Chromotopy

Minimal Models, Maximal Mathematics: Part I

So, we’ve got a topological space, huh? Let’s call it $X$. What are we gonna do with it? Well, the first thing that we might try to do is compute it’s homotopy groups, $\pi_n (X)$. That seems like a pretty normal thing to do, being topologists. Unfortunately for us, computing homotopy groups is (generally) a Pretty Hard Thing. We might try and get around some of this difficulty by only asking for the rational homotopy groups of $X$, which we obtain by taking $\pi_q (X) \otimes \mathbb{Q}$. This effectively kills the torsion in $\pi_q (X)$, and shows us only the rank of $\pi_q (X)$. We’ve thrown away some of the information contained in $\pi_q (X)$ by doing this, but the end result is that we get something that can often be computed.

How, you ask? Well, one way that this can be done is via the theory of minimal models, which was introduced by Sullivan in the 70’s. We’re getting paid by the theorem, so let’s get to it! As usual, there’s a bit of setup involved in starting to talk about this business, but it’ll pay off in the end.

In the following, all algebras are over $\mathbb{R}.$ Let $A = \oplus_{i \geq 0} A^i$ be a differential graded commutative algebra, with the differential $d$ being an antiderivation of degree 1 (it moves things up a degree):

$$d(a \cdot b ) = (da) \cdot b + (-1)^{\mid a\mid} a \cdot db$$

with graded commutativity:

$$a \cdot b = (-1)^{\mid a\mid \cdot \mid b\mid} b \cdot a.$$

An example of this is the De Rham complex of differential forms on a smooth manifnew, with the product being the wedge product and the differential being the exterior derivative.

Definition. We say that such an algebra is free if it satisfies no relations besides associativity and graded commutativity. We write $\Lambda (x_1, \dots, x_n)$ for the free algebra generated by $x_1, \dots, x_n.$ An element in $A$ is said to be decomposable if it is a sum of products of positive elements in $A$ (degree > 0).

Now, we introduce minimal models, which will be our main objects of study.

Definition. A differential graded algebra $M$ is a minimal model for $A$ if:

1. $M$ is free;
2. there is a chain map $f : M \rightarrow A$ that induces an isomorphism on cohomology;
3. the differential of a generator is either zero or decomposable.

Definition. A differential graded algebra $A$ is 1-connected if $H^0 (A)= \mathbb{R}$ and $H^1 (A) = 0$.

Note that the De Rham cohomology of any connected, simply-connected manifnew will have this property. The nice thing about such 1-connected differential algebras is that they have minimal models. That is:

Proposition. Let $A$ be a graded, differential $\mathbb{R}$-algebra that is 1-connected. Then A has a minimal model.

The proof is easy enough, and you’d probably get the idea from the fact that they’re called “minimal models”. We’re going to construct the minimal model $M$ for $A$ by taking free algebras and adding in exactly the extra stuff that we need in order for $M$ to have the same cohomology as $A$ (hence the “minimal”).

Let’s start off trying to get $H^2(M)$ right. That’s not too hard: take $a_1, \dots, a_n$ to be generators of $H^2 (A)$, and define $M_2 = \Lambda (a_1, \dots, a_n)$, setting $\mid a_i \mid = 2$ and $d a_i = 0$. The map from $M_2$ to $A$ is given by

$$f: M_2 \rightarrow A$$ $$a_i \mapsto a_i.$$

As of now, $f$ induces an isomorphism on cohomology (there are no elements in $H^1 (A)$ by hypothesis, and the generators of $H^2 (A)$ are all represented). Moreover, $f$ is an injection in degree 3, as there are no elements in $M_2$ of degree 3. This is going to be the base for an inductive construction of the minimal model. That is, we’ll show that for any $n$ there is a minimal free algebra $M_n$ and a chain map $f : M_n \rightarrow A$ so that

1. the algebra $M_n$ has no elements in degree 1 and no generators in any degree greater than $n$
2. the map $f$ induces an isomorphism in cohomology in dimensions less than $n + 1$ and an injection in dimension $n +1$.

Suppose that this is true for $n = q-1$. By assumption, we’ve got exact sequences

$$0 \rightarrow H^q (M_{q-1}) \rightarrow H^q (A) \rightarrow \mathrm{coker} \, H^q (f) \rightarrow 0$$

and

$$0 \rightarrow \mathrm{ker} \, H^{q+1} (f) \rightarrow H^{q+1} (M_{q+1}) \rightarrow H^{q+1} (A).$$

The first comes from the hypothesis that $f$ is an injection in degree $q$, and the second is simply a rephrasing of what it means to be in $\mathrm{ker} \, H^{q+1} (f).$ The idea is to introduce elements into $M_{q-1}$ in order to kill $\mathrm{coker} \, H^q (f)$ and $\mathrm{ker} \, H^{q+1} (f)$. If we do this, the above short exact sequences give us an isomorphism and injection, respectively (which is exactly what we want for the induction).

Let $\{ [b_i] \}$ be a basis of $\mathrm{coker} \, H^q (f)$, and define

$$M_q = M_{q-1} \otimes \Lambda (b_i, \xi_j), \; \; \dim b_i = \dim \xi_j = q.$$

$M_q$ is a free minimal algebra, if we define the differential

$$d(m \otimes 1 ) = ( dm ) \otimes 1,$$ $$d(1 \otimes b_i) = 0,$$ $$d(1 \otimes \xi_j) = x_j \otimes 1.$$

We extend $f$ to $M_q$ by

$$f(m \otimes 1 ) = f(m)$$ $$f(1 \otimes b_i) = b_i$$ $$f(1 \otimes xi_j) = \alpha_j$$

where $\alpha_j$ is an element of $A$ such that $f(x_j) = d\alpha_j$ (which exists, by $x_j$ being a basis of $\mathrm{ker} H^{q+1}(f)$.

Checking that this new $f$ satisfies the properties required in the induction step is a simple exercise, and is pretty believable after a moment’s thought (we’ve killed the appropriate elements in the short exact sequences written earlier). Because the cohomology of $A$ is assumed to be finite dimensional, the $f$’s constructed become isomorphisms after the stage corresponding to the maximal dimension in $H^{*} (A)$ is reached, giving us a minimal model for $A$.

So, we’ve got ourselves some minimal models. What do we do with them? It seems like we need to know the cohomology of $X$ in order to build the minimal model, so one might wonder if we actually get any information out of it. As promised earlier, though, we actually get some homotopy-theoretic information out of a minimal model, as shown by the following theorem of Sullivan:

Theorem. Let $X$ be a simply connected manifnew and $M$ its minimal model. Then the dimension of the vector space $\pi_q (M) \otimes \mathbb{Q}$ is the number of generators of the minimal model $M$ in dimension $q$.

That’s great (even though it’s actually a bit difficult to prove - see Sullivan’s paper Infinitesimal Computations in Topology for a proof and a huge amount of other stuff)! It turns out that the minimal model lets us detect the non-torsion parts of the homotopy groups, and in a totally computable fashion. Let’s do an example to illustrate the usefulness of this setup.

Proposition. The homotopy groups of an odd sphere $S^{2n+1}$ are torsion except in dimension $2n+1$, where it is infinite cyclic; for an even sphere $S^{2n}$, the exceptional dimensions are $2n$ and $4n -1$.

We’ll show this by simply exhibiting minimal models for spheres. Let’s start with an odd-dimensional sphere. We have that the De Rham cohomology of $S^{2n+1}$ is the exterior algebra on one generator of degree $2n+1$ (corresponding to the volume form of $S^{2n+1}$. So, we have that the cohomology is its own minimal model. By the above theorem, the only homotopy groups of $S^{2n+1}$ with free part are $\pi_{2n+1}$.

For an even-dimensional sphere the situation is slightly more complicated - but not much more. The De Rham cohomology of the sphere $S^{2n}$ is not an exterior algebra because we do not have that anti-symmetry kills the element corresponding to the volume form. That is, if $\omega$ is the volume form, we need to specify that $\omega \wedge \omega = 0$ - this is not forced by the exterior algebra structure. We start off with $\Lambda(x)$, the exterior algebra on one generator $x$ with the degree of $x$ being $2n$ (corresponding to the volume form of the sphere) and set $dx = 0$. We need that $x^2$ (corresponding to $\omega \wedge \omega$) be 0 in the cohomology of the minimal model, so we stick in another generator $y$ (in degree $4n -1$, as $x^2$ has degree $4n$ and $d$ moves us up in degree by 1) and define the differential so that

$$dy = x^2$$

This kills $x^2$ in cohomology, and because the degree of $y$ is odd, $y^2$ is 0 automatically and we do not need to add in any more generators. This gives us a minimal model for the cohomology of the even sphere, and the map from this model to the De Rham complex of the sphere is given by

$$x \mapsto \text{volume form of the sphere}$$ $$y \mapsto 0$$

Again, by the above theorem, we have that the only homotopy groups with positive rank are $\pi_{2n}$ and $\pi_{4n-1}$. Fancy. A good exercise is to do the same computation for the complex projective spaces $\mathbb{CP}^n$ - it’s basically the same deal as with the sphere.

We can get even crazier though, even though I’m not going to do it here. The wedge of the spheres $S^n$ and $S^m$ is the union of $S^n$ and $S^m$ with one point in common, written $S^n \vee S^m$. Using the above theorem, it’s not too tough (with a bit of knowledge of differential topology) to compute some of the rational homotopy groups of, say, $S^2 \vee S^2$ – something that would be difficult to approach otherwise.

Computing rational homotopy groups is not all that you can do with this setup, however! It turns out (by another theorem of Sullivan) that if we have a suitably nice manifnew $X$, any homomorphism from $\pi_n(X) \rightarrow \mathbb{R}$ is realized by integrating the pullbacks of some appropriately chosen differential forms and their primitives. We’ll discuss this more next time!

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