Confusion about the (Grothendieck–Poincaré) double dual of reflexive differentials vs usual differentials on a normal Cohen–Macaulay scheme

Question

$\DeclareMathOperator\Hom{Hom}$Let $\mathcal{A}$ be an abelian category, my question is about the case when $\mathcal{A}$ is the category of quasi-coherent sheaves on a scheme $X$. There is a fully faithful embedding of $\mathcal{A}$ into the derived category $\mathcal{D}(\mathcal{A})$ given by sending an object $A\to A[0]$, a complex centred in degree zero. The inverse (defined on complexes quasi-isomorphic to those centered in degree zero) is given by $H^0$.

Now in particular, if I am not mistaken this means that for $A,B\in \mathcal{A}$ there is an isomorphism in $\mathcal{D}(\mathcal{A})$

$$\Hom_{\mathcal{A}}(A, B)[0]\cong \Hom_{\mathcal{D}(\mathcal{A})}(A[0], B[0])$$

Now to my confusion, suppose that $X$ is a normal projective Cohen–Macaulay scheme with dualizing complex $K^{\bullet}=\omega_X[d_X]$ where $\omega_X$ is a dualising sheaf. Let $A$ and $A'$ be non isomorphic sheaves such that $$\Hom_{\mathcal{A}}(A, \omega_X)\cong \Hom_{\mathcal{A}}(A', \omega_X),$$ for example this happens if $A=\Omega^i_X$ is not reflexive and $A'=A^{\vee \vee}$ is the double dual. ($\omega_X$ is in this context reflexive and can be taken to be $(\Omega^{n}_X)^{\vee \vee}$)

We then have $$\Hom_{\mathcal{A}}(\Hom_{\mathcal{A}}(A, \omega_X),\omega_X)[0]\cong \Hom_{\mathcal{A}}(\Hom_{\mathcal{A}}(A', \omega_X),\omega_X)[0]$$

Which by the argument above implies $$\Hom_{\mathcal{D}(\mathcal{A})}(\Hom_{\mathcal{D}(\mathcal{A})}(A[0], \omega_X[d_X]),\omega_X[d_X])\cong \Hom_{\mathcal{D}(\mathcal{A})}(\Hom_{\mathcal{D}(\mathcal{A})}(A'[0],\omega_X[d_X] ),\omega_X[d_X])$$

Which since $\omega_X[d_X]=K^{\bullet}$ is a dualizing complex gives

$A[0]\cong A'[0]$ in the derived category.

This seems to be a contradiction. What went wrong?

Update: Duality theory says that

$$R\Hom(R\Hom(A[0], \omega_X[d_X]),\omega_X[d_X])\cong A[0]$$ and $$\Hom_{\mathcal{D}(\mathcal{A})}=H^0R\Hom$$ therefore the above question should be changed into: $A$ and $A'$ are not isomorphic, but
$$\Hom_{\mathcal{A}}(A, \omega_X)\cong \Hom_{\mathcal{A}}(A', \omega_X)$$

then why does this not imply that the following objects are isomorphic in the derived category

$$A[0]\cong R\Hom(R\Hom(A[0], \omega_X[d_X]),\omega_X[d_X])\cong R\Hom(R\Hom(A'[0], \omega_X[d_X]),\omega_X[d_X])\cong A'[0]$$

Answer 1

I believe you are confusing local duality (on a Cohen–Macaulay loal ring) and global duality on a scheme. On a scheme $X$ with dualizing complex $\mathcal{K}^{\bullet}$, Serre–Grothendieck duality is stated as:

$$ \mathrm{Hom}_{D(X)}(F,G \otimes \mathcal{K}^{\bullet}) \simeq \mathrm{Hom}_{D(X)}(G,F)^*$$ for any $F,G \in D^{perf}(X)$, and where $^*$ is the dual as vector spaces over the ground field.

There are two issues with your example:

• From the correct version of Serre–Grothendieck duality for a scheme $X$ stated above, it is clear that you can not deduce that $A' \simeq A$.

• Furhtermore, it is not clear to me the $\Omega_{X}$ is a perfect object if $X$ is singular. Hence Grothendieck–Serre duality doesn't necessarily apply to $\Omega_X$.

If you want to work in $D(A)$ where $A$ is a local ring, then the correct version of Grothendieck–Serre duality takes into account the higher $\operatorname{Ext}$. As $\Omega_X$ is supposed to be not locally free in your example, the isomorphism: $$ \operatorname{\mathcal{H}om}_{A}(\Omega_{A}, A) \simeq \operatorname{\mathcal{H}om}_{A}(\Omega^{**}_A,A)$$ will not imply that $\Omega_{A}$ is isomorphic to $\Omega^{**}_A$ (namely because the objects $\mathcal{R}Hom_{D(A)}(\Omega_A^{**}, \mathcal{K})$ and $\mathcal{R}Hom_{D(A)}(\Omega_A, \mathcal{K})$ are not isomorphic).

Chapter 22 of "Introduction algébrique à la géométrie projective" by Peskine explains very clearly, in my opinion, the local duality on a CM local ring.