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).

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.