$\DeclareMathOperator\Hom{Hom}$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

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

  • 3
    $\begingroup$ For a dualizing complex, its not true that the double dual with respect to $Hom_{D(A)}(-,\omega_X)$ is isomorphic to the identity. For that to be true you need to consider the full $RHom_A(-,\omega_X)$. $\endgroup$
    – the L
    Commented Feb 8, 2022 at 19:50
  • $\begingroup$ Thank you! I edited the question! $\endgroup$
    – Adam
    Commented Feb 8, 2022 at 20:58
  • 1
    $\begingroup$ I only answer to the edit in this comment. Simply because $H^0(F) \simeq H^0(G)$ does not imply that $F \simeq G$. By the way, I hope I don't sound too patronizing by saying this, but you should probably read an introductory book on derived categories and homological algebra. Homological Algebra by Gelfand and Manin seems appropriate as far as I can deduce from your questions. $\endgroup$
    – Libli
    Commented Feb 8, 2022 at 21:18

1 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.

  • $\begingroup$ I was thinking about \cite[\href{stacks.math.columbia.edu/tag/0A89}{Tag 0A89}]{stacks-project} which I thought to be valid in my setting? $\endgroup$
    – Adam
    Commented Feb 8, 2022 at 20:31
  • 1
    $\begingroup$ The comment below you question made by @theL is the answer to this issue. You need to take the full $\mathcal{R}Hom_{D(A)}(\bullet, \mathcal{K})$ to get an anti-involution. Then $\mathcal{R}Hom_{D(A)}(\Omega_A, \mathcal{K}) \neq \mathcal{R}Hom_{D(A)}(\Omega^{**}_{A}, \mathcal{K})$ beacause of the higher $Ext$. $\endgroup$
    – Libli
    Commented Feb 8, 2022 at 21:05

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