Formula for compositions of Steenrod squares that produce a form in the top degree

Question

On a smooth $d$ dimensional compact connected manifold $M$, for an $\mathbb{Z}_2$-valued $(d-j)$-cocycle $x_{d-j}$ we have the formula ${\text{Sq}}^{j} (x_{d-j}) = u_{j} \cup x_{d-j}$. Here $u_j \in H^{j}(M,{\mathbb{Z}}_2)$ is the $j$th Wu class and ${\text{Sq}}$ is the Steenrod squaring operation. When composed with the pairing with the $\mathbb{Z}_2$-fundamental class of $M$, this gives a ${\mathbb{Z}}_2$-linear map $H^{d-j}(M,{\mathbb{Z}_2})\rightarrow {\mathbb{Z}}_2$.

Now consider ${\text{Sq}}^{j_1} {\text{Sq}}^{j_2} (x_{d-j_1-j_2})$. This also gives a ${\mathbb{Z}}_2$-linear map $H^{d-j_1-j_2}(M,{\mathbb{Z}_2})\rightarrow {\mathbb{Z}}_2$ by pairing with the $\mathbb{Z}_2$-fundamental class of $M$, i.e. it defines an element of ${\text{Hom}}(H^{d-j_1-j_2}(M,{\mathbb{Z}_2}),{\mathbb{Z}}_2)$. This is isomorphic to $H_{j_1+j_2}(M,{\mathbb{Z}_2}) \cong H^{j_1+j_2}(M,{\mathbb{Z}_2})$. (This last isomorphism holds assuming the homology groups are all finitely generated). Thus there exists some cohomology class $v_{j_1,j_2} \in H^{j_1+j_2}(M,{\mathbb{Z}_2})$ such that ${\text{Sq}}^{j_1} {\text{Sq}}^{j_2} (x_{d-j_1-j_2}) = v_{j_1,j_2} \cup x_{d-j_1-j_2}$. Is there a formula for the cohomology class $v_{j_1,j_2}$ in terms of the characteristic classes of $M$?

I'm guessing the answer is no, since I haven't seen any such formula. I'm curious if my reasoning above is flawed in any way.

Answer 1

$\DeclareMathOperator\Sq{Sq}$Let $X$ be a space and $x,y\in H^*(X;\mathbb{Z}/2)$. For $b\geq0$ consider the element $\Sq^b(x)\cdot y$. Using the Cartan formula write this as $$\Sq^b(x)\cdot y=\Sq^b(x\cdot y)+\sum_{1\leq i_1\leq b}^b\Sq^{b-i_1}(x)\Sq^{i_1}(y).$$ This formula iterates, and each term in the sum can be written in the form $$\Sq^{b-i_1}(x)\Sq^{i_1}(y)=\Sq^{b-i_1}(x\cdot\Sq^{i_1}(y))+\sum_{1\leq i_2\leq b-i_1}^{b-i_1}\Sq^{b-i_1-i_2}(x)\Sq^{i_2}\Sq^{i_1}(y).$$ Thus $$\Sq^b(x)\cdot y=\Sq^b(x\cdot y)+\sum_{1\leq i_1\leq b}\Sq^{b-i_1}(x\cdot \Sq^{i_1}(y))+\sum_{2\leq i_1+i_2\leq b}\Sq^{b-i_1-i_2}(x)\Sq^{i_2}\Sq^{i_1}(y).$$ Induction yields the formula

$$\Sq^b(x)\cdot y=\Sq^b(xy)+\sum_{j=1,\dots,b}\sum_{j\leq i_1+\dotsb+i_j\leq b}\Sq^{b-i_1-\dotsb-i_j}(x\cdot (\Sq^{i_j}\dotsm\Sq^{i_1}y)).$$

Now assume that $X$ is a connected $n$-manifold and $x\in H^{n-a-b}(X;\mathbb{Z}/2)$. Let $v_i\in H^i(X;\mathbb{Z}/2)$ be the $i$th Wu class. Then $$\begin{aligned}\Sq^a\Sq^b(x)&=\Sq^b(x)\cdot v_a\\ &=\Sq^b(xv_a)+\sum_{j=1,\dots,b}\sum_{j\leq i_1+\dotsb+i_j\leq b}\Sq^{b-i_1-\dotsb-i_j}(x\cdot (\Sq^{i_j}\dotsm\Sq^{i_1}v_a))\\ &=xv_av_b+\sum_{j=1,\dots,b}\sum_{j\leq i_1+\dotsb+i_j\leq b}x\cdot v_{b-i_1-\dotsb-i_j}(\Sq^{i_j}\dotsm\Sq^{i_1}v_a).\end{aligned}$$ Thus $$\Sq^a\Sq^bx=x\cdot \left(v_av_b+\sum_{j=1,\dots,b}\sum_{j\leq i_1+\dotsb+i_j\leq b}v_{b-i_1-\dotsb-i_j}(\Sq^{i_j}\dotsm\Sq^{i_1}v_a)\right).$$ This gives

$$v_{a,b}=v_av_b+\sum_{j=1,\dots,b}\sum_{j\leq i_1+\dotsb+i_j\leq b}v_{b-i_1-\dotsb-i_j}(\Sq^{i_j}\dotsm\Sq^{i_1}v_a)$$ where $i_1,\dots,i_j\geq1$ and $v_0=1$.

Answer 2

To add onto the very nice explicit answer of Tyrone, here's another perspective which you might find interesting: Both Stiefel-Whitney classes and Wu classes can be generalized to arbitrary $\mathbb{F}_2$ cohomology operations.

Indeed, for a vector bundle $E$ over $X$ with Thom class $u\in H^n(\mathrm{Th}(E))$, Stiefel-Whitney classes are uniquely characterized by $\mathrm{Sq}^i(u) = w_iu$, so for any cohomology operation $\theta$, you could define $w_\theta$ by $\theta(u) = w_\theta u$. Of course, the Cartan formula together with the formulas for $\mathrm{Sq}^iw_j$ gives you a way to express $w_\theta$ as polynomial in the $w_i$ for $\theta$ any composite of Steenrod operations.

On a closed connected manifold, we may also define generalized Wu classes through Poincaré duality, as you've suggested, by letting $v_\theta \cup\alpha = \theta \alpha$ whenever this expression lands in the top degree cohomology. So classical Wu classes are $v_j = v_{\mathrm{Sq}^j}$, and you're asking about $v_{\mathrm{Sq}^{j_1}\mathrm{Sq}^{j_2}}$.

A careful reading of the proof of the Wu formula then proves the completely analogous relation $$ w_\theta = \sum \theta'(v_{\theta''}), $$ where the sum ranges over terms in $\Delta(\theta) = \sum \theta'\otimes \theta''$, and $\Delta$ denotes the comultiplication in the Hopf algebra structure on the Steenrod algebra (or said differently, the operations in the right hand side are characterized by a Cartan formula $\theta(xy) = \sum \theta'(x)\theta''(y)$).

Since $\Delta(\theta) = \theta\otimes 1 + 1\otimes \theta + \text{ mixed terms}$, one can reverse the generalized Wu formula above to express $v_\theta = w_\theta + \ldots$, with remaining terms depending on $v_{\theta''}$ of strictly smaller degree. This allows you to express the $v_\theta$ as polynomial in Stiefel-Whitney classes inductively.