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I'm reading various books and some notes and here is my question.

Let $(\mathbb{C}^*)^2$ act on $\mathbb{C}^4$ by

$$(\lambda_1,\lambda_2).(x_1, x_2, y_1,y_2)=(\lambda_1 x_1, \lambda_2 x_2, \lambda_1^{-1}y_1,\lambda_2^{-1}y_2).$$

Then $\mathbb{C}^4/\!\!/(\mathbb{C}^*)^2=\mathbb{C}^2$, given by $Spec(\mathbb{C}[x_1 y_1, x_2 y_2])$.

Now for $\chi:(\mathbb{C}^*)^2\rightarrow \mathbb{C}^*$ where $\chi(\lambda_1, \lambda_2)=\lambda_1$, this is equivalent to choosing a section of the trivial line bundle $\mathbb{C}^4\times \mathbb{C}$ over $\mathbb{C}^4$ where

$$ (\lambda_1,\lambda_2).(x_1, x_2, y_1,y_2,z)=(\lambda_1 x_1, \lambda_2 x_2, \lambda_1^{-1}y_1,\lambda_2^{-1}y_2,\lambda_1^{-1}z). $$

Then $\mathbb{C}^4/\!\!/_{\chi}(\mathbb{C}^*)^2= Proj(\mathbb{C}[x_1z])=\mathbb{P}_{S_0}^0$.

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$\mathbf{Question \:1}:$ thinking of $\mathbb{C}[x_1z]$ as a graded ring, what are the elements in the degree $0$ piece?

$\mathbf{Question \: 2}:$ Now $(\lambda_1 x_1, \lambda_2 x_2, \lambda_1^{-1}y_1,\lambda_2^{-1}y_2,\lambda_1^{-1}z)$ intersects the zero section if $x_1=0$. Thus unstable points (with respect to this character) are $\{ (0,x_2,y_1,y_2)\}$ and $\chi$-semi stable points in $\mathbb{C}^4$ are $\{ (x_1,0,0,0): x_1 \in\mathbb{C}^*\}$. Is this correct? Why is $(\mathbb{C}^4)^{\chi-ss}$ so small?

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Added:

$\mathbf{Guess}:$ An answer to $\mathbf{Question \:1}$ is that $\mathbb{C}^4/\!\!/_{\chi}(\mathbb{C}^*)^2$ is isomorphic to $Proj(\mathbb{C}[x_1 y_1, x_2 y_2]\oplus (x_1 z) \oplus \ldots)$, which is isomorphic to $\mathbb{C}^2 \times \mathbb{P}^0$, or just $\mathbb{C}^2$. So $\mathbb{C}^4/\!\!/_{\chi}(\mathbb{C}^*)^2 \cong \mathbb{C}^4/\!\!/(\mathbb{C}^*)^2$.

Then something doesn't make sense because $\mathbb{C}^4/\!\!/_{\chi}(\mathbb{C}^*)^2$ is supposed to be isomorphic to $(\mathbb{C}^4)^{ss}/(\mathbb{C}^*)^2$ in the ordinary topological sense.

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