Triviality of Steenrod operation on $\Sigma^{2k}\mathbb{CP}^n$

Question

I was going through this paper by Tanaka. I am actually stuck at Lemma 5.2, page 365, given below also

The argument he gives above works, in particular for $\operatorname{Sq}^{2^r-2^j}$ but I am not sure how to prove the result for any Steenrod operation i.e. a monomial of the form $\operatorname{Sq}^i\operatorname{Sq}^j\dots \operatorname{Sq}^n$ with total degree ${2^r-2^j}$. I am also aware of the fact that Steenrod algebra has generators of the form $\operatorname{Sq}^{2^i}$ as well as the admissible monomials. Will proving the result for these generators suffices? If so, how can we do it?

Any suggestions or hints will be appreciated and will be of great help.

Answer 1

Let me add some details to Nicholas Kuhn's answer (mostly for my own edification).

First, since Steenrod squares are stable cohomology operations, it is enough to show that the maps $\operatorname{Sq}^I : H^{2^j-2k}(\mathbb{CP}^n) \to H^{2^r-2k}(\mathbb{CP}^n)$ must all be zero.

Lemma: The subspace $F_p = \operatorname{span}\{y^m \mid \alpha(m) \leq p\}$ is a sub-$A$-module of $H^*(\mathbb{CP}^n)$; here $A$ denotes the Steenrod algebra.

Proof: We just need to check that $F_p$ is preserved by each Steenrod square. For degree reasons, we have $\operatorname{Sq}^{2\ell+1}(y^m) = 0 \in F_p$, while $\operatorname{Sq}^{2\ell}(y^m) = \binom{m}{\ell}y^{m+\ell}$; see this answer. If $\binom{m}{\ell} \equiv 0 \bmod 2$, then $\operatorname{Sq}^{2\ell}(y^m) = 0 \in F_p$. On the other hand, if $\binom{m}{\ell} \equiv 1 \bmod 2$, then $\alpha(m + \ell) \leq \alpha(m) \leq p$ as pointed out by Tanaka, so $\operatorname{Sq}^{2\ell}(y^m) = y^{m+\ell} \in F_p$. $\quad\square$

Now let $p = \alpha(2^{j-1}-k)$.

Note that $y^{2^{j-1}-k} \in H^{2^j-2k}(\mathbb{CP}^n)$ is an element of $F_p$, while $y^{2^{r-1}-k} \in H^{2^r-2k}(\mathbb{CP}^n)$ is not because $\alpha(2^{r-1}-k) > \alpha(2^{j-1}-k) = p$. By the lemma, we see that $\operatorname{Sq}^I(y^{2^{j-1}-k}) \in F_p$, so $\operatorname{Sq}^I(y^{2^{j-1}-k}) \neq y^{2^{r-1}-k}$ and hence $\operatorname{Sq}^I(y^{2^{j-1}-k}) = 0$, i.e. $\operatorname{Sq}^I : H^{2^j-2k}(\mathbb{CP}^n) \to H^{2^r-2k}(\mathbb{CP}^n)$ is the zero map.

Answer 2

Recall that $H^*(\mathbb CP^{\infty};\mathbb Z/2)= \mathbb Z/2[y]$ with $|y|=2$. Let $F_k \subset \mathbb Z/2[y]$ be the span of $y^m$ such that $\alpha(m)\leq k$. It is well known, and easy to check, that $F_k$ is a sub-$A$-module of $\mathbb Z/2[y]$. Thus Tanaka's inequality $\alpha(2^{r-1}-k) > \alpha(2^{j-1}-k)$ suffices to show his lemma.

(The $F_k$'s also make up the `primitive filtration' of $\mathbb Z/2[y]$.)