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posicDate: Wed, 2 Jan 2002 23:43:01 +0100 (CET)
From: Leonid Positselski
Subject: Re: quadratic Q-algebras
Dear Albert Solomonovich,
please excuse me for not replying to your question for so long.
I moved from Bonn to Stockholm since then, and I suspect that
I lost a part of your fax, and I cannot see what your problem
was from the 2 pages of fax that I have. I would be grateful
if you could resend the fax [...]
The correct construction is sketched below. Let me first say
that it should not depend on commutativity of A_0, because
it must include the following important example. Let M be
a smooth manifold (or an affine algebraic manifold) and E be
a vector bundle over it. Let Diff_E be the ring of differential
operators acting on the sections of E. It has a filtration by
the order of a differential operator such that the associated
graded ring gr_F Diff_E is isomorphic to End(E)\otimes S(T_M)
(the ring of sections of the graded vector bundle of associative
algebras which is the tensor product of the endomorphisms of E
and the symmetric algebra of the tangent bundle to M).
In particular, A_0 = End(E) for A = Diff_E. This is
noncommutative whenever dim E > 1. The dual Q-algebra to this
"nonhomogeneous quadratic algebra over End(E)" is defined up to
equivalence (replacing Q with Q + [a,-]); in order to obtain
a representative in this equivalence class, one has to choose
a connection in the bundle E. Then the corresponding Q-algebra
is End(E) \otimes \Omega^*(M) with derivation Q being the DeRham
differential corresponding to the chosen connection in E and the
element \omega \in End(E)\otimes\Omega^2(M) being the curvature
of this connection. This is the example one should keep in mind.
I continue with your notation: let A be the union of A_k = F_k A,
K = A_0 be the "base ring". I assume that
gr_F A = A_0 + A_1/A_0 + A_2/A_1 + ...
is a quadratic ring, projective and locally finite dimensional
as a left K-module. Let R \subset (A_1/A_0 \otimes_K A_1/A_0) be
the quadratic relations module, i.e., the kernel of the map to
A_2/A_1. In order to construct the dual Q-algebra, I will have
to choose a complementary left K-submodule V to A_0 in A_1.
This corresponds to choosing a connection in the above example.
Changing the submodule V will lead to an equivalent Q-algebra.
The dual Q-algebra will turn out to be a DG-algebra (i.e., with
\omega = 0) if and only if the left ideal generated by V in A
doesn't intersect A_0 ("no scalars in relations"). (If this is
the case, this left ideal is a linear complement to A_0 in A
and therefore A_0 obtains a structure of left A-module.)
First step is the construction of the quadratic algebra B dual
to gr_F A. It depends on the side, so it's actually better
to say that B is the left dual algebra to gr_F A. It will be
a quadratic graded algebra with the same 0-graded component
B_0 = K, projective over B_0 from the right. The right dual to
the left dual gives back the original algebra. By definition,
B_1 = Hom_{K-left}(A_1/A_0, K), B_2 = Hom_{K-left}(R, K),
and the multiplication map B_1 \otimes B_1 \to B_2 is the dual
map to the embedding of R into A_1/A_0 \otimes_K A_1/A_0. This
determines the quadratic algebra B.
Now of course V projects isomorphically to A_1/A_0, which makes
V a K-K-bimodule. Since V was chosen as a left K-submodule in
A_1, the multiplication map K \otimes_Z V \to A_1 lands in V
and coincides with the left action of K on V. This is not so
for the other multiplication map V \otimes_Z K \to A_1, which
coincides with the right action of K on V modulo A_0 only
(Z denotes the ring of integers). Put q(v,k) = \mu(v,k) - vk,
where \mu(v,k) is the multiplication in A_1 and vk denotes
the right action. This defines a map q: V \otimes K \to K.
Let R~ be the lifting (full preimage) of R in A_1/A_0 \otimes_Z
A_1/A_0 (recall that R is a submodule in the tensor product of
the same modules over K). Equivalently (using the identification
of V with A_1/A_0), R~ is the full preimage of A_1 \subset A
under the multiplication map \mu: V \otimes_Z V \to A. Let us
split the map \mu: R~ \to A_1 into two components (e, -h)
according to the isomorphism A_1 = V + K, so that:
e: R~ \to V, h: R~ \to K, \mu = e-h.
The differentials d_0: K=B_0 \to B_1 and d_1: B_1 \to B_2 are
defined in terms of the maps q and e by the formulas:
< v, d_0(k) > = q(v,k)
< r, d_1(b) > = < e(r~), b> - q(< r~,b >),
where < > denote the pairing of V with B_1 and R with B_2, r~ is
any lifting of r in R~ (which has to be chosen for the parts of
the right hand side to make sense, though the left hand side
doesn't depend on it), and < r~,b > is an element of V \otimes_Z K
obtained by coupling the second component of r~ with b. Note
that it's important to include the q-term in the second formula,
though you wouldn't find it anywhere in duality over a field!
The map h actually factors though R and defines the curvature
element \omega \in B_2. This is the whole of the construction
of the dual quadratic Q-algebra, though one has to make quite
some computations in order to see that everything is
well-defined and compatible.
In the next letter I will spell the construction of the duality
functor on modules. There are actually two such constructions,
one sending a left A-module to a left Q-module over B (with zero
curvature F in the sense of your paper, of course) and another
sending right modules to right modules. Note that B was defined
by taking the left dual quadratic algebra, so the two situations
are not at all symmetric! Indeed, the first functor acts by
tensoring its argument with B over K, while the second functor
tensors with Hom_{K-right}(B, K) = K + V + R + ... over K.
Best wishes and happy New Year.
Yours, Leonid.
Organization: Independent University of Moscow
From: "Leonid Positselski"
Date: Fri, 2 Aug 2002 00:37:18 +0400 (MSD)
Reply-To: posic @ mccme.ru
Subject: Re: quadratic Q-algebras
Lines: 78
Dear Albert Solomonovich,
Please excuse me for not replying to your letters for an even
longer period of time in this last case. I was extremely
distracted by my internet-efforts directed against the war in
Chechnya and other personal matters during all of this year
since last August. This is not my normal mode of operation.
In your letter from January 12 you asked me what I consider
an important and interesting question, the answer to which was
always known to me, however. Let me now try to explain you
this situation and how it works. Here is your question:
>> First step is the construction of the quadratic algebra B
>> dual to $gr_F A$. It depends on the side, so it's actually
>> better to say that B is the left dual algebra to $gr_F A$.
>> It will be a quadratic graded algebra with the same 0-graded
>> component $B_0 = K$, projective over $B_0$ from the right.
>> The right dual to the left dual gives back the original
>> algebra. By definition,
>> $$ B_1 = Hom_{K-left}(A_1/A_0, K),
>> B_2 = Hom_{K-left}(R, K), $$
>> and the multiplication map $B_1\otimes B_1 \to B_2$
>> is the dual map to the embedding of $R$ into
>> $A_1/A_0 \otimes_K A_1/A_0$.
WHEN YOU EMBED $R$ INTO $A_1/A_0 \otimes_K A_1/A_0$.
ONE OF THE FACTORS SHOULD BE CONSIDERED AS A LEFT
MODULE AND ANOTHER FACTOR AS A RIGHT ONE. IT SEEMS
THAT THIS CREATES SOME PROBLEMS. MAY BE I AM WRONG
AND THERE EXISTS A TRIVIAL WAY TO OVERCOME THIS
DIFFICULTY, BUT THIS IS THE POINT THAT WORRIES ME.
Here is the answer. You are right that one of the factors
in the tensor product above is considered as a left module
and another factor as a right one. This does not create
any problems, however, as duality (meaning either left or
right duality) for bimodules commutes with bimodule tensor
product just fine.
LEMMA. Let R, S, T be rings; M be an (R,S)-bimodule and
N be an (S,T)-bimodule. Suppose that N is projective and
finitely generated over S (as a left S-module, that is).
Than there is a natural isomorphism of (T,R)-bimodules
Hom_S(N, S) \otimes_S Hom_R(M, R)
\to Hom_R(M\otimes_S N, R).
(all the Hom's are meant as Hom's of left modules).
Proof: the map is defined by the formula
(f\otimes g) (m\otimes n) = g(mf(n)).
In order to prove that this is an isomorphism, one forgets
about the T-module structures (which aren't used in the
construction of the map) and then reduces the question to
the trivial case N=S.
The only difference between this duality result and the
familiar commutative one is that the noncommutative factors
switch their order under duality: simbolically writing,
we have (M\otimes N)^* = N^* \otimes M^*.
The following result about HOMOGENEOUS quadratic duality
over base rings follows immediately from this Lemma.
Let's consider nonnegatively graded algebras B defined
by quadratic relations over B_0 and such that B_1, B_2
are projective, finitely generated right B_0-modules.
Then the category of such algebras is anti-equivalent
(dual) to the category of nonnegatively graded algebras C
satisfying the same conditions from the left. The duality
is given by B_0=C_0, B_1=Hom_{C_0-left}(C_1, C_0), etc.
In the notation used above, algebras C would correspond
to gr_F A.
So please ask more questions, and I sincerely hope that it
will not take me SO long to answer them the next time...
With best wishes,
Leonid Positselski.
Organization: Independent University of Moscow
From: "Leonid Positselski"
Date: Tue, 6 Aug 2002 early in the morning
Subject: Re: quadratic Q-algebras (misprint corrected)
Dear Albert Solomonovich,
My phone numbers in Moscow are (095) 918-7709 (myself) or <...>
(my mother, where I am to be found some days also). You are
of course most welcome to call. I will stay in Moscow until some
September 20, and then from some October 1 I return to Bonn.
Let me repeat the general outline of the theory that we discussed.
Consider graded rings A = A_0 + A_1 + A_2 + etc., such that A is
defined by quadratic relations over A_0 and A_1, A_2, A_3 are
projective finitely generated left modules over A_0. Then under
the homogeneous quadratic duality over a base ring, the category
of such graded rings A is dual (anti-equivalent) to the category
of graded rings B = B_0 + B_1 + B_2 + etc., satisfying the same
conditions except that projectiveness and finite generateness
is from the right rather that from the left. Under this duality,
one has B_0 = A_0 and B_1 = Hom_{A_0-left}(A_1,A_0).
Now the nonhomogeneous duality is not in general an equivalence,
but only a fully faithful functor, or an embedding of a full
subcategory. The category of filtered rings A~ with increasing
filtration F and the associated graded ring A= gr_F A~
satisfying the above conditions on the algebra A is dual
(anti-equivalent) to a full subcategory of the category of
Q-algebras (B, Q, \omega) with an underlying graded ring B
satisfying the above conditions. Here is a result about this
fully faithful functor, which I believe I can prove, though
I've never actually done it, so I call it a "pretheorem".
Pretheorem: If one restricts to graded rings A and B that are
Koszul (in the appropriate sence, which is easy to define)
than the above nonhomogeneous duality functor becomes not only
fully faithful, but actually an equivalence of categories.
This result should be called "the Poincare-Birkhoff-Witt theorem
for Koszul algebras over noncommutative base rings".
Furthermore, there is a duality theory between modules over A~
and (B, Q, \omega) which can be made into an equivalence of
triangulated categories. If you wish, I can elaborate on this
question in a subsequent letter.
Best regards,
Leonid Positselski.
[...]
From: Leonid Positselski
Subject: Re: quadratic Q-algebras
Dear Albert Solomonovich,
please excuse me for not replying to your question for so long.
I moved from Bonn to Stockholm since then, and I suspect that
I lost a part of your fax, and I cannot see what your problem
was from the 2 pages of fax that I have. I would be grateful
if you could resend the fax [...]
The correct construction is sketched below. Let me first say
that it should not depend on commutativity of A_0, because
it must include the following important example. Let M be
a smooth manifold (or an affine algebraic manifold) and E be
a vector bundle over it. Let Diff_E be the ring of differential
operators acting on the sections of E. It has a filtration by
the order of a differential operator such that the associated
graded ring gr_F Diff_E is isomorphic to End(E)\otimes S(T_M)
(the ring of sections of the graded vector bundle of associative
algebras which is the tensor product of the endomorphisms of E
and the symmetric algebra of the tangent bundle to M).
In particular, A_0 = End(E) for A = Diff_E. This is
noncommutative whenever dim E > 1. The dual Q-algebra to this
"nonhomogeneous quadratic algebra over End(E)" is defined up to
equivalence (replacing Q with Q + [a,-]); in order to obtain
a representative in this equivalence class, one has to choose
a connection in the bundle E. Then the corresponding Q-algebra
is End(E) \otimes \Omega^*(M) with derivation Q being the DeRham
differential corresponding to the chosen connection in E and the
element \omega \in End(E)\otimes\Omega^2(M) being the curvature
of this connection. This is the example one should keep in mind.
I continue with your notation: let A be the union of A_k = F_k A,
K = A_0 be the "base ring". I assume that
gr_F A = A_0 + A_1/A_0 + A_2/A_1 + ...
is a quadratic ring, projective and locally finite dimensional
as a left K-module. Let R \subset (A_1/A_0 \otimes_K A_1/A_0) be
the quadratic relations module, i.e., the kernel of the map to
A_2/A_1. In order to construct the dual Q-algebra, I will have
to choose a complementary left K-submodule V to A_0 in A_1.
This corresponds to choosing a connection in the above example.
Changing the submodule V will lead to an equivalent Q-algebra.
The dual Q-algebra will turn out to be a DG-algebra (i.e., with
\omega = 0) if and only if the left ideal generated by V in A
doesn't intersect A_0 ("no scalars in relations"). (If this is
the case, this left ideal is a linear complement to A_0 in A
and therefore A_0 obtains a structure of left A-module.)
First step is the construction of the quadratic algebra B dual
to gr_F A. It depends on the side, so it's actually better
to say that B is the left dual algebra to gr_F A. It will be
a quadratic graded algebra with the same 0-graded component
B_0 = K, projective over B_0 from the right. The right dual to
the left dual gives back the original algebra. By definition,
B_1 = Hom_{K-left}(A_1/A_0, K), B_2 = Hom_{K-left}(R, K),
and the multiplication map B_1 \otimes B_1 \to B_2 is the dual
map to the embedding of R into A_1/A_0 \otimes_K A_1/A_0. This
determines the quadratic algebra B.
Now of course V projects isomorphically to A_1/A_0, which makes
V a K-K-bimodule. Since V was chosen as a left K-submodule in
A_1, the multiplication map K \otimes_Z V \to A_1 lands in V
and coincides with the left action of K on V. This is not so
for the other multiplication map V \otimes_Z K \to A_1, which
coincides with the right action of K on V modulo A_0 only
(Z denotes the ring of integers). Put q(v,k) = \mu(v,k) - vk,
where \mu(v,k) is the multiplication in A_1 and vk denotes
the right action. This defines a map q: V \otimes K \to K.
Let R~ be the lifting (full preimage) of R in A_1/A_0 \otimes_Z
A_1/A_0 (recall that R is a submodule in the tensor product of
the same modules over K). Equivalently (using the identification
of V with A_1/A_0), R~ is the full preimage of A_1 \subset A
under the multiplication map \mu: V \otimes_Z V \to A. Let us
split the map \mu: R~ \to A_1 into two components (e, -h)
according to the isomorphism A_1 = V + K, so that:
e: R~ \to V, h: R~ \to K, \mu = e-h.
The differentials d_0: K=B_0 \to B_1 and d_1: B_1 \to B_2 are
defined in terms of the maps q and e by the formulas:
< v, d_0(k) > = q(v,k)
< r, d_1(b) > = < e(r~), b> - q(< r~,b >),
where < > denote the pairing of V with B_1 and R with B_2, r~ is
any lifting of r in R~ (which has to be chosen for the parts of
the right hand side to make sense, though the left hand side
doesn't depend on it), and < r~,b > is an element of V \otimes_Z K
obtained by coupling the second component of r~ with b. Note
that it's important to include the q-term in the second formula,
though you wouldn't find it anywhere in duality over a field!
The map h actually factors though R and defines the curvature
element \omega \in B_2. This is the whole of the construction
of the dual quadratic Q-algebra, though one has to make quite
some computations in order to see that everything is
well-defined and compatible.
In the next letter I will spell the construction of the duality
functor on modules. There are actually two such constructions,
one sending a left A-module to a left Q-module over B (with zero
curvature F in the sense of your paper, of course) and another
sending right modules to right modules. Note that B was defined
by taking the left dual quadratic algebra, so the two situations
are not at all symmetric! Indeed, the first functor acts by
tensoring its argument with B over K, while the second functor
tensors with Hom_{K-right}(B, K) = K + V + R + ... over K.
Best wishes and happy New Year.
Yours, Leonid.
Organization: Independent University of Moscow
From: "Leonid Positselski"
Date: Fri, 2 Aug 2002 00:37:18 +0400 (MSD)
Reply-To: posic @ mccme.ru
Subject: Re: quadratic Q-algebras
Lines: 78
Dear Albert Solomonovich,
Please excuse me for not replying to your letters for an even
longer period of time in this last case. I was extremely
distracted by my internet-efforts directed against the war in
Chechnya and other personal matters during all of this year
since last August. This is not my normal mode of operation.
In your letter from January 12 you asked me what I consider
an important and interesting question, the answer to which was
always known to me, however. Let me now try to explain you
this situation and how it works. Here is your question:
>> First step is the construction of the quadratic algebra B
>> dual to $gr_F A$. It depends on the side, so it's actually
>> better to say that B is the left dual algebra to $gr_F A$.
>> It will be a quadratic graded algebra with the same 0-graded
>> component $B_0 = K$, projective over $B_0$ from the right.
>> The right dual to the left dual gives back the original
>> algebra. By definition,
>> $$ B_1 = Hom_{K-left}(A_1/A_0, K),
>> B_2 = Hom_{K-left}(R, K), $$
>> and the multiplication map $B_1\otimes B_1 \to B_2$
>> is the dual map to the embedding of $R$ into
>> $A_1/A_0 \otimes_K A_1/A_0$.
WHEN YOU EMBED $R$ INTO $A_1/A_0 \otimes_K A_1/A_0$.
ONE OF THE FACTORS SHOULD BE CONSIDERED AS A LEFT
MODULE AND ANOTHER FACTOR AS A RIGHT ONE. IT SEEMS
THAT THIS CREATES SOME PROBLEMS. MAY BE I AM WRONG
AND THERE EXISTS A TRIVIAL WAY TO OVERCOME THIS
DIFFICULTY, BUT THIS IS THE POINT THAT WORRIES ME.
Here is the answer. You are right that one of the factors
in the tensor product above is considered as a left module
and another factor as a right one. This does not create
any problems, however, as duality (meaning either left or
right duality) for bimodules commutes with bimodule tensor
product just fine.
LEMMA. Let R, S, T be rings; M be an (R,S)-bimodule and
N be an (S,T)-bimodule. Suppose that N is projective and
finitely generated over S (as a left S-module, that is).
Than there is a natural isomorphism of (T,R)-bimodules
Hom_S(N, S) \otimes_S Hom_R(M, R)
\to Hom_R(M\otimes_S N, R).
(all the Hom's are meant as Hom's of left modules).
Proof: the map is defined by the formula
(f\otimes g) (m\otimes n) = g(mf(n)).
In order to prove that this is an isomorphism, one forgets
about the T-module structures (which aren't used in the
construction of the map) and then reduces the question to
the trivial case N=S.
The only difference between this duality result and the
familiar commutative one is that the noncommutative factors
switch their order under duality: simbolically writing,
we have (M\otimes N)^* = N^* \otimes M^*.
The following result about HOMOGENEOUS quadratic duality
over base rings follows immediately from this Lemma.
Let's consider nonnegatively graded algebras B defined
by quadratic relations over B_0 and such that B_1, B_2
are projective, finitely generated right B_0-modules.
Then the category of such algebras is anti-equivalent
(dual) to the category of nonnegatively graded algebras C
satisfying the same conditions from the left. The duality
is given by B_0=C_0, B_1=Hom_{C_0-left}(C_1, C_0), etc.
In the notation used above, algebras C would correspond
to gr_F A.
So please ask more questions, and I sincerely hope that it
will not take me SO long to answer them the next time...
With best wishes,
Leonid Positselski.
Organization: Independent University of Moscow
From: "Leonid Positselski"
Date: Tue, 6 Aug 2002 early in the morning
Subject: Re: quadratic Q-algebras (misprint corrected)
Dear Albert Solomonovich,
My phone numbers in Moscow are (095) 918-7709 (myself) or <...>
(my mother, where I am to be found some days also). You are
of course most welcome to call. I will stay in Moscow until some
September 20, and then from some October 1 I return to Bonn.
Let me repeat the general outline of the theory that we discussed.
Consider graded rings A = A_0 + A_1 + A_2 + etc., such that A is
defined by quadratic relations over A_0 and A_1, A_2, A_3 are
projective finitely generated left modules over A_0. Then under
the homogeneous quadratic duality over a base ring, the category
of such graded rings A is dual (anti-equivalent) to the category
of graded rings B = B_0 + B_1 + B_2 + etc., satisfying the same
conditions except that projectiveness and finite generateness
is from the right rather that from the left. Under this duality,
one has B_0 = A_0 and B_1 = Hom_{A_0-left}(A_1,A_0).
Now the nonhomogeneous duality is not in general an equivalence,
but only a fully faithful functor, or an embedding of a full
subcategory. The category of filtered rings A~ with increasing
filtration F and the associated graded ring A= gr_F A~
satisfying the above conditions on the algebra A is dual
(anti-equivalent) to a full subcategory of the category of
Q-algebras (B, Q, \omega) with an underlying graded ring B
satisfying the above conditions. Here is a result about this
fully faithful functor, which I believe I can prove, though
I've never actually done it, so I call it a "pretheorem".
Pretheorem: If one restricts to graded rings A and B that are
Koszul (in the appropriate sence, which is easy to define)
than the above nonhomogeneous duality functor becomes not only
fully faithful, but actually an equivalence of categories.
This result should be called "the Poincare-Birkhoff-Witt theorem
for Koszul algebras over noncommutative base rings".
Furthermore, there is a duality theory between modules over A~
and (B, Q, \omega) which can be made into an equivalence of
triangulated categories. If you wish, I can elaborate on this
question in a subsequent letter.
Best regards,
Leonid Positselski.
[...]
