Here is a classic problem of geometric invariant theory: let $ G$ be a reductive linear algebraic group such as $ \mathrm{GL}_n$ and let $ \mathfrak{g}$ be its Lie algebra. Determine the invariant functions $ k[\mathfrak{g}]^G$, where $ G$ acts on $ \mathfrak{g}$ via the adjoint action. This problem is motivated by the search for quotients: What is the quotient $ \mathfrak{g}/G$? Here, the action of $ G$ on $ \mathfrak{g}$ is given by the adjoint action. More explicitly, an element $ g\in G$ acts via the differentiation of $ \mathrm{Int}_g$, where $ \mathrm{Int}_g$ is conjugation by $ g$ on $ G$.
For simplicity, we will stay in the realm of varieties over an algebraically closed field $ k$ of characteristic zero.
First, we should ask:
Obviously, it should be a variety. Perhaps the weakest definition is that the quotient should be any variety $ Y$ with a map $ \mathfrak{g}\to Y$, constant along orbits of $ G$, and such that $ Y$ is universal for this property. In other words, if $ Z$ is a variety with a map $ \mathfrak{g}\to Z$ constant along $ G$-orbits, then $ \mathfrak{g}\to Z$ should factor through $ \mathfrak{g}\to Y$. (In other words, in the category of morphisms $ \mathfrak{g}\to *$, the object $ \mathfrak{g}\to Y$ is an initial object.) Certainly, we would like a quotient to satisfy this property. If such a $ Y$ exists, it is unique up to unique isomorphism and is called a categorical quotient.
Quotients
Now, where would we go looking for a quotient like this? One possibility is to consider the problem on the algebraic side. The regular functions $ k[\mathfrak{g}]$ on $ \mathfrak{g}$ also have a right-action of $ G$. Indeed, if $ g\in G$ and $ \varphi\in k[\mathfrak{g}]$ then $ (\varphi*g)(x) = \varphi(gx)$. Now, since the affine $ k$-algebras and the affine varieties are antiequivalent, it makes sense to look at the invariants: $ k[\mathfrak{g}]^G$. If this is an affine $ k$-algebra, then the corresponding variety would be a reasonable candidate for a quotient.
It is a nontrivial result that if a reductive group $ G$ over our algebraically closed field $ k$ acts on an affine variety $ X$ then $ k[X]^G$ is a finitely generated $ k$-algebra and the corresponding variety is in fact a categorical quotient. Here "acts" means there is a morphism of varieties $ G\times X\to X$ satisfying the axionms of a group action.
Calculations
However, this post we actually be concentrated on describing $ k[\mathfrak{g}]^G$ in fairly simple terms. As an example, let us try $ G = \mathrm{GL}_2$. Then $ \mathfrak{g} = \mathfrak{gl}_2$, the Lie algebra of all $ 2\times 2$ matrices, and the adjoint action is given by conjugation. The Lie algebra has $ k[x_1,x_2,x_3,x_4]$ as its coordinate ring, and is isomorphic to affine $ 4$-space.
Let $ A = \left(\begin{smallmatrix} a & b\\ c & d\end{smallmatrix}\right)\in\mathrm{GL}_2$ and let $ X = \left(\begin{smallmatrix} x_1 & x_2\\ x_3 & x_4\end{smallmatrix}\right)\in\mathfrak{gl}_2$. What does $ AXA^{-1}$ look like? This is an elementary matrix calculation, and it ends up being:
If you meditate with this matrix for a bit, you might notice that the trace of this matrix is $ x_1 + x_4$, which is the trace of $ X$. Thus the function $ x_1 + x_4\in k[x_1,x_2,x_3,x_4] = k[\mathfrak{g}]$ is $ G$-invariant. Is there another? By another of course, I mean another algebraically independent from $ x_1+x_4$.
One could try and play around a bit, perhaps trying something nice and familiar like the determinant (hehe), but the computations can get a bit messy. A computer algebra system might help, but what are you supposed to do if you're stuck in a cabin up north with only a candle and a piece of charcoal to occupy your time? You use the Chevalley restriction theorem!
Chevalley Restriction Theorem
We keep the setting of $ G$ being a reductive algebraic group over an algebraically closed field with Lie algebra $ \mathfrak{g}$. Select a maximal torus $ T\subseteq G$. Then its Lie algebra $ \mathfrak{t}$ is a subalgebra $ \mathfrak{t}\subseteq\mathfrak{g}$, and there is a restriction map $ k[\mathfrak{g}]\to k[\mathfrak{t}]$, and hence an induced map
We define the Weyl group $ W = W(\mathfrak{t})$ of $ \mathfrak{t}$ to be the group $ W = N_G(\mathfrak{t})/N_G(\mathfrak{t})$. This group acts on $ \mathfrak{t}$, and hence on $ k[\mathfrak{t}]$, and any regular function invariant under $ G$ is also invariant under $ N_G(\mathfrak{t})$. Hence we obtain a map
There are various generalisations of next theorem, but the formulation here will be enough for our purposes.
This result is amazing. It reduces the problem of determining $ G$-invariant functions on $ \mathfrak{g}$ to $ W$-invariant functions on $ \mathfrak{t}$. The interesting part is that the Weyl group is a finite group (hint: use rigidity of tori), so computing the $ W$-invariant functions in $ k[\mathfrak{t}]$ is already going to be much easier.
(To be honest, I am not aware of the most general forms of this theorem; for instance, it holds in positive characteristic $ p$ as long as $ p \nmid |W|$ and it holds when $k$ isn't algebraically closed if $G$ is split, and there are other variations, so to keep it simple I excluded these, but in the future I may write a more comprehensive post about this.)
Back to $ \mathrm{GL}_2$
So what are the $ \mathrm{GL}_2$-invariant functions in $ k[\mathfrak{gl}_2]$? Consider the maximal torus of diagonal matrices. The first thing is to determine the Weyl group. This is a straightforward matrix computation, which I will leave to the reader. The answer is $ W\cong \mathbb{Z}/2 = \{1,\sigma\}$, and the action of $ \sigma$ on $ \mathfrak{t}$ is given by permuting the diagonal.
The $ k$-algebra representing $ \mathfrak{t}$ is $ k[x,y]$, and so the invariant subalgebra $ k[x,y]^W$ are the symmetric polynomials. It is a standard result that all such polynomials in two variables are generated by $ x + y$ and $ xy$. So we have determined that $ k[\mathfrak{gl}_2]^{\mathrm{GL}_2}$ is isomorphic to a polynomial ring in two variables, generated by homogeneous polynomials. Isn't this a much easier calculation?
This in particular shows that the categorical quotient $ \mathfrak{gl}_2/\mathrm{GL}_2$ is isomorphic to $ \mathbb{A}^2$.