TITLE: Stiefel-Whitney classes of a projective space bundle 
QUESTION [5 upvotes]: Hi! 
Let $\gamma_1$ denote the twisted line bundle over $S^1$ and add a trivial $(2k-1)$-bundle $\mathbb{R}^{2k-1}$. Consider the projective bundle $P(\gamma_1 \oplus \mathbb{R}^{2k-1})$ over $S^1$. Is it true that the first Stiefel-Whitney class of this bundle is $q^*(w_1)$ and all other vanish? Here $q$ denotes the projection from the total space of the projective bundle to $S^1$ and $w_1$ is the first Stiefel-Whitney class of $\gamma_1$. 
EDIT: The question is about the tangential Stiefel-Whitney classes of the total space of $P(\gamma_1 \oplus \mathbb{R}^{2k-1})$.
best regards

REPLY [2 votes]: You are asking whether a loop in the $2k-1$ manifold $P_k=P(\gamma_1\oplus R^{2k-1})$ is orientation reversing if and only if its projection to the base $S^1$ has odd degree. 
$q:P_k\to S^1$ is a fiber bundle with fiber $F_k=RP^{2k-1}$. Since $RP^{2k-1}$ is orientable, a loop in $P_k$ which projects to a nullhomotopic loop in $S^1$ is homotopic into the fiber, hence orientation preserving. Thus $w_1(P_k)$ is either $q^*(w_1)$ or $0$. 
But it isn't zero, since  it isn't for $k=1$ (the Klein bottle is not orientable), and the normal bundle of $P_1\subset P_k$ is trivial.
EDIT: for the higher SW classes, The inclusion $F_k\subset P_k$ takes the SW classes to those of $F_k$ since the normal bundle of $F_k$ is trivial. If I recall correctly, the total SW class of $RP^{2k-1}$ equals $(1+t)^{2k}$, where $t\in H^1(RP^{2k-1};Z/2)$ denotes the generator. So, for example $(1+t)^6=1+t^2 + t^4$ and so for $k=3$, $w_2(P_k)$ and $w_4(P_k)$ are non-trivial.