Is the principal fiber bundle $GL^+(6,\mathbb R)$ over $GL^+(6,\mathbb R)/SL(3,\mathbb C)$ trivial ?
1 Answer
No. One can reduce to the maximal compact subgroups. So the question is, if $U(3)\rightarrow SO(6)\rightarrow SO(6)/U(3)$ is trivial. But $SO(6)/U(3)\cong \mathbb{C}P^3$ (this can be seen by using $Spin(6)\cong SU(4)$, for instance). But $\pi_2(U(3)\times \mathbb{C}P^3) = \pi_2(\mathbb{C}P^3) \cong \mathbb{Z}$ and $\pi_2(SO(6)) \cong \pi_2(SO(3))\cong \pi_2(S^3) = 0$.

$\begingroup$ Only a small question: why is $SO(3)/U(3)$ simply connected? If the inclusion of $U(3)$ lifts to $Spin(6)\cong SU(4)$ as you suggest, then $\pi_1(SO(6))$ should survive in the quotient. $\endgroup$ Nov 10, 2015 at 16:55

$\begingroup$ The double cover of the inclusion $U(3)\rightarrow SO(6)$ is the inclusion $U(3)\rightarrow SU(4)$ and a little diagram chase shows that $\pi_1(U(3))\rightarrow \pi_1(SO(6))$ is surjective. $\endgroup$ Nov 10, 2015 at 18:44