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.

  • $\begingroup$ 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

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.)


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.