diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -1,15 +1,14 @@
\chapter{Low-Dimensional Examples}\label{ch:sl3}
We are, once again, faced with the daunting task of classifying the
-finite-dimensional representations of a given (semisimple) algebra
-\(\mathfrak{g}\). Having reduced the problem a great deal, all its left is
-classifying the irreducible representations of \(\mathfrak{g}\).
-We have encountered numerous examples of irreducible \(\mathfrak{g}\)-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
-irreducible representations by looking at concrete examples.
-Specifically, we will classify the irreducible finite-dimensional representations
-of certain low-dimensional semisimple Lie algebras.
+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.
Throughout the previous chapters, \(\mathfrak{sl}_2(K)\) has afforded us
surprisingly illuminating examples, so it will serve as our first candidate for
@@ -24,276 +23,284 @@ form a basis for \(\mathfrak{sl}_2(K)\) and satisfy
[e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e
\end{align*}
-Let \(V\) be a finite-dimensional irreducible \(\mathfrak{sl}_2(K)\)-module. We
-now turn our attention to the action of \(h\) on \(V\), in particular, to the
+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, to the
eigenspace decomposition
\[
- V = \bigoplus_{\lambda} V_\lambda
+ M = \bigoplus_{\lambda} M_\lambda
\]
-of \(V\) -- where \(\lambda\) ranges over the eigenvalues of \(h\) and
-\(V_\lambda\) is the corresponding eigenspace. At this point, this is nothing
+of \(M\) -- where \(\lambda\) ranges over the eigenvalues of \(h\) 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. For now we will
postpone the discussion about the real reason of why we chose \(h\). Let
-\(\lambda\) be any eigenvalue of \(h\). Notice \(V_\lambda\) is in general not
-a subrepresentation of \(V\). Indeed, if \(v \in V_\lambda\) then
+\(\lambda\) be any eigenvalue of \(h\). Notice \(M_\lambda\) is in general not
+a \(\mathfrak{sl}_3(K)\)-submodule of \(M\). Indeed, if \(m \in M_\lambda\)
+then
\begin{align*}
- h e v & = 2e v + e h v = (\lambda + 2) e v \\
- h f v & = - 2f v + f h v = (\lambda - 2) f v
+ 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*}
-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 \(V_{\lambda - 2}\).
+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 \arrow[bend left=60]{r}
- & V_{\lambda - 2} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}
- & V_{\lambda} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f}
- & V_{\lambda + 2} \arrow[bend left=60]{r} \arrow[bend left=60]{l}{f}
+ & M_{\lambda - 2} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}
+ & M_{\lambda} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f}
+ & M_{\lambda + 2} \arrow[bend left=60]{r} \arrow[bend left=60]{l}{f}
& \cdots \arrow[bend left=60]{l}
\end{tikzcd}
\end{center}
-This implies \(\bigoplus_{k \in \mathbb{Z}} V_{\lambda - 2 k}\) is an
+This implies \(\bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k}\) is a
\(\mathfrak{sl}_2(K)\)-invariant subspace, which goes to show
\[
- V = \bigoplus_{k \in \mathbb{Z}} V_{\lambda - 2 k},
+ M = \bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k},
\]
and the eigenvalues of \(h\) all have the form \(\lambda - 2 k\) for some
-\(k\).
-Even more so, if \(a = \max \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne 0 \}\) and
-\(b = \min \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne 0 \}\) we can see that
+\(k\). Even more so, if \(a = \max \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne
+0 \}\) and \(b = \min \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne 0 \}\) we can
+see that
\[
- \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le k \le b}} V_{\lambda - 2 k}
+ \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le k \le b}} M_{\lambda - 2 k}
\]
-is also an \(\mathfrak{sl}_2(K)\)-invariant subspace, so that the eigenvalues
-of \(h\) form an unbroken string
+is also a \(\mathfrak{sl}_2(K)\)-submodule, so that the eigenvalues of \(h\)
+form an unbroken string
\[
\ldots, \lambda - 4, \lambda - 2, \lambda, \lambda + 2, \lambda + 4, \ldots
\]
around \(\lambda\).
-Our main objective is to show \(V\) is determined by this string of
+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 \(v \in
-V_\lambda\) and consider the set \(\{v, f v, f^2 v, \ldots\}\).
+\(\lambda\) 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 \(\{v, f v, f^2 v, \ldots\}\) is a basis for \(V\). In addition, the
- action of \(\mathfrak{sl}_2(K)\) on \(V\) is given by the formulas
+ 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 v & \overset{e}{\mapsto} k(\lambda + 1 - k) f^{k - 1} v
- & f^k v & \overset{f}{\mapsto} f^{k + 1} v
- & f^k v & \overset{h}{\mapsto} (\lambda - 2 k) f^k v
+ 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 \(f^k v\) lies in \(V_{\lambda - 2 k}\), so that \(\{v,
- f v, f^2 v, \ldots\}\) is a set of linearly independent vectors. Hence it
- suffices to show \(V = K \langle v, f v, f^2 v, \ldots \rangle\), which in
- light of the fact that \(V\) is irreducible is the same as showing \(K
- \langle v, f v, f^2 v, \ldots \rangle\) is invariant under the action of
- \(\mathfrak{sl}_2(K)\).
-
- 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, \(h f^k v = (\lambda - 2 k) f^k v \in K \langle v, f v, f^2 v,
- \ldots \rangle\), which also goes to show one of the formulas in
- (\ref{eq:irr-rep-of-sl2}). Seeing \(e f^k v \in K \langle v, f v, f^2 v,
- \ldots \rangle\) is a bit more complex. Clearly,
+ 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 f v
- & = h v + f e v \\
+ e \cdot (f \cdot m)
+ & = h \cdot m + f \cdot (e \cdot m) \\
\text{(since \(\lambda\) is the right-most eigenvalue)}
- & = h v + f 0 \\
- & = \lambda v
+ & = h \cdot m + f \cdot 0 \\
+ & = \lambda m
\end{split}
\]
Next we compute
\[
\begin{split}
- e f^2 v
- & = (h + fe) f v \\
- & = h f v + f (\lambda v) \\
- & = 2 (\lambda - 1) f v
+ e \cdot (f^2 \cdot m)
+ & = (h + fe) \cdot (f \cdot m) \\
+ & = h \cdot (f \cdot m) + f \cdot (\lambda m) \\
+ & = 2 (\lambda - 1) f \cdot m
\end{split}
\]
- The pattern is starting to become clear: \(e\) sends \(f^k v\) to a multiple
- of \(f^{k - 1} v\). Explicitly, it is not hard to check by induction that
+ 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
\[
- e f^k v = k (\lambda + 1 - k) f^{k - 1} v,
+ e \cdot (f^k \cdot m) = k (\lambda + 1 - k) \cdot f^{k - 1} m,
\]
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 \(V\). In particular, this goes to show\dots
+\(\mathfrak{sl}_2(K)\) on \(M\). In particular, this goes to show\dots
\begin{corollary}
- Every eigenspace of the action of \(h\) on \(V\) is \(1\)-dimensional.
+ Every eigenspace of the action of \(h\) on \(M\) is \(1\)-dimensional.
\end{corollary}
\begin{proof}
- It suffices to note \(\{v, f v, f^2 v, \ldots \}\) is a basis for \(V\)
- consisting of eigenvalues of \(h\) and whose only element in \(V_{\lambda - 2
- k}\) is \(f^k v\).
+ 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 \(V\) form a symmetric, unbroken string of
+ The eigenvalues of \(h\) in \(M\) form a symmetric, unbroken string of
integers separated by intervals of length \(2\) whose right-most value is
- \(\dim V - 1\).
+ \(\dim M - 1\).
\end{corollary}
\begin{proof}
- If \(f^m\) is the lowest power of \(f\) that annihilates \(v\), it follows
+ 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
\[
- 0 = e 0 = e f^m v = m (\lambda + 1 - m) f^{m - 1} v
+ 0 = e \cdot 0 = e \cdot (f^r \cdot m)
+ = r (\lambda + 1 - r) f^{r - 1} \cdot m
\]
- This implies \(\lambda + 1 - m = 0\) -- i.e. \(\lambda = m - 1 \in \mathbb{Z}\). Now
- since \(\{v, f v, f^2 v, \ldots, f^{m - 1} v\}\) is a basis for \(V\), \(m =
- \dim V\). Hence if \(n = \lambda = \dim V - 1\) and the eigenvalues of \(h\)
- are
+ 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
\[
- \ldots, n - 6, n - 4, n - 2, n
+ \ldots, \lambda - 6, \lambda - 4, \lambda - 2, \lambda
\]
To see that this string is symmetric around \(0\), simply note that the
- left-most eigenvalue of \(h\) is precisely \(n - 2 (m - 1) = -n\).
+ left-most eigenvalue of \(h\) is precisely \(\lambda - 2 (r - 1) =
+ -\lambda\).
\end{proof}
Visually, the situation it thus
\begin{center}
\begin{tikzcd}
- V_{-n} \arrow[bend left=60]{r}{e}
- & V_{- n + 2} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f}
- & V_{- n + 4} \arrow[bend left=60]{r} \arrow[bend left=60]{l}{f}
- & \cdots \arrow[bend left=60]{r} \arrow[bend left=60]{l}
- & V_{n - 4} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}
- & V_{n - 2} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f}
- & V_n \arrow[bend left=60]{l}{f}
+ 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 \(V\) and \(W\) are representations
-of \(\mathfrak{sl}_2(K)\), \(v \in V_\lambda\) and \(w \in W_\lambda\) then by
+\(\mathfrak{sl}_2(K)\)-module. Namely, if \(M\) and \(N\) are
+\(\mathfrak{sl}_2(K)\)-modules, \(m \in M_\mu\) and \(n \in N_\mu\) then by
computing
\[
- H (v + w) = Hv + Hw = \lambda(H) \cdot (v + w)
+ h \cdot (m + n) = h \cdot m + h \cdot n = \mu (m + n)
\]
-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 corresponding eigenspaces are
-the direct sums of the eigenspaces of such irreducible components.
+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 \(V\) all have the same parity -- i.e.
+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 \(V\) 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 \(\lambda
-\in \mathfrak{h}^*\). To conclude our analysis all it is left is to show that
-for each \(n\) there is some finite-dimensional irreducible \(V\) whose highest
-weight is \(\lambda\). Surprisingly, we have already encountered such a \(V\).
+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 \(n \ge 0\) there exists a unique irreducible representation of
- \(\mathfrak{sl}_2(K)\) whose left-most eigenvalue of \(h\) is \(n\).
+ 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 \(V = K[x, y]^{(n)}\) 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 \(n\).
+ 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
- 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
+ \(\mathfrak{sl}_2(K)\)-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]^{(n)} & \to \operatorname{Sym}^n K^2 \\
- x^{n - k} y^k & \mapsto e_1^{n - k} \cdot e_2^k
+ 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
- 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
+ \(\mathfrak{sl}_2(K)\)-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*}
- V & \to W \\
- f^k v & \mapsto f^k w
+ 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 \(K[x,
- y]^{(n)} \isoto \operatorname{Sym}^n K^2\) was constructed.
+ 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} 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 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.
+\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 \(\mathfrak{sl}_3(K)\)}\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.
+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, 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
+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 irreducible representation
-\(V\) of \(\mathfrak{sl}_3(K)\) and proceed step by step, at each point asking
-ourselves how we could possibly adapt the framework we laid out for
+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 \(v\) of \(h\) that is
+\(h\) 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 irreducible
-representations of \(\mathfrak{sl}_2(K)\) culminating in
+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 \(V\) 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
+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
\[
\mathfrak{h}
= \left\{
@@ -304,25 +311,25 @@ way to proceed is to consider the subalgebra
\]
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 \(v \in V\) that is
+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
-\(H v = \lambda(H) \cdot v\) for each \(H \in \mathfrak{h}\) and some nonzero
-\(v \in V\). We call such functionals \(\lambda\) \emph{eigenvalues of
-\(\mathfrak{h}\)}, and we say \emph{\(v\) is an eigenvector of
+\(H \cdot m = \lambda(H) m\) 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}
- V = \bigoplus_\lambda V_\lambda
+ M = \bigoplus_\lambda M_\lambda
\end{equation}
where \(\lambda\) ranges over all eigenvalues of \(\mathfrak{h}\) and
-\(V_\lambda = \{ v \in V : H v = \lambda(H) \cdot v, \forall H \in \mathfrak{h}
+\(M_\lambda = \{ m \in M : H \cdot m = \lambda(H) m, \forall H \in \mathfrak{h}
\}\). We should note that the fact that (\ref{eq:weight-module}) holds is not
-at all obvious. This is because in general \(V_\lambda\) is not the eigenspace
+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,
@@ -340,9 +347,9 @@ integral linear combinations of the eigenvalues of the adjoint action of
\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 an irreducible
-\(\mathfrak{sl}_2(K)\)-modules differ from one another by multiples of \(\pm
-2\).
+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
@@ -392,21 +399,23 @@ Visually we may draw
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 \(v \in V_\lambda\) then
+\(H \in \mathfrak{h}\) and \(m \in M_\lambda\) then
\[
\begin{split}
- H (X v)
- & = X (H v) + [H, X] v \\
- & = X (\lambda(H) \cdot v) + (\alpha(H) \cdot X) v \\
- & = (\lambda + \alpha)(H) \cdot X v
+ H \cdot (X \cdot m)
+ & = X \cdot (H \cdot m) + [H, X] \cdot m \\
+ & = X \cdot (\lambda(H) m) + \alpha(H) X \cdot m \\
+ & = (\lambda + \alpha)(H) X \cdot m
\end{split}
\]
-so that \(X\) carries \(v\) to \(V_{\lambda + \alpha}\). In other words,
-\(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors
+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}.
+\newpage
+
For instance \(\mathfrak{sl}_3(K)_{\alpha_1 - \alpha_3}\) will act on the
-adjoint representation of \(\mathfrak{sl}_3(K)\) via
+adjoint \(\mathfrak{sl}_3(K)\)-modules via
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=2]
@@ -430,9 +439,9 @@ 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 an irreducible
- \(\mathfrak{sl}_3(K)\)-module \(V\) differ from one another by integral
- linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of adjoint
+ 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 \(\alpha_i - \alpha_j\) of the adjoint
action of \(\mathfrak{h}\) on \(\mathfrak{sl}_3(K)\).
\end{theorem}
@@ -441,69 +450,72 @@ This is again entirely analogous to the situation we observed in
\(\mathfrak{sl}_2(K)\): it suffices to note that if we fix some eigenvalue
\(\lambda\) of \(\mathfrak{h}\) and let \(i\) and \(j\) vary then
\[
- \bigoplus_{i j} V_{\lambda + \alpha_i - \alpha_j}
+ \bigoplus_{i j} M_{\lambda + \alpha_i - \alpha_j}
\]
- is an invariant subspace of \(V\).
+ 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 \(V\) and eigenvalues of the
+eigenvalues of the action of \(\mathfrak{h}\) on \(M\) and eigenvalues of the
adjoint action of \(\mathfrak{h}\).
\begin{definition}
- Given a representation \(V\) of \(\mathfrak{sl}_3(K)\), we will call the
- nonzero eigenvalues of the action of \(\mathfrak{h}\) on \(V\) \emph{weights
- of \(V\)}. As you might have guessed, we will correspondingly refer to
+ 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
-representation of \(\mathfrak{sl}_3(K)\) deserve some special attention.
+\(\mathfrak{sl}_3(K)\)-module deserve some special attention.
\begin{definition}
- The weights of the adjoint representation of \(\mathfrak{sl}_3(K)\) are
- called \emph{roots of \(\mathfrak{sl}_3(K)\)}. Once again, the expressions
+ 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}
- The lattice \(Q = \mathbb{Z} \langle \alpha_i - \alpha_j : i, j = 1, 2, 3 \rangle\)
- is called \emph{the root lattice of \(\mathfrak{sl}_3(K)\)}.
+ The lattice \(Q = \mathbb{Z} \langle \alpha_i - \alpha_j : i, j = 1, 2, 3
+ \rangle\) is called \emph{the root lattice of \(\mathfrak{sl}_3(K)\)}.
\end{definition}
\begin{corollary}
- The weights of an irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\)
- are all congruent modulo the root lattice \(Q\). In other words, the weights
- of \(V\) all lie in a single \(Q\)-coset \(t \in \mfrac{\mathfrak{h}^*}{Q}\).
+ 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 the representations of
-\(\mathfrak{sl}_2(K)\) instead. Notice that the canonical inclusion
+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 \(V\) as an
+\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
\[
- \bigoplus_{k \in \mathbb{Z}} V_{\lambda - k (\alpha_1 - \alpha_2)},
+ \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k (\alpha_1 - \alpha_2)},
\]
must be invariant under the action of \(E_{1 2}\) and \(E_{2 1}\) for all
-\(\lambda \in \mathfrak{h}^*\). This goes to show \(\bigoplus_k V_{\lambda - k
-(\alpha_1 - \alpha_2)}\) is a \(\mathfrak{sl}_2(K)\)-submodule of \(V\) for all
-weights \(\lambda\) of \(V\). Furthermore, one can easily see that the
-eigenspace of the eigenspace \(\lambda(H) - 2k\) of \(h\) in \(W\) is precisely
-the weight space \(V_{\lambda - k (\alpha_2 - \alpha_1)}\).
+\(\lambda \in \mathfrak{h}^*\). This goes to show \(\bigoplus_k M_{\lambda - k
+(\alpha_1 - \alpha_2)}\) 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 \(\bigoplus_{k \in \mathbb{Z}} M_{\lambda
+- k (\alpha_1 - \alpha_2)}\) associated with the eigenvalue \(\lambda(H) - 2k\)
+is precisely the weight space \(M_{\lambda - k (\alpha_2 - \alpha_1)}\).
+
+\newpage
Visually,
\begin{center}
@@ -539,14 +551,14 @@ In general, we find\dots
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 \(V\), the space
+ \mathfrak{h}^*\) of \(M\), the space
\[
- W = \bigoplus_{k \in \mathbb{Z}} V_{\lambda - k (\alpha_i - \alpha_j)}
+ N = \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k (\alpha_i - \alpha_j)}
\]
is invariant under the action of \(\mathfrak{s}_{i j}\) and
\[
- V_{\lambda - k (\alpha_i - \alpha_j)}
- = W_{\lambda([E_{i j}, E_{j i}]) - 2k}
+ M_{\lambda - k (\alpha_i - \alpha_j)}
+ = N_{\lambda([E_{i j}, E_{j i}]) - 2k}
\]
\end{proposition}
@@ -584,25 +596,25 @@ In general, we find\dots
which ``erases the \(k\)-th row and the \(k\)-th column'' of a matrix is an
isomorphism.
- To see that \(W\) is invariant under the action of \(\mathfrak{s}_{i j}\), it
- 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 -
+ 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 \(m \in M_{\lambda - k
+ (\alpha_i - \alpha_j)}\) to \(E_{i j} \cdot m \in M_{\lambda - (k - 1) (\alpha_i -
+ \alpha_j)}\) and \(E_{j i} \cdot m \in M_{\lambda - (k + 1) (\alpha_i -
\alpha_j)}\), respectively. Moreover,
\[
(\lambda - k (\alpha_i - \alpha_j))([E_{i j}, E_{j i}])
= \lambda([E_{i j}, E_{j i}]) - k (1 - (-1))
= \lambda([E_{i j}, E_{j i}]) - 2 k,
\]
- which goes to show \(V_{\lambda - k (\alpha_i - \alpha_j)} \subset
- W_{\lambda([E_{i j}, E_{j i}]) - 2k}\). On the other hand, if we suppose \(0
- < \dim V_{\lambda - k (\alpha_i - \alpha_j)} < \dim W_{\lambda([E_{i j}, E_{j
+ which goes to show \(M_{\lambda - k (\alpha_i - \alpha_j)} \subset
+ N_{\lambda([E_{i j}, E_{j i}]) - 2k}\). On the other hand, if we suppose \(0
+ < \dim M_{\lambda - k (\alpha_i - \alpha_j)} < \dim N_{\lambda([E_{i j}, E_{j
i}]) - 2 k}\) for some \(k\) we arrive at
\[
- \dim W
- = \sum_k \dim V_{\lambda - k (\alpha_i - \alpha_j)}
- < \sum_k \dim W_{\lambda([E_{i j}, E_{j i}]) - 2k}
- = \dim W,
+ \dim N
+ = \sum_k \dim M_{\lambda - k (\alpha_i - \alpha_j)}
+ < \sum_k \dim N_{\lambda([E_{i j}, E_{j i}]) - 2k}
+ = \dim N,
\]
a contradiction.
\end{proof}
@@ -615,13 +627,13 @@ As a first consequence of this, we show\dots
\end{definition}
\begin{corollary}\label{thm:sl3-weights-fit-in-weight-lattice}
- Every weight \(\lambda\) of \(V\) lies in the weight lattice \(P\).
+ 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 representation of \(\mathfrak{sl}_2(K)\), so it must
- be an integer. Now since
+ in a finite-dimensional \(\mathfrak{sl}_2(K)\)-module, so it must be an
+ integer. Now since
\[
\lambda
\begin{pmatrix}
@@ -672,9 +684,9 @@ denote by \(B\) 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 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
+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.
@@ -711,7 +723,9 @@ along this direction.
0\) and \(f\) is irrational with respect to the lattice \(Q\).
\end{definition}
-The next observation we make is that all others weights of \(V\) must lie in a
+\newpage
+
+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}
@@ -731,15 +745,15 @@ sort of \(\frac{1}{3}\)-cone with apex at \(\lambda\), as shown in
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 \(V\) 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
+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 \(M_{\lambda +
+\alpha_1 - \alpha_2}\), \(M_{\lambda + \alpha_1 - \alpha_3}\) and \(M_{\lambda
+ \alpha_2 - \alpha_3}\) 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 annihilated by all positive root
+ There is a weight vector \(m \in M\) that is annihilated by all positive root
spaces of \(\mathfrak{sl}_3(K)\).
\end{proposition}
@@ -750,20 +764,20 @@ words\dots
respectively.
\end{proof}
-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 for 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
-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 \(V\) in the first place.
+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 \(V_\lambda\) by successively applying \(E_{2 1}\).
+from \(M_\lambda\) by successively applying \(E_{2 1}\).
\begin{center}
\begin{tikzpicture}
\begin{rootSystem}{A}
@@ -786,11 +800,11 @@ from \(V_\lambda\) by successively applying \(E_{2 1}\).
Notice that \(\lambda([E_{1 2}, E_{2 1}]) \in \mathbb{Z}\) is the right-most
eigenvalue of the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_{k \in \mathbb{Z}}
-V_{\lambda + k (\alpha_i - \alpha_j)}\). In particular, \(\lambda([E_{1 2},
-E_{2 1}]\) must be positive. In addition, since the eigenspace of the
+M_{\lambda - k (\alpha_1 - \alpha_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
-\(\bigoplus_{k \in \mathbb{N}} V_{\lambda + k (\alpha_1 - \alpha_2)}\) is
-\(V_{\lambda + k (\alpha_1 - \alpha_2)}\), the weights of \(V\) appearing the
+\(\bigoplus_{k \in \mathbb{N}} M_{\lambda - k (\alpha_1 - \alpha_2)}\) is
+\(M_{\lambda - k (\alpha_1 - \alpha_2)}\), the weights of \(M\) appearing the
string \(\lambda, \lambda + (\alpha_1 - \alpha_2), \ldots, \lambda + k
(\alpha_1 - \alpha_2), \ldots\) must be symmetric with respect to the line
\(B(\alpha_1 - \alpha_2, \alpha) = 0\). The picture is thus
@@ -811,9 +825,9 @@ string \(\lambda, \lambda + (\alpha_1 - \alpha_2), \ldots, \lambda + k
\end{tikzpicture}
\end{center}
-We could apply this same argument to the subspace \(\bigoplus_k V_{\lambda + k
-(\alpha_3 - \alpha_2)}\), so that the weights in this subspace must be
-symmetric with respect to the line \(B(\alpha_3 - \alpha_2, \alpha) = 0\). The
+We could apply this same argument to the subspace \(\bigoplus_k M_{\lambda - k
+(\alpha_2 - \alpha_3)}\), so that the weights in this subspace must be
+symmetric with respect to the line \(B(\alpha_2 - \alpha_3, \alpha) = 0\). The
picture is now
\begin{center}
\begin{tikzpicture}
@@ -831,7 +845,7 @@ picture is now
\draw[very thick] \weight{0}{-4} -- \weight{0}{4}
node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)};
\draw[very thick] \weight{-4}{0} -- \weight{4}{0}
- node[right]{\small\(B(\alpha_3 - \alpha_2, \alpha) = 0\)};
+ node[right]{\small\(B(\alpha_2 - \alpha_3, \alpha) = 0\)};
\end{rootSystem}
\end{tikzpicture}
\end{center}
@@ -868,11 +882,11 @@ of our string, arriving at
\end{center}
We claim all dots \(\mu\) lying inside the hexagon we have drawn must also be
-weights -- i.e. \(V_\mu \ne 0\). Indeed, by applying the same argument to an
-arbitrary weight \(\nu\) in the boundary of the hexagon we get a representation
-of \(\mathfrak{sl}_2(K)\) whose weights correspond to weights of \(V\) lying in
-a string inside the hexagon, and whose right-most weight is precisely the
-weight of \(V\) we started with.
+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
@@ -908,10 +922,10 @@ weight of \(V\) we started with.
\end{tikzpicture}
\end{center}
-By construction, \(\nu\) corresponds to the right-most weight of the
-representation of \(\mathfrak{sl}_2(K)\), so that all dots lying on the dashed
-string must occur in the representation of \(\mathfrak{sl}_2(K)\). Hence they
-must also be weights of \(V\). The final picture is thus
+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
@@ -947,126 +961,120 @@ must also be weights of \(V\). The final picture is thus
\end{tikzpicture}
\end{center}
-This final picture is known as \emph{the weight diagram of \(V\)}. Finally\dots
+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 \(V\) are precisely the elements of the weight lattice \(P\)
+ 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 \(B(\alpha_i - \alpha_j, \alpha) = 0\).
\end{theorem}
-Having found all of the weights of \(V\), the only thing we are missing is an
+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}. In other words, our next goal is
establishing\dots
\begin{theorem}\label{thm:sl3-existence-uniqueness}
- For each pair of positive integers \(n\) and \(m\), there exists precisely
- one finite-dimensional irreducible representation \(V\) of
- \(\mathfrak{sl}_3(K)\) whose highest weight is \(n \alpha_1 - m \alpha_3\).
+ For each pair of non-negative integers \(k\) and \(\ell\), there exists
+ precisely one finite-dimensional simple \(\mathfrak{sl}_3(K)\)-module \(M\)
+ whose highest weight is \(k \alpha_1 - \ell \alpha_3\).
\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 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 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 \(\mathfrak{sl}_3(K)\).
+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 an irreducible \(\mathfrak{sl}_3(K)\)-representation \(V\) and a
- highest weight vector \(v \in V\), \(V\) is spanned by the images of \(v\)
- under successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\).
+ 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 \(V\) is irreducible, it suffices to show that the subspace
- \(W\) spanned by successive applications of \(E_{2 1}\), \(E_{3 1}\) and
- \(E_{3 2}\) to \(v\) 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 \(W\) is spanned by the elements \(E_{2 1}^a E_{3 1}^b E_{3 1}^c v\).
-
- Recall that \(E_{i j}\) maps \(V_\mu\) to \(V_{\mu + \alpha_i - \alpha_j}\).
- In particular, \(E_{2 1}^a E_{3 1}^b E_{3 1}^c v \in V_{\lambda - a (\alpha_1
- - \alpha_2) - b (\alpha_1 - \alpha_3) - c (\alpha_2 - \alpha_3)}\). In other
- words,
+ 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 \(M_{\mu + \alpha_i - \alpha_j}\).
+ In particular, \(E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m \in M_{\lambda - a
+ (\alpha_1 - \alpha_2) - b (\alpha_1 - \alpha_3) - c (\alpha_2 - \alpha_3)}\).
+ In other words,
\[
- H E_{2 1}^a E_{3 1}^b E_{3 1}^c v
+ H E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m
= (\lambda - a (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3))(H)
- \cdot E_{2 1}^a E_{3 1}^b E_{3 1}^c v
- \in W
+ E_{2 1}^a E_{3 1}^b E_{3 1}^c \cdot m
+ \in N
\]
- for all \(H \in \mathfrak{h}\) and \(W\) is stable under the action of
- \(\mathfrak{h}\). On the other hand, \(W\) is clearly stable under the action
- of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\). All it is left is to show \(W\)
+ 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 v
+ E_{1 2} \cdot (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 v \\
+ 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 v \\
+ E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c \cdot m \\
& \phantom{=} \; +
(\lambda - (a - 1) (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}])
- \cdot
- E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c v \\
+ 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 v \\
+ E_{2 1}^{a - 3} E_{3 1}^b E_{3 2}^c \cdot m \\
& \phantom{=} \; +
(\lambda - (a - 1) (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}])
- \cdot
- E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c v \\
+ E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c \cdot m \\
& \phantom{=} \; +
(\lambda - (a - 2) (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}])
- \cdot
- E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c v \\
+ 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 v \\
+ & = E_{2 1}^a E_{1 2} E_{3 1}^b E_{3 2}^c \cdot m \\
& \phantom{=} \; +
(\lambda - (a - 1) (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}])
- \cdot
- E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c v \\
+ E_{2 1}^{a - 1} E_{3 1}^b E_{3 2}^c \cdot m \\
& \phantom{=} \; +
(\lambda - (a - 2) (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}])
- \cdot
- E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c v \\
+ E_{2 1}^{a - 2} E_{3 1}^b E_{3 2}^c \cdot m \\
& \phantom{=} \; + \cdots \\
& \phantom{=} \; +
(\lambda - (a - a) (\alpha_1 - \alpha_2)
- b (\alpha_1 - \alpha_3)
- c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}])
- \cdot
- E_{2 1}^{a - a} E_{3 1}^b E_{3 2}^c v \\
+ E_{2 1}^{a - a} E_{3 1}^b E_{3 2}^c \cdot m \\
\end{split}
\]
- Since \((\lambda - (a - k) (\alpha_1 - \alpha_2) - b (\alpha_1 - \alpha_3)
- - c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}]) \cdot E_{2 1}^{a - k} E_{3
- 1}^b E_{3 2}^c v \in W\) for all \(k\), it suffices to show \(E_{2 1}^a E_{1
- 2} E_{3 1}^b E_{3 2}^c v \in W\). But
+ Since \((\lambda - (a - k) (\alpha_1 - \alpha_2) - b (\alpha_1 - \alpha_3) -
+ c (\alpha_2 - \alpha_3)) ([E_{1 2}, E_{2 1}]) E_{2 1}^{a - k} E_{3 1}^b
+ E_{3 2}^c \cdot m \in N\) 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
\[
\begin{split}
E_{1 2} E_{3 1}^b
@@ -1080,41 +1088,40 @@ Specifically\dots
\end{split},
\]
given \([E_{1 2}, E_{3 1}] = - E_{3 2}\) and \([E_{3 2}, E_{3 1}] = 0\).
- It then follows from the fact \(E_{1 2} v = 0\) that
+ It then follows from the fact \(E_{1 2} \cdot m = 0\) that
\[
- E_{2 1}^a E_{1 2} E_{3 1}^b E_{3 2}^c v
- = E_{2 1}^a E_{3 1}^b E_{3 2}^c E_{1 2} v
- - b E_{2 1}^a E_{3 1}^b E_{3 2}^{c + 1} v
- = - b E_{2 1}^a E_{3 1}^b E_{3 2}^{c + 1} v \in W,
+ E_{2 1}^a E_{1 2} E_{3 1}^b E_{3 2}^c \cdot m
+ = E_{2 1}^a E_{3 1}^b E_{3 2}^c E_{1 2} \cdot m
+ - b E_{2 1}^a E_{3 1}^b E_{3 2}^{c + 1} \cdot m
+ = - b E_{2 1}^a E_{3 1}^b E_{3 2}^{c + 1} \cdot m \in N,
\]
- given that \(E_{1 2}\) and \(E_{3 2}\) commute. Hence \(E_{1 2} E_{2 1}^a
- E_{3 1}^b E_{3 2}^c v \in W\). Similarly,
+ 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,
\[
- E_{1 3} E_{2 1}^a E_{3 1}^b E_{3 2}^c v,
- E_{2 3} E_{2 1}^a E_{3 1}^b E_{3 2}^c v \in W
+ E_{1 3} \cdot (E_{2 1}^a E_{3 1}^b E_{3 2}^c \cdot m),
+ E_{2 3} \cdot (E_{2 1}^a E_{3 1}^b E_{3 2}^c \cdot m) \in N
\]
\end{proof}
The same argument also goes to show\dots
\begin{corollary}\label{thm:irr-component-of-high-vec}
- Given a finite-dimensional representation \(V\) of \(\mathfrak{sl}_3(K)\)
- with highest weight \(\lambda\) and \(v \in V_\lambda\), the subspace spanned
- by successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\) to
- \(v\) is an irreducible subrepresentation whose highest weight is
- \(\lambda\).
+ 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 representation whose highest weight is \(n \alpha_1 - m
-\alpha_2\) is enough for establishing the ``existence'' part of
-Theorem~\ref{thm:sl3-existence-uniqueness}. Moreover, constructing such
-representation turns out to be quite simple.
+finite-dimensional module whose highest weight is \(k \alpha_1 - \ell
+\alpha_3\) 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]
- 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 \(n \alpha_1 - m \alpha_3\).
+ 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 \(k \alpha_1 - \ell \alpha_3\).
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
@@ -1137,7 +1144,7 @@ representation turns out to be quite simple.
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 for \(\{e_1, e_2,
- e_3\}\) then \(H f_i = - \alpha_i(H) \cdot f_i\) for each \(H \in
+ e_3\}\) then \(H \cdot f_i = - \alpha_i(H) f_i\) 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}
@@ -1157,68 +1164,68 @@ representation turns out to be quite simple.
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
+ On the other hand if we fix two \(\mathfrak{sl}_3(K)\)-modules \(N\)
+ and \(L\), by computing
\[
\begin{split}
- H (u \otimes w)
- & = H u \otimes w + u \otimes H w \\
- & = \lambda(H) \cdot u \otimes w + u \otimes \mu(H) \cdot w \\
- & = (\lambda + \mu)(H) \cdot (u \otimes w)
+ H \cdot (n \otimes l)
+ & = H \cdot n \otimes l + n \otimes H \cdot l \\
+ & = \lambda(H) n \otimes l + n \otimes \mu(H) l \\
+ & = (\lambda + \mu)(H) \, (n \otimes l)
\end{split}
\]
- for each \(H \in \mathfrak{h}\), \(u \in U_\lambda\) 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 highest weights of \(\operatorname{Sym}^n K^3\) and
- \(\operatorname{Sym}^m (K^3)^*\) are \(n \alpha_1\) and \(- m \alpha_3\)
- respectively -- with highest weight vectors \(e_1^n\) and \(f_3^m\).
- 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 \(e_1^n \otimes f_3^m\).
+ 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 \(k \alpha_1\) and \(- \ell
+ \alpha_3\) 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
+ \(k e_1 - \ell e_3\) -- 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 \(V\) and \(W\) be two irreducible representations of
- \(\mathfrak{sl}_3(K)\) with highest weight \(\lambda\). By
- Theorem~\ref{thm:sl3-irr-weights-class}, the weights of \(V\) are precisely
- the same as those of \(W\).
+ 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
\[
- H (v + w)
- = H v + H w
- = \mu(H) \cdot v + \mu(H) \cdot w
- = \mu(H) \cdot (v + w)
+ H \cdot (m + n)
+ = H \cdot m + H \cdot n
+ = \mu(H) m + \mu(H) n
+ = \mu(H) (m + n)
\]
- for each \(H \in \mathfrak{h}\), \(v \in V_\mu\) and \(w \in W_\mu\), we can
- see that the weights of \(V \oplus W\) are same as those of \(V\) and \(W\).
- Hence the highest weight of \(V \oplus W\) is \(\lambda\) -- with highest
- weight vectors given by the sum of highest weight vectors of \(V\) and \(W\).
-
- Fix some \(v \in V_\lambda\) and \(w \in W_\lambda\) and consider the
- subrepresentation \(U = \mathcal{U}(\mathfrak{sl}_3(K)) \cdot v + w \subset V
- \oplus W\) generated by \(v + w\). Since \(v + w\) is a highest weight of \(V
- \oplus W\), 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,
+ 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,
\[
- V \cong U \cong W
+ M \cong L \cong N
\]
\end{proof}
-We have 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 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 \(\mathfrak{g}\).
+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
+\(\mathfrak{g}\).