Introduction#
In the previous part, we saw the construction of the Steenrod algebra and how it is a stable cohomology operation. At the end of the blog, we described a few properties of Steenrod operations. Let us list all the characterizations.
- The stable cohomology operations $Sq^i$ for $i \geq 0$.
- For $n \geq 0$, the operations are stable cohomology operations $Sq^n : H^q(X;\mathbb{Z}_2) \to H^{q+n}(X;\mathbb{Z}_2)$.
- The homomorphism $Sq^{0}$ is the identity.
- For $x \in H^{p}(X;\mathbb{Z}_2)$ or $\deg x = p$, we have $Sq^i(x) = 0$ for $p > i$.
- $Sq^n(x) = x^2$ for $\deg x = n$.
- Cartan formula: $Sq^i(xy) = \sum_{j} Sq^j(x) Sq^{i-j}(y)$.
- Adem relation: The Steenrod operations satisfy the following relation (for $i < 2j$):
$$ Sq^i Sq^j = \sum_k \binom{j-k-i}{j-2k} Sq^{i+j-k} Sq^k $$ - $Sq^1$ is the Bockstein associated with the SES $0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0$.
Since we are looking at cohomology mod 2, it’s helpful to examine Pascal’s triangle to derive the Adem relation. For example:
- $Sq^1 Sq^1 = 0$.
- $Sq^1 Sq^2 = Sq^3 \neq Sq^2 Sq^1$.
- $Sq^2 Sq^2 = Sq^3 Sq^1 = Sq^1 Sq^2 Sq^1$.
In general, we define $Sq^{i_1} \cdots Sq^{i_k} = Sq^{i_1, \cdots, i_k}$, and for a tuple $I = (i_1, \cdots, i_k)$, the previous product is denoted as $Sq^I$. The degree of $I$ is defined as the sum of $i_j$’s.
- Definition: $I$ is admissible if $i_j > 2i_{j+1}$.
- If $I$ is admissible, $I$ has excess $e(I) = 2i_1 - \deg I$.
Steenrod Algebra#
- Definition: The Steenrod algebra $\mathcal{A}$ is the free graded $\mathbb{Z}_2$-algebra (not commutative) with generators $Sq^i, i \geq 0$, $\deg Sq^i = i$, subject to the Adem relation.
Theorem: The algebra $\mathcal{A}$ is minimally generated by $Sq^{2^i}: i \geq 0$.
Theorem: ${Sq^I : I \text{ is admissible} }$ is a $\mathbb{Z}_2$-basis of $\mathcal{A}$.
The Cartan formula $Sq^k \mapsto \sum Sq^i \otimes Sq^{k-i}$ induces a coproduct structure on $\mathcal{A}$ that is an algebra homomorphism, making $\mathcal{A}$ a co-commutative Hopf algebra, which means $\mathcal{A}_{\ast}$ is a commutative algebra.
Theorem (Milnor): $\mathcal{A}_{\ast} = \mathbb{Z}_2[\zeta_1, \cdots]$, where $\deg \zeta_i = 2^i - 1$, and
$$ \Delta(\zeta_k) = \sum_{i=0}^k \zeta_{k-i}^{2^i} \otimes \zeta_i $$
Here, $\zeta_k$ is dual to $Sq^{I_k}$, where $I_k = (2^{k-1}, \cdots, 1, 0)$.
Now, we define $\mathcal{A}(k)$ as the subalgebra generated by $Sq^{i}$ for $i \leq 2^k$. For example, the relations in $\mathcal{A}(1)$ can be represented in the following diagram.
Unstable Modules over $\mathcal{A}$#
We define $Sq: H^{\ast}(X,\mathbb{Z}_2) \to H^{\ast}(X,\mathbb{Z}_2)$ by
$$ Sq(x) = \sum Sq^i(x) $$
(note that for every $x$, it is actually a finite sum). The Cartan formula implies that $Sq$ is a ring homomorphism.
Example: We know $H^{\ast}(RP^{\infty}, \mathbb{Z}_2) \simeq \mathbb{Z}_2[x]$ where $\deg x = 1$.
Note that $Sq(x) = x + x^2$. Using the fact that $Sq$ is a ring homomorphism, we get $Sq(x^k) = Sq(x)^k$ and thus $$ Sq^i(x^k) = \binom{k}{i} x^{k+i} $$
Example 2: $\Sigma CP^2$ is not homotopy equivalent to $S^3 \vee S^5$.
We cannot distinguish them by homology/cohomology ring computations. We can view $H^{\ast}(CP^2; \mathbb{Z}_2)$ as an $\mathcal{A}$-module. Let $\deg x = 2$, then $Sq^2(x) = x^2$. Now, by the stability of Steenrod operations, we get $Sq^2(\Sigma x) = \Sigma x^2$. So there is an $Sq^2$ connection between degrees 3 and 5. However, for $S^3 \vee S^5$, there is no Steenrod algebra connection.
- In general, $\Sigma^k CP^2$ is never homotopy equivalent to a wedge of spheres.
- In other words, the Hopf map $\eta: S^3 \to S^2$ never becomes null after suspending. Thus, $\Sigma^k \eta$ represents a non-trivial element in $\pi_1^S$, and in fact, by the Freudenthal suspension theorem, this is the generator of the stable stem 1.
$Sq^n$ is decomposable if $Sq^n = \sum_{k < n} Sq^{I(k)} Sq^k.$
$Sq^n$ is indecomposable iff $n = 2^k$.
Corollary: If $H^{\ast}(X;\mathbb{Z}_2) = \mathbb{Z}_2[x]$, then $\deg x = 2^n$.
The proof is very short. If $n = \deg x$, then $Sq^n(x) = x^2$. If $Sq^n$ is decomposable, then all the lower Steenrod squares are zero. So $Sq^n$ is indecomposable, and thus $n = 2^k$ for some $k$.