## Model Categories, 2

February 26, 2010

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.

The homotopy category of a model category.

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## Algebraic Coding Theory

February 24, 2010

I 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.

Survey of Algebraic Coding Theory

## Model Categories

January 25, 2010

Model 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.)

Introductory Notes on Model Categories

## Steenrod Squares

November 23, 2009

Steenrod squares recover “stable” information from the cup product structure that otherwise disappears. Recall that ${H^*(\Sigma X)}$ has no nontrivial cup products because it is the union of two contractible open sets. This means that we cannot determine information about $\Sigma X$ through its cohomology ring structure. However, we can determine information about the homotopy type of $\Sigma X$ through the cup product structure on $X$, using properties of Steenrod squares.

For each ${i \ge 0}$, there is a mod-2 cohomology operation ${{\mathrm {sq}}^i : H^{n} \rightarrow H^{n+i}}$ satisfying the following properties:

1. ${{\mathrm {sq}}^i \circ f^* = f^* \circ {\mathrm {sq}}^i}$ (naturality)
2. ${{\mathrm {sq}}^i(x+y) = {\mathrm {sq}}^i(x) + {\mathrm {sq}}^i(y)}$
3. ${{\mathrm {sq}}^n(x \smile y) = \sum_i {\mathrm {sq}}^i(x) \smile {\mathrm {sq}}^{n-i}(y)}$ (Cartan formula)
4. ${{\mathrm {sq}}^i \circ \Sigma = \Sigma \circ {\mathrm {sq}}^i}$ (stability)
5. $\displaystyle \mathrm{sq}^i(\alpha) = \alpha^2 \textrm{ if } i = |\alpha|$,
6. $\displaystyle \mathrm{sq}^i(\alpha) = 0 \textrm{ if } i > |\alpha|$,
7. ${{\mathrm {sq}}^0 = \mathrm{id}}$
8. ${{\mathrm {sq}}^1}$ is the Bockstein homomorphism for the short exact sequence of coefficients
$\displaystyle 0 \rightarrow {\mathbb Z}/2 \rightarrow {\mathbb Z}/4 \rightarrow {\mathbb Z}/2 \rightarrow 0.$

If ${I = (i_1,\dots,i_k)}$, we may write

$\displaystyle {\mathrm {sq}}^I := {\mathrm {sq}}^{i_1}{\mathrm {sq}}^{i_2} \dots {\mathrm {sq}}^{i_k}.$

If, for each ${r}$, ${i_r \ge 2 i_{r+1}}$ (counting ${i_{k+1} = 0}$), we call ${I}$ admissible. The excess of an admissible ${I}$ is the quantity

$\displaystyle e(I) = \sum_{r} i_r - 2i_{r+1} = i_1 - \sum_{r > 1} i_r.$

The following example shows how Steenrod squares can use the cup product structure in ${X}$ to tell us about the homotopy type of ${\Sigma X}$.

Proposition 1 Let ${X = S^1 \vee S^2}$. Then ${X}$ is not stably homotopic to ${P := \mathbb{C}P^2}$.

Proof: Take cohomology modulo 2. Let ${x \in H^1(X;{\mathbb Z}/2) = {\mathbb Z}/2}$ be a generator, and let ${y \in H^1(P;{\mathbb Z}/2)}$ be a generator. Then ${x^2 = 0}$ while ${y^2 \ne 0}$ is a generator of ${H^2(P;{\mathbb Z}/2)}$. Therefore cup products alone tell us that ${X}$ and ${P}$ are not homotopic. However, ${H^n(X;{\mathbb Z}) = H^n(P;{\mathbb Z})}$ for every ${n}$, so cohomology groups alone are not sufficient; unfortunately then the cohomology groups ${H^n(\Sigma^k X; {\mathbb Z}) = H^n(\Sigma^k P; {\mathbb Z})}$, 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 ${{\mathrm {sq}}^1(\Sigma^k x) = \Sigma^k({\mathrm {sq}}^1 x) = \Sigma^k(x^2) = 0}$, while ${{\mathrm {sq}}^1(\Sigma^k y) = \Sigma^k(y^2) \ne 0}$. Therefore ${\Sigma^k X}$ and ${\Sigma^k P}$ are not stably homotopic. $\Box$

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 $K(\mathbb{Z}/2,n)$, which will be discussed in a later post. This calculation expresses the cohomology groups $H^m(K(\mathbb{Z}/2,n);\mathbb{Z}/2)$ in terms of $\mathrm{sq}^I$ over admissible sequences $I$ where $e(I) \le n$.

## Fun Proof of Elementary Fact

October 25, 2009

While thinking about a number theory problem, I stumbled into a proof of a classical fact: the classification of primitive pythagorean triples $(x,y,z)$ where $x^2 + y^2 = z^2$. Here is a proof using unique prime factorization in the Gaussian integers.

Claim: If $x^2 + y^2 = z^2$ where $(x,y,z) = (1)$, then there are integers c,d such that $x = (c^2-d^2), y = 2cd$.

Proof: Suppose $x^2 + y^2 = z^2$. Then we have $(x+iy)(x-iy) = z^2$. We know $(x+iy,x-iy) \supset (2x,2y) = (2) = (1+i)^2$ But if $1+i \mid x+iy$, then $2 \mid x^2 + y^2 = z^2$, so then $4 \mid x^2+y^2$ and we must have that $x,y$ are even, which is impossible for primitive triples (x,y,z).  Therefore $(x+iy,x-iy) = (1)$, and so (x+iy) and (x-iy) must both be squares. So, we can write $x+iy = (c+di)^2 = (c^2-d^2) + (2cd)i$ 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, 2009

I’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 $\times^{\mathrm{alg}} : H^p(C^*) \otimes H^q(D^*) \to H^{p+q}((C_* \otimes D_*)^*)$ is defined by $[f] \otimes [g] \mapsto [\sigma \otimes \tau \mapsto f(\sigma)g(\tau)]$.

It is a simple exercise to show this is well-defined; it’s induced from the standard embedding of $C^* \otimes D^*$ into $(C_* \otimes D_*)^*$.

Definition: A diagonal approximation is a natural chain map $\tau : S_*(X) \to S_*(X) \otimes S_*(X)$ satisfying $\tau(\sigma) = \sigma \otimes \sigma$ for 0-simplices $\sigma$.

Since the functor $S_*(-)$ is free with models $\Delta^n$ and $S_*(-) \otimes S_*(-)$ is acyclic on these models, a diagonal approximation exists, and furthermore any two are chain homotopic. A standard choice of $\tau$ is the Alexander-Whitney diagonal approximation given by $\tau(\sigma) = \sum_{p+q=|\sigma|} { }_p\sigma \otimes \sigma_q$. We let ${ }_p\sigma$ denote the “front face” of $\sigma$ consisting of its first p vertices, and $\sigma_q$ denotes its back q vertices. This choice of $\tau$ 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 $\smile : H^*(X) \otimes H^*(X) \to H^*(X)$ is the composition $(\tau)^* \circ \times^\mathrm{alg} : H^*(X) \otimes H^*(X) \to H^*((S_*X \otimes S_*X)^*) \to H^*(X)$, where $\tau$ is any diagonal approximation.

Since any two diagonal approximations are chain homotopic, all choices of $\tau$ give the same cup product. Using the acyclic models theorem, one can easily show associativity and graded commutativity of $\smile$. However, without a choice of $\tau$, 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: $(a \smile b) = (-1)^{|a||b|}(b \smile a)$ for $a,b \in H^*(X)$.

Proof: Consider $T : S_*(X) \otimes S_*(X) \to S_*(X) \otimes S_*(X)$ given by $T(\sigma \otimes \tau) = (-1)^{|\tau||\sigma|} (\sigma \otimes \tau)$.Then $T \circ \tau$ is a diagonal approximation; so by uniqueness of the cup product $a \smile b = (-1)^{|b||a|}(b \smile a)$.

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 $a \smile b = (-1)^{|a||b|} b \smile a$ 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, 2009

The 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 $F$ and $G$ are functors from a category $\mathcal{A}$ to the category of augmented chain complexes, such that $F$ and $G$ are free and acyclic on models $\mathcal{M}$, then there exists a natural transformation $F \to G$ extending the identity map $\mathbb{Z} \to \mathbb{Z}$, which is unique up to natural chain homotopy.

Corollary 2: If $F$ and $G$ 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 $H_0(F) \to H_0(G)$ 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 $S(X)$ denote the singular chain complex of the space $X$. Then the functors $S(X \times Y)$ and $S(X) \otimes S(Y)$ from $\mathrm{Top}^2$ to $\partial \mathfrak{G}$ are naturally chain homotopic. As a result, $H_n(X \times Y) = H_n(S(X) \otimes S(Y))$.

Proof: By definition both functors $S(X \times Y)$ and $S(X) \otimes S(Y)$ are free with models $\{(\Delta^n,\Delta^m) | n,m \ge 0\}$ (in fact, $S(X \times Y)$ is free with models $\{(\Delta^n, \Delta^n)\}$). Since the space $\Delta^n \times \Delta^m$ is contractible, $S(X \times Y)$ is acyclic in positive dimensions.  Since $\Delta^n$ is itself contractible, $S(\Delta^n) \otimes S(\Delta^m)$ is chain homotopic to $S(*) \otimes S(*)$, which is acyclic in positive dimensions as well. Therefore both functors are free and acyclic.

Consider the (natural) map $S_0(X \times Y) \to (S(X) \otimes S(Y))_0 = S_0(X) \otimes S_0(Y)$  taking $\sigma \times \tau \mapsto \sigma \otimes \tau$.  This map is invertible and so induces an isomorphism on $H_0$. 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 $A$ and $B$ are subspaces of $X$ whose interiors cover $X$, then $S(A + B)$ and $S(X)$ are chain homotopic, where $S(A+B)$ is the chain complex of simplices whose image lies completely in $A$ or completely in $B$.

Excision is typically proven through barycentric subdivision, with the geometric intuition being that every simplex in $X$ can be subdivided into smaller simplices, and eventually each simplex will be small enough to be covered either by the interior of $A$ or the interior of $B$. 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 $S_0(A + B)$ and $S_0(X)$ are isomorphic, since the interiors of $A$ and $B$ form an open cover of $X$. One can consider the category of spaces with open coverings, which has objects $(X, \mathcal{U})$, where $\mathcal{U}$ is an open cover of $X$, and whose maps respect these coverings in the natural way.

Then we can associate to $(X,\mathcal{U})$ the chain complex $S(X, \mathcal{U})$ of simplices whose image is contained wholly in one of the open sets of $\mathcal{U}$, and we can also associate to it the standard chain complex $S(X)$. One can show both are acyclic and free with models $(\square^n, \{\square^n\})$, where $\square^n$ is the $n$-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:

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

## Acyclic Models

August 11, 2009

Acyclic 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 $\mathcal{C}$ be a category, and $\mathcal{M}$ be a collection of objects in $\mathcal{C}$, called model objects. Let $T : \mathcal{C} \to \mathrm{Ab}$ be a functor, where $\mathrm{Ab}$ is the category of abelian groups. If there are objects $g_M \in T(M)$ for $M \in \mathcal{M}$ such that $T(A)$ is freely generated by the images $T(f)(g_M)$, then $T$ is said to be free with models $\mathcal{M}$.

For example, the singular $n$-chains $C_n(X)$ on a space $X$ are free with a single model $\Delta^n$.

The Big Theorem

Let $K$ and $L$ be covariant functors on $\mathcal{C}$ with values in the category of chain complexes, and let $f : K \to L$ be a chain map defined in dimensions $< q$. Then if $K_n$ is free with models $\mathcal{M}$ for all $n \ge q$ and $H_n(L(M)) = 0$ for all $n \ge q-1$ and all $M \in \mathcal{M}$, then $f$ extends to a chain map $f' : K \to L$; furthermore, $f'$ is unique up to a chain homotopy $D$ satisfying $D_n = 0$ for every $n < q$.

In words: if $K$ is free with models and $L$ is acyclic on those models, then partially defined maps $K \to L$ can be extended uniquely up to chain homotopy. The interesting corollary is that if $K$ and $L$ 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 $\Delta^n$, and since simplices are contractible $\widetilde{H}_*(\Delta^n) = 0.$ Note that the (unreduced) singular chain complex is not acyclic since $H_0(\Delta^n) = \mathbb{Z}$.

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

References

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

## Foundations

August 10, 2009

I 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 $\LaTeX$, which makes things awfully convenient for mathematical discourse.