$S^{1}$-bundles over $\mathbb{RP}^2$

Question

How many $S^1$-bundles over $\mathbb{RP}^2$ do exist? Is it true that there exist only two bundles - trivial and not?

Answer 1

If we restrict ourself to principal $S^1$ -bundles and denote the isomorphism classes of principal $S^1$-bundles over a space $X$ with $PCB(X)$, we get the following:

$$PCB(X)\cong[X,BS^1]\cong[X,K(2,\mathbb{Z})]\cong H^2(X;\mathbb{Z})$$

In the case of $\mathbb{R}P^2$ we have $H^2(\mathbb{R}P^2;\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$, so there are indeed - up two isomorphism - exactly two principal circle bundles over the real projective plane.

Answer 2

The projectivised tangent bundle $P(T\mathbb{R}P^2)$ is a non-principal circle bundle over $\mathbb{R}P^2$. To see this note that when restricted to a "circle at infinity" in the base (i.e. a homotopically non-trivial circle), this bundle becomes a Klein bottle which is not a principal bundle over $S^1$ (it admits a global section and yet is not trivial).

Answer 3

archipelago has already given a proof that there are two principal $S^1$-bundles over $\mathbb{R}P^2$. The non-trivial one is the mapping torus of the antipodal map of $S^2$. One way to see this is to consider the space $X:=S^1\times S^2/ (\mathbb{Z}/2\mathbb{Z})$ where $\mathbb{Z}/2\mathbb{Z}$ acts diagonally via the antipodal map on both factors. Projection onto the first factor shows $X$ is an $S^2$-bundle over $S^1$ which allows you to easily show that $\pi_1(X)\cong \mathbb{Z}$. On the other hand, projection onto the second factor gives $X$ the structure of an $S^1$-bundle over $\mathbb{R}P^2$. The $S^1$ action on $S^1\times S^2$ which is trivial on $S^2$ and by rotations on $S^1$ descends to $X$, so $X$ is a principal $S^1$-bundle over $\mathbb{R}P^2$. It is non-trivial because the trivial bundle has $\pi_1(S^1\times \mathbb{R}P^2)\cong \mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}.$

With that out of the way, for the rest of this post, I'll only consider non-principal $S^1$ bundles $\xi$ over $\mathbb{R}P^2$. Since $\operatorname{Diff}(S^1)$ deformation retracts to $O(2)$, it follow easily that this is equivalent to asking that the first Stiefel-Whitney class $w_1(\xi)\in H^1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$ be non-trivial.

Decompose $\mathbb{R}P^2 = M\cup D$ where $M$ is a Mobius band (a neighborhood of a homotopically non-trivial $S^1\subseteq \mathbb{R}P^2$ and $D$ is a disk. I'll write $S^1_c\subseteq M$ to represent the circle through the middle of $M$. So, $S^1_c$ generates $\pi_1(M)$, and via the inclusion $M\subseteq \mathbb{R}P^2$, $S^1_c$ also generates $\pi_1(\mathbb{R}P^2)$.

Because $D$ is contractible, any bundle over it is trivial.

On the other hand, $M$ deformation retracts onto $S^1_c$, and there are precisely two $S^1$-bundles over $S^1_c$ (with total spaces either $T^2$ and the Klein bottle $K$). Thus, there are precisely two $S^1$ bundles over $M$. Restricting $\xi$ to the core circle $S^1_c$, we must get a non-trivial bundle: this $S^1\subseteq M\subseteq \mathbb{R}P^2$ generates $\pi_1(\mathbb{R}P^2)$, so $w_1(\xi_{S^1_c})\neq 0$ since $w_1(\xi)\neq 0$ and the induced map $H^1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})\rightarrow H^1(S^1_c;\mathbb{Z}/2\mathbb{Z})$ is an isomorphism.

One fairly concrete way to think about this non-trivial $S^1$ bundle over $M$, is that it's obtained by pulling back the Klein bottle bundle $S^1\rightarrow K\rightarrow S^1_c$ via the projection map $M\rightarrow S^1_c$. Because this projection map is a homotopy equivalence, it follows that $\pi_1(\xi|_M)\cong \pi_1(K)$. Also, since the projection map is a double covering when restricted to $\partial M$, it follows that the bundle on the boundary is trivial (as it must be to glue to the boundary of $\xi|_D$).

From the above, it follows that we can form $\xi$ by gluing $\xi|_D \cong D\times S^1$ to $\xi|_M$ along their boundary $\partial M\cong \partial D\cong S^1$. The gluing is specified by a map $S^1\rightarrow O(2)$. Of course, the image must lie in $SO(2)$, and $\pi_1(SO(2))\cong \mathbb{Z}$, so there are a $\mathbb{Z}$s worth of possibilities.

Given $z\in \mathbb{Z}$, we can realize this gluing the map $\partial D\times S^1\rightarrow \partial M\times S^1$ given by $\alpha_z(\theta,\phi) = (\theta, z\theta + \phi)$. I'll call the resulting total space $E_z$.

Proposition: For $|z|\neq |z'|$, that $\xi_z$ is not isomorphic to $\xi_{z'}$.

Proof: We'll show that the total spaces $E_z$ and $E_{z'}$ have different fundamental groups.

To that end, we'll use Seifert-van Kampen with respect to the decomposition $E_z = E|_D\cup E|_M$.

Let's set up notation.

We'll write $$\pi_1(E|_M)\cong \langle a,b,| a bab^{-1}\rangle,$$ $$\pi_1(E|_{\partial M}) = \langle c,d| cdc^{-1}d^{-1}\rangle,$$ $$\pi_1(E_{D}) = \langle e\rangle,$$ and $$\pi_1(E_{\partial D})\langle f,g | fgf^{-1}g^{-1}\rangle.$$

Now, the inclusion $\xi_{\partial M}\rightarrow \xi_M$ is, under the homotopy equivalent $\xi_M\cong K$, the usual double cover $T^2\rightarrow K$. In paritcular, the induced map sends, say, $c$ to $a$ and $d$ to $b^2$. Having made this choice, it follows that we're thinking of $c$ as coming from the fiber $S^1$ and $d$ coming from the boundary $\partial M\cong S^1$.

Similarly, the inclusion $\xi_{\partial D}\rightarrow \xi_D$ sends, say, $f$ to $e$ and $g$ to $0$. This means we're thinking of $f$ as the fiber circle and $g$ as the boundary $\partial D\cong S^1$.

From the same trick I used in this similar MSE question, the map $\alpha_z$ sends $d$ to $g+zf$, and it sends $c$ to $f$.

Putting this all together via Seifert-van Kampen, we arrive at the following presentation of $\pi_1(E_z)$.

$$\pi_1(E_z) = \langle a,b,e|abab^{-1}, ae^{-1}, b^2(e^z)^{-1}\rangle$$ which is obviously equivalent to $$\langle a,b| abab^{-1}, b^2(a^z)^{-1}\rangle.$$

According to wikipedia, at least when $|z|\geq 2$, this is the presentation of a Prism Manifold, and $\pi_1(E_z)$ has order $4|z|$. Thus, for $|z|\neq |z'|$, the bundle $\xi_z$ are different, as their total spaces are not homotopy equivalent.

When $|z| = 1$, $a=b^2$ so we may rewrite the presentation as $\langle b| b^2bb^2b^{-1}\rangle$, so the order of $\pi_1(E_z) = 4|z|$ in this case as well. Then $z=0$, $\pi_1(E_0) = \langle ab, b| (ab)^2 , b^2\rangle$, so is the infinite dihedral group.$\square$

This leaves open one last question: what happens when $z = -z'$? Well, then $\pi_1(E_z)\cong \pi_1(E_{z'})$ and, according to that same wikipedia article, this implies that $E_z$ and $E_{z'}$ are diffeomorphic. I'm not sure if the bundles are bundle-isomorphic, however.