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Let $G$ be a complex reductive group acting on a complex affine variety $X$ and let $X // G = \operatorname{Spec}\mathbb{C}[X]^G$ be the GIT quotient.

Is there a relationship between the singular locus of $X$ and that of $X // G$?

Of course, $X//G$ can be highly singular while $X$ is smooth. But, for example, I was wondering if (or under what conditions) singular points of $X$ are mapped to singular points of $X // G$.

Edit. Spenser's nice comment below shows that the answer to the latter is no. But perhaps a better and more precise question is: If $X // G$ is non-singular at $y$, is there a non-singular $x \in X$ mapping to $y$? In other words, do all fibres of $X \to X // G$ at non-singular points contain a non-singular point of $X$? I'm willing to assume irreducibility or other nice properties.

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    $\begingroup$ I think that singular points can be mapped to smooth points. Take $xy = 0$ and let $\mathbb{C}^*$ act by $z\cdot(x, y) = (zx, y)$. The quotient is $\mathbb{C}$. $\endgroup$
    – Spenser
    Aug 28 '20 at 21:28
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    $\begingroup$ @JasonStarr I'm trying to understand your counterexample in the 2 x 2 case, and it doesn't seem to work (I may be wrong). The non-invertible matrices are the locus of $ad - bc = 0$, so the only singular point is the zero matrix. Hence, any non-zero nilpotent matrix is a non-singular point of the fibre above zero. What am I missing? $\endgroup$ Aug 29 '20 at 19:02
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    $\begingroup$ I was wrong! There are, indeed, nilpotent matrices that are regular. $\endgroup$ Aug 29 '20 at 19:03
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    $\begingroup$ This is clearly false for finite groups (take the union of the two coordinate axes in $\mathbb A^2$, with the involution that switches the two axes. For a connected example, embed the cyclic group into $\mathbb G_\mathrm{m}$, and consider the induced action. $\endgroup$
    – Angelo
    Aug 30 '20 at 8:08
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    $\begingroup$ @Angelo. That is precisely how I made the normal counterexample below. I started with a normal counterexample for the cyclic group of order 2 and induced an example for the multiplicative group. $\endgroup$ Aug 30 '20 at 16:35
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This is an answer to the revised question. It is the simplest counterexample that I can think of where the reductive group is smooth and connected, where $X$ is normal and affine, and where $Y=X//G$ is smooth, even though there are fibers of the quotient map that are contained in the singular locus of $X$.

Let $Y$ be $\text{Spec}\ k[x,y,z]$, i.e., affine $3$-space. Let $G$ be the multiplicative group of units, $G=\text{Spec}\ k[u,u^{-1}]$. Let $X$ be $\text{Spec}\ k[x,y,z,s,t,t^{-1}]/\langle f \rangle$ where $f$ is the polynomial, $$f=s^2+t(xz-y^2).$$ Let the action of $G$ on $X$ be defined by $$\mu:G\times_{\text{Spec}\ k} X \to X, \ \ \mu(u,(x,y,z,s,t)) = (x,y,z,us,u^2t). $$ The ring of $G$-invariant polynomials is the subring, $$k[X]^G = k[x,y,z].$$ The quotient map is just the usual projection, $$q:X\to Y, \ \ q(x,y,z,s,t) = (x,y,z).$$ For the dense Zariski open $U = D(xz-y^2)\subset Y$, the inverse image $q^{-1}(U)$ is a $G$-torsor over $U$.

The singular locus of $X$ is the single $q$-fiber, $q^{-1}(0,0,0)$. Even though the origin is a smooth point of $Y$, every point of this $q$-fiber is a singular point of $X$.

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