Another work in progress: a paper about nonhomogeneous quadratic duality etc.
Plan:
1. Quadratic algebras over base rings; quadratic duality
2. Koszul algebras over base rings
3. Nonhomogeneous quadratic rings; duality with CDG-algebras
4. Coalgebroids over rings; quasi-differential structures
5. The Poincare-Birkhoff-Witt theorem over a base ring
6. Derived categories D and D'
7. Koszul duality between the derived category of modules and that of quasi-differential comodules
8. General Koszul duality
Currently only sections 1-3 are written.
%%Organization: Independent University of Moscow
%%From: "Leonid Positselski"
%%Date: Sat, 24 Aug 2002 16:19:30 +0400 (MSD)
%%Subject: domega.tex
%%Lines: 451
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\begin{document}
\title{Nonhomogeneous Koszul Duality and \protect{$\D$--$\Omega$}}
\author{Leonid Positselski}
\address{Independent University of Moscow and
MPI f\"ur Mathematik, Bonn}
\email{posic@mccme.ru, posic@mpim-bonn.mpg.de}
\maketitle
I am writing the following notes mainly in order to give some
readable exposition to a number of results about nonhomogenenous
quadratic rings, derived categories of DG-modules, and Koszul
duality which only exist in my mind as of now.
The main goal of this small paper is to establish a general
algebraic framework, sufficient, e.~g., for proving
the equivalence of an appropriately defined derived category of
DG-modules over the de~Rham complex~$\Omega_X$ with the derived
category of $\D_X$\+modules---all of this over an arbitrary
smooth algebraic variety X (of any characteristic, actually)
or, if one wishes, over a smooth complex manifold.
\smallskip
The main reference is my old paper ``Nonhomogeneous Quadratic
Duality and Curvature'' published in \textit{Functional Analysis
and its Applications} vol.~\textbf{27}, \#3 in~1993.
The plan of what follows is here:
\medskip
\begin{enumerate}
\item Quadratic algebras over base rings; quadratic duality
\item Koszul algebras over base rings
\item Nonhomogeneous quadratic rings; duality with CDG-algebras
\item Coalgebroids over rings; quasi-differential structures
\item The Poincare--Birkhoff--Witt theorem over a base ring
\item Derived categories $D$ and~$D'$
\item Koszul duality between the derived category of modules
and that of quasi-differential comodules
\item General Koszul duality
\end{enumerate}
\Section{Quadratic Algebras over Base Rings; Quadratic Duality}
A graded ring $A = A_0 \op A_1 \op A_2 \op\dsb$ is called
\emph{quadratic} if it is generated by~$A_1$ over~$A_0$ with
relations in degree 2 only.
This means, in other words, that if I consider the graded ring
freely generated by $A_0$\+bimodule $A_1$, i.~e., the graded
ring~$T_{A_0,A_1}$ with components
$$
A_1\ot_{A_0} A_1 \ot_{A_0} \dsb \ot_{A_0} A_1,
$$
then the ring~$A$ will be a quotient ring of~$T_{A_0,A_1}$
by an ideal generated by a certain sub-bimodule~$R_A$ in
$A_1\ot_{A_0} A_1$.
I will always restrict myself to quadratic rings~$A$ which
satisfy some projectiveness, or at least flatness, conditions as
either left or right modules over~$A_0$. When I finally get to
Koszul rings, it will be simply that A is either left, or right
flat (or projective) module over~$A_0$. For general quadratic
rings, it is sensible to use somewhat weaker conditions, namely
as follows.
A quadratic ring~$A$ is called 2\+left projectively finite,
if both left $A_0$\+modules $A_1$ and~$A_2$ are projective
and finitely generated. A quadratic ring~$A$ is called 3\+left
projectively finite, if the same applies to~$A_1$, $A_2$,
and~$A_3$. Further conditions of this kind are not sensible
to consider for general quadratic rings.
\begin{proposition}
There is a duality (anti-equivalence) between the categories
of 2\+left projectively finite quadratic rings A and 2\+right
projectively finite quadratic rings B.
This duality sends 3\+left projectively finite quadratic rings
to 3\+right projectively finite quadratic rings.
The duality functor is defined by the formulas $B_0=A_0$,
$B_1 = \Hom_{A_0\lef}(A_1, A_0)$,
$A_1 = \Hom_{B_0\righ}(B_1, B_0)$,
$B_2 = \Hom_{A_0\lef}(R_A, A_0)$,
$R_B = \Hom_{A_0\lef}(A_2, A_0)$, and vise versa.
\end{proposition}
The proof of Proposition is based on the following lemma about
commutation of ``bimodule duality'' with bimodule tensor product:
\begin{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 as
a left $S$\+module.
Then there is a natural isomorphism of $(T,\.R)$\+bimodules
$$
\Hom_{S\lef}(N,\.S) \ot_S \Hom_{R\lef}(M,\.R) \rarrow
\Hom_{R\lef}(M\ot_S N, \,R).
$$
\end{lemma}
\begin{proof}[Proof of Lemma]
The map is defined by the formula $(f\ot g)\.(m\ot n) = g(mf(n))$.
In order to prove that this is an isomorphism, one should forget
about the $T$\+module structures (which aren't used in the
construction of the map) and then reduce the question to
the trivial case $N=S$.
\end{proof}
The only difference between this duality result and the familiar
commutative one is that noncommutative factors switch their order
under duality; simbolically writing, we have
$(M\ot N)^* = N^*\ot M^*$.
\begin{proof}[Proof of Proposition]
Here it remains to notice that for any 2\+left projectively finite
quadratic ring~$A$ the left $A_0$\+module $R_A$ is projective and
finitely generated, and indeed all modules in the exact triple
$$
0 \rarrow R_A \rarrow A_1\ot_{A_0} A_1 \rarrow A_2 \rarrow 0
$$
are left-projective and finitely generated over~$A_0$. So after
applying $\Hom_{A_0\lef}(-, \.A_0)$ we also get an exact triple.
To prove that 3\+left projectively finite rings correspond to
3\+right projectively finite ones, put $S_A = R_A\ot_{A_0} A_1
\cap A_1\ot_{A_0} R_A$ and consider the following exact
quadruple:
$$
0 \rarrow S_A \rarrow R_A\ot_{A_0}A_1 \op A_1\ot_{A_0}R_A
\rarrow A_1\ot_{A_0}A_1\ot_{A_0} A_1 \rarrow A_3 \rarrow 0.
$$
We see that $S_A$~is a left-projective and finitely generated
$A_0$\+module for any 3\+left projectively finite quadratic
algebra~$A$, and after dualization one has
$B_3 = \Hom_{A_0\lef}(S_A, \.A_0)$.
\end{proof}
The analogous arguments in higher degrees don't work for general
quadratic algebras, because the corresponding longer complexes
don't have to be exact anymore.
In fact, they are exact if and only if the quadratic algebra
is Koszul.
\Section{Koszul Algebras over Base Rings}
It is strange, I think, that in the standard nowadays papers on
Koszul algebras (like A.~Beilinson, V.~Ginzburg, V.~Schechtman
"Koszul duality" or A.~Beilinson, V.~Ginzburg, W.~Soergel "Koszul
duality in representation theory") the authors restrict themselves
to the case when $A_0$~is a semi-simple algebra over a field.
In fact, the basic results about Koszul algebras can be proven
for an arbitrary base ring, provided that an appropriate flatness
condition is imposed on~$A$.
\begin{definition}
A graded ring $A = A_0 \op A_1 \op A_2 \op\dsb$ is called
\emph{Koszul} if it flat as either left or right $A_0$\+module
and one has $\Tor^A_{i,j}(A_0,\.A_0) = 0$ for all $i \ne j$.
Here $A_0$~is endowed with left and right $A$\+module structures
via the augmentation map and the second grading~$j$ on the Tor
spaces comes from the grading of the ring~$A$.
A Koszul ring~$A$ is called left/right (finitely) projective
Koszul, if it is a projective (locally finitely generated)
left/right $A_0$\+module.
\end{definition}
\begin{theorem}
If a quadratic ring $A$ is flat as eiher left or right
$A_0$\+module, then it is Koszul if and only if for each
grading\/~$n$ the lattice in the tensor product of\/ $n$~copies
of~$A_1$ over~$A_0$, generated by\/ $n-1$~submodules
$$
A_1^{\ot i-1}\ot_{A_0}R_A\ot_{A_0}A_1^{\ot n-i-1}
= A_1\ot_{A_0} \dsb \ot_{A_0} A_1 \ot_{A_0} R_A
\ot_{A_0} A_1 \ot_{A_0} \dsb \ot_{A_0} A_1
$$
is distributive.
That is, for any three submodules~$X$, $Y$, $Z$ that
can be obtained from the generating submodules by applying
the operations of sum and intersection one should have
$(X + Y) \cap Z = X\cap Z + Y\cap Z$.
Moreover, if $A$~is a left finitely-projective Koszul ring,
then the quadratic dual ring~$B$ to~$A$ is right
finitely-projective Koszul, and vise versa.
\end{theorem}
This theorem can be proven in a way very similar to the classical
proofs of Backelin's distributivity theorem for Koszul algebras
over a field. The only difference is that it is more convenient
to use the definition of Koszulity in terms of the bar-complex
computing the Tor (rather than the Koszul complex) in this case.
\Section{Nonhomogeneous Quadratic Rings; Duality with CDG-algebras}
\subsection{Nonhomogeneous quadratic rings}
Let $A$~be a (3\+left projectively finite) quadratic ring and
$B$~be the dual (3\+right projectively finite quadratic) ring.
Suppose that we are given a filtered algebra~$\tA$ with
an increasing filtration $F_0\tA \sub F_1\tA \sub F_2\tA
\sub\dsb$ such that the associated graded ring $\gr_F \tA$
is identified with~$A$.
We will call such filtered rings \emph{nonhomogeneous quadratic
3\+left projectively finite rings}.
Which additional structure on the dual ring~$B$ corresponds to
the data of such a filtered ring~$\tA$\,?
The following key example illustrates the situation very well.
Let $M$~be a smooth affine algebraic variety (or a smooth real
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 natural filtration by the order of a differential
operator.
The associated graded ring $\gr_F \Diff_\E$ is isomorphic to
the ring of sections $A = \End(\E)\ot_{\O(M)} \Sym(\T_M)$ 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~$M$.
The ring $A$~is projective locally finitely generated and
quadratic over the base ring $A_0 = \End(\E)$.
The dual ring~$B$ is isomorphic to $\End(\E)\ot_{\O(M)}
\bigwedge \T^*(M)$, i.~e., the ring of exterior forms on~$M$ with
values in $\End(\E)$.
The induced additional structure on~$B$ will be defined as a set
of data up to a certain equivalence relation.
Namely, let us choose a connection~$\nabla$ in the vector
bundle~$\E$.
Then we have the curvature element $\omega \in B_2$ and
the de~Rham differential $d_\nabla$ acting in~$B$
so that $(d_\nabla)^2(x) = [\omega, x]$ and $d_\nabla(\omega)=0$.
When we change the connection, the operator~$d_\nabla$ and the
element~$\omega$ are replaced according to the well-known rules:
if $\nabla' = \nabla+\alpha$, then $d_{\nabla'}(x) = d_{\nabla}(x)
+ [\alpha,x]$ and $\omega' = \omega + d_\nabla(\alpha) + \alpha^2$.
This is the situation which we generalize in the following definition.
\subsection{CDG-algebras}
A CDG-ring is a graded ring $B = \bigoplus_i B^i$ endowed with
an odd derivation $d: B^i \to B^{i+1}$ and a ``curvature element''
$\omega\in B^2$ such that $d^2(x) = [\omega,x]$ for all~$x\in B$
and~$d(\omega)=0$.
A morphism of CDG-rings $B \rarrow C$ is a pair $(f,\.\alpha)$,
where $f: B\rarrow C$ is a morphism of graded rings and
$\alpha$~is an element in~$C^1$ such that $d_C(f(x)) = f(d_B(x))+
[\alpha,x]$ (the supercommutator) for all~$x\in B$ and
$f(\omega_B) = \omega_C + d_C(\alpha) - \alpha^2$.
Composition of morphisms is defined by the rule
$(g,\.\beta)\.(f,\.\alpha) = (gf,\. \beta+g(\alpha))$.
Identity morphisms are $\id_{B,d,\omega} = (\id_B,\. 0)$.
The \emph{category of CDG-rings} is hereby defined.
Note that the natural functor from the category of DG-rings
to the category of CDG-rings is faithful, but not
fully faithful.
In other words, two DG-algebras may be isomorphic in
the category of CDG-algebras without being isomorphic
as DG-algebras.
\subsection{Duality functor}
There is a fully faithful duality functor (that is,
a contravariant embedding of full subcategory) from the category
of all nonhomogeneous quadratic 3\+left projectively finite rings
to the category of CDG-rings whose underlying graded ring is
3\+right projectively finite quadratic.
This functor is constructed as follows.
For each nonhomogeneous 3\+left projective quadratic ring let us
choose a complementary left $A_0$\+submodule~$V$ to the submodule
$A_0 = F_0 \tA$ in the left $A_0$\+module $F_1\tA$.
This can be done, because the quotient module $A_1 =
F_1\tA/F_0\tA$ is projective.
I should emphasize that the subset~$V$ will be only a left
$A_0$\+submodule in $\tA$, but not in general a right
$A_0$\+submodule.
Of course, the choice of~$V$ is an arbitrary one, but we will
see that changing this choice leads to a canonically
isomorphic CDG-algebra, so the duality functor is well-defined
(up to a uniquely defined isomorphism of functors, to be precise).
Since $V$~projects isomorphically into $A_1 = F_1\tA/F_0\tA$,
it is endowed with a structure of $(A_0,\.A_0)$\+bimodule.
The map $V\rarrow \tA$ is only a morphism of left $A_0$\+modules,
however.
The right actions of~$A_0$ on~$V$ and~$\tA$ are compatible
modulo~$F_0\tA$ only.
Put $q(v,k) = \mu(v,k)-vk$, where $\mu(v,k)$ is the product
in~$\tA$ and $vk$~denotes the right action of~$A_0$ on~$V$.
This defines a map $q\: V\ot_\Z A_0 \rarrow A_0$.
Let $\tilde R$ be the full preimage of~$R_A$ in $A_1\ot_\Z A_1$
(recall that $R_A$~is a submodule in the tensor product of
the same modules over~$A_0$).
Equivalently (using the identification of $V$ with~$A_1$),
$\tilde R$ is the full preimage of $F_1\tA$ under
the multiplication map $\mu\: V\ot_\Z \nobreak V \rarrow \tA$.
Let us split the map $\mu\: \tilde R \rarrow F_1\tA$ into two
components $(e, -h)$ according to the direct sum decomposition
$F_1\tA = V + F_0\tA$, so that $e\: \tilde R \rarrow V$,
$\,h\: \tilde R \rarrow A_0$, and $\mu = e-h$.
The differentials $d_0\: B_0\rarrow B_1$ and $d_1\: B_1\rarrow
B_2$ are defined in terms of the maps $q$ and~$e$ by the formulas
$$
\< v,\. d_0(k)\> = q(v,k), \qquad
\< r,\. d_1(b)\> = \< e(\tilde r),\. b\> - q(\< \tilde r,\. b\>),
$$
where $\< \ ,\ \>$ denote the pairings of~$V$ with $B_1$ and of~$R$
with $B_2$, $\,\tilde r$ is any lifting of~$r$ in~$\tilde R$
(which has to be chosen for the summands in the right hand side to
make sense, even though the left hand side doesn't depend on it),
and $\< \tilde r,\. b\>$ is an element of $V \ot_\Z A_0$ obtained
by coupling the second component of~$\tilde r$ with $b\in B_1$.
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 through~$R$ and defines
the curvature element $\omega \in B_2$.
This is the whole of the construction of the dual quadratic
CDG-algebra, though one has to make quite some computations
in order to see that everything is well-defined and compatible.
In particular, the 3\+left projectivity will be actually used
in form of the duality between $S_A$ (where some self-consistency
equations on the defining relations of~$\tA$ live) and $B_3$
(where the equations $d^2(x)=[\omega,x]$, $d(\omega)=0$
are verified).
One cannot but feel a sense of mystery about the formulas that
make this strange and beautiful duality work.
\subsection{Augmented duality}
A map of left $A_0$\+modules $a\: \tA\rarrow A_0$ is called
a \emph{left augmentation} of the ring~$\tA$ over its
subring~$A_0$ if its restriction to~$A_0$ is the identity map
and its kernel is a left ideal in~$\tA$.
There is a bijective correspondence between left augmentation
maps $a\: \tA\rarrow A_0$ and structures of left $\tA$\+module
on~$A_0$ compatible with the obvious structure of left
$A_0$\+module on~$A_0$.
The above duality construction sends nonhomogeneous quadratic
algebras endowed with a left augmentation maps to quadratic
DG-algebras.
More precisely, it suffices to take for~$V$ the kernel of
the restriction of~$a$ to $F_1\tA$ to costruct a DG-ring
out of a left-augmented nonhomogeneous quadratic ring.
In other words, there is fully faithful duality functor from
the category of 3\+left projectively finite left-augmented
nonhomogeneus quadratic rings into the category of
nonnegatively graded DG-rings with differential increasing
the grading by~$1$ whose underlying graded ring is a 3\+right
projectively finite quadratic ring.
\Section{Coalgebroids over Rings; Quasi-Differential Structures}
\end{document}
1. Quadratic algebras over base rings; quadratic duality
2. Koszul algebras over base rings
3. Nonhomogeneous quadratic rings; duality with CDG-algebras
4. Coalgebroids over rings; quasi-differential structures
5. The Poincare-Birkhoff-Witt theorem over a base ring
6. Derived categories D and D'
7. Koszul duality between the derived category of modules and that of quasi-differential comodules
8. General Koszul duality
Currently only sections 1-3 are written.
%%Organization: Independent University of Moscow
%%From: "Leonid Positselski"
%%Date: Sat, 24 Aug 2002 16:19:30 +0400 (MSD)
%%Subject: domega.tex
%%Lines: 451
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\begin{document}
\title{Nonhomogeneous Koszul Duality and \protect{$\D$--$\Omega$}}
\author{Leonid Positselski}
\address{Independent University of Moscow and
MPI f\"ur Mathematik, Bonn}
\email{posic@mccme.ru, posic@mpim-bonn.mpg.de}
\maketitle
I am writing the following notes mainly in order to give some
readable exposition to a number of results about nonhomogenenous
quadratic rings, derived categories of DG-modules, and Koszul
duality which only exist in my mind as of now.
The main goal of this small paper is to establish a general
algebraic framework, sufficient, e.~g., for proving
the equivalence of an appropriately defined derived category of
DG-modules over the de~Rham complex~$\Omega_X$ with the derived
category of $\D_X$\+modules---all of this over an arbitrary
smooth algebraic variety X (of any characteristic, actually)
or, if one wishes, over a smooth complex manifold.
\smallskip
The main reference is my old paper ``Nonhomogeneous Quadratic
Duality and Curvature'' published in \textit{Functional Analysis
and its Applications} vol.~\textbf{27}, \#3 in~1993.
The plan of what follows is here:
\medskip
\begin{enumerate}
\item Quadratic algebras over base rings; quadratic duality
\item Koszul algebras over base rings
\item Nonhomogeneous quadratic rings; duality with CDG-algebras
\item Coalgebroids over rings; quasi-differential structures
\item The Poincare--Birkhoff--Witt theorem over a base ring
\item Derived categories $D$ and~$D'$
\item Koszul duality between the derived category of modules
and that of quasi-differential comodules
\item General Koszul duality
\end{enumerate}
\Section{Quadratic Algebras over Base Rings; Quadratic Duality}
A graded ring $A = A_0 \op A_1 \op A_2 \op\dsb$ is called
\emph{quadratic} if it is generated by~$A_1$ over~$A_0$ with
relations in degree 2 only.
This means, in other words, that if I consider the graded ring
freely generated by $A_0$\+bimodule $A_1$, i.~e., the graded
ring~$T_{A_0,A_1}$ with components
$$
A_1\ot_{A_0} A_1 \ot_{A_0} \dsb \ot_{A_0} A_1,
$$
then the ring~$A$ will be a quotient ring of~$T_{A_0,A_1}$
by an ideal generated by a certain sub-bimodule~$R_A$ in
$A_1\ot_{A_0} A_1$.
I will always restrict myself to quadratic rings~$A$ which
satisfy some projectiveness, or at least flatness, conditions as
either left or right modules over~$A_0$. When I finally get to
Koszul rings, it will be simply that A is either left, or right
flat (or projective) module over~$A_0$. For general quadratic
rings, it is sensible to use somewhat weaker conditions, namely
as follows.
A quadratic ring~$A$ is called 2\+left projectively finite,
if both left $A_0$\+modules $A_1$ and~$A_2$ are projective
and finitely generated. A quadratic ring~$A$ is called 3\+left
projectively finite, if the same applies to~$A_1$, $A_2$,
and~$A_3$. Further conditions of this kind are not sensible
to consider for general quadratic rings.
\begin{proposition}
There is a duality (anti-equivalence) between the categories
of 2\+left projectively finite quadratic rings A and 2\+right
projectively finite quadratic rings B.
This duality sends 3\+left projectively finite quadratic rings
to 3\+right projectively finite quadratic rings.
The duality functor is defined by the formulas $B_0=A_0$,
$B_1 = \Hom_{A_0\lef}(A_1, A_0)$,
$A_1 = \Hom_{B_0\righ}(B_1, B_0)$,
$B_2 = \Hom_{A_0\lef}(R_A, A_0)$,
$R_B = \Hom_{A_0\lef}(A_2, A_0)$, and vise versa.
\end{proposition}
The proof of Proposition is based on the following lemma about
commutation of ``bimodule duality'' with bimodule tensor product:
\begin{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 as
a left $S$\+module.
Then there is a natural isomorphism of $(T,\.R)$\+bimodules
$$
\Hom_{S\lef}(N,\.S) \ot_S \Hom_{R\lef}(M,\.R) \rarrow
\Hom_{R\lef}(M\ot_S N, \,R).
$$
\end{lemma}
\begin{proof}[Proof of Lemma]
The map is defined by the formula $(f\ot g)\.(m\ot n) = g(mf(n))$.
In order to prove that this is an isomorphism, one should forget
about the $T$\+module structures (which aren't used in the
construction of the map) and then reduce the question to
the trivial case $N=S$.
\end{proof}
The only difference between this duality result and the familiar
commutative one is that noncommutative factors switch their order
under duality; simbolically writing, we have
$(M\ot N)^* = N^*\ot M^*$.
\begin{proof}[Proof of Proposition]
Here it remains to notice that for any 2\+left projectively finite
quadratic ring~$A$ the left $A_0$\+module $R_A$ is projective and
finitely generated, and indeed all modules in the exact triple
$$
0 \rarrow R_A \rarrow A_1\ot_{A_0} A_1 \rarrow A_2 \rarrow 0
$$
are left-projective and finitely generated over~$A_0$. So after
applying $\Hom_{A_0\lef}(-, \.A_0)$ we also get an exact triple.
To prove that 3\+left projectively finite rings correspond to
3\+right projectively finite ones, put $S_A = R_A\ot_{A_0} A_1
\cap A_1\ot_{A_0} R_A$ and consider the following exact
quadruple:
$$
0 \rarrow S_A \rarrow R_A\ot_{A_0}A_1 \op A_1\ot_{A_0}R_A
\rarrow A_1\ot_{A_0}A_1\ot_{A_0} A_1 \rarrow A_3 \rarrow 0.
$$
We see that $S_A$~is a left-projective and finitely generated
$A_0$\+module for any 3\+left projectively finite quadratic
algebra~$A$, and after dualization one has
$B_3 = \Hom_{A_0\lef}(S_A, \.A_0)$.
\end{proof}
The analogous arguments in higher degrees don't work for general
quadratic algebras, because the corresponding longer complexes
don't have to be exact anymore.
In fact, they are exact if and only if the quadratic algebra
is Koszul.
\Section{Koszul Algebras over Base Rings}
It is strange, I think, that in the standard nowadays papers on
Koszul algebras (like A.~Beilinson, V.~Ginzburg, V.~Schechtman
"Koszul duality" or A.~Beilinson, V.~Ginzburg, W.~Soergel "Koszul
duality in representation theory") the authors restrict themselves
to the case when $A_0$~is a semi-simple algebra over a field.
In fact, the basic results about Koszul algebras can be proven
for an arbitrary base ring, provided that an appropriate flatness
condition is imposed on~$A$.
\begin{definition}
A graded ring $A = A_0 \op A_1 \op A_2 \op\dsb$ is called
\emph{Koszul} if it flat as either left or right $A_0$\+module
and one has $\Tor^A_{i,j}(A_0,\.A_0) = 0$ for all $i \ne j$.
Here $A_0$~is endowed with left and right $A$\+module structures
via the augmentation map and the second grading~$j$ on the Tor
spaces comes from the grading of the ring~$A$.
A Koszul ring~$A$ is called left/right (finitely) projective
Koszul, if it is a projective (locally finitely generated)
left/right $A_0$\+module.
\end{definition}
\begin{theorem}
If a quadratic ring $A$ is flat as eiher left or right
$A_0$\+module, then it is Koszul if and only if for each
grading\/~$n$ the lattice in the tensor product of\/ $n$~copies
of~$A_1$ over~$A_0$, generated by\/ $n-1$~submodules
$$
A_1^{\ot i-1}\ot_{A_0}R_A\ot_{A_0}A_1^{\ot n-i-1}
= A_1\ot_{A_0} \dsb \ot_{A_0} A_1 \ot_{A_0} R_A
\ot_{A_0} A_1 \ot_{A_0} \dsb \ot_{A_0} A_1
$$
is distributive.
That is, for any three submodules~$X$, $Y$, $Z$ that
can be obtained from the generating submodules by applying
the operations of sum and intersection one should have
$(X + Y) \cap Z = X\cap Z + Y\cap Z$.
Moreover, if $A$~is a left finitely-projective Koszul ring,
then the quadratic dual ring~$B$ to~$A$ is right
finitely-projective Koszul, and vise versa.
\end{theorem}
This theorem can be proven in a way very similar to the classical
proofs of Backelin's distributivity theorem for Koszul algebras
over a field. The only difference is that it is more convenient
to use the definition of Koszulity in terms of the bar-complex
computing the Tor (rather than the Koszul complex) in this case.
\Section{Nonhomogeneous Quadratic Rings; Duality with CDG-algebras}
\subsection{Nonhomogeneous quadratic rings}
Let $A$~be a (3\+left projectively finite) quadratic ring and
$B$~be the dual (3\+right projectively finite quadratic) ring.
Suppose that we are given a filtered algebra~$\tA$ with
an increasing filtration $F_0\tA \sub F_1\tA \sub F_2\tA
\sub\dsb$ such that the associated graded ring $\gr_F \tA$
is identified with~$A$.
We will call such filtered rings \emph{nonhomogeneous quadratic
3\+left projectively finite rings}.
Which additional structure on the dual ring~$B$ corresponds to
the data of such a filtered ring~$\tA$\,?
The following key example illustrates the situation very well.
Let $M$~be a smooth affine algebraic variety (or a smooth real
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 natural filtration by the order of a differential
operator.
The associated graded ring $\gr_F \Diff_\E$ is isomorphic to
the ring of sections $A = \End(\E)\ot_{\O(M)} \Sym(\T_M)$ 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~$M$.
The ring $A$~is projective locally finitely generated and
quadratic over the base ring $A_0 = \End(\E)$.
The dual ring~$B$ is isomorphic to $\End(\E)\ot_{\O(M)}
\bigwedge \T^*(M)$, i.~e., the ring of exterior forms on~$M$ with
values in $\End(\E)$.
The induced additional structure on~$B$ will be defined as a set
of data up to a certain equivalence relation.
Namely, let us choose a connection~$\nabla$ in the vector
bundle~$\E$.
Then we have the curvature element $\omega \in B_2$ and
the de~Rham differential $d_\nabla$ acting in~$B$
so that $(d_\nabla)^2(x) = [\omega, x]$ and $d_\nabla(\omega)=0$.
When we change the connection, the operator~$d_\nabla$ and the
element~$\omega$ are replaced according to the well-known rules:
if $\nabla' = \nabla+\alpha$, then $d_{\nabla'}(x) = d_{\nabla}(x)
+ [\alpha,x]$ and $\omega' = \omega + d_\nabla(\alpha) + \alpha^2$.
This is the situation which we generalize in the following definition.
\subsection{CDG-algebras}
A CDG-ring is a graded ring $B = \bigoplus_i B^i$ endowed with
an odd derivation $d: B^i \to B^{i+1}$ and a ``curvature element''
$\omega\in B^2$ such that $d^2(x) = [\omega,x]$ for all~$x\in B$
and~$d(\omega)=0$.
A morphism of CDG-rings $B \rarrow C$ is a pair $(f,\.\alpha)$,
where $f: B\rarrow C$ is a morphism of graded rings and
$\alpha$~is an element in~$C^1$ such that $d_C(f(x)) = f(d_B(x))+
[\alpha,x]$ (the supercommutator) for all~$x\in B$ and
$f(\omega_B) = \omega_C + d_C(\alpha) - \alpha^2$.
Composition of morphisms is defined by the rule
$(g,\.\beta)\.(f,\.\alpha) = (gf,\. \beta+g(\alpha))$.
Identity morphisms are $\id_{B,d,\omega} = (\id_B,\. 0)$.
The \emph{category of CDG-rings} is hereby defined.
Note that the natural functor from the category of DG-rings
to the category of CDG-rings is faithful, but not
fully faithful.
In other words, two DG-algebras may be isomorphic in
the category of CDG-algebras without being isomorphic
as DG-algebras.
\subsection{Duality functor}
There is a fully faithful duality functor (that is,
a contravariant embedding of full subcategory) from the category
of all nonhomogeneous quadratic 3\+left projectively finite rings
to the category of CDG-rings whose underlying graded ring is
3\+right projectively finite quadratic.
This functor is constructed as follows.
For each nonhomogeneous 3\+left projective quadratic ring let us
choose a complementary left $A_0$\+submodule~$V$ to the submodule
$A_0 = F_0 \tA$ in the left $A_0$\+module $F_1\tA$.
This can be done, because the quotient module $A_1 =
F_1\tA/F_0\tA$ is projective.
I should emphasize that the subset~$V$ will be only a left
$A_0$\+submodule in $\tA$, but not in general a right
$A_0$\+submodule.
Of course, the choice of~$V$ is an arbitrary one, but we will
see that changing this choice leads to a canonically
isomorphic CDG-algebra, so the duality functor is well-defined
(up to a uniquely defined isomorphism of functors, to be precise).
Since $V$~projects isomorphically into $A_1 = F_1\tA/F_0\tA$,
it is endowed with a structure of $(A_0,\.A_0)$\+bimodule.
The map $V\rarrow \tA$ is only a morphism of left $A_0$\+modules,
however.
The right actions of~$A_0$ on~$V$ and~$\tA$ are compatible
modulo~$F_0\tA$ only.
Put $q(v,k) = \mu(v,k)-vk$, where $\mu(v,k)$ is the product
in~$\tA$ and $vk$~denotes the right action of~$A_0$ on~$V$.
This defines a map $q\: V\ot_\Z A_0 \rarrow A_0$.
Let $\tilde R$ be the full preimage of~$R_A$ in $A_1\ot_\Z A_1$
(recall that $R_A$~is a submodule in the tensor product of
the same modules over~$A_0$).
Equivalently (using the identification of $V$ with~$A_1$),
$\tilde R$ is the full preimage of $F_1\tA$ under
the multiplication map $\mu\: V\ot_\Z \nobreak V \rarrow \tA$.
Let us split the map $\mu\: \tilde R \rarrow F_1\tA$ into two
components $(e, -h)$ according to the direct sum decomposition
$F_1\tA = V + F_0\tA$, so that $e\: \tilde R \rarrow V$,
$\,h\: \tilde R \rarrow A_0$, and $\mu = e-h$.
The differentials $d_0\: B_0\rarrow B_1$ and $d_1\: B_1\rarrow
B_2$ are defined in terms of the maps $q$ and~$e$ by the formulas
$$
\< v,\. d_0(k)\> = q(v,k), \qquad
\< r,\. d_1(b)\> = \< e(\tilde r),\. b\> - q(\< \tilde r,\. b\>),
$$
where $\< \ ,\ \>$ denote the pairings of~$V$ with $B_1$ and of~$R$
with $B_2$, $\,\tilde r$ is any lifting of~$r$ in~$\tilde R$
(which has to be chosen for the summands in the right hand side to
make sense, even though the left hand side doesn't depend on it),
and $\< \tilde r,\. b\>$ is an element of $V \ot_\Z A_0$ obtained
by coupling the second component of~$\tilde r$ with $b\in B_1$.
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 through~$R$ and defines
the curvature element $\omega \in B_2$.
This is the whole of the construction of the dual quadratic
CDG-algebra, though one has to make quite some computations
in order to see that everything is well-defined and compatible.
In particular, the 3\+left projectivity will be actually used
in form of the duality between $S_A$ (where some self-consistency
equations on the defining relations of~$\tA$ live) and $B_3$
(where the equations $d^2(x)=[\omega,x]$, $d(\omega)=0$
are verified).
One cannot but feel a sense of mystery about the formulas that
make this strange and beautiful duality work.
\subsection{Augmented duality}
A map of left $A_0$\+modules $a\: \tA\rarrow A_0$ is called
a \emph{left augmentation} of the ring~$\tA$ over its
subring~$A_0$ if its restriction to~$A_0$ is the identity map
and its kernel is a left ideal in~$\tA$.
There is a bijective correspondence between left augmentation
maps $a\: \tA\rarrow A_0$ and structures of left $\tA$\+module
on~$A_0$ compatible with the obvious structure of left
$A_0$\+module on~$A_0$.
The above duality construction sends nonhomogeneous quadratic
algebras endowed with a left augmentation maps to quadratic
DG-algebras.
More precisely, it suffices to take for~$V$ the kernel of
the restriction of~$a$ to $F_1\tA$ to costruct a DG-ring
out of a left-augmented nonhomogeneous quadratic ring.
In other words, there is fully faithful duality functor from
the category of 3\+left projectively finite left-augmented
nonhomogeneus quadratic rings into the category of
nonnegatively graded DG-rings with differential increasing
the grading by~$1$ whose underlying graded ring is a 3\+right
projectively finite quadratic ring.
\Section{Coalgebroids over Rings; Quasi-Differential Structures}
\end{document}