# Wu relation for Stiefel-Whitney classes and Steenrod square

Wu relation for Stiefel-Whitney classes and Steenrod square is given by the following cochain equation $$Sq^n(x_{d-n})=u_n \smile x_{d-n} + \text{d} y_{d-1} \ \text{ mod } 2$$ on a $d$-dimensional manifold $M^d$, where $x_{d-n} \in Z^{d-n}(M^d;\mathbb{Z_2})$ is a cocycle and $y_{d-1} \in C^{d-1}(M^d;\mathbb{Z_2})$ is a cochain.

Also $u_n \in Z^{n}(M^d;\mathbb{Z_2})$ is the Wu class cocycle, which can be expressed in terms of Stiefel-Whitney classes: $u_1=w_1$, $u_2=w_2+w_1^2$, $u_3=w_1w_2$, etc.

What is the explicit expression of $y_{d-1}$ in terms of $x_{d-n}$ and Stiefel-Whitney classes?

A proper answer to the quenstion may need to include a part about choosing a natural cocycle for a triangulation of $M^d$ representing a Stiefel-Whitney class on $M^d$. The triangulation of $M^d$ can contain some decorations, such as the branching structure, Kasteleyn orientations, etc. See https://arxiv.org/abs/1703.10937
• I thought the Wu classes satisfied $\operatorname{Sq}^n(x_{d-n}) = u_n\cup x_{d-n}$. This is what nLab seems to suggest. Do you want $x_{d-n}$ and $y_{d-1}$ to be cochains instead? – Michael Albanese May 29 '17 at 19:20
• I am sorry. $x_{d-n}$ is a cocycle, I update the question. – Xiao-Gang Wen May 29 '17 at 20:00
• Even then, the cocyle representing a Stiefel-Whitney class depends on some "presentation" of $TM$, e.g., an $O(d)$-valued group cocycle or a classifying map to $BO(d)$ or something else. To get a well-defined problem, it might make sense to add this kind of data to your description. – Sebastian Goette May 30 '17 at 5:58