Is the principal fiber bundle $GL^+(6,\mathbb R)$ over $GL^+(6,\mathbb R)/SL(3,\mathbb C)$ trivial ?


1 Answer 1


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. $\endgroup$
    – user_11437
    Nov 10, 2015 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.