It is well known that $SU(3)$ is the unique, non-trivial, principal $SU(2)$- bundle over $S^5$. To my knowledge the way this is proven is by using the following fact:

If $G$ is a Lie group and $H$ a closed subgroup, then $G \to G/H$ is a (locally trivial) principal $H$-bundle.

In our case, $SU(3)$ acts transitively on $S^5 \subseteq \mathbb C^3$, the stabilizer of any point is $SU(2)$, and we use orbit-stabilizer to conclude the desired property. The remaining part is a simple computation of

$$[S^5,BSU(2)] = \pi_5(BSU(2)) = \pi_4(\Omega BSU(2)) = \pi_4(SU(2)) = \pi_4(S^3) = \mathbb Z/2\mathbb Z.$$
And $SU(3)$ is not the trivial one since $\pi_4(SU(3)) \neq \pi_4(SU(2))$.

Unfortunately, I am interested in writing down an explicit trivialization for which this method does not seem to provide any insight. In particular, I would like to trivialize $SU(2) \hookrightarrow SU(3) \rightarrow S^5$ over two open sets given by removing distinct points. More precisely, if $N_\pm=(\pm1,0,0) \subseteq \mathbb C^3$, let's trivialize over $U_\pm = S^5 \setminus\{N_\pm\}$.

**Attempt One:** The naive procedure to begin with is to view $SU(3)$ as the collection of orthonormal bases $\{(e_1,e_2,e_3)\}$ for $\mathbb C^3$. Let us then take the projection $SU(3) \to S^5$ to be $p(e_1,e_2,e_3) = e_1$. The problem is then to map $(e_2,e_3)$ to some element in $SU(2)$ given that $e_1 \neq N_\pm$. I cannot see what that map should be though.

**Attempt Two** My second attempt is to emulate the solution to this overflow post, which would have me define the set $X$ as the set of pairs $(v,A)$ where $v \in S^5 \subseteq \mathbb C^3$ and $A \in (v^\perp)^1 \cong S^3 \cong SU(2)$, the unit sphere in $v^\perp$. This is certainly a principal $SU(2)$-bundle, and while I have yet to show it is non-trivial I believe it is as such. As the non-trivial $SU(2)$-bundle is unique up to isomorphism, there is an $SU(2)$-equivariant diffeomorphism $X \to SU(3)$, but I am not certain as how to find such an isomorphism. Any suggestions are appreciated.

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