The quotient manifold theorem says:
Let $G$ be a Lie group which atcs smoothly, freely and properly on a
smooth manifold $M$. Then the orbit space is a topological manifold, and has a unique smooth structure with the property that the quotient map $\pi: M \to M/G$ is a smooth submersion.
I wonder what happens if $M$ has an additional structure which $G$ respects. For example if $M$ is a complex manifold and $G$ acts holomorphically.
I suspect that then $M/G$ also has this property and $\pi$ respects it.
I read the proof of the quotient manifold theorem in Lee – Introduction to smooth manifolds (Theorem 21.10). What I need is covered in the econd to last paragraph:
Let $(U,\phi)$ and $(\tilde U, \tilde \phi)$ be two adapted charts for $M/G$. First consider the case in which the two adapted charts are both centered at the same point $p\in M$. Write the adapted coordinates as $(x,y)$ and $(\tilde x, \tilde y)$. THe fact that the coordinates are adapted to the $G$-action means that two points with the same $y$-coordinate are in the same orbit, and therefore also have the same $\tilde y$-coordinate. This means that the transition map between these coordinates can be written $(\tilde x, \tilde y) = (A(x,y), B(y))$, where $A$ and $B$ are smooth maps defined on some neighborhood of the origin. The transition map $\tilde \eta \circ \eta^{-1}$ is just $\tilde y = B(y)$, which is clearly smooth.
It seems that the existence of a $G$-adapted atlas ensures the equality $(\tilde x, \tilde y) = (A(x,y), B(y))$. $A$ and $B$ are smooth, probably because the $G$-adapted atlas is smooth.
So the theorem should generalise if we can find a $G$-adapted atlas which is holomorphic, symplectic etc.
Is this correct? And, can we find such an atlas?
Even if this approach does not work, does there exist such a generalisation and if yes, where can I find it?
Best Answer
The quotient manifolds does not always inherits the structure, unless $G$ is discrete.
Consider $M=T^2=S^1\times S^1$. The action of $S^1$ on $T^2$ defined by $g.(x,y)=(x+g,y)$ is proper and free, this actions preserves the symplectic structure of $T^2$, but the quotient manifold $T^2/S^1=S^1$ is not symplectic since it has an odd dimension.