# 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)

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

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. Commented Jun 10, 2016 at 16:16
• @Riccardo: the definition is on pp.83, one page above, although here I summarize it concisely into one sentence. Commented Jun 10, 2016 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 $(B\det)_*(f_\eta) = w_1(\eta) = 0$.

• Very clear answer, JHF! Many thanks! I really wish Cohen can include at least some of your explanation (that something is zero if and only if the bundle is orientable which can then be transferred to the first Stiefel-Whitney class)! Commented Jun 19, 2016 at 2:05
• Why exact sequence of groups induce exact sequence between sets of maps? Commented Jul 24, 2017 at 12:47
• @kp9r4d This is standard; see, e.g., here.
– JHF
Commented Jul 24, 2017 at 14:35