Continuing my discussion of Model Categories (again, mostly through Hovey’s book, with my own reorganization and interpretations.) I’m having my notes follow the lectures I’ve been giving in seminar. They should be readable, but questions are welcome.

## Model Categories, 2

February 26, 2010## Algebraic Coding Theory

February 24, 2010I just found a survey paper I wrote for 6.UAP. Unfortunately, it’s just a draft, and I’ve lost my backup with the .tex file in it, so it’s missing pieces.

In a nutshell: the project was to do some reading on results about error correcting codes. In particular, we can consider error-correcting codes an independent subsets of vertices of hamming graphs, whose vertices are given by bit-strings, with two distinct vertices adjacent whenever the hamming distance between these bit-strings is small. These basic concepts are introduced, and bounds on sizes of independent sets of vertices are discussed. This was written for an audience with a basic understanding of linear programming and linear algebra, so it should be accessible to ambitious undergraduates, for example.

## Model Categories

January 25, 2010Model categories abstract key map lifting and extension properties so that homotopy theory can be performed in categories other than Top.

I’m preparing a short introduction for a seminar I will be taking this term. Here are my notes as of the moment. (Does anyone know nice ways of converting LaTeX to wordpress? Doing things by hand can be really annoying.)

## Steenrod Squares

November 23, 2009Steenrod squares recover “stable” information from the cup product structure that otherwise disappears. Recall that has no nontrivial cup products because it is the union of two contractible open sets. This means that we cannot determine information about through its cohomology ring structure. However, we can determine information about the homotopy type of through the cup product structure on , using properties of Steenrod squares.

For each , there is a mod-2 cohomology operation satisfying the following properties:

- (naturality)
- (Cartan formula)
- (stability)
- ,
- ,
- is the Bockstein homomorphism for the short exact sequence of coefficients

If , we may write

If, for each , (counting ), we call *admissible*. The *excess* of an admissible is the quantity

The following example shows how Steenrod squares can use the cup product structure in to tell us about the homotopy type of .

Proposition 1Let . Then is not stably homotopic to .

*Proof:* Take cohomology modulo 2. Let be a generator, and let be a generator. Then while is a generator of . Therefore cup products alone tell us that and are not homotopic. However, for every , so cohomology groups alone are not sufficient; unfortunately then the cohomology groups , and the cup product structure of the cohomology rings are trivial.

However, Steenrod squares allow us to recover some information from the cup product; indeed , while . Therefore and are not stably homotopic.

It turns out that Steenrod squares generate all mod-2 cohomology operations, which can be seen through the calculation of the cohomology of Eilenberg-Maclane spaces , which will be discussed in a later post. This calculation expresses the cohomology groups in terms of over admissible sequences where .

## Fun Proof of Elementary Fact

October 25, 2009While thinking about a number theory problem, I stumbled into a proof of a classical fact: the classification of primitive pythagorean triples where . Here is a proof using unique prime factorization in the Gaussian integers.

**Claim**: If where , then there are integers c,d such that .

**Proof**: Suppose . Then we have . We know But if , then , so then and we must have that are even, which is impossible for primitive triples (x,y,z). Therefore , and so (x+iy) and (x-iy) must both be squares. So, we can write for some integers c and d, as desired.

The above turns out the be pretty standard, but I still find it amusing and elegant.

## Cup Products

August 20, 2009I’ve always found cup products to be somewhat mysterious. The cup product on a cohomology is a *natural* product in terms of maps of spaces, but it doesn’t mandate a canonical definition like you might expect. In fact, the cup product descends from a map on the cochain level that is quite difficult to reason with, and it is only upon passing to cohomology that it gains many of its nicer features.

The cup product on cohomology is the composition of a *cohomology cross product, *which is simple and easy to understand, and a *diagonal approximation*, which is easy to understand in low dimensions.

**Definition**: The algebraic cohomology cross product is defined by .

It is a simple exercise to show this is well-defined; it’s induced from the standard embedding of into .

**Definition**: A *diagonal approximation* is a natural chain map satisfying for 0-simplices .

Since the functor is free with models and is acyclic on these models, a diagonal approximation exists, and furthermore any two are chain homotopic. A standard choice of is the *Alexander-Whitney diagonal approximation* given by . We let denote the “front face” of consisting of its first *p* vertices, and denotes its back *q* vertices. This choice of is clearly a diagonal approximation, but what is its geometric significance? Not much as far as I can tell — the big advantage of the Alexander-Whitney map, in my opinion, is that it is a clean algebraic device, which is (relatively) easy to calculate, and as such provides for clean computations.

Oh, and before I forget:

**Definition** *(Cup Product)*: The *cup product * is the composition , where is any diagonal approximation.

Since any two diagonal approximations are chain homotopic, all choices of give the same cup product. Using the acyclic models theorem, one can easily show associativity and graded commutativity of . However, without a choice of , it’s not clear how to do any computations with this thing, you have to either use naturality or drop down to the cochain level and the Alexander-Whitney map to know more.

**Proposition**: for .

*Proof*: Consider given by .Then is a diagonal approximation; so by uniqueness of the cup product .

Abstract nonsense is great because it simplifies arguments by exploiting symmetry. However, sometimes you need to get to the nitty-gritty just a little bit to do some actual computations. Try proving *without *using the acyclic models theorem, just using the Alexander-Whitney map directly — I doubt it would be a fun endeavor.

## Acyclic Models, the Eilenberg-Zilber Theorem, and Excision

August 13, 2009The acyclic models theorem, in the full generality that I stated in my previous post, is not usually the form in which it is used. More generally:

**Corollary** **1**: If and are functors from a category to the category of augmented chain complexes, such that and are free and acyclic on models , then there exists a natural transformation extending the identity map , which is unique up to natural chain homotopy.

**Corollary 2: **If and are functors to the category of (non-augmented) chain complexes which are free, and acyclic in dimensions above zero, then any natural chain map between them that restricts to an isomorphism on is a chain homotopy equivalence.

Corollaries 1 and 2 above are effectively restatements of the same powerful idea. The acyclic models theorem reduces equality arguments in homology to proving that appropriate chain complexes are free and acyclic.

**Theorem (Eilenberg-Zilber):** Let denote the singular chain complex of the space . Then the functors and from to are naturally chain homotopic. As a result, .

*Proof:* By definition both functors and are free with models (in fact, is free with models ). Since the space is contractible, is acyclic in positive dimensions. Since is itself contractible, is chain homotopic to , which is acyclic in positive dimensions as well. Therefore both functors are free and acyclic.

Consider the (natural) map taking . This map is invertible and so induces an isomorphism on . Therefore the two functors are chain homotopic. QED.

And it is, remarkably, that simple, although I spent quite a while convincing myself that I hadn’t missed any steps (and I still hope I haven’t). It just turns out that the acyclic models theorem is incredibly powerful; especially in the proof of the excision theorem.

And, yes, I have done some weasely magic by assuming that homology is invariant over homotopic spaces; although that’s a standard result, the Eilenberg-Zilber theorem does yield a *very* simple proof that homotopic maps induce isomorphisms on homology — one could hypothetically just compute the homology of simplices directly if one wanted to be a purist.

The excision theorem is proven in a similar way; at the core of excision is the claim that if and are subspaces of whose interiors cover , then and are chain homotopic, where is the chain complex of simplices whose image lies completely in or completely in .

Excision is typically proven through *barycentric subdivision*, with the geometric intuition being that every simplex in can be subdivided into smaller simplices, and eventually each simplex will be small enough to be covered either by the interior of or the interior of . However, you could check in Hatcher’s book to see that this argument is far less simple than you would like it to be, taking up a few pages to hammer out all of the algebraic (and geometric!) details.

Instead, the acyclic models theorem opens up another avenue of attack: it is painfully clear that and are isomorphic, since the interiors of and form an open cover of . One can consider the category of *spaces with open coverings*, which has objects , where is an open cover of , and whose maps respect these coverings in the natural way.

Then we can associate to the chain complex of simplices whose image is contained wholly in one of the open sets of , and we can also associate to it the standard chain complex . One can show both are acyclic and free with models , where is the -cube, without too much trouble (I will omit the details, but it’s a basic lebesgue number /compactness argument). Therefore, these functors are naturally chain homotopic, and excision is proved.

(Covering bases: the well-definedness of derived functors is not an acyclic models argument so much as it is nearly identical to the *proof* of acyclic models. Sorry about that. )

Next I will be moving into a brief look at cohomology, cup products, and the power of acyclic models in that context for proving things like associativity and commutativity.

**References**:

- R. Schon,
*Acyclic Models and Excision*. Proceedings of the American Mathematical Society, 1976.

## Acyclic Models

August 11, 2009Acyclic Models is a powerful technique for constructing maps between chain complexes. This method can be used to show the equivalence of chain complexes, and thereby construct isomorphisms of homology theories for spaces. Four elementary uses of acyclic models are:

- Excision
- Eilenberg-Zilber theorem
- Construction of Derived functors (for example, Ext and Tor)
- Commutativity of the Cup Product

**The Setup**

Let be a category, and be a collection of objects in , called *model objects*. Let be a functor, where is the category of abelian groups. If there are objects for such that $T(A)$ is freely generated by the images , then is said to be *free with models* .

For example, the singular -chains on a space are free with a single model .

**The Big Theorem**

Let and be covariant functors on with values in the category of chain complexes, and let be a chain map defined in dimensions . Then if is free with models for all and for all and all , then extends to a chain map ; furthermore, is unique up to a chain homotopy satisfying for every .

In words: if is free with models and is acyclic on those models, then partially defined maps can be extended uniquely up to chain homotopy. The interesting corollary is that if and are *both* free and acyclic on the same models, then they are chain equivalent, provided an isomorphism in dimension zero.

This can be particularly useful because the reduced singular chain complex of a space is free with models , and since simplices are contractible Note that the (unreduced) singular chain complex is *not* acyclic since .

In my next post, I will sketch a proof of the excision theorem in singular homology, using the method of Acyclic Models.

**References**

- Samuel Eilenberg and Saunders MacLane, “Acyclic models,”
*American Journal of Mathematics,*vol. 75 (1953), pp. 189-199

## Foundations

August 10, 2009I set this up to discuss ideas from mathematics that I find intriguing or am learning about at the moment. My focus is in algebraic topology, so notes will mostly be in that domain. WordPress is nice for this because it has support for , which makes things awfully convenient for mathematical discourse.