Integral Stiefel-Whitney classes and stable almost complex structure

Question

Question: Let $\xi$ be a real vector bundle over a space $B$. Suppose $\xi$ admits stable almost complex structure, i.e., $\xi\oplus \epsilon^k$ admits almost complex structure for some $k\geq 0$, where $\varepsilon$ denotes a trivial real vector bundle. Is it true that every integral Stiefel–Whitney class $W_j(\xi)\in H^j(B;\Bbb Z)$ of $\xi$ vanishes?

My attempt: Consider a complex vector bundle $\omega$ over $B$ such that $\omega_\Bbb R=\xi\oplus \varepsilon_k$, where $\omega_\Bbb R$ denotes the realification of $\omega$. Now, it is known that (see Proposition 3.8. of Allen Hatcher's Vector bundles and K-theory) $w_{2j}(\omega_\Bbb R)$ is the image of $c_j(\omega)$ under the coefficient homomorphism $\rho_{2j}\colon H^{2j}(B;\Bbb Z)\to H^{2j}(B;\Bbb Z_2)$ and $w_{2j+1}(\omega_\Bbb R)=0$ for all $j\geq 0$. Therefore, $$w_{2j+1}(\xi)=w_{2j+1}(\xi\oplus \varepsilon_k)=w_{2j+1}(\omega_\Bbb R)=0$$ for all $j\geq 0$. Further, $$w_{2j}(\xi)=w_{2j}(\xi\oplus \varepsilon_k)=w_{2j}(\omega_\Bbb R)=\rho_{2j}\big(c_j(\omega)\big)$$ for all $j\geq 1$. Thus, $$\mathrm W_{2j}(\xi)= \beta_{2j-1}\big(w_{2j-1}(\xi)\big)= \beta_{2j-1}\big(0\big)=0$$ and $$\mathrm W_{2j+1}(\xi)= \beta_{2j}\big(w_{2j}(\xi)\big)= \beta_{2j}\rho_{2j}\big(c_j(\omega)\big)=0$$ as $\beta_{2j}\rho_{2j}=0$ for all $j\geq 1$, where $\beta_{n-1}\colon H^{n-1}(B;\Bbb Z_2)\to H^n(B;\Bbb Z)$ is the $(n-1)$-th Bockstein of $0\to \Bbb Z\xrightarrow{\times 2}\Bbb Z\to \Bbb Z/2\Bbb Z\to 0$ for each $n\geq1$.

Am I doing something wrong? The paper Obstructions to the existence of almost complex structures by Massey, W. S. says that the vanishing of the odd integral Stiefel-Whitney classes is a well-known necessary condition for the existence of almost complex structure.

Thank you for reading.

Answer 1

Your argument is correct. Massey only mentions the vanishing of the odd integral Stiefel-Whitney classes because, for an orientable bundle $\xi$, we have $W_{2j+1}(\xi) = 0 \implies W_{2j+2}(\xi) = 0$.

To see this, note that $\rho_{2j+1}(W_{2j+1}(\xi)) = \rho_{2j+1}\beta_{2j}(w_{2j}(\xi)) = \operatorname{Sq}^1(w_{2j}(\xi))$ where $\operatorname{Sq}^1 : H^{2j}(B; \mathbb{Z}_2) \to H^{2j+1}(B; \mathbb{Z}_2)$ is the first Steenrod square. By Wu's formula, we have $\operatorname{Sq}^1(w_{2j}(\xi)) = w_1(\xi)w_{2j}(\xi) + w_{2j+1}(\xi)$, so $\rho_{2j+1}(W_{2j+1}(\xi)) = w_{2j+1}(\xi)$ if $\xi$ is orientable. Therefore, if $W_{2j+1}(\xi) = 0$, then $w_{2j+1}(\xi) = 0$, and hence $W_{2j+2}(\xi) = 0$ as you observed.