# Derived Categories 5: Some Examples

Today we’ll just unravel some toy cases, because understanding these examples is really important, but they tend to be done in generality where the intuition goes away. The first fact is something that we’ll need to use to prove a structure theorem for objects in the derived category of a curve.

As usual, let’s fix ${X/k}$ a smooth, projective variety over a field and let ${D(X)}$ be the bounded derived category of coherent sheaves on ${X}$. In fact, our first fact will work for the derived category of any abelian category. It says that if we take an object (i.e. complex) ${A}$ whose cohomology vanishes for all ${n>i}$, then we can always put it into an exact triangle ${B\rightarrow A\rightarrow H^i(A)[-i]}$.

Let’s be extremely rough and vague to motivate this at first. Given any morphism ${B\stackrel{f}{\rightarrow}A}$ just from the axioms of a triangulated category we can always complete this to an exact triangle ${B\rightarrow A\rightarrow cone(f)}$. Roughly speaking the cone “behaves like a quotient.” I’ll remind you of the exact definition in a second. Our other motivating idea is that we always have a “truncation” process. I’ll chop the very last part of the complex off and then use the “inclusion” map. If the cone is like a quotient I’ll kill everything except that part that I chopped off, and I’ll just be left with the cohomology.

We won’t prove this in general, but we’ll do it in the case of a very small complex to reacquaint ourselves with the cone construction and quasi-isomorphisms. This is one of those cases where this toy case is actually no less general than the general proof when you see what is going on.

Recall that a morphism of complexes ${f: A^\bullet \rightarrow B^\bullet}$ is a collection of morphisms ${f^i: A^i\rightarrow B^i}$ commuting with the differentials: ${d_B\circ f^i=f^{i+1}\circ d_A}$. The complex denoted ${cone(f)=C^\bullet}$ is given by ${C^i=A^{i+1}\bigoplus B^i}$. The differentials are a little tricky, but sort of the only obvious thing that actually makes it into a complex: ${d_C^i=\left(\begin{matrix} -d_A & 0 \\ f^{i+1} & d_B\end{matrix}\right).}$

For simplicity, let’s just assume that our complex has the following form ${0\rightarrow A^0\rightarrow A^1\rightarrow A^2 \rightarrow A^3 \rightarrow 0}$. We’ll assume that cohomology at the second spot is the highest non-zero cohomology. We’ll truncate there and then form the cone. The truncated complex just looks like ${0\rightarrow A^0\rightarrow A^1\rightarrow \ker (d^2)\rightarrow 0}$. Our morphism, ${f}$, of complexes in this easy example at the three different types of places is either the identity morphism, inclusion morphism, or ${0}$ morphism.

The cone complex of this morphism is (I’ll be redundant and put in the zeros for clarity):

$\displaystyle 0\rightarrow A^0\oplus 0 \rightarrow A^1\oplus A^0\rightarrow ker(d^2)\oplus A^1 \rightarrow 0 \oplus A^2\rightarrow 0 \oplus A^3 \rightarrow 0$

One important thing to notice is that at ${A^0\oplus 0}$ is sitting in degree ${-1}$ by definition. By construction the last few maps are again just the original ${d}$ maps. Thus ${H^3(cone(f))=H^3(A^\bullet)=0}$ by assumption. Also, ${H^2(cone(f))=H^2(A^\bullet)}$. If we can show that all other cohomology vanishes, then we will have shown that ${cone(f)}$ is quasi-isomorphic to the single term complex ${H^2(A^\bullet)[-2]}$. This will show the lemma.

This part just follows by “fun with the cone construction.” Note that the first differential is ${(a,0)\mapsto (-d(a), a)}$. If this element is ${(0,0)}$, then ${a=0}$. Thus the kernel is trivial and we get ${H^{-1}(A^\bullet)=0}$. The next one is even more fun. The map is ${(a,b)\mapsto (-d(a), a+d(b))}$. The image of the previous is in the kernel of the next by just doing the maps successively ${(a,0)\mapsto (-d(a), a)\mapsto (d^2(a), -d(a)+d(a))=(0,0)}$. Now we see why that minus sign was important to make the cone a complex.

To check that the image is equal to the kernel, now suppose ${(a,b)}$ is in the kernel. In particular, this means ${a=-d(b)}$ by looking at the second term. The claim is that ${(a,b)}$ is the image of ${(b,0)}$. Well, ${(b,0)\mapsto (-d(b), b)=(a,b)}$. Thus ${H^0(cone(f))=0}$. And now you get the point. There is only one more to check, but it follows the same way.

Of course you could do this with cohomology bounded below with the other truncation functor where you replace the lowest term with a cokernel. In general, this shows an incredibly useful proposition: Given any complex in the bounded derived category of an abelian category there is a finite filtration by truncation ${0\rightarrow A_k \rightarrow A_{k+1} \rightarrow \cdots \rightarrow A_j\rightarrow A^\bullet}$ where the “quotient” (i.e. cone) at each step is ${A_{n-1}\rightarrow A_{n}\rightarrow H^{n}(A^\bullet)[-n]}$. This notation is meant to reflect that ${H^n(A^\bullet)=0}$ if ${n}$ is not in the range ${[k,j]}$.

Here is a remarkable consequence of this proposition. Let ${C/k}$ be a smooth projective curve. Any object ${\mathcal{F}^\bullet\in D(C)}$ is isomorphic to the direct sum of its cohomology sheaves ${\bigoplus_i \mathcal{H}^i(\mathcal{F}^\bullet)[-i]}$. In particular, every object in the derived category is a direct sum of coherent sheaves.

Now we see the power of being able to filtrate by cohomology, because it will allow us to make an induction argument. We prove this by inducting on the length of the complex. The base case is by definition. Suppose the result is true for a length ${k-1}$ complex. Suppose ${\mathcal{F}^\bullet}$ has length ${k}$. By shifting, we suppose ${\mathcal{H}^i(\mathcal{F}^\bullet)\neq 0}$ only in the range ${1\leq i \leq k}$. Consider the first step of the filtration ${\mathcal{E}^\bullet \rightarrow \mathcal{F}^\bullet \rightarrow \mathcal{H}^{k}(\mathcal{F}^\bullet)[-k]\rightarrow \mathcal{E}^\bullet [1]}$.

If this triangle splits we are done, because then the middle term will be a direct sum of the outer terms and by the inductive hypothesis ${\mathcal{E}^\bullet}$ splits as a direct sum of its cohomology. We may as well write it this way

$\displaystyle \mathcal{E}^\bullet\simeq \bigoplus_{i=1}^{k-1} \mathcal{H}^i(\mathcal{E}^\bullet)[-i].$

A sufficient condition to show the triangle splits is to check that ${Hom_{D(C)}(\mathcal{H}^{k}(\mathcal{F}^\bullet)[-k], \mathcal{E}^\bullet [1])=0}$. But by what we just wrote this is the same as

$\displaystyle Hom_{D(C)}(\mathcal{H}^k(\mathcal{F}^\bullet), \bigoplus_{i=1}^{k-1}\mathcal{H}^i(\mathcal{E}^\bullet)[k+1-i]) \simeq \bigoplus_{i=1}^{k-1}Ext_C^{k+1-i}(\mathcal{H}^k(\mathcal{F}^\bullet), \mathcal{H}^i(\mathcal{E}^\bullet))$

Since smooth curves have homological dimension ${1}$ and the exponent is always strictly larger than ${1}$ the Ext groups all vanish. This shows the triangle splits and hence any object in the derived category of a smooth curve splits as a sum of its cohomology sheaves. This exact same proof also shows that for any abelian category with homological dimension less than or equal to ${1}$ the objects of the derived category split as a sum of their cohomology.