How can $SO(4) \cong SO(3) \times S^3$ if the fiber bundle $SO(3) \rightarrow SO(4) \rightarrow S^3$ does not have a global section?

Question

I am under the impression that if I had a product $X = A \times B$ the induced fibration $A \rightarrow X \rightarrow B$ would have a global section, namely $\sigma: B \rightarrow X$ taking $b$ to $(b, a_0)$ for any $a_0 \in A$. Furthermore, I have seen (e.g. here) that $SO(4) \cong SO(3) \times S^3$ as topological spaces. However, it seems that the fiber bundle $SO(3) \rightarrow SO(4) \rightarrow S^3$ doesn't have a global section (see e.g. here; the second answer shows that there are local sections, which is sort of where I got this impression from).

I am not sure what to make of this contradiction. My best guess at what's going on is that the "canonical" fiber bundle in the question linked above doesn't have a section, though one could write down a separate, perhaps less natural bundle $SO(3) \rightarrow SO(4) \rightarrow S^3$ that does have a section. In this case, what is this bundle, and what is the section? I am interested in using a global section to show that $SO(4)$ is homeomorphic to $SO(3) \times S^3$.

EDIT: It just occurs to me that perhaps in the general case the fiber bundle $SO(n) \rightarrow SO(n+1) \rightarrow S^n$ does not have a section (which would be consistent with the fact that in general the product does not seem to hold), but when $n = 3$ it happens to. In this case, can someone write down an explicit section $S^3 \rightarrow SO(4)$?

Answer 1

The fact that there is such a diffeomorphism of manifolds would not contradict the non-existence of a section. The diffeomorphism needn't preserve the projection map in any sense. Your first paragraph is simply false. I provide an example in a post-script.

However, yes, this map has a section. Consider $S^3$ as the unit quaternions sitting in $\Bbb H = \Bbb R^4$. The map $L: S^3 \to SO(4)$ is the map $\Bbb H \to \Bbb H$ given as $L_z(w) = zw$.

I assume your projection is the map $\pi(A) = A(1)$, writing $1 \in S^3$ for the unit. Then $\pi(L_z) = L_z(1) = z*1 = z$, so $L_z$ is a section.

The octonions give a section $S^7 \to SO(8)$ in a similar manner, though unlike $S^3$, the image of $S^7$ is not a subgroup --- it in fact generates $SO(8)$.

Remark. These bundles $SO(n) \to SO(n+1) \to S^n$ are known as the frame bundle of $S^n$; its fibers over $x$ is the space of orthonormal bases of $T_x S^n$. This bundle has a global section if and only if $TS^n$ is trivializable; the global section is a choice of trivialization. So the above remarks (applied to $\Bbb R, \Bbb C, \Bbb H, \Bbb O$) show that $S^0, S^1, S^3, S^7$ have trivializable tangent bundle. It is in fact a remarkable theorem of algebraic topology that this is the full list.

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As example, first let me talk about abelian groups. Set $G = \oplus_{n \geq 0} \oplus_{m \geq 0} \Bbb Z/2^m$, an infinite copy of each cyclic 2-group and let $H = \oplus_{n \geq 0} \Bbb Z/2$ I will label an element in the $n$th copy of something as $(x,n)$. Observe that the map $f: G \to G$ which sends $(x,n) \in \Bbb Z/2^m$ to $(2x, n) \in \Bbb Z/2^{m+1}$ has cokernel isomorphic to $H$ via the obvious projection $p$; of course this sequence does not split (if the direct sum of sequences splits so do the individual sequences, and none of the component sequences here split). Further observe that there is an isomorphism $q: G \oplus H \to G$, basically doing a little juggling with a bijection $\Bbb N \times \{0,1\} \to \Bbb N$. Then the sequence $$0 \to G \xrightarrow{q^{-1}f} G \oplus H \xrightarrow{gq} H \to 0$$ does not split.

Now apply the classifying space functor to obtain a fiber bundle $BG \to BG \times BH \to BH$ which does not admit a global section.