Let $X$ be a smooth complex affine variety, let $G$ be a complex reductive group acting on $X$. Suppose that the stabilizer $G\_x$ of a point $x\in X$ is reductive and connected. Let $\varphi: X\to X//G$ be the GIT quotient. I would like to understand the germ of $X//G$ at the point $\varphi(x)$, and in particular understand if the following is correct:


**Guess.** The analytic germ of $X//G$ at $\varphi(x)$ is isomorphic to the analytic germ at $0$ of the GIT quotient of $T\_X(x)/T\_{O\_x}(x)$ by the linear action of $G\_x$. 


**Question.** Is the above guess correct? If yes, is there a reference for this statement? In particular, how to prove this statement in the case when $x$ is fixed by $G$?



Understanding a germ of a GIT quotient
For a fixed point your guess is right and one doesn't need Luna's slice theorem to prove it: Let $T$ be the tangent space in $x$ and let $\mathfrak m\_x\subset\mathbb C[X]$ be the maximal ideal. Then the canonical surjective linear map $\mathfrak m\_x\to\mathfrak m\_x/\mathfrak m\_x^2\cong T^\*$ has a $G$-equivariant section giving rise to a $G$-linear map $T^\*\hookrightarrow\mathbb C[X]$ and therefore to a morphism $X\to T$ which maps $x$ to $0$ and is étale in $x$. Now use the fact that invariants commute with completions (by complete reducibility) to see that $X//G\to T//G$ is étale in the image of $x$. As Jason indicated this implies an isomorphism of analytic neighborhoods.


It is true that the above argument is part of Luna's slice theorem but it is actually only the "easy" part. In fact, Luna's slice theorem is not just a statement about étale or formal neighborhoods but has a global aspect, as well. One consequence is, e.g., that the unstable set with respect to $x$ (i.e., the set of points having $x$ in their orbit closure) is globally isomorphic to the unstable set of $0$ in $T$. 


Concerning your more general guess: One has to be very careful with Luna's slice theorem: it requires that the orbit of $x$ is closed. This is much stronger than $G\_x$ being reductive. So, your guess is ok if $Gx$ is closed in $X$ (which is trivial if $x$ is a fixed point). 


Otherwise, there is the following counterexample. Let $G=SL(2,\mathbb C)$, let $X$ be the space of $5$-forms, and $x=\xi\eta(\xi+\eta)^3\in X$. It is easily seen that $G\_x=1$. So $V:=T\_x(X)/T\_{\mathcal O\_x}(x)=\mathbb C^3$.


Let $N\subset X$ be the set of unstable $5$-forms. Hilbert has proved that these are those $5$-forms having a linear factor of multiplicity $\ge3$. So $x\in N$. Now, it is well known that $X//G$ is singular in the image of $0$. In particular, it is not isomorphic to $V//G\_x=V$.


The problem of this example is that the codimension of $N$ in $X$ is $2$. Thus, $S\cap N$ is nontrivial and maps to $0$ where $S\cong V$ is a slice to the orbit of $x$.


**Edit**: Concerning the equivalence of étale and analytic singularities see [this MO thread](https://mathoverflow.net/questions/51530/analytic-vs-formal-vs-%C3%A9tale-singularities).


