Why the combinatorial second Stiefel-Whitney class is a cocycle?

Question

From the book "Spin geometry" by Lawson&Michaelson Appendix A or this literature we know that there is a nice combinatorial way to interpret the second SW class by the transition functions of a principal bundle. The basic idea is, we firstly take a good covering, where the principal bundle is trivial on each open sets. Then on each two-fold intersection we have the transition functions $$g_{\alpha\beta}: U_\alpha\bigcap U_\beta\rightarrow \mathrm{SO}(n)$$ Lifting each map to $\overline{g}_{\alpha\beta}:U_\alpha\bigcap U_{\beta}\rightarrow \mathrm{Spin}(n)$ gives a cochain on each three-fold intersection $$w_{\alpha\beta\gamma}=\overline{g}_{\alpha\beta}\overline{g}_{\beta\gamma}\overline{g}_{\gamma\alpha}:U_{\alpha}\bigcap U_{\beta} \bigcap U_{\gamma}\rightarrow \mathbb{Z}_2$$ On both references above they just claim that $w$ is a cocycle. But I just wonder is the argument really so trivial?

Answer 1

Let $U_{ijk} = U_i \cap U_j \cap U_k$ for simplicity.

The Cech $2-$cochain $\omega:U_{ijk} \rightarrow\mathbb{Z}_{2}$ must be single valued on the intersection. Also, one has $$U_{ijk}=U_{\sigma(i) \sigma(j)\sigma(j)}$$ for any permutation $\sigma$ of $\{i, j, k\}$.

Thus, $\omega$ must be totally symmetric. i.e. $$\omega_{ijk}=\omega_{ikj}=\omega_{jik}=\omega_{jki}=\omega_{kji}=\omega_{kij}.$$ Furthermore, since $w$ takes value in $\mathbb{Z}_{2}$ group, one has $\omega_{ijk}=\omega_{ijk}^{-1}$.

Therefore, one has $\bar{g}_{ij}\bar{g}_{jk}\bar{g}_{ki}=\bar{g}_{ik}\bar{g}_{kj}\bar{g}_{ji}$, where $\bar{g}_{ij}=\bar{g}_{ji}^{-1}$.

Using the above equations for $\bar{g}_{ij}$, one has $$\begin{split} (\delta w)_{abcd}&=w_{abc}w_{abd}w_{acd}w_{bcd}\\ &=(\bar{g}_{ab}\bar{g}_{bc}\bar{g}_{ca})(\bar{g}_{ab}\bar{g}_{bd}\bar{g}_{da})(\bar{g}_{ac}\bar{g}_{cd}\bar{g}_{da})(\bar{g}_{bc}\bar{g}_{cd}\bar{g}_{db})\\ &=\bar{g}_{ac}\bar{g}_{cb}\bar{g}_{ba}\bar{g}_{ab}\bar{g}_{bd}\bar{g}_{da}\bar{g}_{ad}\bar{g}_{dc}\bar{g}_{ca}\bar{g}_{bd}\bar{g}_{dc}\bar{g}_{bc}\\ &=\bar{g}_{ac}(\bar{g}_{cb}\bar{g}_{bd}\bar{g}_{dc})\bar{g}_{ca}\bar{g}_{bd}\bar{g}_{dc}\bar{g}_{bc}\\ &=\bar{g}_{ac}w_{cbd}\bar{g}_{ca}w_{bdc}\\ &=w_{cbd}w_{bdc}=1. \end{split}$$