Difference between $\mathfrak{g}/\!/G$ and $G/\!/G$

Question

I am studying a GIT quotient and I have a question that may be very silly.

Let $G$ be a connected reductive group and $\mathfrak{g}$ its Lie algebra. Then are there some differences between $\mathfrak{g}/\!/G$ and $G/\!/G$?

I started this question since I want to find each dimension when $G=U_I$, which is the maximal unipotent radical of a parabolic subgroup in $\operatorname{GL}_n$. Maybe I am wrong, I guess that $\dim(U_I/\!/U_I)=n-1$, but Sevostyanova - The algebra of invariants for the adjoint action of the unitriangular group says that $\dim( \mathfrak{u}_I /\!/ U_I)$ is larger than $n-1$.

In summary, the followings are my questions:

1. Are there some well-known differences between $\mathfrak{g}/\!/G$ and $G/\!/G$?
2. Is there a formula or theorem between dimensions of $\mathfrak{g}/\!/G$ and $G/\!/G$?
3. $\dim U_I/\!/U_i$ is not $n-1$?

I appreciate any comments for these!

Answer 1

$\newcommand{\mmod}{/\!\!/}\newcommand{\fr}[1]{\mathfrak{#1}}$If $G$ is a torus, then $G$ acts trivially on $\fr{g}$, so that $\fr{g}\mmod G \cong \fr{g}$. Similarly, the adjoint action of $G$ on $G$ is trivial, so that $G\mmod G \cong G$, which is of course not isomorphic to $\fr{g}$. Suppose instead that $G$ is semisimple and simply-connected (everything over an algebraically closed field $k$ of characteristic zero, to be safe), and let $W$ be the Weyl group. Then $\fr{g}\mmod G \cong \fr{t}\mmod W$ by Chevalley restriction, and $G\mmod G \cong T\mmod W$ by the Pittie–Steinberg theorem. Both $\fr{t}\mmod W$ and $T\mmod W$ are both isomorphic to an affine space of dimension $\operatorname{rank}(G)$; for $\fr{t}\mmod W$, this is Chevalley–Shepard–Todd, and for $T\mmod W$, this is Theorem 6.1 of Steinberg's "Regular elements of semisimple algebraic groups". So they are indeed isomorphic. For example, say $G = \operatorname{SL}_2$; then $\fr{t}\mmod W \cong \operatorname{Spec}(k[x^2])$ and $T\mmod W \cong \operatorname{Spec}(k[y + y^{-1}])$, and the isomorphism sends $x^2\mapsto y + y^{-1}$.

I don't know what happens if $G$ is instead unipotent, but going through the references of the paper you linked, I found Sevostyanova - The field of invariants of the adjoint action of the unitriangular group in the nilradical of a parabolic subalgebra. In the introduction, the author writes that if $N$ is the subgroup of $\mathrm{GL}_n(k)$ of upper triangular matrices with $1$s on the diagonal and $\fr{n}$ is its Lie algebra, then $k[\fr{n}]^N$ is a polynomial algebra on $n-1$ variables. (So $\fr{n}\mmod N$ is an affine space of dimension $n-1$.)

Answer 2

If you are in characteristic zero and $G$ is unipotent, then the exponential map is a $G$-equivariant isomorphism of algebraic varieties $\mathfrak{g} \longrightarrow G$, and therefore $\mathfrak{g}/\!/G \cong G/\!/G$ in this case.

Also, in characteristic zero but now letting $G$ be any algebraic group, the exponential map is a $G$-equivariant isomorphism from the completion of $\mathfrak{g}$ at $0$ to the completion of $G$ at $e$. I think the completions of $\mathfrak{g}/\!/G$ and $G/\!/G$ at the corresponding maximal ideals should be isomorphic, but Friedrich Knop has left a skeptical comment below, so I'll think about it more.

$\DeclareMathOperator\PSL{PSL}$In comments (1 2) to skd's answer, I pointed out that $\PSL_3/\!/\PSL_3$ is the singular variety $\operatorname{Spec} k[x,y,z]/\langle xy=z^3 \rangle$. Here we have $x=e_1^3$, $y=e_2^3$ and $z=e_1 e_2$ where $e_1$ and $e_2$ are the first and second elementary symmetric functions of the eigenvalues of a matrix in $SL_3$. Note that the identity corresponds to $e_1=e_2=3$, $(x,y,z) = (27,27,9)$, which is a smooth point of $\operatorname{Spec} k[x,y,z]/\langle xy=z^3 \rangle$; the singularity is at $e_1=e_2=0$, $(x,y,z)=0$, which corresponds to a matrix with eigenvalues $(1,\omega, \omega^2)$ for a primitive root of unity $\omega$. So the completion is just a power series ring in two variables.