# Wu relation for Steenrod square and Stiefel-Whitney cocycles

Given a simplicial complex with a branching structure, we can compute the Stiefel-Whitney cocycles $$w_n$$. (see R. Z. Goldstein and E. C. Turner, Proc. Amer. Math. Soc. 58, 339 (1976))

Wu relation for Stiefel-Whitney cocycles 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 simplicial complex with a branching structure and no boundary, 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 cocycles: $$u_1=w_1$$, $$u_2=w_2+w_1^2$$, $$u_3=w_1w_2$$, etc.

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

A branching structure of a simplicial complex is given by assigning arrows to the links which do not form loops around each triangle. The branching structure gives rise to a local order of vertices for every simplex. Recently, a cochain level Adem relations in the mod 2 Steenrod algebra is obtained https://arxiv.org/abs/2006.09354

Updated question

• 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? May 29, 2017 at 19:20
• I am sorry. $x_{d-n}$ is a cocycle, I update the question. May 29, 2017 at 20:00
• You seem to be working at the level of cochains. What is your definition of Wu class in this case? (As far as I can tell they are generally defined only at the level of cohomology classes). May 29, 2017 at 20:02
• I defined Wu class in terms of Stiefel-Whitney classes, and I assume there is way to pick the representative cocycle. May 29, 2017 at 20:06
• 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. May 30, 2017 at 5:58

Let me attempt an answer. Given a $$d$$-manifold and a cocycle $$x$$ in some presentation (e.g., a triangulation), there exists a $$d-1$$-cocycle $$y$$ such that $$\operatorname{Sq}^n(x)=u_n\cup x+\delta y\;.$$ Obviously, we can also choose one for every manifold and $$x$$. The OPs question is how to construct a "canonical choice" of $$y$$. A sensible property of such a canonical choice would be that it is locally computable from, e.g., a triangulation. In any case, it should be compatible with the symmetries of, e.g., the triangulation. I believe that this is not possible.
To illustrate this, let me stick to the simplest case of a $$1$$-cocycle $$x$$ of a $$2$$-manifold, $$x\cup x = \operatorname{Sq}^1(x) = u_1 \cup x +\delta y= \omega_1 \cup x+\delta y\;,$$ and let me work entirely in terms of triangulations, with $$x$$ represented by a $$1$$-cycle. Local combinatorial formulas which compute a $$d-i$$-cycle representing the $$i$$th the Stiefel-Whitney class can be found in . Local combinatorial formulas for the cup product are given in , however those are for cocycles, not cycles. If we want a formula for cycles we have to use the intuition of the cup product as intersections, and an analogue of the branching structure on the dual cellulation to shift one of the cycles to a Poincare-dual cocycle. For two 1-cycles $$x$$ and $$y$$ in $$2$$ dimensions, it suffices to have a dual orientation of each edge, shift $$y$$ according to that dual orientation and let $$x\cup y$$ be the vertices where $$x$$ intersects with the shifted $$y$$. A 1-cycle representing $$\omega_1$$ in this case is given by all edges separating a left-handed and a right-handed triangle.
Now, imagine a triangulation of the real projective plane by gluing a triangulation of a disk with a triangulation of the Moebius strip, both with $$2n$$ edges at their boundary, such that both the disk and Moebius strip triangulations are invariant under cyclic shifts by an arbitrary number of boundary edges, and the Moebius strip triangulation is invariant under reflection at its "core". Take a 1-cycle $$x$$ consisting of a single line which is symmetric under a cyclic shift by $$n$$ edges on the boundary (i.e., a rotation by $$\pi$$). Due to the rotation symmetry at the boundary, $$x$$ will intersect itself in the center of the disk. Due to the reflection symmetry of the Moebius strip, $$\omega_1$$ will consist of its core and intersect $$x$$ there. So $$x\cup x+\omega_1\cup x$$ consists of the center of the disk plus the intersection of $$x$$ with the core of the Moebius strip. The whole setup (triangulation plus $$x$$) has a $$\pi$$-rotation symmetry, but any choice of $$y$$ must break this symmetry.
A physical way to see this is to view $$x$$ as a dynamical field and $$(-1)^{x\cup x}$$ and $$(-1)^{\omega_1\cup x}$$ as two unoriented real lattice TQFTs. If there was a locally computable formula $$y[x]$$, we could build an invertible domain wall between the two, which along a 1-cycle $$s$$ is defined by $$(-1)^{y[x]\cup s}$$. Such an invertible domain wall cannot exist however, as the two axiomatic unoriented TQFTs corresponding to the two lattice TQFTs are non-isomorphic.
The vector space of both of the latter on the circle is 2-dimensional, corresponding to configurations with even and odd $$x$$. One generating cobordism of unoriented TQFT is the annulus with oppositely oriented boundary circles . The amplitude of $$(-1)^{x\cup x}$$ on this cobordism for the two odd $$x$$ configurations is $$-1$$ as due to the opposite orientations, the second copy of $$x$$ is shifted in different directions on the two boundary circles such that $$x$$ must intersect with its shifted copy in the interior. On the other hand, the same amplitude for $$(-1)^{\omega_1\cup x}$$ is $$1$$ since the annulus is orientable and $$\omega_1$$ is locally trivializable. So the full matrices associated to the cobordism are the the Pauli $$Z$$ matrix for $$(-1)^{x\cup x}$$ and the identity for $$(-1)^{\omega_1\cup x}$$, which are not isomorphic due to different eigenvalues.
I'd also like to remark that a canonical choice of $$y$$ in general would imply a canonical choice of trivialization of $$\omega_1^2+\omega_2$$ on 2-manifolds, which would imply a canonical choice of Pin$$^-$$ structure for each 2-manifold.