Steenrod Operation and Steenrod Algebra

Introduction

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In the previous part, we discussed the construction of Steenrod operations and their interpretation as stable cohomology operations. We concluded by listing several fundamental properties. Let us now summarize them clearly before exploring the structure they generate.

• The stable cohomology operations are denoted $Sq^i$ for $i \geq 0$.
• For each $n \geq 0$,
  $$ Sq^n : H^q(X; \mathbb{Z}_2) \longrightarrow H^{q+n}(X; \mathbb{Z}_2) $$ is a stable cohomology operation.
• $Sq^{0}$ acts as the identity.
• For $x \in H^{p}(X; \mathbb{Z}_2)$ (i.e., $\deg x = p$), we have $Sq^i(x) = 0$ if $i > p$.
• $Sq^n(x) = x^2$ whenever $\deg x = n$.
• Cartan formula:
  $$ Sq^i(xy) = \sum_{j} Sq^j(x) , Sq^{i-j}(y) $$
• Adem relation: For $i < 2j$,
  $$ Sq^i Sq^j = \sum_k \binom{j - k - 1}{i - 2k} , Sq^{i+j-k} Sq^k $$
• $Sq^1$ coincides with the Bockstein homomorphism associated to the short exact sequence
  $$ 0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0. $$

Small Examples

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Since we are working over $\mathbb{Z}_2$, the binomial coefficients are reduced mod 2, and Pascal’s triangle provides intuition for the Adem relations:

• $Sq^1 Sq^1 = 0$.
• $Sq^1 Sq^2 = Sq^3 \neq Sq^2 Sq^1$.
• $Sq^2 Sq^2 = Sq^3 Sq^1 = Sq^1 Sq^2 Sq^1$.

In general, for a tuple $I = (i_1, \dots, i_k)$ we define
$$ Sq^I = Sq^{i_1} \cdots Sq^{i_k}, $$ and say that the degree of $I$ is $\deg I = i_1 + \cdots + i_k$.

• Definition: A sequence $I$ is admissible if $i_j \ge 2 i_{j+1}$ for all $j$.
• The excess of an admissible sequence is $e(I) = 2i_1 - \deg I$.

The Steenrod Algebra

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The collection of all Steenrod squares forms an algebraic structure:

Definition:
The Steenrod algebra $\mathcal{A}$ is the free graded (noncommutative) $\mathbb{Z}_2$-algebra generated by elements $Sq^i$ for $i \ge 0$, with grading $\deg Sq^i = i$, subject to the Adem relations.

Key Structural Theorems

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• Minimal Generators:
  $\mathcal{A}$ is generated by ${ Sq^{2^i} : i \ge 0 }$.

• Basis Theorem:
  The set ${ Sq^I : I \text{ admissible} }$ forms a $\mathbb{Z}_2$-basis of $\mathcal{A}$.

Hopf Algebra Structure

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The Cartan formula implies that the assignment $$ Sq^k \longmapsto \sum_{i+j=k} Sq^i \otimes Sq^j $$ is an algebra homomorphism. Hence, $\mathcal{A}$ acquires a coproduct and becomes a cocommutative Hopf algebra. Its dual, $\mathcal{A}_\ast$, is a commutative algebra.

Milnor’s Theorem:

$$\mathcal{A}_\ast = \mathbb{Z}_2[\zeta_1, \zeta_2, \ldots], \quad \deg \zeta_i = 2^i - 1,$$ with coproduct

$$\Delta(\zeta_k) = \sum_{i=0}^k \zeta_{k-i}^{2^i} \otimes \zeta_i.$$

Here, $\zeta_k$ is dual to $Sq^{I_k}$, where $I_k = (2^{k-1}, \cdots, 1, 0)$.

For concreteness, we often denote by $\mathcal{A}(k)$ the subalgebra generated by ${ Sq^i : i \le 2^k }$.
For instance, the relations in $\mathcal{A}(1)$ can be visualized in the diagram on the right.

Unstable Modules over $\mathcal{A}$

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For any space $X$, the total Steenrod square
$$ Sq = \sum_i Sq^i $$ acts on $H^\ast(X; \mathbb{Z}_2)$, and by the Cartan formula this action is multiplicative, making $H^\ast(X; \mathbb{Z}_2)$ an unstable $\mathcal{A}$-module.

Example 1: $\mathbb{R}P^\infty$

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We know $$ H^\ast(\mathbb{R}P^\infty; \mathbb{Z}_2) \cong \mathbb{Z}_2[x], \quad \deg x = 1. $$

Since $Sq(x) = x + x^2$, by multiplicativity we obtain $$ Sq(x^k) = Sq(x)^k = (x + x^2)^k, $$ and hence $$ Sq^i(x^k) = \binom{k}{i} x^{k+i}. $$

Example 2: Detecting Nontrivial Maps via Steenrod Squares

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Consider $\Sigma \mathbb{C}P^2$ and $S^3 \vee S^5$.
Though these spaces have the same cohomology rings, they are not homotopy equivalent.

Indeed, $H^\ast(\mathbb{C}P^2; \mathbb{Z}_2) \cong \mathbb{Z}_2[x]$ with $\deg x = 2$, and $Sq^2(x) = x^2$.
By stability, $$ Sq^2(\Sigma x) = \Sigma x^2, $$ showing an $Sq^2$-link between degrees 3 and 5.
No such link exists for $S^3 \vee S^5$, proving that $\Sigma \mathbb{C}P^2$ cannot be a wedge of spheres.

In fact:

• For any $k$, $\Sigma^k \mathbb{C}P^2$ is not homotopy equivalent to a wedge of spheres.
• Equivalently, the Hopf map $\eta: S^3 \to S^2$ never becomes null after suspension.
  By Freudenthal’s suspension theorem, $\Sigma^k \eta$ represents the generator of the first stable stem, $\pi_1^S$.

Indecomposables and Consequences

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A Steenrod square $Sq^n$ is called decomposable if it can be written as $$ Sq^n = \sum_{k < n} Sq^{I(k)} Sq^k. $$ It is indecomposable otherwise.

Fact: $Sq^n$ is indecomposable if and only if $n = 2^k$.

This gives an elegant criterion on polynomial cohomology algebras:

Corollary:
If $H^\ast(X; \mathbb{Z}_2) \cong \mathbb{Z}_2[x]$, then necessarily $\deg x = 2^n$ for some $n$.

Proof:
If $\deg x = n$, then $Sq^n(x) = x^2$.
If $Sq^n$ were decomposable, the lower operations would vanish on $x$.
Thus, $Sq^n$ must be indecomposable, forcing $n = 2^k$.