I am reading "Dirac Operator in Riemannian Geometry" by T. Friedrich. Trying to figure out when $\mathbb{C}P^n$ has spin structure, he write that (the total space of) the frame bundle $R$ of its tangent space is: $$ R = SU(n+1) \times_{\sigma} SO(2n) $$ where $SU(n+1)$ act transitively on $\mathbb{C}P^n \simeq SU(n+1)/S(U(n) \times U(1))$, with $S(U(n) \times U(1))$ the stabiliser of the point $[0:\dots:0:1]$, and $\sigma$ its isotropy representation defined for $B \in U(n)$ by : $$ \sigma \colon S(U(n)×U(1)) \to U(n) \subset SO(2n), \quad \begin{pmatrix} B & 0 \\ 0 & (det B)^{-1} \end{pmatrix} \mapsto (detB)\ B $$ This may be a stupid question but where this expression for $R$ commes from ? I don't understand it. If it where a trivial frame bundle (which it is not indeed) it $R$ should be the $SO(2n)$-principal bundle $(SU(n+1)/S(U(n) \times U(1))) \times SO(2n)$ which looks close...

It must have something to do with the fact that $\mathbb{C}P^n$ is a homogeneous space $G/H$ and its the tangent bundle is isomorphic to $G \times_H \mathfrak{g}/\mathfrak{h} $ where the action of $H$ on the left on $\mathfrak{g}/\mathfrak{h}$ is induced by the adjoint representation of $G$, but I can't explicit it.

And what is the meaning here of the product $\times_{\sigma}$ ? For me it should means for $g \in S(U(n)×U(1)$ the class $[Ag,\,\rho(g)^{-1}S]$ with $A \in SU(n+1)$ and $S \in SO(2n)$.

And by the way, why the factor $(det B)$ in front of the formula for $\sigma$ since $B$ is already in $U(n)$ ? Or is it $(detB)^{-1}\ B$ so that the image is in $SU(n)$ ?