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
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).
Best Answer
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.