Generalizing a property of the two sphere

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.

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.