Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
Incorporated hidden statements into the statement of a theorem
Incoporated the explicit formulas for the action of sl2 in an irrreducible representation to the theorem where they first show up
1 file changed, 29 insertions, 48 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | sections/sl2-sl3.tex | 77 | 29 | 48 |
diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex @@ -81,9 +81,22 @@ eigenvalues. To do so, we suppose without any loss in generality that V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\). \begin{proposition}\label{thm:basis-of-irr-rep} - The set \(\{v, f v, f^2, \ldots\}\) is a basis for \(V\). + The set \(\{v, f v, f^2, \ldots\}\) is a basis for \(V\). In addition, the + action of \(\mathfrak{sl}_2(K)\) on \(V\) is given by the formulas + \begin{equation}\label{eq:irr-rep-of-sl2} + \begin{aligned} + f^k v & \overset{e}{\mapsto} k(n + 1 - k) f^{k - 1} v + & f^k v & \overset{f}{\mapsto} f^{k + 1} v + & f^k v & \overset{h}{\mapsto} (n - 2 k) f^k v + \end{aligned} + \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 @@ -95,8 +108,9 @@ 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, \(h f^k v = (\lambda - 2 k) f^k v \in K \langle v, f v, f^2, v, - \ldots \rangle\). Seeing \(e f^k v \in K \langle v, f v, f^2 v, \ldots - \rangle\) is a bit more complex. Clearly, + \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, \[ \begin{split} e f v @@ -120,19 +134,14 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\). The pattern is starting to become clear: \(e\) sends \(f^k v\) to a multiple of \(f^{k - 1} v\). Explicitly, it's not hard to check by induction that \[ - e f^k v = k (\lambda + 1 - k) f^{k - 1} v + e f^k v = k (\lambda + 1 - k) f^{k - 1} v, \] + which which is the first formula of (\ref{eq:irr-rep-of-sl2}). \end{proof} -\begin{note} - For this last formula to work we fix the convention that \(f^{-1} v = 0\) -- - which is to say \(e v = 0\). -\end{note} - -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 \(\mathfrak{sl}_2(K)\) on \(V\). In other -words\dots +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 \begin{corollary} \(V\) is completely determined by the right-most eigenvalue \(\lambda\) of @@ -175,8 +184,7 @@ Other important consequences of proposition~\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 - proposition~\ref{thm:basis-of-irr-rep} that + from the formulas in (\ref{eq:irr-rep-of-sl2}) that \[ 0 = e 0 = e f^m v = m (\lambda + 1 - m) f^{m - 1} v \] @@ -208,39 +216,12 @@ 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 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 -\(\lambda\). - -We now know every irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\) -has the form -\begin{center} - \begin{tikzcd} - \cdots \arrow[bend left=60]{r} - & V_{n - 6} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l} - & V_{n - 4} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f} - & V_{n - 2} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f} - & V_n \arrow[bend left=60]{l}{f} - \end{tikzcd} -\end{center} -where \(V_{n - 2 k}\) is the 1-dimensional eigenspace of \(h\) associated to -\(n - 2 k\) and \(n = \dim V - 1\). Even more so, we explicitly know -\[ - V = \bigoplus_{k = 0}^n K f^k v -\] -and -\begin{equation}\label{eq:irr-rep-of-sl2} - \begin{aligned} - f^k v & \overset{e}{\mapsto} k(n + 1 - k) f^{k - 1} v - & f^k v & \overset{f}{\mapsto} f^{k + 1} v - & f^k v & \overset{h}{\mapsto} (n - 2 k) f^k v - \end{aligned} -\end{equation} - -To conclude our analysis all it's left is to show that for each \(n\) such -irreducible \(V\) does indeed exist. Surprisingly, we have already encountered -such a \(V\). +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 \(\lambda\). +To conclude our analysis all it's 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 \(V\). \begin{theorem}\label{thm:sl2-exist-unique} For each \(n \ge 0\) there exists a unique irreducible representation of