# GIT and singularities

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.

• 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 $\mathbb{C}$. Aug 28 '20 at 21:28
• @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? Aug 29 '20 at 19:02
• I was wrong! There are, indeed, nilpotent matrices that are regular. Aug 29 '20 at 19:03
• 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. Aug 30 '20 at 8:08
• @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. Aug 30 '20 at 16:35

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 $$X$$.