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\begin{document}

\begin{slide*}
\mbox{}
\vspace{2cm}

\begin{center}
{\large \bfseries Rigid Dualizing Complexes 
via Differential Graded Algebras}

Lecture Notes
%\footnote[1]{written: 14 Jan 2007}

\vspace{3ex}
Amnon Yekutieli\\
Ben Gurion University, ISRAEL

\texttt{\small http://www.math.bgu.ac.il/$\sim$amyekut}

\vspace{1ex}
{\tiny written: 14 Jan 2007}
\end{center}

\newslide
\vspace{2ex}

Here is the plan of my lecture:
\begin{enumerate}
\item[1.] Dualizing Complexes: Overview
\item[2.] Rigid Complexes and DG Algebras
\item[3.] Properties of Rigid Complexes
\item[4.] Rigid Dualizing Complexes
\item[5.] Rigid Complexes and CM Homomorphisms
\end{enumerate}



\newslide
\section{Dualizing Complexes: Overview}

Let $A$ be a noetherian commutative ring. Denote by 
$\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} A)$
the derived category of bounded complexes of $A$-modules 
with finitely generated cohomology modules. 

\begin{dfn} (Grothendieck \cite{RD}) \label{dfn1}
A {\em dualizing complex} over $A$ is a complex 
$R \in \msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} A)$
satisfying the two conditions:
\begin{enumerate}
\rmitem{i} $R$ has finite injective dimension.
\rmitem{ii} The canonical morphism
$A \to \opn{RHom}_{A}(R, R)$ 
is an isomorphism.
\end{enumerate}
\end{dfn}

Condition (i) means that there is an integer $d$ such that
$\opn{Ext}^i_A(M, R) = 0$ for all $i > d$ and all modules $M$.

\begin{exa} \label{exa0}
If $\K$ is a regular noetherian ring of finite Krull dimension 
(say a field, or the ring of integers $\mbb{Z}$) then 
\[ R := \K \in \msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} \K) \] 
is a dualizing complex. 
\end{exa}

Dualizing complexes over commutative rings are part of 
Grothendieck's duality theory in algebraic geometry, which 
was developed in \cite{RD}. This duality theory deals with 
dualizing complexes on schemes and relations between them.

In this lecture I will explain a new approach to dualizing 
complexes over commutative rings, due to James Zhang 
and myself (see \cite{YZ4} and \cite{YZ5}).
Specifically, I'll talk about existence and 
uniqueness of {\em rigid dualizing complexes}. 

The purpose of rigidity is to eliminate automorphisms, and to
make the dualizing complexes functorial. 

In a sequel paper \cite{Ye2} we use the technique of
{\em perverse coherent sheaves} to construct 
rigid dualizing complexes on schemes, 
and we reproduce almost all of the geometric 
Grothendieck duality theory. 
But that's a subject for a separate lecture.

Related work in noncommutative algebraic geometry (where 
rigid dualizing complexes were first introduced) can be found in 
\cite{VdB, YZ1, YZ2, YZ3}. 


\newslide
\section{Rigid Complexes and DG Algebras}

By default all rings considered in this talk are commutative. 

Let me start with a discussion of rigidity for algebras over a 
field. Suppose 
$\K$ is a field, $B$ is a $\K$-algebra, and
$M \in \msf{D}(\cat{Mod} B)$. 

According to Van den Bergh 
\cite{VdB} a {\em rigidifying isomorphism} for $M$ is an 
isomorphism
\begin{equation} \label{eqn1}
\rho : M \iso \opn{RHom}_{B \otimes_{\K} B}(B, M \otimes_{\K} M)
\end{equation}
in $\msf{D}(\cat{Mod} B)$. 

\newslide
Now suppose $A$ is any ring. 

Trying to write $A$ instead of $\K$ in
formula (\ref{eqn1}) does not make sense:
instead of $M \otimes_{A} M$ we must take the derived tensor 
product
$M \otimes^{\mrm{L}}_{A} M$; but then there is no obvious way to 
make $M \otimes^{\mrm{L}}_{A} M$ into a complex of 
$B \otimes_{A} B$ - modules. 

The problem is torsion: $B$ might 
fail to be a flat $A$-algebra. 

This is where {\em differential graded algebras} (DG algebras) 
enter the picture. 

\newslide
A DG algebra is a graded ring
$\til{A} = \boplus_{i \in \mbb{Z}} \til{A}^i$, 
together with a graded derivation
$\mrm{d} : \til{A} \to \til{A}$ of degree $1$, satisfying 
$\mrm{d} \circ \mrm{d} = 0$. 

A DG algebra quasi-isomorphism is a homomorphism 
$f : \til{A} \to \til{B}$ respecting degrees, multiplications and 
differentials, and such that 
$\mrm{H}(f) : \mrm{H} \til{A} \to \mrm{H} \til{B}$
is an isomorphism (of graded algebras). 

%\newslide
We shall only consider {\em super-commutative non-positive} DG 
algebras. Super-comm- ut\-ative means that 
$a b = (-1)^{i j} b a$ and $c^2 = 0$
for all $a \in \til{A}^i$, $b \in \til{A}^j$ and 
$c \in \til{A}^{2i + 1}$. Non-positive means that
$\til{A} = \boplus_{i \leq 0} \til{A}^i$.

We view a ring $A$ as a DG algebra concentrated in degree $0$. 
Given a DG algebra homomorphism $A \to \til{A}$ we say that 
$\til{A}$ is a DG $A$-algebra.

\newslide
Let $A$ be a ring. 
A {\em semi-free} DG $A$-algebra is a DG $A$-algebra 
$\til{A}$, such that after forgetting the differential $\til{A}$
is isomorphic, as graded $A$-algebra, to a super-polynomial algebra
on some graded set of variables. 

\begin{dfn}
Let $A$ be a ring and $B$ an $A$-algebra. 
A {\em semi-free DG algebra resolution of $B$ relative to $A$} 
is a quasi-isomorphism  $\til{B} \to B$ of DG 
$A$-algebras, where $\til{B}$ is a 
semi-free DG $A$-algebra.
\end{dfn}

Such resolutions always exist, and they are unique up to 
quasi-isomorphism.  

\begin{exa}
Take $A := \mbb{Z}$ and $B = \mbb{Z} / (6)$.
Define $\til{B}$ to be the super-polynomial algebra
$\mbb{Z}[\xi]$ on the variable $\xi$ of degree $-1$. 
So $\til{B} = \mbb{Z} \oplus \mbb{Z} \xi$
as free $\mbb{Z}$-module, and $\xi^2 = 0$.
Let $\mrm{d}(\xi) := 6$. Then $\til{B} \to \mbb{Z} / (6)$ is a 
semi-free DG algebra resolution of $\mbb{Z} / (6)$ relative to 
$\mbb{Z}$.
\end{exa}

\newslide
For a DG algebra $A$ one has the category $\cat{DGMod} \til{A}$ 
of DG $\til{A}$-modules. It is analogous to the category of 
complexes of 
modules over a ring, and by a similar process of inverting 
quasi-isomorphisms we obtain the derived category 
$\til{\cat{D}}(\cat{DGMod} \til{A})$; see \cite{Ke},
\cite{Hi}. 

For a ring $A$ (a DG algebra concentrated in degree 
$0$) we have
\[ \til{\cat{D}}(\cat{DGMod} A) = \msf{D}(\cat{Mod} A) , \]
the usual derived category.

It is possible to derive functors of DG modules, again in analogy 
to $\msf{D}(\cat{Mod} A)$.

An added feature is that for a 
quasi- iso\-morph\-ism $\til{A} \to \til{B}$ 
the restriction of scalars functor
\[ \til{\cat{D}}(\cat{DGMod} \til{B}) \to
\til{\cat{D}}(\cat{DGMod} \til{A}) \]
is an equivalence.

\newslide
Getting back to our original problem, suppose $A$ is a ring and 
$B$ is an $A$-algebra. Choose a semi-free DG algebra resolution 
$\til{B} \to B$ relative to $A$. 
For $M \in \msf{D}(\cat{Mod} B)$ define
\[ \opn{Sq}_{B / A} M :=
\mrm{RHom}_{\til{B} \otimes_{A} \til{B}}(B, 
M \otimes^{\mrm{L}}_{A} M) \]
in $\msf{D}(\cat{Mod} B)$.

\begin{thm}\tup{(\cite{YZ4})}
The functor
\[ \opn{Sq}_{B / A} : \msf{D}(\cat{Mod} B) \to
\msf{D}(\cat{Mod} B) \]
is independent of the resolution $\til{B} \to B$.
\end{thm}

\newslide
The functor $\opn{Sq}_{B / A}$, called the {\em squaring 
operation}, is nonlinear. In fact, given
a morphism $\phi : M \to M$ in $\msf{D}(\cat{Mod} B)$ and an 
element $b \in B$ one has
\begin{equation} \label{eqn2}
\opn{Sq}_{B / A}(b \phi) = b^2 \opn{Sq}_{B / A}(\phi) 
\end{equation}
in
\[ \opn{Hom}_{\msf{D}(\cat{Mod} B)}(\opn{Sq}_{B / A} M,
\opn{Sq}_{B / A} M) . \]

\newslide
\begin{dfn}
Let $B$ be a noetherian $A$-algebra, and let
$M$ be a complex in 
$\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} B)$ that has finite flat 
dimension over $A$. Assume
\[ \rho : M \iso \opn{Sq}_{B / A} M \]
is an isomorphism in $\msf{D}(\cat{Mod} B)$. Then the pair
$(M, \rho)$ is called a {\em rigid complex over $B$ relative to 
$A$}.
\end{dfn}

%\newslide
\begin{dfn} 
Say $(M, \rho)$ and $(N, \sigma)$ are rigid complexes over $B$ 
relative to $A$. A morphism $\phi : M \to N$ in
$\msf{D}(\cat{Mod} B)$ is called a {\em 
rigid morphism relative to $A$} if the diagram
\[ \begin{CD}
M @>{\rho}>> \opn{Sq}_{B / A} M \\
@V{\phi}VV @VV{\opn{Sq}_{B / A} (\phi)}V \\
N  @>{\sigma}>> \opn{Sq}_{B / A} N
\end{CD} \]\
is commutative. 
\end{dfn}

\newslide
We denote by 
$\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} B)_{\mrm{rig} / A}$
the category of rigid complexes over $B$ relative to $A$. 

\begin{exa}
Take $M = B := A$. Then
\[ \opn{Sq}_{A / A} A = 
\mrm{RHom}_{A \otimes_{A} A}(A, A \otimes_{A} A) = A , \]
and we interpret this as the tautological rigidifying isomorphism
\[ \rho^{\mrm{tau}} : A \iso \opn{Sq}_{A / A} A . \]
The {\em tautological rigid complex} is
\[ (A, \rho^{\mrm{tau}}) \in 
\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} A)_{\mrm{rig} / A} . \]
\end{exa}



%\newslide
\section{Properties of Rigid Complexes}

The first property of rigid complexes explains their name.

\begin{thm} \tup{(\cite{YZ4})}
Let $A$ be a ring, $B$ a noetherian $A$-algebra, and 
\[ (M, \rho) \in 
\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} B)_{\mrm{rig} / A} . \]
Assume the canonical homomorphism 
\[ B \to \opn{Hom}_{\msf{D}(\cat{Mod} B)}(M, M) \]
is bijective. Then the only automorphism of $(M, \rho)$
in 
\[ \msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} B)_{\mrm{rig} / A} \]
is the identity $\bsym{1}_M$. 
\end{thm}

\newslide
The proof is very easy: an automorphism $\phi$ of $M$ 
has to be of the form $\phi = b\, \bsym{1}_M$ for some invertible 
element $b \in B$. If $\phi$ is rigid then $b = b^2$
(cf.\ formula (\ref{eqn2})), and hence 
$b = 1$. 

\bigskip
\[ \spadesuit \qquad \spadesuit \qquad \spadesuit \]
\bigskip

We find it convenient to denote ring homomorphisms by $f^*$ etc. 
Thus a ring homomorphism $f^* : A \to B$ corresponds to the 
morphism of schemes
\[ f : \opn{Spec} B \to \opn{Spec} A . \]


\newslide
Let $A$ be a noetherian ring. 
Recall that an $A$-algebra $B$ is called 
essentially finite type if it is a localization of some finitely 
generated $A$-algebra. 

We say that $B$ is {\em essentially smooth}
(resp.\ {\em essentially \'etale}) over $A$ if it is essentially 
finite type and formally smooth (resp.\ formally \'etale).

\begin{exa}
If $A'$ is a localization of $A$ then $A \to A'$ is essentially 
\'etale. If $B = A[t_1, \ldots, t_n]$ is a polynomial algebra then 
$A \to B$ is smooth, and hence also essentially smooth.
\end{exa}

\newslide
Let $A$ be a noetherian ring and $f ^* : A \to B$ an 
essentially smooth homomorphism. Then $\Omega^1_{B / A}$ is a 
finitely generated projective $B$-module. 

Let
\[ \opn{Spec} B = \coprod_i \opn{Spec} B_i \]
be the decomposition into connected components, and for every $i$ 
let $n_i$ be the rank of $\Omega^1_{B_i / A}$. We define a functor
\[ f^{\sharp} : \msf{D}(\cat{Mod} A) \to \msf{D}(\cat{Mod} B) \]
by
\[ f^{\sharp} M := \bigoplus_i \, 
\Omega^{n_i}_{B_i / A}[n_i] \otimes_A M . \]

\newslide
Recall that a ring homomorphism $f^* : A \to B$ is called finite 
if $B$ is a finitely generated $A$-module. Given such a finite 
homomorphism we define a functor
\[ f^{\flat} : \msf{D}(\cat{Mod} A) \to \msf{D}(\cat{Mod} B) \]
by
\[ f^{\flat} M := \opn{RHom}_A(B, M) . \]

\newslide
\begin{thm}\tup{(\cite{YZ4})} \label{thm0.1}
Let $A$ be a noetherian ring, let $B, C$ be essentially finite 
type $A$-algebras, let $f^* : B \to C$ be an $A$-algebra
homomorphism, and let 
\[ (M, \rho) \in 
\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} B)_{\mrm{rig} / A} . \]
\begin{enumerate}
\rmitem{1} If $f^*$ is finite and $f^{\flat} M$ has finite flat 
dimension over $A$, then $f^{\flat} M$ has an induced rigidifying 
isomorphism 
\[ f^{\flat}(\rho) : f^{\flat} M  \iso \opn{Sq}_{C / A}
f^{\flat} M . \]
\rmitem{2} If $f^*$ is essentially smooth
then $f^{\sharp} M$ has an induced rigidifying 
isomorphism 
\[ f^{\sharp}(\rho) : f^{\sharp} M  \iso \opn{Sq}_{C / A}
f^{\sharp} M . \]
\end{enumerate}
\end{thm}



\newslide
\section{Rigid Dualizing Complexes}

Let $\K$ be a noetherian regular ring of finite Krull dimension. 
We denote by $\cat{EFTAlg} / \K$ the category of essentially finite type 
$\K$-algebras. 

\begin{dfn}
A {\em rigid dualizing complex} over $A$ relative to $\K$ is a 
rigid complex $(R_A, \rho_A)$ such that $R_A$ is a dualizing 
complex. 
\end{dfn}

\newslide
\begin{thm} \tup{(\cite{YZ5})} \label{thm4}
Let $\K$ be a regular finite dimensional noetherian ring, and
let $A$ be an essentially finite type $\K$-algebra.
\begin{enumerate}
\rmitem{1} The algebra $A$ has a rigid dualizing complex 
$(R_A, \rho_A)$, which is unique up to a unique rigid isomorphism.
%
\rmitem{2} Given a finite homomorphism $f^* : A \to B$, there is a 
unique rigid isomorphism
$f^{\flat}(R_A, \rho_A) \iso (R_B, \rho_B)$.
%
\rmitem{3} Given an essentially smooth homomorphism $f^* : A \to B$ , 
there is a unique rigid isomorphism
$f^{\sharp}(R_A, \rho_A) \iso (R_B, \rho_B)$.
\end{enumerate}
\end{thm}

\newslide
Here is how the rigid dualizing complex $(R_A, \rho_A)$ is 
obtained. We begin with the tautological 
rigid complex 
\[ (\K, \rho^{\mrm{tau}}) \in
\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} \K)_{\mrm{rig} / \K} , \]
which is dualizing. 
Now the structural homomorphism $\K \to A$ can be factored into
\[ \K \xar{f^*} B \xar{g^*} C \xar{h^*} A , \]
where $f^*$ is smooth ($B$ is a polynomial algebra over $\K$);
$g^*$ is finite (a surjection); 
and $h^*$ is also smooth (a localization). 
Then
\[ (R_A, \rho_A) := h^{\sharp}\, g^{\flat}\, f^{\sharp}
(\K, \rho^{\mrm{tau}}) \in
\msf{D}^{\mrm{b}}_{\mrm{f}}(\cat{Mod} A)_{\mrm{rig} / \K} . \]

\newslide
\begin{dfn}
Given a homomorphism $f^* : A \to B$ in $\cat{EFTAlg} / \K$,
define the {\em twisted inverse image functor}
\[ f^! : \msf{D}^{+}_{\mrm{f}}(\cat{Mod} A) \to
\msf{D}^{+}_{\mrm{f}}(\cat{Mod} B) \]
by the formula
\[ f^! M :=  
\opn{RHom}_B \bigl( B \otimes^{\mrm{L}}_{A} 
\opn{RHom}_A(M, R_A), R_B \bigr) . \]
\end{dfn}

It is not hard to show that the assignment $f^* \mapsto f^!$ 
is a $2$-functor from the category $\cat{EFTAlg} / \K$ to the $2$-category
$\cat{Cat}$ of all categories. 

One can show, using Theorem \ref{thm4}, that this operation has very 
good properties. For instance, when $f^*$ is finite, 
then there is a functorial, nondegenerate trace morphism
\[ \opn{Tr}_f :  f^{!} M \to M . \]



\newslide
\section{Rigid Complexes and CM Homomorphisms}

In this final section I'll talk about the relation between rigid 
complexes and Cohen-Macaulay homomorphisms. 

\begin{dfn}
A ring $A$ is called {\em tractable} if there is an 
essentially finite type homomorphism $\K \to A$, for some 
regular noetherian ring of finite Krull dimension $\K$.
\end{dfn}

The homomorphism $\K \to A$ is {\em not} part of the structure -- 
there is no preferred $\K$. 
``Most commutative noetherian rings we know'' are tractable...

\newslide
\begin{thm}\tup{(\cite{YZ5}, \cite{Ye2})} 
Let $A$ be a tractable ring, and let $B$ be an essentially finite 
type $A$-algebra of finite flat dimension
\tup{(}e.g.\ $B$ is flat over $A$\tup{)}.
\begin{enumerate}
\rmitem{1} There exists a unique 
\tup{(}up to unique rigid isomorphism\tup{)} rigid 
complex $R_{B/A}$ over $B$ relative to $A$, which is nonzero on each 
connected component of $\opn{Spec} B$.
\rmitem{2} If $A$ is a Gorentein ring \tup{(}e.g.\ a regular ring\tup{)} 
then $R_{B/A}$ is a dualizing complex over $B$ 
\end{enumerate} 
\end{thm}

\newslide
Let $f^* : A \to B$ be a finite type flat homomorphism of relative dimension 
$n$; namely the fibers of $f: \opn{Spec} B \to \opn{Spec} A$
are all equidimensional of dimension $n$. 

Recall that $f^*$ is called a {\em Cohen-Macaulay} homomorphism
if the fibers of $f$ are all $n$-dimensional Cohen-Macaulay schemes. 

\begin{thm}\tup{(\cite{Ye2})} 
Let $A$ be a tractable ring, and let $f^* : A \to B$ be a finite 
type flat homomorphism of relative dimension $n$.
Then the following conditions are equivalent:
\begin{enumerate}
\rmitem{i} $f^*$ is a Cohen-Macaulay homomorphism.
\rmitem{ii} 
$\mrm{H}^i R_{B / A} = 0$ for all $i \neq -n$, and the $B$-module
\[ \bsym{\omega}_{B / A} := \mrm{H}^{-n} R_{B / A} \]
is flat over $A$.
\end{enumerate}
\end{thm}

\newslide
The module $\bsym{\omega}_{B / A}$ is the {\em dualizing module} of $B$ 
relative to $A$. 

Note that the complex 
$\bsym{\omega}_{B / A}[n]$ is rigid relative to $A$, but in general 
it is not a dualizing complex (cf.\ part (2) of previous theorem.) 
Still the fibers of $\bsym{\omega}_{B / A}[n]$ are dualizing 
complexes -- this can be seen by taking $A'$ to be a field in the 
next result.

\bigskip
\[ \spadesuit \qquad \spadesuit \qquad \spadesuit \]
\bigskip

Let me end with a ``rigid'' version of Conrad's base change theorem \cite{Co}:

\newslide
\begin{thm}\tup{(\cite{Ye2})} 
Let 
\[ \begin{CD}
A @>{}>> B \\
@V{}VV @VV{}V \\
A'  @>{}>> B'
\end{CD} \]
be a cartesian diagram of rings, i.e.\ 
\[ B' \cong A' \otimes_A B , \]
with $A$ and $A'$ tractable rings. Assume 
$A \to B$ is a Cohen-Macaulay homomorphism. \tup{(}There isn't any  
restriction on the homomorphism $A \to A'$.\tup{)} Then:
\begin{enumerate}
\rmitem{1} $A' \to B'$ is a Cohen-Macaulay homomorphism.
\rmitem{2} There is a unique isomorphism of $B'$-modules
\[ \bsym{\omega}_{B' / A'} \cong A' \otimes_A \bsym{\omega}_{B / A}  \]
which respects rigidity.
\end{enumerate}
\end{thm}





\end{slide*}


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    Springer, 1970.
\bibitem[AJL]{AJL} L.\ Alonso, A.\ Jerem\'{\i}as and J.\ Lipman,
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        in ``Studies in Duality on Noetherian Formal Schemes 
        and Non-Noetherian Ordinary Schemes,'' Contemp.\ Math.\ 
        \textbf{244}, Amer.\ Math.\ Soc., 1999, 3-90.    
\bibitem[Be]{Be} K. Behrend,
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