The Steenrod algebra and its dual

I’ve been reading Milnor’s paper “The Steenrod algebra and its dual,” and want to talk a little about it today. The starting point of this story is the theory of cohomology operations. Given a cohomology theory ${h^*}$ on spaces (or just CW complexes; one can always Kan extend to all spaces), one can consider cohomology operations on ${h^*}$ . Most interesting for our purposes are the stable cohomology operations.

A stable cohomology operation of degree ${k}$ will be a collection of homomorphisms ${h^m(X) \rightarrow h^{m+k}(X)}$ for each ${m}$ , which are natural in the space ${X}$ , and which commute with the suspension isomorphisms. If we think of ${h^*}$ as represented by a spectrum ${E}$ , so that ${h^*(X) = [X, E]}$ is a representable functor (in the stable homotopy category), then a stable cohomology operation comes from a homotopy class of maps ${E \rightarrow E}$ of degree ${k}$ .

A stable cohomology operation is additive, because it comes from a spectrum map, and the stable homotopy category is additive. Moreover, the set of all stable cohomology operations becomes a graded ring under composition. It is equivalently the graded ring ${[E, E]}$ .

The case where ${E}$ is an Eilenberg-MacLane spectrum, and ${h^*}$ ordinary cohomology, is itself pretty interesting. First off, one has to work in finite characteristic—in characteristic zero, there are no nontrivial stable cohomology operations. In fact, the only (possibly unstable) natural transformations ${H^*(\cdot, \mathbb{Q}) \rightarrow H^*(\cdot, \mathbb{Q})}$ come from taking iterated cup products because ${H^*(K(\mathbb{Q}, n))}$ can be computed, via the spectral sequence, to be a free graded-commutative algebra over ${\mathbb{Q}}$ generated by the universal element. These aren’t stable, so the only stable one has to be zero. So we will work with coefficients ${\mathbb{Z}/p}$ for ${p}$ a prime.

Here the algebra of stable cohomology operations is known and has been known since the 1950’s; it’s called the Steenrod algebra ${\mathcal{A}^*}$ . In fact, all unstablecohomology operations are themselves known. Let me state the result for ${p=2}$ .

Steenrod had constructed squaring operations

$\displaystyle \mathrm{Sq}^i: H^*(\cdot, \mathbb{Z}/2) \rightarrow H^{* +i}(\mathbb{Z}/2) .$

These are natural transformations, which have the following properties:

${\mathrm{Sq}^0}$ is the identity operation.
${\mathrm{Sq}^i}$ on a cohomology class ${x}$ of dimension ${n}$ vanishes for ${i > n}$ . For ${i = n}$ , ${\mathrm{Sq}^i}$ acts by the cup square on ${x}$ .
The Steenrod squares behave well with respect to the cohomology cross (and thus cup) product: ${\mathrm{Sq}^i(a \times b) = \sum_{j + k = i} \mathrm{Sq}^j a \times \mathrm{Sq}^k b}$ .
${\mathrm{Sq}^1}$ is the Bockstein connecting homomorphism associated to the short exact sequence ${0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0}$ .
${\mathrm{Sq}^i}$ commutes with suspension (and thus is a homomorphism).

By the last property, the Steenrod squares are stable cohomology operations. One can explicitly construct these using an acyclic models argument. The intuition is that the Steenrod squares measure the deviation from which the cochain algebra ${C^*(X; \mathbb{Z}/2)}$ , which has an associative and homotopy commutative product given by the Alexander-Whitney formula, cannot be “strictified” to a commutative algebra. In fact, one should think of ${C^*(X; \mathbb{Z}/2)}$ as an example of an ${E_\infty}$ -algebra: these are those dgas whose product is commutative up to coherent homotopy. In general, one cannot replace an ${E_\infty}$ -algebra by a cdga, and the Steenrod squares measure this failure. However, the Steenrod squares do not behave the same way on general ${E_\infty}$ algebras as they do on spaces. I’ll come back to this.

To determine unstable cohomology operations, one reduces by Yoneda’s lemma to determining the mod ${2}$ cohomology of the spaces ${K(\mathbb{Z}/2, n)}$ . This was done by Serre by use of his spectral sequence. In fact, Serre showed that the cohomology of ${K(\mathbb{Z}/2, n)}$ is a polynomial ring on the iterated Steenrod squares of the universal class ${\iota_n}$ (except those which have to vanish for dimensional reasons). The corollary of this result is:

Theorem 1 The algebra of stable cohomology operations in ${\mathbb{Z}/2}$ -cohomology is generated by the ${\mathrm{Sq}^i}$ .

One might ask what the relations are. These are the Adem relations, for which I’ll refer you to Wikipedia. The point of the Adem relations is that, whenever you have an expression ${\mathrm{Sq}^i \mathrm{Sq}^j}$ where ${i \leq 2j}$ , then you can write as a sum of smaller terms. So the basis for the Steenrod algebra is given by sequences ${\mathrm{Sq}^{i_1}\mathrm{Sq}^{i_2}\dots \mathrm{Sq}^{i_k}}$ where ${i_j \geq 2 i_{j-1}}$ . It turns out that the Adem relations are exactly the relations one has to impose. That is, the Steenrod algebra ${\mathcal{A}^*}$ is the quotient of the free associative algebra on the ${\mathrm{Sq}^i}$ modulo the Adem relations (and ${\mathrm{Sq}^0 =1}$ ).

It’s somewhat interesting that establishing ${\mathrm{Sq}^0 = 1}$ is actually nontrivial. In fact, this is false for the action of the Steenrod algebra on general ${E_\infty}$ -algebras (and one also gets Steenrod squares in negative degrees). So ${\mathrm{Sq}^0 = 1}$ is actually special to topological spaces! One way to see this is to note that on ${H^0}$ , ${\mathrm{Sq}^0}$ should act by the cup square. For a cohomology algebra mod two, any element in degree zero squares to itself, but that is not true for the cohomology of a general ${E_\infty}$ -algebra.

OK, so we have ${\mathcal{A}^*}$ . We know, moreover, that there is a natural action

$\displaystyle \mathcal{A}^* \otimes H^*(X) \rightarrow H^*(X)$

for any space ${X}$ (where cohomology is all mod ${2}$ ). Now one might ask how a general Steenrod operation acts with respect to the ring structure of ${H^*(X)}$ . This is the source of the additional structure of ${\mathcal{A}^*}$ .

Milnor showed that ${\mathcal{A}^*}$ is a connected Hopf algebra (a graded one). In other words, there is a comultiplication map

$\displaystyle \Delta: \mathcal{A}^* \rightarrow \mathcal{A}^* \otimes \mathcal{A}^*$

This is required to satisfy the condition that ${\Delta}$ be a homomorphism of graded algebras, and also ${\Delta(a) = a \otimes 1 + 1 \otimes a + \text{smaller terms}}$ for any homogeneous ${a}$ . Finally, the requirement is that in degree zero, ${\mathcal{A}^*}$ is generated by the unit element. This ensures that the map given by projecting to the zeroth coordinate is a counit, so we do have a Hopf algebra in an ordinary sense.

How do we define ${\Delta}$ ? Milnor does it to make the following diagram commute for any spaces ${X, Y}$ :

Here the left vertical map is the composite of the comultiplication and the braiding. The right vertical map is the action of the Steenrod algebra on ${H^*(X \times Y)}$ . The bottom horizontal arrow is the composite of the actions of the Steenrod algebras on both cohomology rings. The top horizontal arrow is the cohomology cross product.

In particular, this means that if ${a \in \mathcal{A}^*}$ , then ${a}$ acts on the cup product ${x \cup y}$ of two cohomology classes by applying ${\Delta a}$ to ${x \otimes y}$ and taking the cup product of them. That is, if ${\Delta a = \sum a_i \otimes b_i}$ , then ${a( x \cup y) = \sum a_i(x) \cup b_i (y)}$ .

Theorem 2 (Milnor) There is a unique homomorphism ${\Delta}$ which satisfies the above conditions, and makes ${\mathcal{A}^*}$ into a connected Hopf algebra.

Milnor shows this result as follows. He considers the set of all ${a}$ for which there exists some ${\sum a_i \otimes b_i}$ in ${\mathcal{A}^* \otimes \mathcal{A}^*}$ which do satisfy the above conditions. This set includes the operations ${\mathrm{Sq}^i}$ , by the Cartan product formula. It is also easily checked to be an algebra, and that the choice of ${\sum a_i \otimes b_i}$ is unique (by working over an Eilenberg-MacLane space). This pretty much implies the theorem.

Note that ${\mathcal{A}^*}$ is cocommutative and coassociative (while it’s associative, but not commutative). This follows by chasing through the diagrams because the cup product is commutative and associative. Since we’re working mod ${2}$ , we don’t even have to worry about signs.

So, now that ${\mathcal{A}^*}$ is a connected Hopf algebra, we can take its dual and get a new connected Hopf algebra.

Definition 3 ${\mathcal{A}_*}$ is the dual Hopf algebra of ${\mathcal{A}^*}$ and is called the dual Steenrod algebra.

The nice thing about ${\mathcal{A})_*}$ is that, since the Steenrod algebra was coassociative and cocommutative, its dual ${\mathcal{A}_*}$ is actually commutative and associative. One of Milnor’s results in this paper is that ${\mathcal{A}_*}$ is actually a polynomial ring.

Climbing Mount Bourbaki