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Three filtrations on the Grothendieck ring of a scheme

Today’s blog post is inspired by the Chern classes seminar that we are conducting, but it takes a rather big leap from where I left off. Everything I say is based on SGA6 and Fulton–Lang’s Riemann–Roch algebra (which is a true gem I discovered).

The lambda-ring associated to a scheme

If you have a sufficiently nice scheme X (the buzzwords are connected, noetherian and having an ample line bundle) and let’s even assume that \dim(X)=d. Then we have a Grothendieck group \mathrm{K}^0(X), obtained by considering the category of locally free sheaves on X and modding out exact sequences of locally free sheaves. Or, as is explained in SGA6 IV, one can take the (triangulated) category of perfect complexes and mod out triangles, where SGA6 II addresses the issues of being globally (and not just locally) isomorphic to a complex of locally free sheaves.

The tensor product of locally free sheaves induces a multiplication, hence we obtain a Grothendieck ring.

So far so good. Now we can turn this into a lambda-ring. The problem with this extra structure on a ring is that writing down all the axioms is boring. It helps if you know where the structure actually comes from: besides direct sums and tensor products we also have the exterior product of locally free sheaves. If we were to take globally free sheaves, we can prove that, just as in linear algebra, we have a decomposition

\bigwedge^n(\mathcal{E})\cong\bigoplus_{i=0}^n\left( \bigwedge^i(\mathcal{E}')\otimes\bigwedge^{n-i}(\mathcal{E}'') \right)

if 0\to\mathcal{E}'\to\mathcal{E}\to\mathcal{E}''\to 0 is an exact sequence of globally free sheaves. In the case that the exact sequence is only one of locally free sheaves there is no global decomposition, but there exists a filtration whose quotients are exactly the terms in the decomposition above. If you feel like it: do Hartshorne, exercise II.5.16(d). Hence, in the Grothendieck ring we have a sum decomposition

\left[ \bigwedge^n(\mathcal{E}) \right]=\sum_{i+j=n}\left[ \bigwedge^i(\mathcal{E}')\otimes\bigwedge^j(\mathcal{E}'') \right]

which says that exterior powers are compatible with the equivalence relation. Now that we know where the intuition comes from, we can formalise things.

Definition. Let A be a commutative ring. Then a lambda-ring structure consists of endomorphisms \lambda^i\colon A\to A for all i\in\mathbb{N}, such that

  1. \lambda^0(x)=1 for all x\in A;
  2. \lambda^1(x)=x for all x\in A;
  3. \lambda^n(x+y)=\sum_{i=0}^n\lambda^i(x)\lambda^{k-i}(y) for all x,y\in A.

So we have just formalised the notion of exterior products. Moreover we have a notion of rank for a locally free sheaf, which by linearity induces a surjective ring homomorphism r\colon A\to\mathbb{Z}.

We are not quite there yet, because actually we are working with special lambda-rings. The splitting principle in intersection theory, which allows us to treat vector bundles as if they are iterated extensions of line bundles, gives extra data. Namely we have expressions for \bigwedge^n\left( \bigwedge^m\mathcal{E} \right) and \bigwedge^n\left( \mathcal{E}\otimes\mathcal{F} \right) in terms of the Chern roots of the original bundles. We will not need these at the moment, but maybe I will write something about this later.

Three filtrations

The title suggests that there are three filtrations on this ring. It’s about time I say which, and how they relate.

The topological filtration

The easiest filtration is the topological filtration. In the introduction SGA6 it is defined in a special case, for the presentation of the Grothendieck group via coherent sheaves, by just saying that \mathrm{F}_{\mathrm{top}}^i\mathrm{K}^0(X) is generated by coherent sheaves whose support has codimension bigger than i. In Fulton–Lang a more formal definition is given in terms of (global) perfect complexes. The idea is to use cohomological support of complexes and codimension, but I don’t feel like spelling out details.

One can prove that it is compatible with the ring structure, and under the assumption that \dim(X)=d it is a filtration of length \leq d.

The lambda-filtration

The hardest filtration is the lambda-filtration, or Grothendieck filtration, or gamma-filtration. I’m not sure whether my intuition is entirely correct, but it seems that the motivation is to use the splitting principle to obtain a finer filtration than the topological filtration.

To define this we put a different lambda-structure on our lambda-ring, by considering endomorphisms \gamma^i\colon\mathrm{K}^0(X)\to\mathrm{K}^0(X) defined by

\sum_{i\in\mathbb{N}}\lambda^i(x))\left( \frac{t}{1-t} \right)^i=\sum_{i\in\mathbb{N}}\gamma^i(x)t^i

i.e. we consider a formal power series ring over our lambda-ring, we choose a different generator for it and write out the change in coefficients.

We then set

  1. \mathrm{F}_\lambda^0\mathrm{K}^0(X)=\mathrm{K}^0(X);
  2. \mathrm{F}_\lambda^1\mathrm{K}^0(X)=\mathrm{ker}(r), i.e. those elements which are of (formal) rank 0;
  3. \mathrm{F}_\lambda^n\mathrm{K}^0(X) is generated by the expressions \prod_{i=1}^k\gamma^{r_i}(x_i), for x_i\in\mathrm{F}^1\mathrm{K}^0(X) and \sum_{i=1}^kr_i\geq n.

This induces a descending sequence of ideals in the Grothendieck ring. One can prove that \mathrm{F}_{\mathrm{top}}^1\mathrm{K}^0(X)=\mathrm{F}_\lambda^1\mathrm{K}^0(X) (Fulton–Lang, V.3.5), and the proof boils down to the silly fact that [\Sigma\mathcal{E}^\bullet]=-[\mathcal{E}^\bullet], if we use the presentation of the Grothendieck group via complexes.

In general we have that the lambda-filtration is finer than the topological filtration (Fulton–Lang, V.3.9), i.e. \mathrm{F}_\lambda^n\mathrm{K}^0(X)\subseteq\mathrm{F}_{\mathrm{top}}^n\mathrm{K}^0(X), and the proof of this fact uses the splitting principle, which reduces the statement to line bundles, then we use the equality in degree 1 and the fact that \mathrm{F}_{\mathrm{top}}^i is compatible with the ring structures.

One of the main results (SGA6 VII) is that the associated graded rings are, up to torsion, isomorphic!

The third filtration

I was aware of the previous two filtrations in SGA6, but hidden inside SGA6 exposé VI.6.5 (not IV, as indicated in Fulton–Lang) there is a third filtration. This blog post has gone on for long enough now and the filtration has properties similar to the topological filtration (but it is not functorial), hence I won’t discuss it. But reading SGA6 and Fulton–Lang was fun (in some perverse way, maybe), so maybe I will come back to it later.

Only five smooth projective toric Fano surfaces

The past few weeks I’ve been dabbling in some toric geometry, and the following result in toric geometry puzzled me when I first read it:

The only smooth projective toric Fano surfaces are \mathbb{P}^2,\mathbb{P}^1\times\mathbb{P}^1,\mathrm{Bl}_1\mathbb{P}^2,\mathrm{Bl}_2\mathbb{P}^2,\mathrm{Bl}_3\mathbb{P}^2.

If you are as ignorant about toric geometry as I am (i.e. you’ve looked at Cox–Little–Schenck’s marvellous book but not much more than that) this statement might be a bit puzzling. Because as you might know: the del Pezzo surfaces (a shorthand term for smooth projective Fano varieties of dimension 2) are classified, and these are exactly the surfaces \mathbb{P}^2,\mathbb{P}^1\times\mathbb{P}^1,\mathrm{Bl}_1\mathbb{P}^2,\ldots,\mathrm{Bl}_9\mathbb{P}^2, where the points are chosen sufficiently general.

So why did this fact puzzle me? Well: there is a perfectly good notion of blowing up a toric variety. Hence why do we not get \mathrm{Bl}_4\mathbb{P}^2 as a toric del Pezzo surface? Surely it is del Pezzo, and via a blow-up we can realise it as a toric surface, right?!

No! Think about this for a while. If you don’t know this: a toric blow-up is not an arbitrary blow-up, it is a blow-up in a torus-invariant point. This is really the crucial point.

In the case of \mathbb{P}^2 there are 3 torus-invariant divisors, and they form a triangle of \mathbb{P}^1‘s in \mathbb{P}^2. If you do a toric blow-up you blow up in one of the intersections. You get a square of \mathbb{P}^1‘s in \mathrm{Bl}_1\mathbb{P}^2, hence there are now 4 intersections to choose from for your toric blow-up, but only two of them come from \mathbb{P}^2. If you blow-up one of those two original points it was as if you performed a simultaneous blow-up, to get \mathrm{Bl}_2\mathbb{P}^2, but if you pick one of the other 2 points, which lie on the exceptional divisor, you are getting something different.

Hence, this is why you cannot get \mathrm{Bl}_4\mathbb{P}^2 as a toric del Pezzo surface: it is not toric! It is all completely trivial, but it was a nice thing to realise for me, and what is better than sharing your ignorance with the world?

What are Chern classes?

This year we are continuing ANAGRAMS, the local student seminar (previously announced on my weblog). The first series of lectures we are starting with will be about Chern classes, and I will try to blog more regularly about the things we discuss. Hence in this post I will outline the motivation and interpretation of Chern classes (the gory details will not be discussed here).

Characteristic classes

People like vector bundles. They are easy-to-understand coherent sheaves, we have some obvious candidates (tangent and cotangent bundles, line bundles) and there are many interesting constructions for them (duals, direct sums, tensor products, symmetric products, exterior products, …), but nevertheless they carry significant amounts of geometric information. Even if you don’t like vector bundles as such, some seemingly totally non-vector bundleish questions such as “how many lines lie on a cubic surface” can be rephrased in terms of vector bundles and then we can actually solve them!

So what is there to know about vector bundles? We know that locally they are just copies of the structure sheaf, so the main question to ask about your vector bundle is whether it is globally (non)trivial. Continuing this train of thought one should ask, whenever a vector bundle is non-trivial, whether we can say in which sense it is non-trivial. This admittedly vague statement will be made more precise (but not too precise) in the next section.

So how can we quantify information about vector bundles? Using characteristic classes! So what are they? Depending on which context you are working in (differential geometry, complex geometry, algebraic geometry) you have different cohomology theories associated to the space on which you are considering vector bundles. There is Betti cohomology, the Chow ring, de Rham cohomology, … The characteristic class of your vector bundle is then an element in this cohomology theory, which then measures something about the vector bundle: how non-trivial it is, and in which sense it is non-trivial. If you are lucky then you can interpret the characteristic class of your bundle, and draw an interesting conclusion.

To summarise: pick your favourite cohomology ring associated to your variety, this is a graded ring. Then the characteristic classes of your vector bundle (there are possibly multiple non-zero classes) all lie in one particular piece of this graded ring.

Chern classes

Let X be a smooth projective variety over some algebraically closed field (one can get by with much less). If one takes the Chow ring \mathrm{CH}^\bullet(X) for the receptacle of this characteristic classes mumbo-jumbo one obtains Chern classes. The Chow ring is defined in terms of cycles on X, and hence our Chern classes describe something in terms of the subvarieties: \mathrm{c}_i(\mathcal{F}) is an element of \mathrm{CH}^i(X), hence defines some cycle of codimension i.

One can interpret these Chern classes in terms of degeneracy loci. Recall that we were trying to measure in which sense a vector bundle is non-trivial, and for ease of statement we will assume that our vector bundle is globally generated. If we denote r=\mathrm{rk}(\mathcal{F}), then for any k=1,\dotsc,r we can consider global sections s_1,\dotsc,s_{r-k+1}. If we evaluate these in a point we get r-k+1 vectors of length r, hence we can ask whether these are linearly (in)dependent. The degeneracy locus of our set of global sections is then exactly the set of points in which the evaluations become linearly dependent, hence they degenerate.

Of course one has to choose these sections sufficiently generally in order to make the codimension of the degeneracy locus correct, but one can prove that if the codimension is correct and the vector bundle is globally generated, then the degeneracy locus is independent of the choice! This can then be one way of defining the Chern classes of a vector bundle.

Example

During the series of lectures I intend to have people discuss some “numbers you should recognise”. Hence you have to explain why e.g. the number 2875 is interesting from the viewpoint of an enumerative geometer. As the toy example to introduce vector bundles in enumerative geometry I will do the number 27, which is as everyone should know, the number of lines on a smooth cubic surface.

How can we phrase the question

How many lines lie on a cubic surface?

in terms of vector bundles, and therefore apply Chern class machinery to try and solve it? The Grassmannian G=\mathrm{Grass}(2,4) parametrising lines in \mathbb{P}^3 or planes in \mathbb{A}^4 is a smooth projective variety (more specifically: it is a quadric hypersurface in \mathbb{P}^5) that comes equipped with a tautological bundle \mathcal{F} (not to be confused with the canonical bundle, which is the highest exterior power of the cotangent bundle).

This tautological bundle has rank 2, and is a subbundle of the trivial vector bundle \mathcal{O}_G^{\oplus 4}, obtained by taking as the fiber in each point the 2-dimensional subspace (of the 4-dimensional vector space) that corresponds to the point of the Grassmannian. The inclusion \mathcal{F}\hookrightarrow\mathcal{O}_G^{\oplus 4} dualises to a surjection \mathcal{O}_G^{\oplus 4,\vee}\twoheadrightarrow\mathcal{F}^\vee, which corresponds to restricting a linear function on \mathbb{P}^3 to the line. Taking dth symmetric powers of this surjection corresponds to restricting a homogeneous degree d polynomial on \mathbb{P}^3 to the line.

Now let X be the hypersurface defined by a cubic, to this corresponds a f, a section of \mathrm{Sym}^3(\mathcal{O}_G^{\oplus 4}) which is mapped to \mathrm{Sym}^3(\mathcal{F}^\vee) by the symmetric cube of the map from the previous paragraph. Hence we get a global section of the symmetric cube of our tautological bundle, which describes exactly the set of lines in \mathbb{P}^3 on which f vanishes!

Now, assuming this set is finite, we can apply some intersection theory machinery (see section 14.1 of Fulton’s bible) which says that inside the correct Chow group the zero locus of our global section is described by the top Chern class of our vector bundle! Hence it is completely determined in terms of Chern classes, and if we can compute this particular Chern class (and its degree) we know exactly how big this set is. To see why this is the case: observe that

\mathrm{rk}(\mathrm{Sym}^3(\mathcal{F}^\vee))={2+3-1 \choose 3}=4=\mathrm{dim}(G),

hence the top Chern class of this vector bundle lands precisely where it should land.

To determine this top Chern class \mathrm{c}_4(\mathrm{Sym}^3\mathcal{F}^\vee) one has to study how Chern classes behave with respect to symmetric powers and express it in terms of \mathcal{F}^\vee only, which gives

\mathrm{c}_4(\mathrm{Sym}^3\mathcal{F}^\vee)=9\mathrm{c}_2(\mathcal{F}^\vee)^2+18\mathrm{c}_1(\mathcal{F}^\vee)^2\mathrm{c}_2(\mathcal{F}^\vee).

Now we have to interpret the values in the right-hand side, which can either be done using general Schubert calculus on a Grassmannian variety (the moduli interpretation of these varieties gives strong tools to compute things on these varieties), or in this small case one can resort to explicit interpretations. Either way, the products turn out to be 1 in both cases, hence there are 27 lines on a cubic surface.

This is certainly not the only thing Chern classes are good for, but I think it’s a nice example of showing how constructions with vector bundles are reflected by operations on Chern classes, and how enumerative geometry can be rephrased in terms of intersection theory.

Why homotopy colimits in triangulated categories are defined the way they are

This post is one big triviality. But I only realised now why homotopy colimits in triangulated categories are defined the way they are defined.

Just think of triangulated categories as a bit wonky abelian categories, and in (non-wonky) abelian categories a colimit can be constructed from having coproducts and coequalisers. In a triangulated category you don’t have coequalisers / cokernels, but cones play this role. Hence one takes a particular cone of a direct sum that simulates a coequaliser, and one is done.

Of course, everything dualises to homotopy limits.

Why did no-one tell me this?!

Each statement is to be interpreted under suitable conditions on the objects in sight, e.g. existence of direct sums etc. etc.

Comparison of topologies

Last year I made a graph comparing a bunch of Grothendieck topologies, and I had been toying with the idea of making an interactive web version of it. Two weeks ago Shane Kelly published a preprint Points in algebraic geometry, while Stefan Schroër tackled in a preprint that appeared on the same day the intriguingly hard case of Points in the fppf topology.

The article by Kelly contains a comparison graph, just like the one I made last year (mine emphasised the “usual” topologies more, and had less topologies), but also describes what type of rings act as stalks for this topology, and gave short definitions of each of the topologies in use. Hence I had all the data to make a complete comparison!

There is a very preliminary draft available. I’ll be away from the internet for 2 weeks, hence if you are looking for a summer programming project and feel like contributing, there is the GitHub repository. I want to:

  1. Improve the layout (I’m a horrible designer, as you can tell, and my SVG skills are rather sketchy).
  2. Add explanations on how to use the comparison graph.
  3. Fix all the TODOs, the code is riddled with them…
  4. Add more data (you don’t have to be a programmer for this): I am especially looking for improved comparison results, more topologies, … I know there is a fpuo and a l' topology out there. Are there more topologies in use?
  5. Add the canonical and extensive topology to the graph, and indicate these properties in the “passport” as well.
  6. I want a tighter integration with the Stacks project, preferably all the result that are used should turn up there, this requires both writing for the Stacks project and providing the functionality in the comparison graph.
  7. Right now everything is for \mathrm{Sch}/S, the category of separated schemes of finite type over S a separated noetherian scheme: I don’t know what goes wrong if we take a different category, hence
    • I would like to be able to switch between categories, and suitably update the comparison graph.
    • I have no clue regarding the actual results that should be there…

New feature for the Stacks project: slogans

Update from the Stacks project: there is a new feature, namely slogans. Johan de Jong has discussed this idea before, and right now they are implemented on the website *and* there is a nice web page (compatible with your smartphone, I hope) to suggest slogans. Moreover, there are numerous small bug fixes (and hopefully not too many new bugs).

Hence: first suggest slogans and then see them appear on the Stacks project website.

I would like to thank Johan and Grietje Commelin for providing an excellent place to stay during the programmingathon Johan (Commelin, having multiple Johans makes for confusing mail conversations) and I did. And special thanks to their 2.5 year old daughter Hannah for the great conversations one can have with toddlers during breaks :).

Notes on stratification of triangulated categories

The last week of June I attended a summer school on derived categories in Nantes. Really interesting summer school, and a truly brilliant city. During the long and boring train ride home I decided to TeX up my notes for Henning Krause‘s lecture series on stratification of triangulated categories. The result of that is now proofread and corrected (thanks Henning!), and can be found on my personal webpage (or his, or hopefully the summer school’s page soon).

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