Proj as a quotient of an affine variety by $\mathbb C^*$ I wonder how good is the quotient of $\operatorname{Spec} A$ by $\mathbb C^*$ given by $\operatorname{Proj} A$.
First, an action of the group $\mathbb C^*$ on an complex affine variety $\operatorname{Spec} A$ is the same as $\mathbb Z$-grading on its function ring $A$. Suppose also that there are only positive gradings, so it is $\mathbb N$-grading. Then there is a rational map $\operatorname{Spec} A \dashrightarrow \operatorname{Proj} A$ which is defined on the open subset $(\operatorname{Spec} A)_{ss}$ whose complement is given by the irrelevant ideal $\sum_{i>0} A_i$.
Second, it is well-known that for a reductive group $G$ the variety $\operatorname{Spec} B^G$ classifies closed $G$-orbits in $\operatorname{Spec} B$. But $\operatorname{Proj} A$ is locally given by $\operatorname{Spec} (A_f)_0$ for a homogeneous element $f \in A_k$, so I wonder whether it classifies the closed orbits in $(\operatorname{Spec} A)_{ss}$? As the group is $\mathbb C^*$, every orbit is either one point or one-dimensional. Maybe $(\operatorname{Spec} A)_{ss}$ contains only one-dimensional orbits, so all of them are closed?
If nothing this simple holds, is there a simple counter-example, or a good reference?
 A: Let's write $X:=\operatorname{Spec}(A)\subseteq\Bbb A^n\newcommand{\sms}{\mathrm{ss}}$ where I also assume that the coordinates of $\Bbb A^n$ are such that $\Bbb C^\times$ acts by a character on it (we can always choose coordinates in this way). Since you insisted that the induced grading vanishes in negative degrees, this implies that $t\in\Bbb C^\times$ acts on a point $x=(x_1,\ldots,x_n)\in X$ as $t.(x_1,\ldots,x_n)=(t^{k_1}x_1,\ldots,t^{k_n}x_n)$ for certain natural numbers $k_i\ge 0$. 
From this you can easily construct examples where $0$ is not in the closure of $\Bbb C^\times.x$ (i.e. $x\in X_\sms$), but the orbit of $x$ is not closed:
Choose $n=2$, $X=\Bbb A^2$, $k_1=0$, $k_2=1$, then $x=(1,1)$ is semistable because $t.(1,1)=(1,t)$ and zero is not in the closure of these points, but the point $y=(1,0)$ is in the closure of the orbit, but not the orbit itself. Hence, going from $X$ to $X_\sms$ does not guarantee in general that you have only closed orbits left, even for an action of $\Bbb C^\times$.
I also hope that the first paragraph might make things easier to work with for you, because the situation is still quite controllable.
