# How to compute the tangent space of a quotient by a finite group

Let $I\subseteq R:=\mathbb C[x_0,\ldots,x_n]$ be a homogeneous ideal defining a subscheme $X\subseteq\Bbb P^n$. As in my previous question, the permutation group $\mathfrak S_{n+1}$ acts on $R$ by permuting the variables, inducing an action on $\Bbb P^n$. and there is a subgroup $G\subseteq\mathfrak S_{n+1}$ which leaves $X$ invariant. I would like to (practically) compute the dimension of the Zariski Tangent space of $Y:=X/\!/G$ at a point $y$. Here, $Y$ is defined as $\operatorname{Proj}(R^G/I^G)$.

Let $\pi\colon X\twoheadrightarrow X/\!/G = Y$ be the projection and $\pi(x)=y$. Computing the tangent space of $X$ at some point $x$ is rather "easy" in the sense that I can do it practically: I can (in a computer algebra system) provide generators for $I$ and then compute the kernel of the Jacobian. However, I am not sure how to compute the tangent space of $Y$ at $y$ efficiently.

There are computer algebra methods to compute generators of the invariant ring $R^G$ and also generators for $I^G$ in terms of these, but I have tried them and my examples seem too large for them to handle, even though $R^G$ is a polynomial ring by the Chevalley–Shephard–Todd Theorem.

Maybe I am doing something wrong, but mostly I think it might be easier to compute just the tangent space. Note however that the differential $d\pi\colon \mathcal T_{X,x}\to \mathcal T_{Y,y}$ does not have to be surjective: Given a (homogeneous) prime ideal $\mathfrak p\subseteq R/I$, we only have $(\mathfrak p^G)^2\subseteq(\mathfrak p^2)^G$, usually equality will not hold, so $\mathfrak p^G / (\mathfrak p^G)^2 \to \mathfrak p / \mathfrak p^2$ is not generally injective. Even if something like this could be said however, I do not know what the map $d\pi$ even looks like.

The smallest case that interests me is $n=36$, where $\Bbb P^n$ can be though of as (classes of) $6\times 6$ matrices and $G\cong\mathfrak S_3$ acts by permuting the first $3$ rows of a matrix. I will not go into detail about $X$ unless requested, I guess it is enough to know that it is given by explicit equations and computing a Groebner basis for the ideal $I$ is presumably out of reach.

Let's assume $G$ acts effectively. Since $G$ is finite all orbits are closed. Take an affine open set $U\subset X$ that contains $x$, and assume $x$ is a smooth point.

Then using the Luna Slice Theorem on $U$ you can deduce that $T_{\pi(x)}(U//G)$ is isomorphic to $T_0(T_x(U)//Stab_G(x))$ since the tangent space to the orbit of $x$ is trivial. However, $T_{\pi(x)}(U//G)=T_{\pi(x)}(X//G)$ and $T_0(T_x(U)//Stab_G(x))=T_0(T_x(X)//Stab_G(x))$.

So $$T_{\pi(x)}(X//G)=T_0(T_x(X)//Stab_G(x))$$ at smooth points $x$.

Independent of this, if the action is free (which it generically is), then $\pi$ is étale and so the tangent spaces are all isomorphic.

When $x$ is singular and the action is not free, there are examples where $\pi(x)$ remains singular and examples where it becomes smooth. A simple example of the latter is $x^2-y^2=0$ with $\{\pm 1\}$ acting by $(x,y)\mapsto (x,-y)$, which has a smooth quotient.

On the other hand, if your finite group is acting linearly then I just recalled a result in (Chapter 5, Section 5):

N. Bourbaki, Groupes et Algbbres de Lie, chaps. 4-6, Act. Sci. Ind. 1337, Hermann, Paris, 1968

that implies that if $x$ is a closed point then $\pi(x)$ is smooth if and only if the stabilizer action on the local slice (from Luna's Slice Theorem applied to an affine open containing $x$) is generated by pseudoreflections.

See here for a great exposition of Luna' Slice Theorem by Drézet. I learned about the above fact about pseudoreflections in Richardson's paper in Vol. 57, No. 1 Duke Mathematical Journal (1988) Conjugacy classes of $n$-tuples in Lie Algebras and Algebraic Groups starting around page 14.

• Thanks a bunch! Unfortunately, I am interested in the singular points of $X$, namely the points where two components meet. Can something be said when $x$ is not smooth? Also: $G$ in my case does not act freely. However, its action on codimension one points is free. – Jesko Hüttenhain Apr 20 '16 at 5:45
• Generally speaking the action is free on a Zariski open set and the projection is étale at any point where the action is free. So at singularities in $X$ where the action is free you still get an isomorphism of tangent spaces. – Sean Lawton Apr 20 '16 at 12:18
• Interesting, but my goal is to show that each component of $X$ is smooth, and I want to show that $Y=X/\!/G$ is smooth in order to achieve this. For this, I (think I) need to compute the dimension of the tangent space of $Y$ at $\pi(x)$ for any $x$, especially the worst of them. – Jesko Hüttenhain Apr 20 '16 at 12:20
• It sounds like from here mathoverflow.net/questions/236405/… that you have an isomorphism on components of $X$ with $X//G$. So it sounds like restricting to a component might make the intersection points smooth in the component and then the tangent space computation above applies in that component and may allow a dimension count. – Sean Lawton Apr 20 '16 at 12:55
• It's a chicken and egg problem: I have this isomorphism only if $Y$ is normal. The only way I know how to prove that is by showing it is smooth. – Jesko Hüttenhain Apr 20 '16 at 13:03