# Is there a nontrivial fiber or principal bundle over $S^3$?

Is there a nontrivial fiber or principal bundle over $S^3$?I know that, by a paper of Steenrod,see the link below, every sphere bundle on 3- sphere is trivial but what about arbitrary fiber bundle?

Triviality/non-triviality of line/circle bundle over $S^3$

Isomorphism classes of principal $G$-bundles over $S^3$ are classified by $\pi_3(BG)\cong \pi_2(G)$. So there is a nontrivial principal $G$-bundle over $S^3$ if and only if $G$ has non-vanishing second homotopy group.
As all Lie groups have vanishing second homotopy group $G$, any fiber bundle over $S^3$ with structure group a Lie group is trivial, in particular $S^3$ has no nontrivial vector bundles over it.
To get a nontrivial example, pick your favorite topological group $G$ with $\pi_2(G)\neq 0$ and a nontrivial element in that group. Pulling back the universal $G$-bundle gives you a principal $G$-bundle over $S^3$ which is nontrivial.
A good source of examples are diffeomorphism groups $Diff(M)$ of closed manifolds $M$, which are not finite dimensional manifolds (in particular Lie groups) in general, so may have nonvanishing second homotopy group. By the above argument $\pi_2(Diff(M))$ classifies principal $Diff(M)$-bundles and hence (via the associated bundle construction) also fiber bundles with fiber $M$ and structure group $Diff(M)$, i.e. smooth $M$-bundles over $S^3$.
As an example, Hatcher calculated the homotopy type of $Diff(S^1\times S^2)$: It is the one of $O(2)\times O(3)\times \Omega SO(3)$, so we arrive at $$\pi_2(O(2)\times O(3)\times \Omega SO(3))=\pi_2(\Omega SO(3))\cong\pi_3(SO(3))\cong\pi_3(S^3)\cong\mathbb Z.$$ Hence, smooth $S^1\times S^2$-bundles over $S^3$ are classified by the integers.
• +1 for a cool answer with great facts and examples! Couldn't we also use the clutching construction to say directly that the $F$ bundles over $S^3$ are classified by $\pi_2(F)$. So taking $F=S^2$ (or $F=S^2 \times S^1$ for your example) the smooth $S^2$ (or $S^1 \times S^2$) bundles are classified by integers $\pi_{3-1}(F) \cong \pi_2(S^2) \cong \mathbb{Z}$? Commented Apr 24, 2022 at 0:43
• @IanGershonTeixeira I do not think that will do! Clutching works for the associated principal bundle, which in the above example is a $\rm{Diff}(S^1 \times S^2)$ bundle. Hence, you do need the Hatcher's result to conclude the classification. Commented Apr 18, 2023 at 6:18