lie-algebras-and-their-representations

Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules

Name	Size	Mode
..
sections/sl2-sl3.tex	55845B	-rw-r--r--

\chapter{Representations of \(\mathfrak{sl}_2(K)\) \& \(\mathfrak{sl}_3(K)\)}\label{ch:sl3}

We are, once again, faced with the daunting task of classifying the
finite-dimensional modules of a given (semisimple) algebra \(\mathfrak{g}\).
Having reduced the problem a great deal, all its left is classifying the simple
\(\mathfrak{g}\)-modules. We have encountered numerous examples of simple
\(\mathfrak{g}\)-modules over the previous chapter, but we have yet to subject
them to any serious scrutiny. In this chapter we begin a systematic
investigation of simple modules by looking at concrete examples. Specifically,
we will classify the simple finite-dimensional modules of certain
low-dimensional semisimple Lie algebras: \(\mathfrak{sl}_2(K)\) and
\(\mathfrak{sl}_3(K)\).

The reason why we chose \(\mathfrak{sl}_2(K)\) is a simple one: throughout the
previous chapters \(\mathfrak{sl}_2(K)\) has afforded us surprisingly
illuminating examples. We begin our analysis by recalling that the elements
\begin{align*}
  e & = \begin{pmatrix} 0 & 1 \\ 0 &  0 \end{pmatrix} &
  f & = \begin{pmatrix} 0 & 0 \\ 1 &  0 \end{pmatrix} &
  h & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{align*}
form a basis for \(\mathfrak{sl}_2(K)\) and satisfy
\begin{align*}
  [e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e
\end{align*}

Let \(M\) be a finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module. We now
turn our attention to the action of \(h\) on \(M\), in particular, we
investigate the subspace \(\bigoplus_{\lambda} M_\lambda \subset M\) -- where
\(\lambda\) ranges over the eigenvalues of \(h\!\restriction_M\) and
\(M_\lambda\) is the corresponding eigenspace.

At this point, this is nothing short of a gamble: why look at the eigenvalues
of \(h\)? The short answer is that, as we shall see, this will pay off. We will
postpone the discussion about the real reason of why we chose \(h\), but for
now we may notice that, perhaps surprisingly, the action \(h\!\restriction_M\)
of \(h\) on a finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module \(M\)
is always a diagonalizable operator.

Let \(\lambda\) be any eigenvalue of \(h\!\restriction_M\). Notice
\(M_\lambda\) is in general not a \(\mathfrak{sl}_2(K)\)-submodule of \(M\).
Indeed, if \(m \in M_\lambda\) then the identities
\begin{align*}
  h \cdot (e \cdot m) &=  2e \cdot m + e h \cdot m = (\lambda + 2) e \cdot m \\
  h \cdot (f \cdot m) &= -2f \cdot m + f h \cdot m = (\lambda - 2) f \cdot m
\end{align*}
follow. In other words, \(e\) sends an element of \(M_\lambda\) to an element
of \(M_{\lambda + 2}\), while \(f\) sends it to an element of \(M_{\lambda -
2}\). Visually, we may draw
\begin{center}
  \begin{tikzcd}
    \cdots          \rar[bend left=60]                          &
    M_{\lambda - 2} \rar[bend left=60]{e} \lar[bend left=60]    &
    M_{\lambda}     \rar[bend left=60]{e} \lar[bend left=60]{f} &
    M_{\lambda + 2} \rar[bend left=60]    \lar[bend left=60]{f} &
    \cdots                                \lar[bend left=60]
  \end{tikzcd}
\end{center}

This implies \(\bigoplus_\lambda M_\lambda\) is a
\(\mathfrak{sl}_2(K)\)-submodule, so that \(\bigoplus_\lambda M_\lambda\) is
either \(0\) or the entirety of \(M\) -- recall that \(M\) is simple. Since
\(M\) is finite dimensional, \(h\!\restriction_M\) has at least one eigenvalue
and therefore
\[
  M = \bigoplus_\lambda M_\lambda
\]

Even more so, we have seen that for any eigenvalue \(\lambda \in K\) of
\(h\!\restriction_M\), \(\bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k}\) is a
\(\mathfrak{sl}_2(K)\)-invariant subspace, which goes to show
\[
  M = \bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k},
\]
and the eigenvalues of \(h\) all have the form \(\lambda - 2 k\) for some
\(k\). By the same token, if \(a\) is the greatest \(k \in \mathbb{Z}\) such
that \(V_{\lambda - 2 k} \ne 0\) and, likewise, \(b\) is the smallest \(k \in
\mathbb{Z}\) such that \(V_{\lambda - 2 k} \ne 0\) then
\[
  M = \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le k \le b}}
      M_{\lambda - 2 k}
\]

The eigenvalues of \(h\) thus form an unbroken string
\[
  \ldots, \lambda - 4, \lambda - 2, \lambda, \lambda + 2, \lambda + 4, \ldots
\]
around \(\lambda\). Our main objective is to show \(M\) is determined by this
string of eigenvalues. To do so, we suppose without any loss in generality that
\(\lambda\) is the right-most eigenvalue of \(h\), fix some nonzero \(m \in
M_\lambda\) and consider the set \(\{m, f \cdot m, f^2 \cdot m, \ldots\}\).

\begin{proposition}\label{thm:basis-of-irr-rep}
  The set \(\{m, f \cdot m, f^2 \cdot m, \ldots\}\) is a basis for \(M\). In
  addition, the action of \(\mathfrak{sl}_2(K)\) on \(M\) is given by the
  formulas
  \begin{equation}\label{eq:irr-rep-of-sl2}
    \begin{aligned}
        f^k \cdot m & \overset{e}{\mapsto} k(\lambda + 1 - k) f^{k - 1} \cdot m
      & f^k \cdot m & \overset{f}{\mapsto} f^{k + 1} \cdot m
      & f^k \cdot m & \overset{h}{\mapsto} (\lambda - 2 k) f^k \cdot m
    \end{aligned}
  \end{equation}
\end{proposition}

\begin{proof}
  First of all, notice \(f^k \cdot m\) lies in \(M_{\lambda - 2 k}\), so that
  \(\{m, f \cdot m, f^2 \cdot m, \ldots\}\) is a set of linearly independent
  vectors. Hence it suffices to show \(M = K \langle m, f \cdot m, f^2 \cdot m,
  \ldots \rangle\), which in light of the fact that \(M\) is simple is the same
  as showing \(K \langle m, f \cdot m, f^2 \cdot m, \ldots \rangle\) is
  invariant under the action of \(\mathfrak{sl}_2(K)\).

  The fact that \(h \cdot (f^k \cdot m) \in K \langle m, f \cdot m, f^2 \cdot
  m, \ldots \rangle\) follows immediately from our previous assertion that
  \(f^k \cdot m \in M_{\lambda - 2 k}\) -- indeed, \(h \cdot (f^k \cdot m) =
  (\lambda - 2 k) f^k \cdot m \in K \langle m, f \cdot m, f^2 \cdot m, \ldots
  \rangle\), which also goes to show one of the formulas in
  (\ref{eq:irr-rep-of-sl2}). Seeing \(e \cdot (f^k \cdot m) \in K \langle m, f
  \cdot m, f^2 \cdot m, \ldots \rangle\) is a bit more complex. Clearly,
  \[
    \begin{split}
      e \cdot (f \cdot m)
      & = h \cdot m + f \cdot (e \cdot m) \\
      \text{(since \(\lambda\) is the right-most eigenvalue)}
      & = h \cdot m + f \cdot 0 \\
      & = \lambda m
    \end{split}
  \]

  Next we compute
  \[
    \begin{split}
      e \cdot (f^2 \cdot m)
      & = (h + fe) \cdot (f \cdot m) \\
      & = h \cdot (f \cdot m) + f \cdot (\lambda m) \\
      & = 2 (\lambda - 1) f \cdot m
    \end{split}
  \]

  The pattern is starting to become clear: \(e\) sends \(f^k \cdot m\) to a
  multiple of \(f^{k - 1} \cdot m\). Explicitly, it is not hard to check by
  induction that
  \[
    e \cdot (f^k \cdot m) = k (\lambda + 1 - k) \cdot f^{k - 1} m,
  \]
  which which is the first formula of (\ref{eq:irr-rep-of-sl2}).
\end{proof}

The significance of Proposition~\ref{thm:basis-of-irr-rep} should be
self-evident: we have just provided a complete description of the action of
\(\mathfrak{sl}_2(K)\) on \(M\). In particular, this goes to show\dots

\begin{corollary}
  Every eigenspace of the action of \(h\) on \(M\) is \(1\)-dimensional.
\end{corollary}

\begin{proof}
  It suffices to note \(\{m, f \cdot m, f^2 \cdot m, \ldots \}\) is a basis for
  \(M\) consisting of eigenvalues of \(h\) and whose only element in
  \(M_{\lambda - 2 k}\) is \(f^k \cdot m\).
\end{proof}

\begin{corollary}\label{thm:sl2-find-weights}
  The eigenvalues of \(h\) in \(M\) form a symmetric, unbroken string of
  integers separated by intervals of length \(2\) whose right-most value is
  \(\dim M - 1\).
\end{corollary}

\begin{proof}
  If \(f^r\) is the lowest power of \(f\) that annihilates \(m\), it follows
  from the formulas in (\ref{eq:irr-rep-of-sl2}) that
  \[
    0 = e \cdot 0 = e \cdot (f^r \cdot m)
    = r (\lambda + 1 - r) f^{r - 1} \cdot m
  \]

  This implies \(\lambda + 1 - r = 0\) -- i.e. \(\lambda = r - 1 \in
  \mathbb{Z}\). Now since \(\{m, f \cdot m, f^2 \cdot m, \ldots, f^{r - 1}
  \cdot m\}\) is a basis for \(M\), \(r = \dim V\). Hence if \(\lambda =
  \dim V - 1\) then the eigenvalues of \(h\) are
  \[
    \ldots, \lambda - 6, \lambda - 4, \lambda - 2, \lambda
  \]

  To see that this string is symmetric around \(0\), simply note that the
  left-most eigenvalue of \(h\) is precisely \(\lambda - 2 (r - 1) =
  -\lambda\).
\end{proof}

Visually, the situation it thus
\begin{center}
  \begin{tikzcd}
    M_{-\lambda}      \rar[bend left=60]{e}                       &
    M_{- \lambda + 2} \rar[bend left=60]{e} \lar[bend left=60]{f} &
    M_{- \lambda + 4} \rar[bend left=60]    \lar[bend left=60]{f} &
    \cdots            \rar[bend left=60]    \lar[bend left=60]    &
    M_{\lambda - 4}   \rar[bend left=60]{e} \lar[bend left=60]    &
    M_{\lambda - 2}   \rar[bend left=60]{e} \lar[bend left=60]{f} &
    M_\lambda                               \lar[bend left=60]{f}
  \end{tikzcd}
\end{center}

Corollary~\ref{thm:sl2-find-weights} can be used to find the eigenvalues of the
action of \(h\) on an arbitrary finite-dimensional
\(\mathfrak{sl}_2(K)\)-module. Namely, if \(M\) and \(N\) are
\(\mathfrak{sl}_2(K)\)-modules, \(m \in M_\mu\) and \(n \in N_\mu\) then by
computing
\[
  h \cdot (m + n) = h \cdot m + h \cdot n = \mu (m + n)
\]
we can see that \((M \oplus N)_\mu = M_\mu + N_\mu\). Hence the set of
eigenvalues of \(h\) in a \(\mathfrak{sl}_2(K)\)-module \(M\) is the union of
the sets of eigenvalues in its simple components, and the corresponding
eigenspaces are the direct sums of the eigenspaces of such simple components.

In particular, if the eigenvalues of \(M\) all have the same parity -- i.e.
they are either all even integers or all odd integers -- and the dimension of
each eigenspace is no greater than \(1\) then \(M\) must be simple, for if \(N,
L \subset M\) are submodules with \(M = N \oplus L\) then either \(N_\lambda =
0\) for all \(\lambda\) or \(L_\lambda = 0\) for all \(\lambda \in
\mathfrak{h}^*\). To conclude our analysis all it is left is to show that for
each \(\lambda \in \mathbb{Z}\) with \(\lambda \ge 0\) there is some
finite-dimensional simple \(M\) whose highest weight is \(\lambda\).
Surprisingly, we have already encountered such a \(M\).

\begin{theorem}\label{thm:sl2-exist-unique}
  For each \(\lambda \ge 0\), \(\lambda \in \mathbb{Z}\), there exists a unique
  simple \(\mathfrak{sl}_2(K)\)-module whose left-most eigenvalue of \(h\) is
  \(\lambda\).
\end{theorem}

\begin{proof}
  Let \(M = K[x, y]^{(\lambda)}\) be the \(\mathfrak{sl}_2(K)\)-module of
  homogeneous polynomials of degree \(\lambda\) in two variables, as in
  Example~\ref{ex:sl2-polynomial-subrep}. A simple calculation shows \(M_{n - 2
  k} = K x^{\lambda - k} y^k\) for \(k = 0, \ldots, \lambda\) and \(M_\mu = 0\)
  otherwise. In particular, the right-most eigenvalue of \(M\) is \(\lambda\).
  Alternatively, one can readily check that if \(K^2\) is the natural
  \(\mathfrak{sl}_2(K)\)-module, then \(M = \operatorname{Sym}^\lambda K^2\)
  satisfies the relations of (\ref{eq:irr-rep-of-sl2}). Indeed, the map
  \begin{align*}
    K[x, y]^{(\lambda)} & \to     \operatorname{Sym}^\lambda K^2 \\
             x^k y^\ell & \mapsto e_1^k \cdot e_2^\ell
  \end{align*}
  is an isomorphism.

  Either way, by the previous observation that a finite-dimensional
  \(\mathfrak{sl}_2(K)\)-module whose eigenvalues all have the same parity and
  whose corresponding eigenspace are all \(1\)-dimensional must be simple,
  \(M\) is simple. As for the uniqueness of \(M\), it suffices to notice that
  if \(N\) is a finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module with
  right-most eigenvalue \(\lambda\) and \(n \in N_\lambda\) is nonzero then
  relations (\ref{eq:irr-rep-of-sl2}) imply the map
  \begin{align*}
              M & \to     N           \\
    f^k \cdot m & \mapsto f^k \cdot n
  \end{align*}
  is an isomorphism -- this is, in effect, precisely how the isomorphism \(K[x,
  y]^{(\lambda)} \isoto \operatorname{Sym}^\lambda K^2\) was constructed.
\end{proof}

Our initial gamble of studying the eigenvalues of \(h\) may have seemed
arbitrary at first, but it payed off: we have \emph{completely} described
\emph{all} simple \(\mathfrak{sl}_2(K)\)-modules. It is not yet clear, however,
if any of this can be adapted to a general setting. In the following section we
shall double down on our gamble by trying to reproduce some of these results
for \(\mathfrak{sl}_3(K)\), hoping this will somehow lead us to a general
solution. In the process of doing so we will find some important clues on why
\(h\) was a sure bet and the race was fixed all along.

\section{Representations of \(\mathfrak{sl}_{2 + 1}(K)\)}\label{sec:sl3-reps}

The study of representations of \(\mathfrak{sl}_2(K)\) reminds me of the
difference between the derivative of a function \(\mathbb{R} \to \mathbb{R}\)
and that of a smooth map between manifolds: it is a simpler case of something
greater, but in some sense it is too simple of a case, and the intuition we
acquire from it can be a bit misleading in regards to the general setting. For
instance, I distinctly remember my Calculus I teacher telling the class ``the
derivative of the composition of two functions is not the composition of their
derivatives'' -- which is, of course, the \emph{correct} formulation of the
chain rule in the context of smooth manifolds.

The same applies to \(\mathfrak{sl}_2(K)\). It is a simple and beautiful
example, but unfortunately the general picture, modules of arbitrary semisimple
algebras, lacks its simplicity. The general purpose of this section is to
investigate to which extent the framework we developed for
\(\mathfrak{sl}_2(K)\) can be generalized to other semisimple Lie algebras. Of
course, the algebra \(\mathfrak{sl}_3(K)\) stands as a natural candidate for
potential generalizations: \(\mathfrak{sl}_3(K) = \mathfrak{sl}_{2 + 1}(K)\)
after all.

Our approach is very straightforward: we will fix some simple
\(\mathfrak{sl}_3(K)\)-module \(M\) and proceed step by step, at each point
asking ourselves how we could possibly adapt the framework we laid out for
\(\mathfrak{sl}_2(K)\). The first obvious question is one we have already asked
ourselves: why \(h\)?  More specifically, why did we choose to study its
eigenvalues and is there an analogue of \(h\) in \(\mathfrak{sl}_3(K)\)?

The answer to the former question is one we will discuss at length in the next
chapter, but for now we note that perhaps the most fundamental property of
\(h\) is that \emph{there exists an eigenvector \(m\) of \(h\) that is
annihilated by \(e\)} -- that being the generator of the right-most eigenspace
of \(h\). This was instrumental to our explicit description of the simple
\(\mathfrak{sl}_2(K)\)-modules culminating in
Theorem~\ref{thm:sl2-exist-unique}.

Our first task is to find some analogue of \(h\) in \(\mathfrak{sl}_3(K)\), but
it is still unclear what exactly we are looking for. We could say we are
looking for an element of \(M\) that is annihilated by some analogue of \(e\),
but the meaning of \emph{some analogue of \(e\)} is again unclear. In fact, as
we shall see, no such analogue exists and neither does such element. Instead,
the actual way to proceed is to consider the subalgebra
\[
  \mathfrak{h}
  = \left\{
    X \in
    \begin{pmatrix} K & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K \end{pmatrix}
    : \operatorname{Tr}(X) = 0
    \right\}
\]

The choice of \(\mathfrak{h}\) may seem like an odd choice at the moment, but
the point is we will later show that there exists some \(m \in M\) that is
simultaneously an eigenvector of each \(H \in \mathfrak{h}\) and annihilated by
half of the remaining elements of \(\mathfrak{sl}_3(K)\). This is exactly
analogous to the situation we found in \(\mathfrak{sl}_2(K)\): \(h\)
corresponds to the subalgebra \(\mathfrak{h}\), and the eigenvalues of \(h\) in
turn correspond to linear functions \(\lambda : \mathfrak{h} \to k\) such that
\(H \cdot m = \lambda(H) m\) for each \(H \in \mathfrak{h}\) and some nonzero
\(m \in M\). We call such functionals \(\lambda\) \emph{eigenvalues of
\(\mathfrak{h}\)}, and we say \emph{\(m\) is an eigenvector of
\(\mathfrak{h}\)}.

Once again, we will pay special attention to the eigenvalue decomposition
\begin{equation}\label{eq:weight-module}
  M = \bigoplus_\lambda M_\lambda
\end{equation}
where \(\lambda\) ranges over all eigenvalues of \(\mathfrak{h}\) and
\(M_\lambda = \{ m \in M : H \cdot m = \lambda(H) m, \forall H \in \mathfrak{h}
\}\). We should note that the fact that (\ref{eq:weight-module}) holds is not
at all obvious. This is because in general \(M_\lambda\) is not the eigenspace
associated with an eigenvalue of any particular operator \(H \in
\mathfrak{h}\), but instead the eigenspace of the action of the entire algebra
\(\mathfrak{h}\). Fortunately for us, (\ref{eq:weight-module}) always holds,
but we will postpone its proof to the next chapter.

Next we turn our attention to the remaining elements of \(\mathfrak{sl}_3(K)\).
In our analysis of \(\mathfrak{sl}_2(K)\) we saw that the eigenvalues of \(h\)
differed from one another by multiples of \(2\). A possible way to interpret
this is to say \emph{the eigenvalues of \(h\) differ from one another by
integral linear combinations of the eigenvalues of the adjoint action of
\(h\)}. In English, since
\begin{align*}
  \operatorname{ad}(h) e & = 2 e  &
  \operatorname{ad}(h) f & = -2 f &
  \operatorname{ad}(h) h & = 0,
\end{align*}
the eigenvalues of the adjoint actions of \(h\) are \(0\) and \(\pm 2\), and
the eigenvalues of the action of \(h\) on a simple
\(\mathfrak{sl}_2(K)\)-module differ from one another by integral multiples of
\(2\).

In the case of \(\mathfrak{sl}_3(K)\), a simple calculation shows that if \([H,
X]\) is scalar multiple of \(X\) for all \(H \in \mathfrak{h}\) then all but
one entry of \(X\) are zero. Hence the eigenvectors of the adjoint action of
\(\mathfrak{h}\) are \(E_{i j}\) and its eigenvalues are \(\epsilon_i -
\epsilon_j\), where
\[
  \epsilon_i
  \begin{pmatrix}
    a_1 &   0 &   0 \\
      0 & a_2 &   0 \\
      0 &   0 & a_3
  \end{pmatrix}
  = a_i
\]

Visually we may draw

\begin{figure}[h]
  \centering
  \begin{tikzpicture}[scale=2]
    \begin{rootSystem}{A}
      \filldraw[black] \weight{0}{0} circle (.5pt);
      \node[black, above right] at \weight{0}{0} {\small$0$};
      \wt[black]{-1}{2}
      \wt[black]{-2}{1}
      \wt[black]{1}{1}
      \wt[black]{-1}{-1}
      \wt[black]{2}{-1}
      \wt[black]{1}{-2}
      \node[above] at \weight{-1}{2}  {$\epsilon_2 - \epsilon_3$};
      \node[left]  at \weight{-2}{1}  {$\epsilon_2 - \epsilon_1$};
      \node[right] at \weight{1}{1}   {$\epsilon_1 - \epsilon_3$};
      \node[left]  at \weight{-1}{-1} {$\epsilon_3 - \epsilon_1$};
      \node[right] at \weight{2}{-1}  {$\epsilon_1 - \epsilon_2$};
      \node[below] at \weight{1}{-2}  {$\epsilon_3 - \epsilon_1$};
      \node[black, above] at \weight{1}{0}  {$\epsilon_1$};
      \node[black, above] at \weight{-1}{1} {$\epsilon_2$};
      \node[black, above] at \weight{0}{-1} {$\epsilon_3$};
      \filldraw[black] \weight{1}{0}  circle (.5pt);
      \filldraw[black] \weight{-1}{1} circle (.5pt);
      \filldraw[black] \weight{0}{-1} circle (.5pt);
    \end{rootSystem}
  \end{tikzpicture}
\end{figure}

If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) on
\(\mathfrak{sl}_3(K)\) associated to \(\alpha\) by
\(\mathfrak{sl}_3(K)_\alpha\) and fix some \(X \in \mathfrak{sl}_3(K)_\alpha\),
\(H \in \mathfrak{h}\) and \(m \in M_\lambda\) then
\[
  \begin{split}
    H \cdot (X \cdot m)
    & = X \cdot (H \cdot m) + [H, X] \cdot m \\
    & = X \cdot (\lambda(H) m) + \alpha(H) X \cdot m \\
    & = (\lambda + \alpha)(H) X \cdot m
  \end{split}
\]
so that \(X\) carries \(m\) to \(M_{\lambda + \alpha}\). In other words,
\(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(M\) by translating vectors
between eigenspaces}.

For instance \(\mathfrak{sl}_3(K)_{\epsilon_1 - \epsilon_3}\) will act on the
adjoint \(\mathfrak{sl}_3(K)\)-modules via
\begin{figure}[h]
  \centering
  \begin{tikzpicture}[scale=2]
    \begin{rootSystem}{A}
      \wt[black]{0}{0}
      \wt[black]{-1}{2}
      \wt[black]{-2}{1}
      \wt[black]{1}{1}
      \wt[black]{-1}{-1}
      \wt[black]{2}{-1}
      \wt[black]{1}{-2}
      \draw[-latex, black] \weight{-1.9}{1.1} -- \weight{-1.1}{1.9};
      \draw[-latex, black] \weight{-.9}{-.9} -- \weight{-.1}{-.1};
      \draw[-latex, black] \weight{0.1}{0.1} -- \weight{.9}{.9};
      \draw[-latex, black] \weight{1.1}{-1.9} -- \weight{1.9}{-1.1};
    \end{rootSystem}
  \end{tikzpicture}
\end{figure}

This is again entirely analogous to the situation we observed in
\(\mathfrak{sl}_2(K)\). In fact, we may once more conclude\dots

\begin{theorem}\label{thm:sl3-weights-congruent-mod-root}
  The eigenvalues of the action of \(\mathfrak{h}\) on a simple
  \(\mathfrak{sl}_3(K)\)-module \(M\) differ from one another by integral
  linear combinations of the eigenvalues \(\epsilon_i - \epsilon_j\) of the adjoint
  action of \(\mathfrak{h}\) on \(\mathfrak{sl}_3(K)\).
\end{theorem}

\begin{proof}
  This proof goes exactly as that of the analogous statement for
  \(\mathfrak{sl}_2(K)\): it suffices to note that if we fix some eigenvalue
  \(\lambda\) of \(\mathfrak{h}\) and let \(i\) and \(j\) vary then
  \[
    \bigoplus_{i j} M_{\lambda + \epsilon_i - \epsilon_j}
  \]
  is an invariant subspace of \(M\).
\end{proof}

To avoid confusion we better introduce some notation to differentiate between
eigenvalues of the action of \(\mathfrak{h}\) on \(M\) and eigenvalues of the
adjoint action of \(\mathfrak{h}\).

\begin{definition}\index{weights}
  Given a \(\mathfrak{sl}_3(K)\)-module \(M\), we will call the \emph{nonzero}
  eigenvalues of the action of \(\mathfrak{h}\) on \(M\) \emph{weights of
  \(M\)}. As you might have guessed, we will correspondingly refer to
  eigenvectors and eigenspaces of a given weight by \emph{weight vectors} and
  \emph{weight spaces}.
\end{definition}

It is clear from our previous discussion that the weights of the adjoint
\(\mathfrak{sl}_3(K)\)-module deserve some special attention.

\begin{definition}\index{weights!roots}
  The weights of the adjoint \(\mathfrak{sl}_3(K)\)-module are called
  \emph{roots of \(\mathfrak{sl}_3(K)\)}. Once again, the expressions
  \emph{root vector} and \emph{root space} are self-explanatory.
\end{definition}

Theorem~\ref{thm:sl3-weights-congruent-mod-root} can thus be restated as\dots

\begin{definition}\index{weights!root lattice}
  The lattice \(Q = \mathbb{Z} \langle \epsilon_i - \epsilon_j : i, j = 1, 2, 3
  \rangle\) is called \emph{the root lattice of \(\mathfrak{sl}_3(K)\)}.
\end{definition}

\begin{corollary}
  The weights of a simple \(\mathfrak{sl}_3(K)\)-module \(M\) are all congruent
  modulo the root lattice \(Q\). In other words, the weights of \(M\) all lie
  in a single \(Q\)-coset \(\xi \in \mfrac{\mathfrak{h}^*}{Q}\).
\end{corollary}

At this point we could keep playing the tedious game of reproducing the
arguments from the previous section in the context of \(\mathfrak{sl}_3(K)\).
However, it is more profitable to use our knowledge of
\(\mathfrak{sl}_2(K)\)-modules instead. Notice that the canonical inclusion
\(\mathfrak{gl}_2(K) \to \mathfrak{gl}_3(K)\) -- as described in
Example~\ref{ex:gln-inclusions} -- restricts to an injective homomorphism
\(\mathfrak{sl}_2(K) \to \mathfrak{sl}_3(K)\). In other words,
\(\mathfrak{sl}_2(K)\) is isomorphic to the image \(\mathfrak{s}_{1 2} = K
\langle E_{1 2}, E_{2 1}, [E_{1 2}, E_{2 1}] \rangle \subset
\mathfrak{sl}_3(K)\) of the inclusion \(\mathfrak{sl}_2(K) \to
\mathfrak{sl}_3(K)\). We may thus regard \(M\) as a
\(\mathfrak{sl}_2(K)\)-module by restricting to \(\mathfrak{s}_{1 2}\).

Our first observation is that, since the root spaces act by translation, the
subspace
\[
  \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k (\epsilon_1 - \epsilon_2)},
\]
must be invariant under the action of \(E_{1 2}\) and \(E_{2 1}\) for all
\(\lambda \in \mathfrak{h}^*\). This goes to show \(\bigoplus_k M_{\lambda - k
(\epsilon_1 - \epsilon_2)}\) is a \(\mathfrak{sl}_2(K)\)-submodule of \(M\) for all
weights \(\lambda\) of \(M\). Furthermore, one can easily see that the
eigenspace of the action of \(h\) on \(\bigoplus_{k \in \mathbb{Z}} M_{\lambda
- k (\epsilon_1 - \epsilon_2)}\) associated with the eigenvalue \(\lambda(H) - 2k\)
is precisely the weight space \(M_{\lambda - k (\epsilon_2 - \epsilon_1)}\).

Visually,
\begin{center}
  \begin{tikzpicture}
    \begin{rootSystem}{A}
      \node at \weight{-4}{2} (l) {};
      \node at \weight{-2}{1} (a) {};
      \node at \weight{0}{0}  (b) {};
      \node at \weight{2}{-1} (c) {};
      \node at \weight{4}{-2} (r) {};
      \draw \weight{-3}{1.5} -- \weight{3}{-1.5};
      \draw[dotted] \weight{-3}{1.5} -- (l);
      \draw[dotted] \weight{3}{-1.5} -- (r);
      \foreach \i in {-1, 0, 1}{\wt[black]{-2*\i}{\i}}
      \draw[-latex] (l) to[bend left=40] (a);
      \draw[-latex] (a) to[bend left=40] (b);
      \draw[-latex] (b) to[bend left=40] (c);
      \draw[-latex] (c) to[bend left=40] (r);
      \draw[-latex] (r) to[bend left=40] (c);
      \draw[-latex] (c) to[bend left=40] (b);
      \draw[-latex] (b) to[bend left=40] (a);
      \draw[-latex] (a) to[bend left=40] (l);
      \node[above right] at (b) {\small\(\lambda\)};
      \node[above right=2pt] at \weight{-3}{1.5} {\small\(E_{1 2}\)};
      \node[below left=2pt]  at \weight{-3}{1.5} {\small\(E_{2 1}\)};
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

In general, we find\dots

\begin{proposition}
  Given \(i < j\), the subalgebra \(\mathfrak{s}_{i j} = K \langle E_{i j},
  E_{j i}, [E_{i j}, E_{j i}] \rangle\) is isomorphic to
  \(\mathfrak{sl}_2(K)\). In addition, given a weight \(\lambda \in
  \mathfrak{h}^*\) of \(M\), the space
  \[
    N = \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k (\epsilon_i - \epsilon_j)}
  \]
  is invariant under the action of \(\mathfrak{s}_{i j}\) and
  \[
    M_{\lambda - k (\epsilon_i - \epsilon_j)}
    = N_{\lambda([E_{i j}, E_{j i}]) - 2k}
  \]
\end{proposition}

\begin{proof}
  In effect, if \(i \ne k \ne j\) then \(\mathfrak{s}_{i j}\) is the subalgebra
  of matrices whose \(k\)-th row and \(k\)-th column are nil. For instance, if
  \(i = 1\) and \(j = 3\) then
  \[
    \mathfrak{s}_{1 3}
    = \begin{pmatrix} K & 0 & K \\ 0 & 0 & 0 \\ K & 0 & K \end{pmatrix}
      \cap \mathfrak{sl}_3(K)
  \]

  In this case, the map
  \begin{align*}
    \mathfrak{s}_{1 3} & \to \mathfrak{sl}_2(K) \\
    \begin{pmatrix} a & 0 & b \\ 0 & 0 & 0 \\ c & 0 & -a \end{pmatrix}
    & \mapsto
    \begin{pmatrix}
                  a &    \tm{topA}{0} &              b \\
      \tm{leftA}{0} &               0 & \tm{rightA}{0} \\
                  c & \tm{bottomA}{0} &             -a
    \end{pmatrix}
    = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}
    \DrawVLine[black]{topA}{bottomA}
    \DrawHLine[black]{leftA}{rightA}
  \end{align*}
  is an isomorphism of Lie algebras. In general, the map
  \begin{align*}
    \mathfrak{s}_{i j} & \to     \mathfrak{sl}_2(K) \\
    E_{i j}            & \mapsto e                  \\
    E_{j i}            & \mapsto f                  \\
    [E_{i j}, E_{j i}] & \mapsto h
  \end{align*}
  which ``erases the \(k\)-th row and the \(k\)-th column'' of a matrix is an
  isomorphism.

  To see that \(N\) is invariant under the action of \(\mathfrak{s}_{i j}\), it
  suffices to notice \(E_{i j}\) and \(E_{j i}\) map \(m \in M_{\lambda - k
  (\epsilon_i - \epsilon_j)}\) to \(E_{i j} \cdot m \in M_{\lambda - (k - 1) (\epsilon_i -
  \epsilon_j)}\) and \(E_{j i} \cdot m \in M_{\lambda - (k + 1) (\epsilon_i -
  \epsilon_j)}\), respectively. Moreover,
  \[
    (\lambda - k (\epsilon_i - \epsilon_j))([E_{i j}, E_{j i}])
    = \lambda([E_{i j}, E_{j i}]) - k (1 - (-1))
    = \lambda([E_{i j}, E_{j i}]) - 2 k,
  \]
  which goes to show \(M_{\lambda - k (\epsilon_i - \epsilon_j)} \subset
  N_{\lambda([E_{i j}, E_{j i}]) - 2k}\). On the other hand, if we suppose \(0
  < \dim M_{\lambda - k (\epsilon_i - \epsilon_j)} < \dim N_{\lambda([E_{i j}, E_{j
  i}]) - 2 k}\) for some \(k\) we arrive at
  \[
    \dim N
    = \sum_k \dim M_{\lambda - k (\epsilon_i - \epsilon_j)}
    < \sum_k \dim N_{\lambda([E_{i j}, E_{j i}]) - 2k}
    = \dim N,
  \]
  a contradiction.
\end{proof}

As a first consequence of this, we show\dots

\begin{definition}\index{weights!weight lattice}
  The lattice \(P = \mathbb{Z} \langle \epsilon_1, \epsilon_2, \epsilon_3 \rangle\)
  is called \emph{the weight lattice of \(\mathfrak{sl}_3(K)\)}.
\end{definition}

\begin{corollary}\label{thm:sl3-weights-fit-in-weight-lattice}
  Every weight \(\lambda\) of \(M\) lies in the weight lattice \(P\).
\end{corollary}

\begin{proof}
  It suffices to note \(\lambda([E_{i j}, E_{j i}])\) is an eigenvalue of \(h\)
  in a finite-dimensional \(\mathfrak{sl}_2(K)\)-module, so it must be an
  integer. Now since
  \[
    \lambda
    \begin{pmatrix}
      a & 0 & 0     \\
      0 & b & 0     \\
      0 & 0 & -a -b
    \end{pmatrix}
    =
    \lambda
    \begin{pmatrix}
      a & 0 & 0  \\
      0 & 0 & 0  \\
      0 & 0 & -a
    \end{pmatrix}
    +
    \lambda
    \begin{pmatrix}
      0 & 0 & 0  \\
      0 & b & 0  \\
      0 & 0 & -b
    \end{pmatrix}
    =
    a \lambda([E_{1 3}, E_{3 1}]) + b \lambda([E_{2 3}, E_{3 2}]),
  \]
  which is to say \(\lambda = \lambda([E_{1 3}, E_{3 1}]) \epsilon_1 +
  \lambda([E_{2 3}, E_{3 2}]) \epsilon_2 \in P\).
\end{proof}

There is a clear parallel between the case of \(\mathfrak{sl}_3(K)\) and that
of \(\mathfrak{sl}_2(K)\), where we observed that the eigenvalues of the action
of \(h\) all lied in the lattice \(P = \mathbb{Z}\) and were congruent modulo
the sublattice \(Q = 2 \mathbb{Z}\).

Among other things, this last result goes to show that the diagrams we have
been drawing are in fact consistent with the theory we have developed. Namely,
since all weights lie in the rational span of \(\{\epsilon_1, \epsilon_2,
\epsilon_3\}\), we may as well draw them in the Cartesian plane. In fact, the
attentive reader may notice that \(\kappa(E_{1 2}, E_{2 3}) = - \sfrac{1}{2}\),
so that the angle -- with respect to the Killing form \(\kappa\) -- between the
root vectors \(E_{1 2}\) and \(E_{2 3}\) is precisely the same as the angle
between the points representing their roots \(\epsilon_1 - \epsilon_2\) and
\(\epsilon_2 - \epsilon_3\) in the Cartesian plane. Since \(\epsilon_1 - \epsilon_2\)
and \(\epsilon_2 - \epsilon_3\) span \(\mathfrak{h}^*\), this implies the diagrams
we've been drawing are given by an isometry \(\mathbb{Q} P \isoto
\mathbb{Q}^2\), where \(\mathbb{Q} P\) is endowed with the bilinear form
defined by \((\epsilon_i - \epsilon_j, \epsilon_k - \alpha_\ell) \mapsto \kappa(E_{i
j}, E_{k \ell})\) -- which we denote by \(\kappa\) as well.

To proceed we once more refer to the previously established framework: next we
saw that the eigenvalues of \(h\) form an unbroken string of integers symmetric
around \(0\). To prove this we analyzed the right-most eigenvalues of \(h\) and
their eigenvectors, providing an explicit description of the simple
\(\mathfrak{sl}_2(K)\)-modules in terms of these vectors. We may
reproduce these steps in the context of \(\mathfrak{sl}_3(K)\) by fixing a
direction in the plane an considering the weight lying the furthest in that
direction. 

For instance, let's say we fix the direction
\begin{center}
  \begin{tikzpicture}[scale=2]
    \begin{rootSystem}{A}
      \wt[black]{0}{0}
      \wt[black]{-1}{2}
      \wt[black]{-2}{1}
      \wt[black]{1}{1}
      \wt[black]{-1}{-1}
      \wt[black]{2}{-1}
      \wt[black]{1}{-2}
      \draw[-latex, black, thick] \weight{-1.5}{-.5} -- \weight{1.5}{.5};
    \end{rootSystem}
  \end{tikzpicture}
\end{center}
and let \(\lambda\) be the weight lying the furthest in this direction.

Its easy to see what we mean intuitively by looking at the previous picture,
but its precise meaning is still allusive. Formally this means we will choose a
linear functional \(f : \mathbb{Q} P \to \mathbb{Q}\) and pick the weight that
maximizes \(f\). To avoid any ambiguity we should choose the direction of a
line irrational with respect to the root lattice \(Q\) -- for if \(f\) is not
irrational there may be multiple choices the ``weight lying the furthest''
along this direction.

\begin{definition}
  We say that a root \(\alpha\) is positive if \(f(\alpha) > 0\) -- i.e. if it
  lies to the right of the direction we chose. Otherwise we say \(\alpha\) is
  negative. Notice that \(f(\alpha) \ne 0\) since by definition \(\alpha \ne
  0\) and \(f\) is irrational with respect to the lattice \(Q\).
\end{definition}

The next observation we make is that all others weights of \(M\) must lie in a
sort of \(\frac{1}{3}\)-cone with apex at \(\lambda\), as shown in
\begin{center}
  \begin{tikzpicture}
    \AutoSizeWeightLatticefalse
    \begin{rootSystem}{A}
      \weightLattice{3}
      \fill[gray!50,opacity=.2] (hex cs:x=5,y=-7) -- (hex cs:x=1,y=1) --
      (hex cs:x=-7,y=5) arc (150:270:{7*\weightLength});
      \draw[black, thick] (hex cs:x=5,y=-7) -- (hex cs:x=1,y=1) --
      (hex cs:x=-7,y=5);
      \filldraw[black] (hex cs:x=1,y=1) circle (1pt);
      \node[above right=-2pt] at (hex cs:x=1,y=1) {\small\(\lambda\)};
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

Indeed, if this is not the case then, by definition, \(\lambda\) is not the
weight placed the furthest in the direction we chose. Given our previous
assertion that the root spaces of \(\mathfrak{sl}_3(K)\) act on the weight
spaces of \(M\) via translation, this implies that \(E_{1 2}\), \(E_{1 3}\) and
\(E_{2 3}\) all annihilate \(M_\lambda\), or otherwise one of \(M_{\lambda +
\epsilon_1 - \epsilon_2}\), \(M_{\lambda + \epsilon_1 - \epsilon_3}\) and \(M_{\lambda
+ \epsilon_2 - \epsilon_3}\) would be nonzero -- which contradicts the hypothesis
that \(\lambda\) lies the furthest in the direction we chose. In other
words\dots

\begin{proposition}\label{thm:sl3-mod-is-highest-weight}
  There is a weight vector \(m \in M\) that is annihilated by all positive root
  spaces of \(\mathfrak{sl}_3(K)\).
\end{proposition}

\begin{proof}
  It suffices to note that the positive roots of \(\mathfrak{sl}_3(K)\) are
  precisely \(\epsilon_1 - \epsilon_2\), \(\epsilon_1 - \epsilon_3\) and \(\epsilon_2 -
  \epsilon_3\), with root vectors \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\),
  respectively.
\end{proof}

\index{weights!highest weight}
We call \(\lambda\) \emph{the highest weight of \(M\)}, and we call any nonzero
\(m \in M_\lambda\) \emph{a highest weight vector}. Going back to the case of
\(\mathfrak{sl}_2(K)\), we then constructed an explicit basis for our simple
module in terms of a highest weight vector, which allowed us to provide an
explicit description of the action of \(\mathfrak{sl}_2(K)\) in terms of its
standard basis, and finally we concluded that the eigenvalues of \(h\) must be
symmetrical around \(0\). An analogous procedure could be implemented for
\(\mathfrak{sl}_3(K)\) -- and indeed that's what we will do later down the line
-- but instead we would like to focus on the problem of finding the weights of
\(M\) in the first place.

We will start out by trying to understand the weights in the boundary of
previously drawn cone. As we have just seen, we can get to other weight spaces
from \(M_\lambda\) by successively applying \(E_{2 1}\).
\begin{center}
  \begin{tikzpicture}
    \begin{rootSystem}{A}
      \node at \weight{3}{1}  (a) {};
      \node at \weight{1}{2}  (b) {};
      \node at \weight{-1}{3} (c) {};
      \node at \weight{-3}{4} (d) {};
      \node at \weight{-5}{5} (e) {};
      \draw \weight{3}{1} -- \weight{-4}{4.5};
      \draw[dotted] \weight{-4}{4.5} -- \weight{-5}{5};
      \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
      \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
      \draw[-latex] (a) to[bend left=40] (b);
      \draw[-latex] (b) to[bend left=40] (c);
      \draw[-latex] (c) to[bend left=40] (d);
      \draw[-latex] (d) to[bend left=40] (e);
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

Notice that \(\lambda([E_{1 2}, E_{2 1}]) \in \mathbb{Z}\) is the right-most
eigenvalue of the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_{k \in \mathbb{Z}}
M_{\lambda - k (\epsilon_1 - \epsilon_2)}\). In particular, \(\lambda([E_{1 2},
E_{2 1}])\) must be positive. In addition, since the eigenspace of the
eigenvalue \(\lambda([E_{1 2}, E_{2 1}]) - 2k\) of the action of \(h\) on
\(\bigoplus_{k \in \mathbb{N}} M_{\lambda - k (\epsilon_1 - \epsilon_2)}\) is
\(M_{\lambda - k (\epsilon_1 - \epsilon_2)}\), the weights of \(M\) appearing the
string \(\lambda, \lambda + (\epsilon_1 - \epsilon_2), \ldots, \lambda + k
(\epsilon_1 - \epsilon_2), \ldots\) must be symmetric with respect to the line
\(\kappa(\epsilon_1 - \epsilon_2, \alpha) =  0\). The picture is thus
\begin{center}
  \begin{tikzpicture}
    \AutoSizeWeightLatticefalse
    \begin{rootSystem}{A}
      \setlength{\weightRadius}{2pt}
      \weightLattice{4}
      \draw[thick] \weight{3}{1} -- \weight{-3}{4};
      \wt[black]{0}{0}
      \node[above left] at \weight{0}{0} {\small\(0\)};
      \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
      \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
      \draw[very thick] \weight{0}{-4} -- \weight{0}{4}
      node[above]{\small\(\kappa(\epsilon_1 - \epsilon_2, \alpha) = 0\)};
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

We could apply this same argument to the subspace \(\bigoplus_k M_{\lambda - k
(\epsilon_2 - \epsilon_3)}\), so that the weights in this subspace must be
symmetric with respect to the line \(\kappa(\epsilon_2 - \epsilon_3, \alpha) = 0\).
The picture is now
\begin{center}
  \begin{tikzpicture}
    \AutoSizeWeightLatticefalse
    \begin{rootSystem}{A}
      \setlength{\weightRadius}{2pt}
      \weightLattice{4}
      \draw[thick] \weight{3}{1} -- \weight{-3}{4};
      \draw[thick] \weight{3}{1} -- \weight{4}{-1};
      \wt[black]{0}{0}
      \wt[black]{4}{-1}
      \node[above left] at \weight{0}{0} {\small\(0\)};
      \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
      \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
      \draw[very thick] \weight{0}{-4} -- \weight{0}{4}
      node[above]{\small\(\kappa(\epsilon_1 - \epsilon_2, \alpha) = 0\)};
      \draw[very thick] \weight{-4}{0} -- \weight{4}{0}
      node[right]{\small\(\kappa(\epsilon_2 - \epsilon_3, \alpha) = 0\)};
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

Needless to say, we could keep applying this method to the weights at the ends
of our string, arriving at
\begin{center}
  \begin{tikzpicture}
    \AutoSizeWeightLatticefalse
    \begin{rootSystem}{A}
      \setlength{\weightRadius}{2pt}
      \weightLattice{5}
      \draw[thick] \weight{3}{1} -- \weight{-3}{4};
      \draw[thick] \weight{3}{1} -- \weight{4}{-1};
      \draw[thick] \weight{-3}{4} -- \weight{-4}{3};
      \draw[thick] \weight{-4}{3} -- \weight{-1}{-3};
      \draw[thick] \weight{1}{-4} -- \weight{4}{-1};
      \draw[thick] \weight{-1}{-3} -- \weight{1}{-4};
      \wt[black]{-4}{3}
      \wt[black]{-3}{1}
      \wt[black]{-2}{-1}
      \wt[black]{-1}{-3}
      \wt[black]{1}{-4}
      \wt[black]{2}{-3}
      \wt[black]{3}{-2}
      \wt[black]{4}{-1}
      \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
      \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
      \draw[very thick] \weight{-5}{5} -- \weight{5}{-5};
      \draw[very thick] \weight{0}{-5} -- \weight{0}{5};
      \draw[very thick] \weight{-5}{0} -- \weight{5}{0};
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

We claim all dots \(\mu\) lying inside the hexagon we have drawn must also be
weights -- i.e. \(M_\mu \ne 0\). Indeed, by applying the same argument to an
arbitrary weight \(\nu\) in the boundary of the hexagon we get a
\(\mathfrak{sl}_2(K)\)-module whose weights correspond to weights of \(M\)
lying in a string inside the hexagon, and whose right-most weight is precisely
the weight of \(M\) we started with.
\begin{center}
  \begin{tikzpicture}
    \AutoSizeWeightLatticefalse
    \begin{rootSystem}{A}
      \setlength{\weightRadius}{2pt}
      \weightLattice{5}
      \draw[thick] \weight{3}{1}   -- \weight{-3}{4};
      \draw[thick] \weight{3}{1}   -- \weight{4}{-1};
      \draw[thick] \weight{-3}{4}  -- \weight{-4}{3};
      \draw[thick] \weight{-4}{3}  -- \weight{-1}{-3};
      \draw[thick] \weight{1}{-4}  -- \weight{4}{-1};
      \draw[thick] \weight{-1}{-3} -- \weight{1}{-4};
      \wt[black]{-4}{3}
      \wt[black]{-3}{1}
      \wt[black]{-2}{-1}
      \wt[black]{-1}{-3}
      \wt[black]{1}{-4}
      \wt[black]{2}{-3}
      \wt[black]{3}{-2}
      \wt[black]{4}{-1}
      \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
      \node[above right=-2pt] at \weight{1}{2} {\small\(\nu\)};
      \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
      \draw[very thick] \weight{-5}{5} -- \weight{5}{-5};
      \draw[very thick] \weight{0}{-5} -- \weight{0}{5};
      \draw[very thick] \weight{-5}{0} -- \weight{5}{0};
      \draw[dashed, thick] \weight{1}{2} -- \weight{-2}{-1};
      \wt[black]{1}{2}
      \wt[black]{-2}{-1}
      \wt[black]{0}{1}
      \wt[black]{-1}{0}
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

By construction, \(\nu\) corresponds to the right-most weight of a
\(\mathfrak{sl}_2(K)\)-module, so that all dots lying on the dashed string must
occur in \(\mathfrak{sl}_2(K)\)-module. Hence they must also be weights of
\(M\). The final picture is thus
\begin{center}
  \begin{tikzpicture}
    \AutoSizeWeightLatticefalse
    \begin{rootSystem}{A}
      \setlength{\weightRadius}{2pt}
      \weightLattice{5}
      \draw[thick] \weight{3}{1} -- \weight{-3}{4};
      \draw[thick] \weight{3}{1} -- \weight{4}{-1};
      \draw[thick] \weight{-3}{4} -- \weight{-4}{3};
      \draw[thick] \weight{-4}{3} -- \weight{-1}{-3};
      \draw[thick] \weight{1}{-4} -- \weight{4}{-1};
      \draw[thick] \weight{-1}{-3} -- \weight{1}{-4};
      \wt[black]{-4}{3}
      \wt[black]{-3}{1}
      \wt[black]{-2}{-1}
      \wt[black]{-1}{-3}
      \wt[black]{1}{-4}
      \wt[black]{2}{-3}
      \wt[black]{3}{-2}
      \wt[black]{4}{-1}
      \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
      \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
      \draw[very thick] \weight{-5}{5} -- \weight{5}{-5};
      \draw[very thick] \weight{0}{-5} -- \weight{0}{5};
      \draw[very thick] \weight{-5}{0} -- \weight{5}{0};
      \wt[black]{-2}{2}
      \wt[black]{0}{1}
      \wt[black]{-1}{0}
      \wt[black]{0}{-2}
      \wt[black]{1}{-1}
      \wt[black]{2}{0}
    \end{rootSystem}
  \end{tikzpicture}
\end{center}

\index{weights!weight diagrams}
This final picture is known as \emph{the weight diagram of \(M\)}. Finally\dots

\begin{theorem}\label{thm:sl3-irr-weights-class}
  The weights of \(M\) are precisely the elements of the weight lattice \(P\)
  congruent to \(\lambda\) module the sublattice \(Q\) and lying inside hexagon
  with vertices the images of \(\lambda\) under the group generated by
  reflections across the lines \(\kappa(\epsilon_i - \epsilon_j, \alpha) = 0\).
\end{theorem}

Having found all of the weights of \(M\), the only thing we are missing is an
existence and uniqueness theorem analogous to
Theorem~\ref{thm:sl2-exist-unique}. It is clear from the symmetries of the
locus of weights found in Theorem~\ref{thm:sl3-irr-weights-class} that if
\(\lambda \in P\) is the highest weight of some finite-dimensional simple
\(\mathfrak{sl}_3(K)\)-module \(M\) then \(\lambda\) lies in the cone
\(\mathbb{N} \langle \epsilon_1, - \epsilon_3 \rangle\). What's perhaps more
surprising is the fact that this condition is sufficient for the existence of
such a \(M\). In other words, our next goal is establishing\dots

\begin{definition}\index{weights!dominant weight}
  An element \(\lambda \in P\) is called \emph{dominant} if it lies in the cone
  \(\mathbb{N} \langle \epsilon_1, - \epsilon_3 \rangle\).
\end{definition}

\begin{theorem}\label{thm:sl3-existence-uniqueness}
  For each dominant \(\lambda \in P\), there exists precisely one
  finite-dimensional simple \(\mathfrak{sl}_3(K)\)-module \(M\) whose highest
  weight is \(\lambda\).
\end{theorem}

To proceed further we once again refer to the approach we employed in the case
of \(\mathfrak{sl}_2(K)\): next we showed in
Proposition~\ref{thm:basis-of-irr-rep} that any simple
\(\mathfrak{sl}_2(K)\)-module is spanned by the images of its highest weight
vector under \(f\). A more abstract way of putting it is to say that a simple
\(\mathfrak{sl}_2(K)\)-module \(M\) of is spanned by the images of its highest
weight vector under successive applications of the action of half of the root
spaces of \(\mathfrak{sl}_2(K)\). The advantage of this alternative formulation
is, of course, that the same holds for \(\mathfrak{sl}_3(K)\).
Specifically\dots

\begin{proposition}\label{thm:sl3-positive-roots-span-all-irr-rep}
  Given a simple \(\mathfrak{sl}_3(K)\)-module \(M\) and a highest weight
  vector \(m \in M\), \(M\) is spanned by the images of \(m\) under successive
  applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\).
\end{proposition}

\begin{proof}
  Given the fact \(M\) is simple, it suffices to show that the subspace \(N\)
  spanned by successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3
  2}\) to \(m\) is stable under the action of \(\mathfrak{sl}_3(K)\). In
  addition, since \([E_{2 1}, E_{3 1}] = [E_{3 1}, E_{3 2}] = 0\) and \([E_{2
  1}, E_{3 2}] = - E_{3 1}\), all successive product of \(E_{2 1}\), \(E_{3
  1}\) and \(E_{3 2}\) in \(\mathcal{U}(\mathfrak{sl}_3(K))\) can be written as
  \(E_{2 1}^a E_{3 1}^b E_{3 1}^c\) for some \(a\), \(b\) and \(c\), so that
  \(N\) is spanned by the elements \(E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m\).

  Recall that \(E_{i j}\) maps \(M_\mu\) to \(M_{\mu + \epsilon_i - \epsilon_j}\).
  In particular, \(E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m \in M_{\lambda - a
  (\epsilon_1 - \epsilon_2) - b (\epsilon_1 - \epsilon_3) - c (\epsilon_2 - \epsilon_3)}\).
  In other words,
  \[
    H E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m
    = (\lambda - a (\epsilon_1 - \epsilon_2)
               - b (\epsilon_1 - \epsilon_3)
               - c (\epsilon_2 - \epsilon_3))(H)
      E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m
      \in N
  \]
  for all \(H \in \mathfrak{h}\) and \(N\) is stable under the action of
  \(\mathfrak{h}\). On the other hand, \(N\) is clearly stable under the action
  of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\). All it is left is to show \(N\)
  is stable under the action of \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\).

  We begin by analyzing the case of \(E_{1 2}\). We have
  \[
    \begin{split}
      E_{1 2} E_{2 1}^a E_{3 1}^b E_{3 2}^c \cdot m
      & = ([E_{1 2}, E_{2 1}] + E_{2 1} E_{1 2})
          E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c \cdot m \\
      & = E_{2 1} ([E_{1 2}, E_{2 1}] + E_{2 1} E_{1 2})
          E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \phantom{=} \; +
          (\lambda - (a - 1) (\epsilon_1 - \epsilon_2)
                   - b (\epsilon_1 - \epsilon_3)
                   - c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}])
          E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c \cdot m \\
      & = E_{2 1}^2 ([E_{1 2}, E_{2 1}] + E_{2 1} E_{1 2})
          E_{2 1}^{a - 3} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \phantom{=} \; +
          (\lambda - (a - 1) (\epsilon_1 - \epsilon_2)
                   - b (\epsilon_1 - \epsilon_3)
                   - c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}])
          E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \phantom{=} \; +
        (\lambda - (a - 2) (\epsilon_1 - \epsilon_2)
                   - b (\epsilon_1 - \epsilon_3)
                   - c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}])
          E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \; \; \vdots \\
      & = E_{2 1}^a E_{1 2} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \phantom{=} \; +
          (\lambda - (a - 1) (\epsilon_1 - \epsilon_2)
                   - b (\epsilon_1 - \epsilon_3)
                   - c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}])
          E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \phantom{=} \; +
        (\lambda - (a - 2) (\epsilon_1 - \epsilon_2)
                   - b (\epsilon_1 - \epsilon_3)
                   - c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}])
          E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c \cdot m \\
      & \phantom{=} \; \; \, \vdots \\
      & \phantom{=} \; +
        (\lambda - (a - a) (\epsilon_1 - \epsilon_2)
                   - b (\epsilon_1 - \epsilon_3)
                   - c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}])
          E_{2 1}^{a - a} E_{3 1}^b E_{3 2}^c \cdot m \\
    \end{split}
  \]

  Since \((\lambda - (a - k) (\epsilon_1 - \epsilon_2) - b (\epsilon_1 - \epsilon_3) -
  c (\epsilon_2 - \epsilon_3)) ([E_{1 2}, E_{2 1}]) E_{2 1}^{a - k} E_{3 1}^b
  E_{3 2}^c \cdot m \in N\) for all \(k\), it suffices to show \(E_{2 1}^a E_{1
  2} E_{3 1}^b E_{3 2}^c \cdot m \in N\). But
  \[
    \begin{split}
      E_{1 2} E_{3 1}^b
      & = (E_{3 1} E_{1 2} - E_{3 2}) E_{3 1}^{b - 1} \\
      & = E_{3 1} E_{1 2} E_{3 1}^{b - 1}
        - E_{3 1} E_{3 2} E_{3 1}^{b - 1} \\
      & = E_{3 1} (E_{3 1} E_{1 2} - E_{3 2}) E_{3 1}^{b - 2}
        - E_{3 2} E_{3 1}^b \\
      & \; \; \vdots \\
      & = E_{3 1}^b  E_{1 2} - b E_{3 2} E_{3 1}^b \\
    \end{split},
  \]
  given \([E_{1 2}, E_{3 1}] = - E_{3 2}\) and \([E_{3 2}, E_{3 1}] = 0\).
  It then follows from the fact \(E_{1 2} \cdot m = 0\) that
  \[
    E_{2 1}^a E_{1 2} E_{3 1}^b E_{3 2}^c \cdot m
    = E_{2 1}^a E_{3 1}^b E_{3 2}^c E_{1 2} \cdot m
    - b E_{2 1}^a E_{3 1}^b E_{3 2}^{c + 1} \cdot m
    = - b E_{2 1}^a E_{3 1}^b E_{3 2}^{c + 1} \cdot m \in N,
  \]
  given that \(E_{1 2}\) and \(E_{3 2}\) commute. Hence \(E_{1 2} \cdot (E_{2
  1}^a E_{3 1}^b E_{3 2}^c \cdot m) \in N\). Similarly,
  \[
    E_{1 3} \cdot (E_{2 1}^a E_{3 1}^b E_{3 2}^c \cdot m),
    E_{2 3} \cdot (E_{2 1}^a E_{3 1}^b E_{3 2}^c \cdot m) \in N
  \]
\end{proof}

The same argument also goes to show\dots

\begin{corollary}\label{thm:irr-component-of-high-vec}
  Given a finite-dimensional \(\mathfrak{sl}_3(K)\)-module \(M\) with highest
  weight \(\lambda\) and \(m \in M_\lambda\), the subspace spanned by
  successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\) to \(m\)
  is a simple submodule whose highest weight is \(\lambda\).
\end{corollary}

This is very interesting to us since it implies that finding \emph{any}
finite-dimensional module whose highest weight is \(\lambda\) is enough for
establishing the ``existence'' part of
Theorem~\ref{thm:sl3-existence-uniqueness}. Moreover, constructing such a
module turns out to be quite simple.

\begin{proof}[Proof of existence]
  Take \(\lambda = k \epsilon_1 - \ell \epsilon_3 \in P\) with \(k, \ell \ge 0\),
  so that \(\lambda\) is dominant. Consider the natural
  \(\mathfrak{sl}_3(K)\)-module \(K^3\). We claim that the highest weight of
  \(\operatorname{Sym}^k K^3 \otimes \operatorname{Sym}^\ell (K^3)^*\) is
  \(\lambda\).

  First of all, notice that the weight vector of \(K^3\) are the canonical
  basis elements \(e_1\), \(e_2\) and \(e_3\), whose corresponding weights are
  \(\epsilon_1\), \(\epsilon_2\) and \(\epsilon_3\) respectively. Hence the weight
  diagram of \(K^3\) is
  \begin{center}
    \begin{tikzpicture}[scale=2]
      \AutoSizeWeightLatticefalse
      \begin{rootSystem}{A}
        \weightLattice{2}
        \wt[black]{1}{0}
        \wt[black]{-1}{1}
        \wt[black]{0}{-1}
        \node[right] at \weight{1}{0}  {$\epsilon_1$};
        \node[above left] at \weight{-1}{1} {$\epsilon_2$};
        \node[below left] at \weight{0}{-1} {$\epsilon_3$};
      \end{rootSystem}
    \end{tikzpicture}
  \end{center}
  and \(\epsilon_1\) is the highest weight of \(K^3\).

  On the one hand, if \(\{f_1, f_2, f_3\}\) is the dual basis for \(\{e_1, e_2,
  e_3\}\) then \(H \cdot f_i = - \epsilon_i(H) f_i\) for each \(H \in
  \mathfrak{h}\), so that the weights of \((K^3)^*\) are precisely the
  opposites of the weights of \(K^3\). In other words,
  \begin{center}
    \begin{tikzpicture}[scale=2]
      \AutoSizeWeightLatticefalse
      \begin{rootSystem}{A}
        \weightLattice{2}
        \wt[black]{-1}{0}
        \wt[black]{1}{-1}
        \wt[black]{0}{1}
        \node[left]        at \weight{-1}{0} {$-\epsilon_1$};
        \node[below right] at \weight{1}{-1} {$-\epsilon_2$};
        \node[above right] at \weight{0}{1}  {$-\epsilon_3$};
      \end{rootSystem}
    \end{tikzpicture}
  \end{center}
  is the weight diagram of \((K^3)^*\) and \(\epsilon_3\) is the highest weight
  of \((K^3)^*\).

  On the other hand if we fix two \(\mathfrak{sl}_3(K)\)-modules \(N\) and
  \(L\), by computing
  \[
    \begin{split}
      H \cdot (n \otimes l)
      & = H \cdot n \otimes l + n \otimes H \cdot l \\
      & = \lambda(H) n \otimes l + n \otimes \mu(H) l \\
      & = (\lambda + \mu)(H) \, (n \otimes l)
    \end{split}
  \]
  for each \(H \in \mathfrak{h}\), \(n \in N_\lambda\) and \(l \in L_\mu\) we
  can see that the weights of \(N \otimes L\) are precisely the sums of the
  weights of \(N\) with the weights of \(L\).

  This implies that the highest weights of \(\operatorname{Sym}^k K^3\) and
  \(\operatorname{Sym}^\ell (K^3)^*\) are \(k \epsilon_1\) and \(- \ell
  \epsilon_3\) respectively -- with highest weight vectors \(e_1^k\) and
  \(f_3^\ell\). Furthermore, by the same token the highest weight of
  \(\operatorname{Sym}^k K^3 \otimes \operatorname{Sym}^\ell (K^3)^*\) must be
  \(\lambda = k e_1 - \ell e_3\) -- with highest weight vector \(e_1^k \otimes
  f_3^\ell\).
\end{proof}

The ``uniqueness'' part of Theorem~\ref{thm:sl3-existence-uniqueness} is even
simpler than that.

\begin{proof}[Proof of uniqueness]
  Let \(M\) and \(N\) be two simple \(\mathfrak{sl}_3(K)\)-modules with highest
  weight \(\lambda\). By Theorem~\ref{thm:sl3-irr-weights-class}, the weights
  of \(M\) are precisely the same as those of \(N\).

  Now by computing
  \[
    H \cdot (m + n)
    = H \cdot m + H \cdot n
    = \mu(H) m + \mu(H) n
    = \mu(H) (m + n)
  \]
  for each \(H \in \mathfrak{h}\), \(m \in M_\mu\) and \(n \in N_\mu\), we can
  see that the weights of \(M \oplus N\) are same as those of \(M\) and \(N\).
  Hence the highest weight of \(M \oplus N\) is \(\lambda\) -- with highest
  weight vectors given by the sum of highest weight vectors of \(M\) and \(N\).

  Fix some \(m \in M_\lambda\) and \(n \in N_\lambda\) and consider the
  submodule \(L = \mathcal{U}(\mathfrak{sl}_3(K)) \cdot m + n \subset M \oplus
  N\) generated by \(m + n\). Since \(m + n\) is a highest weight of \(M \oplus
  N\), it follows from corollary~\ref{thm:irr-component-of-high-vec} that \(L\)
  is simple. The projection maps \(\pi_1 : L \to M\), \(\pi_2 : L \to N\),
  being nonzero homomorphism between simple \(\mathfrak{sl}_3(K)\)-modules,
  must be isomorphism. Finally,
  \[
    M \cong L \cong N
  \]
\end{proof}

We have been very successful in our pursue for a classification of the simple
modules of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\), but so far we
have mostly postponed the discussion on the motivation behind our methods. In
particular, we did not explain why we chose \(h\) and \(\mathfrak{h}\), and
neither why we chose to look at their eigenvalues. Apart from the obvious fact
we already knew it would work a priory, why did we do all that? In the
following chapter we will attempt to answer this question by looking at what we
did in the last chapter through more abstract lenses and studying the
representations of an arbitrary finite-dimensional semisimple Lie algebra
\(\mathfrak{g}\).