Let $\Sigma$ be a two-dimensional connected smooth manifold without boundary. (Feel free to assume it is compact and orientable.)


Let $\mathcal{X}(\Sigma)$ denote the smooth vector fields on $\Sigma$ and let $\operatorname{Diff}(\Sigma)$ denote the group of diffeomorphisms of $\Sigma$. Clearly $\operatorname{Diff}(\Sigma)$ acts on $\mathcal{X}(\Sigma)$.


I would expect naively that the space of orbits $\mathcal{M}:=\mathcal{X}(\Sigma)/\operatorname{Diff}(\Sigma)$ would be finite-dimensional. Is this actually the case?


**Question**


What can one say about $\mathcal{M}$ in general?


Any references where I could read about this question would be appreciated.



Smooth vector fields on a surface modulo diffeomorphisms
This is not finite dimensional. For example, consider non-vanishing vector fields on $T^2=\mathbb R^2/\mathbb Z^2$, transversal to vertical circles. Any such field defines a self diffeo $S^1\to S^1$ on a vertical circle, called the return map. Suppose that such a diffeo $\varphi$ has $n$ fixed points $x\_i$ (they correspond to closed orbits of the field). Then for each fixed point $x\_i\in S^1$ we have a linear map on the tangent space $d\varphi: T\_{x\_i}S^1\to T\_{x\_i}S^1$ given by multiplication by $\alpha\_i\in \mathbb R^\*$. Such $\alpha\_i$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $\alpha\_i$'s are independent, we see that the dimension is not finite.


If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $1$-forms. Now the spaces of *minimal* closed $1$-forms on a closed genus $g$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $1$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)


