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 $\mathbb{C}[a_{11}, a_{22}, a_{21}a_{12}]$, so the GIT quotient is $\mathbb{C}^3$. 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.

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?

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?

For $\mathrm{SL}_2$, the invariant ring is $\mathbb{C}[a_{11}, a_{22}, a_{21}a_{12}]$, so the GIT quotient is $\mathbb{C}^3$. 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.

What does the situation look like for general $G$? This should be a well-known exercise. I appreciate a reference or explanation.

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 $\mathbb{C}[a_{11}, a_{22}, a_{21}a_{12}]$, so the GIT quotient is $\mathbb{C}^3$. 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.