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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

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Could you tell us what you mean by SW-classes of bundles of projective spaces? –  Mark Grant Feb 22 '12 at 9:43
The OP might ask about the first SW class of the total space of the bundle. –  BS. Feb 22 '12 at 10:21
I should have been more precise here. My question is about the tangential SW-classes of the total space of the bundle $P(\gamma \oplus \mathbb{R}^{2k-1}$. –  rotkaeppchen Feb 22 '12 at 11:29

2 Answers 2

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 $(x+y)y^{2k-1}=0$.]

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Thanks for taking time to answer. However, i cannot quite follow your reasoning as I am not into spectral sequences. What is $y$? Anyway, I came as far as this: The tangential SW-classes of any projective bundle $P(\xi \rightarrow M)$ can be computed as $\Bigl(\sum \limits_{j=0}^k (1+a)^j q^*(v_{k-j})\Bigr) \cup q^*(w(M))$ where $k=rk(\xi)$, $a$ is the first SW-class of the canonical line bundle in $q^*(\xi)$ and $v_j$ are the SW-classes of $\xi$. Hence in our case $w(TP(V))=(1+a)^{2m-1}(1+a+q^*(v_1))$. Now my guess was that $a=q^*(v_1)(=x)$ which would make things simple. What do you think? –  rotkaeppchen Feb 22 '12 at 22:34
Your $a$ is my $w_1(L)$, but I don't understand your formula for $w(TP(\xi))$. I would rather say it's $q^*(w(M)) q^*(w(\xi)) (1+a)^{-1}$. This said, I admit my spectral sequence calculation might be a bit bugged, not beeing careful enough with filtration/gradation issues to give the correct algebra structure. Instead, it is preferable to use the Leray-Hirsch theorem, saying that $H^*(P(\xi))$ has basis $(1,a,a^2,dots,a^{k-1})$ as a $H^*(M)$-module (mod 2 coeffs everywhere), and the multiplicative structure is given by the relation $\sum_{0\leq j\leq k} w_j(\xi)a^{k-j}=0$. –  BS. Feb 23 '12 at 10:50
(cont'd) See chapter 3 in Allen Hatcher's "Vector bundles and K-theory" chapter for this. Actually, it is the way he defines Stiefel-Whitney classes. In your case, this gives $H^*(P(\xi))\simeq \mathbb{F}_2[x,a]/(x^2,(x+a)a^{2k-1})$, and $w(TP(\xi))=(1+x)/(1+a)=1+(x+a)/(1+a)=1+(x+a)(1+a\dots +a^{2k-2})$. Note that $a$ can't be $x$, since $x^2=0$. –  BS. Feb 23 '12 at 11:01
Sorry for the late answer. Thank you for posting the corrected computation. The formula cited above can be found in page 517. 15.4 Our formulas are not very different. Mine can be manipulated to $w(TP(V))=(1+a+x)(1+a+\ldots+a^{2m-1})$. Are you sure that the last summand is $a^{2m-2}$ and that $1$ is outside the brackets? Finally, i wonder if all SW-numbers of $V$ vanish? –  rotkaeppchen Mar 2 '12 at 20:20
I discovered a mistake in my calculation. I still believe that the total SW-class is $(1+a)^{2m-1}(1+a+x)$. I agree, with your structure for $H^*(P(\xi))$, however i do not understand your formula for $w(TP(\xi))$ and i cannot find it in Hatcher's notes. Can you please give more details on that? –  rotkaeppchen Mar 5 '12 at 7:51

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.

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Thanks for this point of view. My calculation also yields that the first SW-class is $q^*(v_1)$. –  rotkaeppchen Mar 2 '12 at 22:23
EDIT (typo): I completely agree with you here. I wonder if it is possible to build the monomial $a^{2k}$ with the SW-classes of $P_k$? –  rotkaeppchen Mar 5 '12 at 15:11

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