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.