# Frame bundle of $\mathbb{C}P^n$ as homogeneous space

I am reading "Dirac Operator in Riemannian Geometry" by T. Friedrich, where he writes that (the total space of) the frame bundle $$R$$ of the tangent space of $$\mathbb{C}P^n$$ 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 does this expression for $$R$$ come from ?

Locally this frame bundle is a $$SO(2n)$$-principal bundle (a $$GL(2n)$$-principal bundle reduced to $$SO(2n)$$ from metric and orientability) isomorphic to $$\mathbb{C}P^n \times SO(2n) \simeq SU(n+1)/S(U(n) \times U(1))\ \times \ SO(2n).$$

Quotienting via the action of $$S(U(n) \times U(1)$$ on $$SU(n+1)$$ on the right (right multiplication) and on the left on $$SO(2n)$$ via $$\sigma(g)^{-1}$$ acting by conjugaison makes it global ? Is it the correct meaning of $$\times_{\sigma}$$ ?

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 $$Ad(h^{-1}$$ of $$H$$, and it is probably a generic formula for homogeneous space but I can't figure it out.

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 $$SU(n)$$ ?

• One thing confuses me. Let's fix a notation for the inclusion $\iota: GL(n, \mathbb{C}) \to GL(2n, \mathbb{R})$. We have in general $\det \iota(A) = |\det A|^2.$ So for the image of your map we get $\det [\det(B)\iota(B)] = \det(B)^{2n}\det\iota(B) = \det(B)^{2n}|\det(B)|^2 = \det(B)^{2n}$. So the image is not inside $SO(2n)$. Apr 26, 2023 at 16:33
• @Vit : This expression for sigma from Friedrich's book seems correct to me : it doesn't say that $(det B) B \in SO(2n)$, but that $(det B) B \in U(n)$ when $B \in U(n)$, and you can inject $U(n)$ in $SO(2n)$ Apr 27, 2023 at 8:53
• Thanks for clarification. Apr 27, 2023 at 18:48

There's a more general description of frame bundles on homogeneous spaces here: if you take $$G/H$$ and give it a $$G$$-invariant Riemannian metric, then $$H$$ preserves the identity coset, and so acts on the tangent space there $$T_{eH}G/H=\mathfrak{g}/\mathfrak{h}$$, preserving the metric $$g$$; actually this is induced by the adjoint action. Thus, we have a map $$H\to SO(\mathfrak{g}/\mathfrak{h},g)$$ and the frame bundle will always be isomorphic to $$G\times^HSO(\mathfrak{g}/\mathfrak{h},g)$$, where $$\times^H$$ is the balanced product given by $$G\times SO(\mathfrak{g}/\mathfrak{h},g)$$ modulo the $$H$$ action by $$h\cdot (g,v)=(gh^{-1},hv)$$.
The homomorphism $$\sigma$$ is just an explicit form of the action on this quotient of Lie algebras. The product $$(\det B)B$$ can be thought of this way: the unit coset is identified with the line $$(0,...,0,*)$$. We can identify the tangent space with $$\mathbb{C}^{n-1}$$ by sending $$(a_1,\dots, a_{n-1})$$ to the velocity vector of $$(a_1t,\dots,a_{n-1}t,1)$$. Thus, acting by $$B$$ sends this to $$(tB\mathbf{a},(\det B)^{-1})$$. To rewrite this as desired, we have to multiply through by $$\det B$$, since in the $$\mathbb{CP}^{n-1}$$, this is the same as $$(t(\det B)B\mathbf{a},1)$$. This is a coordinatish way of saying that the tangent bundle on $$\mathbb{CP}^{n-1}$$ is $$\mathrm{Hom}(L,\mathbb{C}^n/L)$$ where $$L$$ is a the tautological line bundle.
• OK, I have overlooked that you called the $G$-invariant metric $g$ (so not an element of $G$ as I had understood !). So OK for $SO(\mathfrak{g}/\mathfrak{h},g)$ Apr 27, 2023 at 9:48