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

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

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

  • $\begingroup$ 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? $\endgroup$ – rotkaeppchen Feb 22 '12 at 22:34
  • $\begingroup$ 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 $\sum_{0\leq j\leq k} w_j(\xi)a^{k-j}=0$. $\endgroup$ – BS. Feb 23 '12 at 10:50
  • $\begingroup$ (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 $x^2=0$. $\endgroup$ – BS. Feb 23 '12 at 11:01
  • $\begingroup$ Sorry for the late answer. Thank you for posting the corrected computation. The formula cited above can be found in maths.ed.ac.uk/~aar/papers/bh1.pdf 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? $\endgroup$ – rotkaeppchen Mar 2 '12 at 20:20
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ Thanks for this point of view. My calculation also yields that the first SW-class is $q^*(v_1)$. $\endgroup$ – rotkaeppchen Mar 2 '12 at 22:23
  • $\begingroup$ 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 $P_k$? $\endgroup$ – rotkaeppchen Mar 5 '12 at 15:11

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