Euler class of a principal $SO(2)$-bundle over a lens space

Question

Note that for a manifold $X$, isomorphism classes of principal $SO(2)$-bundles over $X$ are classified by their Euler classes in $H^2(Z;\Bbb Z)$. Now consider a lens space $L(p,q)$; we have $H^2(L(p,q);\Bbb Z)=\Bbb Z_p$. Suppose there is a principal $SO(2)$-bundle (or equivalently a principal $S^1$-bundle) $P\to L(p,q)$ such that $P$ is homeomorphic to $S^3\times S^1$. Then should the Euler class of the bundle be a unit (i.e. a generator) of $H^2(L(p,q);\Bbb Z)=\Bbb Z_p$? This seems true, according to p.338 (second paragraph) of this paper: https://bpb-us-e2.wpmucdn.com/faculty.sites.uci.edu/dist/3/246/files/2011/03/23_PseudoFreeOrbifolds.pdf, but I can't see why.

Answer 1

We may construct principal $S^1$-bundles over $L(p,q)$ as follows.

We have a $\Bbb Z_p$ action on $S^3\times S^1$ generated by $((z_1,z_2),z)\mapsto ((e^{i2\pi/p}z_1,e^{i2\pi q/p}z_2),e^{i2\pi k/p}z)$, where $1\le k\le p-1$. The projection $S^3\times S^1\to S^3$ induces $\tilde{p}:(S^3\times S^1)/\Bbb Z_p\to S^3/\Bbb Z_p=L(p,q)$ by passing to the orbit space, which is a $S^1$-bundle. This bundle is principal, being the unit sphere bundle of a complex line bundle $(S^3\times \Bbb C)/\Bbb Z_p\to L(p,q)$. To emphasize the importance of $k$ in this construction, we will use the notation $S^3\times_{\Bbb Z_p}\Bbb C(k)$ and $S^3\times_{\Bbb Z_p}S^1(k)$ to denote the total space of the complex line bundle and its unit sphere bundle respectively.

We now investigate the total space.

• If $\gcd(k,p)=d\neq 1$, then $S^3\times_{\Bbb Z_p}S^1(k)\not\cong S^3\times S^1$. Consider the subgroup $\Bbb Z_d$ of $\Bbb Z_p$ acting on $S^3\times S^1$ by $$((z_2,z_2),z_2)\mapsto ((e^{i2\pi(p/d)/p}z_1,e^{2\pi q(p/d)/p}z_2),e^{i2\pi k(p/d)/p}z)=((e^{i2\pi/d}z_1,e^{i2\pi q/d}z_2),z)$$ This action is trivial on $S^1$ factor, so we deduce that the quotient space $L(d,q)\times S^1$ is a $p/d$-sheeted cover of $S^3\times_{\Bbb Z_p}S^1(k)$. Hence, $\pi_1(S^3\times_{\Bbb Z_p}S^1(k))$ contains a subgroup isomorphic to $\Bbb Z\times \Bbb Z_d$, so $S^3\times_{\Bbb Z_p}S^1(k)\not\cong S^3\times S^1$.
• If $\gcd(k,p)=1$, then $S^3\times_{\Bbb Z_p}S^1(k)$ is essentially the mapping torus of a rotation of $S^3$. Since this is orientation preserving, we have a homotopy between this self-homeomorphism and the identity through homeomorphisms (given by a path in $SO(4)$ connecting the identity and this rotation). Therefore, $S^3\times_{\Bbb Z_p}S^1(k)$ is homeomorphic to the mapping torus of the identity on $S^3$, which is $S^3\times S^1$. A reference of a general result about homeomorphic mapping tori can be found here.

We note that the complex line bundle $S^3\times_{\Bbb Z_p}\Bbb C(k)\to L(p,q)$ is the $k$-fold tensor product of $\xi:S^3\times_{\Bbb Z_p}\Bbb C(1)\to L(p,q)$. We can use Gysin sequence to show that $c_1(\xi)=e(\xi)$ generates $H^2(L(p,q);\Bbb Z)$, so $e(\xi^{\otimes k})=c_1(\xi^{\otimes k})=kc_1(\xi)=ke(\xi)$. So in fact, we get all possible isomorphism classes of principal $S^1$-bundles over $L(p,q)$ using the construction above.

So, if the total space $P$ is homeomorphic to $S^3\times S^1$, then up to isomorphism we are in the situation of $\gcd(k,p)=1$, so the Euler class is $ke(\xi)$ which is a generator of $H^2$.