Question

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

Answer 1

The situation seems more complicated.

In fact, let $V$ be the vector bundle $\gamma_1 \oplus \mathbb{R}^{2k-1}$ over $S^1$.

Then $TP(V)$ is isomorphic (via the choice of a connection) to $q^*TS^1 \oplus q^*V / L$, where $L\subset q^*V$ is the tautological bundle.

Hence the total Stiefel-Whitney class of $TP(V)$ is $q^* w(V)\cup w(L)^{-1}$ in the algebra $H^*(TP(V),\mathbb{F}_2)$.

This algebra is isomorphic to $\mathbb{F}_2[x,y]/(x^2,y^{2k})$, [EDIT: as a module over $H^*(S^1)$] since the $\mathbb{F}_2$-cohomology spectral sequence of $P(V)\to S^1$ necessarily has zero differentials on the $E_2$ page. Here $x=q^*(w_1(\gamma_1))$. Note that $\pi_1(S^1)$ acts trivially on $H^*(P^{2k-1},\mathbb{F}_2) \simeq \mathbb{F}_2[y]/(y^{2k})$. [EDIT : $y\in H^1(P(V ))$ is a class that restricts to the generator of $H^1$ of any fiber. But this doesn't characterize it : one may add $x$ to it. Hence the algebra structure must be determined by other means. See the comments].

But $w_1(L)\in H^1(P(V),\mathbb{F}_2)\simeq Hom(\pi_1(P(V)),\mathbb{F}_2)$ is easily checked to be $x+y$ : first note that $\pi_1(V)\simeq \mathbb{Z}\times \mathbb{Z}/2$, then that $L$ is non trivial along the section of $q$ given by $P(\gamma_1)$. Hence the $x$ summand. The $y$ summand comes from restriction to a fibre. [EDIT : here I may precise a choice of $y$. It is Poincaré dual to the "hyperplane section" $S^1\times P(\mathbb{R}^{2k-1})$ in $P(V)$. But this doesn't determine the multiplicative structure.]

[EDIT : The following calculation was wrong, due to a wrong algebra structure. See the comments for calculations with the correct one, given by $x^2=0$ and $(x+y)y^{2k-1}=0$.]

Answer 2

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.