# Construction of Dualizing sheaf

I was going through the construction of dualizing sheaf given in Hartshorne [III, 7, Lemma 7.4]. The proof apparently omits lots of details. In particular it does not mention any of the $i^* , i_*, i^{!}$ that should appear, so almost none of the things he writes are actually defined. I was trying to clean up the proof and I noticed the following more general algebraic lemma.

Lemma: Let $\mathcal C, \mathcal D$ be abelian categories and $\mathcal D$ has enough injectives. $F : \mathcal C \to \mathcal D$ and $G:\mathcal D \to \mathcal C$ are additive functors such that

(i) $F$ is left exact and sends nonzero objects to nonzero objects.

(ii) $G$ is right adjoint to $F$.

(ii) For some fixed object $P$ of $\mathcal D$, we have $(R^i (F \circ G) )(P) = 0$ for all $i < r$ for some fixed $r \in \mathbb N$.

Then it follows that for any object $A$ of $\mathcal C$, $\text{Ext}^i (F(A), P) = 0$ for $i < r$ and there is a functorial isomorphism $\text{Ext}^r (F(A), P) \cong \text{Hom}(A, (R^r (G))(P))$.

Proof: We choose an injective resolution of $P$ in $\mathcal D$, $0 \to P \to I^{\bullet}$. Define $J^{\bullet} := G(I^{\bullet})$. Since $F$ is left exact and left adjoint it is in fact exact. By adjointness, $J^{\bullet}$ is a complex of injective objects. Now $F(h^iJ^{\bullet}) = h^i( F(J^{\bullet}))= h^i (F \circ G )(I^{\bullet}) = R^i(F \circ G)= 0$ for $i<r$ by hypothesis. So $J^{\bullet}$ is exact till $r$-th step. Therefore we can write $J^{\bullet}$ as $J_1^{\bullet} \oplus J_2^{\bullet}$, where $J_1^{\bullet}$ is injective, exact and nonzero only in degrees $\le r$ and $J_2$ is also injective and nonzero only in degrees $\ge r$. Now $\text{Ext}^r(F(A), P) = h^r (\text{Hom}(F(A), I^{\bullet})= h^r(\text{Hom}(A, J^{\bullet}))$ $=h^r(\text{Hom}(A, J_2^{\bullet})) =$ $\text{Ker}((\text{Hom}(A, J_2^r) \to \text{Hom}(A, J_2^{r+1}))$ $= \text{Hom}(A, \text{Ker}(J_2^{r} \to J_2^{r+1}))$. But $\text{Ker}(J_2^{r} \to J_2^{r+1}) = h^r(J^{\bullet})= h^r(G(I^{\bullet}))= (R^rG)(P)$.

Now to prove the existence of dualizing sheaf for a closed immersion $j : X \to \mathbb{P}^n_k$, we take $\mathcal C = \mathfrak {Qco} (X), \mathcal D = \mathfrak {Qco} (\mathbb P ^n _k), F = j_*, G = j^{!}$. Then $(R^i (j_* \circ j^{!})) (\omega_{\mathbb{P}^n_k}) = \mathscr{Ext}^i(j_* \mathscr{O}_X, \omega_{\mathbb{P}^n_k}) = 0$ for $i < r:= n - \dim X$. Therefore, it follows from lemma that $\text{Ext}^r(j_*\mathscr{F}, \omega_{\mathbb{P}^n_k}) = \text{Hom}_{X}(\mathscr{F}, (R^{r} (j^!))(\omega_{\mathbb{P}^n_k}) )$. Now $(R^{r} (j^!))(\omega_{\mathbb{P}^n_k})$ is same as the quasicoherent $\mathcal{O}_X$- module corresponding to the quasicoherent $j_* \mathcal{O}_X$ module $\mathscr{Ext}^r (j_* \mathcal O _{X}, \omega_{\mathbb{P}^n_k})$, which we write by $\mathscr{Ext}^r (j_* \mathcal O _{X}, \omega_{\mathbb{P}^n_k})^{\sim}$.

## Questions:

(i)Does the lemma follow quickly from some spectral sequence (which I don't properly know) argument? Is there any reference for the lemma?

(ii)In these notes (http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec24_dualizing.pdf), the dualizing sheaf is defined to be $j^*(\mathscr{Ext}^r (j_* \mathcal{O}_X, \omega_{\mathbb{P}^n_k}))$. Is it the same as $\mathscr{Ext}^r (j_* \mathcal{O}_X, \omega_{\mathbb{P}^n_k})^{\sim}$?

I don't really understand some parts of the proof in that note, in particular it talks about isomorphism of sheaves in two different topological spaces, which should really be related by a push forward.

• What, exactly, are you asserting is bogus in Hartshorne's proof? – Hoot May 18 '16 at 15:16
• I'm not saying anything is bogus. I'm just saying they are very improperly written. Equations like $\omega_X ^0 = \mathscr{Ext}_P^r (\mathcal O _X, \omega_P)$ doesn't make too much of sense and at least to me it is not obvious how to interpret them. I am just trying to clean things up and state things properly. Also I am trying to separate out the algebraic part of the proof as stated in the lemma. Using the generalized lemma, the proof immediately follows. – Shubhodip Mondal May 18 '16 at 15:21
• I take your point about $i^!$. I'm not sure that Hartshorne ever does this. Does this tag from the Stacks project help? I haven't actually looked at your lemma. – Hoot May 18 '16 at 15:52
• As for your second question: note that $\mathscr Ext^r_P(\mathcal O_X, \omega_P)$ is supported on $X$, since $\mathcal O_X$ is. That is, every local section of $\mathcal I_X$ kills it. Thus, it is of no harm to confuse it with its restriction to $X$. – R. van Dobben de Bruyn May 18 '16 at 15:54
• I think that there may not be a spectral sequence argument giving you the lemma 'for free'. The problem is that you're composing the left exact functor $F(-)$ with the left exact but contravariant functor $\operatorname{Hom}(-,P)$ (so it takes right exact sequences to left exact sequences). So certainly Grothendieck's spectral sequence does not apply here. (You might have been aware of this already.) – R. van Dobben de Bruyn May 18 '16 at 15:58

I am not sure I see the point of your worry. As $i_*$ is exact, there is really no harm in ignoring it. Hartshorne actually says this earlier and states that this "abuse of notation" will be used. As far as I can tell, $i^*$ is only used (that is "should be used, but not marked") when a sheaf supported on $X$, but considered as one on $\mathbb P^n$ is restricted to $X$. This is just using the previous principle. In other words, Hartshorne (and pretty much everyone else) adopts the maxim that

If the sheaf $\mathscr F$ on a scheme $Z$ is supported on the closed subscheme $X\subseteq Z$, then it may be considered a sheaf on $X$.

This is about the same as when a module $M$ over the ring $A$ has annihilator $I\subseteq A$, then $M$ maybe considered a module over $A/I$.

As for your lemma, it seems to me that it is indeed relatively straightforward from a Grothendieck spectral sequence argument: As you point out, (i) and (ii) imply that $F$ is actually exact (and of course, $i_*$ is, so you could even start with this). Then (i) and (iii) imply that you actually have
(iii*) $R^i G (P) = 0$ for all $i < r$ for some fixed $r \in \mathbb N$. Then considering the fact that $$R\text{Hom}(F(A), \_\_)\cong R\text{Hom}(A, G(\_\_)),$$ and applying GSS to the RHS, and pluggin in $P$ you see that by (iii*) $$E^{p,q}_2=R^p\text{Hom}(A, R^qG(P))=0$$ as long as $q<r$ (and of course it is $0$ for $p<0$), so the first non-zero term from the SW ("the lower left corner") of the spectral sequence computing the RHS is $\text{Hom}(A, R^rG(P))$. This proves both statements.

• Thanks for your answer! I guess my lack of experience with these notations at that time made me insist on writing all the proper notations everywhere. Also, thanks for the proof of the lemma. Writing it in terms of composition of derived functor and using the isomorphism theorem gives a clean way to see the result as well. But I believe for this we have to prove that for an injective $I$, $G(I)$ is acyclic for $\text{Hom} (A, \_\_)$. For this I was considering the functor in second variable and using adjointness. However, this assumes the existence of enough projectives in the category. – Shubhodip Mondal Dec 24 '16 at 23:38
• I assume there is a way to fix this, since one needs this hypothesis in your spectral sequence argument as well. Also, sorry for the late response. – Shubhodip Mondal Dec 24 '16 at 23:49
• @ShubhodipMondal: yes, you are perfectly correct, I neglected that issue. I suppose you need to add that as an assumption to your lemma. As for the application, it seems to me that you only need sheaf-Hom and being acyclic is a local property, so to check this condition you can restrict to an affine open on which there are enough projectives. What do you think? – Sándor Kovács Dec 25 '16 at 1:19
• That does seem fine for the application. But I just looked at my previous proof (which never talks about projectives) and I think one can fix this easily without that assumption. For any injective $I$, $\text{Hom}( \_\_, G(I)) = \text{Hom}(F( \_\_), I)$. Since $F$ is exact this implies that $G(I)$ is also an injective. Does this look fine? – Shubhodip Mondal Dec 25 '16 at 9:22
• Yeah, that's an even better way. :) – Sándor Kovács Dec 26 '16 at 0:37