Let $M$ be a smooth manifold with tangent bundle $TM$ and cotangent bundle $TM^*$ and $\psi\in TM^*$ a one-form. We denote the quotient manifold of $M$ by the free and proper $G$-action $\varphi$ as $\bar M=M/G$. The projection map $\pi:M\mapsto \bar M$ is a smooth submersion. I would like to know how to project the one-form $\psi$ on $\bar M$ i.e. define a one-form $\bar \psi\in T\bar M^*$ by projecting $\psi$ with $\pi$ and eventually to know the conditions $\psi$ has to satisfy in order for $\bar \psi$ to be well-defined. In the case of a vector field, the pushforward furnishes a natural way in order to define vectors at each point of $\bar M$ but as the pullback operates backwards, it doesn't seem suited to define one-forms from $M$ to $\bar M$. Does anyone has any suggestions or references ? Thank you very much.

Edit: More precisely, I'm interested in a manifold $M$ quotiented by the action defined by the flow of a given vector field so the group $G=(R,+)$ which is not discrete and the projection map $\pi:M\mapsto \bar M=M/G$ is not an immersion. I dispose of a one-form or covector field on $M$, $\psi:TM\mapsto C^\infty$ and would like to know the conditions on $\psi$ in order for $\psi$ to induce a well-defined one-form field $\bar \psi:T\bar M\mapsto C^\infty$ on the quotient manifold $\bar M$.

  • $\begingroup$ For reference: Take a look at Kobayashi-Numizu "Foundations of Differential Geometry", volume I, they discuss such matters in great detail. $\endgroup$ – Moishe Kohan Aug 8 '13 at 11:45
  • $\begingroup$ @studiosus Thanks for the reference, I did look into it but I can't seem to find the right passage. Do you know where it is discussed, by any chance ? $\endgroup$ – user88737 Aug 8 '13 at 14:45

I'm assuming by a projection you mean a $\tilde \psi$ that pulls back to $\psi$.

First let's consider the pointwise problem: project a single tangent covector $\psi_p \in TM^*$. I'll omit the explicit $p$ dependence of the linear maps $d\pi$ and $\pi^*$.

I guess this is the most obvious observation: The pullback $\pi^*$ is the adjoint of the differential, so it is injective because $d\pi$ is surjective and thus the projection is unique if it exists. Likewise, it is surjective when the differential is injective; so we have a necessary and sufficient condition for every $\psi_p \in TM^*$ to have a projection $\tilde\psi_{\pi (p)}$: we need $\pi$ to be an immersion. This happens when $G$ acts discontinuously.

If the action is not discontinuous, the $\psi_p$ that have projections are only those in the image of $\pi^*$. In particular I believe this means that if $G$ contains a one-parameter subgroup $\phi : \mathbb{R} \times M \to M$ with corresponding flow field $X_p = d/dt(\phi_t (p))$ then we need $\psi_p(X)=0$.

Now if we want to a project a field $\psi \in \Gamma (TM^*)$ we need the above conditions to be satisfied at each point; but this not sufficient: we could get different values of $\tilde \psi_{\pi(p)}$ by applying the argument above to $\psi_p$ and $\psi_{q}$ where $p,q$ share an orbit (so they will project to the same point in the quotient).

This problem arises in general, e.g. when projecting functions or vector fields. To resolve this we have to check that the field $\psi$ is preserved by the $G$-action; i.e. for every $g \in G$ we need $L_g^*\psi=\psi$ where $L_g:M \to M$ is the diffeomorphism $x \mapsto g \cdot x$.

  • $\begingroup$ @Georges: I assumed by the OP's notation $\psi \in TM^*$ that he meant a single cotangent vector rather than a field. $\endgroup$ – Anthony Carapetis Aug 8 '13 at 11:28
  • $\begingroup$ I did mean a field, sorry for the lack of clarity. $\endgroup$ – user88737 Aug 8 '13 at 11:55
  • $\begingroup$ Anthony: George's answer is now deleted, so you should add few more words to yours. $\endgroup$ – Moishe Kohan Aug 8 '13 at 12:17
  • $\begingroup$ @studiosus: Thanks, I revised my answer. $\endgroup$ – Anthony Carapetis Aug 8 '13 at 12:43

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