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
git clone: git://git.pablopie.xyz/lie-algebras-and-their-representations
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Commit: 9254cd7294b0453294327a77019176db372a91bd
Parent: f6e2ffa60d3ef2f76c92416b34631804fff08886
Author: Pablo <pablo-escobar@riseup.net>
Date: Sun, 27 Nov 2022 18:19:18 +0000
Revised the third chapter for a third time!
Diffstat

1 file changed, 39 insertions, 62 deletions
Status	File Name	N° Changes	Insertions	Deletions
Modified	sections/sl2-sl3.tex	101	39	62
diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -11,7 +11,7 @@ irreducible representations by looking at concrete examples.
 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
+Throughout the previous chapters, \(\mathfrak{sl}_2(K)\) has afforded us
 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*}
@@ -92,11 +92,6 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
   \end{equation}
 \end{proposition}
 
-\begin{note}
-  For these formulas to work we fix the convention that \(f^{-1} v = 0\) --
-  which reflects the fact that \(e v = 0\).
-\end{note}
-
 \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
@@ -144,30 +139,7 @@ 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
 
 \begin{corollary}
-  \(V\) is completely determined by the right-most eigenvalue \(\lambda\) of
-  \(h\).
-\end{corollary}
-
-\begin{proof}
-  If \(W\) is an irreducible representation of \(\mathfrak{sl}_2(K)\) whose
-  right-most eigenvalue of \(h\) is \(\lambda\) and \(w \in W_\lambda\) is
-  nonzero, consider the linear isomorphism
-  \begin{align*}
-    T : V     & \to     W      \\
-        f^k v & \mapsto f^k w
-  \end{align*}
-
-  We claim \(T\) is an intertwining operator. Indeed, the explicit calculations
-  of \(e f^k v\) and \(h f^k v\) from the previous proof imply
-  \begin{align*}
-    T e & = e T & T f & = f T & T h & = h T
-  \end{align*}
-\end{proof}
-
-Other important consequences of proposition~\ref{thm:basis-of-irr-rep} are\dots
-
-\begin{corollary}
-  Every \(h\) eigenspace is 1-dimensional.
+  Every eigenspace of the action of \(h\) on \(V\) is 1-dimensional.
 \end{corollary}
 
 \begin{proof}
@@ -191,7 +163,7 @@ Other important consequences of proposition~\ref{thm:basis-of-irr-rep} are\dots
 
   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 \(n = \lambda = \dim V - 1\) then the eigenvalues of \(h\)
+  \dim V\). Hence if \(n = \lambda = \dim V - 1\) and the eigenvalues of \(h\)
   are
   \[
     \ldots, n - 6, n - 4, n - 2, n
@@ -303,7 +275,7 @@ of \(h\). This was instrumental to our explicit description of the irreducible
 representations of \(\mathfrak{sl}_2(K)\) culminating in
 theorem~\ref{thm:sl2-exist-unique}.
 
-Our fist task is to find some analogue of \(h\) in \(\mathfrak{sl}_3(K)\), but
+Our first task is to find some analogue of \(h\) in \(\mathfrak{sl}_3(K)\), but
 it's still unclear what exactly we are looking for. We could say we're 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
@@ -341,21 +313,21 @@ at all obvious. This is because in general \(V_\lambda\) 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 section.
+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, the eigenvalues of the adjoint actions of \(h\) are \(0\)
-and \(\pm 2\) since
+\(h\)}. In English, since
 \begin{align*}
   \operatorname{ad}(h) e & = 2 e  &
   \operatorname{ad}(h) f & = -2 f &
   \operatorname{ad}(h) h & = 0,
 \end{align*}
-and the eigenvalues of the action of \(h\) on an irreducible
+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 \(\pm
 2\).
 
@@ -413,10 +385,10 @@ If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) on
     H (X v)
     & = X (H v) + [H, X] v \\
     & = X (\lambda(H) \cdot v) + (\alpha(H) \cdot X) v \\
-    & = (\alpha + \lambda)(H) \cdot X v
+    & = (\lambda + \alpha)(H) \cdot X v
   \end{split}
 \]
-so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words,
+so that \(X\) carries \(v\) to \(V_{\lambda + \alpha}\). In other words,
 \(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors
 between eigenspaces}.
 
@@ -446,9 +418,9 @@ This is again entirely analogous to the situation we observed in
 
 \begin{theorem}\label{thm:sl3-weights-congruent-mod-root}
   The eigenvalues of the action of \(\mathfrak{h}\) on an irreducible
-  \(\mathfrak{sl}_3(K)\)-representation \(V\) differ from one another by
-  integral linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of
-  adjoint action of \(\mathfrak{h}\) on \(\mathfrak{sl}_3(K)\).
+  \(\mathfrak{sl}_3(K)\)-module \(V\) differ from one another by integral
+  linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of adjoint
+  action of \(\mathfrak{h}\) on \(\mathfrak{sl}_3(K)\).
 \end{theorem}
 
 \begin{proof}
@@ -666,12 +638,13 @@ As a first consequence of this, we show\dots
 \end{proof}
 
 There's a clear parallel between the case of \(\mathfrak{sl}_3(K)\) and that of
-\(\mathfrak{sl}_2(K)\), where we observed that the weights 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've been
-drawing are in fact consistent with the theory we've developed. Namely, since
-all weights lie in the rational span of \(\{\alpha_1, \alpha_2, \alpha_3\}\),
-we may as well draw them in the Cartesian plane.
+\(\mathfrak{sl}_2(K)\), 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've been drawing are in fact consistent with the
+theory we've developed. Namely, since all weights lie in the rational span of
+\(\{\alpha_1, \alpha_2, \alpha_3\}\), we may as well draw them in the Cartesian
+plane.
 
 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
@@ -680,7 +653,11 @@ 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 plane an considering the weight lying the furthest in that
-direction. For instance, let's say we fix the direction
+direction. 
+
+\newpage 
+
+For instance, let's say we fix the direction
 \begin{center}
   \begin{tikzpicture}[scale=2.5]
     \begin{rootSystem}{A}
@@ -769,7 +746,7 @@ 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
-weight spaces from \(V_\lambda\) by successively applying \(E_{1 2}\).
+weight spaces from \(V_\lambda\) by successively applying \(E_{2 1}\).
 \begin{center}
   \begin{tikzpicture}
     \begin{rootSystem}{A}
@@ -786,10 +763,6 @@ weight spaces from \(V_\lambda\) by successively applying \(E_{1 2}\).
       \draw[-latex] (b) to[bend left=40] (c);
       \draw[-latex] (c) to[bend left=40] (d);
       \draw[-latex] (d) to[bend left=40] (e);
-      \draw[-latex] (e) to[bend left=40] (d);
-      \draw[-latex] (d) to[bend left=40] (c);
-      \draw[-latex] (c) to[bend left=40] (b);
-      \draw[-latex] (b) to[bend left=40] (a);
     \end{rootSystem}
   \end{tikzpicture}
 \end{center}
@@ -846,6 +819,8 @@ picture is now
   \end{tikzpicture}
 \end{center}
 
+\newpage
+
 Needless to say, we could keep applying this method to the weights at the ends
 of our string, arriving at
 \begin{center}
@@ -983,9 +958,10 @@ 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
+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
@@ -1107,10 +1083,11 @@ course, that the same holds for \(\mathfrak{sl}_3(K)\). Specifically\dots
 The same argument also goes to show\dots
 
 \begin{corollary}\label{thm:irr-component-of-high-vec}
-  Given a 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 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
+  \(\lambda\).
 \end{corollary}
 
 This is very interesting to us since it implies that finding \emph{any}
@@ -1179,9 +1156,9 @@ representation turns out to be quite simple.
   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 \(\operatorname{Sym}^n K^3\) and
+  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 maximal weight vectors \(e_1^n\) and \(f_3^m\).
+  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 \(e_1^n \otimes f_3^m\).