# Quotient Manifold Theorem for Riemannian, Complex, Symplectic, ... Manifolds

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?

• In the complex setting it suffices to assume that $G$ is a complex Lie group and the action $G\times M\to M$ is holomorphic (and also proper and free, of course). Jun 16, 2019 at 5:01

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.
• Oh, I see. So what if $G$ is discrete? Does the quotient inherit the structure? Jun 15, 2019 at 12:44
• Yes it inherits the structure since an atlas of $M/G$ can be defined by $(p(U_i)_{i\in I})$ such that the restriction of the quotient map $p$ to $U_i$ is a diffeomorphism onto its image, and $(U_i)$ is an atlas of $M$ such that the transition functions preserve the structure. Jun 15, 2019 at 13:09
• That's nice. Is it because $(p|_{U_i})^{-1}\circ p|_{U_j}$ is the translation by a group element and thus it is a holomorphic, symplectic,... map? Jun 15, 2019 at 15:58