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?