# 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.

# This Week’s Finds in Arithmetic Geometry

It has been going around the math blogosphere that in honor of John Baez’s 20 year anniversary of doing This Week’s Finds we all do one in our area. So here’s a brief This Week’s Finds in Arithmetic Geometry. Hopefully this will raise awareness of the blog that essentially pioneered math/physics blogging (and if you’re into arithmetic geometry some papers you might not have caught yet).

Since the content and style of Baez’s This Week’s Finds vary so much, I’ll just copy what Jordan Ellenberg did here and give some papers posted the last week that caught my attention.

The fields of definition of branched Galois covers of the projective line by Hilaf Hasson caught my attention because I just met him and saw him speak on this exact topic a few weeks ago at the Joint Meetings. Although the results are certainly interesting in themselves, the part that a blog audience might appreciate is the set of corollaries to the results.

Recall that a major open problem in number theory is the “Inverse Galois Problem” which asks which groups arise as Galois groups. I even posted an elementary proof that if you don’t care what your fields are, then any finite group arises as ${Gal(L/K)}$. In general (for example if you force ${K=\mathbb{Q}}$), then the problem is extremely hard and wide open.

If you haven’t seen this type of thing before, then it might be surprising, but you can actually use geometry to study this question. This is exactly the type of result that Hilaf gets.

Next is New derived autoequivalences of Hilbert schemes and generalised Kummer varieties by Andreas Krug. This topic is near and dear to me because I study derived categories in the arithmetic setting. I haven’t taken a look at this paper in any depth, but I’ll just point out why these types of things are important.

In the classification of varieties one often tries to study the problem up to some type of birational equivalence otherwise it would be too difficult. Often times birational varieties are derived equivalent, but not the other way around. So one could think of studying varieties up to derived equivalence as a slightly looser classification.

When trying to figure out what two varieties that are derived equivalent have in common, a typical sticking point is that you need to know certain automorphisms of the derived category (i.e. autoequivalence) exist to get nice cohomological properties or something. When papers constructing new autoequivalences come out it always catches my attention because I want to know if the method used transfers to situations I work in.

Lastly, we’ll do La conjecture de Tate entière pour les cubiques de dimension quatre sur un corps fini by François Charles and Alena Pirutka. If you don’t know what the Tate conjecture is, then lots of people refer to it as the Hodge conjecture in positive characteristic.

If you’ve ever seen any cohomology theory, then you should be at least passingly familiar with the idea that certain sub-objects (subvarieties or submanifolds etc) can be realized as classes in the cohomology. Sometimes this is due to construction and sometimes it is a major theorem.

The particular case of the Tate conjecture says the following. Consider the relatively easy to prove fact. If you take a cycle on your variety ${X/k}$, then the cohomology class it maps to (in ${\ell}$-adic cohomology) will be invariant under the natural Galois action ${Gal(\overline{k}/k)}$ (because it is defined over ${k}$!). The Tate conjecture is that any Galois invariant cohomology class actually comes from one of these cycles.

The fact that mathematicians can have honest arguments over whether or not the Tate conjecture or the Hodge conjecture (a million dollar problem!) is harder just gives credence to the fact that it is darned hard. If you weren’t convinced, then just consider that this paper is proving the Tate conjecture in the particular case of smooth hypersurfaces of degree ${3}$ in ${\mathbb{P}^5}$ just for the cohomology classes of degree ${4}$. People consider this progress, and they should.