lie-algebras-and-their-representations

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -918,7 +918,7 @@ define\dots
 \begin{example}\label{ex:sl2-polynomial-subrep}
   Let \(K[x, y]\) be the \(\mathfrak{sl}_2(K)\)-module as in
   example~\ref{ex:sl2-polynomial-rep}. Since \(e\), \(f\) and \(h\) all
-  preserve the degree of monomials, the space \(K_n[x, y] = \bigoplus_{k = 0}^n
+  preserve the degree of monomials, the space \(K[x, y]^{(n)} = \bigoplus_{k = 0}^n
   K x^{n - k} y^k\) of homogeneous polynomials of degree \(n\) is a
   finite-dimensional subrepresentation of \(K[x, y]\).
 \end{example}

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -12,7 +12,7 @@ Specifically, we'll classify the irreducible finite-dimensional representations
 of certain low-dimensional semisimple Lie algebras.
 
 Throughout the previous chapters \(\mathfrak{sl}_2(K)\) has afforded us
-surprisingly elucidating examples, so it will serve as our first candidate for
+surprisingly illuminating examples, so it will serve as our first candidate for
 low-dimensional algebra. We begin our analysis by recalling that the elements
 \begin{align*}
   e & = \begin{pmatrix} 0 & 1 \\ 0 &  0 \end{pmatrix} &
@@ -45,7 +45,7 @@ a subrepresentation of \(V\). Indeed, if \(v \in V_\lambda\) then
 
 In other words, \(e\) sends an element of \(V_\lambda\) to an element of
 \(V_{\lambda + 2}\), while \(f\) sends it to an element of \(V_{\lambda - 2}\).
-Hence
+Visually, we may draw
 \begin{center}
   \begin{tikzcd}
     \cdots \arrow[bend left=60]{r}
@@ -55,18 +55,18 @@ Hence
     & \cdots \arrow[bend left=60]{l}
   \end{tikzcd}
 \end{center}
-and \(\bigoplus_{n \in \ZZ} V_{\lambda + 2 n}\) is an
-\(\mathfrak{sl}_2(K)\)-invariant subspace. This implies
+
+This implies \(\bigoplus_{k \in \ZZ} V_{\lambda - 2 k}\) is an
+\(\mathfrak{sl}_2(K)\)-invariant subspace, which goes to show
 \[
-  V = \bigoplus_{n \in \ZZ} V_{\lambda + 2 n},
+  V = \bigoplus_{k \in \ZZ} V_{\lambda - 2 k},
 \]
-so that the eigenvalues of \(h\) all have the form \(\lambda + 2 n\) for some
-\(n\) -- since \(V_\mu = 0\) for all \(\mu \notin \lambda + 2 \ZZ\).
-
-Even more so, if \(a = \min \{ n \in \ZZ : V_{\lambda + 2 n} \ne 0 \}\) and
-\(b = \max \{ n \in \ZZ : V_{\lambda + 2 n} \ne 0 \}\) we can see that
+and the eigenvalues of \(h\) all have the form \(\lambda - 2 k\) for some
+\(k\).
+Even more so, if \(a = \max \{ k \in \ZZ : V_{\lambda - 2 k} \ne 0 \}\) and
+\(b = \min \{ k \in \ZZ : V_{\lambda - 2 k} \ne 0 \}\) we can see that
 \[
-  \bigoplus_{\substack{n \in \ZZ \\ a \le n \le b}} V_{\lambda + 2 n}
+  \bigoplus_{\substack{k \in \ZZ \\ a \le n \le b}} V_{\lambda - 2 k}
 \]
 is also an \(\mathfrak{sl}_2(K)\)-invariant subspace, so that the eigenvalues
 of \(h\) form an unbroken string
@@ -80,9 +80,9 @@ eigenvalues. To do so, we suppose without any loss in generality that
 \(\lambda\) is the right-most eigenvalue of \(h\), fix some nonzero \(v \in
 V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
 
-\begin{theorem}\label{thm:basis-of-irr-rep}
+\begin{proposition}\label{thm:basis-of-irr-rep}
   The set \(\{v, f v, f^2, \ldots\}\) is a basis for \(V\).
-\end{theorem}
+\end{proposition}
 
 \begin{proof}
   First of all, notice \(f^k v\) lies in \(V_{\lambda - 2 k}\), so that \(\{v,
@@ -94,8 +94,9 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
 
   The fact that \(h f^k v \in K \langle v, f v, f^2 v, \ldots \rangle\) follows
   immediately from our previous assertion that \(f^k v \in V_{\lambda - 2 k}\)
-  -- indeed, \(h f^k v = (\lambda - 2 k) f^k v\). Seeing \(e f^k v \in K
-  \langle v, f v, f^2 v, \ldots \rangle\) is a bit more complex. Clearly,
+  -- indeed, \(h f^k v = (\lambda - 2 k) f^k v \in K \langle v, f v, f^2, v,
+  \ldots \rangle\). Seeing \(e f^k v \in K \langle v, f v, f^2 v, \ldots
+  \rangle\) is a bit more complex. Clearly,
   \[
     \begin{split}
       e f v
@@ -128,7 +129,7 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
   which is to say \(e v = 0\).
 \end{note}
 
-Theorem~\ref{thm:basis-of-irr-rep} may seem unrelated to our problem at first,
+Proposition~\ref{thm:basis-of-irr-rep} may seem unrelated to our problem at first,
 but its significance lies in the fact that we have just provided a complete
 description of the action of \(\mathfrak{sl}_2(K)\) in \(V\). In other
 words\dots
@@ -154,7 +155,7 @@ words\dots
   \end{align*}
 \end{proof}
 
-Other important consequences of theorem~\ref{thm:basis-of-irr-rep} are\dots
+Other important consequences of proposition~\ref{thm:basis-of-irr-rep} are\dots
 
 \begin{corollary}
   Every \(h\) eigenspace is 1-dimensional.
@@ -166,7 +167,7 @@ Other important consequences of theorem~\ref{thm:basis-of-irr-rep} are\dots
   k}\) is \(f^k v\).
 \end{proof}
 
-\begin{corollary}
+\begin{corollary}\label{thm:sl2-find-weights}
   The eigenvalues of \(h\) in \(V\) form a symmetric, unbroken string of
   integers separated by intervals of length \(2\) whose right-most value is
   \(\dim V - 1\).
@@ -175,7 +176,7 @@ Other important consequences of theorem~\ref{thm:basis-of-irr-rep} are\dots
 \begin{proof}
   If \(f^m\) is the lowest power of \(f\) that annihilates \(v\), it follows
   from the formula for \(e f^k v\) obtained in the proof of
-  theorem~\ref{thm:basis-of-irr-rep} that
+  proposition~\ref{thm:basis-of-irr-rep} that
   \[
     0 = e 0 = e f^m v = m (\lambda + 1 - m) f^{m - 1} v
   \]
@@ -192,6 +193,26 @@ Other important consequences of theorem~\ref{thm:basis-of-irr-rep} are\dots
   left-most eigenvalue of \(h\) is precisely \(n - 2 (m - 1) = -n\).
 \end{proof}
 
+Corollary~\ref{thm:sl2-find-weights} can be used to find the eigenvalues of the
+action of \(h\) in an arbitrary finite-dimensional
+\(\mathfrak{sl}_2(K)\)-module. Namely, if \(V\) and \(W\) are representations
+of \(\mathfrak{sl}_2(K)\), \(v \in V_\lambda\) and \(w \in W_\lambda\) then by
+computing
+\[
+  H (v + w) = Hv + Hw = \lambda(H) \cdot (v + w)
+\]
+we can see that \((V \oplus W)_\lambda = V_\lambda + W_\lambda\). Hence the set
+of eigenvalues of \(h\) in a representation \(V\) is the union of the sets of
+eigenvalues in its irreducible components, and the correspoding eigenspaces are
+the direct sums of the eigenspaces of such irreducible components.
+
+In particular, if the eigenvalues of \(V\) 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 greather than \(1\) then \(V\) must be irreducible, for
+if \(U, W \subset V\) are subrepresentations with \(V = W \oplus U\) then
+either \(W_\lambda = 0\) for all \(\lambda\) or \(U_\lambda = 0\) for all
+\(\lambda\).
+
 We now know every irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\)
 has the form
 \begin{center}
@@ -218,8 +239,8 @@ and
 \end{equation}
 
 To conclude our analysis all it's left is to show that for each \(n\) such
-\(V\) does indeed exist and is irreducible. Surprisingly, we have already
-encountered such a \(V\).
+irreducible \(V\) does indeed exist. Surprisingly, we have already encountered
+such a \(V\).
 
 \begin{theorem}\label{thm:sl2-exist-unique}
   For each \(n \ge 0\) there exists a unique irreducible representation of
@@ -227,31 +248,33 @@ encountered such a \(V\).
 \end{theorem}
 
 \begin{proof}
-  Let \(V = K_n[x, y]\) be as in example~\ref{ex:sl2-polynomial-subrep}. A
-  simple calculation shows \(V_{n - 2 k} = K x^{n - k} y^k\) for \(k = 0,
-  \ldots, n\) and \(V_\lambda = 0\) otherwise. In particular, the right-most
-  eigenvalue of \(V\) is \(n\). Moreover, \(V\) must be irreducible, for if
-  \(V\) could be decomposed as the sum of at least two irreducible
-  representations we would either find an eigenspace of dimension greater than
-  \(1\) or find an eigenvalue whose difference with \(n\) is odd.
+  Let \(V = K[x, y]^{(n)}\) be the \(\mathfrak{sl}_2(K)\)-module of homogeneous
+  polynomials of degree \(n\) in two variables, as in
+  example~\ref{ex:sl2-polynomial-subrep}. A simple calculation shows \(V_{n - 2
+  k} = K x^{n - k} y^k\) for \(k = 0, \ldots, n\) and \(V_\lambda = 0\)
+  otherwise. In particular, the right-most eigenvalue of \(V\) is \(n\).
   Alternatively, one can readily check that if \(K^2\) is the natural
   representation of \(\mathfrak{sl}_2(K)\), then \(V = \operatorname{Sym}^n
   K^2\) satisfies the relations of (\ref{eq:irr-rep-of-sl2}). Indeed, the map
   \begin{align*}
-        K_n[x, y] & \to     \operatorname{Sym}^n K^2 \\
+        K[x, y]^{(n)} & \to     \operatorname{Sym}^n K^2 \\
     x^{n - k} y^k & \mapsto e_1^{n - k} \cdot e_2^k
   \end{align*}
   is an isomorphism.
 
-  As for the uniqueness of \(V\), it suffices to notice that if \(W\) is a
-  finite-dimensional irreducible representation of \(\mathfrak{sl}_2(K)\) with
-  right-most eigenvector \(w\) then relations (\ref{eq:irr-rep-of-sl2}) imply
-  the map
+  Either way, by the previous observation that a finite-dimensional
+  representation whose eigenvalues all have the same parity and whose
+  corresponding eigenspace are all 1-dimensional must be irreducible, \(V\) is
+  irreducible. As for the uniqueness of \(V\), it suffices to notice that if
+  \(W\) is a finite-dimensional irreducible representation of
+  \(\mathfrak{sl}_2(K)\) with right-most eigenvector \(w\) then relations
+  (\ref{eq:irr-rep-of-sl2}) imply the map
   \begin{align*}
         V & \to     W     \\
     f^k v & \mapsto f^k w
   \end{align*}
-  is an isomorphism.
+  is an isomorphism -- this is, in effect, precisely how the isomorphism \(K[x,
+  y]^{(n)} \isoto \operatorname{Sym}^n K^2\) was constructed.
 \end{proof}
 
 Our initial gamble of studying the eigenvalues of \(h\) may have seemed
@@ -259,10 +282,9 @@ arbitrary at first, but it payed off: we've \emph{completely} described
 \emph{all} irreducible representations of \(\mathfrak{sl}_2(K)\). 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 the results of this section for \(\mathfrak{sl}_3(K)\), hoping this
-will \emph{somehow} lead us to a general solution. In the process of doing so
-we'll learn a bit more why \(h\) was a sure bet and the race was fixed all
-along.
+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'll find some important
+clues on why \(h\) was a sure bet and the race was fixed all along.
 
 \section{Representations of \(\mathfrak{sl}_3(K)\)}\label{sec:sl3-reps}
 
@@ -270,21 +292,20 @@ The study of representations of \(\mathfrak{sl}_2(K)\) reminds me of the
 difference the derivative of a function \(\RR \to \RR\) and that of a smooth
 map between manifolds: it's a simpler case of something greater, but in some
 sense it's 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
+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's a simple and beautiful
-example, but unfortunately the general picture -- representations of arbitrary
-semisimple algebras -- lacks its simplicity, and, of course, much of this
-complexity is hidden in the case of \(\mathfrak{sl}_2(K)\).  The general
-purpose of this section is to investigate to which extent the framework used in
-the previous section to classify the representations of \(\mathfrak{sl}_2(K)\)
-can be generalized to other semisimple Lie algebras, and the algebra
-\(\mathfrak{sl}_3(K)\) stands as a natural candidate for potential
-generalizations: \(3 = 2 + 1\) after all.
+example, but unfortunately the general picture, representations 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'll fix some irreducible representation
 \(V\) of \(\mathfrak{sl}_3(K)\) and proceed step by step, at each point asking
@@ -347,14 +368,15 @@ 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, the eigenvalues of of the adjoint actions of \(h\) are
-\(\pm 2\) since
+\(0\) and \(\pm 2\) since
 \begin{align*}
-  [h, f] & = -2 f &
-  [h, e] & = 2 e
+  \operatorname{ad}(h) e & = 2 e  &
+  \operatorname{ad}(h) f & = -2 f &
+  \operatorname{ad}(h) h & = 0,
 \end{align*}
 and the eigenvalues of the action of \(h\) in an irreducible
-\(\mathfrak{sl}_2(K)\)-representation differ from one another by multiples of
-\(\pm 2\).
+\(\mathfrak{sl}_2(K)\)-modules differ from one another by multiples of \(\pm
+2\).
 
 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
@@ -489,8 +511,7 @@ Theorem~\ref{thm:sl3-weights-congruent-mod-root} can thus be restated as\dots
 \begin{corollary}
   The weights of an irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\)
   are all congruent module the root lattice \(Q\). In other words, the weights
-  of \(V\) all lie in a single \(Q\)-coset of \(t \in
-  \mfrac{\mathfrak{h}^*}{Q}\).
+  of \(V\) all lie in a single \(Q\)-coset \(t \in \mfrac{\mathfrak{h}^*}{Q}\).
 \end{corollary}
 
 At this point we could keep playing the tedious game of reproducing the
@@ -503,7 +524,7 @@ example~\ref{ex:gln-inclusions} -- restricts to an injective homomorphism
 \(\mathfrak{sl}_2(K)\) 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 consider regard \(V\) as an
+\mathfrak{sl}_3(K)\). We may thus regard \(V\) as an
 \(\mathfrak{sl}_2(K)\)-module by restricting to \(\mathfrak{s}_{1 2}\).
 
 Our first observation is that, since the root spaces act by translation, the
@@ -601,7 +622,7 @@ In general, we find\dots
   suffices to notice \(E_{i j}\) and \(E_{j i}\) map \(v \in V_{\lambda - k
   (\alpha_i - \alpha_j)}\) to \(E_{i j} v \in V_{\lambda - (k - 1) (\alpha_i -
   \alpha_j)}\) and \(E_{j i} v \in V_{\lambda - (k + 1) (\alpha_i -
-  \alpha_j)}\). Moreover,
+  \alpha_j)}\), respectively. Moreover,
   \[
     (\lambda - k (\alpha_i - \alpha_j))([E_{i j}, E_{j i}])
     = \lambda([E_{i j}, E_{j i}]) - k (1 - (-1))
@@ -677,7 +698,7 @@ symmetric around \(0\). To prove this we analyzed the right-most eigenvalue of
 \(h\) and its eigenvector, providing an explicit description of the irreducible
 representation of \(\mathfrak{sl}_2(K)\) in terms of this vector. We may
 reproduce these steps in the context of \(\mathfrak{sl}_3(K)\) by fixing a
-direction in the place an considering the weight lying the furthest in that
+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=2.5]
@@ -715,7 +736,7 @@ multiple choices the ``weight lying the furthest'' along this direction.
   0\) and \(f\) is irrational with respect to the lattice \(Q\).
 \end{definition}
 
-The first observation we make is that all others weights of \(V\) must lie in a
+The next observation we make is that all others weights of \(V\) must lie in a
 sort of \(\frac{1}{3}\)-plane with corners at \(\lambda\), as shown in
 \begin{center}
   \begin{tikzpicture}
@@ -733,35 +754,37 @@ sort of \(\frac{1}{3}\)-plane with corners at \(\lambda\), as shown in
 \end{center}
 
 Indeed, if this is not the case then, by definition, \(\lambda\) is not the
-furthest weight along the line we chose. Given our previous assertion that the
-root spaces of \(\mathfrak{sl}_3(K)\) act on the weight spaces of \(V\) via
-translation, this implies that \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\) all
-annihilate \(V_\lambda\), or otherwise one of \(V_{\lambda + \alpha_1 -
-\alpha_2}\), \(V_{\lambda + \alpha_1 - \alpha_3}\) and \(V_{\lambda + \alpha_2
-- \alpha_3}\) would be nonzero -- which contradicts the hypothesis that
-\(\lambda\) lies the furthest along the direction we chose. In other words\dots
-
-\begin{theorem}
+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 \(V\) via translation, this implies that \(E_{1 2}\), \(E_{1 3}\) and
+\(E_{2 3}\) all annihilate \(V_\lambda\), or otherwise one of \(V_{\lambda +
+\alpha_1 - \alpha_2}\), \(V_{\lambda + \alpha_1 - \alpha_3}\) and \(V_{\lambda
++ \alpha_2 - \alpha_3}\) would be nonzero -- which contradicts the hypothesis
+that \(\lambda\) lies the furthest in the direction we chose. In other
+words\dots
+
+\begin{proposition}
   There is a weight vector \(v \in V\) that is killed by all positive root
   spaces of \(\mathfrak{sl}_3(K)\).
-\end{theorem}
+\end{proposition}
 
 \begin{proof}
   It suffices to note that the positive roots of \(\mathfrak{sl}_3(K)\) are
   precisely \(\alpha_1 - \alpha_2\), \(\alpha_1 - \alpha_3\) and \(\alpha_2 -
-  \alpha_3\).
+  \alpha_3\), with root vectors \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\),
+  respectively.
 \end{proof}
 
-We call \(\lambda\) \emph{the highest weight of \(V\)}, and we call any \(v \in
-V_\lambda\) \emph{a highest weight vector}. Going back to the case of
+We call \(\lambda\) \emph{the highest weight of \(V\)}, and we call any nonzero
+\(v \in V_\lambda\) \emph{a highest weight vector}. Going back to the case of
 \(\mathfrak{sl}_2(K)\), we then constructed an explicit basis of our
 irreducible representations 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
+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'll do later
 down the line -- but instead we would like to focus on the problem of finding
-the weights of \(V\) for the moment.
+the weights of \(V\) in the first place.
 
 We'll start out by trying to understand the weights in the boundary of
 \(\frac{1}{3}\)-plane previously drawn. As we've just seen, we can get to other
@@ -953,7 +976,7 @@ must also be weights of \(V\). The final picture is thus
   \end{tikzpicture}
 \end{center}
 
-Finally\dots
+This final picture is known as \emph{the weight diagram of \(V\)}. Finally\dots
 
 \begin{theorem}\label{thm:sl3-irr-weights-class}
   The weights of \(V\) are precisely the elements of the weight lattice \(P\)
@@ -974,14 +997,14 @@ establishing\dots
 \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 theorem~\ref{thm:basis-of-irr-rep}
-that any irreducible representation of \(\mathfrak{sl}_2(K)\) is spanned by the
-images of its highest weight vector under \(f\). A more abstract way of putting
-it is to say that an irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\)
-is spanned by the images of its highest weight vector under successive
-applications by 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
+of \(\mathfrak{sl}_2(K)\): next we showed in
+proposition~\ref{thm:basis-of-irr-rep} that any irreducible representation of
+\(\mathfrak{sl}_2(K)\) is spanned by the images of its highest weight vector
+under \(f\). A more abstract way of putting it is to say that an irreducible
+representation \(V\) of \(\mathfrak{sl}_2(K)\) is spanned by the images of its
+highest weight vector under successive applications by 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{theorem}\label{thm:irr-sl3-span}
   Given an irreducible \(\mathfrak{sl}_3(K)\)-representation \(V\) and a
@@ -989,8 +1012,9 @@ for \(\mathfrak{sl}_3(K)\). Specifically\dots
   under successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\).
 \end{theorem}
 
+% TODO: Add a proof? We can hind the induction in "..." fairly easily
 The proof of theorem~\ref{thm:irr-sl3-span} is very similar to that of
-theorem~\ref{thm:basis-of-irr-rep}: we use the commutator relations of
+proposition~\ref{thm:basis-of-irr-rep}: we use the commutator relations of
 \(\mathfrak{sl}_3(K)\) to inductively show that the subspace spanned by the
 images of a highest weight vector under successive applications of \(E_{2 1}\),
 \(E_{3 1}\) and \(E_{3 2}\) is invariant under the action of
@@ -1011,14 +1035,14 @@ theorem~\ref{thm:sl3-existence-uniqueness}. Moreover, constructing such
 representation turns out to be quite simple.
 
 \begin{proof}[Proof of existence]
-  Consider the natural representation \(V = K^3\) of \(\mathfrak{sl}_3(K)\). We
-  claim that the highest weight of \(\operatorname{Sym}^n V \otimes
-  \operatorname{Sym}^m V^*\) is \(n \alpha_1 - m \alpha_3\).
-
-  First of all, notice that the eigenvectors of \(V\) are the canonical basis
-  vectors \(e_1\), \(e_2\) and \(e_3\), whose eigenvalues are \(\alpha_1\),
-  \(\alpha_2\) and \(\alpha_3\) respectively. Hence the weight diagram of \(V\)
-  is
+  Consider the natural representation \(K^3\) of \(\mathfrak{sl}_3(K)\). We
+  claim that the highest weight of \(\operatorname{Sym}^n K^3 \otimes
+  \operatorname{Sym}^m (K^3)^*\) is \(n \alpha_1 - m \alpha_3\).
+
+  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
+  \(\alpha_1\), \(\alpha_2\) and \(\alpha_3\) respectively. Hence the weight
+  diagram of \(K^3\) is
   \begin{center}
     \begin{tikzpicture}[scale=2.5]
       \AutoSizeWeightLatticefalse
@@ -1033,12 +1057,12 @@ representation turns out to be quite simple.
       \end{rootSystem}
     \end{tikzpicture}
   \end{center}
-  and \(\alpha_1\) is the highest weight of \(V\).
+  and \(\alpha_1\) is the highest weight of \(K^3\).
 
   On the one hand, if \(\{f_1, f_2, f_3\}\) is the dual basis of \(\{e_1, e_2,
   e_3\}\) then \(H f_i = - \alpha_i(H) \cdot f_i\) for each \(H \in
-  \mathfrak{h}\), so that the weights of \(V^*\) are precisely the opposites of
-  the weights of \(V\). In other words,
+  \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=2.5]
       \AutoSizeWeightLatticefalse
@@ -1053,8 +1077,8 @@ representation turns out to be quite simple.
       \end{rootSystem}
     \end{tikzpicture}
   \end{center}
-  is the weight diagram of \(V^*\) and \(\alpha_3\) is the highest weight of
-  \(V^*\).
+  is the weight diagram of \((K^3)^*\) and \(\alpha_3\) is the highest weight
+  of \((K^3)^*\).
 
   On the other hand if we fix two \(\mathfrak{sl}_3(K)\)-representations \(U\)
   and \(W\), by computing
@@ -1066,16 +1090,16 @@ representation turns out to be quite simple.
       & = (\lambda + \mu)(H) \cdot (u \otimes w)
     \end{split}
   \]
-  for each \(H \in \mathfrak{h}\), \(u \in U_\lambda\) and \(w \in W_\lambda\)
-  we can see that the weights of \(U \otimes W\) are precisely the sums of the
+  for each \(H \in \mathfrak{h}\), \(u \in U_\lambda\) and \(w \in W_\mu\) we
+  can see that the weights of \(U \otimes W\) are precisely the sums of the
   weights of \(U\) with the weights of \(W\).
 
-  This implies that the maximal weights of \(\operatorname{Sym}^n V\) and
-  \(\operatorname{Sym}^m V^*\) are \(n \alpha_1\) and \(- m \alpha_3\)
+  This implies that the maximal weights of \(\operatorname{Sym}^n K^3\) and
+  \(\operatorname{Sym}^m (K^3)^*\) are \(n \alpha_1\) and \(- m \alpha_3\)
   respectively -- with maximal weight vectors \(e_1^n\) and \(f_3^m\).
-  Furthermore, by the same token the highest weight of \(\operatorname{Sym}^n V
-  \otimes \operatorname{Sym}^m V^*\) must be \(n e_1 - m e_3\) -- with highest
-  weight vector \(e_1^n \otimes f_3^m\).
+  Furthermore, by the same token the highest weight of \(\operatorname{Sym}^n
+  K^3 \otimes \operatorname{Sym}^m (K^3)^*\) must be \(n e_1 - m e_3\) -- with
+  highest weight vector \(e_1^n \otimes f_3^m\).
 \end{proof}
 
 The ``uniqueness'' part of theorem~\ref{thm:sl3-existence-uniqueness} is even
@@ -1105,16 +1129,12 @@ simpler than that.
   \oplus W\), it follows from corollary~\ref{thm:irr-component-of-high-vec}
   that \(U\) is irreducible. The projection maps \(\pi_1 : U \to V\), \(\pi_2 :
   U \to W\), being nonzero homomorphism between irreducible representations of
-  \(\mathfrak{sl}_3(K)\) must be isomorphism. Finally,
+  \(\mathfrak{sl}_3(K)\), must be isomorphism. Finally,
   \[
     V \cong U \cong W
   \]
 \end{proof}
 
-The situation here is analogous to that of the previous section, where we saw
-that the irreducible representations of \(\mathfrak{sl}_2(K)\) are given by
-symmetric powers of the natural representation.
-
 We've been very successful in our pursue for a classification of the
 irreducible representations of \(\mathfrak{sl}_2(K)\) and
 \(\mathfrak{sl}_3(K)\), but so far we've mostly postponed the discussion on the