The Derived Category 4: A Nice Spanning Class

Recall that we are assuming that ${X/k}$ is a smooth projective variety. Let’s also say it is of dimension ${n}$. We’re going to be lazy (i.e. sane) and all functors will be derived when talking about the derived category even though the ${\mathbf{L}}$ and ${\mathbf{R}}$ will be omitted. Our derived category always has some (auto-) functors. For example, we definitely have the shift functor that comes with any triangulated category ${[k]}$ just by shifting where the sheaves occur in the complex.

Also, given any coherent sheaf we have the functor ${\mathcal{F}\otimes - : D(X)\rightarrow D(X)}$. In particular, we could tensor with the canonical bundle and shift by ${n}$. This functor is so useful it has a name and notation ${S_X(\mathcal{E})=\mathcal{E}\otimes \omega_X [n]}$ (again, ${\mathcal{E}}$ is any object of ${D(X)}$ and hence a complex even though I didn’t write ${\mathcal{E}^\bullet}$). We call this the Serre functor.

This name just comes from the fact that the generalized form of Serre duality for the derived category says that there is a functorial isomorphism

$\displaystyle \eta: Hom_{D(X)}(\mathcal{E}, \mathcal{F})\stackrel{\sim}{\rightarrow} Hom_{D(X)}(\mathcal{F}, S_X(\mathcal{E}))^*.$

Notice that if ${\mathcal{E}}$ and ${\mathcal{F}}$ are honest sheaves sitting in degree ${0}$ we can use that ${Ext^i(\mathcal{E}, \mathcal{F})=Hom_{D(X)}(\mathcal{E}, \mathcal{F}[i])}$ to derive the special case ${Ext^i(\mathcal{E}, \mathcal{F})\simeq Ext^{n-i}(\mathcal{F}, \mathcal{E}\otimes \omega_X)^*}$ which is the standard form of Serre duality given in classic texts like Hartshorne.

For the rest of today let’s look at a very important concept from triangulated categories. One might wonder how much we can know about certain triangulated categories just from knowing certain special classes of objects. A collection of objects ${\Omega}$ is called a spanning class for a triangulated category ${\mathcal{D}}$ if the following hold:

If ${Hom(A, B[i])=0}$ for all ${A\in \Omega}$ and ${i\in \mathbf{Z}}$, then ${B\simeq 0}$.

If ${Hom(B[i], A)=0}$ for all ${A\in \Omega}$ and ${i\in \mathbf{Z}}$, then ${B\simeq 0}$.

It is not in general true that these two conditions are equivalent, but it is easy to check that Serre duality for ${D(X)}$ will allow us to only have to check one of the conditions. The idea of spanning classes (which may not come up for awhile) is that you can check certain properties just on these objects to get properties on the whole category. For example, one can use this idea to prove necessary and sufficient conditions for a Fourier-Mukai transform to be fully-faithful.

Since our triangulated category ${D(X)}$ is somehow built out of ${X}$, to any (closed) point of ${X}$ we have a natural object associated to it that we’ll call ${k(x)}$. This is just the skyscraper sheaf at the point ${x}$. One hope would be that the set of objects of this form is a spanning class. This intuitively makes sense, because checking a property on this class in the derived category is sort of like checking a property on “points” of variety. It is indeed the case that this forms a spanning class.

Suppose ${\mathcal{F}}$ is a non-trivial object of ${D(X)}$. We’ll check the second condition. This says that we must produce some closed point ${x}$ and some integer ${i}$ so that ${Hom(\mathcal{F}, k(x)[i])\neq 0}$ (well, almost, we used Serre duality again to flip the i over to the other side). We will use the standard local-to-global spectral sequence

$\displaystyle E_2^{p,q}=Ext^p(\mathcal{H}^{-q}(\mathcal{E}), \mathcal{G})\Rightarrow Ext^{p+q}(\mathcal{E}, \mathcal{G}).$

If we plug in ${\mathcal{E}=\mathcal{F}}$ and ${\mathcal{G}=k(x)}$ we get

$\displaystyle E_2^{p,q}=Hom(\mathcal{H}^{-q}(\mathcal{F}), k(x)[p])\Rightarrow Hom(\mathcal{F}, k(x)[p+q]).$

Let ${m}$ be the maximal ${m}$ such that ${\mathcal{H}^m(\mathcal{F})\neq 0}$. The sheaf itself is assumed non-trivial, so there exists ${m}$ with that sheaf non-zero, but ${X}$ is regular so there are only finitely many non-zero and hence such an ${m}$ exists. We will now argue that ${E_2^{0, -m}=E_{\infty}^{0, -m}}$ by showing that all differentials with source and target ${E_r^{0,-m}}$ for any ${r}$ must be trivial.

On the one hand, ${E_2^{p,q}}$ is the ${p}$-th Ext group between coherent sheaves, so when ${p<0}$ it always vanishes. This means that any differential with target ${E_r^{0,-m}}$ must be trivial. On the other hand, our choice of ${m}$ maximal implies that any differential with source ${E_r^{0, -m}}$ is trivial.

Now ${\mathcal{H}^m(\mathcal{F})}$ is non-trivial, so in particular it has non-trivial support which is a closed set and hence contains some closed point ${x}$. This tells us that ${E_{\infty}^{0, -m}=E_2^{0, -m}=Hom(\mathcal{H}^m(\mathcal{F}), k(x))\neq 0}$. But this says that ${Hom(\mathcal{F}, k(x)[-m])\neq 0}$ which is what we set out to prove and hence the collection of skyscraper sheaves of closed points do form a spanning set.

Nondegenerate Hodge de Rham

Let’s construct the example today as a quick post. First, we’ll need a theorem that will be used to show that HdR doesn’t degenerate. Let ${Z}$ be a smooth variety over an algebraically closed field of positive characteristic with the property that ${\mathrm{Pic}^\tau(Z)\simeq \alpha_p}$ as a group scheme. Then HdR does not degenerate for ${Z}$. We’ll use a lot of things we haven’t talked about or proved, but the purpose of this post is to give the example and a flavor of the why it is true. Later we’ll come back and look at the parts that go into it more carefully.

Here is how this is proved. It is well-known that for HdR to degenerate it must be the case that all global ${1}$-forms are closed. So we assume this is the case otherwise we are done. Now Oda has a theorem that says when we are in this case ${H^1_{dR}(Z/k)\simeq D(pPic^\tau(Z))}$ where ${D(-)}$ is the Dieudonne module. By assumption that ${\mathrm{Pic}^\tau(Z)\simeq \alpha_p}$ we get that the first de Rham Betti number ${h^1_{dR}=1}$.

Since ${H^1(Z, \mathcal{O}_Z)}$ is isomorphic to the tangent space to the Picard scheme we get that ${1=\dim H^1(Z, \mathcal{O}_Z)}$, but also ${\alpha_p}$ gives a global ${1}$-form on the Picard scheme, so ${\dim H^0(Z, \Omega^1)\geq 1}$. Thus we have a contradiction if HdR degenerates and hence this does not happen.

This gives our example if we can come up with something that satisfies those properties. Let ${R}$ be the extension of ${W(k)}$ of ramification index ${2}$. Let ${S=\textrm{Spec}(R)}$ and ${G}$ be a finite flat group scheme over ${S}$ such that ${G_0\simeq \alpha_p}$. It is a theorem of Tate and Oort that if there are elements ${a}$ and ${c}$ in ${\frak{m}_R}$ such that ${ac=p}$, then such a ${G}$ exists.

Now it is a theorem of Raynaud that there is a projective space ${P}$ over ${S}$ with a linear action of ${G}$ which contains a relative complete intersection surface ${Y}$ which is stabilized by ${G}$ and such that ${G}$ acts freely on ${Y}$ and ${Y=X/G}$ is smooth over ${S}$. It follows that ${\textrm{Pic}^\tau(X_0)\simeq G_0^D\simeq \alpha_p}$. Thus for any characteristic we have an example of a smooth variety that lifts to characteristic ${0}$ over a very small ramified extension of ${W(k)}$ but the HdR spectral sequence does not degenerate! We’ll try to unpack this better next time.

When does Hodge-de Rham Degenerate?

I wrote up half of the follow up to the last post almost immediately, but then got really stuck. I wasn’t sure how to avoid the technical details and still have it be a worthwhile post. I’ve tried twice to do it, and now I’ve decided it just isn’t worth going down that road right now. Instead I’ve thought of something else to do which I find quite amazing and shocking. It also fits in really well with the past series of posts. In this post I’ll overview the example and why it is shocking. Next time I’ll give the construction. Then we’ll end with a closer examination of what is going on and why it shouldn’t be shocking that it exists.

Let’s recall some facts and theorems about the Hodge-de Rham spectral sequence. We’ve thought quite a bit about this over the past six months or so. Let ${X}$ be a smooth projective variety over a field of characteristic ${0}$. Deligne and Illusie came up with a pretty crazy way to prove that HdR degenerates at ${E_1}$. First off, degeneration is somewhat rare in positive characteristic, but they decided to reduce mod ${p}$ and prove that if ${Y}$ is a scheme over a perfect field of characteristic bigger than ${\dim Y}$ and ${Y}$ admits a lift to ${W_2(k)}$, then HdR SS degenerates at ${E_1}$. Since this always happens for varieties that came from characteristic ${0}$, it is true for ${Y}$. To finish it off we can extrapolate back.

The moral of this story is that we have degeneration in characteristic ${0}$, and we have degeneration in positive characteristic if we have even just a single step in trying to lift to characteristic ${0}$. Note that there are varieties that lift to ${W_2(k)}$ but do not lift to ${W_3(k)}$, and there are varieties that admit a formal lift by being able to lift each stage from ${W_n(k)}$ to ${W_{n+1}(k)}$ all the way up and then don’t algebraize to get an actual lift to characteristic ${0}$. Both of these types of varieties have HdR degenerate at ${E_1}$! They don’t have to lift to get the degeneration. In some sense they merely need to exhibit some sort of “evidence” that a lift is possible.

Lastly, we have a theorem that relates degeneration of HdR of the generic and closed fibers in a flat family over a DVR. It says that if ${\frak{X}\rightarrow S=\mathrm{Spec}(R)}$ is a flat family over a DVR and ${R/m\simeq k}$ is of characteristic ${p}$ and ${\mathrm{Frac}(R)=K}$ is of characteristic ${0}$, then if ${\dim H^q(X_0, \Omega^p)=\dim H^q(X_\eta, \Omega^p)}$ then HdR degenerates at ${E_1}$ for ${X_0}$. In other words, if we have a lift of ${X_0}$ to characteristic ${0}$ (over ANY ring, not necessarily ${W(k)}$), and the Hodge numbers match the lifted Hodge numbers this guarantees degeneration. This argument is quite simple and just involves counting dimensions and upper-semicontinuity, so we actually didn’t need characteristic ${0}$. It can be generalized to say if HdR degenerates on the generic fiber (of whatever characteristic) then it degenerates on the special fiber.

Let’s just recap here. HdR degenerates for characteristic ${0}$ things. Things that merely exhibit evidence of a lift to characteristic ${0}$ by lifting to ${W_2(k)}$ have HdR degenerate (this includes things that provably can’t be lifted all the way to characteristic ${0}$). We can relate degeneration of HdR on the generic fiber to degeneration on the special fiber of a flat family. One might make the guess that if you have an actual honest lift to characteristic ${0}$ (rather than just “evidence”) that HdR must degenerate. You may even think you have a proof by just saying lift it ${\frak{X}\rightarrow S}$ and now since the generic fiber degenerates the special fiber must also.

You would be wrong. William Lang came up with an example of a smooth variety (over any characteristic that you want) that lifts to characteristic ${0}$ but has non-degenerate HdR spectral sequence. This should be sort of shocking in light of the above theorems and “proof” outline. I’m going to partially give away the punchline now. There are two things going on here. First, we needed to be able to say that the Hodge numbers matched up after doing the lifting to say that degeneration of one implied degeneration of the other. This example will not have that property. The other thing going on is that it seems that degeneration of HdR somehow really has something to do with liftability to ${W_2(k)}$ and is not something that is merely about characteristic ${0}$. This example will be a lift over a ramified (of degree ${2}$, so it is as close to ${W(k)}$ as possible) extension of ${W(k)}$.

Hodge and de Rham Cohomology Revisited

I was going to talk about how the moduli of K3 surfaces is stratified by height in positive characteristic and some of the cool properties of this (for instance, “most” K3 surfaces have height 1). Instead I’m going to shift gears a little. We’ve talked about ${\ell}$-adic étale cohomology, Witt cohomology, cohomology on any site you want to put on ${X}$, de Rham cohomology, and we’ve implicitly used Hodge theory in places. Secretly we’ve been heading straight towards cyrstalline cohomology. I think it might be neat to start a series of posts on how each of these relate to eachother and then really motivate the need for crystalline stuff.

Away from this blog I’ve been thinking about degeneration of the Hodge-de Rham Spectral Sequence a lot. Suppose for a minute we’re in the nicest situation possible. We have a smooth variety ${X}$ over ${\mathbb{C}}$. This means we can look at the ${\mathbb{C}}$-points and get an actual complex manifold. We defined the algebraic de Rham cohomology awhile ago to be ${H^i_{dR}(X/\mathbb{C}):=\mathbf{H}^i(\Omega_{X/\mathbb{C}}^\cdot)}$ the hypercohomology of the complex ${0\rightarrow \mathcal{O}_X\rightarrow \Omega^1\rightarrow \Omega^2\rightarrow \cdots}$. Since we’re in this nice case, this actually agrees perfectly with the standard singular cohomology on the manifold with coefficients in ${\mathbb{C}}$ (and by the de Rham theorem, the standard de Rham cohomology).

On a complex manifold we also have a nice working notion of Hodge theory. The Hodge numbers are $h^{ij}=\dim_{\mathbb{C}}H^j(X, \Omega^i)$ which we would normally derive through the Dolbeault resolution. We also have a Hodge decomposition ${H_{dR}^j(X/\mathbb{C})=\bigoplus_{p+q=j} H^q(X, \Omega^p)}$.

How do we see this using fancy language? Well, merely from the fact that de Rham cohomology is defined as the hypercohomology of a complex, we get the spectral sequence arising from hypercohomology. Without doing any work we can just check what this spectral sequence is and we find ${E_1^{ij}=H^j(X, \Omega^i)\Rightarrow H_{dR}^{i+j}(X/\mathbb{C})}$. This is because the ${E_1^{ij}}$ terms come from resolving each individual part of the complex which by definition just gives sheaf cohomology of the ${\Omega^i}$.

Of course, there was nothing special about ${X}$ being over ${\mathbb{C}}$, we could just as easily be over an arbitrary field and all of this still works. There is a great theorem that says that this spectral sequence degenerates at ${E_1}$ if ${X}$ is smooth over a characteristic ${0}$ field. There are several known proofs, some more analytic and some more algebraic. The coolest one is certainly by Deligne and Illusie.

They prove this preliminary result that if ${X}$ is smooth over a field ${k}$ of characteristic ${p}$ where ${p>\dim(X)}$ and ${X}$ has a lift to ${W_2(k)}$, then the Hodge-de Rham spectral sequence degenerates at ${E_1}$. Maybe we’ll talk about how this is done some other day, but if you know about the Cartier isomorphism then it is related to that. Using this side result that seems to be about as unrelated to the characteristic ${0}$ case as possible they then amazingly prove the characteristic ${0}$ case by reducing to positive characteristic and using a Lefschetz principle type argument.

Now despite the fact that H-dR degenerating being the norm in characteristic ${0}$, it turns out to be not so much the case in positive characteristic, so it is really shocking that to prove the characteristic ${0}$ case they moved themselves to this situation where it was likely not to degenerate. But we’ll get a better intuition later for why this wasn’t as risky as it sounds. Namely that since it came from characteristic ${0}$, there wasn’t going to be a problem lifting it back so the lifting to ${W_2(k)}$ was not a problem. It seems that the obstruction to being able to do this is almost exactly the failure of degeneracy. Recall that every K3 surface lifts to characteristic ${0}$, so (if you don’t know the proof of this) you’d expect the H-dR SS to degenerate at ${E_1}$. It might be a fun exercise for you to try to figure out why this is (very important hint: there are no global vector fields on a K3 so ${h^{1,0}=0}$).

Before ending this post it should be pointed out that all of this can be done in the relative setting as well. We actually originally defined de Rham cohomology purely in the relative setting without thinking about it over a field like we did today. Suppose ${\pi: X\rightarrow S}$ is a smooth scheme. The relative H-dR SS is given by ${E_1^{ij}=H^j(X, \Omega^i_{X/S})\Rightarrow \mathbf{R}^{i+j}\pi_*(\Omega_{X/S}^\cdot)=H_{dR}^{i+j}(X/S)}$.

We’ll continue with this next time, but I’ll just leave you with the thought that you can basically formulate for any class of schemes you want a large open problem by asking yourself whether or not the HdR SS degenerates at ${E_1}$ or at all.

Gauss-Manin Connection 2

It’s probably been awhile since you read the first post in this series, so I’ll quickly remind you of the key point. ${S}$ is a smooth scheme over a field ${k}$. We fixed connection ${\rho}$ on ${\mathcal{E}}$. Then given a derivation ${\delta}$ corresponding to ${D: \Omega\rightarrow \mathcal{O}_S}$, then for any element of ${\mathcal{D}er_k(\mathcal{O}_S)}$, the sheaf of germs of ${k}$-derivations we can compose the maps we have and we get ${\overline{D}\in \mathcal{E}nd_k(\mathcal{E})}$.
So every connection gives a map ${\mathcal{D}er_k(\mathcal{O}_S)\rightarrow \mathcal{E}nd_k(\mathcal{E})}$, ${\delta\mapsto \overline{D}}$ and we had a relation between ${D}$ and ${\overline{D}}$, and any such map satisfying the relation comes from a connection.

Now we’ll go back to the construction of the Gauss-Manin connection from last time. We haven’t actually checked that it is a connection or that it is flat. Recall that it is just one of the maps we get from the spectral sequence associated to the filtration of the complex ${\Omega_{X/k}^\cdot}$. Now the filtration is compatible with taking wedge products (${F^i\wedge F^j\subset F^{i+j}}$) and the functors ${\mathbf{R}^q\pi_*}$ are multiplicative, so we have a product structure on the terms of the spectral sequence as follows.

If we take sections of the sheaves over an open, then ${E^{p,q}_r\times E^{p',q'}_r\rightarrow E^{p+p', q+q'}_r}$ by ${(e,e')\mapsto e\cdot e'}$. If you want the actual construction see Godement. The product satisfies a few important properties. We have a type of anti-commutativity ${e\cdot e'=(-1)^{(p+q)(p'+q')}e'\cdot e}$. Also we know how it behaves under the differential map: ${d_r(e\cdot e')=d_r(e)\cdot e'+ (-1)^{p+q}e\cdot d_r(e')}$.

In particular, let’s look at what this product rule for the differential is for the Gauss-Manin map. For ${\nabla=d_1^{0,q}:E_1^{0, q}\rightarrow E_1^{1,q}}$ which is really mapping ${\mathcal{H}^q_{dR}(X/S)\rightarrow \Omega_{S/k}\otimes \mathcal{H}^q_{dR}(X/S)}$, the differential is really just ${d_{S/k}\otimes Id}$. Thus that rule says that ${\nabla(\omega\cdot e)=d\omega\cdot e+(-1)^0\omega\cdot \nabla(e)}$. So it is a connection!

The curvature is easily seen to be ${d_1^{1,q}\circ d_1^{0,q}}$ and since the ${d_1}$‘s are maps of a complex we get that it is ${0}$, and hence ${\nabla}$ is flat and hence the Gauss-Manin connection is integrable. We’ve now proved the theorem that any smooth ${k}$-morphism of smooth ${k}$-schemes gives rise to a canonical integrable connection on the relative de Rham cohomology sheaves that is compatible with the cup product.

If you want a more explicit way to see what the map is see the paper, but it is kind of tedious since writing out how it appears in the spectral sequence you will quickly find that it is the connecting homomorphism when taking the long exact sequence after applying the functor ${\mathbf{R}^q\pi_*}$ to the exact sequence ${0\rightarrow \mathrm{gr}^{p+1}\rightarrow F^p/F^{p+2}\rightarrow \mathrm{gr}^p\rightarrow 0}$.

This is the third post on this topic, and I haven’t given you a reason to care yet. Here’s why we should care. One would hope (via a conjecture of Grothendieck) that there is some sort of relative de Rham Leray Spectral Sequence: ${E_2^{p,q}=\mathbf{H}^p(S, \Omega_{S/k}^\cdot \otimes_{\mathcal{O}_S} \mathcal{H}^q(X/S))\Rightarrow H_{dR}^{p+q}(X/k)}$. For the ${E_2}$-term to make any sort of sense we needed ${\Omega_{S/k}^\cdot \otimes_{\mathcal{O}_S} \mathcal{H}^q(X/S)}$ to be a complex, and since the Gauss-Manin connection is integrable it is a complex. Also, ${H_{dR}^{p+q}(X/k)}$ is defined to be ${\mathbf{H}^{p+q}(X, \Omega_{X/k}^\cdot)}$.

It turns out that when ${S}$ is affine such a spectral sequence exists. In case you’re wondering, affineness is needed for a nice proof of this because it makes certain cohomologies vanish. Deligne has proved it in a more complicated way when ${S}$ is not affine (but still with our standing assumptions). This is of great importance in proving every K3 surface lifts from characteristic ${p}$ to characteristic ${0}$.

The Grothendieck Spectral Sequence

Well, I meant to do lots more examples building up some more motivation for how powerful spectral sequences can be in some simple cases. But I’m just running out of steam on posting about them. Since we’ve done spectral sequences associated to a double complex, we may as well do the Grothendieck Spectral Sequence, then I might move on to another topic for a bit (I admit it is sort of sad to not prove the Kunneth formula using a SS).

I haven’t scoured the blogs to see whether these topics have been done yet, but I’m thinking about either basics on abelian varieties a la Mumford, or some curve theory, possibly building slowly to and culminating in Riemann-Roch.

In any case, we have the tools to do the Grothendieck Spectral Sequence (GSS) quite easily. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories with enough injectives. Let $\mathcal{A}\stackrel{G}{\to}\mathcal{B}\stackrel{F}{\to}\mathcal{C}$ be functors (and $FG:\mathcal{A}\to\mathcal{C}$ the composition). Suppose that $F$ and $G$ are left exact and for every injective $J\in\mathcal{A}$ we have $G(J)$ is acyclic. This just means that $R^iF(J)=0$ for all positive $i$.

Then there exists a spectral sequence (the GSS) with $E_2^{pq}\simeq (R^pF)(R^qG)(X)\Rightarrow R^{p+q}(FG)(X)$ with differential $d_{r}:E_r^{pq}\to E_r^{p+r, q-r+q}$.

The proof of this is just to resolve $X$ using the injectives that we know exist. This gives us a double complex. From a double complex, the way to get the $E_2$ term is to take vertical then horizontal homology, or horizontal and then vertical. Both of these will converge to the same thing. One way completed collapses to the “0 row” due to the fact that the exact sequence remained exact after applying the functor except at the 0 spot. Thus it stabilizes at this term and writing it out, you see that it is exactly $R^{p+q}(FG)(X)$. Taking homology in the other order gives us exactly $(R^pF)(R^qG)(X)$ by definition of a derived functor. This completes the proof.

I probably should write the diagram out for clarity, but really they are quite a pain to make and import into wordpress. The entire outline of the proof is here, so if you’re curious about the details, just carefully fill in what everything is from the previous posts.

This is quite a neat spectral sequence. It is saying that just by knowing the derived functors of $F$ and $G$ you can get to the derived functors of the composition of them. There are two important spectral sequence consequences of this one. They are the Leray SS and the Lyndon-Hochschild-Serre SS. The later computes group cohomolgy.

I promised early on to do the Leray SS for all the algebraic geometers out there. The Leray SS gives a way to compute sheaf cohomology. Let $\mathcal{A}=Sh(X)$ and $\mathcal{B}=Sh(Y)$ be the category of sheaves of abelian groups on X and Y. Let $\mathcal{C}=Ab$ the category of abelian groups. Let $f:X\to Y$ be a continuous map, then we have the functor $F=f_*$ and the two global section functors $\Gamma_X$ and $\Gamma_Y$.

Applying the GSS to these functors, we get that $H^p(Y, R^qf_*\mathcal{F})\Rightarrow H^{p+q}(X, \mathcal{F})$.

There are a few things to verify to make sure that the GSS applies, and we need the fact that $\Gamma_Y\circ f_*=\Gamma_X$. It would also be nice to have an example to see that this is useful. So maybe I’ll do those two things next time.

Spectral Sequence of a Double Complex

I knew my posting would be sparser this quarter, but this was a pretty large gap. I sort of lost motivation for posting, since my class moved on from spectral sequences to group cohomology, so I wasn’t constantly reminded about things to post about. Now the class is coming back around to spectral sequences in order to compute the group cohomology, so I’m sort of in the mood again.

We’ve mostly been looking at how to get a spectral sequence out of a filtered complex, and last time I promised I’d talk about convergence. Since my posting is so sparse, and I only wanted to give an idea about it rather than a proof of all the details of the convergence, I’m going to keep it brief in order to post about something more important.

The theorem is that if we have a filtration ${\{K^p\}}$ of a complex ${K}$ and the following three conditions are satisfied

1. ${K^p=0}$ when ${p<0}$
2. ${H_{p+q}(K^p/K^{p-1})=0}$ when ${q\leq -1}$
3. ${K=\cup K^p}$

then if ${r\geq \max\{p, q+1\}+1}$, then ${E^r_{p,q}\simeq E^\infty_{p,q}}$. Where we define ${E^\infty_{p,q}:= F_{p,q}/F_{p-1, q+1}}$ where ${F_{p,q}:=im (H_{p+q}(K^p)\rightarrow H_{p+q}(K))}$.

Basically what this says is that under some “mild" conditions that were naturally satisfied in our examples, the spectral sequence converges. And not only does it converge, but it converges to an associated graded object of the homology of the original complex.

The fact that it converges you can work out without much trouble if you've followed the previous posts. It basically amounts to noticing the first two conditions force all non-zero ${E^r}$ terms to be in the “first quadrant" when ${r>1}$. Since the differential maps get longer and longer as ${r}$ increases, paying attention to some specific ${E^*_{p,q}}$, the differential pointing to this term will have to come from the fourth quadrant, and eventually the differential will point to the second quadrant. But we said that these two quadrants were filled with 0 groups. Thus our sequence we are taking homology of is ${0\rightarrow E^r_{p,q}\rightarrow 0}$ for all ${r}$ after some point. So they stabilize. Figuring out exactly what these things are that they stabilize to is far more tedious and I won't do it.

The next topic that should be addressed is how to get a spectral sequence out of a double complex. This is actually much more useful it turns out, since there are two ways to do it that will both converge to the same thing which will give us lots of information.

First, let's recall that a double complex ${C_{p,q}}$ is just sets of complexes where the ${q}$ index the rows and the ${p}$ index the columns. So for a fixed ${q}$, we get ${\cdots \leftarrow C_{p-1, q}\leftarrow C_{p, q}\leftarrow \cdots }$ a chain complex with differential, say ${d^h}$ for horizontal. Likewise we get them going down in columns for any fixed ${p}$ with differential ${d^v}$ for vertical. The differentials are compatible in the sense that ${d^h\circ d^v+d^v\circ d^h=0}$. These also have to fit together into what we call the total complex ${(Tot C)_n=\bigoplus_{p+q=n} C_{p,q}}$ where the differential is ${d^h+d^v: C_{p,q}\rightarrow C_{p-1, q}\bigoplus C_{p,q-1}}$.

Now if we consider just the single complex ${Tot C}$, and put a filtration on it, we'll have a spectral sequence as before. One natural filtration is at the ${p}$th filtered part to chop everything from the ${p}$-th column to the right off and replace it with ${0}$. Notationally, this is ${(F_pC)_n=\bigoplus_{j+q=n \ \text{and} \ j\leq p} C_{j, q}}$.

The other natural one is to chop everything from the ${p}$-th row and up off and replace it with ${0}$. We'll notate this with a prime on the left ${F_pC}$. Both of these certainly satisfy ${\cdots \subset F_pC\subset F_{p+1}C\subset \cdots}$ since from ${p}$ to ${p+1}$ everything stays exactly the same except the ${p+1}$ column of ${C_{p,q}}$ gets stuck on. We also satisfy ${\cup F_pC=Tot C}$. We do need to assume that ${C_{p,q}=0}$ when ${p<0}$ or ${q<0}$ to get all the convergence conditions above, but this isn't too bad since most naturally arising double complexes (that we'll look at) are zero before some point and we can just shift indices.

The spectral sequence we get has ${E^1_{p,q}=H_{p+q}(F_pC/F_{p-1}C)}$. Note that for the first spectral sequence the ${E^2}$ term is computed by first taking vertical homology, and then horizontal homology: ${E^2_{p,q}\simeq H_p^h(H_q^v(C))}$ and for the other filtration it is ${E^2_{p,q}\simeq H_q^v(H^h_p(C))}$.

As usual, the formal aspect of this might seem overwhelming and confusing right now, so I'll hold off on working out what the maps actually are and let this absorb a little. It should become more clear what is going on when we do some examples.

Spectral Sequence Arrows Revisited

I don’t want to do much today. In fact, we’ll probably not cover any new ground, but start to fill in what is going on a little better. Notice that in order to work with the spectral sequence last time, we didn’t actually have to figure out what the ${i, j,}$ or ${k}$ or ${d}$ maps were. We only needed that they existed. The main thing of today is to work out all these sequences. Some are exact. Some are chain complexes. Some are commutative diagrams. Most of them fit together in a really organized way, other parts are just related. It is useful to see how these fit together, since sometimes you can get by without the maps as long as you know it is part of a commutative exact diagram.

Recall that our current situation is that given a filtered chain complex ${\cdots K^{p-1}\subset K^p\subset \cdots \subset K}$ we get an exact couple ${(D_{p,q}, E_{p,q})}$ where ${D_{p,q}= H_{p+q}(K^p)}$ and ${E_{p,q}=H_{p+q}(K^p/K^{p-1})}$ by taking the long exact sequence in homology associated to the short exact sequence ${0\rightarrow K^{p-1}\rightarrow K^p\rightarrow K^p/K^{p-1}\rightarrow 0}$.

Two times ago we saw exactly what the ${i_{p,q}}$, ${j_{p,q}}$ and ${k_{p,q}}$ maps were.

Let’s develop them again, but being more careful as to how they all fit together. Well ${i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})}$ and it comes from the induced maps on homology from the injection ${K^p\hookrightarrow K^{p+1}}$. So we get infinite strings of ${i_{p,q}}$ for fixed ${q}$ that relate as injections ${p}$ changes. i.e we’ll get as part of a large diagram columns that look like ${\cdots \rightarrow H_{p+q}(K^{p-2})\rightarrow H_{p+q}(K^{p-1})\rightarrow H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})\rightarrow\cdots}$.

The ${j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})}$ are induced on homology from the projection map ${K^p\rightarrow K^p/K^{p-1}}$.

The ${k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})}$ comes from the snake lemma. It is the boundary map in the long exact sequence.

Note that these last two will fit into rows: ${\cdots \rightarrow H_{p+q}(K^p)\stackrel{j}{\rightarrow} H_{p+q}(K^p/K^{p-1})\stackrel{k}{\rightarrow} H_{p+q-1}(K^{p-1})\stackrel{j}{\rightarrow} H_{p+q-1}(K^{p-1}/K^{p-2})\rightarrow\cdots}$.

Thus we can form a large commutative diagram from these long exact sequences, and figure out how to extrapolate the exact couple from it. Below is the diagram relating what was just said:

(Sorry about the image. I’ll figure this out eventually. For now you can click on it and zoom to get the full size).

The labelled arrows are the ${i,j,k}$ maps of just a single exact couple. Note that there are infinitely many exact couples going on here, and they are all related. Note also that the exact couples don’t just go across or down.

Now let’s figure out the differentials and how to get the spectral sequence. Well, take the ${H_{p+q}(K^p/K^{p-1})}$ spot. Recall that in the exact couple, this is the ${E}$ term which is where our chain complex comes from. Then doing ${j\circ k}$ is the ${d}$ map. Note this is the really the ${d^1}$ map. So our page of ${E^1_{p,q}}$ comes from this row only.

Take homology with respect to this, and due to the nature of the exact couples, we’ll get another page of these things. Now we are on the ${E^2_{p,q}}$ page of groups. The ${d^2}$ map now is to go right, then up, then right. This is the composition ${k\circ i^{-1} \circ j: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-2}/K^{p-3})}$.

Take homology again. In general the ${d^r}$ map is going right, then up ${r-1}$ times then right one more time. This gives us exactly what we said last time, ${d^r_{p,q}: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}$.

Hopefully this helps clarify some of the inner workings of what is going on here. Next time we’ll talk about why the conditions from last time imply that the spectral sequence converges. We won’t prove it, but there are some nice concepts behind why it works that should be enlightening.

Serre Spectral Sequence

Today we’re going to back up and not worry about the tediousness of the last post. Let’s try to form some motivation and see how these things work even if we have to just assume some of the theory works for now. I think it will be helpful. So why care about spectral sequences? Well, one thing is that sometimes it is really hard to compute say ${H_n(C)}$, but it isn’t so bad to do ${H_n(C\otimes A)}$ for some ${A}$‘s. There is no good way to convert the information you get out of ${H_n(C\otimes A)}$ to information about ${H_n(C)}$, but this is exactly what the Bockstein spectral sequence does.

Today, I’ll talk about the Serre spectral sequence. This will help us calculate the homology of the loop space of an ${n}$-sphere. There is a nice proof that Tor is independent of which resolution you take (i.e. ${Tor(A, B)}$ can be calculated by resolving ${A}$ or ${B}$). There is a nice proof of the Snake Lemma using spectral sequences. There is a proof of the K\:{u}nneth formula. For the algebraic geometers, the Leray spectral sequence tells us information about sheaf cohomology. Anyway, there are tons of examples hitting many areas of math of uses of spectral sequences.

Right now we’ll not worry about the technical details of when and how a spectral sequence converges, but recall that the situation we ended on was that given a complex ${K}$, we could form a spectral sequence from a filtration ${\cdots \subset K^p\subset K^{p+1}\subset \cdots \subset K}$. We’ll just accept for now that if we have ${K^p=0}$ for ${p\max(p, q+1)}$ then ${E^r_{p,q}\simeq E^\infty_{p,q}}$. This ${E^\infty}$ term is the associated graded group of a filtration of ${H_*(K)}$.

This should be exciting. It means that if we can formulate a statement that reads, “There is a spectral sequence with ${E^2_{p,q}=}$ (fill in the blank), differential map ${d^r: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}$, that converges strongly to ${H_{p+q}}$(fill in again).” Then we can just chase around differentials and hope that everything collapses fast (lots of times these things only have a few terms or even ${E^2=E^\infty}$). If this happens then we can just read off what the known things are, and we’ll have figured out new information. Note that being able to say such a statement is not bad. We only need the mere existence to begin work, and taking those conditions above for granted, we actually can formulate lots of statements such as these.

Let ${\Omega S^n}$ be the loop space of ${S^n}$, namely the space of loops based at the north pole. Let ${PS^n}$ be the path space, the space of paths starting at the north pole. Note here that our standard tools of algebraic topology are not very useful in trying to calculate ${H_q(\Omega S^n)}$. But we know that ${PS^n}$ is contractible using the obvious map of retracting along all the paths simultaneously. And we also know something very useful, that there is a spectral sequence with ${E^2_{pq}=H_p(S^n)\otimes_\mathbb{Z} H_q(\Omega S^n)}$, differentials ${d^r : E^r_{pq}\rightarrow E^r_{p-r, q+r-1}}$ converging strongly to ${H_{p+q}(PS^n)}$.

So recall that we get an entire page of groups. The ${E^2}$ groups (${r=2}$). Since ${H_p(S^n)}$ is ${0}$ for all ${p}$ except ${p=0, n}$, and is ${\mathbb{Z}}$ in those spots. We also know that ${H_0(\Omega S^n)=\mathbb{Z}}$. This allows us to fill in the whole ${q=0}$ row, and all ${p}$ columns are completely ${0}$ except in ${p=0}$ and ${p=n}$.

The differential map ${d^r}$ goes to the left ${r}$ and up ${r-1}$.

Examine the bottom row. Due to all the ${0}$‘s in columns not ${p=0}$ or ${n}$, all the differential maps that don’t land in the ${p=0}$ column must be 0. Thus there is only one possible non-zero differential coming out of there. Namely, the ${d^n_{n,0}}$ map. It lands in the ${E^2_{0, n-1}=H_{n-1}(\Omega S^n)}$ spot. Now the complex of ${d^n}$ maps is ${\cdots \rightarrow 0\rightarrow \mathbb{Z}\stackrel{d^n_{n,0}}{\rightarrow} H_{n-1}(\Omega S^n)\rightarrow 0\rightarrow \cdots }$. Recall that to get to the different ${r}$ values in ${E^r_{p,q}}$ you take homology with respect to the ${d^r}$ maps. Since ${E^\infty_{p,q}=H_{p+q}(PS^n)=0}$ for ${p+q\neq 0}$ we must have ${d^n_{n, 0}}$ an isomorphism in order for it to vanish. This gives us a ${\mathbb{Z}}$ in the 0, ${n-1}$ spot and hence another in the ${n}$, ${n-1}$ spot. We can keep repeating this argument giving a ${\mathbb{Z}}$ in the ${p=0, n}$ and ${q=}$ multiples of ${n-1}$ spots.

Since the differentials are all ${0}$ before ${r=n}$, nothing changes for ${E^2=E^3=\cdots E^n}$. Then at ${r=n}$ we get isomorphisms except at the ${E^2_{0,0}}$ position, so everything vanishes except that position which stays a ${\mathbb{Z}}$. Thus ${H_q(\Omega S^n)=\begin{cases} \mathbb{Z} \ if \ (n-1)|q \\ 0 \ else\end{cases}}$.

This is a pretty typical situation. Hopefully that gives a better overview of the process so that going back to the general case will have some sort of reference. If there are questions, feel free to ask. Maybe I’ll do another example before going back to theory.

Exact Couples

Today we begin a long set of posts on spectral sequences. I must first off say that I only learned what a spectral sequence was about 2 weeks ago, so I am far from knowledgable about them. This means if you have questions, I may not know the answers. I’m also going to go incredibly slowly. I want to carefully develop everything and give lots of motivating examples. Lastly, all of this is coming from the notes I’ve taken from John Palmieri’s class this quarter, so any great insights should be attributed to him and mistakes should be attributed to myself.

Like I’ve already said, I’m not familiar with the literature, so I’m not sure what the standard development is. For people who know some of this already, we’re going to define the spectral sequence associated to an exact couple. Then anytime we want a spectral sequence, we just need to figure out the exact couple it comes from. This is a different approach than say Weibel.

In general, this is going to work in any abelian category, but we’ll use the category of ${A}$-modules just to keep it less scary for those that haven’t worked with abelian categories.

Our first definition is that of an exact couple. This is a diagram of ${A}$-modules, ${D}$ and ${E}$ such that:

is exact in every spot. Then we immediately get a map ${d: E\rightarrow E}$ by ${d=j\circ k}$. This may look strange, since we’re not just going around the diagram. We sort of jump, but this is fine since at this stage we aren’t identifying ${i(D)\subset D}$ or anything. Now this map has a nice property, ${d\circ d=j\circ k\circ j\circ k=j\circ (k\circ j)\circ k=j\circ (0)\circ k=0}$ by exactness that middle term is ${0}$ making the whole thing ${0}$.

This means that ${\cdots \stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\rightarrow \cdots}$ is actually a chain complex. Thus we can take homology: $\displaystyle {H(E, d)=\frac{ker d}{im d}}$.

Now take ${D'= im i}$. Define the map ${i': D'\rightarrow D'}$ by ${i'=i\big|_{D'}}$. Take ${E'=H(E, d)}$. We will now think of ${j': D'\rightarrow E'}$ as ${j\circ i^{'-1}}$, but to be precise we pick an element whose preimage is ${x=i^{-1}(y)}$, and take an equivalence class, so ${j'(x)=[j(y)]}$. Lastly define ${k': E'\rightarrow D'}$ by ${k'([z])=k(z)}$. Now we have made lots of choices, but as usual in homological things it turns out by examination that everything is well-defined. It is also the case that with these new maps, the corresponding diagram is exact at every place. We call this the derived exact couple.

Since the derived exact couple is an exact couple, there is nothing stopping us from defining the derived exact couple of the derived exact couple. Then do it again, and again, … . Obviously, if we aren’t careful we’ll lose track of where we are, so choosing notation is essential. Suppose ${(D, E)}$ form an exact couple. Then let ${D^1=D}$ and ${E^1=E}$. Next let ${D^2=D'}$ and ${E^2=E'}$. This inductively gives us that ${D^{r+1}=(D^r)'}$ and ${E^{r+1}=(E^r)'}$. In each of these we get three maps which tell us that each stage is an exact couple ${i^r: D^r\rightarrow D^r}$, also ${j^r: D^r\rightarrow E^r}$, and ${k^r: E^r\rightarrow D^r}$.

Since the triangle is exact, defining (which technically already happened in the definition of ${E^{r+1}}$) ${d^r: E^r\rightarrow E^r}$ by ${d^r=j^r\circ k^r}$ we get a chain complex and can take homology. We say that ${\{(E^r, d^r)\}_{\{r\geq 1\}}}$ is the spectral sequence associated to the exact couple. Note that ${E^{r+1}=H(E^r, d^r)}$.

This is probably incredibly useless for people wanting to know what a spectral sequence is and how it is used. But in general we are going to have a lot more information floating around than just this abstract exact couple. More specifically, we have to get the exact couple somehow, and the process of forming it will give us things to work with. The other thing is that we want to know that ${\displaystyle \lim_{r\rightarrow\infty}(E^r, d^r)}$ converges in some sense to something, and hopefully we have information about what this “${E^\infty}$” term is.

I don’t want to go through an entire example today, but we’ll set up the situation for the example (for those who have seen this, it will be the Serre Spectral Sequence).

Suppose we have a chain complex ${K}$. Then given a filtration ${\cdots \subset K^n\subset K^{n+1}\subset \cdots \subset K}$ (not necessarily starting at the 0 complex or ending at the total complex, but the ${j}$th grade of any ${K^n}$ should be a submodule of the ${j}$th grade of ${K}$ and the submodules should respect the differentials). Then we will get a short exact sequence of chain complexes ${0\rightarrow K^{P-1}\rightarrow K^P\rightarrow K^P/K^{P-1}\rightarrow 0}$. This is just because it works trivially on each grade.

Well, our old standby says that a short exact sequence of chain complexes induces a long exact sequence in homology, so ${\cdots \rightarrow H_n(K^{P-1})\stackrel{i}{\rightarrow} H_n(K^P)\stackrel{j}{\rightarrow} H_n(K^P/K^{P-1})\stackrel{k}{\rightarrow} H_{n-1}(K^{P-1})\rightarrow \cdots}$.

From here we define our exact couple. Let ${D_{p,q}=H_{p+q}(K^p)}$ and ${E_{p,q}=H_{p+q}(K^p/K^{p-1})}$.

Now use the maps given in the long exact sequence.

So ${i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})}$ which is really ${D_{p,q}\rightarrow D_{p+1, q-1}}$.

Our ${j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})}$ which maps ${D_{p,q}\rightarrow E_{p,q}}$.

Lastly the map ${k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})}$ which maps ${E_{p,q}\rightarrow D_{p-1, q}}$.

Lastly, the differential ${d_{p,q} = j_{p-1, q}\circ k_{p,q}: E_{p,q}\rightarrow E_{p-1, q}}$.

Don’t worry if this seems horribly confusing. We’ll look at a very concrete example next time in which we’ll see that often everyting stays stationary or goes to 0 leaving us with very little. I’ll just try to summarize a little of this filtered chain complex example. We are getting infinitely many exact couples all mingled together somehow. If we think of a ${(p,q)}$-plane, then at each integer valued place we get a group ${E_{p,q}}$. The differential maps move us to the “left” on place. So we have rows of chain complexes for each fixed ${q}$. Remember that when we form the spectral sequence this is really only the ${E^1_{p,q}}$ part. We can think of this term as an entire “sheet” or “page” of groups related to each other. We get a page ${E^r_{p,q}}$ for each ${r\geq 1}$.

Some things that would be nice (and to get you thinking about why this might be useful) is if for each fixed pair ${(p,q)}$ the groups stabilize, i.e. ${E^N_{p,q}=E^{N+1}_{p,q}=\cdots}$. Then we would have a notion of ${E^\infty_{p,q}}$. It would be nice to know how this limit term relates to the original chain complex, or filtration. It would be nice if the spectral sequence limit were somehow independent of the filtration, that way taking a limit of different filtrations would give us information we might not know. Like I said earlier, don’t worry if this post was confusing. It just needed to be done. After some examples, this will probably not seem so bad.