This is the third in a series of posts on the Steenrod algebra.

We’ll need to determine the structure of the dual Steenrod algebra ${\mathcal{A}_*}$ described earlier. To do so, let’s start by defining certain generators ${\xi_i}$ in it. For ${i > 0}$, we let ${\xi_i}$ be the dual element to ${\mathrm{Sq}^{2^{i-1}} \mathrm{Sq}^{2^{i-2}} \dots \mathrm{Sq}^{1} \in \mathcal{A}}$, with respect to the Serre-Cartan basis. In other words, ${\xi_i}$ evaluates to ${1}$ on the element ${\mathrm{Sq}^{2^{i-1}} \dots \mathrm{Sq}^1}$, but evaluates to zero on ${\mathrm{Sq}^I}$ when ${I}$ is any admissible sequence.

We want to prove:

Theorem 4 (Milnor) ${\mathcal{A}_*}$ is the polynomial ring ${\mathbb{Z}/2[\xi_1, \xi_2, \dots, ]}$.

In this post, I’ll describe the proof of Milnor’s theorem, and then give the computation of the comultiplication map on $\mathcal{A}_*$ (which one should expect to be nontrivial, given that the dual product map on the Steenrod algebra is complicated). As a result, this post will be somewhat lightly computational (certainly more so than usual for this blog). After this, I’m hoping to get to applications of these computations to more theoretical material.

Proof of Milnor’s theorem

For a sequence ${J = (j_1, \dots, j_k)}$, we let ${\xi^J = \xi_1^{j_1} \dots \xi^{j_k}_k \in \mathcal{A}_*}$. To prove the theorem, we will consider how the products ${\xi^J}$ act on the Cartan-Serre basis elements ${\mathrm{Sq}^I \in \mathcal{A}}$ (for $I$ admissible, it is a theorem of Serre that these form a basis for the Steenrod algebra). Although the ${\xi_j}$ were dual elements to certain ${\mathrm{Sq}^I}$, the products ${\xi^J}$ are not dual to simple elements in the same way. Nonetheless, we will see that one gets an upper-triangular matrix with ones on the diagonal, from which the result will follow.

Let’s start by determining how exactly an element ${\xi_n}$ acts on an element of the Steenrod algebra. We know how it acts on ${\mathrm{Sq}^I}$ for ${I}$ an admissible sequence, but we’ll also need to know how it acts without the admissibility hypothesis. We will need the following alternative description of the functional ${\xi_n}$:

Lemma 5 Consider ${\mathbb{RP}^\infty}$ and identify all its higher cohomologies with ${\mathbb{Z}/2}$ in the unique way. Let ${x \in H^1(\mathbb{RP}^\infty, \mathbb{Z}/2)}$ be a generator; then ${\xi_n}$ can be described by

$\displaystyle \xi_n(a) = ax , \quad a \in \mathcal{A}^{2^{n}-1}.$

Here ${ax}$ lives in some cohomology of ${\mathbb{RP}^\infty}$, which we identify with ${\mathbb{Z}/2}$. So, to evaluate ${\xi_n}$ on some element ${a}$ of the Steenrod algebra (of the appropriate degree), we make ${a}$ act on the nonzero element of ${H^1(\mathbb{RP}^\infty, \mathbb{Z}/2)}$, and determine whether we get zero or not.

Proof: In fact, the claim is the following. If ${I}$ is an admissible sequence, then ${\mathrm{Sq}^I x = 0}$ unless ${I = (2^n, 2^{n-1}, \dots, 1)}$ for some ${n}$, in which case we just get ${x^{2^n}}$. This is pretty easy to see by induction. Namely, we have to use the fact that the total squaring operation ${\mathrm{Sq} = \sum \mathrm{Sq}^i}$ is a homomorphism, and it sends ${x \mapsto \mathrm{Sq}^0 x + \mathrm{Sq}^1 x = x + x^2}$. More generally, ${\mathrm{Sq} x^{2^k} = x^{2^k} + x^{2^{k+1}}}$ because ${\mathrm{Sq}}$ is a homomorphism and commutes with squaring. So, if we’re starting with ${x}$, and want to apply a product of Steenrod operations and get something nonzero at the end, the only way we can do that is if the sequence of Steenrod operations looks like ${\mathrm{Sq}^{2^n}, \mathrm{Sq}^{2^{n-1}}, \dots, \mathrm{Sq}^1}$ (possibly with zeros interpersed). But for an admissible sequence, the only possibility is the one listed. $\Box$

We’ve in fact shown something a bit stronger. Namely, we have seen:

Corollary 6 If ${I}$ is any sequence (not necessarily admissible), then ${\xi_n( \mathrm{Sq}^I) = 0}$ unless ${I}$ is the sequence ${(2^{n-1}, 2^{n-2}, \dots, 1)}$ with zeros interspersed.

This is now clear, because we know that ${\xi_n}$ acts by evaluating the Steenrod operation on ${x}$, and because we have seen how things of the form ${\mathrm{Sq}^I}$ act on ${x}$ in the preceding proof.

With this in mind, we can prove Milnor’s theorem. Namely, we’re going to need to show that as ${J}$ ranges over all sequences of nonnegative integers (well, finite sequences), the terms ${\xi^J \in \mathcal{A}_*}$ form a basis for the dual Steenrod algebra. To do this, we’ll need to know how the elements ${\xi^J}$ act on the admissible elements ${\mathrm{Sq}^I}$ in the Steenrod algebra. We want to show that the ${\xi^J}$ at least approximate a dual basis.

To do this, we’ll need a pairing between the ${I}$‘s and ${J}$‘s; here recall that ${I}$ ranges over admissible sequences, while ${J}$ ranges over sequences of nonnegative integers. Thus, we define the map ${\gamma}$ which sends an admissible sequence ${(a_1, \dots, )}$ to ${(a_1 - 2a_2, a_2 - 2a_3, \dots, )}$. One easily sees that ${\gamma}$ establishes a bijection between sequences of nonnegative integers, all but finitely whose elements are zero, and admissible sequences of nonnegative integers.

With this notation, we have:

Lemma 7 (The basic computation) For any sequences ${I, J}$, if ${I = \gamma^{-1}(J)}$, then ${\xi^J ( \mathrm{Sq}^I) = 1}$. If ${I < \gamma^{-1}(J)}$, then ${\xi^J (\mathrm{Sq}^I ) = 0}$.

I haven’t yet told you what the ordering means. The ordering is a lexicographic order from the right. We can do this, since the sequences terminate after finitely many terms. But it is a little funny that ${(2, 3) < (235, 4)}$, for instance. However, it is a perfectly good order, which is preserved under addition, for instance. Note also that in this lemma, we haven’t assumed ${I}$ admissible. In fact, for the proof of Milnor’s theorem, we will only need need the case ${I}$ admissible, but for the lemma itself, this result will be useful.

So let’s prove the lemma. When ${J}$ is the zero sequence, there is nothing to prove. Let’s verify what happens when ${J}$ is the sequence ${(0, 0, \dots, 0, 1, 0, \dots)}$ where the unique ${1}$ is in the ${n}$th place. In this case, the corresponding ${\gamma^{-1}(J)}$ is the sequence ${(2^{n-1}, 2^{n-2}, \dots, 1, 0, \dots)}$.

The first statement, that if ${I = \gamma^{-1}(J)}$ then ${\xi^J (\mathrm{Sq}^I) = 1}$, is part of the definition. The second statement requires a tiny bit more work. Namely, if ${I < \gamma^{-1}(J)}$, then there must be less than ${n}$ terms, as the ${n}$th term must be zero. But we know that ${\zeta_n}$ on any sequence ${I}$ with fewer than ${n}$ terms is zero, by the previous corollary. So we’re done in this case.

Now we need to do this in general. Here we can use induction. Namely, let’s consider the sequence ${J}$, and pick the last nonzero term. That is, we write ${J = J_1 + J_2}$ where ${J_2}$ is the sequence ${(0, 0, \dots, 1)}$ where the ${1}$ is in, say, the ${n}$th place. We will assume that the result is true for ${J_1}$ and ${J_2}$ (and for anything replacing ${I}$ that is appropriate).

Then ${\xi^J = \xi^{J_1} \xi^{J_2}}$, and if ${m}$ and ${m^*}$ are the multiplication and comultiplications, we have

$\displaystyle ( \xi^{J_1}\xi^{J_2}) (\mathrm{Sq}^I) = m( \xi^{J_1} \otimes \xi^{J_2}) ( \mathrm{Sq}^I) = (\xi^{J_1}\otimes \xi^{J_2}) ( m^* \mathrm{Sq}^I).$

In other words, we have to write out ${m^* \mathrm{Sq}^I}$. But this is easy; it’s ${\sum_{I_1 + I_2 = I } \mathrm{Sq}^{I_1} \otimes \mathrm{Sq}^{I_2}}$ by the explicit description of the comultiplication on the Steenrod algebra ${\mathcal{A}}$.

So we need to evaluate the pairing

$\displaystyle \sum_{I_1 + I_2 = I} ( \xi^{J_1} \otimes \xi^{J_2})(\mathrm{Sq}^{I_1} \otimes \mathrm{Sq}^{I_2}).$

Now, we’re assuming ${I_1 + I_2 \leq J_1 + J_2}$. Thus either ${I_1 \leq I_2}$ or ${J_1 \leq J_2}$. If we have strict inequality in either case, then we get zero by the inductive hypothesis. If we have equality in both cases, then we get ${1}$ by the inductive hypothesis again. This proves the lemma.

With this lemma, we want to prove Milnor’s theorem. This is now linear algebra, since the lexicographic order is a total ordering. We have seen that the ${\xi^J}$, expressed in terms of the dual basis of the ${\mathrm{Sq}^I}$, are given by a triangular matrix with ones on the diagonal. Thus the ${\xi^J}$ have to form a basis of ${\mathcal{A}_*}$.

The comultiplication on $\mathcal{A}_*$

Finally, we want to work out the comultiplication on the dual Steenrod algebra ${\mathcal{A}_*}$. Here, we will again use the real projective space ${\mathbb{RP}^\infty}$ to help in the calculations. Namely, let’s recall the following. If ${X}$ is any space, then there is an action map

$\displaystyle \mathcal{A} \otimes H^*(X; \mathbb{F}_2) \rightarrow H^*(X; \mathbb{F}_2) .$

Each element ${a \in \mathcal{A}}$ induces an endomorphism of ${H^*(X; \mathbb{F}_2)}$. If all the homology groups of ${X}$ are finite-dimensional, then we can dualize and obtain maps ${H_*(X; \mathbb{F}_2) \rightarrow H_*(X; \mathbb{F}_2)}$ for each ${a \in \mathcal{A}}$. This gives a map

$\displaystyle \mathcal{A} \otimes H_*(X; \mathbb{F}_2) \rightarrow H_*(X; \mathbb{F}_2).$

One can check that this is a coalgebra homomorphism. That is, since the homology groups form a coalgebra, and ${\mathcal{A}}$ is (as a Hopf algebra), the previous assertion makes sense. This is mostly an exercise in chasing some diagrams and keeping straight exactly what the distinguishing property the codiagonal map of ${\mathcal{A}}$ had (and the fact that the coalgebra structure in homology is dual to the algebra structure in homology); I’ll refer the reader to Milnor’s paper or Mosher-Tangora’s book on cohomology operations for details.

Anyway, dualizing again, we get a map

$\displaystyle \psi: H^*(X; \mathbb{F}_2) \rightarrow H^*(X; \mathbb{F}_2) \hat{\otimes} \mathcal{A}_*$

where the tensor product is completed because infinite sums may occur. As the dual to a coalgebra homomorphism, this is an algebra homomorphism. If one unwinds the definitions, this is what the map ${\psi}$ does. The map ${\psi}$ sends a cohomology class ${x}$ to an element ${\sum y \otimes \zeta \in H^*(X; \mathbb{F}_2) \hat{\otimes} \mathcal{A}_*}$. This element has the following property. If you want to evaluate a Steenrod operation on ${x}$, you can just plug it into this element, and pair it with the relevant elements in ${\mathcal{A}_*}$.

In particular, we find the following.

Proposition 8 If ${x}$ is the generator of ${H^1(\mathbb{RP}^\infty; \mathbb{F}_2)}$, then ${\psi(x) = \sum_n x^{2^n} \otimes \xi_n}$.

This comes from how the Steenrod operations can act on ${x}$, as we saw earlier.

Now we need another property of ${\psi}$. Namely, if we apply ${\psi}$ twice, we get a map ${H^*(X; \mathbb{F}_2) \rightarrow H^*(X; \mathbb{F}_2) \hat{\otimes} \mathcal{A}_* \otimes \mathcal{A}_*}$. But we could also apply ${\psi}$ and then the comultiplication map on ${\mathcal{A}_*}$. That is, we have a diagram

This is commutative. Again, this is another assertion that one has to check using some diagram-chasing, but it is essentially linear algebra. The dual assertion is simply that the action of ${\mathcal{A}^*}$ on the homology is an algebra action.

OK, now the point is that if we keep track of this diagram, and use the map ${\psi}$, we can completely determine the effect of the comultiplication homomorphism on ${\mathcal{A}_*}$ just by using what we know about ${\mathbb{RP}^\infty}$. Indeed, we know that ${\psi(x)}$ (for ${x}$ as before, in ${H^1(\mathbb{RP}^\infty, \mathbb{F}_2)}$) is ${\sum_n x^{2^n} \otimes \xi_n}$. Now we can apply ${1 \otimes \psi}$ to this. Note that we know

$\displaystyle \psi(x^{2^n}) = \psi(x)^{2^n} = \sum_k x^{2^{n+k}} \otimes \xi_k^{2^n} ,$

because ${\psi}$ is a homomorphism. Consequently, we get

$\displaystyle (1 \otimes \psi) \circ \psi(x) = \sum_m \sum_k x^{2^{m+k}} \otimes \xi_k^{2^m} \otimes \xi_m.$

But this is also equal to

$\displaystyle ( \Delta \otimes 1) \circ \psi(x) = \sum_n \Delta(\xi_n) \otimes x^{2^n}.$

Now we just have to compare terms of ${x^{2^n}}$ in the above two terms, and equate them. We find:

Proposition 9 (Coalgebra structure) The coalgebra map on ${\mathcal{A}_*}$ is given by

$\displaystyle \Delta \xi_n = \sum_{k} \xi_k^{2^{n-k}} \otimes \xi_{n-k}.$

This was all a bit computational, but ultimately the Steenrod algebra turns out to be huge. Namely, as I want to explain in a later post, given any appropriate spectrum ${E}$, representing a homology theory ${E_*}$, and given any spectra ${X, Y}$, there is a spectral sequence that gives information about at least the ${E}$-part of homotopy classes of maps ${[X, Y]}$. (${E}$-part is not defined; I mean something like 2-torsion when ${E}$ is an Eilenberg-MacLane spectrum mod 2.) To get this, though, we’ll need the structure of ${E_*(X), E_*(Y)}$ as comodules over ${E_*(E)}$, which is somehow approximately “dual” to the space of cohomology operations. In the important case where ${E}$ is the mod 2 Eilenberg-MacLane spectrum, we’ll get information about, say, the 2-component of stable homotopy groups once we know how the dual Steenrod algebra coacts on homology. And these are precisely the relevant computations, to do that.