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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    $\begingroup$ No more so than "Complex simple Lie algebra" :) $\endgroup$ Oct 21 '13 at 22:30
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    $\begingroup$ I think the title should start with "Local". $\endgroup$ Oct 22 '13 at 0:05
  • $\begingroup$ Added "local" in the title. $\endgroup$
    – Qfwfq
    Oct 22 '13 at 9:44
  • $\begingroup$ I haven't worked out the details, but here's a start (I think): Let $U = S^5\setminus \{N_+\}$. Define $f:SU(2)\times U\rightarrow \pi^{-1}(U)$ first by mapping $(A, N_-)$ to the block diagonal matrix diag(-1, A'). where $A'$ denotes $A$ with rows swapped (so $diag(-1,A')$ has $det = 1$). Now, given any $p\in U$, there is a unique geodesic $\gamma$ connecting $N_-$ and $p$. If $P_\gamma(A')$ denote the parallel translation of $A'$ (thought of as a pair of orthonormal vectors in $T_{N_-} S^5$, then define $f(A,p)$ as the matrix with column 1 = p and column 2 and 3 = $P_\gamma(A')$. $\endgroup$ Oct 22 '13 at 14:33
  • $\begingroup$ @JasonDeVito Thanks for the reply. I had been playing around with something similar, minus the parallel transport. I'll see if I can't grind out some details. $\endgroup$ Oct 22 '13 at 17:43

The problem seems to be one of systematically extending $z=(z_1,z_2,z_3)\in S^5$ to an orthonormal basis of $\mathbb{C}^3$. To this end, let $e_1,e_2,e_3$ be the standard basis vectors for $\mathbb{C}^3$. For $j\in\{1,2,3\}$, let $$U_j:=\{z\in S^5:\{z,e_1,e_2,e_3\}\setminus\{e_j\}\text{ is a basis of }\mathbb{C}^3\}.$$ This is an open cover of $S^5$.

Taking determinants, we find that $U_j=\{z\in S^5:z_j\neq 0\}$. Perhaps we can trivialize your bundle over the three open sets $U_j$. For $z\in U_j$, I would suggest using Gram-Schmidt on the basis $\{z,e_1,e_2,e_3\}\setminus\{e_j\}$ to obtain $\varphi_j(z):=[z\text{ }f_{1j}(z)\text{ }f_{2j}(z)]\in SU(3)$, where $f_{1j},f_{2j}:U_j\rightarrow\mathbb{C}^3$. The sections $\varphi_j$ should be computable and they should allow you to trivialize the bundle over the open sets $U_j$.

I realize that I have not given local trivializations over the complements of the north and south poles. Still, it is some local trivialization.


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