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Let $G=\mathbb R^*$ act on $X=\mathbb R^3\setminus\{0\}$ by pointwise multiplication. That is for any $t\in\mathbb G$ and $(x_1,x_2,x_3)\in X$ we have $$t\cdot(x_1,x_2,x_3)=(tx_1,tx_2,tx_3)$$

Is the resulting quotient a smooth manifold?

I know from this answer that I only need to check that the action is smooth, free and proper. Smoothness and freeness are easy.

How do I show that the action is proper?

I have to show that the map $\theta:G\times X\to X\times X$ given by $\theta(g,x)=(gx,x)$ is proper (inverse image of compact set is compact) or equivalently - for each pair of pints $x,y\in X$ there are neighbourhoods $V_x$ and $V_y$ of $x$ and $y$ respectively such that $H=\{\ g\in G\ |\ gV_x\cap V_y\neq\emptyset\ \}$ is relatively compact in $G$. But I am unable to do this. Any help would be appreciate.

Thank you.

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    $\begingroup$ Do you understand that the action is a scaling? So it $x, y$ does not lie in the same line, $gx$ will never be $y$, So in this case you can choose small neigborhood $V_x, V_y$ so that $\{ g\in G : gV_x \cap V_y \neq \emptyset\} = \emptyset$. $\endgroup$ – user99914 Jun 2 '16 at 7:08
  • $\begingroup$ @John Ma, Yes I see! But what about when $y=gx$ for some $g$? Can I choose neighbourhoods so that $H=\{g\}$? $\endgroup$ – R_D Jun 2 '16 at 8:04
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    $\begingroup$ No you cannot, but you need only that $H$ is bounded and bounded away from $0$. Just pick any small $V_x$, and $V_y$. $\endgroup$ – user99914 Jun 2 '16 at 9:57
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    $\begingroup$ I think you can directly use that in a compact subset of $X\times X$ both components are bounded in norm above and below (i.e. bounded away from zero), This readily implies that the pre-image under $\theta$ is bounded in norm in both directions, which implies that $\theta$ is proper. Just as a sideremark: For the example in question, it is probably easier to directly construct local charts via inhomogeneous coordinates on projective space than the verify the properties of the action. $\endgroup$ – Andreas Cap Jun 2 '16 at 10:44
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The following reformulation of properness is often very useful in practice.

Proposition. (Lee, Introduction to smooth manifolds, Proposition 21.5) Let $M$ be a manifold and $G$ a Lie group acting continuously on $M$. The following are equivalent.

  1. The action is proper.
  2. If $(p_i)$ is a sequence in $M$ and $(g_i)$ is a sequence in $G$ such that both $(p_i)$ and $(g_i\cdot p_i)$ converge, then a subsequence of $(g_i)$ converges.

Now, we can easily prove that your action is proper by proving that the second item in this proposition holds.

Let $(x_n,y_n,z_n)\in\Bbb R^3-\{0\}$ and $t_n\in\Bbb R^*$ be such that $(x_n,y_n,z_n)$ and $(t_nx_n,t_ny_n,t_nz_n)$ converge. Then, $$\lim_{n\to\infty}(x_n,y_n,z_n)=(x,y,z)\in\Bbb R^3-\{0\}.$$ Since $(x,y,z)\in\Bbb R^3-\{0\}$, one of the coordinates is non-zero, and hence we may assume without loss of generality that $x\neq 0$. Hence $x_n\to x\neq 0$ as $n\to\infty$. In particular, $x_n\neq 0$ for all $n$ large enough, so $$\lim_{n\to\infty}\frac{1}{x_n}=\frac{1}{x}.\tag{1}$$ (By removing enough of the first elements of the sequence.) Now, we are given that $(t_nx_n,t_ny_n,t_nz_n)$ also converges, so $$\lim_{n\to\infty}t_nx_n=a,\tag{2}$$ for some $a\in\Bbb R$. Putting $(1)$ and $(2)$ together, we get $$\lim_{n\to\infty}t_n=\lim_{n\to\infty}t_nx_n\frac{1}{x_n}=\left(\lim_{n\to\infty}t_nx_n\right)\left(\lim_{n\to\infty}\frac{1}{x_n}\right)=\frac{a}{x}\in\Bbb R.$$ The second equality is justified since both limit exist. So $t_n$ converges in $\Bbb R$. The last thing we need to check is that the limit is actually in $\Bbb R^*$. But this must be the case as otherwise $$\lim_{n\to\infty}(t_nx_n,t_ny_n,t_nz_n)=(0\cdot x,0\cdot y,0\cdot z)=(0,0,0)\notin\Bbb R^3-\{0\}$$ (because $(x_n,y_n,z_n)$ converges to $(x,y,z)$). This shows that the action is proper and hence the quotient $(\Bbb R^3-\{0\})/\Bbb R^*$ is a manifold.

(The manifold that you constructed with this quotient is called a real projective space, and is usually denoted $\Bbb{RP}^3$.)

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