# The first Stiefel-Whitney class is zero if and only if the bundle is orientable

Ralph Cohen's notes on the topology of fiber bundles pp.84 (theorem 3.3) says that it follows immediately from the definition of the first Stiefel-Whitney class of real vector bundles (pp.83)

\begin{equation} (BSO(n) \to BO(n)) \in \text{Prin}_{O(1)}(BO(n)) \cong [BO(n),BO(1)] \cong H^1(BO(n);\mathbb{Z}_2) \implies w_1(\eta) = f_{\eta}^*(w_1) \in H^1(X;\mathbb{Z}_2) \end{equation}

and that a bundle has an $SO(n)$ structure if and only if it is orientable. But I am still stuck with how to apply the orientability to the definition. Could someone please point that out for me?

• where did you find this chain of implications/isomorphisms ? on cohen at page 84 I wasn't able to find it. – Riccardo Jun 10 '16 at 16:16
• @Riccardo: the definition is on pp.83, one page above, although here I summarize it concisely into one sentence. – PhysicsMath Jun 10 '16 at 18:37

Consider the short exact sequence of groups (note that I use $O(1) \cong \mathbb{Z}_2$) $$SO(n) \rightarrow O(n) \xrightarrow{\det} \mathbb{Z}_2.$$ This induces an exact sequence $$[X, BSO(n)] \rightarrow [X, BO(n)] \xrightarrow{(B\det)_*} [X, B\mathbb{Z}_2].$$
The real rank $n$ bundle $\eta$ is classified by the map $f_\eta \in [X, BO(n)]$. We want to show that it lifts to an element in $[X, BSO(n)]$. Then it will be a $SO(n)$-bundle and hence orientable. By exactness, to show that it lifts is to show that $(B\det)_* (f_\eta) = 0$.
Consider the commutative square $$\begin{array}{ccc} [BO(n),BO(n)] & \rightarrow & [BO(n), B\mathbb{Z}_2] \cong H^1(BO(n);\mathbb{Z}_2) \\ \downarrow & & \downarrow \\ [X, BO(n)] & \rightarrow & [X, B\mathbb{Z}_2] \cong H^1(X;\mathbb{Z}_2) \end{array}$$ The vertical maps are $f_\eta^*$ and the horizontal maps are $(B\det)_*$. Trace the image of the identity $\operatorname{id}_{BO(n)}$ through the square. By definition of the Stiefel-Whitney class from Cohen's notes, $(B\det)_*(\operatorname{id}_{BO(n)}) = w_1 \in H^1(BO(n);\mathbb{Z}_2)$, so going right and then down gives $$f_\eta^*(B\det)_*(\operatorname{id}_{BO(n)}) = f_\eta^*(w_1) =: w_1(\eta).$$ On the other hand, going down and then right gives $$(B\det)_*f_\eta^*(\operatorname{id}_{BO(n)}) = (B\det)_* (f_\eta).$$
So $\eta$ is orientable iff $(B\det)_*(f_\eta) = w_1(\eta) = 0$.