A Mind for Madness

Musings on art, philosophy, mathematics, and physics


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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.


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Sheaf of Witt Vectors

I was going to go on to prove a bunch of purely algebraic properties of the Witt vectors, but honestly this is probably only interesting to you if you are a pure algebraist. From that point of view, this ring we’ve constructed should be really cool. We already have the ring of {p}-adic integers, and clearly {W_{p^\infty}} directly generalizes it. They have some nice ring theoretic properties, especially {W_{p^\infty}(k)} where {k} is a perfect field of characteristic {p}.

Unfortunately it would take awhile to go through and prove these things, and it would just be tedious algebra. Let’s actually see why algebraic geometers and number theorists care about the Witt vectors. First, we’ll need a few algebraic facts that we haven’t talked about. For today, we’re going to fix a prime {p} and we have an {\mathbf{important}} notational change: when I write {W(A)} I mean {W_{p^\infty}(A)}, which means I’ll also write {(a_0, a_1, \ldots)} when I mean {(a_{p^0}, a_{p^1}, \ldots)} and I’ll write {W_n(A)} when I mean {W_{p^n}(A)}. This shouldn’t cause confusion as it is really just a different way of thinking about the same thing, and it is good to get used to since this is the typical way they appear in the literature (on the topics I’ll be discussing).

There is a cool application by thinking about these functors as representable by group schemes or ring schemes, but we’ll delay that for now in order to think about cohomology of varieties in characteristic {p} and hopefully relate it back to de Rham stuff from a month or so ago.

In addition to the fixed {p}, we will assume that {A} is a commutative ring with {1} and of characteristic {p}.

We have a shift operator {V: W_n(A)\rightarrow W_{n+1}(A)} that is given on elements by {(a_0, \ldots, a_{n-1})\mapsto (0, a_0, \ldots, a_{n-1})}. The V stands for Verschiebung which is German for “shift”. Note that this map is additive, but is not a ring map.

We have the restriction map {R: W_{n+1}(A)\rightarrow W_n(A)} given by {(a_0, \ldots, a_n)\mapsto (a_0, \ldots, a_{n-1})}. This one is a ring map as was mentioned last time.

Lastly, we have the Frobenius endomorphism {F: W_n(A)\rightarrow W_n(A)} given by {(a_0, \ldots , a_{n-1})\mapsto (a_0^p, \ldots, a_{n-1}^p)}. This is also a ring map, but only because of our necessary assumption that {A} is of characteristic {p}.

Just by brute force checking on elements we see a few relations between these operations, namely that {V(x)y=V(x F(R(y)))} and {RVF=FRV=RFV=p} the multiplication by {p} map.

Now on to the algebraic geometry part of all of this. Suppose {X} is a variety defined over an algebraically closed field of characteristic {p}, say {k}. Then we can form the sheaf of Witt vectors on {X} as follows. Notice that all the stalks of the structure sheaf {\mathcal{O}_x} are local rings of characteristic {p}, so it makes sense to define the Witt rings {W_n(\mathcal{O}_x)} for any postive {n}. Now just form the natural sheaf {\mathcal{W}_n} that has as its stalks {(\mathcal{W}_{n})_x=W_n(\mathcal{O}_x)}.

Note that forgetting ring structure and thinking only as a sheaf of sets we have that {\mathcal{W}_n} is just {\mathcal{O}^n}, and when {n=1} it is actually isomorphic as a sheaf of rings. For larger {n} the addition and multiplication is defined in that strange way, so we no longer get an isomorphism of rings. Using our earlier operations and the isomorphism for {n=1}, we can use the following sequences to extract information.

When {n\geq m} we have the exact sequence {0\rightarrow \mathcal{W}_m\stackrel{V}{\rightarrow} \mathcal{W}_n\stackrel{R}{\rightarrow}\mathcal{W}_{n-m}\rightarrow 0}. If we take {m=1}, then we get the sequence {0\rightarrow \mathcal{O}_X\rightarrow \mathcal{W}_n\rightarrow \mathcal{W}_{n-1}\rightarrow 0}. This will be useful later when trying to convert cohomological facts about {\mathcal{O}_X} to {\mathcal{W}}.

We could also define {H^q(X, \mathcal{W}_n)} as sheaf cohomology because we can think of {\mathcal{W}_n} just as a sheaf of abelian groups. Let {\Lambda=W(k)}, then since {\mathcal{W}_n} are {\Lambda}-modules annihilated by {p^n\Lambda}, we get that {H^q(X, \mathcal{W}_n)} are also {\Lambda}-modules annihilated by {p^n\Lambda}. Next time we’ll talk about some other fundamental properties of the cohomology of these sheaves.


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Connections and Curvature … on Schemes! … in Characteristic p?!

I feel bad about my absence. I lasted posted during winter break, and now winter quarter is completely over. I kept meaning to do a series on “well-known” algebraic geometry results and constructions that don’t appear with any amount of thoroughness in the references. I thought it would be good to get that information out there. Unfortunately, I had already written these things down into a notebook and just couldn’t motivate myself to type something up that I already had. Anyway, one thing led to another and I didn’t do any posts. I’m not sure why I’m trying to justify my absence with an excuse.

Recently I’ve been typing up a translation of Deligne’s argument (written down by Illusie) that every K3 surface in characteristic {p>0} lifts to characteristic {0}. I’m not to the point of trying to understand it, but I wanted a typed version, so that when I get the background material (namely crystalline cohomology!) and go to understand it, I can just fill in the details into my typed notes quickly and easily. I also was curioius as to the overall format of the argument.

This led me to the 1968 paper by Katz and Oda called On the Differentiation of de Rham Cohomology Classes with Respect to Parameters. The next few posts will be about the main result from this paper. It is really quite amazing.

First, some definitions. We’ll always be working with a smooth scheme {S} over a field {k} (no assumptions here!). Let {\mathcal{E}} be a quasi-coherent sheaf of {\mathcal{O}_S}-modules. We’ll write {\Omega} for {\Omega^1_{S/k}} and unless otherwise noted, all tensor products will be over {\mathcal{O}_S}. We say that {\nabla} is a connection on {\mathcal{E}} if it is a homomorphism {\nabla: \mathcal{E}\rightarrow \Omega\otimes \mathcal{E}} that satisfies the “Leibniz rule”.

In other words, {\nabla(fg)=f\nabla(g)+df\otimes g}. This is the standard shorthand meaning {\nabla(U): \mathcal{E}(U)\rightarrow \Omega(U)\otimes \mathcal{E}(U)} satisfies the rule where {f\in \mathcal{O}_S(U)}, {g\in \mathcal{E}(U)} and {df} is the image of {f} under the universal map {\mathcal{O}_S\rightarrow \Omega}.

Given a connection {\rho}, we get homomorphisms for all {i}, {\rho_i: \Omega^i\otimes\mathcal{E}\rightarrow \Omega^{i+1}\otimes \mathcal{E}}. These are given by {\rho_i(\omega\otimes e)=d\omega \otimes e+(-1)^i \omega\wedge \rho(e)}.

The notation is just the one that makes sense: {\rho(e)\in \Omega\otimes \mathcal{E}}, so it looks like {\tau\otimes \epsilon}. So we define {\omega\wedge \rho(e)=\omega\wedge(\tau\otimes \epsilon)} to be {(\omega\wedge \tau)\otimes \epsilon\in \Omega^{i+1}\otimes \mathcal{E}}.

Now we define the curvature of the connection {K:\mathcal{E}\rightarrow \Omega^2\otimes \mathcal{E}} to be the map {\rho_1\circ \rho}. The curvature is related to the other {\rho_i} by an easy check {\rho_{i+1}\circ \rho_i(\omega\otimes e)=\omega\wedge K(e)}.

This gives some sort of meaning to the curvature now. If the curvature is {0}, then the natural de Rham-like sequence we get from a connection by stringing together the {\rho_i} as follows {0\rightarrow \mathcal{E}\stackrel{\rho}{\rightarrow} \Omega\otimes\mathcal{E}\stackrel{\rho_1}{\rightarrow} \Omega^2\otimes \mathcal{E}\rightarrow \cdots} is an honest complex that we can take cohomology with respect to, since {\rho_{i+1}\circ \rho_i=0}.

When this happens we call the connection {\rho} integrable. Now let {\mathcal{D}er_k(\mathcal{O}_S)} be the sheaf of germs of {k}-derivations of {\mathcal{O}_S} into itself. From the fact that the module of differentials is a representing object, we get that as a sheaf of {\mathcal{O}_S}-modules, {\mathcal{D}er_k(\mathcal{O}_S)\simeq \mathcal{H}om_{\mathcal{O}_S}(\Omega, \mathcal{O}_S)}.

Let {\mathcal{E}nd_k(\mathcal{E})} be the sheaf of germs of {k}-linear endomorphisms of {\mathcal{E}}. Given any connection {\rho} on {\mathcal{E}} we get an induced {\mathcal{O}_S}-linear map {\mathcal{D}er_k(\mathcal{O}_S)\rightarrow \mathcal{E}nd_k(\mathcal{E})} as follows. Let {\delta} be a derivation, then it corresponds to a map {D: \Omega\rightarrow \mathcal{O}_S}.

So consider the composition {\overline{D}:\mathcal{E}\rightarrow \Omega\otimes \mathcal{E}\rightarrow \mathcal{O}_S\otimes \mathcal{E}\simeq \mathcal{E}}, where the first is the connection and the second is {D\otimes Id}. This gives the map {\mathcal{D}er_k(\mathcal{O}_S)\rightarrow \mathcal{E}nd_k(\mathcal{E})} as {\delta\mapsto \overline{D}}.

Lastly for today, note that we get a nice relation between {D} and {\overline{D}} as follows {\overline{D}(fe)=D(f)e+f\overline{D}(e)} and that any map {\mathcal{D}er_k(\mathcal{O}_S)\rightarrow \mathcal{E}nd_k(\mathcal{E})} satisfying this relation comes from a unique connection on {\mathcal{E}}.

Today was just a bunch of notation and definitions, but next time it should get more interesting.


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The Cohomology Computation

Alright, I’m in a sort of tough spot. Yesterday I started typing this up, but I just don’t have the motivation. There are lots of tedious details that no one is going to read and will not come up in our study after this. It is all incredibly standard chasing Fourier coefficients around, so I’m not going to do it. This post will be an outline in how one would go about doing it, and I even may provide quick ideas behind it, but if it will be weeks before I continue on if I don’t just get through this. Someday, if it seems important, I’ll come back and fill it in. Or if someone comments and really wants to see one particular part, at least I’ll have motivation that someone is going to read it.

Here goes. Recall some of my conventions. X is a compact complex Lie group. We let A=\Gamma(X, \mathcal{C}) the global sections of the sheaf of C^\infty-functions on X. This was important to the Dolbeaut resolution. \overline{T} are the \mathbb{C}-antilinear functionals. We use \bigwedge^q to mean \bigwedge^q(\overline{T}).

We’ve shown that A\otimes_\mathbb{C}\bigwedge^q is isomorphic to \Gamma(X, \mathcal{C}^{0,q}) as complexes and hence we have the iso H^q(X, \mathcal{O}_X)\simeq H^q(A\otimes_\mathbb{C}\bigwedge).

Now we want to show that we actually have an induced isomorphism on cohomology from the inclusion map i: \bigwedge \hookrightarrow A\otimes \bigwedge. We’ll do this by comparing Fourier series. So we set up a normalized measure on X say \mu. By integrating functions against this measure we get linear function \mu_{\wedge} (I called this something different last time).

Now we need the lemma that for all \omega\in A\otimes \bigwedge^q we have \mu_\wedge (\overline{\partial} \omega)=0. This follows from periodicity and translation invariance of the vector field that comes up when you go to write it down.

The next step is to define the “Fourier coefficients” of a function. Now if we choose any integer valued function from the lattice defining X, say \lambda. Then it extends to an \mathbb{R}-linear function on the tangent space at the identity, V. We then have the exponentiation map \displaystyle v\mapsto e^{2\pi i \lambda(v)}. This respects the lattice and hence descends to $\latex X$. Call this map c_\lambda.

Define the \mathbb{C}-linear function A\to \mathbb{C} by Q_\lambda(f)=\int_X c_{-\lambda}fd\mu. Don’t be intimidated here. This is just the standard Fourier coefficient when your in a familiar situation. i.e. f=\sum e_\lambda \otimes Q_\lambda(f).

Now choose a Hermitian inner product, which gives us a norm to work with. Here things will become less detailed. One can next prove a lemma that the map f\to \{Q_\lambda(f)\}_\lambda is an isomorphism A\to the space of maps decreasing at infinity faster than \|\lambda \|^{-n} for all n.

Then do a computation to see that Q_\lambda (\overline{\partial}\omega)=(-1)^p 2\pi i \left(Q_\lambda(\omega)\wedge \overline{C}(\lambda)\right).

Lastly we want to get back to showing that the inclusion is a homotopy equivalence. Thus use the Hermitian inner product to define a map \lambda^*\in Hom_\mathbb{C}(\overline{T}, \mathbb{C}) for every \lambda by \displaystyle \lambda^*(x)=\frac{\langle x, \overline{C}(\lambda)\rangle}{2\pi i \|\overline{C}(\lambda)\|^2}.

We need to define for \omega\in A\otimes \bigwedge^p a map k(\omega) which will only be defined in terms of its Fourier coefficients. We define Q_\lambda(k(\omega))=(-1)^p \lambda^* \neg Q_\lambda(\omega) if \lambda\neq 0 and the coefficient is 0 if lambda is 0. Now it has all been set up so that comparing Fourier coefficients on \overline{\partial} k + k \overline{\partial} we get exactly the same ones as in id_{A\otimes\wedge} - i \mu_\wedge. Thus we are done by the magic of Fourier coefficients being unique.

That last computation I left out requires use of things such as “Cartan’s magic formula” and breaking it into two cases. Anyway, for not doing any details, I think this is a pretty thorough outline and filling any of the details you don’t believe or would like to know shouldn’t be too hard.


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Cohomology of Abelian Varieties II

Two posts in the same week! Before we get started today, we need to introduce one more piece of new notation. Let {\overline{T}} be the {\mathbb{C}}-antilinear maps {V\rightarrow \mathbb{C}}. Our goal is to prove that {H^q(X, \mathcal{O}_X)\simeq \bigwedge^q\overline{T}} and {H^q(X, \Omega^p)\simeq \bigwedge^pT\otimes \bigwedge^q\overline{T}}.

To do the calculation we will use the Dolbeault resolution: {0\rightarrow \mathcal{O}_X\rightarrow \mathcal{C}^{0,0}\rightarrow \mathcal{C}^{0,1}\rightarrow \mathcal{C}^{0,2}\rightarrow\cdots}. This is an acyclic resolution of the structure sheaf, and so is fine to use for the calculation of cohomology. The first upper index of {\mathcal{C}} refers to the degree of the {\mathbb{C}}-linear part and the second upper index refers to the degree of the {\mathbb{C}}-antilinear part. The map of the chain complex is {\overline{\partial}}.

Let’s examine the complex a little more closely. Define {\phi_{p,q}: \mathcal{C}\otimes (\bigwedge^pT\otimes \bigwedge^q\overline{T})\rightarrow \mathcal{C}^{p,q}} by {\sum f_i\otimes \alpha_i\mapsto \sum f_i\omega_{\alpha_i}}. Where we define {\omega_{\alpha}} to be the natural translation invariant {(p,q)}-form associated to {\alpha\in \wedge^pT\otimes \wedge^q\overline{T}} by left-invariantizing.

Note that {\omega_{\alpha\wedge\beta}=\omega_\alpha\wedge \omega_\beta}. Thus to prove that all {\omega_\alpha} are {\overline{\partial}}-closed (which we’ll denote {d} from now on for simplicity), we only need to check this for {(1,0)} and {(0,1)} forms. Now {exp: V\rightarrow X} is a local iso, so we also only need to check {d(exp^*(\omega_\alpha))=0}. But {exp^*(\omega_\alpha)=d\alpha}, so {d} of this expression, is {(d\circ d)(\alpha)=0}.

This gives us that our map {\phi_{0,q}} is an iso {\Gamma(X, \mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T} \rightarrow \Gamma(X, \mathcal{C}^{0,q})}. The map {\overline{\partial}} is defined to be {\overline{\partial}(f\otimes \alpha)=\overline{\partial}(f)\otimes \alpha}. So since the {\omega_\alpha} are closed, the iso commutes with the differential and we get these are actually isomorphic as chain complexes. Thus computing cohomology of one is equivalent to computing cohomology of the other.

Explicitly, we know that {H^q(X, \mathcal{O}_X)\simeq H^q(X, \Gamma(X,\mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T})}.

Since this notation is cumbersome, let {A=\Gamma(X, \mathcal{C})} and {\bigwedge^*=\bigwedge^*\overline{T}}.

Let {i: \bigwedge \rightarrow A\otimes_\mathbb{C} \bigwedge} be the inclusion. We want to show this gives an iso {\bigwedge^q\stackrel{\sim}{\rightarrow} H^q(X, A\otimes \bigwedge^*)}. This is precisely the goal given at the start of the post.

Now {V} is a vector space so we have a natural Euclidean measure. Let {\mu} be the measure on {X} induced from this that is normalized so that {\mu(X)=1}. (For those familiar, this is just the unique translation invariant Haar measure, and {X} is compact, so is finite and can be normalized). Define the linear map represented by {\mu} by {S: A\rightarrow \mathbb{C}}. It is just {S(f)=\int_X fd\mu}. If {W} is any {\mathbb{C}}-vector space, denote by {S_W} the map {A\otimes_\mathbb{C} W\rightarrow W}. In particular, we have {S_\wedge: A\otimes \bigwedge^*\rightarrow \bigwedge^*} so {S_\wedge\circ i=id_\wedge}.

We have a good ways to go yet, so next time we’ll pick up with the lemma that {S_\wedge(\overline{\partial}\omega)=0} for all {\omega\in A\otimes \bigwedge}.


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Cohomology of Abelian Varieties

Hopefully I’ll start updating more than once a month. Since it’s been awhile and the previous post was tangent to what we’re actually doing, I’ll recap some notation. {X} will be a compact complex (connected) Lie group of dimension {g}. We showed that we have an analytic isomorphism {X\simeq (S^1)^{2g}\simeq (\mathbb{R}/\mathbb{Z})^{2g}}. Let {V=T_0X} (note that I’ll assume {0} is the identity).

Under the exponential map {exp: V\rightarrow X} (which we showed was a local isomorphism), we have that {V} is the universal covering space of {X}. We showed that {ker(exp)=U} is a lattice. Now the title of this post will seem a little silly to experts out there in cohomology, since we know that topologically these things are all tori. We’ll go through the details anyway.

First, we’ll show that {H^r(X, \mathbb{Z})\simeq } the group of alternating {r}-forms {U\times \cdots \times U\rightarrow \mathbb{Z}}. We proceed by induction.

By general covering space theory {exp^{-1}(0)=U=\pi_1(X, 0)}. Thus we get the base case {H^1(X,\mathbb{Z})\simeq Hom(U=\pi_1(X), \mathbb{Z})}. Now we’ll want to show that the cup product induces the isomorphism {\bigwedge^r\left(H^1(X,\mathbb{Z})\right)\rightarrow H^r(X,\mathbb{Z})}. By the {r=1} case this proves the statement.

We first reduce to the case of showing it is true for {S^1}. Since {X} is just a product of {S^1}‘s, if we show that if the statement is true for {X_1} and {X_2}, then it is also true for the product we can make the reduction. (For simplicitly, coefficients are in {\mathbb{Z}}, but I’ll omit that). Since we only need to apply this in the case where {X_1} or {X_2} is finite product of {S^1}‘s, we can also assume that the cohomologies are finitely generated and that they two spaces are connected for simplicity.

First, by the K\”{u}nneth formula: {H^1(X_1\times X_2)\simeq \left(H^1(X_1)\otimes_\mathbb{Z}H^0(X_2)\right)\bigoplus \left(H^0(X_1)\otimes_\mathbb{Z}H^1(X_2)\right)}, but the spaces are connected, so {H^0(X_i)=\mathbb{Z}}. Thus {\bigwedge^r(H^1(X_1\times X_2))\simeq \bigwedge^r(H^1(X_1)\oplus H^1(X_2))}
{\simeq \displaystyle\sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}.

But now our inductive hypothesis is that for all p less than r, {\bigwedge^p(H^1(X_i))\simeq H^p(X_i)}. Thus we get {\displaystyle \sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}
{\displaystyle \simeq \sum_{p+q=r} H^p(X_1)\otimes H^q(X_2)}
{\simeq H^r(X_1\times X_2)}. In other words, stringing all these isos together we get the iso we wanted. So we’ve reduced to the case of showing the statement for {S^1}, which follows immediately since {H^n(S^1)=0} for {n>1}.

Note that if you have basic facts about singular cohomology at your disposal, this isn’t at all surprising. But let’s look at sheaf cohomology instead. This will require us to look at the Hodge structure which could be interesting. We won’t go very far today, but let’s at least get a few things out of the way.

Let {\Omega^p} be the sheaf of holomorphic {p}-forms on {X}. We’d like to compute {H^r(X, \Omega^p)}. Let {T=Hom(V, \mathbb{C})}, i.e. the (complex) cotangent space at the identity to {X}. As with vectors and vector fields, every {p}-covector, i.e. element of {\bigwedge^pT} can be extended uniquely to a left invariant {p}-form by pulling back along the left multiplication by {-x} map. We’ll denote the correspondence {\alpha\mapsto \omega_\alpha}. This map defines an isomorphism of sheaves {\mathcal{O}_X\otimes_\mathbb{C} \bigwedge^pT\stackrel{\sim}{\rightarrow} \Omega^p}.

This says that {\Omega^p} is a free sheaf of {\mathcal{O}_X}-modules. Now take global sections to get that {\Gamma(X, \Omega^p)\simeq \Gamma(X, \mathcal{O}_X\otimes \bigwedge^pT)\simeq \bigwedge^pT}, since the global sections of {\mathcal{O}_X} are constants. Thus the only global sections of {\Omega^p} are the {p}-forms that are invariant under left translation. Thus this isomorphism reduces our calculation to {H^r(X, \Omega^p)\simeq H^r(X, \mathcal{O}_X)\otimes_\mathbb{C} \bigwedge^pT}. So we’ll start in on that next time.


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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.


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Generalized HT90

I officially promise this is my last post on Hilbert’s Theorem 90, but because of that it is going to go really fast for those who have not seen group cohomology (it is really cool, so I couldn’t pass it up).

An abelian group is a G-module (G a group) if for all \sigma\in G and a\in A there is a unique element \sigma(a)\in A satisfying two conditions: \sigma(a+b)=\sigma(a)+\sigma(b), and (\sigma\tau)(a)=\sigma(\tau(a)) for all \sigma, \tau\in G and a,b\in A.

Just check any algebra text or here for more information on modules.

Now define an n-cochain of G over A to be a a function of n variables from G into A. If n=0 it is just an element of A. C^n(G, A) is the set of all n-cochains, and can be made into a group by the operation (f+g)(\sigma_1, \ldots, \sigma_n)=f(\sigma_1, \ldots, \sigma_n)+g(\sigma_1, \ldots, \sigma_n).

We can also get from C^n(G, A) to C^{n+1}(G, A) (which is what people who know about cohomology were hoping for), by the function \delta:

(\delta f)(\sigma_1, \ldots, \sigma_{n+1})=\sigma_1(f(\sigma_2, \ldots, \sigma_{n+1}))+

\sum_{i=1}^n(-1)^i f(\sigma_1, \ldots, \sigma_i\sigma_{i+1}, \ldots, \sigma_{n+1})+(-1)^{n+1}f(\sigma_1, \ldots, \sigma_n).

OK. That looks bad, but really in some sense it is the natural choice. I’ll leave it to you to check that this is both a homomorphism and that \delta\delta f=0 (i.e. we have a chain complex).

Now if we label them \delta_0: A\to C^1(G, A)

\delta_1: C^1(G, A)\to C^2(G, A) and

\delta_n: C^n(G, A)\to C^{n+1}(G, A). Then we form \displaystyle H^n(G, A)=\frac{\ker\delta_n}{im\delta_{n-1}}=\frac{Z^n(G, A)}{B^n(G, A)}. We call the elements of Z^n(G, A) the n-cocycles and the elements of B^n(G, A) the n-coboundaries.

So if you don’t like that, we can scratch it now, since in HT90 we only care about H^1(G, A), so let’s take a closer look at that. We can completely classify what the elements of Z^1(G, A) look like. For any f\in Z^1(G, A) and any \sigma, \tau\in G we need (\delta f)(\sigma, \tau)=\sigma(f(\tau))-f(\sigma\tau)+f(\sigma)=0. Which is to say that f(\sigma\tau)=\sigma(f(\tau))+f(\sigma).

Now let’s classify what B^1(G,A) looks like. Well, if g\in B^1(G, A), then g=\delta h for some h\in A=C^0(G,A). So g(\sigma)=(\delta h)(\sigma)=\sigma(a)-a for some a\in A. Well, I think you might be able to see the previous formulation of the theorem coming from unravelling these definitions.

Theorem statement: If K/F is a finite Galois extension and G=Gal(K/F), then H^1(G, K^\times) is trivial.

Proof: Let a be a cocycle. Then let \alpha: K\to K by c\mapsto \sum_{\sigma\in G}a(\sigma)\sigma(c). As in the last post, \alpha is not 0 by linear independence. So let c\in K such that \alpha(c)\neq 0 and set b=\alpha(c).

Then for any \tau\in G we have

\displaystyle\tau(b)=\sum_{\sigma\in G}\tau(a(\sigma)\sigma(c))

\displaystyle =\sum_{\sigma\in G}\tau(a\sigma)(\tau\sigma)(c)

\displaystyle =\sum_{\sigma\in G}a(\tau)^{-1}a(\tau\sigma)(\tau\sigma)(c). Now we use that a is a cocycle (in the kernel) to continue the equality as

\displaystyle = a(\tau)^{-1}\sum_{\sigma\in G}a(\tau\sigma)(\tau\sigma)(c)

=a(\tau)^{-1}b.

Aha, so a(\tau)=b\tau(b)^{-1} is a coboundary! Thus every cocycle is a coboundary, so the quotient is trivial.

Test your understanding by now trying to prove the other formulation as a corollary to this (remember you assume that G is cyclic in that version and have to relate it back to the norm).


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Hilbert’s Theorem 90…the math

So now that I’ve presented the origins of Hilbert’s Theorem 90, I thought it might be good to actually present it mathematically with a proof.

Statement: Suppose K is a finite Galois extension of F with G=Gal(K/F) cyclic, say G=<\sigma > of order n. If a\in K, then N_{K/F}(a)=1 if and only if a=b/\sigma(b) for some b\in K.

Let’s parse this a little before the proof. First off let’s define the notation N_{K/F}(a). Notice that we can view K as a vector sapce over F, so t_a: K\to K by u\mapsto au is a linear transformation. So we want a norm, so we define N_{K/F}(a)=\det (t_a). i.e. the determinant of the matrix that represents that linear transformation. Thus, we get all our nice results from linear algebra like independence of basis choice. If we fight with definitions and linear algebra for a bit we also get that if the extension is Galois, \displaystyle N_{K/F}(a)=\prod_{\phi\in Gal(K/F)} \phi (a). This is left as an (not required) exercise to familiarize yourself with the terms before moving on. Note that we can determine Tr_{K/F}(a) similarly to be the trace of t_a.

Proof of the theorem: Let’s do the backwards way first (I’m going to suppress the norm notation since we know where it is happening). Suppose a=b/\sigma(b) for some b\in K. Then let’s simply use the result (that you proved!) to see that \displaystyle N(a)=\left(\frac{b}{\sigma(b)}\right)\left(\frac{\sigma(b)}{\sigma^2(b)}\right)\cdots \left(\frac{\sigma^{n-1}(b)}{\sigma^n(b)}\right)=1 since \sigma has order n.

Now for the forwards direction. Suppose N(a)=1. Again using a standard linear algebra argument we know that 1, \sigma, \ldots, \sigma^{n-1} are K-linearly independent (in fact, any set of field automorphisms of K are K-linearly independent over the vector space of all functions from K to K).

Thus, \displaystyle \phi = a\cdot 1+(a\sigma(a))\sigma + (a\sigma(a)\sigma^2(a))\sigma^2+\cdots + \left(\prod_{i=0}^{n-1}\sigma^i (a)\right)\sigma^{n-1} from K to K is not 0. By the thing you proved the coefficient on \sigma^{n-1} is N(a) which is 1. More importantly, there is some c\in K such that \phi(c)\neq 0, so let’s define b=\phi(c).

But \sigma(b)=\frac{1}{a}(b-ac)+\sigma^n(c)=\frac{1}{a}(b-ac)+c=\frac{b}{a}. Thus a=b/\sigma(b). And we are done.

Does this proof remind anyone of Lagrange resolvents? Eh, whatever, I won’t dig for a connection now.

I also mentioned that there is a generalization of this.

Statement: If K/F is a finite Galois extension and G=Gal(K/F), then H^1(G, K^\times) is trivial. I’m debating whether or not to parse and prove this version next time, or to just drop the Hilbert Theorem 90 posts. Suggestions?


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Property of the Plane

I was reading the blog topological musings, and the category talk reminded me of this neat thing I posted on AoPS awhile ago.

By using the Mayer-Vietoris sequence in cohomology, we determine whether or not \mathbb{R}^2 can be written as the union of two open connected sets, U and V, such that U\cap V is disconnected.

Well, it seems that we are concerned with H^0 stuff, since that tells us the number of connected components.

The sequence gives 0\to H^0(\mathbb{R}^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)\to 0, a nice short exact sequence in which we mostly know they are connected, so

0\to \mathbb{R}\to \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V)\to 0, basically the question reduces to: Is it possible for the second to last term to have dimension greater than 1, but the sequence to remain exact?

Well, no, since the exactness tells us the dimension of H^0(U\cap V) is the dimension of \frac{\mathbb{R}\oplus\mathbb{R}}{\mathbb{R}}\cong \mathbb{R}.

So we have a nifty result: the plane cannot be broken into two connected parts in which the intersection of these two parts is disconnected.

Challenge for the readers: This seems extremely simple and obvious. Find a proof that does not require advanced techniques such as cohomology.

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