# Counterexamples to gluing complexes of sheaves

Note: I asked the question below last week on MathSE but received no answer.

Background:

I have read the claim that perverse sheaves behave more like sheaves than like complexes of sheaves. This refers to the fact that they can be glued.

For instance, suppose that $X$ is a complex analytic space and $P_1^{\bullet}, P_2^{\bullet}$ are perverse sheaves defined on the open sets $U_1, U_2$ respectively. Then if we are given an isomorphism $\alpha_{12}: P_1^{\bullet}|_{U_1 \cap U_2} = P_2^{\bullet}|_{U_1 \cap U_2},$ then there exists a unique (up to canonical isomorphism) perverse sheaf $P^{\bullet}$ defined on $U_1 \cup U_2$ with the property that there are isomorphisms $\beta_i:P^{\bullet}|_{U_i} \to P_i^{\bullet}$ with ${\beta_2}|_{U_1 \cap U_2}={(\alpha_{12} \circ \beta_1) }|_{U_1 \cap U_2}.$

More generally, if one has an open cover $\{U_i\}$ of $X$ and perverse sheaves $P_i^{\bullet}$ of with isomorphisms on the overlaps $U_i \cap U_j$ satisfying the co-cycle condition, then this data glue in the usual way.

My question:

What goes wrong if one tries to glue ordinary complexes of sheaves? Are there counterexamples showing that the gluing property cannot hold in $C^{\bullet}(Sh(X))$ (the category of sheaves on $X$) or in $D^b(X)$ (the bounded derived category of sheaves on $X$)?

• Just out of curiosity, because it came to mind: have you tried the simplest things possible, like trying to glue complexes of vector bundles and skyscraper sheaves on $\mathbb P^1$? – Tabes Bridges Nov 28 '17 at 5:49
• Your description of how to glue is off. The statement is that the choice of an isomorphism $P_1 \vert_{U_1 \cap U_2} \cong P_2 \vert_{U_1 \cap U_2}$ defines a unique sheaf $P$ on $U_1 \cup U_2$. Different choices of isomorphism will lead to different $P$'s. – Dan Petersen Nov 28 '17 at 7:14
• Think of when you have a vector bundle on $S^2$, covered by two "caps" homeomorphic to $\mathbb R^2$. Your vector bundle has to be trivial on both $U_1$ and $U_2$, so if your statement were literally true every vector bundle on $S^2$ would be trivial. The correct statement is that a vector bundle on $S^2$ is completely determined by a "clutching function", which is exactly a choice of an isomorphism between two trivial vector bundles on $U_1 \cap U_2$. – Dan Petersen Nov 28 '17 at 7:14
• Math.SE copy: Counterexamples to gluing complexes of sheaves – Martin Sleziak Nov 28 '17 at 7:18
• I think you can glue complexes of sheaves with on-the-nose isomorphisms. The trouble is that you can't glue them with only quasi-isos in the derived category, and if you are obtaining your original complexes of sheaves by resolution, you won't get on-the-nose isomorphisms. I asked Adeel Khan this exact question yesterday, by chance, and that is what he said. You can, however, glue things like cotangent complexes if you move to a fully derived setting like derived schemes since you also have all the higher coherences that are missing in the derived category. – Harry Gindi Nov 28 '17 at 14:25

## 3 Answers

This doesn't quite answer your question, but it might be useful.

First of all, note that it is extremely unbelievable (please, correct me if I am wrong) that you can glue objects of (bounded) derived category, if you cannot glue morphisms. One reason is that you usually construct gluing "by induction on number of open subsets". But you can't do this induction, if you can't glue morphisms.

Also, from category theoretic point of view it is more reasonable to ask for a functor $R:GluingData(\{U_i\}_{i\in I}) \to D^b(X)$ from the category of gluing data (in bounded derived category) with respect to an open covering $\{U_i\}_{i\in I}$ to the bounded derived category $D^b(X)$.

With all these said, let me show that you can't glue morphisms in (bounded) derived categories. Namely, I will show that morphisms are not uniquely defined by gluing data.

Pick any noetherian scheme X with a non-split short exact sequence of vector bundles $$0 \to \mathcal E' \to \mathcal E \to \mathcal E'' \to 0.$$ [An explicit example of this will be a scheme $\mathbf P^1_k$ and an extension $$0 \to \mathcal O(-2) \to \mathcal O(-1)^{\oplus 2} \to \mathcal O \to 0.]$$ This extension defines a class $\xi \in Ext^1(\mathcal E'', \mathcal E')=Hom_{D^b(X)}(\mathcal E'', \mathcal E'[1])$. So, we can think of $\xi$ as a morphism in the derived category. Now choose any cover of $X$ by open affine subschemes $\{U_i\}_{i\in I}$. The restriction of $\xi$ onto each open $U_i$ is equal to $0$ because we don't have any higher Ext groups between locally free sheaves on affine schemes. Thus $\xi|_{U_i}=0|_{U_i}$ for each $i$, but $\xi \neq 0$. So there is no chance to glue morphisms in (bounded) derived categories.

UPDATE: Actually, it seems that you can glue two objects of bounded derived categories. In order to make everything precise let me say what I mean by "bounded derived category". For me $D^b(X)$ will actually mean $D^b(Qcoh(X))$ and I will denote by $D^b_{coh}(X)$ the bounded derived category of coherent sheaves.

Please, check it carefully. There might be a mistake. The result looks quite mysterious for me.

Claim: Let $X$ be any noetherian scheme and let $U,V$ be two open subschemes of $X$. Given two objects $K\in D^b_{coh}(U), L \in D^b_{coh}(V)$ and an isomorphism $c:K|_{U\cap V} \to L|_{U\cap V}$ we can construct an element $F\in D^b_{coh}(X)$ s.t. $F|_{U}\cong K$ and $F|_{V} \cong L$.

Proof: Let us denote the intesection $U\cap V$ by $W$ and consider the derived pushforward functors $$\mathbf Rj_{U}:D^b(U) \to D^b(X), \mathbf Rj_{V}:D^b(V) \to D^b(X)\text{ and }\mathbf Rj_{W}:D^b(W) \to D^b(X).$$

Note that there is a natural morphism $$b:\mathbf Rj_{V}(L) \to \mathbf Rj_{W}(L|_W)$$ that is adjoint to the identity morphism $$\mathbf Rj_{V}(L)|_{W} \xrightarrow{\mathbf 1} L|_W.$$ Also, there is a natural morphism $$a:\mathbf Rj_{U}(K) \to \mathbf Rj_{W}(L|_W)$$ that is adjoint to the composition $$\mathbf Rj_{U}(K)|_{W} \xrightarrow{\mathbf 1} K|_{W} \xrightarrow{c} L|_{W}.$$ Taking sum of these morphisms we obtain a map $$\phi:\mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L) \xrightarrow{(a,b)} \mathbf Rj_{W}(L|_W).$$ By the TR3 axiom of triangulated category we can find an object $F\in D^b(X)$ such that the following triangle is distinguished $$F\xrightarrow{(f_1,f_2)} \mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L) \to \mathbf Rj_{W}(L|_W) \to F[1] (*).$$ I claim that $F|_{U}$ is isomorphic to $K$ (and the same for $V$ and $L$). Indeed, restrict the triangle $(*)$ on $V$: $$F|_{U}\to K\oplus \mathbf Rj_{V}(L)|_{U} \xrightarrow{a|_{U}, b|_{U}} \mathbf Rj_{W}(L|_W)|_{U} \to F|_{U}[1] (*').$$ But $b|_{U}:\mathbf Rj_{V}(L)|_{U} \to Rj_{W}(L|_W)|_{U}$ is an isomorphism (basically, by definition of $b$). Thus the triangle $(*')$ has a section. The general result about triangulated categories implies that $f_1|_{U}: F|_{U} \to K$ is an isomorphism.

An argument is a bit trickier for $V$ because we a priori know only that $a|_{W}$ is an isomorphism, not $a|_{V}$. In other words, the natural map $$f_2|_{V}:F|_{V} \to L$$ is an isomorphism on $W$. Hence, the support of its cone is concentrated on $Y:=V-W$. Therefore there is a scheme structure on $Y$ s.t. $Cone(a|_{V})=\mathbf Ri_Y(Q)$ for some element $Q\in D^b(Y)$ (where $i_{Y}:Y\to V$ is the natural closed immersion).

In order to show that $f_2|_{V}$ is actually an isomorphism, let us recall Mayer–Vietoris distinguished triangle $$F \to \mathbf Rj_{U}((F)|_{U})\oplus \mathbf Rj_{V}((F)|_{V}) \to \mathbf Rj_{W}((F)|_{W}) \to F[1] (**).$$

Observe that there is a morphism of distinguished triangles induced by a pair $(\mathbf Rj_{U}(f_1|_{U}), \mathbf Rj_{V}(f_2|_{V}))$ (I don't know how to type commutative diagrams here) $$F \to \mathbf Rj_{U}((F)|_{U})\oplus \mathbf Rj_{V}((F)|_{V}) \to \mathbf Rj_{W}((F)|_{W}) \to F[1] (**) \\ \downarrow \\ F\to \mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L) \to \mathbf Rj_{W}(L|_W) \to F[1] (*).$$ The first map $F \to F$ is an isomorphism by definition, the third map $\mathbf R j_W F|_W \to \mathbf R j_W L|_{W}$ is an isomorphism since $f_1|_{W}$ and $f_2|_{W}$ are isomorphisms. Thus we conclude that the second morphism $$\mathbf Rj_{U}((F)|_{U})\oplus \mathbf Rj_{V}((F)|_{V}) \to \mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L)$$ is an isomorphism. In particular, $\mathbf Rj_{V}(f_2|_{V})$ is an isomorphism. Apply this result to a distinguished triangle $$F|_{V} \xrightarrow{f_2|_{V}} L \to \mathbf Ri_{Y}(Q) \to F|_{V}[1]$$ to obtain the result that $$0\cong \mathbf Rj_{V} \mathbf Ri_{Y}(Q) \cong \mathbf Ri_{Y\to X}Q.$$ On the other hand, $\mathbf Li_{Y\to X}\mathbf Ri_{Y\to X} Q \cong Q$ since $i_{Y \to X}$ is a finite map. Thus we see that $$Q\cong \mathbf Li_{Y\to X}\mathbf Ri_{Y\to X} Q \cong 0.$$ So, the cone $Cone(f_2|_{V}) \cong 0$. In other words, $f_2|_{V}$ is an isomorphism. Therefore, we see that $K|_{V} \cong L$.

The last thing to check is that $K\in D^b_{Coh}(X)$, but it is clear since $K|_{U}\in D^b_{Coh}(U)$ and $K|_{V}\in D^b_{Coh}(V)$. So, we are done!

• +1 Nice answer. – Ben Lim Dec 1 '17 at 22:05

One can ask whether the derived category forms a stack (of triangulated categories). The answer is no:

Let $S^2$ denote the two-sphere (or $\mathbb{P}^1$ if you prefer) and let $k$ denote a ring of coefficients (e.g. $\mathbb{Z}$). Let $k_{S^2}$ denote the constant sheaf on $S^2$. Consider a map $\alpha: k_{S^2}[-2] \to k_{S^2}$ given by a nontrivial element of $Hom(k_{S^2}[-2],k_{S^2}) = H^2(S^2)$. For any proper subset $U \subset S^2$ the restrction map $H^2(S^2) \to H^2(U)$ is zero. Thus if the derived category of complexes of sheaves were a stack then $\alpha$ would be zero, but it isn't! (Of course $S^2$ isn't special ... you just need non-trivial cohomology.)

Now let me get to your question. Suppose that we have $F$ on $U$, $G$ on $V$ and an isomorphism $F_{U \cap V} = H = G_{U \cap V}$. Then if C denotes the cocone of the complex

$C \to F \oplus G \to H \stackrel{[1]}{\to}$

(I have omitted lots of $j_*$'s) then $C$ will have the property you want, but it won't be defined up to canonical isomorphism. I guess you'll also get intro trouble if you have more complicated covers, because of non-canonicity of cones.

Also, note that you won't be able to get all objects in this way. For example, I think it is impossible to construct the cone of $\alpha$ (where $\alpha$ is as in the first paragraph), by glueing of things defined on proper open subsets.

I'm not sure if you are still interested in this question. Actually for an open cover $$\{U_i\}$$ and complexes of sheaves on each $$U_i$$, we could give the "higher" descent data and "higher" cocycle conditions in terms of twisted complexes, which was first introduced by O’Brian, Toledo and Tong in late 1970's and 1980's. I have written a paper to describe twisted complexes as a dg-enhancement of the derived category of sheaves on the space (sorry for the self-citation). Later in a joint paper with Block and Holstein we proved that twisted complexes indeed give the homotopy limit of cosimplicial diagrams of dg-categories coming from geometry (we can consider homotopy limit as a more formal way to say descent data).

I would like to mention that in a recent preprint, the authors generalized the main result in the paper of Block, Holstein and me to arbitrary cosimplicial diagrams of dg-categories.