Given smooth action of $\Bbb R$ on $M = \Bbb R^2 \setminus \{(0; 0)\}$ : $g \cdot (m_1, m_2) = (e^tm_1, e^{-t}m_2)$
The $\Bbb R$-action stabilizes the subset $S = [0,+\infty) \times (0,+\infty)$. Prove that
The action is free.
$S/\Bbb R$, with the quotient topology, is homeomorphic to $[0,+\infty)$ (with the Euclidean topology).
The quotient topology on $M/\Bbb R$ is not Hausdorff.
Prove that $M/\Bbb R$ becomes a manifold upon removing four well-chosen points
I already solved the first 2 points, which might be useful for the other two. I am having trouble with those. How can I show that? in the last point first I have to make it into a Hausdorff space I guess.
Edit:
About the orbits, to find them I did this:
Let say I want to find the orbit (class) of $(a,b )$ \in $M$ So I fix $(a,b)$, I want to find all the $(x,y)$ such that $(x,y) \sim (a,b)$ which means that there must exist some $g \in G=\Bbb R$ such that $g\cdot(x,y)= (a,b)$ that is $(x e^t, y e^-t)=(a,b)$ so I get two equations: $xe^t=a$ and $y e^{-t }=b$ Analyzing by cases I find these cases:
$a=0$, $b \neq 0, b > 0: (x,y)= (0,be^t)$ that is all the point of the form $(0 , x)$ with $x > 0$ In the same way
$a=0, b \neq 0, b <0:$ the class is $(0,x)$ with $x < 0$
$b=0, a \neq 0, a >0:$ the class is $(x,0)$ with $x > 0$
$b=0, a \neq 0, a <0: $ the class is $(x,0)$ with $x < 0$
And for the case $a\neq 0$ and $b\neq 0$ I have from $xe^t=a$ and $y e^{-t }=b$ that $xy=ab$, so those are hyperbolas. for every value of the non-zero constant $ab$ Am I right that these are all the orbits?
Then what would the neighborhoods of (0,1) will look like? I think like this: the +y semiaxis plus a continuous of hyperbolas next to it, that I think at the end you get the whole region of the I quadrant under a certain hyperbola withouth including the origin and the semiaxis x+.
Is all of this correct? What should I do next?
