So next, I would like to describe the action of the Steenrod squares in a more general context (which is pure homological algebra). This more general approach to Steenrod operations seems to be due to Peter May, but I learned it from these lecture notes by Jacob Lurie.

As I said earlier, the Steenrod squares will act on the cohomology of any ${E_\infty}$-algebra. What is an ${E_\infty}$-algebra? It’s supposed to be a dga with a multiplication law that is coherent and associative up to coherent homotopy. Or, equivalently, an algebra over an ${E_\infty}$-operad in chain complexes. Unfortunately, I don’t know too many simple ones. For instance, the standard ${E_\infty}$-operad in spaces, the infinite little cubes operad, is huge: the individual terms have to be contractible spaces large enough to admit a free action of the symmetric group.

But we’ll need less structure than an ${E_\infty}$-algebra structure. We’ll need a multiplication law on a chain complex ${X}$ which is commutative up to coherent homotopy, but without quite all the coherence that one would need for an ${E_\infty}$-algebra.

What should this be? Well, a multiplication law would be a map ${X \otimes X \rightarrow X}$, and to say that it is commutative would be to say that it is ${\Sigma_2}$-equivariant where ${\Sigma_2}$ acts by permutation on ${X \otimes X}$ and trivially on ${X}$. In other words, it would be a map ${(X \otimes X)_{\Sigma_2} \rightarrow X}$ where ${(X \otimes X)_{\Sigma_2}}$ denotes the coinvariants. This is generally going to be too strong: most of the homotopy commutative multiplication laws we deal with will not be rectifiable to strictly commutative ones.

Instead, we’ll use maps not out of the coinvariants of ${(X \otimes X)_{\Sigma_2}}$, but out of the homotopy coinvariants. To get this, one first tensors ${X \otimes X}$ with a complex of free ${\Sigma_2}$-modules which is contractible as a complex of vector spaces, and then takes the coinvariants.

More generally:

Definition 1 If ${Y}$ is a ${\Sigma_2}$-equivariant complex of ${\mathbb{F}_2}$-vector spaces, then we define the homotopy coinvariants ${(h Y)_{\Sigma_2}}$ as the coinvariants of ${(Y \otimes E \Sigma_2)_{\Sigma_2}}$. Here ${E \Sigma_2}$ is any acyclic resolution of the complex ${\mathbb{F}_2[0]}$ (just ${\mathbb{F}_2}$ in degree zero, nothing elsewhere) consisting of free ${\mathbb{F}_2[\Sigma_2]}$-modules.

For instance, we could take ${E \Sigma_2}$ as the complex

$\displaystyle \dots \rightarrow \mathbb{F}_2[\mathbb{Z}_2] \rightarrow \mathbb{F}_2[\mathbb{Z}_2] \rightarrow \dots \rightarrow \mathbb{F}_2[\mathbb{Z}_2] \rightarrow 0,$

where each map is multiplication by ${1 + \sigma }$ (for ${\sigma \in \mathbb{Z}_2}$ the non-unital element). This corresponds to projective space as a ${K(\mathbb{Z}_2, 1)}$.

This is analogous to the situation in algebraic topology. Given a space with a ${\Sigma_2}$-action, call it ${X}$, we can define the homotopy coinvariants of ${X}$ as ${(X \times E \Sigma_2)/\Sigma_2}$. Once again, the point is to take the product with a freely acted upon but contractible ${\Sigma_2}$-space.

The benefit of taking homotopy coinvariants rather than plain coinvariants is that it is homotopy invariant, for one thing (though that’s not the most relevant to us). That is, if ${X \rightarrow Y}$ is a weak equivalence of ${\Sigma_2}$-equivariant complexes, then ${(h X)_{\Sigma_2} \rightarrow (h Y)_{\Sigma_2}}$ is a weak equivalence (i.e. quasi-isomorphism) as well. To see this, we need only note that ${X \otimes E\Sigma_2 \rightarrow Y \otimes E \Sigma_2}$ is a weak equivalence of freely acted upon ${\Sigma_2}$-equivariant complexes. Taking coinvariants preserves weak equivalences for freely acted upon ${\Sigma_2}$-complexes. Essentially, this is because taking coinvariants is a left Quillen functor, and should preserves weak equivalences (quasi-isomorphisms) between cofibrant (free) ${\Sigma_2}$-complexes.

Definition 2 coherently commutative multiplication law on a complex ${X}$ will be a map ${\mathrm{h}\mathrm{Sym}^2 X \rightarrow X}$.

Consider the model for ${\mathrm{h}\mathrm{Sym}^2}$ given by the explicit example of ${E \Sigma_2}$ above, i.e. the one resembling real projective space. This means that there is a multiplication map ${m: X \otimes X \rightarrow X}$ (coming from the identity element in ${(E \Sigma_2)_0}$) which is not required to be commutative. However, ${m}$ and the twist ${m \circ T: X \otimes X \rightarrow X}$ are required to be homotopic to one another; that’s because the identity plus the non-identity in ${(E \Sigma_2)_0}$ are cohomologous.

Example 1 The claim is that there is a natural coherently commutative multiplication on the cochain algebra ${C^*(X, \mathbb{F}_2)}$ of ${\mathbb{F}_2}$-cochains on a topological space ${X}$. One may see this as follows. First, one shows that there is a coherently cocommutative comultiplication on ${C_*(X, \mathbb{F}_2)}$. To do this, we need to get a ${\Sigma_2}$-equivariant map

$\displaystyle C_*(X, \mathbb{F}_2) \otimes E \Sigma_2 \rightarrow C_*(X, \mathbb{F}_2) \otimes C_*(X, \mathbb{F}_2).$

Both are augmented functors from spaces to cochains. One may apply here the method of “acyclic models” (with the ring ${\mathbb{F}_2[\mathbb{Z}_2]}$) to deduce that, since the left side is free and the right side is acyclic on the models given by the simplices, there is a natural equivariant map. Moreover, this map is unique up to natural equivariant chain homotopy.Dualizing now gives the required homotopy coherently commutative multiplication on ${C^*(X, \mathbb{F}_2)}$.

We are going to make the Steenrod operations act on any complex with a coherently commutative multiplication. More generally, we will define the Steenrod operations

$\displaystyle H^*(C) \rightarrow H^{* + i}( \mathrm{h}\mathrm{Sym}^2 C)$

for any complex ${C}$. For a complex ${C}$ equipped with a multiplication ${\mathrm{h}\mathrm{Sym}^2 C \rightarrow C}$, one gets the Steenrod operations to act on ${H^*(C)}$.

To get these, note that giving an element ${c}$ of ${H^n(C)}$ is the same as giving a homotopy class of maps ${\mathbb{F}_2[-n] \rightarrow C}$. Here ${\mathbb{F}_2[-n]}$ is the complex with ${\mathbb{F}_2}$ in degree ${n}$ and zero elsewhere (and zero differentials, obviously). Given such a map, we get a homotopy class of maps ${\mathrm{h}\mathrm{Sym}^2 \mathbb{F}_2[-n] \rightarrow \mathrm{h}\mathrm{Sym}^2 C}$.

If we used ordinary coinvariants instead of homotopy coinvariants, we would be getting the map ${\mathrm{Sym}^2 \mathbb{F}_2[-n] \rightarrow \mathrm{Sym}^2 C}$ or ${\mathbb{F}_2[-2n] \rightarrow \mathrm{Sym}^2 C}$. In other words, we would be sending the element ${c}$ to ${c \otimes c}$; this is essentially “squaring.”

But when we take the homotopy symmetric square, there are additional terms. Namely, ${\mathrm{h}\mathrm{Sym}^2 \mathbb{F}_2[-n]}$ will contain many additional terms. To see this, we note that ${\mathrm{h}\mathrm{Sym}^2 \mathbb{F}_2[-n]}$ can be taken as ${\mathbb{F}_[-2n] \otimes (E\mathbb{Z}_2)_{\Sigma_2}}$. Taking the coinvariants of ${E\mathbb{Z}_2}$ produces the complex which is ${\mathbb{Z}/2}$ in all nonpositive dimensions and with zero differentials. So ${\mathrm{h}\mathrm{Sym}^2 \mathbb{F}_2[-n]}$ is a complex which has a unique nonzero element ${e_k}$ in all dimensions ${k \leq 2n}$, and what we get is a map

$\displaystyle \mathrm{h}\mathrm{Sym}^2 \mathbb{F}_2[-n] \rightarrow \mathrm{h}\mathrm{Sym}^2 C.$

Each ${e_k, k \leq 2n}$, maps to a well-defined element in ${H^k(\mathrm{h}\mathrm{Sym}^2 C)}$. If ${k = 2n - i}$, then we have maps

$\displaystyle \mathrm{Sq}^{n-i} : H^n(C) \rightarrow H^{2n-i}(\mathrm{h}\mathrm{Sym}^2 C), \quad i \geq 0$

which are well-defined, and which are called the Steenrod operations. Thus, on ${H^n(C)}$, we have ${\mathrm{Sq}^n, \mathrm{Sq}^{n-1}, \dots}$ (possibly in infinitely many negative degrees). Consequently, if we have a coherently commutative multiplication ${\mathrm{h}\mathrm{Sym}^2 C \rightarrow C}$, then we get maps

$\displaystyle \mathrm{Sq}^{n-i}: H^n(C) \rightarrow H^{2n-i}(C).$

The point now is to prove things about the Steenrod operations using this general theory.

We start by noting that the operation ${\mathrm{Sq}^n}$ is just the cup square. Namely, if ${c \in H^n(C)}$, then we can consider ${c \otimes c \in H^{2n}(\mathrm{h}\mathrm{Sym}^2 C)}$ by representing ${c}$ by a cycle ${x \in C_n}$ and considering ${x \otimes x \otimes \text{something}}$ in ${C \otimes C \otimes E \mathbb{Z}_2}$, and then taking the image of that. The choices one makes are not relevant in cohomology.

One should finally note that these operations are zero (well, except for the one which is squaring) for a strictly commutative dga. That’s because the map from homotopy coinvariants factors through the ordinary symmetric square. Consequently the map $\mathrm{hSym}^2 \mathbb{F}_2[-2n] \to C$ factors through the ordinary symmetric square $\mathrm{hSym}^2 \mathbb{F}_2[-2n]$ and that means there’s nothing interesting. Since the Steenrod operations are natural over complexes with a coherently commutative multiplication law , this observation shows that not every coherently commutative multiplication law can be rectified to a strictly commutative one: the Steenrod operations provide obstructions. (Actually, I haven’t given an example where the Steenrod operations are trivial, but this is not too hard, and there are plenty of interesting examples in topology.)

Anyway, so Lurie’s notes cover all this, and he shows how one can derive the properties of the Steenrod operations one usually proves in the topological case purely from homological algebra. Not everything works the same (e.g. ${\mathrm{Sq}^0}$ is not necessarily the identity), but the Adem relations still hold. I’m going to go back to the structure of the Steenrod algebra in the case of topological spaces, though, because I have other applications in mind that I would like to blog about in the future: namely, those related to stable homotopy theory.