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 $SU(n)$ ?

  • $\begingroup$ 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 $SO(2n)$. $\endgroup$ Apr 26, 2023 at 16:33
  • 1
    $\begingroup$ @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 $SO(2n)$ $\endgroup$
    – ychemama
    Apr 27, 2023 at 8:53
  • $\begingroup$ Thanks for clarification. $\endgroup$ Apr 27, 2023 at 18:48

1 Answer 1


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.

  • $\begingroup$ 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 $SO(\mathfrak{g}/\mathfrak{h},g)$ $\endgroup$
    – ychemama
    Apr 27, 2023 at 9:48

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