2] 
              = (μ yFx+μ yF̅-y^2u̅)⊗ŷ∧D̂∧Ê, 
              
    
          [(S_r+1+S_1D)⊗ŷ∧D̂,(S_r+1+S_1D)⊗ŷ∧D̂]
              = 0,
              
    
          [(S_r+1+S_1D)⊗ŷ∧D̂,ω_2]
              = 0,
              
    
          [ω_2,ω_2]
              = 0.
          .

Here u∈ S_r, λ∈, v∈ S_r+1, w∈ S_1 and
μ∈ and u̅∈ S_r-1 are such that u=λ y^r+xu̅.

Let us first recall from <cit.> how one can compute the
Gerstenhaber bracket in the standard complex _A^e( A,A). If
f:A^⊗ q→ A is a q-cochain in the standard complex
_A^e( A,A), which we identify as usual with
(A^⊗∙,A), and p≥ q, we denote _p(f):A^⊗
p→ A^p-q+1 the p-cochain in the same complex such that
  
    _p(f)(a_1⊗⋯⊗ a_p) 
    
        = ∑_i=1^p-q+1(-1)^(q-1)(i-1)
            a_1⊗⋯⊗ a_i-1⊗ f(a_i⊗⋯⊗ a_i+q-1)
            ⊗ a_i+q⊗⋯⊗ a_p.

If now α and β are a p- and a q-cocycle in the standard
complex, the Gerstenhaber composition ♢ (which is usually written
simply ∘) of α and β is the (p+q-1)-cochain
  
    α♢β = α∘_p+q-1(β)

and the Gerstenhaber bracket is the graded commutator for this composition,
so that
  
    [α,β] = α♢β
        -(-1)^(p-1)(q-1)β♢α.

Next, if α and β are now a p- and a q-cochain 
in the complex _A^e(,A), we
can lift them to a p-cochain α̃=α∘ψ_p and a
q-cochain β̃=β∘ψ_q in the
standard complex _A^e( A,A), and the Gerstenhaber bracket of the
classes of α and β is then represented by the
(p+q-1)-cochain [α̃,β̃]∘ϕ_p+q-1. This is
the computation we have to do in order to compute brackets
in ^∙(A), except that in some favorable circumstances we can
take advantage of the compatibility of the bracket with the product to cut
down the work. We do this in several steps.

  *  Since the morphism ψ is normalized and ^0(A) is spanned by 1∈, it
follows immediately that 
  
    [^0(A),^∙(A)]=0.

  *  The Gerstenhaber bracket on ^1(A) is induced by the commutator
of derivations. From Proposition prop:partial we have a basis
of ^1(A) whose elements are classes of certain derivations, and it is
immediate to check that those derivations commute, so that 
  
    [^1(A),^1(A)]=0.

  *  We know that the subspace S_r⊗D̂∧Ê of ^2(A) is
^1(A)⌣^1(A). Since ^∙(A) is a Gerstenhaber algebra
and we now that (<ref>) holds, it follows that
  
    [^1(A),S_r⊗D̂∧Ê] = 0.

For exactly the same reasons we also have that
  
    [S_r⊗D̂∧Ê,S_r⊗D̂∧Ê]=0.

  *  Let α=u⊗D̂+λ⊗Ê, with u∈ S_r and
λ∈. If β=(v+wD)⊗ŷ∧D̂, with v∈
S_r+1 and w∈ S_1, one can compute that
(α̃♢β̃)∘ϕ=uw⊗ŷ∧D̂
and that (β̃♢α̃)∘ϕ=0: it follows from
this that
  
    [α,(v+wD)⊗ŷ∧D̂] = uw⊗ŷ∧D̂.

On the other hand, we have
(ω̃_2♢α̃)∘ϕ=0 and
  
    [α̃,ω̃_2]∘ϕ   = (α̃♢ω̃_2)∘ϕ
         = (yu-λ y^r+1)⊗x̂∧D̂
           +λ yF̅⊗ŷ∧D̂
       = [t][.75]
            (
            (μ-λ)yF_x+μ y F̅-y^2u̅)⊗ŷ∧D̂
    
           - δ^1(
                     ((μ-λ)F̅-yu̅)E⊗D̂
                     + (λ-μ)y⊗x̂)

with u̅∈ S_r-1 and μ∈ chosen so that u=μ y^r+xu̅.

Finally, if v∈ S_r+1 and w∈ S_1, using the compatibility
of the bracket and the product and what we know so far we see that
  
    [α,(v+wD)⊗ŷ∧D̂∧Ê]
           = [α,1⊗ E⌣ (v+wD)⊗ŷ∧D̂] 
       = 1⊗ E⌣ [α,(v+wD)⊗ŷ∧D̂] 
       = 1⊗ E⌣ uw⊗ŷ∧D̂
       = uw⊗ŷ∧D̂∧Ê

and, similarly, that
  
    [α,ω_3]
           = [α,ω_2⌣1⊗Ê]
         = [α,ω_2]⌣1⊗Ê
           + ω_2⌣[α,1⊗Ê] 
       = ((μ-λ)yF_x+μ y F̅-y^2u̅)
             ⊗ŷ∧D̂∧Ê.

  *  Let u∈ S_r. If v∈ S_r+1 and w∈ S_1, we have
  
    [u⊗D̂∧Ê,(v+wD)⊗ŷ∧D̂]
          = [u⊗D̂⌣1⊗Ê,(v+wD)⊗ŷ∧D̂] 
        = [t][.8]
            [u⊗D̂,(v+wD)⊗ŷ∧D̂]⌣1⊗Ê
    
            +
            u⊗D̂⌣[1⊗Ê,(v+wD)⊗ŷ∧D̂] 
            
        = uw⊗ŷ∧D̂⌣ 1⊗Ê
          = uw⊗ŷ∧D̂∧Ê.

Similarly, 
  
    [u⊗D̂∧Ê,ω_2]
            = [u⊗D̂⌣1⊗Ê,ω_2] 
        = [u⊗D̂,ω_2]⌣1⊗Ê
            + u⊗D̂⌣[1⊗Ê,ω_2] 
        = (
            μ yF_x+μ y F̅-y^2u̅)⊗ŷ∧D̂∧Ê.

if u=μ y^r+xu̅ with μ∈ and u̅∈ S_r-1.


  *  Let now α=(v+wD)⊗ŷ∧D̂ and
β=(s+tD)⊗ŷ∧D̂, with v, s∈ S_r+1 and
w, t∈ S_1. We claim that
(α̃♢β̃)∘ϕ=0, so that, by symmetry, we
have [α̃,β̃]∘ϕ=0. To verify our claim, we
compute:
#1#1
    1|x∧ y∧ E|1   ϕ[x,y,E]^⊗ 5_3(β̃) 0; 
    1|x∧ D∧ E|1   ϕ[x,D,E]^⊗ 5_3(β̃) 0; 
    1|x∧ y∧ D|1   ϕ[t][.7]
          1|x|y|D|1 
          - 1|x|D|y|1 
          + 1|D|x|y|1 
    
          - 1|D|y|x|1 
          + 1|y|D|x|1 
          - 1|y|x|D|1 
          + S^⊗ 5
       _3(β̃)
          1|(s+tD)|x|1-1|x|(s+tD)|1
          
       ψ
            - s1|x∧ s2|s3
            - t1|x∧ t2|t3D
            - t|x∧ D|1
          
       α
            0;
          
    1|y∧ D∧ E|1   ϕ[t][.7]
          1|y|D|E|1 
          - 1|y|E|D|1 
          + 1|E|y|D|1 
    
          - 1|E|D|y|1 
          + 1|D|E|y|1 
          - 1|D|y|E|1 
          + [x,y,E]^⊗ 5
       _3(β̃)
          1|(s+tD)|E|1 - 1|E|(s+tD)|1 
          
       ψ
          s1|s2∧ E|s3 + t1|t2∧ E|t3D + t|D∧ E|1
          
       α
          0.

  *  Let again α=(v+wD)⊗ŷ∧D̂, with v∈
S_r+1 and w∈ S_1, and let us compute that
(ω̃_2♢α̃)∘ϕ_3=-w(yD-y^r+1E)⊗x̂∧ŷ∧D̂.
  
    1|x∧ y∧ z|1
            ϕ_3[x,y,E]^⊗5_2(α̃)
         0 
    
       1|x∧ D∧ E|1
            ϕ_3[x,D,E]^⊗ 5_3(α̃)
         0 
    
      1|x∧ y∧ D|1
           ϕ_3[t][.7]
          1|x|y|D|1 
          - 1|x|D|y|1 
          + 1|D|x|y|1 
    
          - 1|D|y|x|1 
          + 1|y|D|x|1 
          - 1|y|x|D|1 
          + S^⊗ 5
       _3(α̃)
          1|(v+wD)|x|1 + 1|x|(v+wD)|1
          
       ψ_2
          -v1|x∧ v 2|v3 
          -w1|x∧ w 2|w3 D
          -w|x∧ D|1
          
       ω_2
          -w(yD-y^r+1E)
          
    
      1|y∧ D∧ E|1
           ϕ_3[t][.7]
          1|y|D|E|1 
          - 1|y|E|D|1 
          + 1|E|y|D|1 
    
          - 1|E|D|y|1 
          + 1|D|E|y|1 
          - 1|D|y|E|1 
          + [x,y,E]^⊗ 5
       _3(α̃)
          1|(v+wD)|E|1 - 1|E|(v+wD)|1 
          
       ψ_2
          v1|v2∧ E|v3 + w1|w2∧ E|w3D + w|D∧ E|1
          
       ω_2
          0.

Similarly, we have that
(α̃♢ω̃_2)∘ϕ_3=y(v+wD)⊗x̂∧ŷ∧D̂:
  
    1|x∧ y∧ z|1
            ϕ_3[x,y,E]^⊗5
       _2(ω̃_2)
           0 
    
       1|x∧ D∧ E|1
            [t][.7]
           ϕ_3
           1|x|D|E|1 
           - 1|x|E|D|1 
           + 1|E|x|D|1 
    
           - 1|E|D|x|1 
           + 1|D|E|x|1 
           - 1|D|x|E|1 
           
       _3(ω̃_2)
           -1|E|(yD-y^r+1E)|1+1|(yD-y^r+1)|E|1
           
       ψ_2
           -1|y∧ E|D-y|D∧ E|1+∑_i=0^ry^i|y∧ E|y^r-i
       α
           0
           
    
      1|x∧ y∧ D|1
           [t][.7]
          ϕ_3
          1|x|y|D|1 
          - 1|x|D|y|1 
          + 1|D|x|y|1 
    
          - 1|D|y|x|1 
          + 1|y|D|x|1 
          - 1|y|x|D|1 
          + S^⊗ 5
       [t][.7]
          _3(ω̃_2)
          -1|x|yF̅ E|x|1 - 1|(yD-y^r+1E)|y|1 
    
          +1|yF̅E|x|1+1|y|(yD-y^r+1E)|1
          
       [t][.7]
          ψ_2
          y|y∧ D|1-y^r+1|y∧ E|1 
    
          -(yF̅E)1|x∧(yF̅E)2|(yF̅E)3
          
       α
          -y(v+wD)
          
    
      1|y∧ D∧ E|1
           [t][.7]
          ϕ_3
          1|y|D|E|1 
          - 1|y|E|D|1 
          + 1|E|y|D|1 
    
          - 1|E|D|y|1 
          + 1|D|E|y|1 
          - 1|D|y|E|1 
          + [x,y,E]^⊗ 5
       _3(ω̃_2)
          -1|E|yF̅E|1 + 1|yF̅E|E|1 
          
       ψ_2
          (yF̅E)1|(yF̅E)2∧ E|(yF̅E)3
          
       α
          0.

It follows from this that
  
    [ω̃_2,α̃]∘ϕ_3
            = -w (yD-y^r+1E)⊗x̂∧ŷ∧D̂
            + y(v+wD)⊗x̂∧ŷ∧D̂
       = (yv+y^r+1E)⊗x̂∧ŷ∧D̂

and, as we say in p:3-cocycles, this is a coboundary.


  *  The one computation that remains is that of the bracket of ω_2
with itself, which is represented by the 3-cocycle
  
    [ω̃_2,ω̃_2]∘ϕ_3
        = 2(ω̃_2♢ω̃_2)∘ϕ_3
        = 2y^2F̅E⊗x̂∧ŷ∧D̂,

as can be seen from the following calculation:
  
    1|x∧ y∧ z|1
            ϕ_3[x,y,E]^⊗5
       _2(ω̃_2)
           0 
    
       1|x∧ D∧ E|1
            [t][.7]
           ϕ_3
           1|x|D|E|1 
           - 1|x|E|D|1 
           + 1|E|x|D|1 
    
           - 1|E|D|x|1 
           + 1|D|E|x|1 
           - 1|D|x|E|1 
           
       _3(ω̃_2)
           -1|E|(yD-y^r+1E)|1+1|(yD-y^r+1)|E|1
           
       ψ_2
           -1|y∧ E|D-y|D∧ E|1+∑_i=0^ry^i|y∧ E|y^r-i
       ω_2
           0
           
    
      1|x∧ y∧ D|1
           [t][.7]
          ϕ_3
          1|x|y|D|1 
          - 1|x|D|y|1 
          + 1|D|x|y|1 
    
          - 1|D|y|x|1 
          + 1|y|D|x|1 
          - 1|y|x|D|1 
          + S^⊗ 5
       [t][.7]
          _3(ω̃_2)
          -1|x|yF̅ E|x|1 - 1|(yD-y^r+1E)|y|1 
    
          +1|yF̅E|x|1+1|y|(yD-y^r+1E)|1
          
       [t][.7]
          ψ_2
          y|y∧ D|1-y^r+1|y∧ E|1 
    
          -(yF̅E)1|x∧(yF̅E)2|(yF̅E)3
          
       ω_2
          -y^2F̅E
          
    
      1|y∧ D∧ E|1
           [t][.7]
          ϕ_3
          1|y|D|E|1 
          - 1|y|E|D|1 
          + 1|E|y|D|1 
    
          - 1|E|D|y|1 
          + 1|D|E|y|1 
          - 1|D|y|E|1 
          + [x,y,E]^⊗ 5
       _3(ω̃_2)
          -1|E|yF̅E|1 + 1|yF̅E|E|1 
          
       ψ_2
          (yF̅E)1|(yF̅E)2∧ E|(yF̅E)3
          
       ω_2
          0.

Now the 3-cocycle (<ref>) is a coboundary, again by what we saw
in p:3-cocycles, so that the class of ω_2 has bracket-square
zero.

This completes the proof of the proposition.



§ HOCHSCHILD HOMOLOGY, CYCLIC HOMOLOGY AND K-THEORY


  
 For completeness, we determine the rest of the `usual'
homological invariants of our algebraA. Recall that our ground
fieldis of characteristic zero.

prop:hom
The inclusion T=[E]→ A induces an isomorphism in Hochschild
homology and in cyclic homology. In particular, there are isomorphisms of
vector spaces
  
    _i(A) ≅
            T,    if i=0 or i=1;
    
            0,    if i≥2;      _i(A) ≅
            T,    if i=0;
    _i(),    if i>0.

On the other hand, the inclusion → A induces an
isomorphism in periodic cyclic homology and in higher K-theory.

As we know, the algebra A is _0-graded and for each n∈_0 its
homogeneous component A_n of degree n is the eigenspace corresponding
to the eigenvalue n of the derivation (E):A→ A. On one hand, this
grading of A induces as usual an _0-grading on the Hochschild
homology _∙(A) of A; on the other, the derivation (E)
induces a linear map L_(E):_∙(A)→_∙(A) as
in <cit.>*4.1.4 and, in fact, for all n∈_0 the
homogeneous component _∙(A)_n of degree n for that grading
coincides with the eigenspace corresponding to the eigenvalue n
of L_(E). As the derivation (E) is inner, it follows from
<cit.>*Proposition 4.1.5 that the map L_(E) is actually the
zero map and this tells us in our situation that _∙(A)_n=0 for
all n≠0. Of course, this means that _∙(A)=_∙(A)_0
and, since A is non-negatively graded, it is immediate that the 0th
homogeneous component _∙(A)_0 coincides with the Hochschild
homology _∙(A_0) of A_0 and that the map
_∙(A_0)→_∙(A) induced by the inclusion
A_0↪ A is an isomorphism.
Now, in the notation of <cit.>*Theorem 4.1.13, this tells us that
≈_∙(A)=0 so that by that theorem we also have
≈_∙(A)=0: this means precisely that the inclusion
A_0↪ A induces an isomorphism
_∙(A_0)→_∙(A) in cyclic homology. Together with the
well-known computation of the Hochschild homology of a polynomial
ring and that of the cyclic homology of symmetric algebras
<cit.>*Theorem 3.2.5, this proves the first claim of the statement.

In the proof of the lemma of p:P we constructed an increasing
filtration F on the algebra A with F_-1A=0 and such that the
corresponding graded algebra is the commutative polynomial ring A=[x,y,D,E] with generators x and y in degree 0 and D and E
in degree 1. In particular, both A and its subalgebra _0A of
degree 0 have finite global dimension. It follows from a theorem of
D. Quillen <cit.>*p. 117, Theorem 7 that the inclusion
[x,y]=F_0A→ A induces an isomorphism K_i([x,y])→ K_i(A) in
K-theory for all i≥0. Similarly, the theorem of J. Block
<cit.>*Theorem 3.4 tells us that that inclusion induces an
isomorphism HP_∙([x,y])→ HP_∙(A) in periodic cyclic
homology. As the inclusion →[x,y] induces an isomorphism in
K-theory and in periodic cyclic homology, we see that the second claim of
the proposition holds.



§ THE CALABI–YAU PROPERTY


  
 The enveloping algebraA^eofAis a bimodule over
itself, with left and right actions▹and◃given by `outer' and `inner' multiplication, respectively, so that ifa⊗ b,c⊗ dande⊗ fare elementary tensors
inA^e, we have 
  
    a⊗ b ▹ c⊗ d ◃ e⊗ f
        = ace⊗ fdb.

From this bimodule structure we obtain a duality functor
  
    _A^e(,A^e) : A^e→A^e.

On the other hand, using the anti-automorphismτ:A^e→ A^esuch thatτ(a⊗ b)=b⊗ afor alla,b∈ A, we can turn
a rightA^e-moduleMinto a leftA^e-module, with actionu▹ m=m◃τ(u)for allu∈ A^eand allm∈
M. In this way, we obtain an isomorphism of categoriesτ^*:A^e→A^e. We denote()^∨:A^e→A^ethe compositionτ^*∘_A^e(,A^e).

Let nowWbe a finite dimensional vector space,
letW^*be the vector space dual toW, and viewA⊗ W⊗ AandA⊗ W^*⊗ Aas leftA^e-modules using the usual
`exterior' action. There is a unique-linear map
  
    Φ:A⊗ W^*⊗ A→ (A⊗ W⊗ A)^∨

such thatΦ(a⊗ϕ⊗ b)(1⊗ w⊗ 1)=ϕ(w)b⊗
aand it is an isomorphism of leftA^e-modules: we will view it in all
that follows as an identification.


The algebra A is twisted Calabi-Yau of dimension 4 with modular
automorphism σ:A→ A such that
  
    σ(x) = x,
            σ(y) = y,
            σ(D) = D + F_y,
            σ(E) = E + r+2.


Let us recall from <cit.> that this means thatAhas a
resolution of finite length by finitely generated projectiveA-bimodules,
that_A^e^i(A,A^e)=0ifi≠4and that_A^e^4(A,A^e)≅
A_σ, theA-bimodule obtained fromAby twisting its right action
using the automorphismσ, so thata▹ x◃
b=axσ(b)for alla,b∈ Aand allx∈ A_σ.


A direct computation shows that there is indeed an automorphism σ
of A as in the statement of the proposition.
We already know that A has a resolution  of length 4 by finitely
generated free A-bimodules, so we need only
compute _A^e^∙(A,A^e), and this is the cohomology of the
complex ^∨ obtained by applying the functor described in p:vee
to . Using the identifications introduced there, this
complex ^∨ is
  
    [column sep=0.95em]
      A⊗ A [r, "d_1^∨"]
                A⊗ V^*⊗ A [r, "d_2^∨"]
                A⊗Λ^2V^*⊗ A [r, "d_3^∨"]
                A⊗Λ^3V^*⊗ A [r, "d_4^∨"]
                A⊗Λ^4V^*⊗ A

with left A^e-linear differentials such that
  
    d_1^∨(1⊗ 1) 
            = - [x,1⊗⊗ 1] 
              - [y,1⊗⊗ 1] 
              - [D,1⊗⊗ 1] 
              - [E, 1⊗⊗ 1];
      
    [.9]
      d_2^∨( 1 ⊗⊗  1) 
            =   [y, 1⊗∧⊗ 1 ] 
              + [D, 1⊗∧⊗ 1 ]
              + [E, 1⊗∧⊗ 1 ] 
    
              + 1⊗∧⊗  1  
              + ∇̃^∧_x(F) ;
      
    [.9]
      d_2^∨( 1 ⊗⊗  1) 
            = - [x, 1⊗∧⊗ 1 ] 
              + [D, 1⊗∧⊗ 1 ] 
              + [E, 1⊗∧⊗ 1 ] 
    
              + 1⊗∧⊗  1
              + ∇̃^∧_y(F) ;
      
    [.9]
      d_2^∨( 1 ⊗⊗  1) 
            = - [x, 1⊗∧⊗ 1 ] 
              - [y, 1⊗∧⊗ 1 ]
              + [E, 1⊗∧⊗ 1 ] 
    
              + r⊗∧⊗  1 ;
      
    
      d_2^∨( 1 ⊗⊗  1) 
            = - [x, 1⊗∧⊗ 1 ] 
              - [y, 1⊗∧⊗ 1 ]
              - [D, 1⊗∧⊗ 1 ] ;
      
    [.9]
        d_3^∨(1 ⊗∧⊗  1) 
            = - [D, 1⊗∧∧⊗ 1]
              - ∇̃^∧∧_y(F) 
    
              - [E, 1⊗∧∧⊗ 1] 
              - 2⊗∧∧⊗ 1 ;
      
    [.9]
      d_3^∨(1 ⊗∧⊗  1) 
            =   [y, 1⊗∧∧⊗ 1]
              - [E, 1⊗∧∧⊗ 1] 
    
              - (r+1)⊗∧∧⊗ 1 ;
      
    [.9]
      d_3^∨(1 ⊗∧⊗  1) 
            =   [y, 1⊗∧∧⊗ 1]
              + [D, 1⊗∧∧⊗ 1] 
    
              + ∇̃^∧∧_x(F) ;
      
    [.9]
      d_3^∨(1 ⊗∧⊗  1) 
            = - [x, 1⊗∧∧⊗ 1]
              - [E, 1⊗∧∧⊗ 1] 
    
              - (r+1)⊗∧∧⊗ 1 ;
      
    [.9]
      d_3^∨(1 ⊗∧⊗  1) 
            = - [x, 1⊗∧∧⊗ 1]
              + [D, 1⊗∧∧⊗ 1] 
    
              + ∇̃^∧∧_y(F) ;
      
    
      d_3^∨(1⊗∧⊗  1) 
            = - [x, 1⊗∧∧⊗ 1] 
              - [y,1⊗∧∧⊗ 1] ;
      
    [.9]
      d_4^∨ (1 ⊗∧∧⊗ 1) 
            =   [E , 1 ⊗∧∧∧⊗ 1] 
    
              + (r+2) ⊗∧∧∧⊗ 1;
      
    
      d_4^∨ (1 ⊗∧∧⊗ 1) 
            = - [D , 1 ⊗∧∧∧⊗ 1]
              - ∇̃^∧∧∧_y(F);
      
    
      d_4^∨( 1⊗∧∧⊗ 1) 
            =   [y, 1⊗∧∧∧⊗ 1] ;
      
    
      d_4^∨ ( 1⊗∧∧⊗ 1) 
            = - [x, 1 ⊗∧∧∧⊗ 1],

where each ∇̃^u_x is the image of ∇^u_x under the map
a⊗ u⊗ b↦ b⊗ u⊗ a, and the same with each
∇̃^u_y.

Let us now identify ⊗_A A_σ with  as vector spaces,
remembering that the bimodule structure on  with this identification is
given by a▹ x◃ b=axσ(b) for all a, b∈
A and all x∈. There is a morphism of complexes of A-bimodules
ψ : ^∨→⊗_AA_σ such that
  
    ψ(1⊗∧∧∧⊗ 1) 
            = 1⊗ 1 ; 
    ψ(1⊗∧∧⊗ 1)
            = -1⊗ x⊗ 1 ; 
    ψ( 1⊗∧∧⊗ 1)
          = 1⊗ y⊗ 1 ; 
    ψ( 1⊗∧∧⊗  1)
          = -1⊗ D⊗ 1 - ξ ; 
    ψ( 1⊗∧∧⊗ 1)
          =  1 ⊗ E⊗  1 ; 
    ψ( 1⊗∧⊗ 1)
          = -1 ⊗  x∧ y ⊗  1 ; 
    ψ( 1 ⊗∧⊗ 1)
          = 1⊗ y∧ E⊗ 1 ; 
    ψ( 1 ⊗∧⊗  1)
          = - 1⊗ x∧ E⊗ 1 ; 
    ψ( 1 ⊗∧⊗  1)
          = 1 ⊗  x∧ D ⊗  1 + x∧ξ ; 
    ψ( 1 ⊗∧⊗  1)
          = - 1⊗  y∧ D⊗ 1 + ζ ; 
    ψ( 1 ⊗∧⊗  1)
          = -1⊗ D∧ E⊗ 1 - ξ∧ E ; 
    ψ( 1 ⊗⊗  1)
          = 1 ⊗  x∧ y∧ D ⊗  1; 
    ψ( 1 ⊗⊗  1)
          = - 1⊗ x∧ y∧ E⊗ 1 ; 
    ψ( 1 ⊗⊗  1)
          = 1⊗ x∧ D∧ E⊗ 1 + x∧ξ∧ E; 
    ψ( 1 ⊗⊗  1)
          = -1⊗ y∧ D∧ E⊗ 1 + ζ∧ E ; 
    ψ( 1⊗   1)
          = 1⊗ x∧ y∧ D∧ E⊗ 1,

where ξ∈ A⊗ V⊗ A and ζ∈ A⊗Λ^2V⊗
A are chosen so that
  
    d_1(ξ)
        = ∇̃_y(F) -1|F_y,
             d_2(ζ)
        =  ξ y - y ξ - 1 | y |F_y - ∇̃_x^x(F) + ∇(F).

That there are elements which satisfy these two conditions follows
immediately from the exactness of the Koszul resolution of S as an
S-bimodule —indeed, the right hand sides of the two conditions are
cycles in that complex— but we can exhibit a specific choice:
if we write F=∑_a+b=r+1c_ax^ay^b, with
c_0, …, c_r-1∈, then we can pick
  
    ξ =∑_a+b=r+1
    s+t+1=b-1(t+1) c_a y^s|y|x^a y^t,
            ζ = ∑_a+b=r+1
    s+t+1=b
    s'+t'+1=a
            c_ax^s'y^s|x∧ y|x^t'y^t.

That these formulas for ψ do indeed define a morphism of complexes
follows from a direct computation and it is easy to see that it is in fact
an isomorphism, as for an appropriate ordering of the bases of the
bimodules involved the matrices for the components of ψ are upper
triangular. Of course, it therefore induces an isomorphism in cohomology
and, since A_σ is A-projective on the left, we conclude that there
are isomorphisms of A-bimodules
  
    H^i(^∨) 
        ≅ H^i(⊗_A A_σ)
        ≅
          A_σ   if i=4;
    
          0    if i>0.

This completes the proof.
[4]




§ AUTOMORPHISMS, ISOMORPHISMS AND NORMAL ELEMENTS


  
 Our next objective is to compute the group of automorphisms of
the algebraA. We start by describing some graded automorphisms ofA. Later
we will see that these are, in fact, all the graded automorphisms of
our algebra, and that together with the exponentials of locally-nilpotent elements they generate the whole group(A).

lemma:iso:gr
If [ a b; c d ]∈_2()
and e∈^× are such that
  
    1/(ad-bc)eQ(ax+by,cx+dy) = Q(x,y),

and v∈ and ϕ_0∈ S_r,
then there is a homogeneous algebra automorphism θ:A→ A such that
  
    θ(x) = ax+by,
            θ(y) = cx+dy,
            θ(E) = E+v

and
  
    θ(D) =
        ϕ_0 - ebF/ax+byE + eD,     if b≠ 0; 
    ϕ_0+eD,     if not.

This is proved by a straightforward calculation. It should be noted that
the quotient appearing in the formula (<ref>) is always a
polynomial.


  
 Recall that a higher derivation ofAis a sequenced=(d_i)_i≥0of linear mapsA→ Asuch thatd_0=𝕀_Aand for alla,b∈ Aand alli≥0we have the higher Leibniz
identity
    d_i(ab) = ∑_s+t=i d_s(a)d_t(b).

It is clear that ifd=(d_i)_i≥0is a higher derivation andm≥0,
then the sequenced^[m]=(d_i^[m])_i≥0with
  
    d_i^[m] =
        
        d_i/m,     if i is divisible by m; 
    
        0,     if not

is also a higher derivation. On the other hand,
ifd=(d_i)_i≥0andd'=(d'_i)_i≥0are higher derivations
ofA, we can construct a new higher derivation(d”_i)_i≥0, which
we denoted∘ d', puttingd”_i = ∑_s+t=i d_s∘ d'_tfor alli≥0. Finally, ifδ:A→ Ais a derivation ofA, then the
sequence(1i!δ^i)_i≥0is a higher
derivation, which we denote byexp(δ); notice that this makes sense
because our ground fieldhas characteristic zero.

We letD(A)be the associative subalgebra of_(A)generated
by(A), and say that two higher derivationsd=(d_i)_i≥0andd'=(d'_i)_i≥0ofAare equivalent, and writed∼ d', if for
alli≥0the mapd_i-d'_iis in the subalgebra
of_(A)generated byD(A)andd_0, …,d_i-1; one can
check that this is indeed an equivalence relation on the set of higher
derivations.



  
 We recall the following very useful lemma from <cit.>:

lemma:ac
If d=(d_i)_i≥0 is a higher derivation of A, then d_i∈ D(A) for
all i≥0.

The result is an easy consequence of the fact that
  
    if d is a higher derivation of A and j≥1, then there
      exists a higher derivation d'=(d'_i)_i≥0 such that d'∼ d,
      d'_i=0 if 1<i<j, and d'_j is an element of (A).

To prove that this holds, let d=(d_i)_i≥0 and suppose there is an j≥1 such
that that d_i=0 if 1<i<j. The higher Leibniz identity implies that
d_j is an element of (A), and then we can consider the higher
derivation exp(-d_j)^[j]. We let d'=(d'_i)_i≥0 be the
composition exp(-d_j)^[j]∘ d. It is immediate that d∼ d'
and a simple computation shows that d'_i=0 if 1<i<j+1. The
claim (<ref>) follows inductively from this.

An element of A commutes with x and with y if and only if it belongs
to S.

The sufficiency of the condition is clear. To prove the necessity, let e∈
A be such that [x,e]=[y,e]=0.
There are an integer m≥0 and elements ϕ_0, …, ϕ_m in the
subalgebra generated by x, y and D in A such that
e=∑_i=0^mϕ_iE^i, and we have
0=[x,e_l]=∑_i=0^mϕ_iτ_1(E^i): this tells us that ϕ_i=0 if
i>0, and that e=ϕ_0. In particular, there are an integer n≥0
and elements ψ_0, …, ψ_n in S such that
e=∑_i=0^nψ_iD^i.
If i≥0 we have [D^i,y]≡ iFD^i-1⊕_j=0^i-2SD^j,
so that
  
    0 = [e,y] = ∑_i=0^nψ_i[D^i,y]
        ≡ nψ_nFD^n-1⊕_j=0^n-2SD^i.

Proceeding by descending induction we see from this that ψ_i=0 if
i>0, so that e=ψ_0∈ S.

If θ:A→ A is an automorphism of A such that for all i≥0 and
all a∈ A_i we have θ(a)∈ a+⊕_j>iA_j, then here
exists an f∈ S, uniquely determined up to the addition of a constant, such that
  
    θ(x) = x,
            θ(y) = y,
            θ(D) = D-Ff_y,
            θ(E) = E-[E,f].

Conversely, every f∈ S determines in this way an automorphism of A
satisfying that condition.

Let θ:A→ A be an automorphism of A as in the statement.
For each j≥0 there is a unique linear map θ_j:A→ A of
degree j such that for each i≥0 and each a∈ A_i the element
θ_j(a) is the (i+j)th homogeneous component of θ(a).
We have that for all a∈ A we have θ_j(a)=0 for j≥0 and
θ(a)=∑_j≥0θ_i(a) and, moreover, the sequence
(θ_j)_j≥0 is a higher derivation of A. In particular, it
follows from Lemma lemma:ac that
  
    θ_i∈ D(A) for all i≥0.

We know, from Proposition prop:hh, that (A)=S_rD̂⊕Ê⊕(A). If u is an irreducible factor of xF,
then (ϕD̂)(uA), Ê(uA) and [a,uA] are all contained
in uA for all ϕ∈ S_r and all a∈ A, and therefore
(<ref>) implies that that θ(uA)⊆ uA. As our
argument also applies to the inverse automorphism θ^-1, we have
θ^-1(uA)⊆ uA and, therefore, θ(uA)=uA. Since all units
of A are in , we see that θ(u)=u. Since of xF has two
linearly independent linear factors, we can conclude that θ(x)=x and
θ(y)=y.

Let θ(E)=E+e_1+⋯+e_l with e_i∈ A_i for each
i∈{1,…,l}. We have
  
    x=θ(x)=[θ(E),θ(x)]=[E,x]+[e_1,x]+⋯+[e_l,x]

and, by looking at homogeneous components, we see that [e_i,x]=0 for all
i∈{1,…,l} Similarly, [e_i,y]=0 for such i, and therefore
Lemma lemma:iso:xy tells us that e_1, …, e _l∈ S.

Suppose now that θ(D)=D+d_r+1+⋯+d_l with d_j∈ A_j for
each j∈{r+1,…,l}. Considering the equality
[θ(E),θ(D)]=rθ(D) we see that d_r+i=1iFe_iy
for each i∈{1,…,l}. Putting
f=-∑_i=1^l1ie_i, we obtain the first part of the lemma. The
second part follows from a direct verification.


  
 The automorphisms described in Proposition prop:iso:id
are precisely the exponentials of the inner derivations corresponding to
locally-nilpotent elements ofA. This  is a consequence of the
following result:

prop:local-ad-nilp
An element of A is locally -nilpotent if and only if it belongs
to S. If f∈ S, then the automorphism exp(f) maps x, y, D
and E to x, y, D-Ff_y and E-[E,f], respectively.

Suppose that e∈ A is a locally -nilpotent element. The kernel
(e) is a factorially closed subalgebra of A, so that whenever
a, b∈ A and (e)(ab)=0 we have (e)(a)=0 or (e)(b)=0;
see <cit.> for the proof of this in the commutative case, which
adapts to ours.

Since [x^iy^jD^kE^l,x]=-x^i+1y^jD^kτ_1(E^l) for all
i, j, k, l≥0, we have [A,x]⊆ xA and from this we see
immediately that [A,xA]⊆ xA. This implies that there is a
sequence (u_k)_k≥0 in A such that (e)^k(x)=xu_k for all
k≥0. Since e is locally -nilpotent, we can consider the
integer k_0=max{k∈_0:(e)^k(x)≠0}, and then we have
0≠ xu_k_0∈(e). As (e) is factorially closed, we see that
(e)(x)=0. In other words, the element e commutes with x.

There are an integer m≥0 and elements ϕ_0, …, ϕ_m in the
subalgebra generated by x, y and D in A such that
e=∑_i=0^mϕ_iE^i, and we have
0=[x,e]=∑_i=0^mϕ_iτ_1(E^i): this tells us that ϕ_i=0 if
i>0, and that e=ϕ_0. In particular, there are an integer n≥0
and elements ψ_0, …, ψ_n in S such that
e=∑_i=0^nψ_iD^i.

An induction shows that [D^i,F]∈ FA for all i≥0, and using this we
see that [e,F]=∑_i=0^nψ_i[D^i,F]∈ FA, from which it follows
that in fact [e,FA]⊆ FA. There is therefore a sequence
(v_i)_i≥0 of elements of A such that (e)^i(F)=Fv_i for all
i≥0. The local nilpotence of the map (e) allows us to consider the integer
  
    i_0=max{i∈_0:(e)^i(F)≠0},

and then 0≠
Fv_i_0∈(e). If ax+by is any of the factors of F, we have
b≠0 and ax+by∈(e): clearly, this implies that y commutes
with e. 

In view of Lemma lemma:iso:xy, we see that e∈ S:
this proves the necessity of the condition
for local -nilpotency given in the lemma. Its sufficiency is a direct
consequence of the fact that the graded algebra associated to the
filtration on A described in p:start is commutative. Finally, the
truth of the last sentence of the proposition can be verified by
an easy computation.


  
 We write_0(A)the set all automorphisms ofAdescribed
in Lemma lemma:iso:gr, and(A)the set of all automorphisms
ofAdescribed in Proposition prop:iso:id; they are subgroups of
the full group of automorphisms(A).


The group (A) is the semidirect product _0(A)⋉(A),
corresponding to the action of _0(A) on (A) given by
  
    θ_0 ·exp(f) = exp(θ^-1(f))

for all θ_0∈_0(A) and f∈ S. The subgroup _0(A) is
precisely the set of automorphisms of A preserving the grading and
(A) is the set of exponentials of locally nilpotent inner
derivations of A.


Notice that the action described in this statement makes sense, asθ_0(S)=Swheneverθ_0belongs to_0(A).


Let θ:A→ A be an automorphism and let us write
θ(E)=e_0+⋯+e_l,
θ(x)=x_0+⋯+x_l,
θ(y)=e_0+⋯+y_l,
θ(D)=d_0+⋯+d_l with e_i, x_i, y_i, d_i∈ A_i for each
i∈{0,…,l}. Since θ is an automorphism, we have
  
    [θ(E),θ(x)] = θ(x),
             [θ(E),θ(y)] = θ(y),
             [θ(E),θ(D)] = rθ(D).

Looking at the degree zero parts of these equalities, and remembering that
A_0 is a commutative ring, wee see x_0=y_0=d_0=0. As θ(x)≠0,
we can consider the number s=min{i∈_0:x_i≠0} and we have
s>0. Looking that the component of degree s of the first equality
in (<ref>), we see that [e_0,x_s]=x_s. This means that the
restriction (e_0):A_s→ A_s has a nonzero fixed vector. Now A_s as
a right [E]-module is free with basis {x^iy^jD^k:i+j+rk=s}, the
map (e_0) is right [E]-linear, and coincides with right
multiplication by -τ_s(e_0) on A_s. Clearly, the existence of nonzero
fixed vector implies that -τ_s(e_0)=1, so that e_0=uE+v for some
u∈^× and v∈ with su=1.
Putting now s'=min{i∈_0:y_i≠0} and
s”=min{u∈_0:d_i≠0} and looking at the components in the least
possible degree in the second and third equations of (<ref>),
we find that s'u=1 and s”u=r. In particular, s=s' and s”=rs.

Suppose for a moment that s>1. As θ(x), θ(y) and
θ(D) are in the ideal (A_s) generated by A_s, the composition
q:A→ A of θ with the quotient map A→ A/(A_s) is a surjection
such that q(A_0)=A/(A_s). This is impossible, as A_0 is a commutative
ring and A/(A_s) is not: we therefore have s=1 and, as a consequence,
u=1.

There exist a, b, c, d∈[E] such that x_1=xa+yb and
y_1=xc+yd. The four elements θ(E), θ(x), θ(y)
and θ(D) generate A and, as θ(D) is in ⊕_i≥
rA_i, the elements x and y are in the subalgebra generated by the first three.
It follows at once that x, y∈ x_1[E]+y_1[E] and, therefore,
that [ a b; c d ]∈_2([E]).

Let us write f∈[E]↦f⃗∈[E] the unique algebra morphism
such that E⃗=E+1.
We have [θ(x),θ(y)]=0 and in degree 2 this tells us that
  
    x^2(ac⃗-a⃗c)
          + xy T
          + y^2(bd⃗-b⃗d)
          = 0,

so that 
  
    ac⃗=a⃗c,
             bd⃗=c⃗d.

Suppose that
a is not constant. As the characteristic of  is zero (and possibly
after replacing  by an algebraic extension, which does not change
anything) there is then a ξ∈ such that
a(ξ)=0 and a⃗(ξ)=a(ξ+1)≠0, and the first
equality in (<ref>) implies that c(ξ)=0. The determinant of
[ a b; c d ] is thus divisible by
E-ξ, and this is impossible. Similarly, we find that all of
b, c, d must be constant.

Since d_r∈ A_r, there exist k≥0, ϕ_0, …, ϕ_k∈ S_r and
h∈[E] such that d_r=∑_i=0^kϕ_iE^i+Dh. The component of
degree r+1 of [θ(D),θ(x)] is
  
    0  
        = [d_r,x_1]
        = - ∑_i=0^k(ax+by)ϕ_iτ_1(E^i)
          - (ax+by)Dτ_1(h)
          + bFh⃗.

We thus see that h is constant, that ϕ_i=0 if
i≥2, and that 
  
    (ax+by)ϕ_1+bhF=0.

If b=0, then ϕ_1=0, and if
instead b≠0, then either h≠0 and we see that ax+by divides F and that
ϕ_1=-bhF/(ax+by), or h=0 and ϕ_1=0. 
In any case, we see that 
  
    d_r =
        ϕ_0 - hbF/ax+byE + hD,     if b≠ 0; 
    ϕ_0+hD,     if not.

Finally, the component of degree r+1 of the equality
[θ(D),θ(y)]=θ(F) tells us that
  
    F(ax+by,cx+dy) = (ad-bc)hxF/ax+by.

It follows now from Lemma lemma:iso:gr that there is a graded automorphism
θ_0:A→ A such that θ_0(x)=ax+by, θ_0(y)=cx+dy,
θ_0(E)=E+v and θ_0(D)=d_r.
The composition θ_0^-1∘θ satisfies the hypothesis of
Proposition prop:iso:id, and then there exists an f∈ S such that
θ=θ_0∘exp(f). This shows that
(A)=_0(A)·(A). Moreover, if θ is a graded
automorphism, then so is exp(f)=θ_0^-1∘θ and,
since it maps E to E-[E,f],
this is possible if and only if f∈, that is, if and only if
exp(f)=𝕀_A; this proves the last claim of the theorem.

Finally, computing the action of both sides of the equation on the generators
of A, we see that
  
    exp(f)∘θ_0 = θ_0∘exp(θ^-1(f))

for all f∈ S and all θ_0∈_0(A), and this tells us that (A) is
indeed a semidirect product _0(A)⋉(A).


  
 As usual, we say that an elementuofAis
normal ifuA=Au. Such an element, since it is not a
zero-divisor, determines an automorphismθ_u:A→ Auniquely by the
condition thatua=θ_u(a)ufor allu∈ A.


Let Q=α_0⋯α_r+1 be a factorization of Q as
a product of linear factors. The set of non-zero normal elements of A is
  
    (A) = 
      {λα_0^i_0⋯α_r+1^i_r+1:
            λ∈^×,i_0,…,i_r+1∈_0}.

This set is the saturated multiplicatively closed subset of A or of S
generated by Q.

A direct computation shows that each of the factors
α_0, …, α_r+1 of Q is normal in A, so the
set (A) is contained in the set of normal elements of A, for the
latter is multiplicatively closed. The set (A) is 
multiplicatively closed and it is saturated because S is closed under
divisors in A, and it is clear that as a saturated multiplicatively
closed it is generated by Q. To conclude the proof, we have to show
that every non-zero normal element of A belongs to (A).

Let u be a normal element in A and let θ_u:A→ A be the
associated automorphism, so that ua=θ_u(a)u for all a∈ A.
There are k, l∈_0 with k≤ l and elements u_k, … u_l∈
A such that u=u_k+⋯+u_l, u_i∈ A_i if k≤ i≤ l, and
u_k≠0≠ u_l. Similarly, there are s, t∈_0 with s≤ t
and elements e_s, …, e_t∈ A such that
θ_u(E)=e_s+⋯+e_t, e_i∈ A_i if s≤ i≤ t, and
e_s≠0≠ e_t. As we have
  
    u_kE+⋯+u_lE
        = uE
        = θ_u(E)u 
        = e_su_k + ⋯ + e_tu_l

with u_kE, u_lE, e_su_k and e_tu_l all non-zero, looking at the
homogeneous components of both sides we see that s=t=0. This means that
θ_u(E)=f(E)∈[E], and therefore the above equality is really of the
form
  
    u_kE + ⋯ + u_lE = f(E)u_k + ⋯ + f(E)u_l.

It follows from this that u_iE=f(E)u_i=u_if(E+i) for all
i∈{k,…,l} and therefore that E=f(E+k) and that E=f(E+l).
Since our ground field has characteristic zero, this is only possible if
k=l: the element u is homogeneous of degree l.

Now, since ua=θ_u(a)u for all a∈ A, the homogeneity of u
implies immediately that θ_u is a homogeneous map. There are
n∈_0 and ϕ_0, …, ϕ_n in the subalgebra of A
generated by x, y and D, such that ϕ_n≠0 and 
u=∑_i=0^nϕ_i E^i. As
θ_u(x) has degree 1, it belongs to S_1 and we have
  
    θ_u(x)∑_i=0^nϕ_iE^i
        = θ_u(x)u
        = ux
        = ∑_i=0^nϕ_iE^ix
        = x∑_i=0ϕ_i(E+1)^i.

Considering only the terms that have E^n as a factor we see that
θ_u(x)=x, and then the equality tells us that in fact
∑_i=0^nϕ_iE^i=∑_i=0ϕ_i(E+1)^i. Looking now at the terms
which have E^n-1 as a factor here we see that moreover n=0, so that
u∈[x,y,D]. There exist then m∈_0 and
ψ_0, …, ψ_m∈ S such that ψ_m≠0 and
u=∑_i=0^mψ_iD^i. As θ_u(y) has degree 1, it belongs
to S_1 and we have
  
    θ_u(y)∑_i=0^mψ_iD^i
        =θ_u(y)u
        = uy
        = ∑_i=0^mψ_iD^iy
        = ∑_i=0^myψ_iD^i + ∑_i=0^mψ_i[D^i,y].

Comparing the terms that have D^m as a factor we conclude that also
θ_u(y)=y.

As θ_u fixes x and y, the element u commutes with x and y,
and Lemma lemma:iso:xy allows us to conclude that u is in S_l.
Moreover, we know that all homogeneous automorphisms of A are those
described in Lemma lemma:iso:gr, so there exist ϕ∈ S_r and
e∈^× such that θ_u(D)=ϕ+eD. We then have that
  
    uD 
        = θ_u(D)u 
        = (ϕ+eD)u
        = ϕ u + euD + eu_yF

and this implies that e=1 and ϕ u+u_yF=0. Suppose now that α
is a linear factor of u and let k∈ and v∈ S be such that
u=α^kv and v is not divisible by α. The last equality
becomes ϕα^kv+kα^k-1α_yvF+α^kv_yF=0 and implies
that α divides α_yF: this means that α is a non-zero
multiple of x or a linear factor of F. As u can be factored as a
product of linear factors, we can therefore conclude that u belongs to
the set described in the statement of the proposition.


  
 There is a close connection between normal elements, the
first Hochschild cohomology space that we computed in
Section <ref> and the modular automorphisms ofA.

prop:conn
Let Q=α_0⋯α_r+1 be a factorization of Q as
a product of linear factors. 

  *  Every linear combination of the derivations
∂_α_0, …, ∂_α_r+1:A→ A
described in Proposition prop:partial is locally nilpotent.


  *  If u=λα_0^i_0⋯α_r+1^i_r+1, with
λ∈^× and i_0, …, i_r+1∈_0, is a normal
element of A, then the automorphism θ_u:A→ A associated to u
is
  
    θ_u = exp(-∑_j=0^r+1i_j∂_α_j).

This automorphism is such that θ_u(f) = f for all f∈ S and
  
    θ_u(δ) = δ + δ(u)/u

for all δ∈().


  *  The modular automorphism σ:A→ A described in
Proposition prop:CY coincides with the automorphism
θ_Q associated to the normal element Q.



  
 Another immediate application of the determination of the set of
normal elements is the classification under isomorphisms of our algebras.


Let  and ' be two central arrangements of lines in Å^2.
The algebras D() and D(') are isomorphic
if and only if the arrangements  and ' are isomorphic.

The sufficiency of the condition being obvious, we prove only its
necessity. We will denote with primes the objects associated to the
arrangement ', so that for example A'=D(') and so on. Moreover, in
view of the sufficiency of the condition we
can suppose without loss of generality that both arrangements 
and ' contain the line with equation x=0.

Let us suppose that there is an isomorphism of algebras ϕ:A→ A'.
Since ϕ maps locally -nilpotent elements to locally
-nilpotent elements, it follows from
Proposition prop:local-ad-nilp that ϕ(S)=S' and therefore that
ϕ restricts to an isomorphism of algebras ϕ:S→ S'. On the other
hand, ϕ also maps normal elements to normal elements, so that ϕ
restricts to a monoid homomorphism ϕ:(A)→(A'). 
Let Q=α_0⋯α_r+1 and Q'=α'_0⋯α'_r'+1 be
the factorizations of Q and of Q' as products of linear factors.
The invertible elements of the monoid (A) are the units of  and
the quotient (A)/^× is the free abelian monoid generated by
(the classes of) α_0, …, α_r+1 and, of course, a similar
statement holds for the other arrangement. Since ϕ induces an
isomorphism (A)/^×→(A')/^× we see, first, that r=r'
and, second, that there are a permutation π of the set {0,…,r+1}
and a function λ:{0,…,r+1}→^× such that
ϕ(α_i)=λ(i)α'_π(i) for all i∈{0,…,r+1}.
As there are at least two lines in each arrangement, this implies that the
restriction ϕ|_S:S→ S' restricts to an isomorphism of vector spaces
ϕ:S_1→ S'_1, so that ϕ|_S is linear, and that ϕ(Q)=Q'.
It is clear that this implies that the arrangements  and ' are
isomorphic.


  
 A simple and final observation that we can make at this point is that our
algebraAand the full algebra(S)of regular differentials operators
ofSare birational, that is, that they have the same fields of
quotients. In fact, the two algebras become isomorphic already after localization
at a single element:

prop:birational
The inclusion A→(S) induces after localization at Q an isomorphism
A[1/Q]→(S)[1/Q] and, in particular, A and (S)
have isomorphic fields of fractions.


That both localizations actually exist follows from the usual
characterization of quotient rings; see, for example, <cit.>*Chapter 2.


Clearly the map A[1/Q]→(S)[1/Q] induced by the
inclusion is injective, and it is surjective since S is contained in its
image as are ∂_y=1FD and ∂_x=1xE-yQD.

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