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Given a compact group $G$ acting freely on a manifold $X$, is there a "nice" way to describe the tangent bundle of the quotient $X/G$ (when it is a manifold)?

In the case the group $G$ is finite, or more generally when its action is properly discontinuous, the projection $p : X \to X/G$ is a local diffeomorphism, and therefore the tangent space $T_xX$ is isomorphic to $T_{p(x)}(X/G)$ via $d_xp$. In fact, I can prove that $T(X/G) \simeq (TX)/G$, for a suitable action of $G$ on $TX$.

My question here is motivated by trying to understand the tangent space of $\mathbb CP^n$. It can be seen as $S^{2n+1}/U(1)$, but here, the action of $U(1)$ is not properly discontinuous, and everything breaks down. $U(1)$ acts on $TS^{2n+1} \simeq \{ (x,v) : x,v \in \mathbb{C}^{n+1} \|x\| = 1, x \bot v \}$ (orthogonality is for the real inner product on $\mathbb{R}^{2n+2} = \mathbb C^{n+1}$) by multiplication on both factors. The quotient is indeed a vector bundle on $\mathbb CP^n$, but it's not the tangent bundle: it doesn't even have the correct dimension.

More specifically, I'm trying to prove that for a complex line $D \subset \mathbb C^{n+1}$, the tangent space of $\mathbb CP^n$ at $D$ is isomorphic to $\mathrm{Hom}(D, D^\bot)$. This is "proven" in "The Topology of Fiber Bundles: Lecture Notes" by Ralph L. Cohen (found online) in section 2.2, but the author merely says that the result is "proved in the same way" as in the real case; but in the real case, $\{\pm 1\}$ acts properly discontinuously on the sphere, and the projection is a local diffeomorphism. This isn't true in the complex case.

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  • $\begingroup$ Look at the space not as a sphere modulo $U(1)$ but as $\mathbb C^n\setminus0$ modulo $\mathbb C^\times$; then the construction does work the same for the reals and the complex numbers. $\endgroup$ Apr 14 '13 at 9:53
  • $\begingroup$ I tried doing that, and I find that the tangent space at $D$ is $\{(x,v) : x \in D \setminus \{0\}, v \in \mathbb C^{n+1} \}$ modulo $(x,v) = (\lambda x, \lambda v)$. This has (complex) dimension $n+1$ (for a fixed $y \in D, y \neq 0$, then it's isomorphic to $\{y\} \times \mathbb C^{n+1}$), which isn't as expected. Where do I go wrong? $\endgroup$ Apr 14 '13 at 10:24
  • $\begingroup$ @Mariano: Also, why would it work for one construction and not the other? What's different about them? $\endgroup$ Apr 14 '13 at 10:41
  • $\begingroup$ Well, it seems to bother you that one group is discrete and the other isn't, so his is just a way to get them both non-discrete! :-) $\endgroup$ Apr 14 '13 at 14:37
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I think the best answer for your general question comes from observing that $\pi:X\to X/G$ is a surjective submersion. Hence for each $x\in X$, you can identify the tangent space $T_{\pi(x)}X/G$ with the quotient of $T_xX$ by the tangent space of the orbit through $x$. The latter is the subspace of $T_xX$ spanned by the fundamental vector fields for the action associated to elements in the Lie algebra $\mathfrak g$ of $G$. Here for $A\in\mathfrak g$, the fundamental vecor field is defined by $\zeta_A(x)=\frac{d}{dt}|_{t=0}x\cdot exp(tA)$. For $\mathbb CP^n$ realized as a quotient of $S^{2n+1}$, this gives you and identification of the tangent space to the complex line spanned by $x$ as the quotient of $x^\perp$ (real orthocomplement) by imaginary multiples of $x$. This is the right space, but not quite the identification that you would like to get.

To get an identification like the one you want, you probably have to realize $\mathbb CP^n$ as a homogeneous space (since this allows you to compare tangent spaces at different points to some extent). The most general version of this is viewing it as a homogeneous space of $GL(n,\mathbb C)$. Then for each line $D\subset\Bbb C^{n+1}$, the map $A\mapsto A(D)$ defines a surjective submersion from $GL(n,\Bbb C)$ onto $\mathbb CP^n$. Hence you get an identification of $T_D\mathbb CP^n$ with the quotient of the tangent space of $GL(n,\mathbb C)$ (which is just the space of complex $n\times n$-matrices) by the Lie algebra of the stabilizer of $D$, which is simply formed by all matrices mapping $D$ to itself. This quotient can be identified with the space of linear maps from $D$ to $\mathbb C^{n+1}/D$, and this is the identification you want. (If you prefer to involve a complex orthocomplement, you can set up a similar picture with $U(n)$ or $SU(n)$ acting on the space of lines.)

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