Let F be a quasi-coherent sheaf on a smooth projective variety X over an algebraically closed field and $D\_X(?)=RHom(?,\omega \_X[n])$ the dualizing functor. Is it the case that $D\_X(D\_X(F))$ is a sheaf? (i.e. $\mathcal H^n(D\_X(D\_X(F)))=0$ for $n\ne 0$). If so is there a convenient reference?



Is the derived double dual of a quasi-coherent sheaf a sheaf?
For coherent sheaves, by using resolutions, one checks that the assertion in question is true: for any perfect complex $K$, one even has $D\_X^2(K) \simeq K$. However, for larger quasicoherent sheaves $F$, the value $D\_X^2(F)$ fails to even be a sheaf; I give an example below in Corollary 5.


Fix a countable set $I$, and a PID $A$ with at least two non-zero irreducible elements $p,q \in A$ such that $(p,q) = (1)$. We begin with the following basic observation that lies at the heart of the argument:


**Lemma 1:** One has $Hom\_A(\prod\_I A,A) \simeq \oplus\_I A$ via the natural map in $D(A)$.


*Proof:* See [Is it true that, as $\Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other?](https://mathoverflow.net/questions/10239/is-it-true-that-as-z-modules-the-polynomial-ring-and-the-power-series-ring-over).


This Lemma shows that $\prod\_I A$ is not projective. More precisely, we get:


**Lemma 2:** The complex $K := RHom\_A(\prod\_I A,A)$ has non-zero $H^1$.


*Proof:* If $k$ is a residue field of $A$ at a maximal ideal (ex: $k = A/(p)$), then $k$ is $A$-perfect, so $- \otimes\_A k$ commutes with all (homotopy) limits and colimits. This gives $K \otimes\_A k \simeq Hom\_k(\prod\_I k, k)$, which is a vector space of uncountable dimension. On the other hand, by Lemma 1, if we assume that $H^1(K) = 0$, then we get $K \otimes\_A k \simeq \oplus\_I k$, which is a vector space of countable dimension; so we conclude $H^1(K) \neq 0$.


Now we globalize. Set $X = \mathbf{P}^1\_k$ over some field $k$. All complexes below are considered as living in the derived category $D(O\_X)$ of all $O\_X$-modules on $X$, and similarly for functors. (I prefer this to using $D\_{qc}(O\_X)$ as the description of products in the latter is more subtle.) The global analogue of Lemma 2 is: 


**Lemma 3:** The complex $K := RHom\_X(\prod\_I O\_X,O\_X)$ has a non-zero $H^1$.


*Proof:* This does not formally follow from Lemma 2 as $\prod\_I O\_X$ is not quasi-coherent. However, we can argue as follows: (a) first show that $H^0(K) := Hom\_X(\prod\_I O\_X,O\_X) \simeq \oplus\_I O\_X$ via the natural map, and (b) show that $K \otimes\_{O\_X} k \simeq Hom\_k(\prod\_I k,k)$ is a vector space of uncountable dimension, where $k$ is the residue field at $0$ viewed as an $O\_X$-complex. Then (a) and (b) immediately imply the claim by the argument of Lemma 2. Moreover, (b) is clear since $k$ is perfect over $O\_X$, so it remains to show (a). Consider the natural map $\eta:\oplus\_I O\_X \to Hom(\prod\_I O\_X,O\_X)$. Clearly $\eta$ is injective. For surjectivity, fix some $f:\prod\_I O\_X \to O\_X$. We want to show that $f$ has finite support, i.e., that $f$ factors through a projection to finitely many components of the product on the left. For any affine open $U \subset X$, the map $f(U)$ has finite support by Lemma 1. A glueing argument then finishes the proof.


Writing a product as the dual of a sum then gives:


**Lemma 4:** The complex $RHom\_X(RHom\_X(\oplus\_I O\_X,O\_X),O\_X)$ has non-zero cohomology in degree $0$ and $1$.


*Proof:* This follows formally from Lemma 3 as $RHom\_X(-,O\_X)$ carries coproducts to products.


Here is the promised example:


**Corollary 5:** On $X = \mathbf{P}^1\_k$, if $D\_X = RHom\_X(-,\omega\_X[1])$, then the functor $D\_X^2$ carries the sheaf $\oplus\_I O\_X$ to a complex with non-zero $H^0$ and $H^1$.


*Proof:* As $\omega\_X$ is invertible, one has $D\_X = RHom\_X(-,O\_X) \otimes \omega\_X[1]$, and hence $D\_X^2 = RHom\_X(RHom\_X(-,O\_X),O\_X)$, so the claim follows form Lemma 4.


