Dualizing Complexes

Question

Let $R$ be a dualizing complex of a Noetherian graded algebra $A$ (not necessary commutative). For any $M\in D_c^b(A)$, there is a natural morphism $$ \theta: R\Gamma_m(M) \to RHom_{A^o}(RHom_A(M,R),R\Gamma_m(R)) $$ in $D^+(A)$. Notations and terminologies we refer to the paper [Dualizing complexex over noncommutative graded algebras, A. Yekutieli], in which the author show that $\theta$ is an isomorphism whenever $R$ is balanced. Nevertheless I wonder whether or not $\theta$ is an isomorphism for arbitrary $R$.

Concerning the natural morphism, one can take $M$ to be a left bounded complex of injective modules, and take $R$ to be a complex of bimodules which are injective over $A$ and $A^{op}$. Thus $\Gamma_m(R)$ is also a complex of bimodules which are injective ofer $A$ and $A^{op}$ because $A$ is Noetherian. The natural morphism $\theta$ is given by the following morphism: $$ \Gamma_m(M) \to Hom_{A^{op}}(Hom_A(M,R),\Gamma_m(R)). $$

Answer 1

For an arbitray dualizing complex (not necessary balanced), I am going to prove the following version of local duality theorem. I will be grateful for any comments and corrections. Notations and concepts are refered to Here.

Theorem: Let $A$ be a connected graded Noetherian algebra, and let $R\in D(A^e)$ be a dualizing complex (not necessary balanced). Then for any $X\in D^b_{fg}(A)$, the natural morphism $$ \theta: R\Gamma_{\mathfrak{m}} (X) \to R\text{Hom}_{A^o}(R\text{Hom}_A(X,R),R\Gamma_{\mathfrak{m}}(R)) $$ is an isomorphism in $D(A)$.

Proof of the Theorem: Firestly choose an injective resolution $X\to I$ in $K^+(A)$, and also choose an injective resolution $R\to E$ in $K^+(A^e)$. Then one knows that $A^o$-modules occures in $\text{Hom}_A(A/A_{\geq n},E)$ and $\Gamma_\mathfrak{m}(E)$ are all injective. So for any $n\geq 1$ we get a commutative diagram in $K(A)$: $$ \begin{array}{ccc} \text{Hom}_A(A/A_{\geq n},I) & \xrightarrow{f_n} & \text{Hom}_{A^o}(\text{Hom}_A(I,E),\text{Hom}_A(A/A_{\geq n},E))\\ \downarrow_{p_n} & & \downarrow_{q_n}\\ \Gamma_\mathfrak{m}(I) & \xrightarrow{g} & \text{Hom}_{A^o}(\text{Hom}_A(I,E),\Gamma_\mathfrak{m}(E)), \end{array} $$ where all $p_n,q_n,f_n,g$ are the canonical jomomorphisms of complexes. Note that $\lim_{n\to \infty} p_n$ and $\lim_{n\to \infty} f_n$ are all quasi-isomorphisms. So to prove that $g$ is a quasi-isomorphism it quivalent to show that $\lim_{n\to\infty}q_n$ is a quasi-isomorphism.

Since $H^i(\text{Hom}_A(I,E))$ is finitely generated and vanishes when $|i|\gg0$, one can choose a resolution $P\xrightarrow{\sim} \text{Hom}_A(I,E)$ in $D(A^o)$ such that $P$ is bounded above and all $P_i$ are finitely generated pojective $A^o$-modules. Again we have a commutative diagram in $K(A)$: \begin{array}{ccc} \text{Hom}_{A^o}(\text{Hom}_A(I,E),\text{Hom}_A(A/A_{\geq n},E)) &\xrightarrow{u_n} & \text{Hom}_{A^o}(P,\text{Hom}_A(A/A_{\geq n},E))\\ \downarrow_{q_n} & & \downarrow_{r_n}\\ \text{Hom}_{A^o}(\text{Hom}_A(I,E),\Gamma_\mathfrak{m}(E)) & \xrightarrow{v} & \text{Hom}_{A^o}(P,\Gamma_\mathfrak{m}(E)), \end{array} Where $r_n,u_n,v$ are the natural homomorphism of complexes. Now note that $\lim_{n\to \infty} u_n$ and $v$ are quasi-isomorphisms. From the fact that $\text{Hom}_{A^o}(P,-)$ commutes with direct limits of complexes with a common lower lound, one obtains that $\lim_{n\to \infty} r_n$ is a quasi-isomorphism, and so is $\lim_{n\to\infty}q_n$ too. Therefore the cononical homomorphism $g$ is a quasi-isomorphism.

At last, the construction of $\theta$ tells that it is an isomorphism in $D^+(A)$.