# Why is the derived tensor product only defined for bounded above derived categories?

In "Residues and Duality" by Hartshorne, the derived tensor $\otimes$ only defined for the bounded above categories (see Chapter I, section 4), that is one has

$$\otimes: D^{-}(X) \times D^{-}(X) \to D(X),$$ where $X$ is a variety, and $D^{-}(X), D(X)$ are derived category of bounded above complex of $\mathcal{O}_X$-modules and derived category of complex of $\mathcal{O}_X$-modules respectively .

The main reason it is defined for the bounded above complex is that Hartshorne used the fact that every sheaf of module has a flat resolution.

However, why cannot one use the injective resolution for the derived tensor, and defined it for bounded below derived cateogry $D^{+}(X)$?

In fact, the later is the case I need (to be precise, I need $F\otimes : D^{+}(X) \to D(X)$, where $F$ is a sheaf of module), and I checked every conditions for the existence of derived functor, it seems that everything is satisfied.

• I don't see how you can use injective resolutions for a left derived functor. – Zhen Lin Mar 6 '14 at 16:20
• That doesn't make sense! You need a resolution by objects $P$ which are acyclic, in this case such that the functor $P\otimes -\$ is exact. – abx Mar 6 '14 at 17:17
• @ZhenLin (and also abx) I know this is weird, but if you have the book I mentioned above, you can look at Corollary 5.3 Page 56. It says: if $A,B$ are abelian categories, and $F$ is an additive functor, assume $A$ has enough injectives, then there exist derived functor $\mathbb{R}^+F : D^{+}(A) \to D(B)$. – Li Yutong Mar 7 '14 at 1:22
• The derived tensor product is a left derived functor. There is no use making it a right derived functor. – Zhen Lin Mar 7 '14 at 7:59
• Could you take a look at my new question below? It contains more information. mathoverflow.net/questions/159668/… – Li Yutong Mar 7 '14 at 13:57

## 2 Answers

I would like to add my 2 euro ¢. In a nutshell: your information is outdated. "Residues and Duality" by Hartshorne was the first available text on derived categories of coherent sheaves and Grothedieck duality but it was published in 1966. In 1988, Spaltenstein showed in ("Resolutions of unbounded complexes", Compositio Math. 65 (1988), no. 2, 121–154) that one can use unbounded homotopically flat resolutions of sheaves to get a derived tensor product of unbounded complexes. Similarly, you may use homotopically injective resolutions of sheaves to derive functors on the right.

A very readable and useful update of duality for coherent sheaves using this developments is Joe Lipman's text "Notes on Derived Functors and Grothendieck Duality" in Springer Lecture Notes, no. 1960 (2009), 1–259.

There is a lot of extra literature on variants for Grotehdieck abelian categories, differential graded algebras and so on.

• @Adeel The paper you mention is a great reference, but it uses model categories, the ones I gave restrict to methods of homological algebra, enough for a lot of purposes. In any case, a general knowledge of model categories is useful too. – Leo Alonso Mar 6 '14 at 18:08
• It should be pointed out that life isn't perfect: you lose some theorems when you extend your complexes to be unbounded and use Spaltenstein's methods. For example, the proper base change theorem no longer holds in the same generality as in the bounded case. (See Higher Topos Theory Remark 6.5.4.3.) That said, if you're working with nice $X$ you should be okay. – Dylan Wilson Mar 6 '14 at 20:11
• @LeoAlonso Thank you so much for your wonderful answer. Suppose I want to define $\otimes F: D^{+}(X) \to D(X)$ where $F$ is a sheaf. For any complex $G^{\bullet} \in D^{+}(X)$, I can first do an injective resolution in bounded below homotopy category, and then do the ordinary tensor with $F$. Then one projects the complex $F \otimes G^{\bullet}$ down to the derived category $D(X)$. Is that correct? – Li Yutong Mar 7 '14 at 1:27
• I am not an expert of these things, but I don't think you get the correct result using an injective resolution. The reason is that injective objects aren't acyclic for the tensor product (if you want it is not true that $\otimes I$ is exact for injective $I$) and this implies that $F\otimes G^\bullet$ depends on the particular representative $F$ and not on its quasi-isomorphism class. – Denis Nardin Mar 7 '14 at 4:01
• @LiYutong You want to use homotopically flat. Makes sense, because you want to define a derived tensor product :) Ideally you might want to use homotopically projective objects for your goal but they don't exist in this setting. Injectivity has nothing to do with tensor products, but Hom and things like that. It's good not to lose sight of the forest from the trees or whatever the expression is. – user36931 Mar 7 '14 at 5:59

You can read about this in many references, for example in Section Tag 06Y7. It also gives some refences.