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Let $G$ be a connected complex reductive group with Lie algebra $\mathfrak{g}$. One knows a lot about the GIT quotient $\mathfrak{g}/\!/G$: the invariant ring is a free polynomial algebra on $\mathrm{rank}([G,G])$ generators, unstable points are the nilpotent elements, polystable points are the semisimple elements, and there are no stable points.

Now let $T$ be a maximal torus of $G$. Then $T$ acts on $\mathfrak{g}$.

Question: what does $\mathfrak{g}/\!/T$ look like? What is the invariant ring, and the unstable/stable points? What does its singularities look like?

I presume this a well-known exercise. I appreciate a reference or explanation.

For $\mathrm{SL}_2$, the invariant ring is $\mathbb{C}[a_{11}, a_{22}, a_{21}a_{12}, a_{11}+a_{22}]$, so the GIT quotient is $\mathbb{C}^2$. Semistable points are matrices which are neither strictly upper triangular nor strictly lower triangular. Stable points are the ones which are neither upper triangular nor lower triangular.

For $G=\mathrm{SL}_3$, it seems the GIT quotient is singular.

Note, we have a canonical map $\mathfrak{g}/\!/T\rightarrow \mathfrak{g}/\!/G$. What do the fibres look like?

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    $\begingroup$ For $\mathfrak{sl}_3$ it is $\mathbb C[a_{11},a_{33},a_{12}a_{21},a_{23}a_{32},a_{13}a_{31}, a_{12}a_{23}a_{31},a_{21}a_{13}a_{32}]$ i. e. $\mathbb C[x,y,z,t,u,v,w]/(ztu-vw)$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 6 at 7:26
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    $\begingroup$ You seem to forget the relation $a_{11} + a_{22} = 0$ for $\mathrm{SL}_2$. If you take it into account, the GIT quotient is $\mathbb{A}^2$. $\endgroup$
    – Sasha
    Commented Feb 6 at 7:45
  • $\begingroup$ Ah yes. I was thinking about GL. It is corrected now. Thanks @Sasha $\endgroup$
    – Dr. Evil
    Commented Feb 6 at 23:38

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