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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).

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  • $\begingroup$ 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). $\endgroup$ – Moishe Kohan Apr 10 at 20:54
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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.

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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.

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