Let $G$ be a real Lie group with a right action on a smooth manifold $X$. Assume the action is free and proper. This implies there is a unique manifold structure on the quotient space $G \backslash X$ such that the quotient map $\pi$ is a submersion, and that $\pi$ is a principal fiber bundle with $G$ is a fiber.

In other words, locally, the quotient map looks like $U \times G \rightarrow U, (u,g) \mapsto u$ for sufficiently small open sets $U \subset G \backslash X$, with $G$ acting by $(u,x).g = (u,xg)$.

Let $\omega \in \Omega^k(X \backslash G)$ be a smooth differential $k$-form on $X \backslash G$. Then $\omega$ pulls back to a differential $k$-form $\pi^{\ast}(\omega)$ on $X$ which is $G$-invariant.

1 . Is $\omega \mapsto \pi^{\ast}(\omega)$ an injective map $\Omega^k(G \backslash X) \rightarrow \Omega^k(X)$?

2 . What is the image of $\pi^{\ast}$? Does it consist of all $G$-invariant differential $k$-forms on $X$?

If $X$ is a trivial principal fiber bundle, i.e. $X = Y \times G$ for a smooth manifold $Y$, then all this seems obvious, but I'm not sure if there are some complications that can arise if $X$ is merely covered by such things.


$\omega\rightarrow \pi^*\omega$ is linear. Suppose that $\pi^*\omega=0$, since $\pi$ is a submersion, for every $x\in G/X, u_1,...,u_k\in T_x(G/X)$ and $y\in\pi^{-1}(x)$, there exists $v\in T_yX$ such that $d\pi_y(v_i)=u_i$, $\pi^*\omega_y(v_1,...,v_k)=\omega_x(u_1,...,u_k)=0$ implies that $\omega=0$ and $\omega\rightarrow \pi^*\omega$ is injective.

Consider $X\times G$ and take any non zero $1$-form $\beta$ invariant on $G$ by the right translations. Write $\alpha_{(x,g)}(u,v)=\beta_g(v)$ is a form invariant by $G$. You cannot have $\alpha=\pi^*\omega$ since $\pi^*\omega$ vanishes on the fibre.

  • $\begingroup$ Okay, thanks. So the injectivity is not a big deal. $\endgroup$ – D_S Jan 7 at 0:19
  • $\begingroup$ The images of $\pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field. $\endgroup$ – Andreas Cap Jan 7 at 11:30

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