I am reading A. Yekutieli's original article on dualizing complexes for noncommutative algebras and I found a problem I cannot solve.

First, some background. We start with a field $k$ and a noetherian $\mathbb N$-graded algebra $A$ and assume that $A_0 = k$. All objects are graded, and we write $\mathcal A$ for the category of graded $A$-bimodules. A dualizing complex is an object $R^\bullet$ in the derived category $\mathcal D(\mathcal A)$ satisfying the following:

  1. Its cohomology modules are finitely generated as left and right $A$-modules.

  2. If we look at $R^\bullet$ as a complex of left $A$-modules, then it has finite injective dimension, and the same holds exchangng left for right.

  3. The obvious morphisms $A \to \mathcal R Hom_{A^\circ}(R^\bullet, R^\bullet)$ and $A^\circ \to \mathcal R Hom_{A}(R^\bullet, R^\bullet)$ are isomorphisms in the category $\mathcal D (\mathcal A)$.

These "obvious morphism" come from the following setup: By 2 we may replace $R^\bullet$ with a bounded complex $I^\bullet$ such that each $I^i$ is injective as left and right $A$-module (but not necessarily as bimodule). Thus for every complex of left $A$-modules $M^\bullet$ there is a map, defined at the level of chain complexes \begin{align*} M^\bullet \to Hom_{A^\circ}(Hom_A(M^\bullet, I^\bullet), I^\bullet) \end{align*} given by simple evaluation. Passing to the derived category, the complex on the r.h.s. is isomorphic to $\mathcal R Hom_{A^\circ}(\mathcal R Hom_{A}(M^\bullet, R^\bullet), R^\bullet)$. Putting $M^\bullet = A$ gives us the morphism of point 3 of the above definition (if $M^\bullet$ is a complex of bimodules then the map respects this structure), and in the paper Yekutieli proves that this map is an isomorphism for $A$ if and only if it is an isomorphism whenever $M^\bullet$ is a bounded complex with finitely generated cohomology modules. So far, so good.

Dualizing complexes, when they exists, are not unique. In order to fix a special class of isomorphisms Yekutieli introduces local cohomology. Set $\Gamma_m(-) = \varinjlim Hom_A(A/A_{\geq n},-)$, where $A_{\geq n}$ denotes the ideal of elements of $A$ of degree at least $n$. This is a left exact functor, with a right derived functor $\mathcal R \Gamma_m$ defined over left bounded complexes of left $A$-modules. Then he states that there is a map similar to that above, \begin{align*} \mathcal R \Gamma_m(M^\bullet) \to \mathcal R Hom_{A^\circ}(\mathcal R Hom_{A}(M^\bullet, R^\bullet), \mathcal R \Gamma_m(R^\bullet)) \end{align*} It is this map I am trying to make sense of. By looking at it simply at the level of complexes, we might feel tempted to replace $M^\bullet$ by an injective resolution $J^\bullet$, reuse our old resolution of $R^\bullet$ called $I^\bullet$, and thus obtain a map \begin{align*} \Gamma_m(J^\bullet) \to Hom_{A^\circ}(Hom_{A}(J^\bullet, I^\bullet), \mathcal \Gamma_m(I^\bullet)) \end{align*} given again by evaluation. This is indeed well-defined, but the problem is that passing to the derived category, it is not immediate, at least to me, that the complex on the r.h.s. is isomorphic to $\mathcal R Hom_{A^\circ}(\mathcal R Hom_{A}(M^\bullet, R^\bullet), \mathcal R \Gamma_m(R^\bullet))$. In order for that to happen, $\Gamma_m(I^\bullet)$ should consist of $A$-module which are injective as right $A$-modules, and I see no reason for that to be true (in the article it is proven that these modules are injective as left $A$-modules).

Any ideas on how to fix this? Have I made a mistake with the definition of this last morphism?


2 Answers 2


I happened to be looking at this very issue a few days ago! Here is my explanation to myself...

Indeed, in the notation above, there is a homomorphism of complexes of graded modules (not derived) $$ \zeta : \Gamma_{\mathfrak{m}} (J) \to \mathrm{Hom}_{A^{\mathrm{op}}} ( \mathrm{Hom}_A ( J, I ), \Gamma_{\mathfrak{m}} (I) ) . $$ And $$ \Gamma_{\mathfrak{m}} (J) \cong \mathrm{R} \Gamma_{\mathfrak{m}} (M) . $$ But now we can choose a quasi-isomorphism $ \Gamma_{\mathfrak{m}} (I) \to K $ where $K$ is a complex of bimodules, injective on the right (and bounded below). We then get a morphism $$ \eta : \mathrm{Hom}_{A^{\mathrm{op}}} ( \mathrm{Hom}_A ( J, I ), \Gamma_{\mathfrak{m}} (I) ) \to \mathrm{Hom}_{A^{\mathrm{op}}} ( \mathrm{Hom}_A ( J, I ), K) \cong \mathrm{RHom}_{A^{\mathrm{op}}} ( \mathrm{RHom}_A ( M, R ), \mathrm{R}\Gamma_{\mathfrak{m}} (R) ) . $$ The morphism we want is $\eta \circ \zeta$. It is functorial.

In case $A$ is noetherian, we know that $ \Gamma_{\mathfrak{m}} (I) $ is already injective on the right (as mentioned in the previous answer), and then $\eta$ is an isomorphism.

I hope this makes sense!

Added 27 Nov 2018:

  1. When $A$ is noetherian connected, and $R$ is a balanced DC, then $\mathrm{R} \Gamma_{\mathfrak{m}}(R)$ is isomorphic to $A^{*}$, and this is graded-injective on both sides.
  2. In the book (in preparation) Derived Categories (prepulication ver 3) there is an improved treatment of local duality in the NC connected graded case -- see Thm 17.2.7.

You can consult [Michel Van den Bergh, Existence theorems for dualizing complexes over noncommutative graded and filtered rings, Lemma 4.1]. Your problem can be solved when $A$ is assumed to be Noetherian.

  • $\begingroup$ Just to clarify, said lemma states that if $E$ is an injective $A$-bimodule then $\Gamma_m(E)$ is a direct limit of injective right modules, which is also injective when $A$ is noetherian. Still, I am curious about Yekutieli's construction, which is done in general for coherent algebras. $\endgroup$ Oct 2, 2013 at 11:38

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