Suppose we have a $2\times 2$ matrix

$$

M = \begin{pmatrix}

x_{11} & x_{12}\\

x_{21} & x_{22}

\end{pmatrix}

$$

with entries in a field $F$. The characteristic polynomial of this matrix is $p(t) := {\rm det}(tI_2 – M) = t^2 – (x_{11} + x_{22})t + x_{11}x_{22} – x_{21}x_{12}$. One might ask: how can we produce a matrix with a given characteristic polynomial? This can be accomplished using the rational canonical form:

$$

t^2 + at + b\mapsto

\begin{pmatrix}

0 & -b\\

1 & -a

\end{pmatrix}.

$$

We can calculate that the characteristic polynomial of this matrix to be $t^2 + at + b$. This map gives a bijection between quadratic monic polynomials in $F[t]$ and matrices of the above form. One way to understand this phenomenon is through algebraic groups. To explain, let's stick with $F$ having characteristic zero, though much of what we do can be done in characteristic $p$ for $p$ sufficiently large as well using very different techniques.

# Algebraic Groups and Representations

We can look at the $2\times 2$ matrices with coefficients in $F$ as the $F$-points of the Lie algebra $\mathfrak{gl}_2$ of $\GL_2$. If we let $F[\mathfrak{gl}_2]$ be the coordinate ring of $\mathfrak{gl}_2$, then the coefficients of the characteristic polynomial, which in this case are the *trace* and the *determinant* (up to sign), generate the algebra $F[\mathfrak{gl}_2]^{\GL_2}$ of invariant functions under the action of $\GL_2$. Here, $\GL_2$ acts on $\mathfrak{gl}_2$ by the action of conjugation. The representation of an algebraic group $G$ on its Lie algebra $\mathfrak{g}$ in general is known as the *adjoint representation*. In the case of the adjoint representation, it is known that $k[\mathfrak{g}]^G$ is a polynomial algebra generated by $m$ invariants where $m$ is the rank of $G$. This is the dimension of the maximal torus of $G$, so that for $\GL_2$, we have verified this: we have two invariants.

The rational canonical form is an example of a Kostant-Weierstrass section or KW-section, as it is known in invariant theory: in this case, an affine subvariety $\cfr\subseteq\gfr$ such that the composition $\cfr\rightarrow\gfr\to \gfr//G$. Here, we have written $\gfr//G$ for the categorical quotient $\Spec(k[\gfr]^G)$, and the map $\gfr\to\gfr//G$ is induced by inclusion.

This question is interesting and subtle, and there are important examples of representations for which this question is believed to be true but not proven. However, in our case where $F$ has characteristic zero and the representation is the adjoint representation, a KW-section was constructed by Kostant. In this setting, it is usually called the Kostant section since he was the first one to notice this phenomenon.

# The Kostant Section

To reiterate our setting, we are looking at a connected reductive algebraic group $G$ over a field $F$ of characteristic zero acting on its Lie algebra $\gfr$. We denote the action of $g\in G$ on $x\in\gfr$ by ${\rm Ad}(g)x$.

The interesting thing about the Kostant section is that its construction actually gives many different possible sections. In order to introduce it, we must first introduce regular elements: we say that $x\in\gfr$ is regular if ${\rm Ad}(G)x$, the orbit of $x$, has maximal dimension. It is a theorem that $x$ is regular if and only if its centraliser in $\gfr$ has minimal dimension. Often this latter criterion is more useful since the centraliser in $\gfr$ is easy to compute.

Next, we introduce $\mathfrak{sl}_2$-triples. An $\mathfrak{sl}_2$-triple in $\gfr$ is a triple $(x,e,f)$ of three elements of $\gfr$ such that:

- $[x,e]=2e$
- $[x,f]=-2f$
- $[e,f]=x$

The point is that the three 'obvious' basis elements

$$

x=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix},

e=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},

f= \begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}

$$

of $\mathfrak{sl}_2$ also satisfy these relations (exercise), and that any such elements span a Lie algebra isomorphic to $\mathfrak{sl}_2$. Kostant's section then is $e + \zfr_\gfr(f)$ where

$$

\zfr_\gfr(f) = \{ y\in \gfr : [y,f] = 0\}

$$

is the centraliser of $f$ in $\gfr$.

# An Example

Let's apply the construction of the Kostant section to our example of $\GL_2$ acting on $\mathfrak{gl}_2$. First of course, we have to find a regular nilpotent element. Of course, in this case, we might as well just take

$$e = \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}.$$

Of course, we need to verify that it is in fact regular:

*Proof.*The computation

$$

\begin{pmatrix}

x & y\\

z & t

\end{pmatrix}

\begin{pmatrix}

0 & 1\\

0 & 0

\end{pmatrix} –

\begin{pmatrix}

0 & 1\\

0 & 0

\end{pmatrix}\begin{pmatrix}

x & y\\

z & t

\end{pmatrix}=

\begin{pmatrix}

-z & x-t\\

0 & z

\end{pmatrix}

$$

shows that $\zfr_\gfr(e)$ is two-dimensional. This is the rank of $\GL_2$, so $e$ is indeed regular.

Next, we need to embed $e$ into an $\mathfrak{sl}_2$-triple. In this case this is easy because the basis of $\mathfrak{sl}_2$ we described above will work. In general, finding the $\mathfrak{sl}_2$-triple is a little more tricky, but usually not really difficult.

We then compute that

$$

\zfr_\gfr(f) = \left\{

\begin{pmatrix}

x & 0\\

z & x

\end{pmatrix} : x,z\in F

\right\}

$$

So, the section is given by the inclusion of

$$

e + \zfr_\gfr(f) = \left\{

\begin{pmatrix}

x & 1\\

z & x

\end{pmatrix} : x,z\in F

\right\}

$$

into $\mathfrak{gl_2}$. Notice that, as expected, there is exactly one matrix of this type one we specify what the trace and determinant must be. Here's a parting question: