- Commit
- d310eab2ea34c3b2e1de91689d21d108440a64f4
- Parent
- 490c4786b2a6050f9aacffbc3f3dce5f440d8fe9
- Author
- Pablo <pablo-escobar@riseup.net>
- Date
Moved the discussion on sl2 and sl3 to a new chapter
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
Diffstat
3 files changed, 1789 insertions, 1785 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | sections/complete-reducibility.tex | 1785 | 0 | 1785 |
| Added | sections/sl2-sl3.tex | 1787 | 1787 | 0 |
| Modified | tcc.tex | 2 | 2 | 0 |
diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex @@ -894,1788 +894,3 @@ finite-dimensional representations of semisimple Lie algebras are far from over. In particular, there is so far no indication on how we could go about understanding the irreducible \(\mathfrak{g}\)-modules. Once more, we begin by investigating a simple case: that of \(\mathfrak{sl}_2(K)\). - -\section{Representations of \(\mathfrak{sl}_2(K)\)}\label{sec:sl2} - -The primary goal of this section is proving\dots - -\begin{theorem}\label{thm:sl2-exist-unique} - For each \(n > 0\), there exists precisely one irreducible representation - \(V\) of \(\mathfrak{sl}_2(K)\) with \(\dim V = n\). -\end{theorem} - -The general approach we'll take is supposing \(V\) is an irreducible -representation of \(\mathfrak{sl}_2(K)\) and then derive some information about -its structure. We begin our analysis by recalling that the elements -\begin{align*} - e & = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & - f & = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} & - h & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} -\end{align*} -form a basis of \(\mathfrak{sl}_2(K)\) and satisfy -\begin{align*} - [e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e -\end{align*} - -This is interesting to us because it implies every subspace of \(V\) invariant -under the actions of \(e\), \(f\) and \(h\) has to be \(V\) itself. Next we -turn our attention to the action of \(h\) in \(V\), in particular, to the -eigenspace decomposition -\[ - V = \bigoplus_{\lambda} V_\lambda -\] -of \(V\) -- where \(\lambda\) ranges over the eigenvalues of \(h\) and -\(V_\lambda\) 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 -- which -conveniently justifies the epigraph of this chapter. 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 -\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 -\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 \(V_{\lambda - 2}\). -Hence -\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} - & \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 -\[ - V = \bigoplus_{n \in \ZZ} V_{\lambda + 2 n}, -\] -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 -\[ - \bigoplus_{\substack{n \in \ZZ \\ a \le n \le b}} V_{\lambda + 2 n} -\] -is also an \(\mathfrak{sl}_2(K)\)-invariant subspace, 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 -eigenvalues. To do so, we suppose without any loss in generality that -\(\lambda\) is the right-most eigenvalue of \(h\), fix some non-zero \(v \in -V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\). - -\begin{theorem}\label{thm:basis-of-irr-rep} - The set \(\{v, f v, f^2, \ldots\}\) is a basis for \(V\). -\end{theorem} - -\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, \(h f^k v = (\lambda - 2 k) f^k v\). 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 - & = h v + f e v \\ - \text{(since \(\lambda\) is the right-most eigenvalue)} - & = h v + f 0 \\ - & = \lambda v - \end{split} - \] - - Next we compute - \[ - \begin{split} - e f^2 v - & = (h + fe) f v \\ - & = h f v + f (\lambda v) \\ - & = 2 (\lambda - 1) f v - \end{split} - \] - - 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 - \[ - e f^k v = k (\lambda + 1 - k) f^{k - 1} v - \] -\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} - -Theorem~\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 - -\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 - non-zero, 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 theorem~\ref{thm:basis-of-irr-rep} are\dots - -\begin{corollary} - Every \(h\) eigenspace is one-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 \(f^k v\). -\end{proof} - -\begin{corollary} - The eigenvalues of \(h\) in \(V\) form a symmetric, unbroken string of - integers separated by intervals of length \(2\) whose right-most value is - \(\dim V - 1\). -\end{corollary} - -\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 - \[ - 0 = e 0 = e f^m v = m (\lambda + 1 - m) f^{m - 1} v - \] - - This implies \(\lambda + 1 - m = 0\) -- i.e. \(\lambda = m - 1 \in \ZZ\). 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 \(h\) - are - \[ - \ldots, n - 6, n - 4, n - 2, n - \] - - To see that this string is symmetric around \(0\), simply note that the - left-most eigenvalue of \(h\) is precisely \(n - 2 (m - 1) = -n\). -\end{proof} - -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 one-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 -\(V\) does indeed exist and is irreducible. In other words\dots - -\begin{theorem}\label{thm:irr-rep-of-sl2-exists} - For each \(n \ge 0\) there exists a (unique) irreducible representation of - \(\mathfrak{sl}_2(K)\) whose left-most eigenvalue of \(h\) is \(n\). -\end{theorem} - -\begin{proof} - The fact the representation \(V\) from the previous discussion exists is - clear from the commutator relations of \(\mathfrak{sl}_2(K)\) -- just look at - \(f^k v\) as abstract symbols and impose the action given by - (\ref{eq:irr-rep-of-sl2}). 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}). To see that \(V\) is irreducible let \(W\) be a - non-zero subrepresentation and take some non-zero \(w \in W\). Suppose \(w = - \alpha_0 v + \alpha_1 f v + \cdots + \alpha_n f^n v\) and let \(k\) be the - lowest index such that \(\alpha_k \ne 0\), so that - \[ - w = \alpha_k f^k v + \cdots + \alpha_n f^n v - \] - - Now given that \(f^m = f^{n + 1}\) annihilates \(v\), - \[ - f w = \alpha_k f^{k + 1} v + \cdots + \alpha_{n - 1} f^n v - \] - - Proceeding inductively we arrive at \(f^{n - k} w = \alpha_k f^n v\), so - that \(f^n v \in W\). Hence \(e^i f^n v = \prod_{k = 1}^i k(n + 1 - k) f^{n - - i} v \in W\) for all \(i = 1, 2, \ldots, n\). Since \(k \ne 0 \ne n + 1 - k\) - for all \(k\) in this range, we can see that \(f^k v \in W\) for all \(k = 0, - 1, \ldots, n\). In other words, \(W = V\). We are done. -\end{proof} - -Our initial gamble of studying the eigenvalues of \(h\) may have seemed -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. - -\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 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 -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. - -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 -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'll 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 -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 -theorem~\ref{thm:irr-rep-of-sl2-exists}. - -Our fist 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 -see, no such analogue exists and neither does such element. Instead, the actual -way to proceed is to consider the subalgebra -\[ - \mathfrak{h} - = \left\{ - X \in - \begin{pmatrix} K & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K \end{pmatrix} - : \operatorname{Tr}(X) = 0 - \right\} -\] - -The choice of \(\mathfrak{h}\) may seem like an odd choice at the moment, but -the point is we'll later show that there exists some \(v \in V\) 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 non-zero -\(v \in V\). We call such functionals \(\lambda\) \emph{eigenvalues of -\(\mathfrak{h}\)}, and we say \emph{\(v\) is an eigenvector of -\(\mathfrak{h}\)}. - -Once again, we'll pay special attention to the eigenvalue decomposition -\begin{equation}\label{eq:weight-module} - V = \bigoplus_\lambda V_\lambda -\end{equation} -where \(\lambda\) ranges over all eigenvalues of \(\mathfrak{h}\) and -\(V_\lambda = \{ v \in V : H v = \lambda(H) \cdot v, \forall H \in \mathfrak{h} -\}\). 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 -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. - -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 of the adjoint actions of \(h\) are -\(\pm 2\) since -\begin{align*} - [h, f] & = -2 f & - [h, e] & = 2 e -\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 -\(\pm 2\). - -In the case of \(\mathfrak{sl}_3(K)\), a simple calculation shows that if \([H, -X]\) is scalar multiple of \(X\) for all \(H \in \mathfrak{h}\) then all but -one entry of \(X\) are zero. Hence the eigenvectors of the adjoint action of -\(\mathfrak{h}\) are \(E_{i j}\) and its eigenvalues are \(\alpha_i - -\alpha_j\), where -\[ - \alpha_i - \begin{pmatrix} - a_1 & 0 & 0 \\ - 0 & a_2 & 0 \\ - 0 & 0 & a_3 - \end{pmatrix} - = a_i -\] - -Visually we may draw - -\begin{figure}[h] - \centering - \begin{tikzpicture}[scale=2.5] - \begin{rootSystem}{A} - \filldraw[black] \weight{0}{0} circle (.5pt); - \node[black, above right] at \weight{0}{0} {\small$0$}; - \wt[black]{-1}{2} - \wt[black]{-2}{1} - \wt[black]{1}{1} - \wt[black]{-1}{-1} - \wt[black]{2}{-1} - \wt[black]{1}{-2} - \node[above] at \weight{-1}{2} {$\alpha_2 - \alpha_3$}; - \node[left] at \weight{-2}{1} {$\alpha_2 - \alpha_1$}; - \node[right] at \weight{1}{1} {$\alpha_1 - \alpha_3$}; - \node[left] at \weight{-1}{-1} {$\alpha_3 - \alpha_1$}; - \node[right] at \weight{2}{-1} {$\alpha_1 - \alpha_2$}; - \node[below] at \weight{1}{-2} {$\alpha_3 - \alpha_1$}; - \node[black, above] at \weight{1}{0} {$\alpha_1$}; - \node[black, above] at \weight{-1}{1} {$\alpha_2$}; - \node[black, above] at \weight{0}{-1} {$\alpha_3$}; - \filldraw[black] \weight{1}{0} circle (.5pt); - \filldraw[black] \weight{-1}{1} circle (.5pt); - \filldraw[black] \weight{0}{-1} circle (.5pt); - \end{rootSystem} - \end{tikzpicture} -\end{figure} - -If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) in -\(\mathfrak{sl}_3(K)\) 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 -\[ - \begin{split} - H (X v) - & = X (H v) + [H, X] v \\ - & = X (\lambda(H) \cdot v) + (\alpha(H) \cdot X) v \\ - & = (\alpha + \lambda)(H) \cdot X v - \end{split} -\] -so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words, -\(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors -between eigenspaces}. - -For instance \(\mathfrak{sl}_3(K)_{\alpha_1 - \alpha_3}\) will act on the -adjoint representation of \(\mathfrak{sl}_3(K)\) via -\begin{figure}[h] - \centering - \begin{tikzpicture}[scale=2.5] - \begin{rootSystem}{A} - \wt[black]{0}{0} - \wt[black]{-1}{2} - \wt[black]{-2}{1} - \wt[black]{1}{1} - \wt[black]{-1}{-1} - \wt[black]{2}{-1} - \wt[black]{1}{-2} - \draw[-latex, black] \weight{-1.9}{1.1} -- \weight{-1.1}{1.9}; - \draw[-latex, black] \weight{-.9}{-.9} -- \weight{-.1}{-.1}; - \draw[-latex, black] \weight{0.1}{0.1} -- \weight{.9}{.9}; - \draw[-latex, black] \weight{1.1}{-1.9} -- \weight{1.9}{-1.1}; - \end{rootSystem} - \end{tikzpicture} -\end{figure} - -This is again entirely analogous to the situation we observed in -\(\mathfrak{sl}_2(K)\). In fact, we may once more conclude\dots - -\begin{theorem}\label{thm:sl3-weights-congruent-mod-root} - The eigenvalues of the action of \(\mathfrak{h}\) in 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 \(\mathfrak{h}\) in \(\mathfrak{sl}_3(K)\). -\end{theorem} - -\begin{proof} - This proof goes exactly as that of the analogous statement for - \(\mathfrak{sl}_2(K)\): it suffices to note that if we fix some eigenvalue - \(\lambda\) of \(\mathfrak{h}\) and let \(i\) and \(j\) vary then - \[ - \bigoplus_{i j} V_{\lambda + \alpha_i - \alpha_j} - \] - is an invariant subspace of \(V\). -\end{proof} - -To avoid confusion we better introduce some notation to differentiate between -eigenvalues of the action of \(\mathfrak{h}\) in \(V\) and eigenvalues of the -adjoint action of \(\mathfrak{h}\). - -\begin{definition} - Given a representation \(V\) of \(\mathfrak{sl}_3(K)\), we'll call the - non-zero eigenvalues of the action of \(\mathfrak{h}\) in \(V\) \emph{weights - of \(V\)}. As you might have guessed, we'll correspondingly refer to - eigenvectors and eigenspaces of a given weight by \emph{weight vectors} and - \emph{weight spaces}. -\end{definition} - -It's clear from our previous discussion that the weights of the adjoint -representation of \(\mathfrak{sl}_3(K)\) 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 - \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{corollary} - The weights of an irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) - are all congruent module the lattice \(Q\) generated by the roots \(\alpha_i - - \alpha_j\) of \(\mathfrak{sl}_3(K)\). -\end{corollary} - -\begin{definition} - The lattice \(Q = \ZZ \langle \alpha_i - \alpha_j : i, j = 1, 2, 3 \rangle\) - is called \emph{the root lattice of \(\mathfrak{sl}_3(K)\)}. -\end{definition} - -To proceed we once more refer to the previously established framework: next we -saw that the eigenvalues of \(h\) formed 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 -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. For instance, let's say we fix the direction -\begin{center} - \begin{tikzpicture}[scale=2.5] - \begin{rootSystem}{A} - \wt[black]{0}{0} - \wt[black]{-1}{2} - \wt[black]{-2}{1} - \wt[black]{1}{1} - \wt[black]{-1}{-1} - \wt[black]{2}{-1} - \wt[black]{1}{-2} - \draw[-latex, black, thick] \weight{-1.5}{-.5} -- \weight{1.5}{.5}; - \end{rootSystem} - \end{tikzpicture} -\end{center} -and let \(\lambda\) be the weight lying the furthest in this direction. - -Its easy to see what we mean intuitively by looking at the previous picture, -but its precise meaning is still allusive. Formally this means we'll choose a -linear functional \(f : \mathfrak{h}^* \to \QQ\) and pick the weight that -maximizes \(f\). To avoid any ambiguity we should choose the direction of a -line irrational with respect to the root lattice \(Q\). For instance if we -choose the direction of \(\alpha_1 - \alpha_3\) and let \(f\) be the rational -projection \(Q \to \QQ \langle \alpha_1 - \alpha_3 \rangle \cong \QQ\) then -\(\alpha_1 - 2 \alpha_2 + \alpha_3 \in Q\) lies in \(\ker f\), so that if a -weight \(\lambda\) maximizes \(f\) then the translation of \(\lambda\) by any -multiple of \(\alpha_1 - 2 \alpha_2 + \alpha_3\) must also do so. In others -words, if the direction we choose is parallel to a vector lying in \(Q\) then -there may be multiple choices the ``weight lying the furthest'' along this -direction. - -\begin{definition} - We say that a root \(\alpha\) is positive if \(f(\alpha) > 0\) -- i.e. if it - lies to the right of the direction we chose. Otherwise we say \(\alpha\) is - negative. Notice that \(f(\alpha) \ne 0\) since by definition \(\alpha \ne - 0\) and \(f\) is irrational with respect to the lattice \(Q\). -\end{definition} - -The first 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} - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \weightLattice{3} - \fill[gray!50,opacity=.2] (hex cs:x=5,y=-7) -- (hex cs:x=1,y=1) -- - (hex cs:x=-7,y=5) arc (150:270:{7*\weightLength}); - \draw[black, thick] (hex cs:x=5,y=-7) -- (hex cs:x=1,y=1) -- - (hex cs:x=-7,y=5); - \filldraw[black] (hex cs:x=1,y=1) circle (1pt); - \node[above right=-2pt] at (hex cs:x=1,y=1) {\small\(\lambda\)}; - \end{rootSystem} - \end{tikzpicture} -\end{center} - -Indeed, if this is not the case then, by definition, \(\lambda\) is not the -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 non-zero -- which contradicts the hypothesis that -\(\lambda\) lies the furthest along the direction we chose. In other words\dots - -\begin{theorem} - There is a weight vector \(v \in V\) that is killed by all positive root - spaces of \(\mathfrak{sl}_3(K)\). -\end{theorem} - -\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 \(\alpha_2 - - \alpha_3\). -\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 -\(\mathfrak{sl}_2(K)\), 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 -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. - -We'll start out by trying to understand the weights in the boundary of -\(\frac{1}{3}\)-plane previously drawn. Since the root spaces act by -translation, the action of \(E_{2 1}\) in \(V_\lambda\) will span a subspace -\[ - W = \bigoplus_k V_{\lambda + k (\alpha_2 - \alpha_1)}, -\] -and by the same token \(W\) must be invariant under the action of \(E_{1 2}\). - -To draw a familiar picture -\begin{center} - \begin{tikzpicture} - \begin{rootSystem}{A} - \node at \weight{3}{1} (a) {}; - \node at \weight{1}{2} (b) {}; - \node at \weight{-1}{3} (c) {}; - \node at \weight{-3}{4} (d) {}; - \node at \weight{-5}{5} (e) {}; - \draw \weight{3}{1} -- \weight{-4}{4.5}; - \draw[dotted] \weight{-4}{4.5} -- \weight{-5}{5}; - \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} - \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; - \draw[-latex] (a) to[bend left=40] (b); - \draw[-latex] (b) to[bend left=40] (c); - \draw[-latex] (c) to[bend left=40] (d); - \draw[-latex] (d) to[bend left=40] (e); - \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} - -What's remarkable about all this is the fact that the subalgebra spanned by -\(E_{1 2}\), \(E_{2 1}\) and \(H = [E_{1 2}, E_{2 1}]\) is isomorphic to -\(\mathfrak{sl}_2(K)\) via -\begin{align*} - E_{2 1} & \mapsto e & - E_{1 2} & \mapsto f & - H & \mapsto h -\end{align*} - -In other words, \(W\) is a representation of \(\mathfrak{sl}_2(K)\). Even more -so, we claim -\[ - V_{\lambda + k (\alpha_2 - \alpha_1)} = W_{\lambda(H) - 2k} -\] - -Indeed, \(V_{\lambda + k (\alpha_2 - \alpha_1)} \subset W_{\lambda(H) - 2k}\) -since \((\lambda + k (\alpha_2 - \alpha_1))(H) = \lambda(H) + k (-1 - 1) = -\lambda(H) - 2 k\). On the other hand, if we suppose \(0 < \dim V_{\lambda + k -(\alpha_2 - \alpha_1)} < \dim W_{\lambda(H) - 2 k}\) for some \(k\) we arrive -at -\[ - \dim W - = \sum_k \dim V_{\lambda + k (\alpha_2 - \alpha_1)} - < \sum_k \dim W_{\lambda(H) - 2k} - = \dim W, -\] -a contradiction. - -There are a number of important consequences to this, of the first being that -the weights of \(V\) appearing on \(W\) must be symmetric with respect to the -the line \(B(\alpha_1 - \alpha_2, \alpha) = 0\). The picture is -thus -\begin{center} - \begin{tikzpicture} - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \setlength{\weightRadius}{2pt} - \weightLattice{4} - \draw[thick] \weight{3}{1} -- \weight{-3}{4}; - \wt[black]{0}{0} - \node[above left] at \weight{0}{0} {\small\(0\)}; - \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} - \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; - \draw[very thick] \weight{0}{-4} -- \weight{0}{4} - node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)}; - \end{rootSystem} - \end{tikzpicture} -\end{center} - -Notice we could apply this same argument to the subspace \(\bigoplus_k -V_{\lambda + k (\alpha_3 - \alpha_2)}\): this subspace is invariant under the -action of the subalgebra spanned by \(E_{2 3}\), \(E_{3 2}\) and \([E_{2 3}, -E_{3 2}]\), which is again isomorphic to \(\mathfrak{sl}_2(K)\), so that the -weights in this subspace must be symmetric with respect to the line -\(B(\alpha_3 - \alpha_2, \alpha) = 0\). The picture is now -\begin{center} - \begin{tikzpicture} - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \setlength{\weightRadius}{2pt} - \weightLattice{4} - \draw[thick] \weight{3}{1} -- \weight{-3}{4}; - \draw[thick] \weight{3}{1} -- \weight{4}{-1}; - \wt[black]{0}{0} - \wt[black]{4}{-1} - \node[above left] at \weight{0}{0} {\small\(0\)}; - \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} - \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; - \draw[very thick] \weight{0}{-4} -- \weight{0}{4} - node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)}; - \draw[very thick] \weight{-4}{0} -- \weight{4}{0} - node[right]{\small\(B(\alpha_3 - \alpha_2, \alpha) = 0\)}; - \end{rootSystem} - \end{tikzpicture} -\end{center} - -In general, given a weight \(\mu\), the space -\[ - \bigoplus_k V_{\mu + k (\alpha_i - \alpha_j)} -\] -is invariant under the action of the subalgebra \(\mathfrak{s}_{\alpha_i - -\alpha_j} = K \langle E_{i j}, E_{j i}, [E_{i j}, E_{j i}] \rangle\), which is -once more isomorphic to \(\mathfrak{sl}_2(K)\), and again the weight spaces in -this string match precisely the eigenvalues of \(h\). Needless to say, we could -keep applying this method to the weights at the ends of our string, arriving at -\begin{center} - \begin{tikzpicture} - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \setlength{\weightRadius}{2pt} - \weightLattice{5} - \draw[thick] \weight{3}{1} -- \weight{-3}{4}; - \draw[thick] \weight{3}{1} -- \weight{4}{-1}; - \draw[thick] \weight{-3}{4} -- \weight{-4}{3}; - \draw[thick] \weight{-4}{3} -- \weight{-1}{-3}; - \draw[thick] \weight{1}{-4} -- \weight{4}{-1}; - \draw[thick] \weight{-1}{-3} -- \weight{1}{-4}; - \wt[black]{-4}{3} - \wt[black]{-3}{1} - \wt[black]{-2}{-1} - \wt[black]{-1}{-3} - \wt[black]{1}{-4} - \wt[black]{2}{-3} - \wt[black]{3}{-2} - \wt[black]{4}{-1} - \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} - \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; - \draw[very thick] \weight{-5}{5} -- \weight{5}{-5}; - \draw[very thick] \weight{0}{-5} -- \weight{0}{5}; - \draw[very thick] \weight{-5}{0} -- \weight{5}{0}; - \end{rootSystem} - \end{tikzpicture} -\end{center} - -We claim all dots \(\mu\) lying inside the hexagon we've 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. -\begin{center} - \begin{tikzpicture} - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \setlength{\weightRadius}{2pt} - \weightLattice{5} - \draw[thick] \weight{3}{1} -- \weight{-3}{4}; - \draw[thick] \weight{3}{1} -- \weight{4}{-1}; - \draw[thick] \weight{-3}{4} -- \weight{-4}{3}; - \draw[thick] \weight{-4}{3} -- \weight{-1}{-3}; - \draw[thick] \weight{1}{-4} -- \weight{4}{-1}; - \draw[thick] \weight{-1}{-3} -- \weight{1}{-4}; - \wt[black]{-4}{3} - \wt[black]{-3}{1} - \wt[black]{-2}{-1} - \wt[black]{-1}{-3} - \wt[black]{1}{-4} - \wt[black]{2}{-3} - \wt[black]{3}{-2} - \wt[black]{4}{-1} - \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} - \node[above right=-2pt] at \weight{1}{2} {\small\(\nu\)}; - \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; - \draw[very thick] \weight{-5}{5} -- \weight{5}{-5}; - \draw[very thick] \weight{0}{-5} -- \weight{0}{5}; - \draw[very thick] \weight{-5}{0} -- \weight{5}{0}; - \draw[gray, thick] \weight{1}{2} -- \weight{-2}{-1}; - \wt[black]{1}{2} - \wt[black]{-2}{-1} - \wt{0}{1} - \wt{-1}{0} - \end{rootSystem} - \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 gray -string must occur in the representation of \(\mathfrak{sl}_2(K)\). Hence they -must also be weights of \(V\). The final picture is thus -\begin{center} - \begin{tikzpicture} - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \setlength{\weightRadius}{2pt} - \weightLattice{5} - \draw[thick] \weight{3}{1} -- \weight{-3}{4}; - \draw[thick] \weight{3}{1} -- \weight{4}{-1}; - \draw[thick] \weight{-3}{4} -- \weight{-4}{3}; - \draw[thick] \weight{-4}{3} -- \weight{-1}{-3}; - \draw[thick] \weight{1}{-4} -- \weight{4}{-1}; - \draw[thick] \weight{-1}{-3} -- \weight{1}{-4}; - \wt[black]{-4}{3} - \wt[black]{-3}{1} - \wt[black]{-2}{-1} - \wt[black]{-1}{-3} - \wt[black]{1}{-4} - \wt[black]{2}{-3} - \wt[black]{3}{-2} - \wt[black]{4}{-1} - \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} - \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; - \draw[very thick] \weight{-5}{5} -- \weight{5}{-5}; - \draw[very thick] \weight{0}{-5} -- \weight{0}{5}; - \draw[very thick] \weight{-5}{0} -- \weight{5}{0}; - \wt[black]{-2}{2} - \wt[black]{0}{1} - \wt[black]{-1}{0} - \wt[black]{0}{-2} - \wt[black]{1}{-1} - \wt[black]{2}{0} - \end{rootSystem} - \end{tikzpicture} -\end{center} - -Another important consequence of our analysis is the fact that \(\lambda\) lies -in the lattice \(P\) generated by \(\alpha_1\), \(\alpha_2\) and \(\alpha_3\). -Indeed, \(\lambda([E_{i j}, E_{j i}])\) is an eigenvalue of \(h\) in a -representation of \(\mathfrak{sl}_2(K)\), so it must be an integer. Now since -\[ - \lambda - \begin{pmatrix} - a & 0 & 0 \\ - 0 & b & 0 \\ - 0 & 0 & -a -b - \end{pmatrix} - = - \lambda - \begin{pmatrix} - a & 0 & 0 \\ - 0 & 0 & 0 \\ - 0 & 0 & -a - \end{pmatrix} - + - \lambda - \begin{pmatrix} - 0 & 0 & 0 \\ - 0 & b & 0 \\ - 0 & 0 & -b - \end{pmatrix} - = - a \lambda([E_{1 3}, E_{3 1}]) + b \lambda([E_{2 3}, E_{3 2}]), -\] -which is to say \(\lambda = \lambda([E_{1 3}, E_{3 1}]) \alpha_1 + -\lambda([E_{2 3}, E_{3 2}]) \alpha_2\), we can see that \(\lambda \in -P\). - -\begin{definition} - The lattice \(P = \ZZ \alpha_1 \oplus \ZZ \alpha_2 \oplus \ZZ \alpha_3\) is - called \emph{the weight lattice of \(\mathfrak{sl}_3(K)\)}. -\end{definition} - -Finally\dots - -\begin{theorem}\label{thm:sl3-irr-weights-class} - The weights of \(V\) 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} - -Once more 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 \(P = \ZZ\) and were congruent modulo the lattice \(Q = 2 \ZZ\). -Having found all of the weights of \(V\), the only thing we're 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 irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) whose highest - weight is \(n \alpha_1 - m \alpha_3\). -\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 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 - -\begin{theorem}\label{thm:irr-sl3-span} - 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 \(E_{3 2}\). -\end{theorem} - -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 -\(\mathfrak{sl}_3(K)\) 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 -\(\mathfrak{sl}_3(K)\) -- please refer to \cite{fulton-harris} for further -details. The same argument also goes to show\dots - -\begin{corollary} - 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 \(\lambda\). -\end{corollary} - -This is very interesting to us since it implies that finding \emph{any} -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. - -\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 - \begin{center} - \begin{tikzpicture}[scale=2.5] - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \weightLattice{2} - \wt[black]{1}{0} - \wt[black]{-1}{1} - \wt[black]{0}{-1} - \node[right] at \weight{1}{0} {$\alpha_1$}; - \node[above left] at \weight{-1}{1} {$\alpha_2$}; - \node[below left] at \weight{0}{-1} {$\alpha_3$}; - \end{rootSystem} - \end{tikzpicture} - \end{center} - and \(\alpha_1\) is the highest weight of \(V\). - - 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, - \begin{center} - \begin{tikzpicture}[scale=2.5] - \AutoSizeWeightLatticefalse - \begin{rootSystem}{A} - \weightLattice{2} - \wt[black]{-1}{0} - \wt[black]{1}{-1} - \wt[black]{0}{1} - \node[left] at \weight{-1}{0} {$-\alpha_1$}; - \node[below right] at \weight{1}{-1} {$-\alpha_2$}; - \node[above right] at \weight{0}{1} {$-\alpha_3$}; - \end{rootSystem} - \end{tikzpicture} - \end{center} - is the weight diagram of \(V^*\) and \(\alpha_3\) is the highest weight of - \(V^*\). - - On the other hand if we fix two \(\mathfrak{sl}_3(K)\)-representations \(U\) - and \(W\), by computing - \[ - \begin{split} - H (u \otimes w) - & = H u \otimes w + u \otimes H w \\ - & = \lambda(H) \cdot u \otimes w + u \otimes \mu(H) \cdot w \\ - & = (\lambda + \mu)(H) \cdot (u \otimes w) - \end{split} - \] - 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 - 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 \(- m \alpha_3\) - 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 \(e_1^n \otimes f_3^m\). -\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 \(W\). - - Now by computing - \[ - H (v + w) - = H v + H w - = \mu(H) \cdot v + \mu(H) \cdot w - = \mu(H) \cdot (v + w) - \] - 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 - irreducible representation \(U = \mathfrak{sl}_3(K) \cdot v + w\) generated - by \(v + w\). The projection maps \(\pi_1 : U \to V\), \(\pi_2 : U \to W\), - being non-zero homomorphism between irreducible representations of - \(\mathfrak{sl}_3(K)\) 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 -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 section 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 \(\mathfrak{g}\). - -\section{Simultaneous Diagonalization \& the General Case} - -At the heart of our analysis of \(\mathfrak{sl}_2(K)\) and -\(\mathfrak{sl}_3(K)\) was the decision to consider the eigenspace -decomposition -\begin{equation}\label{sym-diag} - V = \bigoplus_\lambda V_\lambda -\end{equation} - -This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the -reasoning behind it, as well as the mere fact equation (\ref{sym-diag}) holds, -are harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace -decomposition associated with an operator \(V \to V\) is a very well-known -tool, and this type of argument should be familiar to anyone familiar with -basic concepts of linear algebra. On the other hand, the eigenspace -decomposition of \(V\) with respect to the action of an arbitrary subalgebra -\(\mathfrak{h} \subset \mathfrak{gl}(V)\) is neither well-known nor does it -hold in general: as previously stated, it may very well be that -\[ - \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \subsetneq V -\] - -We should note, however, that this two cases are not as different as they may -sound at first glance. Specifically, we can regard the eigenspace decomposition -of a representation \(V\) of \(\mathfrak{sl}_2(K)\) with respect to the -eigenvalues of the action of \(h\) as the eigenvalue decomposition of \(V\) -with respect to the action of the subalgebra \(\mathfrak{h} = K h \subset -\mathfrak{sl}_2(K)\). Furthermore, in both cases \(\mathfrak{h} \subset -\mathfrak{sl}_n(K)\) is the subalgebra of diagonal matrices, which is Abelian. -The fundamental difference between these two cases is thus the fact that \(\dim -\mathfrak{h} = 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim -\mathfrak{h} > 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The -question then is: why did we choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} > -1\) for \(\mathfrak{sl}_3(K)\)? - -% TODO: Add a note on how irreducible representations of Abelian algebras are -% all one dimensional to the previous chapter -The rational behind fixing an Abelian subalgebra is a simple one: we have seen -in the previous chapter that representations of Abelian -algebras are generally much simpler to understand than the general case. -Thus it make sense to decompose a given representation \(V\) of -\(\mathfrak{g}\) into subspaces invariant under the action of \(\mathfrak{h}\), -and then analyze how the remaining elements of \(\mathfrak{g}\) act on this -subspaces. The bigger \(\mathfrak{h}\) the simpler our problem gets, because -there are fewer elements outside of \(\mathfrak{h}\) left to analyze. - -Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h} -\subset \mathfrak{g}\), which leads us to the following definition. - -\begin{definition} - An subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a Cartan - subalgebra of \(\mathfrak{g}\)} if is self-normalizing -- i.e. \([X, H] \in - \mathfrak{h}\) for all \(H \in \mathfrak{h}\) if, and only if \(X \in - \mathfrak{h}\) -- and nilpotent. Equivalently for reductive \(\mathfrak{g}\), - \(\mathfrak{h}\) is called \emph{a Cartan subalgebra of \(\mathfrak{g}\)} if - it is Abelian, \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in - \mathfrak{h}\) and if \(\mathfrak{h}\) is maximal with respect to the former - two properties. -\end{definition} - -\begin{proposition} - There exists a Cartan subalgebra \(\mathfrak{h} \subset \mathfrak{g}\). -\end{proposition} - -\begin{proof} - Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose - elements act as diagonal operators via the adjoint representation. Indeed, - \(0\) -- the only element of \(0 \subset \mathfrak{g}\) -- is such that - \(\operatorname{ad}(0) = 0\). Furthermore, given a chain of Abelian - subalgebras - \[ - 0 \subset \mathfrak{h}_1 \subset \mathfrak{h}_2 \subset \cdots - \] - such that \(\operatorname{ad}(H)\) is a diagonal operator for each \(H \in - \mathfrak{h}_i\), the subalgebra \(\bigcup_i \mathfrak{h}_i \subset - \mathfrak{g}\) is Abelian, and its elements also act diagonally in - \(\mathfrak{g}\). It then follows from Zorn's lemma that there exists a - subalgebra \(\mathfrak{h}\) which is maximal with respect to both these - properties -- i.e. a Cartan subalgebra. -\end{proof} - -We have already seen some concrete examples. For instance, one can readily -check that every pair of diagonal matrices commutes, so that -\[ - \mathfrak{h} = - \begin{pmatrix} - K & 0 & \cdots & 0 \\ - 0 & K & \cdots & 0 \\ - \vdots & \vdots & \ddots & \vdots \\ - 0 & 0 & \cdots & K - \end{pmatrix} -\] -is an Abelian -- and hence nilpotent -- subalgebra of \(\mathfrak{gl}_n(K)\). A -simple calculation also shows that if \(i \ne j\) then the coefficient of -\(E_{i j}\) in \([E_{i i}, X]\) is the same as the coefficient of \(E_{i j}\) -in \(X\), for all \(X \in \mathfrak{gl}_n(K)\). In particular, if \([E_{i i}, -X]\) is diagonal for all \(i\), then so is \(X\) -- i.e. \(\mathfrak{h}\) is -self-normalizing. Hence \(\mathfrak{h}\) is a Cartan subalgebra of -\(\mathfrak{gl}_n(K)\). - -The intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the -subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of -\(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the -subalgebras described the previous two sections. The remaining question then -is: if \(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(V\) -is a representation of \(\mathfrak{g}\), does the eigenspace decomposition -\[ - V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda -\] -of \(V\) hold? The answer to this question turns out to be yes. This is a -consequence of something known as \emph{simultaneous diagonalization}, which is -the primary tool we'll use to generalize the results of the previous section. -What is simultaneous diagonalization all about then? - -\begin{definition}\label{def:sim-diag} - Given a \(K\)-vector space \(V\), a set of operators \(\{T_j : V \to V\}_j\) - is called \emph{simultaneously diagonalizable} if there is a basis \(\{v_1, - \ldots, v_n\}\) for \(V\) such that \(T_j v_i\) is a scalar multiple of - \(v_i\), for all \(i, j\). -\end{definition} - -\begin{proposition} - Given a \emph{finite-dimensional} vector space \(V\), A set of diagonalizable - operators \(V \to V\) is simultaneously diagonalizable if, and only if all of - its elements commute with one another. -\end{proposition} - -We should point out that simultaneous diagonalization \emph{only works in the -finite-dimensional setting}. In fact, simultaneous diagonalization is usually -framed as an equivalent statement about diagonalizable \(n \times n\) matrices --- where \(n\) is, of course, finite. - -Simultaneous diagonalization implies that to show \(V = \bigoplus_\lambda -V_\lambda\) it suffices to show that \(H\!\restriction_V : V \to V\) is a -diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we -introduce \emph{the Jordan decomposition of an operator} and \emph{the abstract -Jordan decomposition of a semisimple Lie algebra}. - -\begin{proposition}[Jordan] - Given a finite-dimensional vector space \(V\) and an operator \(T : V \to - V\), there are unique commuting operators \(T_s, T_n : V \to V\), with - \(T_s\) diagonalizable and \(T_n\) nilpotent, such that \(T = T_s + T_n\). - The pair \((T_s, T_n)\) is known as \emph{the Jordan decomposition of \(T\)}. -\end{proposition} - -\begin{proposition} - Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are - \(X_s, X_n \in \mathfrak{g}\) such that \(X = X_s + X_n\), \([X_s, X_n] = - 0\), \(\operatorname{ad}(X_s)\) is a diagonalizable operator and - \(\operatorname{ad}(X_n)\) is a nilpotent operator. The pair \((X_s, X_n)\) - is known as \emph{the Jordan decomposition of \(X\)}. -\end{proposition} - -It should be clear from the uniqueness of \(\operatorname{ad}(X)_s\) and -\(\operatorname{ad}(X)_n\) that the Jordan decomposition of -\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) = \operatorname{ad}(X_s) + -\operatorname{ad}(X_n)\). What's perhaps more remarkable is the fact this holds -for \emph{any} finite-dimensional representation of \(\mathfrak{g}\). In other -words\dots - -\begin{proposition}\label{thm:preservation-jordan-form} - Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\) and \(X - \in \mathfrak{g}\). Denote by \(X\!\restriction_V\) the action of \(X\) in - \(V\). Then \(X_s\!\restriction_V = (X\!\restriction)_s\) and - \(X_n\!\restriction_V = (X\!\restriction)_n\). -\end{proposition} - -This last result is known as \emph{the preservation of the Jordan form}, and a -proof can be found in appendix C of \cite{fulton-harris}. We should point out -this fails spectacularly in positive characteristic. Furthermore, the statement -of proposition~\ref{thm:preservation-jordan-form} only makes sense for -\emph{semisimple} Lie algebras -- i.e. the algebras \(\mathfrak{g}\) for which -the abstract Jordan decomposition of \(\mathfrak{g}\) is defined. Nevertheless, -as promised this implies\dots - -\begin{corollary}\label{thm:finite-dim-is-weight-mod} - Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset - \mathfrak{g}\) be a Cartan subalgebra and \(V\) be any finite-dimensional - representation of \(\mathfrak{g}\). Then there is a basis \(\{v_1, \ldots, - v_n\}\) of \(V\) so that each \(v_i\) is simultaneously an eigenvector of all - elements of \(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as - a diagonal matrix in this basis. In other words, there are linear functionals - \(\lambda_i \in \mathfrak{h}^*\) so that - \( - H v_i = \lambda_i(H) \cdot v_i - \) - for all \(H \in \mathfrak{h}\). In particular, - \[ - V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda - \] -\end{corollary} - -\begin{proof} - Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_V - : V \to V\) is a diagonalizable operator. - - If we write \(H = H_s + H_n\) for the abstract Jordan decomposition of \(H\), - we know \(\operatorname{ad}(H_s) = \operatorname{ad}(H)_s\). But - \(\operatorname{ad}(H)\) is a diagonalizable operator, so that - \(\operatorname{ad}(H)_s = \operatorname{ad}(H)\). This implies - \(\operatorname{ad}(H_n) = \operatorname{ad}(H)_n = 0\), so that \(H_n\) is a - central element of \(\mathfrak{g}\). Since \(\mathfrak{g}\) is semisimple, - \(H_n = 0\). Proposition~\ref{thm:preservation-jordan-form} then implies - \((H\!\restriction_V)_n = (H_n)\!\restriction_V = 0\), so \(H\!\restriction_V - = (H\!\restriction_V)_s\) is a diagonalizable operator. -\end{proof} - -We should point out that this last proof only works for semisimple Lie -algebras. This is because we rely heavily on -proposition~\ref{thm:preservation-jordan-form}, as well in the fact that -semisimple Lie algebras are centerless. In fact, -corollary~\ref{thm:finite-dim-is-weight-mod} fails even for reductive Lie -algebras. For a counterexample, consider the algebra \(\mathfrak{g} = K\): the -Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a -\(\mathfrak{g}\)-module is simply a vector space \(V\) endowed with an operator -\(V \to V\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) in -\(V\). In particular, if we choose an operator \(V \to V\) which is \emph{not} -diagonalizable we find \(V \ne \bigoplus_{\lambda \in \mathfrak{h}^*} -V_\lambda\). - -However, corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive -\(\mathfrak{g}\) if we assume that the representation in question is -irreducible, since central elements of \(\mathfrak{g}\) act on irreducible -representations as scalar operators. The hypothesis of finite-dimensionality is -also of huge importance. In the next chapter we will encounter -infinite-dimensional \(\mathfrak{g}\)-modules for which the eigenspace -decomposition \(V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda\) fails. -As a first consequence of corollary~\ref{thm:finite-dim-is-weight-mod} - -\begin{corollary} - The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate. -\end{corollary} - -\begin{proof} - Consider the eigenspace decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus - \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint representation, where - \(\alpha\) ranges over all nonzero eigenvalues of the adjoint action of - \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = \mathfrak{h}\). - - Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h}) - \mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the - other hand, since \(\mathfrak{h}\) is self-normalizing, if \([X, H] = 0 \in - \mathfrak{h}\) for all \(H \in \mathfrak{h}\) then \(X \in \mathfrak{h}\) -- - i.e. \(\mathfrak{g}_0 \subset \mathfrak{h}\). So the eigenspace decomposition - becomes - \[ - \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha - \] - - We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to - \(\mathfrak{g}_\alpha\) with respect to \(B\) for any \(\alpha \ne 0\). - Indeed, given \(X \in \mathfrak{g}_\alpha\) and \(H_1, H_2 \in \mathfrak{h}\) - with \(\alpha(H_1) \ne 0\) we have - \[ - \alpha(H_1) \cdot B(X, H_2) - = B([H_1, X], H_2) - = - B([X, H_1], H_2) - = - B(X, [H_1, H_2]) - = 0 - \] - - Hence the non-degeneracy of \(B\) implies the non-degeneracy of its - restriction. -\end{proof} - -We should point out that the restriction of \(B\) to \(\mathfrak{h}\) is -\emph{not} the Killing form of \(\mathfrak{h}\). In fact, since -\(\mathfrak{h}\) is Abelian, its Killing form is identically zero -- which is -hardly ever a non-degenerate form. - -\begin{note} - Since \(B\) induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\), it - induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in - \(\mathfrak{h}^*\). We denote this form by \(B\). -\end{note} - -We now have most of the necessary tools to reproduce the results of the -previous chapter in a general setting. Let \(\mathfrak{g}\) be a -finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\) -and let \(V\) be a finite-dimensional irreducible representation of -\(\mathfrak{g}\). We will proceed, as we did before, by generalizing the -results about of the previous two sections in order. By now the pattern should -be starting become clear, so we will mostly omit technical details and proofs -analogous to the ones on the previous sections. Further details can be found in -appendix D of \cite{fulton-harris} and in \cite{humphreys}. - -We begin our analysis by remarking that in both \(\mathfrak{sl}_2(K)\) and -\(\mathfrak{sl}_3(K)\), the roots were symmetric about the origin and spanned -all of \(\mathfrak{h}^*\). This turns out to be a general fact, which is a -consequence of the non-degeneracy of the restriction of the Killing form to the -Cartan subalgebra. - -\begin{proposition}\label{thm:weights-symmetric-span} - The eigenvalues \(\alpha\) of the adjoint action of \(\mathfrak{h}\) in - \(\mathfrak{g}\) are symmetrical about the origin -- i.e. \(- \alpha\) is - also an eigenvalue -- and they span all of \(\mathfrak{h}^*\). -\end{proposition} - -\begin{proof} - We'll start with the first claim. Let \(\alpha\) and \(\beta\) be two - eigenvalues of the adjoint action of \(\mathfrak{h}\). Notice - \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha + - \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and \(Y \in - \mathfrak{g}_\beta\) then - \[ - [H [X, Y]] - = [X, [H, Y]] - [Y, [H, X]] - = (\alpha + \beta)(H) \cdot [X, Y] - \] - for all \(H \in \mathfrak{h}\). - - This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X) - \operatorname{ad}(Y)\) is nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then - \[ - (\operatorname{ad}(X) \operatorname{ad}(Y))^n Z - = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ] - \in \mathfrak{g}_{n \alpha + n \beta + \gamma} - = 0 - \] - for \(n\) large enough. In particular, \(B(X, Y) = - \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if - \(- \alpha\) is not an eigenvalue we find \(B(X, \mathfrak{g}_\beta) = 0\) - for all eigenvalues \(\beta\), which contradicts the non-degeneracy of \(B\). - Hence \(- \alpha\) must be an eigenvalue of the adjoint action of - \(\mathfrak{h}\). - - For the second statement, note that if the eigenvalues of \(\mathfrak{h}\) do - not span all of \(\mathfrak{h}^*\) then there is some \(H \in \mathfrak{h}\) - non-zero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is - to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in - \mathfrak{g}\). Another way of putting it is to say \(H\) is an element of - the center \(\mathfrak{z}\) of \(\mathfrak{g}\), which is zero by the - semisimplicity -- a contradiction. -\end{proof} - -Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and -\(\mathfrak{sl}_3(K)\) one can show\dots - -\begin{proposition}\label{thm:root-space-dim-1} - The eigenspaces \(\mathfrak{g}_\alpha\) are all 1-dimensional. -\end{proposition} - -The proof of the first statement of -proposition~\ref{thm:weights-symmetric-span} highlights something interesting: -if we fix some some eigenvalue \(\alpha\) of the adjoint action of -\(\mathfrak{h}\) in \(\mathfrak{g}\) and a eigenvector \(X \in -\mathfrak{g}_\alpha\), then for each \(H \in \mathfrak{h}\) and \(v \in -V_\lambda\) we find -\[ - H (X v) - = X (H v) + [H, X] v - = (\lambda + \alpha)(H) \cdot X v -\] -so that \(X\) carries \(v\) to \(V_{\lambda + \alpha}\). We have encountered -this formula twice in this chapter: again, we find \(\mathfrak{g}_\alpha\) -\emph{acts on \(V\) by translating vectors between eigenspaces}. In other -words, if we denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\) -then\dots - -\begin{theorem}\label{thm:weights-congruent-mod-root} - The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) are - all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\). -\end{theorem} - -% TODOO: Turn this into a proper discussion of basis and give the idea of the -% proof of existance of basis? -To proceed further, as in the case of \(\mathfrak{sl}_3(K)\) we have to fix a -direction in \(\mathfrak{h}^*\) -- i.e. we fix a linear function -\(\mathfrak{h}^* \to \QQ\) such that \(Q\) lies outside of its kernel. This -choice induces a partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of -roots of \(\mathfrak{g}\) and once more we find\dots - -\begin{definition} - The elements of \(\Delta^+\) and \(\Delta^-\) are called \emph{positive} and - \emph{negative roots}, respectively. The subalgebra \(\mathfrak{b} = - \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is - called \emph{the Borel subalgebra associated with \(\mathfrak{h}\)}. -\end{definition} - -\begin{theorem} - There is a weight vector \(v \in V\) that is killed by all positive root - spaces of \(\mathfrak{g}\). -\end{theorem} - -% TODO: Here we may take a weight of maximal height, but why is it unique? -% TODO: We don't really need to talk about height tho, we may simply take a -% weight that maximizes B(gamma, lambda) in QQ -% TODOO: Either way, we need to move this to after the discussion on the -% integrality of weights -\begin{proof} - It suffices to note that if \(\lambda\) is the weight of \(V\) lying the - furthest along the direction we chose and \(V_{\lambda + \alpha} \ne 0\) for - some \(\alpha \in \Delta^+\) then \(\lambda + \alpha\) is a weight that is - furthest along the direction we chose than \(\lambda\), which contradicts the - definition of \(\lambda\). -\end{proof} - -Accordingly, we call \(\lambda\) \emph{the highest weight of \(V\)}, and we -call any \(v \in V_\lambda\) \emph{a highest weight vector}. The strategy then -is to describe all weight spaces of \(V\) in terms of \(\lambda\) and \(v\), as -in theorem~\ref{thm:sl3-irr-weights-class}, and unsurprisingly we do so by -reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). Namely, we -show\dots - -\begin{proposition}\label{thm:distinguished-subalgebra} - Given a root \(\alpha\) of \(\mathfrak{g}\) the subspace - \(\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{- \alpha} - \oplus [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) is a subalgebra - isomorphic to \(\mathfrak{sl}_2(k)\). -\end{proposition} - -\begin{corollary}\label{thm:distinguished-subalg-rep} - For all weights \(\mu\), the subspace - \[ - V_\mu[\alpha] = \bigoplus_k V_{\mu + k \alpha} - \] - is invariant under the action of the subalgebra \(\mathfrak{s}_\alpha\) - and the weight spaces in this string match the eigenspaces of \(h\). -\end{corollary} - -The proof of proposition~\ref{thm:distinguished-subalgebra} is very technical -in nature and we won't include it here, but the idea behind it is simple: -recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both -1-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) -is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}] -\ne 0\) and that no generator of \([\mathfrak{g}_\alpha, \mathfrak{g}_{- -\alpha}] \ne 0\) is annihilated by \(\alpha\), so that by adjusting scalars we -can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in -\mathfrak{g}_{- \alpha}\) such that \(H_\alpha = [E_\alpha, F_\alpha]\) -satisfies -\begin{align*} - [H_\alpha, F_\alpha] & = -2 F_\alpha & - [H_\alpha, E_\alpha] & = 2 E_\alpha -\end{align*} - -The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely -determined by this condition, but \(H_\alpha\) is. The second statement of -corollary~\ref{thm:distinguished-subalg-rep} imposes a restriction on the -weights of \(V\). Namely, if \(\mu\) is a weight, \(\mu(H_\alpha)\) is an -eigenvalue of \(h\) in some representation of \(\mathfrak{sl}_2(K)\), so -that\dots - -\begin{proposition} - The weights \(\mu\) of an irreducible representation \(V\) of - \(\mathfrak{g}\) are so that \(\mu(H_\alpha) \in \ZZ\) for each \(\alpha \in - \Delta\). -\end{proposition} - -Once more, the lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha) -\in \ZZ, \forall \alpha \in \Delta \}\) is called \emph{the weight lattice of -\(\mathfrak{g}\)}, and we call the elements of \(P\) \emph{integral}. Finally, -another important consequence of theorem~\ref{thm:distinguished-subalgebra} -is\dots - -\begin{corollary} - If \(\alpha \in \Delta^+\) and \(T_\alpha : \mathfrak{h}^* \to - \mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to - \(\alpha\) with respect to the Killing form, - corollary~\ref{thm:distinguished-subalg-rep} implies that all \(\nu \in P\) - lying inside the line connecting \(\mu\) and \(T_\alpha \mu\) are weights -- - i.e. \(V_\nu \ne 0\). -\end{corollary} - -\begin{proof} - It suffices to note that \(\nu \in V_\mu[\alpha]\) -- see appendix D of - \cite{fulton-harris} for further details. -\end{proof} - -\begin{definition} - We refer to the group \(\mathcal{W} = \langle T_\alpha : \alpha \in \Delta^+ - \rangle \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl group of - \(\mathfrak{g}\)}. -\end{definition} - -This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we -found that the weights of the irreducible representations were symmetric with -respect to the lines \(K \alpha\) with \(B(\alpha_i - \alpha_j, \alpha) = 0\). -Indeed, the same argument leads us to the conclusion\dots - -\begin{theorem}\label{thm:irr-weight-class} - The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) with - highest weight \(\lambda\) are precisely the elements of the weight lattice - \(P\) congruent to \(\lambda\) modulo the root lattice \(Q\) lying inside the - convex hull of the image of \(\lambda\) under the action of the Weyl group - \(\mathcal{W}\). -\end{theorem} - -Now the only thing we are missing for a complete classification is an existence -and uniqueness theorem analogous to theorem~\ref{thm:sl2-exist-unique} and -theorem~\ref{thm:sl3-existence-uniqueness}. Lo and behold\dots - -\begin{definition} - An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all - \(\alpha \in \Delta^+\) is referred to as an \emph{integral dominant weight - of \(\mathfrak{g}\)}. -\end{definition} - -\begin{theorem}\label{thm:dominant-weight-theo} - For each dominant integral \(\lambda \in P\) there exists precisely one - irreducible finite-dimensional representation \(V\) of \(\mathfrak{g}\) whose - highest weight is \(\lambda\). -\end{theorem} - -Fix some dominant integral \(\lambda \in P\). The ``uniqueness'' part of the -theorem follows at once from the argument used for \(\mathfrak{sl}_3(K)\). The -``existence'' part is more nuanced. Our first instinct is, of course, to try to -generalize the proof used for \(\mathfrak{sl}_3(K)\). The issue is that our -proof relied heavily on our knowledge of the roots of \(\mathfrak{sl}_3(K)\). -Instead, we need a new strategy for the general setting. To that end, we -introduce a special class of \(\mathfrak{g}\)-modules, known as \emph{Verma -modules}. - -\begin{definition}\label{def:verma} - The \(\mathfrak{g}\)-module \(M(\lambda) = - \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K v^+\), where the action of - \(\mathfrak{b}\) in \(K v^+\) is given by \(H v^+ = \lambda(H) \cdot v^+\) - for all \(H \in \mathfrak{h}\) and \(X v^+ = 0\) for \(X \in - \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma - module of weight \(\lambda\)} -\end{definition} - -We should point out that, unlike most representations we've encountered so far, -Verma modules are \emph{highly infinite-dimensional}. Indeed, the dimension of -\(M(\lambda)\) is the same as the codimension of \(\mathcal{U}(\mathfrak{b})\) -in \(\mathcal{U}(\mathfrak{g})\), which is always infinite. Nevertheless, -\(M(\lambda)\) turns out to be quite well behaved. For instance, by -construction \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\) -- where -\(v^+ = 1 \otimes v^+ \in M(\lambda)\) is as in definition~\ref{def:verma}. -Moreover, we find\dots - -\begin{proposition}\label{thm:verma-is-weight-mod} - The weight spaces decomposition - \[ - M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu - \] - holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in - \mathfrak{h}^*\) and \(\dim M(\lambda) = 1\). Finally, \(\lambda\) is the - highest weight of \(M(\lambda)\), with highest weight vector given by \(v^+ = - 1 \otimes v^+ \in M(\lambda)\) as in definition~\ref{def:verma}. -\end{proposition} - -\begin{proof} - The Poincaré-Birkhoff-Witt theorem implies that \(M(\lambda)\) is spanned by - the vectors \(F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+\) for - \(\alpha_i \in \Delta^-\) and \(F_{\alpha_i} \in \mathfrak{g}_{\alpha_i}\) as - in the proof of proposition~\ref{thm:distinguished-subalgebra}. But - \[ - \begin{split} - H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+ - & = ([H, F_{\alpha_1}] + F_{\alpha_1} H) - F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\ - & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v^+ - + F_{\alpha_1} ([H, F_{\alpha_2}] + F_{\alpha_2} H) - F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\ - & \;\; \vdots \\ - & = (\alpha_1 + \cdots + \alpha_n)(H) \cdot - F_{\alpha_1} \cdots F_{\alpha_n} v^+ - + F_{\alpha_1} \cdots F_{\alpha_n} H v^+ \\ - & = (\lambda + \alpha_1 + \cdots + \alpha_n)(H) \cdot - F_{\alpha_1} \cdots F_{\alpha_n} v^+ \\ - & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v^+ - \in M(\lambda)_{\lambda + \alpha_1 + \cdots + \alpha_n} - \end{split} - \] - - Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*} - M(\lambda)_\mu\), as desired. In fact we have established - \[ - M(\lambda) - \subset - \bigoplus_{\substack{k_i \in \ZZ \\ k_i \ge 0}} - M(\lambda)_{\lambda + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n} - \] - where \(\{\alpha_1, \ldots, \alpha_m\} = \Delta^-\), so that all weights of - \(M(\lambda)\) have the form \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots + - k_n \cdot \alpha_n\). - - This already gives us that the weights of \(M(\lambda)\) are bounded by - \(\lambda\) -- in the sense that no weight of \(M(\lambda)\) is ``higher'' - than \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that - \(v^+\) is nonzero weight vector. Clearly \(v^+ \in V_\lambda\). The - Poincaré-Birkhoff-Witt theorem implies - \[ - M(\lambda) - \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right) - \otimes_{\mathcal{U}(\mathfrak{b})} K v^+ - \cong \bigoplus_i \mathcal{U}(\mathfrak{b}) - \otimes_{\mathcal{U}(\mathfrak{b})} K v^+ - \cong \bigoplus_i K v^+ - \ne 0 - \] - as \(\mathcal{U}(\mathfrak{b})\)-modules, so \(v^+ \ne 0\) -- for if this was - not the case we would find \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+ - = 0\). Hence \(V_\lambda \ne 0\) and therefore \(\lambda\) is the highest - weight of \(M(\lambda)\), with highest weight vector \(v^+\). - - To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only - finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots - F_{\alpha_n}^{k_n}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots - + k_n \cdot \alpha_n\). Since \(M(\lambda)_\mu\) is spanned by the images of - \(v^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In - particular, there is a single monomials \(F_{\alpha_1}^{k_1} - F_{\alpha_2}^{k_2} \cdots F_{\alpha_n}^{k_n}\) such that \(\lambda = \lambda - + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n\) -- which is, of course, - the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim V_\lambda = 1\). -\end{proof} - -\begin{example}\label{ex:sl2-verma} - If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K - h\) and \(\mathfrak{b} = K e \oplus K h\). If \(\lambda \in - \mathfrak{h}^*\) is the map \(h \mapsto 2\) then \(M(\lambda) = - \bigoplus_{k \ge 0} K f^k v^+\), and the action of \(\mathfrak{sl}_2(K)\) in - \(M(\lambda)\) is given by - \begin{align*} - f^{k + 1} v^+ & \overset{e}{\mapsto} (2 - k (k - 1)) f^k v^+ & - f^{k + 1} v^+ & \overset{f}{\mapsto} f^{k + 2} v^+ & - f^{k + 1} v^+ & \overset{h}{\mapsto} - 2 k f^{k + 1} v^+ & - \end{align*} - - In the language of the diagrams used in section~\ref{sec:sl2}, we write - % TODO: Add a label to the righ of the diagram indicating that the top arrows - % are the action of e and the bottom arrows are the action of f - \begin{center} - \begin{tikzcd} - \cdots \arrow[bend left=60]{r}{-10} - & M(\lambda)_{-6} \arrow[bend left=60]{r}{-4} \arrow[bend left=60]{l}{1} - & M(\lambda)_{-4} \arrow[bend left=60]{r}{0} \arrow[bend left=60]{l}{1} - & M(\lambda)_{-2} \arrow[bend left=60]{r}{2} \arrow[bend left=60]{l}{1} - & M(\lambda)_0 \arrow[bend left=60]{r}{2} \arrow[bend left=60]{l}{1} - & M(\lambda)_2 \arrow[bend left=60]{l}{1} - \end{tikzcd} - \end{center} - where \(M(\lambda)_{2 - 2 k} = K f^k v\). In this case, unlike we have see in - section~\ref{sec:sl2}, the string of weight spaces to left of the diagram is - infinite. -\end{example} - -What's interesting to us about all this is that we've just constructed a -\(\mathfrak{g}\)-module whose highest weight is \(\lambda\). This is not a -proof of theorem~\ref{thm:dominant-weight-theo}, however, since \(M(\lambda)\) -is neither irreducible nor finite-dimensional. Nevertheless, we can use -\(M(\lambda)\) to construct an irreducible representation of \(\mathfrak{g}\) -whose highest weight is \(\lambda\). - -\begin{proposition}\label{thm:max-verma-submod-is-weight} - Every subrepresentation \(V \subset M(\lambda)\) is the direct sum of its - weight spaces. In particular, \(M(\lambda)\) has a unique maximal - subrepresentation \(N(\lambda)\) and a unique irreducible quotient - \(\sfrac{M(\lambda)}{N(\lambda)}\). -\end{proposition} - -\begin{proof} - Let \(V \subset M(\lambda)\) be a subrepresentation and take any nonzero \(v - \in V\). Because of proposition~\ref{thm:verma-is-weight-mod}, we know there - are \(\mu_1, \ldots, \mu_n \in \mathfrak{h}^*\) and nonzero \(v_i \in - M(\lambda)_{\mu_i}\) such that \(v = v_1 + \cdots + v_n\). We want to show - \(v_i \in V\) for all \(i\). - - Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\). - Then - \[ - v_1 - - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_3 - - \cdots - - \frac{(\mu_n - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_n - = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v - \in V - \] - - Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By - applying the same procedure again we get - \begin{multline*} - v_1 - - - \frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)} - {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_4 - - \cdots - - \frac{(\mu_n - \mu_3)(H_3) \cdot (\mu_n - \mu_1)(H_2)} - {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_n \\ - = - \left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right) - \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v - \in V - \end{multline*} - - By applying the same procedure over and over again we can see that \(v_1 = X - v \in V\) for some \(X \in \mathcal{U}(\mathfrak{g})\). Furthermore, if we - reproduce all this for \(v_2 + \cdots + v_n = v - v_1 \in V\) we get that - \(v_2 \in V\). Now by applying the same procedure over and over we find - \(v_1, \ldots, v_n \in V\). Hence - \[ - V = \bigoplus_\mu V_\mu = \bigoplus_\mu M(\lambda)_\mu \cap V - \] - - Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\), \(V\) is a proper - subrepresentation then \(v^+ \notin V\). Hence any proper submodule lies in - the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum - \(N(\lambda)\) of all such submodules is still proper. In fact, this implies - \(N(\lambda)\) is the unique maximal subrepresentation of \(M(\lambda)\) and - \(\sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible quotient. -\end{proof} - -\begin{example}\label{ex:sl2-verma-quotient} - If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we - can see from example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge - 3} K f^k v^+\), so that \(\sfrac{M(\lambda)}{N(\lambda)}\) is the - \(3\)-dimensional irreducible representation of \(\mathfrak{sl}_2(K)\) -- - i.e. the finite-dimensional irreducible representation with highest weight - \(\lambda\) constructed in section~\ref{sec:sl2}. -\end{example} - -This last example is particularly interesting to us, since it indicates that -the finite-dimensional irreducible representations of \(\mathfrak{sl}_2(K)\) as -quotients of Verma modules. This is because the quotient -\(\sfrac{M(\lambda)}{N(\lambda)}\) in example~\ref{ex:sl2-verma-quotient} -happened to be finite-dimensional. As it turns out, this is always the case for -semisimple \(\mathfrak{g}\). Namely\dots - -\begin{proposition}\label{thm:verma-is-finite-dim} - If \(\lambda\) is dominant integral then the unique irreducible quotient of - \(M(\lambda)\) is finite-dimensional. -\end{proposition} - -The proof of proposition~\ref{thm:verma-is-finite-dim} is very technical and we -won't include it here, but the idea behind it is to show that the set of -weights of \(\sfrac{M(\lambda)}{N(\lambda)}\) is stable under the natural -action of the Weyl group \(\mathcal{W}\) in \(\mathfrak{h}^*\). One can then -show that the every weight of \(V\) is conjugate to a single dominant integral -weight of \(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant -integral weights of such irreducible quotient is finite. Since \(W\) is -finitely generated, this implies the set of weights of the unique irreducible -quotient of \(M(\lambda)\) is finite. But each weight space is -finite-dimensional. Hence so is the irreducible quotient. - -We refer the reader to \cite[ch. 21]{humphreys} for further details. What we -are really interested in is\dots - -\begin{corollary} - There is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose - highest weight is \(\lambda\). -\end{corollary} - -\begin{proof} - Let \(V = \sfrac{M(\lambda)}{N(\lambda)}\). It suffices to show that its - highest weight is \(\lambda\). We have already seen that \(v^+ \in - M(\lambda)_\lambda\) is a highest weight vector. Now since \(v\) lies outside - of the maximal subrepresentation of \(M(\lambda)\), the projection \(v^+ + - N(\lambda) \in V\) is nonzero. - - % TODO: Why is V_mu = M(lambda)_mu + N(lambda)? Turn this into a proposition? - We now claim that \(v^+ + N(\lambda) \in V_\lambda\). Indeed, - \[ - H (v^+ + N(\lambda)) - = H v^+ + N(\lambda) - = \lambda(H) \cdot (v^+ + N(\lambda)) - \] - for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(V\), with - weight vector \(v^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is - the highest weight of \(V\), for if this was not the case we could find a - weight \(\mu\) of \(M(\lambda)\) which is higher than \(\lambda\). -\end{proof} - -% TODO: Write a conclusion and move this to the next chapter
diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex @@ -0,0 +1,1787 @@ +\chapter{Representations of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\)} + +\section{Representations of \(\mathfrak{sl}_2(K)\)}\label{sec:sl2} + +The primary goal of this section is proving\dots + +\begin{theorem}\label{thm:sl2-exist-unique} + For each \(n > 0\), there exists precisely one irreducible representation + \(V\) of \(\mathfrak{sl}_2(K)\) with \(\dim V = n\). +\end{theorem} + +The general approach we'll take is supposing \(V\) is an irreducible +representation of \(\mathfrak{sl}_2(K)\) and then derive some information about +its structure. We begin our analysis by recalling that the elements +\begin{align*} + e & = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & + f & = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} & + h & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} +\end{align*} +form a basis of \(\mathfrak{sl}_2(K)\) and satisfy +\begin{align*} + [e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e +\end{align*} + +This is interesting to us because it implies every subspace of \(V\) invariant +under the actions of \(e\), \(f\) and \(h\) has to be \(V\) itself. Next we +turn our attention to the action of \(h\) in \(V\), in particular, to the +eigenspace decomposition +\[ + V = \bigoplus_{\lambda} V_\lambda +\] +of \(V\) -- where \(\lambda\) ranges over the eigenvalues of \(h\) and +\(V_\lambda\) 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 -- which +conveniently justifies the epigraph of this chapter. 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 +\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 +\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 \(V_{\lambda - 2}\). +Hence +\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} + & \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 +\[ + V = \bigoplus_{n \in \ZZ} V_{\lambda + 2 n}, +\] +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 +\[ + \bigoplus_{\substack{n \in \ZZ \\ a \le n \le b}} V_{\lambda + 2 n} +\] +is also an \(\mathfrak{sl}_2(K)\)-invariant subspace, 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 +eigenvalues. To do so, we suppose without any loss in generality that +\(\lambda\) is the right-most eigenvalue of \(h\), fix some non-zero \(v \in +V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\). + +\begin{theorem}\label{thm:basis-of-irr-rep} + The set \(\{v, f v, f^2, \ldots\}\) is a basis for \(V\). +\end{theorem} + +\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, \(h f^k v = (\lambda - 2 k) f^k v\). 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 + & = h v + f e v \\ + \text{(since \(\lambda\) is the right-most eigenvalue)} + & = h v + f 0 \\ + & = \lambda v + \end{split} + \] + + Next we compute + \[ + \begin{split} + e f^2 v + & = (h + fe) f v \\ + & = h f v + f (\lambda v) \\ + & = 2 (\lambda - 1) f v + \end{split} + \] + + 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 + \[ + e f^k v = k (\lambda + 1 - k) f^{k - 1} v + \] +\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} + +Theorem~\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 + +\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 + non-zero, 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 theorem~\ref{thm:basis-of-irr-rep} are\dots + +\begin{corollary} + Every \(h\) eigenspace is one-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 \(f^k v\). +\end{proof} + +\begin{corollary} + The eigenvalues of \(h\) in \(V\) form a symmetric, unbroken string of + integers separated by intervals of length \(2\) whose right-most value is + \(\dim V - 1\). +\end{corollary} + +\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 + \[ + 0 = e 0 = e f^m v = m (\lambda + 1 - m) f^{m - 1} v + \] + + This implies \(\lambda + 1 - m = 0\) -- i.e. \(\lambda = m - 1 \in \ZZ\). 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 \(h\) + are + \[ + \ldots, n - 6, n - 4, n - 2, n + \] + + To see that this string is symmetric around \(0\), simply note that the + left-most eigenvalue of \(h\) is precisely \(n - 2 (m - 1) = -n\). +\end{proof} + +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 one-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 +\(V\) does indeed exist and is irreducible. In other words\dots + +\begin{theorem}\label{thm:irr-rep-of-sl2-exists} + For each \(n \ge 0\) there exists a (unique) irreducible representation of + \(\mathfrak{sl}_2(K)\) whose left-most eigenvalue of \(h\) is \(n\). +\end{theorem} + +\begin{proof} + The fact the representation \(V\) from the previous discussion exists is + clear from the commutator relations of \(\mathfrak{sl}_2(K)\) -- just look at + \(f^k v\) as abstract symbols and impose the action given by + (\ref{eq:irr-rep-of-sl2}). 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}). To see that \(V\) is irreducible let \(W\) be a + non-zero subrepresentation and take some non-zero \(w \in W\). Suppose \(w = + \alpha_0 v + \alpha_1 f v + \cdots + \alpha_n f^n v\) and let \(k\) be the + lowest index such that \(\alpha_k \ne 0\), so that + \[ + w = \alpha_k f^k v + \cdots + \alpha_n f^n v + \] + + Now given that \(f^m = f^{n + 1}\) annihilates \(v\), + \[ + f w = \alpha_k f^{k + 1} v + \cdots + \alpha_{n - 1} f^n v + \] + + Proceeding inductively we arrive at \(f^{n - k} w = \alpha_k f^n v\), so + that \(f^n v \in W\). Hence \(e^i f^n v = \prod_{k = 1}^i k(n + 1 - k) f^{n - + i} v \in W\) for all \(i = 1, 2, \ldots, n\). Since \(k \ne 0 \ne n + 1 - k\) + for all \(k\) in this range, we can see that \(f^k v \in W\) for all \(k = 0, + 1, \ldots, n\). In other words, \(W = V\). We are done. +\end{proof} + +Our initial gamble of studying the eigenvalues of \(h\) may have seemed +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. + +\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 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 +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. + +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 +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'll 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 +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 +theorem~\ref{thm:irr-rep-of-sl2-exists}. + +Our fist 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 +see, no such analogue exists and neither does such element. Instead, the actual +way to proceed is to consider the subalgebra +\[ + \mathfrak{h} + = \left\{ + X \in + \begin{pmatrix} K & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K \end{pmatrix} + : \operatorname{Tr}(X) = 0 + \right\} +\] + +The choice of \(\mathfrak{h}\) may seem like an odd choice at the moment, but +the point is we'll later show that there exists some \(v \in V\) 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 non-zero +\(v \in V\). We call such functionals \(\lambda\) \emph{eigenvalues of +\(\mathfrak{h}\)}, and we say \emph{\(v\) is an eigenvector of +\(\mathfrak{h}\)}. + +Once again, we'll pay special attention to the eigenvalue decomposition +\begin{equation}\label{eq:weight-module} + V = \bigoplus_\lambda V_\lambda +\end{equation} +where \(\lambda\) ranges over all eigenvalues of \(\mathfrak{h}\) and +\(V_\lambda = \{ v \in V : H v = \lambda(H) \cdot v, \forall H \in \mathfrak{h} +\}\). 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 +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. + +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 of the adjoint actions of \(h\) are +\(\pm 2\) since +\begin{align*} + [h, f] & = -2 f & + [h, e] & = 2 e +\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 +\(\pm 2\). + +In the case of \(\mathfrak{sl}_3(K)\), a simple calculation shows that if \([H, +X]\) is scalar multiple of \(X\) for all \(H \in \mathfrak{h}\) then all but +one entry of \(X\) are zero. Hence the eigenvectors of the adjoint action of +\(\mathfrak{h}\) are \(E_{i j}\) and its eigenvalues are \(\alpha_i - +\alpha_j\), where +\[ + \alpha_i + \begin{pmatrix} + a_1 & 0 & 0 \\ + 0 & a_2 & 0 \\ + 0 & 0 & a_3 + \end{pmatrix} + = a_i +\] + +Visually we may draw + +\begin{figure}[h] + \centering + \begin{tikzpicture}[scale=2.5] + \begin{rootSystem}{A} + \filldraw[black] \weight{0}{0} circle (.5pt); + \node[black, above right] at \weight{0}{0} {\small$0$}; + \wt[black]{-1}{2} + \wt[black]{-2}{1} + \wt[black]{1}{1} + \wt[black]{-1}{-1} + \wt[black]{2}{-1} + \wt[black]{1}{-2} + \node[above] at \weight{-1}{2} {$\alpha_2 - \alpha_3$}; + \node[left] at \weight{-2}{1} {$\alpha_2 - \alpha_1$}; + \node[right] at \weight{1}{1} {$\alpha_1 - \alpha_3$}; + \node[left] at \weight{-1}{-1} {$\alpha_3 - \alpha_1$}; + \node[right] at \weight{2}{-1} {$\alpha_1 - \alpha_2$}; + \node[below] at \weight{1}{-2} {$\alpha_3 - \alpha_1$}; + \node[black, above] at \weight{1}{0} {$\alpha_1$}; + \node[black, above] at \weight{-1}{1} {$\alpha_2$}; + \node[black, above] at \weight{0}{-1} {$\alpha_3$}; + \filldraw[black] \weight{1}{0} circle (.5pt); + \filldraw[black] \weight{-1}{1} circle (.5pt); + \filldraw[black] \weight{0}{-1} circle (.5pt); + \end{rootSystem} + \end{tikzpicture} +\end{figure} + +If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) in +\(\mathfrak{sl}_3(K)\) 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 +\[ + \begin{split} + H (X v) + & = X (H v) + [H, X] v \\ + & = X (\lambda(H) \cdot v) + (\alpha(H) \cdot X) v \\ + & = (\alpha + \lambda)(H) \cdot X v + \end{split} +\] +so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words, +\(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors +between eigenspaces}. + +For instance \(\mathfrak{sl}_3(K)_{\alpha_1 - \alpha_3}\) will act on the +adjoint representation of \(\mathfrak{sl}_3(K)\) via +\begin{figure}[h] + \centering + \begin{tikzpicture}[scale=2.5] + \begin{rootSystem}{A} + \wt[black]{0}{0} + \wt[black]{-1}{2} + \wt[black]{-2}{1} + \wt[black]{1}{1} + \wt[black]{-1}{-1} + \wt[black]{2}{-1} + \wt[black]{1}{-2} + \draw[-latex, black] \weight{-1.9}{1.1} -- \weight{-1.1}{1.9}; + \draw[-latex, black] \weight{-.9}{-.9} -- \weight{-.1}{-.1}; + \draw[-latex, black] \weight{0.1}{0.1} -- \weight{.9}{.9}; + \draw[-latex, black] \weight{1.1}{-1.9} -- \weight{1.9}{-1.1}; + \end{rootSystem} + \end{tikzpicture} +\end{figure} + +This is again entirely analogous to the situation we observed in +\(\mathfrak{sl}_2(K)\). In fact, we may once more conclude\dots + +\begin{theorem}\label{thm:sl3-weights-congruent-mod-root} + The eigenvalues of the action of \(\mathfrak{h}\) in 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 \(\mathfrak{h}\) in \(\mathfrak{sl}_3(K)\). +\end{theorem} + +\begin{proof} + This proof goes exactly as that of the analogous statement for + \(\mathfrak{sl}_2(K)\): it suffices to note that if we fix some eigenvalue + \(\lambda\) of \(\mathfrak{h}\) and let \(i\) and \(j\) vary then + \[ + \bigoplus_{i j} V_{\lambda + \alpha_i - \alpha_j} + \] + is an invariant subspace of \(V\). +\end{proof} + +To avoid confusion we better introduce some notation to differentiate between +eigenvalues of the action of \(\mathfrak{h}\) in \(V\) and eigenvalues of the +adjoint action of \(\mathfrak{h}\). + +\begin{definition} + Given a representation \(V\) of \(\mathfrak{sl}_3(K)\), we'll call the + non-zero eigenvalues of the action of \(\mathfrak{h}\) in \(V\) \emph{weights + of \(V\)}. As you might have guessed, we'll correspondingly refer to + eigenvectors and eigenspaces of a given weight by \emph{weight vectors} and + \emph{weight spaces}. +\end{definition} + +It's clear from our previous discussion that the weights of the adjoint +representation of \(\mathfrak{sl}_3(K)\) 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 + \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{corollary} + The weights of an irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) + are all congruent module the lattice \(Q\) generated by the roots \(\alpha_i + - \alpha_j\) of \(\mathfrak{sl}_3(K)\). +\end{corollary} + +\begin{definition} + The lattice \(Q = \ZZ \langle \alpha_i - \alpha_j : i, j = 1, 2, 3 \rangle\) + is called \emph{the root lattice of \(\mathfrak{sl}_3(K)\)}. +\end{definition} + +To proceed we once more refer to the previously established framework: next we +saw that the eigenvalues of \(h\) formed 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 +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. For instance, let's say we fix the direction +\begin{center} + \begin{tikzpicture}[scale=2.5] + \begin{rootSystem}{A} + \wt[black]{0}{0} + \wt[black]{-1}{2} + \wt[black]{-2}{1} + \wt[black]{1}{1} + \wt[black]{-1}{-1} + \wt[black]{2}{-1} + \wt[black]{1}{-2} + \draw[-latex, black, thick] \weight{-1.5}{-.5} -- \weight{1.5}{.5}; + \end{rootSystem} + \end{tikzpicture} +\end{center} +and let \(\lambda\) be the weight lying the furthest in this direction. + +Its easy to see what we mean intuitively by looking at the previous picture, +but its precise meaning is still allusive. Formally this means we'll choose a +linear functional \(f : \mathfrak{h}^* \to \QQ\) and pick the weight that +maximizes \(f\). To avoid any ambiguity we should choose the direction of a +line irrational with respect to the root lattice \(Q\). For instance if we +choose the direction of \(\alpha_1 - \alpha_3\) and let \(f\) be the rational +projection \(Q \to \QQ \langle \alpha_1 - \alpha_3 \rangle \cong \QQ\) then +\(\alpha_1 - 2 \alpha_2 + \alpha_3 \in Q\) lies in \(\ker f\), so that if a +weight \(\lambda\) maximizes \(f\) then the translation of \(\lambda\) by any +multiple of \(\alpha_1 - 2 \alpha_2 + \alpha_3\) must also do so. In others +words, if the direction we choose is parallel to a vector lying in \(Q\) then +there may be multiple choices the ``weight lying the furthest'' along this +direction. + +\begin{definition} + We say that a root \(\alpha\) is positive if \(f(\alpha) > 0\) -- i.e. if it + lies to the right of the direction we chose. Otherwise we say \(\alpha\) is + negative. Notice that \(f(\alpha) \ne 0\) since by definition \(\alpha \ne + 0\) and \(f\) is irrational with respect to the lattice \(Q\). +\end{definition} + +The first 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} + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \weightLattice{3} + \fill[gray!50,opacity=.2] (hex cs:x=5,y=-7) -- (hex cs:x=1,y=1) -- + (hex cs:x=-7,y=5) arc (150:270:{7*\weightLength}); + \draw[black, thick] (hex cs:x=5,y=-7) -- (hex cs:x=1,y=1) -- + (hex cs:x=-7,y=5); + \filldraw[black] (hex cs:x=1,y=1) circle (1pt); + \node[above right=-2pt] at (hex cs:x=1,y=1) {\small\(\lambda\)}; + \end{rootSystem} + \end{tikzpicture} +\end{center} + +Indeed, if this is not the case then, by definition, \(\lambda\) is not the +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 non-zero -- which contradicts the hypothesis that +\(\lambda\) lies the furthest along the direction we chose. In other words\dots + +\begin{theorem} + There is a weight vector \(v \in V\) that is killed by all positive root + spaces of \(\mathfrak{sl}_3(K)\). +\end{theorem} + +\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 \(\alpha_2 - + \alpha_3\). +\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 +\(\mathfrak{sl}_2(K)\), 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 +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. + +We'll start out by trying to understand the weights in the boundary of +\(\frac{1}{3}\)-plane previously drawn. Since the root spaces act by +translation, the action of \(E_{2 1}\) in \(V_\lambda\) will span a subspace +\[ + W = \bigoplus_k V_{\lambda + k (\alpha_2 - \alpha_1)}, +\] +and by the same token \(W\) must be invariant under the action of \(E_{1 2}\). + +To draw a familiar picture +\begin{center} + \begin{tikzpicture} + \begin{rootSystem}{A} + \node at \weight{3}{1} (a) {}; + \node at \weight{1}{2} (b) {}; + \node at \weight{-1}{3} (c) {}; + \node at \weight{-3}{4} (d) {}; + \node at \weight{-5}{5} (e) {}; + \draw \weight{3}{1} -- \weight{-4}{4.5}; + \draw[dotted] \weight{-4}{4.5} -- \weight{-5}{5}; + \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} + \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; + \draw[-latex] (a) to[bend left=40] (b); + \draw[-latex] (b) to[bend left=40] (c); + \draw[-latex] (c) to[bend left=40] (d); + \draw[-latex] (d) to[bend left=40] (e); + \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} + +What's remarkable about all this is the fact that the subalgebra spanned by +\(E_{1 2}\), \(E_{2 1}\) and \(H = [E_{1 2}, E_{2 1}]\) is isomorphic to +\(\mathfrak{sl}_2(K)\) via +\begin{align*} + E_{2 1} & \mapsto e & + E_{1 2} & \mapsto f & + H & \mapsto h +\end{align*} + +In other words, \(W\) is a representation of \(\mathfrak{sl}_2(K)\). Even more +so, we claim +\[ + V_{\lambda + k (\alpha_2 - \alpha_1)} = W_{\lambda(H) - 2k} +\] + +Indeed, \(V_{\lambda + k (\alpha_2 - \alpha_1)} \subset W_{\lambda(H) - 2k}\) +since \((\lambda + k (\alpha_2 - \alpha_1))(H) = \lambda(H) + k (-1 - 1) = +\lambda(H) - 2 k\). On the other hand, if we suppose \(0 < \dim V_{\lambda + k +(\alpha_2 - \alpha_1)} < \dim W_{\lambda(H) - 2 k}\) for some \(k\) we arrive +at +\[ + \dim W + = \sum_k \dim V_{\lambda + k (\alpha_2 - \alpha_1)} + < \sum_k \dim W_{\lambda(H) - 2k} + = \dim W, +\] +a contradiction. + +There are a number of important consequences to this, of the first being that +the weights of \(V\) appearing on \(W\) must be symmetric with respect to the +the line \(B(\alpha_1 - \alpha_2, \alpha) = 0\). The picture is +thus +\begin{center} + \begin{tikzpicture} + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \setlength{\weightRadius}{2pt} + \weightLattice{4} + \draw[thick] \weight{3}{1} -- \weight{-3}{4}; + \wt[black]{0}{0} + \node[above left] at \weight{0}{0} {\small\(0\)}; + \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} + \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; + \draw[very thick] \weight{0}{-4} -- \weight{0}{4} + node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)}; + \end{rootSystem} + \end{tikzpicture} +\end{center} + +Notice we could apply this same argument to the subspace \(\bigoplus_k +V_{\lambda + k (\alpha_3 - \alpha_2)}\): this subspace is invariant under the +action of the subalgebra spanned by \(E_{2 3}\), \(E_{3 2}\) and \([E_{2 3}, +E_{3 2}]\), which is again isomorphic to \(\mathfrak{sl}_2(K)\), so that the +weights in this subspace must be symmetric with respect to the line +\(B(\alpha_3 - \alpha_2, \alpha) = 0\). The picture is now +\begin{center} + \begin{tikzpicture} + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \setlength{\weightRadius}{2pt} + \weightLattice{4} + \draw[thick] \weight{3}{1} -- \weight{-3}{4}; + \draw[thick] \weight{3}{1} -- \weight{4}{-1}; + \wt[black]{0}{0} + \wt[black]{4}{-1} + \node[above left] at \weight{0}{0} {\small\(0\)}; + \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} + \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; + \draw[very thick] \weight{0}{-4} -- \weight{0}{4} + node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)}; + \draw[very thick] \weight{-4}{0} -- \weight{4}{0} + node[right]{\small\(B(\alpha_3 - \alpha_2, \alpha) = 0\)}; + \end{rootSystem} + \end{tikzpicture} +\end{center} + +In general, given a weight \(\mu\), the space +\[ + \bigoplus_k V_{\mu + k (\alpha_i - \alpha_j)} +\] +is invariant under the action of the subalgebra \(\mathfrak{s}_{\alpha_i - +\alpha_j} = K \langle E_{i j}, E_{j i}, [E_{i j}, E_{j i}] \rangle\), which is +once more isomorphic to \(\mathfrak{sl}_2(K)\), and again the weight spaces in +this string match precisely the eigenvalues of \(h\). Needless to say, we could +keep applying this method to the weights at the ends of our string, arriving at +\begin{center} + \begin{tikzpicture} + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \setlength{\weightRadius}{2pt} + \weightLattice{5} + \draw[thick] \weight{3}{1} -- \weight{-3}{4}; + \draw[thick] \weight{3}{1} -- \weight{4}{-1}; + \draw[thick] \weight{-3}{4} -- \weight{-4}{3}; + \draw[thick] \weight{-4}{3} -- \weight{-1}{-3}; + \draw[thick] \weight{1}{-4} -- \weight{4}{-1}; + \draw[thick] \weight{-1}{-3} -- \weight{1}{-4}; + \wt[black]{-4}{3} + \wt[black]{-3}{1} + \wt[black]{-2}{-1} + \wt[black]{-1}{-3} + \wt[black]{1}{-4} + \wt[black]{2}{-3} + \wt[black]{3}{-2} + \wt[black]{4}{-1} + \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} + \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; + \draw[very thick] \weight{-5}{5} -- \weight{5}{-5}; + \draw[very thick] \weight{0}{-5} -- \weight{0}{5}; + \draw[very thick] \weight{-5}{0} -- \weight{5}{0}; + \end{rootSystem} + \end{tikzpicture} +\end{center} + +We claim all dots \(\mu\) lying inside the hexagon we've 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. +\begin{center} + \begin{tikzpicture} + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \setlength{\weightRadius}{2pt} + \weightLattice{5} + \draw[thick] \weight{3}{1} -- \weight{-3}{4}; + \draw[thick] \weight{3}{1} -- \weight{4}{-1}; + \draw[thick] \weight{-3}{4} -- \weight{-4}{3}; + \draw[thick] \weight{-4}{3} -- \weight{-1}{-3}; + \draw[thick] \weight{1}{-4} -- \weight{4}{-1}; + \draw[thick] \weight{-1}{-3} -- \weight{1}{-4}; + \wt[black]{-4}{3} + \wt[black]{-3}{1} + \wt[black]{-2}{-1} + \wt[black]{-1}{-3} + \wt[black]{1}{-4} + \wt[black]{2}{-3} + \wt[black]{3}{-2} + \wt[black]{4}{-1} + \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} + \node[above right=-2pt] at \weight{1}{2} {\small\(\nu\)}; + \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; + \draw[very thick] \weight{-5}{5} -- \weight{5}{-5}; + \draw[very thick] \weight{0}{-5} -- \weight{0}{5}; + \draw[very thick] \weight{-5}{0} -- \weight{5}{0}; + \draw[gray, thick] \weight{1}{2} -- \weight{-2}{-1}; + \wt[black]{1}{2} + \wt[black]{-2}{-1} + \wt{0}{1} + \wt{-1}{0} + \end{rootSystem} + \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 gray +string must occur in the representation of \(\mathfrak{sl}_2(K)\). Hence they +must also be weights of \(V\). The final picture is thus +\begin{center} + \begin{tikzpicture} + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \setlength{\weightRadius}{2pt} + \weightLattice{5} + \draw[thick] \weight{3}{1} -- \weight{-3}{4}; + \draw[thick] \weight{3}{1} -- \weight{4}{-1}; + \draw[thick] \weight{-3}{4} -- \weight{-4}{3}; + \draw[thick] \weight{-4}{3} -- \weight{-1}{-3}; + \draw[thick] \weight{1}{-4} -- \weight{4}{-1}; + \draw[thick] \weight{-1}{-3} -- \weight{1}{-4}; + \wt[black]{-4}{3} + \wt[black]{-3}{1} + \wt[black]{-2}{-1} + \wt[black]{-1}{-3} + \wt[black]{1}{-4} + \wt[black]{2}{-3} + \wt[black]{3}{-2} + \wt[black]{4}{-1} + \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}} + \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)}; + \draw[very thick] \weight{-5}{5} -- \weight{5}{-5}; + \draw[very thick] \weight{0}{-5} -- \weight{0}{5}; + \draw[very thick] \weight{-5}{0} -- \weight{5}{0}; + \wt[black]{-2}{2} + \wt[black]{0}{1} + \wt[black]{-1}{0} + \wt[black]{0}{-2} + \wt[black]{1}{-1} + \wt[black]{2}{0} + \end{rootSystem} + \end{tikzpicture} +\end{center} + +Another important consequence of our analysis is the fact that \(\lambda\) lies +in the lattice \(P\) generated by \(\alpha_1\), \(\alpha_2\) and \(\alpha_3\). +Indeed, \(\lambda([E_{i j}, E_{j i}])\) is an eigenvalue of \(h\) in a +representation of \(\mathfrak{sl}_2(K)\), so it must be an integer. Now since +\[ + \lambda + \begin{pmatrix} + a & 0 & 0 \\ + 0 & b & 0 \\ + 0 & 0 & -a -b + \end{pmatrix} + = + \lambda + \begin{pmatrix} + a & 0 & 0 \\ + 0 & 0 & 0 \\ + 0 & 0 & -a + \end{pmatrix} + + + \lambda + \begin{pmatrix} + 0 & 0 & 0 \\ + 0 & b & 0 \\ + 0 & 0 & -b + \end{pmatrix} + = + a \lambda([E_{1 3}, E_{3 1}]) + b \lambda([E_{2 3}, E_{3 2}]), +\] +which is to say \(\lambda = \lambda([E_{1 3}, E_{3 1}]) \alpha_1 + +\lambda([E_{2 3}, E_{3 2}]) \alpha_2\), we can see that \(\lambda \in +P\). + +\begin{definition} + The lattice \(P = \ZZ \alpha_1 \oplus \ZZ \alpha_2 \oplus \ZZ \alpha_3\) is + called \emph{the weight lattice of \(\mathfrak{sl}_3(K)\)}. +\end{definition} + +Finally\dots + +\begin{theorem}\label{thm:sl3-irr-weights-class} + The weights of \(V\) 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} + +Once more 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 \(P = \ZZ\) and were congruent modulo the lattice \(Q = 2 \ZZ\). +Having found all of the weights of \(V\), the only thing we're 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 irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) whose highest + weight is \(n \alpha_1 - m \alpha_3\). +\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 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 + +\begin{theorem}\label{thm:irr-sl3-span} + 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 \(E_{3 2}\). +\end{theorem} + +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 +\(\mathfrak{sl}_3(K)\) 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 +\(\mathfrak{sl}_3(K)\) -- please refer to \cite{fulton-harris} for further +details. The same argument also goes to show\dots + +\begin{corollary} + 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 \(\lambda\). +\end{corollary} + +This is very interesting to us since it implies that finding \emph{any} +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. + +\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 + \begin{center} + \begin{tikzpicture}[scale=2.5] + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \weightLattice{2} + \wt[black]{1}{0} + \wt[black]{-1}{1} + \wt[black]{0}{-1} + \node[right] at \weight{1}{0} {$\alpha_1$}; + \node[above left] at \weight{-1}{1} {$\alpha_2$}; + \node[below left] at \weight{0}{-1} {$\alpha_3$}; + \end{rootSystem} + \end{tikzpicture} + \end{center} + and \(\alpha_1\) is the highest weight of \(V\). + + 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, + \begin{center} + \begin{tikzpicture}[scale=2.5] + \AutoSizeWeightLatticefalse + \begin{rootSystem}{A} + \weightLattice{2} + \wt[black]{-1}{0} + \wt[black]{1}{-1} + \wt[black]{0}{1} + \node[left] at \weight{-1}{0} {$-\alpha_1$}; + \node[below right] at \weight{1}{-1} {$-\alpha_2$}; + \node[above right] at \weight{0}{1} {$-\alpha_3$}; + \end{rootSystem} + \end{tikzpicture} + \end{center} + is the weight diagram of \(V^*\) and \(\alpha_3\) is the highest weight of + \(V^*\). + + On the other hand if we fix two \(\mathfrak{sl}_3(K)\)-representations \(U\) + and \(W\), by computing + \[ + \begin{split} + H (u \otimes w) + & = H u \otimes w + u \otimes H w \\ + & = \lambda(H) \cdot u \otimes w + u \otimes \mu(H) \cdot w \\ + & = (\lambda + \mu)(H) \cdot (u \otimes w) + \end{split} + \] + 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 + 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 \(- m \alpha_3\) + 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 \(e_1^n \otimes f_3^m\). +\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 \(W\). + + Now by computing + \[ + H (v + w) + = H v + H w + = \mu(H) \cdot v + \mu(H) \cdot w + = \mu(H) \cdot (v + w) + \] + 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 + irreducible representation \(U = \mathfrak{sl}_3(K) \cdot v + w\) generated + by \(v + w\). The projection maps \(\pi_1 : U \to V\), \(\pi_2 : U \to W\), + being non-zero homomorphism between irreducible representations of + \(\mathfrak{sl}_3(K)\) 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 +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 section 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 \(\mathfrak{g}\). + +\section{Simultaneous Diagonalization \& the General Case} + +At the heart of our analysis of \(\mathfrak{sl}_2(K)\) and +\(\mathfrak{sl}_3(K)\) was the decision to consider the eigenspace +decomposition +\begin{equation}\label{sym-diag} + V = \bigoplus_\lambda V_\lambda +\end{equation} + +This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the +reasoning behind it, as well as the mere fact equation (\ref{sym-diag}) holds, +are harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace +decomposition associated with an operator \(V \to V\) is a very well-known +tool, and this type of argument should be familiar to anyone familiar with +basic concepts of linear algebra. On the other hand, the eigenspace +decomposition of \(V\) with respect to the action of an arbitrary subalgebra +\(\mathfrak{h} \subset \mathfrak{gl}(V)\) is neither well-known nor does it +hold in general: as previously stated, it may very well be that +\[ + \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \subsetneq V +\] + +We should note, however, that this two cases are not as different as they may +sound at first glance. Specifically, we can regard the eigenspace decomposition +of a representation \(V\) of \(\mathfrak{sl}_2(K)\) with respect to the +eigenvalues of the action of \(h\) as the eigenvalue decomposition of \(V\) +with respect to the action of the subalgebra \(\mathfrak{h} = K h \subset +\mathfrak{sl}_2(K)\). Furthermore, in both cases \(\mathfrak{h} \subset +\mathfrak{sl}_n(K)\) is the subalgebra of diagonal matrices, which is Abelian. +The fundamental difference between these two cases is thus the fact that \(\dim +\mathfrak{h} = 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim +\mathfrak{h} > 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The +question then is: why did we choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} > +1\) for \(\mathfrak{sl}_3(K)\)? + +% TODO: Add a note on how irreducible representations of Abelian algebras are +% all one dimensional to the previous chapter +The rational behind fixing an Abelian subalgebra is a simple one: we have seen +in the previous chapter that representations of Abelian +algebras are generally much simpler to understand than the general case. +Thus it make sense to decompose a given representation \(V\) of +\(\mathfrak{g}\) into subspaces invariant under the action of \(\mathfrak{h}\), +and then analyze how the remaining elements of \(\mathfrak{g}\) act on this +subspaces. The bigger \(\mathfrak{h}\) the simpler our problem gets, because +there are fewer elements outside of \(\mathfrak{h}\) left to analyze. + +Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h} +\subset \mathfrak{g}\), which leads us to the following definition. + +\begin{definition} + An subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a Cartan + subalgebra of \(\mathfrak{g}\)} if is self-normalizing -- i.e. \([X, H] \in + \mathfrak{h}\) for all \(H \in \mathfrak{h}\) if, and only if \(X \in + \mathfrak{h}\) -- and nilpotent. Equivalently for reductive \(\mathfrak{g}\), + \(\mathfrak{h}\) is called \emph{a Cartan subalgebra of \(\mathfrak{g}\)} if + it is Abelian, \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in + \mathfrak{h}\) and if \(\mathfrak{h}\) is maximal with respect to the former + two properties. +\end{definition} + +\begin{proposition} + There exists a Cartan subalgebra \(\mathfrak{h} \subset \mathfrak{g}\). +\end{proposition} + +\begin{proof} + Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose + elements act as diagonal operators via the adjoint representation. Indeed, + \(0\) -- the only element of \(0 \subset \mathfrak{g}\) -- is such that + \(\operatorname{ad}(0) = 0\). Furthermore, given a chain of Abelian + subalgebras + \[ + 0 \subset \mathfrak{h}_1 \subset \mathfrak{h}_2 \subset \cdots + \] + such that \(\operatorname{ad}(H)\) is a diagonal operator for each \(H \in + \mathfrak{h}_i\), the subalgebra \(\bigcup_i \mathfrak{h}_i \subset + \mathfrak{g}\) is Abelian, and its elements also act diagonally in + \(\mathfrak{g}\). It then follows from Zorn's lemma that there exists a + subalgebra \(\mathfrak{h}\) which is maximal with respect to both these + properties -- i.e. a Cartan subalgebra. +\end{proof} + +We have already seen some concrete examples. For instance, one can readily +check that every pair of diagonal matrices commutes, so that +\[ + \mathfrak{h} = + \begin{pmatrix} + K & 0 & \cdots & 0 \\ + 0 & K & \cdots & 0 \\ + \vdots & \vdots & \ddots & \vdots \\ + 0 & 0 & \cdots & K + \end{pmatrix} +\] +is an Abelian -- and hence nilpotent -- subalgebra of \(\mathfrak{gl}_n(K)\). A +simple calculation also shows that if \(i \ne j\) then the coefficient of +\(E_{i j}\) in \([E_{i i}, X]\) is the same as the coefficient of \(E_{i j}\) +in \(X\), for all \(X \in \mathfrak{gl}_n(K)\). In particular, if \([E_{i i}, +X]\) is diagonal for all \(i\), then so is \(X\) -- i.e. \(\mathfrak{h}\) is +self-normalizing. Hence \(\mathfrak{h}\) is a Cartan subalgebra of +\(\mathfrak{gl}_n(K)\). + +The intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the +subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of +\(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the +subalgebras described the previous two sections. The remaining question then +is: if \(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(V\) +is a representation of \(\mathfrak{g}\), does the eigenspace decomposition +\[ + V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda +\] +of \(V\) hold? The answer to this question turns out to be yes. This is a +consequence of something known as \emph{simultaneous diagonalization}, which is +the primary tool we'll use to generalize the results of the previous section. +What is simultaneous diagonalization all about then? + +\begin{definition}\label{def:sim-diag} + Given a \(K\)-vector space \(V\), a set of operators \(\{T_j : V \to V\}_j\) + is called \emph{simultaneously diagonalizable} if there is a basis \(\{v_1, + \ldots, v_n\}\) for \(V\) such that \(T_j v_i\) is a scalar multiple of + \(v_i\), for all \(i, j\). +\end{definition} + +\begin{proposition} + Given a \emph{finite-dimensional} vector space \(V\), A set of diagonalizable + operators \(V \to V\) is simultaneously diagonalizable if, and only if all of + its elements commute with one another. +\end{proposition} + +We should point out that simultaneous diagonalization \emph{only works in the +finite-dimensional setting}. In fact, simultaneous diagonalization is usually +framed as an equivalent statement about diagonalizable \(n \times n\) matrices +-- where \(n\) is, of course, finite. + +Simultaneous diagonalization implies that to show \(V = \bigoplus_\lambda +V_\lambda\) it suffices to show that \(H\!\restriction_V : V \to V\) is a +diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we +introduce \emph{the Jordan decomposition of an operator} and \emph{the abstract +Jordan decomposition of a semisimple Lie algebra}. + +\begin{proposition}[Jordan] + Given a finite-dimensional vector space \(V\) and an operator \(T : V \to + V\), there are unique commuting operators \(T_s, T_n : V \to V\), with + \(T_s\) diagonalizable and \(T_n\) nilpotent, such that \(T = T_s + T_n\). + The pair \((T_s, T_n)\) is known as \emph{the Jordan decomposition of \(T\)}. +\end{proposition} + +\begin{proposition} + Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are + \(X_s, X_n \in \mathfrak{g}\) such that \(X = X_s + X_n\), \([X_s, X_n] = + 0\), \(\operatorname{ad}(X_s)\) is a diagonalizable operator and + \(\operatorname{ad}(X_n)\) is a nilpotent operator. The pair \((X_s, X_n)\) + is known as \emph{the Jordan decomposition of \(X\)}. +\end{proposition} + +It should be clear from the uniqueness of \(\operatorname{ad}(X)_s\) and +\(\operatorname{ad}(X)_n\) that the Jordan decomposition of +\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) = \operatorname{ad}(X_s) + +\operatorname{ad}(X_n)\). What's perhaps more remarkable is the fact this holds +for \emph{any} finite-dimensional representation of \(\mathfrak{g}\). In other +words\dots + +\begin{proposition}\label{thm:preservation-jordan-form} + Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\) and \(X + \in \mathfrak{g}\). Denote by \(X\!\restriction_V\) the action of \(X\) in + \(V\). Then \(X_s\!\restriction_V = (X\!\restriction)_s\) and + \(X_n\!\restriction_V = (X\!\restriction)_n\). +\end{proposition} + +This last result is known as \emph{the preservation of the Jordan form}, and a +proof can be found in appendix C of \cite{fulton-harris}. We should point out +this fails spectacularly in positive characteristic. Furthermore, the statement +of proposition~\ref{thm:preservation-jordan-form} only makes sense for +\emph{semisimple} Lie algebras -- i.e. the algebras \(\mathfrak{g}\) for which +the abstract Jordan decomposition of \(\mathfrak{g}\) is defined. Nevertheless, +as promised this implies\dots + +\begin{corollary}\label{thm:finite-dim-is-weight-mod} + Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset + \mathfrak{g}\) be a Cartan subalgebra and \(V\) be any finite-dimensional + representation of \(\mathfrak{g}\). Then there is a basis \(\{v_1, \ldots, + v_n\}\) of \(V\) so that each \(v_i\) is simultaneously an eigenvector of all + elements of \(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as + a diagonal matrix in this basis. In other words, there are linear functionals + \(\lambda_i \in \mathfrak{h}^*\) so that + \( + H v_i = \lambda_i(H) \cdot v_i + \) + for all \(H \in \mathfrak{h}\). In particular, + \[ + V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda + \] +\end{corollary} + +\begin{proof} + Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_V + : V \to V\) is a diagonalizable operator. + + If we write \(H = H_s + H_n\) for the abstract Jordan decomposition of \(H\), + we know \(\operatorname{ad}(H_s) = \operatorname{ad}(H)_s\). But + \(\operatorname{ad}(H)\) is a diagonalizable operator, so that + \(\operatorname{ad}(H)_s = \operatorname{ad}(H)\). This implies + \(\operatorname{ad}(H_n) = \operatorname{ad}(H)_n = 0\), so that \(H_n\) is a + central element of \(\mathfrak{g}\). Since \(\mathfrak{g}\) is semisimple, + \(H_n = 0\). Proposition~\ref{thm:preservation-jordan-form} then implies + \((H\!\restriction_V)_n = (H_n)\!\restriction_V = 0\), so \(H\!\restriction_V + = (H\!\restriction_V)_s\) is a diagonalizable operator. +\end{proof} + +We should point out that this last proof only works for semisimple Lie +algebras. This is because we rely heavily on +proposition~\ref{thm:preservation-jordan-form}, as well in the fact that +semisimple Lie algebras are centerless. In fact, +corollary~\ref{thm:finite-dim-is-weight-mod} fails even for reductive Lie +algebras. For a counterexample, consider the algebra \(\mathfrak{g} = K\): the +Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a +\(\mathfrak{g}\)-module is simply a vector space \(V\) endowed with an operator +\(V \to V\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) in +\(V\). In particular, if we choose an operator \(V \to V\) which is \emph{not} +diagonalizable we find \(V \ne \bigoplus_{\lambda \in \mathfrak{h}^*} +V_\lambda\). + +However, corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive +\(\mathfrak{g}\) if we assume that the representation in question is +irreducible, since central elements of \(\mathfrak{g}\) act on irreducible +representations as scalar operators. The hypothesis of finite-dimensionality is +also of huge importance. In the next chapter we will encounter +infinite-dimensional \(\mathfrak{g}\)-modules for which the eigenspace +decomposition \(V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda\) fails. +As a first consequence of corollary~\ref{thm:finite-dim-is-weight-mod} + +\begin{corollary} + The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate. +\end{corollary} + +\begin{proof} + Consider the eigenspace decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus + \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint representation, where + \(\alpha\) ranges over all nonzero eigenvalues of the adjoint action of + \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = \mathfrak{h}\). + + Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h}) + \mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the + other hand, since \(\mathfrak{h}\) is self-normalizing, if \([X, H] = 0 \in + \mathfrak{h}\) for all \(H \in \mathfrak{h}\) then \(X \in \mathfrak{h}\) -- + i.e. \(\mathfrak{g}_0 \subset \mathfrak{h}\). So the eigenspace decomposition + becomes + \[ + \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha + \] + + We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to + \(\mathfrak{g}_\alpha\) with respect to \(B\) for any \(\alpha \ne 0\). + Indeed, given \(X \in \mathfrak{g}_\alpha\) and \(H_1, H_2 \in \mathfrak{h}\) + with \(\alpha(H_1) \ne 0\) we have + \[ + \alpha(H_1) \cdot B(X, H_2) + = B([H_1, X], H_2) + = - B([X, H_1], H_2) + = - B(X, [H_1, H_2]) + = 0 + \] + + Hence the non-degeneracy of \(B\) implies the non-degeneracy of its + restriction. +\end{proof} + +We should point out that the restriction of \(B\) to \(\mathfrak{h}\) is +\emph{not} the Killing form of \(\mathfrak{h}\). In fact, since +\(\mathfrak{h}\) is Abelian, its Killing form is identically zero -- which is +hardly ever a non-degenerate form. + +\begin{note} + Since \(B\) induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\), it + induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in + \(\mathfrak{h}^*\). We denote this form by \(B\). +\end{note} + +We now have most of the necessary tools to reproduce the results of the +previous chapter in a general setting. Let \(\mathfrak{g}\) be a +finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\) +and let \(V\) be a finite-dimensional irreducible representation of +\(\mathfrak{g}\). We will proceed, as we did before, by generalizing the +results about of the previous two sections in order. By now the pattern should +be starting become clear, so we will mostly omit technical details and proofs +analogous to the ones on the previous sections. Further details can be found in +appendix D of \cite{fulton-harris} and in \cite{humphreys}. + +We begin our analysis by remarking that in both \(\mathfrak{sl}_2(K)\) and +\(\mathfrak{sl}_3(K)\), the roots were symmetric about the origin and spanned +all of \(\mathfrak{h}^*\). This turns out to be a general fact, which is a +consequence of the non-degeneracy of the restriction of the Killing form to the +Cartan subalgebra. + +\begin{proposition}\label{thm:weights-symmetric-span} + The eigenvalues \(\alpha\) of the adjoint action of \(\mathfrak{h}\) in + \(\mathfrak{g}\) are symmetrical about the origin -- i.e. \(- \alpha\) is + also an eigenvalue -- and they span all of \(\mathfrak{h}^*\). +\end{proposition} + +\begin{proof} + We'll start with the first claim. Let \(\alpha\) and \(\beta\) be two + eigenvalues of the adjoint action of \(\mathfrak{h}\). Notice + \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha + + \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and \(Y \in + \mathfrak{g}_\beta\) then + \[ + [H [X, Y]] + = [X, [H, Y]] - [Y, [H, X]] + = (\alpha + \beta)(H) \cdot [X, Y] + \] + for all \(H \in \mathfrak{h}\). + + This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X) + \operatorname{ad}(Y)\) is nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then + \[ + (\operatorname{ad}(X) \operatorname{ad}(Y))^n Z + = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ] + \in \mathfrak{g}_{n \alpha + n \beta + \gamma} + = 0 + \] + for \(n\) large enough. In particular, \(B(X, Y) = + \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if + \(- \alpha\) is not an eigenvalue we find \(B(X, \mathfrak{g}_\beta) = 0\) + for all eigenvalues \(\beta\), which contradicts the non-degeneracy of \(B\). + Hence \(- \alpha\) must be an eigenvalue of the adjoint action of + \(\mathfrak{h}\). + + For the second statement, note that if the eigenvalues of \(\mathfrak{h}\) do + not span all of \(\mathfrak{h}^*\) then there is some \(H \in \mathfrak{h}\) + non-zero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is + to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in + \mathfrak{g}\). Another way of putting it is to say \(H\) is an element of + the center \(\mathfrak{z}\) of \(\mathfrak{g}\), which is zero by the + semisimplicity -- a contradiction. +\end{proof} + +Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and +\(\mathfrak{sl}_3(K)\) one can show\dots + +\begin{proposition}\label{thm:root-space-dim-1} + The eigenspaces \(\mathfrak{g}_\alpha\) are all 1-dimensional. +\end{proposition} + +The proof of the first statement of +proposition~\ref{thm:weights-symmetric-span} highlights something interesting: +if we fix some some eigenvalue \(\alpha\) of the adjoint action of +\(\mathfrak{h}\) in \(\mathfrak{g}\) and a eigenvector \(X \in +\mathfrak{g}_\alpha\), then for each \(H \in \mathfrak{h}\) and \(v \in +V_\lambda\) we find +\[ + H (X v) + = X (H v) + [H, X] v + = (\lambda + \alpha)(H) \cdot X v +\] +so that \(X\) carries \(v\) to \(V_{\lambda + \alpha}\). We have encountered +this formula twice in this chapter: again, we find \(\mathfrak{g}_\alpha\) +\emph{acts on \(V\) by translating vectors between eigenspaces}. In other +words, if we denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\) +then\dots + +\begin{theorem}\label{thm:weights-congruent-mod-root} + The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) are + all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\). +\end{theorem} + +% TODOO: Turn this into a proper discussion of basis and give the idea of the +% proof of existance of basis? +To proceed further, as in the case of \(\mathfrak{sl}_3(K)\) we have to fix a +direction in \(\mathfrak{h}^*\) -- i.e. we fix a linear function +\(\mathfrak{h}^* \to \QQ\) such that \(Q\) lies outside of its kernel. This +choice induces a partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of +roots of \(\mathfrak{g}\) and once more we find\dots + +\begin{definition} + The elements of \(\Delta^+\) and \(\Delta^-\) are called \emph{positive} and + \emph{negative roots}, respectively. The subalgebra \(\mathfrak{b} = + \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is + called \emph{the Borel subalgebra associated with \(\mathfrak{h}\)}. +\end{definition} + +\begin{theorem} + There is a weight vector \(v \in V\) that is killed by all positive root + spaces of \(\mathfrak{g}\). +\end{theorem} + +% TODO: Here we may take a weight of maximal height, but why is it unique? +% TODO: We don't really need to talk about height tho, we may simply take a +% weight that maximizes B(gamma, lambda) in QQ +% TODOO: Either way, we need to move this to after the discussion on the +% integrality of weights +\begin{proof} + It suffices to note that if \(\lambda\) is the weight of \(V\) lying the + furthest along the direction we chose and \(V_{\lambda + \alpha} \ne 0\) for + some \(\alpha \in \Delta^+\) then \(\lambda + \alpha\) is a weight that is + furthest along the direction we chose than \(\lambda\), which contradicts the + definition of \(\lambda\). +\end{proof} + +Accordingly, we call \(\lambda\) \emph{the highest weight of \(V\)}, and we +call any \(v \in V_\lambda\) \emph{a highest weight vector}. The strategy then +is to describe all weight spaces of \(V\) in terms of \(\lambda\) and \(v\), as +in theorem~\ref{thm:sl3-irr-weights-class}, and unsurprisingly we do so by +reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). Namely, we +show\dots + +\begin{proposition}\label{thm:distinguished-subalgebra} + Given a root \(\alpha\) of \(\mathfrak{g}\) the subspace + \(\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{- \alpha} + \oplus [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) is a subalgebra + isomorphic to \(\mathfrak{sl}_2(k)\). +\end{proposition} + +\begin{corollary}\label{thm:distinguished-subalg-rep} + For all weights \(\mu\), the subspace + \[ + V_\mu[\alpha] = \bigoplus_k V_{\mu + k \alpha} + \] + is invariant under the action of the subalgebra \(\mathfrak{s}_\alpha\) + and the weight spaces in this string match the eigenspaces of \(h\). +\end{corollary} + +The proof of proposition~\ref{thm:distinguished-subalgebra} is very technical +in nature and we won't include it here, but the idea behind it is simple: +recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both +1-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) +is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}] +\ne 0\) and that no generator of \([\mathfrak{g}_\alpha, \mathfrak{g}_{- +\alpha}] \ne 0\) is annihilated by \(\alpha\), so that by adjusting scalars we +can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in +\mathfrak{g}_{- \alpha}\) such that \(H_\alpha = [E_\alpha, F_\alpha]\) +satisfies +\begin{align*} + [H_\alpha, F_\alpha] & = -2 F_\alpha & + [H_\alpha, E_\alpha] & = 2 E_\alpha +\end{align*} + +The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely +determined by this condition, but \(H_\alpha\) is. The second statement of +corollary~\ref{thm:distinguished-subalg-rep} imposes a restriction on the +weights of \(V\). Namely, if \(\mu\) is a weight, \(\mu(H_\alpha)\) is an +eigenvalue of \(h\) in some representation of \(\mathfrak{sl}_2(K)\), so +that\dots + +\begin{proposition} + The weights \(\mu\) of an irreducible representation \(V\) of + \(\mathfrak{g}\) are so that \(\mu(H_\alpha) \in \ZZ\) for each \(\alpha \in + \Delta\). +\end{proposition} + +Once more, the lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha) +\in \ZZ, \forall \alpha \in \Delta \}\) is called \emph{the weight lattice of +\(\mathfrak{g}\)}, and we call the elements of \(P\) \emph{integral}. Finally, +another important consequence of theorem~\ref{thm:distinguished-subalgebra} +is\dots + +\begin{corollary} + If \(\alpha \in \Delta^+\) and \(T_\alpha : \mathfrak{h}^* \to + \mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to + \(\alpha\) with respect to the Killing form, + corollary~\ref{thm:distinguished-subalg-rep} implies that all \(\nu \in P\) + lying inside the line connecting \(\mu\) and \(T_\alpha \mu\) are weights -- + i.e. \(V_\nu \ne 0\). +\end{corollary} + +\begin{proof} + It suffices to note that \(\nu \in V_\mu[\alpha]\) -- see appendix D of + \cite{fulton-harris} for further details. +\end{proof} + +\begin{definition} + We refer to the group \(\mathcal{W} = \langle T_\alpha : \alpha \in \Delta^+ + \rangle \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl group of + \(\mathfrak{g}\)}. +\end{definition} + +This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we +found that the weights of the irreducible representations were symmetric with +respect to the lines \(K \alpha\) with \(B(\alpha_i - \alpha_j, \alpha) = 0\). +Indeed, the same argument leads us to the conclusion\dots + +\begin{theorem}\label{thm:irr-weight-class} + The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) with + highest weight \(\lambda\) are precisely the elements of the weight lattice + \(P\) congruent to \(\lambda\) modulo the root lattice \(Q\) lying inside the + convex hull of the image of \(\lambda\) under the action of the Weyl group + \(\mathcal{W}\). +\end{theorem} + +Now the only thing we are missing for a complete classification is an existence +and uniqueness theorem analogous to theorem~\ref{thm:sl2-exist-unique} and +theorem~\ref{thm:sl3-existence-uniqueness}. Lo and behold\dots + +\begin{definition} + An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all + \(\alpha \in \Delta^+\) is referred to as an \emph{integral dominant weight + of \(\mathfrak{g}\)}. +\end{definition} + +\begin{theorem}\label{thm:dominant-weight-theo} + For each dominant integral \(\lambda \in P\) there exists precisely one + irreducible finite-dimensional representation \(V\) of \(\mathfrak{g}\) whose + highest weight is \(\lambda\). +\end{theorem} + +Fix some dominant integral \(\lambda \in P\). The ``uniqueness'' part of the +theorem follows at once from the argument used for \(\mathfrak{sl}_3(K)\). The +``existence'' part is more nuanced. Our first instinct is, of course, to try to +generalize the proof used for \(\mathfrak{sl}_3(K)\). The issue is that our +proof relied heavily on our knowledge of the roots of \(\mathfrak{sl}_3(K)\). +Instead, we need a new strategy for the general setting. To that end, we +introduce a special class of \(\mathfrak{g}\)-modules, known as \emph{Verma +modules}. + +\begin{definition}\label{def:verma} + The \(\mathfrak{g}\)-module \(M(\lambda) = + \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K v^+\), where the action of + \(\mathfrak{b}\) in \(K v^+\) is given by \(H v^+ = \lambda(H) \cdot v^+\) + for all \(H \in \mathfrak{h}\) and \(X v^+ = 0\) for \(X \in + \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma + module of weight \(\lambda\)} +\end{definition} + +We should point out that, unlike most representations we've encountered so far, +Verma modules are \emph{highly infinite-dimensional}. Indeed, the dimension of +\(M(\lambda)\) is the same as the codimension of \(\mathcal{U}(\mathfrak{b})\) +in \(\mathcal{U}(\mathfrak{g})\), which is always infinite. Nevertheless, +\(M(\lambda)\) turns out to be quite well behaved. For instance, by +construction \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\) -- where +\(v^+ = 1 \otimes v^+ \in M(\lambda)\) is as in definition~\ref{def:verma}. +Moreover, we find\dots + +\begin{proposition}\label{thm:verma-is-weight-mod} + The weight spaces decomposition + \[ + M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu + \] + holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in + \mathfrak{h}^*\) and \(\dim M(\lambda) = 1\). Finally, \(\lambda\) is the + highest weight of \(M(\lambda)\), with highest weight vector given by \(v^+ = + 1 \otimes v^+ \in M(\lambda)\) as in definition~\ref{def:verma}. +\end{proposition} + +\begin{proof} + The Poincaré-Birkhoff-Witt theorem implies that \(M(\lambda)\) is spanned by + the vectors \(F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+\) for + \(\alpha_i \in \Delta^-\) and \(F_{\alpha_i} \in \mathfrak{g}_{\alpha_i}\) as + in the proof of proposition~\ref{thm:distinguished-subalgebra}. But + \[ + \begin{split} + H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+ + & = ([H, F_{\alpha_1}] + F_{\alpha_1} H) + F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\ + & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v^+ + + F_{\alpha_1} ([H, F_{\alpha_2}] + F_{\alpha_2} H) + F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\ + & \;\; \vdots \\ + & = (\alpha_1 + \cdots + \alpha_n)(H) \cdot + F_{\alpha_1} \cdots F_{\alpha_n} v^+ + + F_{\alpha_1} \cdots F_{\alpha_n} H v^+ \\ + & = (\lambda + \alpha_1 + \cdots + \alpha_n)(H) \cdot + F_{\alpha_1} \cdots F_{\alpha_n} v^+ \\ + & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v^+ + \in M(\lambda)_{\lambda + \alpha_1 + \cdots + \alpha_n} + \end{split} + \] + + Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*} + M(\lambda)_\mu\), as desired. In fact we have established + \[ + M(\lambda) + \subset + \bigoplus_{\substack{k_i \in \ZZ \\ k_i \ge 0}} + M(\lambda)_{\lambda + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n} + \] + where \(\{\alpha_1, \ldots, \alpha_m\} = \Delta^-\), so that all weights of + \(M(\lambda)\) have the form \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots + + k_n \cdot \alpha_n\). + + This already gives us that the weights of \(M(\lambda)\) are bounded by + \(\lambda\) -- in the sense that no weight of \(M(\lambda)\) is ``higher'' + than \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that + \(v^+\) is nonzero weight vector. Clearly \(v^+ \in V_\lambda\). The + Poincaré-Birkhoff-Witt theorem implies + \[ + M(\lambda) + \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right) + \otimes_{\mathcal{U}(\mathfrak{b})} K v^+ + \cong \bigoplus_i \mathcal{U}(\mathfrak{b}) + \otimes_{\mathcal{U}(\mathfrak{b})} K v^+ + \cong \bigoplus_i K v^+ + \ne 0 + \] + as \(\mathcal{U}(\mathfrak{b})\)-modules, so \(v^+ \ne 0\) -- for if this was + not the case we would find \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+ + = 0\). Hence \(V_\lambda \ne 0\) and therefore \(\lambda\) is the highest + weight of \(M(\lambda)\), with highest weight vector \(v^+\). + + To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only + finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots + F_{\alpha_n}^{k_n}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots + + k_n \cdot \alpha_n\). Since \(M(\lambda)_\mu\) is spanned by the images of + \(v^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In + particular, there is a single monomials \(F_{\alpha_1}^{k_1} + F_{\alpha_2}^{k_2} \cdots F_{\alpha_n}^{k_n}\) such that \(\lambda = \lambda + + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n\) -- which is, of course, + the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim V_\lambda = 1\). +\end{proof} + +\begin{example}\label{ex:sl2-verma} + If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K + h\) and \(\mathfrak{b} = K e \oplus K h\). If \(\lambda \in + \mathfrak{h}^*\) is the map \(h \mapsto 2\) then \(M(\lambda) = + \bigoplus_{k \ge 0} K f^k v^+\), and the action of \(\mathfrak{sl}_2(K)\) in + \(M(\lambda)\) is given by + \begin{align*} + f^{k + 1} v^+ & \overset{e}{\mapsto} (2 - k (k - 1)) f^k v^+ & + f^{k + 1} v^+ & \overset{f}{\mapsto} f^{k + 2} v^+ & + f^{k + 1} v^+ & \overset{h}{\mapsto} - 2 k f^{k + 1} v^+ & + \end{align*} + + In the language of the diagrams used in section~\ref{sec:sl2}, we write + % TODO: Add a label to the righ of the diagram indicating that the top arrows + % are the action of e and the bottom arrows are the action of f + \begin{center} + \begin{tikzcd} + \cdots \arrow[bend left=60]{r}{-10} + & M(\lambda)_{-6} \arrow[bend left=60]{r}{-4} \arrow[bend left=60]{l}{1} + & M(\lambda)_{-4} \arrow[bend left=60]{r}{0} \arrow[bend left=60]{l}{1} + & M(\lambda)_{-2} \arrow[bend left=60]{r}{2} \arrow[bend left=60]{l}{1} + & M(\lambda)_0 \arrow[bend left=60]{r}{2} \arrow[bend left=60]{l}{1} + & M(\lambda)_2 \arrow[bend left=60]{l}{1} + \end{tikzcd} + \end{center} + where \(M(\lambda)_{2 - 2 k} = K f^k v\). In this case, unlike we have see in + section~\ref{sec:sl2}, the string of weight spaces to left of the diagram is + infinite. +\end{example} + +What's interesting to us about all this is that we've just constructed a +\(\mathfrak{g}\)-module whose highest weight is \(\lambda\). This is not a +proof of theorem~\ref{thm:dominant-weight-theo}, however, since \(M(\lambda)\) +is neither irreducible nor finite-dimensional. Nevertheless, we can use +\(M(\lambda)\) to construct an irreducible representation of \(\mathfrak{g}\) +whose highest weight is \(\lambda\). + +\begin{proposition}\label{thm:max-verma-submod-is-weight} + Every subrepresentation \(V \subset M(\lambda)\) is the direct sum of its + weight spaces. In particular, \(M(\lambda)\) has a unique maximal + subrepresentation \(N(\lambda)\) and a unique irreducible quotient + \(\sfrac{M(\lambda)}{N(\lambda)}\). +\end{proposition} + +\begin{proof} + Let \(V \subset M(\lambda)\) be a subrepresentation and take any nonzero \(v + \in V\). Because of proposition~\ref{thm:verma-is-weight-mod}, we know there + are \(\mu_1, \ldots, \mu_n \in \mathfrak{h}^*\) and nonzero \(v_i \in + M(\lambda)_{\mu_i}\) such that \(v = v_1 + \cdots + v_n\). We want to show + \(v_i \in V\) for all \(i\). + + Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\). + Then + \[ + v_1 + - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_3 + - \cdots + - \frac{(\mu_n - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_n + = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v + \in V + \] + + Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By + applying the same procedure again we get + \begin{multline*} + v_1 + - + \frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)} + {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_4 + - \cdots - + \frac{(\mu_n - \mu_3)(H_3) \cdot (\mu_n - \mu_1)(H_2)} + {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_n \\ + = + \left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right) + \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v + \in V + \end{multline*} + + By applying the same procedure over and over again we can see that \(v_1 = X + v \in V\) for some \(X \in \mathcal{U}(\mathfrak{g})\). Furthermore, if we + reproduce all this for \(v_2 + \cdots + v_n = v - v_1 \in V\) we get that + \(v_2 \in V\). Now by applying the same procedure over and over we find + \(v_1, \ldots, v_n \in V\). Hence + \[ + V = \bigoplus_\mu V_\mu = \bigoplus_\mu M(\lambda)_\mu \cap V + \] + + Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\), \(V\) is a proper + subrepresentation then \(v^+ \notin V\). Hence any proper submodule lies in + the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum + \(N(\lambda)\) of all such submodules is still proper. In fact, this implies + \(N(\lambda)\) is the unique maximal subrepresentation of \(M(\lambda)\) and + \(\sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible quotient. +\end{proof} + +\begin{example}\label{ex:sl2-verma-quotient} + If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we + can see from example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge + 3} K f^k v^+\), so that \(\sfrac{M(\lambda)}{N(\lambda)}\) is the + \(3\)-dimensional irreducible representation of \(\mathfrak{sl}_2(K)\) -- + i.e. the finite-dimensional irreducible representation with highest weight + \(\lambda\) constructed in section~\ref{sec:sl2}. +\end{example} + +This last example is particularly interesting to us, since it indicates that +the finite-dimensional irreducible representations of \(\mathfrak{sl}_2(K)\) as +quotients of Verma modules. This is because the quotient +\(\sfrac{M(\lambda)}{N(\lambda)}\) in example~\ref{ex:sl2-verma-quotient} +happened to be finite-dimensional. As it turns out, this is always the case for +semisimple \(\mathfrak{g}\). Namely\dots + +\begin{proposition}\label{thm:verma-is-finite-dim} + If \(\lambda\) is dominant integral then the unique irreducible quotient of + \(M(\lambda)\) is finite-dimensional. +\end{proposition} + +The proof of proposition~\ref{thm:verma-is-finite-dim} is very technical and we +won't include it here, but the idea behind it is to show that the set of +weights of \(\sfrac{M(\lambda)}{N(\lambda)}\) is stable under the natural +action of the Weyl group \(\mathcal{W}\) in \(\mathfrak{h}^*\). One can then +show that the every weight of \(V\) is conjugate to a single dominant integral +weight of \(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant +integral weights of such irreducible quotient is finite. Since \(W\) is +finitely generated, this implies the set of weights of the unique irreducible +quotient of \(M(\lambda)\) is finite. But each weight space is +finite-dimensional. Hence so is the irreducible quotient. + +We refer the reader to \cite[ch. 21]{humphreys} for further details. What we +are really interested in is\dots + +\begin{corollary} + There is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose + highest weight is \(\lambda\). +\end{corollary} + +\begin{proof} + Let \(V = \sfrac{M(\lambda)}{N(\lambda)}\). It suffices to show that its + highest weight is \(\lambda\). We have already seen that \(v^+ \in + M(\lambda)_\lambda\) is a highest weight vector. Now since \(v\) lies outside + of the maximal subrepresentation of \(M(\lambda)\), the projection \(v^+ + + N(\lambda) \in V\) is nonzero. + + % TODO: Why is V_mu = M(lambda)_mu + N(lambda)? Turn this into a proposition? + We now claim that \(v^+ + N(\lambda) \in V_\lambda\). Indeed, + \[ + H (v^+ + N(\lambda)) + = H v^+ + N(\lambda) + = \lambda(H) \cdot (v^+ + N(\lambda)) + \] + for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(V\), with + weight vector \(v^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is + the highest weight of \(V\), for if this was not the case we could find a + weight \(\mu\) of \(M(\lambda)\) which is higher than \(\lambda\). +\end{proof} + +% TODO: Write a conclusion and move this to the next chapter +
diff --git a/tcc.tex b/tcc.tex @@ -26,6 +26,8 @@ \input{sections/complete-reducibility} +\input{sections/sl2-sl3} + \input{sections/mathieu} \printbibliography