11
$\begingroup$

Is every $ \mathbb{R}P^{2n} $ bundle over the circle trivial?

Are there exactly two $ \mathbb{R}P^{2n+1} $ bundles over the circle?

This is a cross-post of (part of) my MSE question

https://math.stackexchange.com/questions/4349052/diffeomorphisms-of-spheres-and-real-projective-spaces

which has been up for a couple weeks and got 8 upvotes and some nice comments but no answers.

My intuition for thinking both answer are yes is that there are exactly 2 sphere bundles over the circle. The trivial one and then the non-trivial (and non-orientable) one which can be realized as the mapping torus of an orientation reversing map of the sphere. So importing that intuition to projective spaces then the orientable $ \mathbb{R}P^{2n+1} $ should have a nontrivial (and non orientable) bundle over the circle while the non orientable $ \mathbb{R}P^{2n} $ should have only the trivial bundle. For $ n=1 $ this checks out since that projective space is orientable and thus we have exactly two bundles over the circle (the trivial one=the 2 torus and the nontrivial one=the Klein bottle).

$\endgroup$

2 Answers 2

8
$\begingroup$

Your answer is correct if appropriately understood, but it's a little subtle. Here I should note that I'm interpreting your question as a purely homotopy theoretic one (in particular ignoring smooth structure), and that by "bundle" you mean "Serre bundle". If you care about smooth bundles, see Tom Goodwillie's answer.

The way I know to do this involves a little algebraic topology.

First, it's always the case that $X$-bundles over a circle (for $X$ some topological space) are classified by the group $\pi_0(\mathrm{Aut}(X)),$ where $\mathrm{Aut}(X)$ is the group of maps $X\to X$ which are invertible up to homotopy. Thus a good technique to carry out this classification is to first find all self-maps $X\to X$ up to homotopy, then see which ones are invertible.

Now we want to specialize to the case $X = RP^n,$ for some $n$. However it turns out that it's much easier to classify maps not into $RP^n$ itself, but rather into $RP^\infty.$ Namely, maps from some space $X$ to $RP^\infty$ up to homotopy are always classified by $H^1(X, \mathbb{Z}/2)$ (equivalently, this is group homomorphisms from $\pi_1(X)$ into $\mathbb{Z}/2$). Now the difference between $RP^n$ and $RP^\infty$ isn't actually too terrible. Namely, the CW approximation theorem tells us that any map $RP^n\to RP^\infty$ actually can be chosen up to homotopy such that it factorizes through $RP^n\subset RP^\infty.$ So we can compute $$\pi_0\mathrm{Maps}(RP^n, RP^\infty) \cong H^1(RP^n, \mathbb{Z}/2)\cong \mathbb{Z}/2.$$ The element $0\in \mathbb{Z}/2$ corresponds to the trivial map $$*:RP^n\to RP^n$$ mapping everything to a basepoint in $RP^\infty$ and the nontrivial element $1\in \mathbb{Z}/2$ corresponds to the identity map, $$id:RP^n\to RP^n\subset RP^\infty.$$ And any other map $RP^n\to RP^n$ will be homotopy equivalent to one of these as an element of $\mathrm{Maps}(RP^n, RP^\infty)$. However there's a catch: while any map $RP^n\to RP^n$ is homotopy equivalent to one of the maps $id, *$ as a map to $RP^\infty$, the homotopy between the two maps might not live in $RP^n$. So a priori, there can be multiple homotopy classes of self-maps of $RP^n$ homotopic to one of these maps in $RP^\infty$ (and indeed, sometimes there are). To get a handle on how badly maps to $RP^\infty$ are undercounting, you can apply the CW approximation theorem again to see that any homotopy between two maps $RP^n\to RP^\infty$ will factorize, up to homotopy, through $$RP^{n+1}\subset RP^\infty.$$ The "error" of such a homotopy existing in $RP^n$ will be classified by a map to the quotient, $$RP^n\times [0,1]\to RP^{n+1}/RP^n\cong S^{n+1},$$ taking both $RP^n\times \{0\}, RP^n\times \{1\}$ to the basepoint, in other words, a based map from the space $$RP^n_+\wedge S^1 = \Sigma(RP^n_+)$$ (you can think of this as the ordinary suspension $\Sigma(RP^n)$ with the two suspension points identified) to $S^{n+1}.$ Let's write $$D: = \pi_0\text{Maps}(RP^n_+\wedge S^1, S^n)$$ for the set of possible such "defects" of a homotopy in $RP^{n+1}$ restricting to $RP^n$.

Based maps from a closed $n+1$-dimensional manifold to $S^{n+1}$ are classified by ordinary $H^{n+1},$ and so we have $$D \cong H^{n+1}(\Sigma(RP^n_+)) \cong H^n(RP^n) \cong \begin{cases} \mathbb{Z}/2, & n\text{ even}\\ \mathbb{Z}, & n\text{ odd}. \end{cases}$$

Thus for each of the maps $*, id: RP^n\to RP^n,$ there can be at worst $D$ worth of distinct other homotopy classes of self-maps $RP^n\to RP^n$ homotopic to it as maps to $CP^{\infty}.$ Since a map homotopic to $*$ cannot be an automorphism (it would have to induce the trivial map on $H^1$ which cannot come from an automorphism), we can restrict our attention to maps homotopic in $CP^\infty$ to $id:RP^n\to RP^n.$ A priori, there could have been elements of $D$ which are not realizable as "defects" of homotopies between maps $RP^n\to RP^n$, but in this case we don't run into this problem: indeed, every element of $D$ occurs as the defect of some homotopy $RP^n\times [0,1]\to RP^{n+1}$ between the identity $id:RP^n\to RP^n$ and another map. Namely, recall that we have realized $D \cong H^n(RP^n)$ as a cyclic group, either $\mathbb{Z}$ or $\mathbb{Z}/2$ (depending on parity). Let $\alpha$ be a generator of this group. Then every element of $D$ can be written $k\alpha$ for some $k\in \mathbb{Z}$. By doing a calculation, you can see that each element $k\alpha\in D$ is realized as the defect of the homotopy $$[0,1]\cdot RP^n\to RP^{n+1}$$ induced by the map $[0,1]\times S^n\to S^{n+1}$ given by rotating $S^n$ in a circle around some ($n-1$-dimensional) axis inside $S^{n+1},$ by an angle of $$k\cdot \pi.$$ Now if $n$ is even, we see the resulting "new" map $RP^n\to RP^n$ is once again the identity. If $n$ is odd, the new map is induced from the "reflection" map given by $$\sigma:(x_1,x_2,\dots, x_n)\mapsto (-x_1,x_2, \dots, x_n)$$ (in some coordinates). Thus from what we've done so far, there can be at most two homotopy invertible self-maps up to homotopy $$RP^n\to RP^n$$ for any $n$, namely $id$ and $\sigma.$ It remains to check whether the induced two self-maps $RP^n\to RP^n$ are homotopic to each other. When $n$ is odd, they cannot be homotopic to each other since $RP^n$ is orientable, and $\sigma$ reverses orientation (so $\sigma$ can be distinguished from $id$ by looking at action on $H^n$). But when $n$ is even, the map $\sigma$ is homotopy equivalent via sphere rotations to the map $(x_1,x_2\dots, x_n)\mapsto (-x_1,-x_2\dots, -x_n),$ and this induces a homotopy between $\sigma$ and $id$ as maps $RP^n\to RP^n.$ Thus we have $$\pi_0(\mathrm{Aut}(RP^n)) \cong \begin{cases} \{id\}, & n \text{ even}\\ \{id, \sigma\}, & n \text{ odd}. \end{cases} $$ As mentioned, $RP^n$-bundles on $S^1$ are classified by the same data, so your guess is correct.

$\endgroup$
10
  • 3
    $\begingroup$ You are apparently classifying up to fibre homotopy equivalence. As Tom Goodwillie points out, the classification up to diffeomorphism is different. $\endgroup$ Jan 23, 2022 at 13:43
  • $\begingroup$ Right. Tom Goodwillie's answer came while I was writing mine, and I added a disclaimer. $\endgroup$ Jan 23, 2022 at 13:44
  • $\begingroup$ And the MSE question had "diffeomorphism" in the title. $\endgroup$ Jan 23, 2022 at 13:58
  • 1
    $\begingroup$ As I say it is Corollary 6 in "Coverings of fibrations" by Becker and Gottlieb. $\endgroup$ Jan 23, 2022 at 14:42
  • 1
    $\begingroup$ Three different questions are possible: classification of smooth fiber bundles up to fiber-preserving diffeomorphism, classification of topological fiber bundles up to fiber-preserving homeomorphism, classification of fibrations up to fiber homotopy equivalence. I addressed the first, Vaintrob addressed the third. $\endgroup$ Jan 23, 2022 at 18:11
8
$\begingroup$

No. Every smooth bundle over $S^1$ with fiber $M$ is the mapping torus of some diffeomorphism $f:M\to M$. Isomorphism classes of bundles correspond to conjugacy classes in the group of isotopy classes of diffeomorphisms.

There are already surprises when $M$ is $S^n$. For example, an exotic $7$-sphere can be realized as the union of two copies of $D^7$ glued by a diffeomorphism of $S^6$ that is not isotopic to the identity, and this leads to examples.

There are related examples with $M=P^n$.

$\endgroup$
9
  • 1
    $\begingroup$ For r=11mod16 the quotient of $\pi_0 Diff^+(RP^r)$ modulo concordance has been computed in Wells:The concordance group of real projective space, jstor.org/stable/1996838?seq=19#metadata_info_tab_contents $\endgroup$
    – ThiKu
    Jan 23, 2022 at 13:49
  • $\begingroup$ Or at least it has been proven to equal something else. $\endgroup$
    – ThiKu
    Jan 23, 2022 at 13:51
  • $\begingroup$ I am a little uneducated about exotic smooth structures. Although I am interested in both perspectives, I think the answer by Dmitry Vaintrob about classification up to homeomorphism is more what I am looking for, so I have accepted it. You are right that the MSE question had diffeomorphism in the title. Sorry if that caused any confusion. And thanks for the answer, maybe this will motivate me to look into exotic smooth structures more! $\endgroup$ Jan 23, 2022 at 14:22
  • 1
    $\begingroup$ @IanGershonTeixeira: perhaps your confusion comes from ignoring the structure group. One can classify up to fiber homotopy equivalence (as in Vaintrob's answer), fiber preserving homeomorphism, fiber preserving diffeomorphism (as in Goodwillie's answer), and also one can look at linear $S^n$ bundles over the circle, which have the structure group $O(n+1)$. Since $\pi_1(BO(n+1))\cong \pi_0(O(n+1))\cong\mathbb Z_2$ there are two linear $S^n$ bundles over $S^1$. $\endgroup$ Jan 23, 2022 at 14:35
  • 2
    $\begingroup$ @IanGershonTeixeira: yes, there is a non-orientable circle bundle over $RP^2$ whose total space is diffeomorphic to $RP^3\# RP^3$. See e.g. p.459 in Scott's "The geometries of 3-manifolds", homepages.warwick.ac.uk/~masgar/Teach/2021_3MFDS/References/…. It is probably not hard to describe the bundle explicitly. $\endgroup$ Jan 23, 2022 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.