# An example of a quotient which is NOT a manifold

I want to

find a smooth action of $$\mathbb R$$ on a smooth n-manifold M such that the orbit space isn't a smooth manifold.

So far I only know that such a result can't happen for finite groups acting freely (Quotient Manifold Theorem).

• A more interesting question would be about quotients $Q$ which are Hausdorff but are not manifolds or, more precisely, do not admit smooth manifold structures making the quotient map $M\to Q$ smooth (which is what one normally likes to have in this setting). – Moishe Kohan Apr 10 at 20:54

The group $$\mathbb R$$ acts on the smooth $$2$$-manifold $$\mathbb R^2$$ by the formula $$t \cdot (x,y) = (e^t x, e^{-t}y)$$. The orbits of this action are:
• The origin $$(0,0)$$
• The positive $$x$$-axis
• The positive $$y$$-axis
• The negative $$x$$-axis
• The negative $$y$$-axis
• Each one of the two sheets of each 2-sheeted hyperbola of the form $$xy = c$$, for all $$c \ne 0$$.
It follows from this description that the quotient space is not Hausdorff, and therefore the orbit space is not a smooth manifold. For example, the equivalence class of the origin $$(0,0)$$ forms a one-point subset of the quotient that is not closed, because every open ball in $$\mathbb R^2$$ containing $$(0,0)$$ has nontrivial intersection with each positive and negative axis. Also, every neighborhood of the positive $$x$$-axis and every neighborhood of the positive $$y$$-axis have nonempty intersection, because for every open ball around $$(1,0)$$ in $$\mathbb R^2$$ and for every open ball around $$(0,1)$$ in $$\mathbb R^2$$ there exists $$c>0$$ such that those two balls each intersect the $$xy=c$$ curve in the 1st quadrant.
Consider $$T^2$$ the two dimensional torus, it is the quotient of $$\mathbb{R}^2$$ by $$t_1(x,y)=(x+1,y)$$ and $$t_2(x,y)=(x,y+1)$$, consider the action of $$\mathbb{R}$$ induced on $$T^2$$ by $$f_t(x,y)=(x+t\alpha,y)$$ where $$\alpha$$ is irrational, its orbits are dense and the quotient space is not Haussdorff.