There are many circle bundles over the sphere $ S^2 $ (in fact infinitely many) but all of them are principal.

Do there exist any other manifolds besides $ S^2 $ for which (nontrivial bundles exist and) all circle bundle are principal?

I believe that $ S^2 $ is the only surface with this property.

For example $ \mathbb{R}P^2 $ admits infinitely many non principal circle bundles, as does the Klein bottle $ K^2 $. And the torus $ T^2 $ admits two different non principal circle bundles. And the plane $ \mathbb{R}^2 $ is contractible so all circle bundles are principal but also all circle bundles are trivial (indeed it is true for all $ n $ that all circle bundles over $ \mathbb{R}^n $ are trivial, in fact even all fiber bundles are trivial see A fiber bundle over Euclidean space is trivial.).

Also note that the Klein bottle $ K^2 $ is a non principal circle bundle over $ S^1 $. So the circle does not have this property.

Higher spheres $ S^n $ for $ n \geq 3 $ are other examples where this property fails, since every circle bundle is trivial, this follows from the fact that $ \pi_{n-1}(S^1)=0 $ for $ n\geq 3 $ see for example https://math.stackexchange.com/a/269761/758507.


1 Answer 1


First, you didn't state the structure group of the bundle, but $Diff(S^1)$ deformation retracs to $O(2)\subseteq Diff(S^1)$, so we can assume that you mean $O(2)$. Said another way, all $S^1$-bundles over a manifold are the unit circle bundles in a rank $2$ vector bundle.

Also, for ease of writing, I'll write $G$ for the group $\mathbb{Z}/2\mathbb{Z}$.

Claim 1: For a manifold $M$, all circle bundles over $M$ are principal iff $H^1(M;G) = 0$.

Proof: Assume first that $H^1(M;G)$ is trivial. And choose a circle bundle $E\rightarrow M$. As mentioned above, we may assume the bundle is linear, so we can talk about characteristic classes. By assumption, $w_1(E)\in H^1(M;G) = 0$, so the bundle is orientable. This mean the structure group is actually $Diff^+(S^1)$, which deformation retracts to $SO(2)$. But this, then, means the bundle is principal.

Conversely, assume that $0\neq x \in H^1(M;G)$. Since $\mathbb{R}P^\infty$ is a $K(G,1)$, there must be a map $\phi:M\rightarrow \mathbb{R}P^\infty$ for which $\phi^\ast:H^1(\mathbb{R}P^\infty;G)\rightarrow H^1(M;G)$ has image $x$. Pulling back the tautological real rank $1$ bundle over $\mathbb{R}P^\infty$ along $\phi$, we obtain a rank $1$ vector bundle $E$ over $M$ with $w_1(E) = x\neq 0$.

Now, consider the vector bundle $E\oplus 1$ where $1$ denotes a trivial rank $1$-bundle. By the Whitney sum formula, $w_1(E\oplus 1) = w_1(E) = x$. Finally, the unit sphere bundle in $E\oplus 1$ gives a circle bundle over $M$ with $w_1$ non-trivial. Hence this bundle is not orientable and thus not principal. $\square$.

From claim 1, it follows that among closed surfaces, $S^2$ is the only one for which all circle bundles are principal.

Claim 2: For a manifold $M$, there is a non-trivial circle bundle over $M$ iff $H^2(M;\mathbb{Z})\neq 0$.

Proof: Such bundles are classified by their Euler class, an element of $H^2(M)$. In more detail, any element $x\in H^2(M)$ corresponds to a map to $K(\mathbb{Z},2) = \mathbb{C}P^\infty$. Pulling back the tautological rank 1 complex vector bundle over $\mathbb{C}P^\infty$ gives a circle bundle over $M$ with Euler class $x$. Moreover, all principal $S^1$ bundles arise from this construction since $BS^1 = \mathbb{C}P^\infty$. $\square$

Combining claim 1 and claim 2, $M$ meets your requirements iff $H^1(M;G) = 0$ and $H^2(M;\mathbb{Z})\neq 0$.

Now getting examples is easy. The following are all examples of what you want:

  1. If $H^1(N;G) = 0$, then $N\times S^2$ is an example. More generally, if $M$ is any example of what you want, then $M\times N$ is an example.
  2. $\mathbb{C}P^n$ for every $n$
  3. Any non-abelian connected compact Lie group mod its maximal torus.
  4. Any simply connected closed $4$-manifold except $S^4$.
  5. If $M$ has the property and $H^1(N;G) = 0$, and $\dim M = \dim N \geq 4$, then $M\sharp N$ has the property.
  6. Any lens space $L_{p,q}$ with $p$ odd.

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