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- 98de47f460c1b89a39f8ddaa05fe2fd0cad2b42b
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- c120e70fe65f0dcc3981ea289f1f55931a1ce3bd
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- Pablo <pablo-escobar@riseup.net>
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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
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3 files changed, 1699 insertions, 1699 deletions
| Status | Name | Changes | Insertions | Deletions |
| Deleted | sections/lie-algebras.tex | 1 file changed | 0 | 1698 |
| Added | sections/semisimple-algebras.tex | 1 file changed | 1698 | 0 |
| Modified | tcc.tex | 2 files changed | 1 | 1 |
diff --git a/sections/lie-algebras.tex /dev/null @@ -1,1698 +0,0 @@ -\chapter{Semisimple Lie Algebras \& their Representations}\label{ch:lie-algebras} - -\epigraph{Nobody has ever bet enough on a winning horse.}{\textit{Some -gambler}} - -I guess we could simply define semisimple Lie algebras as the class of -Lie algebras whose representations are completely reducible, but this is about -as satisfying as saying ``the semisimple are the ones who won't cause us any -trouble''. Who are the semisimple Lie algebras? Why does complete reducibility -holds for them? - -\section{Semisimplicity \& Complete Reducibility} - -Let \(K\) be an algebraicly closed field of characteristic \(0\). -There are multiple equivalent ways to define what a semisimple Lie algebra is, -the most obvious of which we have already mentioned in the above. Perhaps the -most common definition is\dots - -\begin{definition}\label{thm:sesimple-algebra} - A Lie algebra \(\mathfrak g\) over \(k\) is called \emph{semisimple} if it - has no non-zero solvable ideals -- i.e. subalgebras \(\mathfrak h\) with - \([\mathfrak h, \mathfrak g] \subset \mathfrak h\) whose derived series - \[ - \mathfrak h - \supseteq [\mathfrak h, \mathfrak h] - \supseteq [[\mathfrak h, \mathfrak h], [\mathfrak h, \mathfrak h]] - \supseteq - [ - [[\mathfrak h, \mathfrak h], [\mathfrak h, \mathfrak h]], - [[\mathfrak h, \mathfrak h], [\mathfrak h, \mathfrak h]] - ] - \supseteq \cdots - \] - converges to \(0\) in finite time. -\end{definition} - -\begin{example} - The Lie algebras \(\mathfrak{sl}_n(K)\) and \(\mathfrak{sp}_{2 n}(K)\) are both - semisimple -- see the section of \cite{kirillov} on invariant bilinear forms - and the semisimplicity of classical Lie algebras. -\end{example} - -A popular alternative to definition~\ref{thm:sesimple-algebra} is\dots - -\begin{definition}\label{def:semisimple-is-direct-sum} - A Lie algebra \(\mathfrak g\) is called semisimple if it is the direct sum of - simple Lie algebras -- i.e. non-Abelian Lie algebras \(\mathfrak s\) whose - only ideals are \(0\) and \(\mathfrak s\). -\end{definition} - -% TODO: Remove the reference to compact algebras -I suppose this last definition explains the nomenclature, but what does any of -this have to do with complete reducibility? Well, the special thing about -semisimple Lie algebras is that they are \emph{compact algebras}. -Compact Lie algebras are, as you might have guessed, \emph{algebras that come -from compact groups}. In other words\dots - -\begin{theorem} - Every representation of a semisimple Lie algebra is completely reducible. -\end{theorem} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -% TODO: Move this to the introduction -%By the same token, most of other aspects of representation -%theory of compact groups must hold in the context of semisimple algebras. For -%instance, we have\dots -% -%\begin{lemma}[Schur] -% Let \(V\) and \(W\) be two irreducible representations of a complex -% semisimple Lie algebra \(\mathfrak{g}\) and \(T : V \to W\) be an -% intertwining operator. Then either \(T = 0\) or \(T\) is an isomorphism. -% Furthermore, if \(V = W\) then \(T\) is scalar multiple of the identity. -%\end{lemma} -% -%\begin{corollary} -% Every irreducible representation of an Abelian Lie group is 1-dimensional. -%\end{corollary} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -% TODO: Turn this into a proper proof -Alternatively, one could prove the same statement in a purely algebraic manner -by showing the first Lie algebra cohomology group \(H^1(\mathfrak{g}, V) = -\operatorname{Ext}^1(K, V)\) vanishes for all \(V\), as do \cite{kirillov} and -\cite{lie-groups-serganova-student} in their proofs. More precisely, one can -show that there is a natural bijection between \(H^1(\mathfrak{g}, \operatorname{Hom}(V, -W))\) and isomorphism classes of the representations \(U\) of \(\mathfrak{g}\) -such that there is an exact sequence -\begin{center} - \begin{tikzcd} - 0 \arrow{r} & V \arrow{r} & U \arrow{r} & W \arrow{r} & 0 - \end{tikzcd} -\end{center} - -This implies every exact sequence of \(\mathfrak{g}\)-representations splits -- -which, if you recall theorem~\ref{thm:complete-reducibility-equiv}, is -equivalent to complete reducibility -- if, and only if \(H^1(\mathfrak{g}, -\operatorname{Hom}(V, W)) = 0\) for all \(V\) and \(W\). - -% TODO: Comment on the geometric proof by Weyl -%The algebraic approach has the -%advantage of working for Lie algebras over arbitrary fields, but in keeping -%with our principle of preferring geometric arguments over purely algebraic one -%we'll instead focus in the unitarization trick. What follows is a sketch of its -%proof, whose main ingredient is\dots - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -% TODO: Move this to somewhere else: the Killing form is not needed for this -% proof -%\section{The Killing Form} -% -%\begin{definition} -% Given a -- either real or complex -- Lie algebra, its Killing form is the -% symmetric bilinear form -% \[ -% K(X, Y) = \Tr(\ad(X) \ad(Y)) -% \] -%\end{definition} -% -%The Killing form certainly deserves much more attention than what we can -%afford at the present moment, but what's relevant to us is the fact that -%theorem~\ref{thm:compact-form} can be deduced from an algebraic condition -%satisfied by the Killing forms of complex semisimple algebras. Explicitly\dots -% -%\begin{theorem}\label{thm:killing-form-is-negative} -% If \(\mathfrak g\) is semisimple then there exists a semisimple real Lie -% algebra \(\mathfrak{g}_\RR\) whose complexification is precisely \(\mathfrak -% g\) and whose Killing form is negative-definite. -%\end{theorem} -% -%The proof of theorem~\ref{thm:killing-form-is-negative} is combinatorial in -%nature and it can be found in chapter 26 of \cite{fulton-harris}. What we're -%interested in at the moment is showing it implies -%theorem~\ref{thm:compact-form}. We'll start out by showing\dots -% -%\begin{lemma} -% If \(\mathfrak{g}_\RR\) is a real Lie algebra with negative-definite Killing -% form and \(G\) is its simply connected form then \(\mfrac{G}{Z(G)}\) is -% compact. -%\end{lemma} -% -%\begin{proof} -% Let \(G\) be the simply connected form of \(\mathfrak{g}_\RR\). Consider the -% the adjoint action \(\Ad : G \to \Aut(\mathfrak{g}_\RR)\). -% -% We'll start by point out that given \(g \in G\), -% \[ -% \begin{split} -% K(X, Y) -% & = \Tr(\ad(X) \ad(Y)) \\ -% & = \Tr(\Ad(g) (\ad(X) \ad(Y)) \Ad(g)^{-1}) \\ -% & = \Tr((\Ad(g) \ad(X) \Ad(g)^{-1}) (\Ad(g) \ad(Y) \Ad(g)^{-1})) \\ -% \text{(because \(\Ad(g)\) is a homomorphism)} -% & = \Tr(\ad(\Ad(g) X) \ad(\Ad(g) Y)) \\ -% & = K(\Ad(g) X, \Ad(g) Y)) -% \end{split} -% \] -% -% Now since \(K\) is negative-definite, \(\Ad(g)\) is an orthogonal operator. -% Hence \(\Ad(G)\) is a closed subgroup of \(\operatorname{O}(n)\) -- where \(n -% = \dim \mathfrak{g}_\RR\). Notice \(Z(G) = \ker \Ad\). Indeed, if \(\Ad(g) = -% \Id\) by corollary~\ref{thm:lie-group-morphism-at-identity} -% \(h \mapsto g h g^{-1}\) is the identity map -- i.e. \(g \in Z(G)\). It then -% follows from the fact that \(\operatorname{O}(n)\) is compact that -% \[ -% \mfrac{G}{Z(G)} -% = \mfrac{G}{\ker \Ad} -% \cong \Ad(G) -% \] -% is compact. -%\end{proof} -% -%We should point out that this last trick can also be used to prove that -%\(\mathfrak{g}_\RR\) is the direct sum of simple algebras. Indeed, if -%\(\mathfrak{g}_\RR\) is not simple then, by definition, it has a proper -%subalgebra \(\mathfrak h\). We can then consider its orthogonal complement -%\(\mathfrak{h}^\perp\) under the Killing form, so that \(\mathfrak{h}^\perp\) -%is a subalgebra and \(\mathfrak{g}_\RR = \mathfrak{h} \oplus -%\mathfrak{h}^\perp\). Now by induction on the dimension of \(\mathfrak{g}_\RR\) -%we see that theorem~\ref{thm:killing-form-is-negative} implies the -%characterization of definition~\ref{def:semisimple-is-direct-sum}. -% -%To conclude this dubious attempt at a proof, we refer to a theorem by Hermann -%Weyl, whose proof is beyond the scope of this notes as it requires calculating -%the Ricci curvature of \(G\) \footnote{The Ricci curvature is a tensor related -%to any given connection in a manifold. In this proof we're interested in the -%Ricci curvature of the Riemannian connection of \(\widetilde H\) under the -%metric given by the pullback of the unique bi-invariant metric of \(H\) along -%the covering map \(\widetilde H \to H\).} -- for a proof please refer to -%theorem 3.2.15 of \cite{gorodski}. What's interesting about this theorem is it -%implies\dots -% -%\begin{theorem}[Weyl] -% If \(H\) is a compact connected Lie group with discrete center then its -% universal cover \(\widetilde H\) is also compact. -%\end{theorem} -% -%\begin{proof}[Proof of theorem~\ref{thm:compact-form}] -% Let \(\mathfrak{g}_\RR\) be a semisimple real form of \(\mathfrak g\) with -% negative-definite Killing form. Because of the previous lemma, we already -% know \(\mfrac{G}{Z(G)}\) is compact and centerless. Hence by Weyl's theorem -% it suffices to show \(Z(G) = \ker \Ad\) is discrete -- so that the universal -% cover of \(\mfrac{G}{Z(G)}\) is \(G\). -% -% To do so, we consider its Lie algebra \(\mathfrak z = \ker \ad\) -- also -% known as the center of \(\mathfrak{g}_\RR\). Notice \(\mathfrak z\) is an -% ideal. In fact, \(\mathfrak z\) is a solvable ideal of \(\mathfrak{g}_\RR\) -% -- indeed, \([\mathfrak z, \mathfrak z] = 0\). This implies \(\mathfrak z = -% 0\) and therefore \(Z(G)\) is a 0-dimensional Lie group -- i.e. a discrete -% group. We are done. -%\end{proof} -% -%This results can be generalized to a certain extent by considering the exact -%sequence -%\begin{center} -% \begin{tikzcd} -% 0 \arrow{r} & -% \Rad(\mathfrak g) \arrow{r} & -% \mathfrak g \arrow{r} & -% \mfrac{\mathfrak g}{\Rad(\mathfrak g)} \arrow{r} & -% 0 -% \end{tikzcd} -%\end{center} -%where \(\Rad(\mathfrak g)\) is the sum of all solvable ideals of \(\mathfrak -%g\) -- i.e. a maximal solvable ideal -- for arbitrary complex \(\mathfrak g\). -%This implies we can deduce information about the representations of \(\mathfrak -%g\) by studying those of its semisimple part \(\mfrac{\mathfrak -%g}{\Rad(\mathfrak g)}\). In practice though, this isn't quite satisfactory -%because the exactness of this last sequence translates to the -%underwhelming\dots -% -%\begin{theorem}\label{thm:semi-simple-part-decomposition} -% Every irreducible representation of \(\mathfrak g\) is the tensor product of -% an irreducible representation of its semisimple part \(\mfrac{\mathfrak -% g}{\Rad(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak -% g\). -%\end{theorem} -% -%We say that this isn't satisfactory because -%theorem~\ref{thm:semi-simple-part-decomposition} is a statement about -%\emph{irreducible} representations of \(\mathfrak g\). This may sound a bit -%unfair, as theorem~\ref{thm:semi-simple-part-decomposition} does lead to a -%complete classification of a large class of representations of \(\mathfrak g\) -%-- those that are the direct sum of irreducible representations -- but the -%point is that these may not be all possible representations if \(\mathfrak g\) -%is not semisimple. That said, we can finally get to the classification itself. -%Without further ado, we'll start out by highlighting a concrete example of the -%general paradigm we'll later adopt: that of \(\sl_2(\CC)\). -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -% TODO: This shouldn't be considered underwelming! The primary results of this -% notes are concerned with irreducible representations of reducible Lie -% algebras -This results can be generalized to a certain extent by considering the exact -sequence -\begin{center} - \begin{tikzcd} - 0 \arrow{r} & - \operatorname{Rad}(\mathfrak g) \arrow{r} & - \mathfrak g \arrow{r} & - \mfrac{\mathfrak g}{\operatorname{Rad}(\mathfrak g)} \arrow{r} & - 0 - \end{tikzcd} -\end{center} -where \(\operatorname{Rad}(\mathfrak g)\) is the sum of all solvable ideals of \(\mathfrak -g\) -- i.e. a maximal solvable ideal -- for arbitrary \(\mathfrak g\). -This implies we can deduce information about the representations of \(\mathfrak -g\) by studying those of its semisimple part \(\mfrac{\mathfrak -g}{\operatorname{Rad}(\mathfrak g)}\). In practice though, this isn't quite satisfactory -because the exactness of this last sequence translates to the -underwhelming\dots - -\begin{theorem}\label{thm:semi-simple-part-decomposition} - Every irreducible representation of \(\mathfrak g\) is the tensor product of - an irreducible representation of its semisimple part \(\mfrac{\mathfrak - g}{\operatorname{Rad}(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak - g\). -\end{theorem} - -\section{Representations of \(\mathfrak{sl}_2(K)\)} - -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 pointing out 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 \(\lambda\). - -% TODO: We should clarify what right-most means in the context of an arbitrary -% field -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. - -% TODO: This doesn't make any sence in field other than C -In practice this means we'll choose a linear functional \(f : \mathfrak{h}^* -\to \RR\) 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 projection \(Q \to \RR \langle \alpha_1 - -\alpha_3 \rangle \cong \RR\) 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. - -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. - -\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} - -% TODO: Rewrite this: we haven't chosen any line -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 \(\langle \alpha_1 - \alpha_2, \alpha \rangle = 0\). 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\(\langle \alpha_1 - \alpha_2, \alpha \rangle=0\)}; - \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 \(\langle \alpha_3 - -\alpha_2, \alpha \rangle = 0\). 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\(\langle \alpha_1 - \alpha_2, \alpha \rangle=0\)}; - \draw[very thick] \weight{-4}{0} -- \weight{4}{0} - node[right]{\small\(\langle \alpha_3 - \alpha_2, \alpha \rangle=0\)}; - \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 \(\langle \alpha_i - \alpha_j, \alpha \rangle = - 0\). -\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: Rewrite this: we haven't dealt with finite groups at all -The rational behind fixing an Abelian subalgebra is one we have already -encountered when dealing with finite groups: representations of Abelian groups -and 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. - -% TODO: Remove or adjust the comment on maximal tori -% TODO: Turn this into a proper definition -% TODO: Also define the associated Borel subalgebra -Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h} -\subset \mathfrak{g}\). When \(\mathfrak{g}\) is semisimple, these coincide -with the so called \emph{Cartan subalgebras} of \(\mathfrak{g}\) -- i.e. -self-normalizing nilpotent subalgebras. A simple argument via Zorn's lemma is -enough to establish the existence of Cartan subalgebras for semisimple -\(\mathfrak{g}\): it suffices to note that if -\[ - \mathfrak{h}_1 - \subset \mathfrak{h}_2 - \subset \cdots - \subset \mathfrak{h}_n - \subset \cdots -\] -is a chain of Abelian subalgebras, then their sum is again an Abelian -subalgebra. Alternatively, one can show that every compact Lie group \(G\) -contains a maximal tori, whose Lie algebra is therefore a maximal Abelian -subalgebra of \(\mathfrak{g}\). - -That said, we can easily compute 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 subalgebra of \(\mathfrak{gl}_n(K)\). A simple calculation then shows -that if \(X \in \mathfrak{gl}_n(K)\) commutes with every diagonal matrix \(H \in -\mathfrak{h}\) then \(X\) is a diagonal matrix, so that \(\mathfrak{h}\) 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{proposition} - Let \(\mathfrak{g}\) be a Lie algebra, \(\mathfrak{h} \subset \mathfrak{g}\) - be an Abelian 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 is some linear functional - \(\lambda \in \mathfrak{h}^*\) so that - \[ - H v_i = \lambda(H) \cdot v_i - \] - for all \(H \in \mathfrak{h}\) and all \(i\). -\end{proposition} - -% TODO: h is not semisimple. Fix this proof -\begin{proof} - We claim \(\mathfrak{h}\) is semisimple. Indeed, if \(\{H_1, \ldots, H_m\}\) - is basis of \(\mathfrak{h}\) then - \[ - \mathfrak{h} \cong \bigoplus_i K H_i - \] - as vector spaces. Usually this is simply a linear isomorphism, but since - \(\mathfrak{h}\) is Abelian this is an isomorphism of Lie algebras -- here - \(K H_i\) represents the 1-dimensional subalgebra spanned by \(H_i\), which - is isomorphic to the trivial Lie algebra \(K\). Each \(K H_i\) is simple, - so \(\mathfrak{h}\) is isomorphic to a direct sum of simple algebras -- i.e. - \(\mathfrak{h}\) is semisimple. - - Hence - \[ - V - = \operatorname{Res}_{\mathfrak h}^{\mathfrak g} V - \cong \bigoplus V_i, - \] - as representations of \(\mathfrak{h}\), where each \(V_i\) is an irreducible - representation of \(\mathfrak{h}\). Since \(\mathfrak{h}\) is Abelian, it - follows from Schur's lemma that each \(V_i\) is 1-dimensional. Say \(V_i = - k v_i\) and consider the basis \(\{v_1, \ldots, v_n\}\) of \(V\). Now the - assertion that each \(v_i\) is an eigenvector of all elements of - \(\mathfrak{h}\) is equivalent to the statement that each \(K v_i\) is - stable under the action of \(\mathfrak{h}\). -\end{proof} - -As promised, this implies\dots - -\begin{corollary} - Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra - and \(\mathfrak{h}\) be a Cartan subalgebra of \(\mathfrak{g}\). Given a - finite-dimensional representation \(V\) of \(\mathfrak{g}\), - \[ - V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda - \] -\end{corollary} - -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 following theorem. - -% TODO: Add a refenrence to a proof (probably Humphreys) -% TODO: Changed the notation for the Killing form so that there is no conflict -% with the notation for the base field -% TODO: Clarify the meaning of "non-degenerate" -% TODO: Move this to before the analysis of sl3 -\begin{theorem} - If \(\mathfrak g\) is semisimple then its Killing form \(K\) is - non-degenerate. Furthermore, the restriction of \(K\) to \(\mathfrak{h}\) is - non-degenerate. -\end{theorem} - -\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, \(K(X, Y) = \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). - Now if \(- \alpha\) is not an eigenvalue we find \(K(X, \mathfrak{g}_\beta) = - 0\) for all eigenvalues \(\beta\), which contradicts the non-degeneracy of - \(K\). 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} - -% TODO: Rewrite this: the concept of direct has no sence in the general setting -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 \(\mathfrak{h}^* \to -\RR\) 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{theorem} - There is a weight vector \(v \in V\) that is killed by all positive root - spaces of \(\mathfrak{g}\). -\end{theorem} - -\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 \(W = \langle T_\alpha : \alpha \in \Delta^+ \rangle - \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl group of - \(\mathfrak{g}\)}. -\end{definition} - -% TODO: Note that this is the line orthogonal to alpha_i - alpha_j with respect -% to the Killing form -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 \(\langle \alpha_i - \alpha_j, \alpha \rangle = 0\). 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 - \(W\). -\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{theorem}\label{thm:dominant-weight-theo} - For each \(\lambda \in P\) such that \(\lambda(H_\alpha) \ge 0\) for - all positive roots \(\alpha\) there exists precisely one irreducible - representation \(V\) of \(\mathfrak{g}\) whose highest weight is \(\lambda\). -\end{theorem} - -\begin{note} - An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all - \(\alpha \in \Delta^+\) is usually referred to as an \emph{integral - dominant weight of \(\mathfrak{g}\)}. -\end{note} - -Unsurprisingly, our strategy is to copy what we did in the previous section. -The ``uniqueness'' part of the theorem follows at once from the argument used -for \(\mathfrak{sl}_3(K)\), and the proof of existence of can once again be reduced -to the proof of\dots - -\begin{theorem}\label{thm:weak-dominant-weight} - There exists \emph{some} -- not necessarily irreducible -- finite-dimensional - representation of \(\mathfrak{g}\) whose highest weight is \(\lambda\). -\end{theorem} - -The trouble comes when we try to generalize the proof of -theorem~\ref{thm:weak-dominant-weight} we used for the case when \(\mathfrak{g} -= \mathfrak{sl}_3(K)\). 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. - -% TODO: Add further details. turn this into a proper proof? -Alternatively, one could construct a potentially infinite-dimensional -representation of \(\mathfrak{g}\) whose highest weight is some fixed dominant -integral weight \(\lambda\) by taking the induced representation -\(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} V_\lambda = \mathcal{U}(\mathfrak{g}) -\otimes_{\mathcal{U}(\mathfrak{b})} V_\lambda\), where \(\mathfrak{b} = -\mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha \subset -\mathfrak{g}\) is the so called \emph{Borel subalgebra of \(\mathfrak{g}\)}, -\(\mathcal{U}(\mathfrak{g})\) denotes the \emph{universal enveloping algebra -of \(\mathfrak{g}\)} and \(\mathfrak{b}\) acts on \(V_\lambda = K v\) via \(H -v = \lambda(H) \cdot v\) and \(X v = 0\) for \(X \in \mathfrak{g}_\alpha\), as -does \cite{humphreys} in his proof. The fact that \(v\) is annihilated by all -positive root spaces guarantees that the maximal weight of \(V\) is at most -\(\lambda\), while the Poincare-Birkhoff-Witt \cite{humphreys} theorem -guarantees that \(v = 1 \otimes v \in V\) is a non-zero weight vector of -\(\lambda\) -- so that \(\lambda\) is the highest weight of \(V\). -The challenge then is to show that the irreducible component of \(v\) in \(V\) -is finite-dimensional -- see chapter 20 of \cite{humphreys} for a proof. -
diff --git /dev/null b/sections/semisimple-algebras.tex @@ -0,0 +1,1698 @@ +\chapter{Semisimple Lie Algebras \& their Representations}\label{ch:lie-algebras} + +\epigraph{Nobody has ever bet enough on a winning horse.}{\textit{Some +gambler}} + +I guess we could simply define semisimple Lie algebras as the class of +Lie algebras whose representations are completely reducible, but this is about +as satisfying as saying ``the semisimple are the ones who won't cause us any +trouble''. Who are the semisimple Lie algebras? Why does complete reducibility +holds for them? + +\section{Semisimplicity \& Complete Reducibility} + +Let \(K\) be an algebraicly closed field of characteristic \(0\). +There are multiple equivalent ways to define what a semisimple Lie algebra is, +the most obvious of which we have already mentioned in the above. Perhaps the +most common definition is\dots + +\begin{definition}\label{thm:sesimple-algebra} + A Lie algebra \(\mathfrak g\) over \(k\) is called \emph{semisimple} if it + has no non-zero solvable ideals -- i.e. subalgebras \(\mathfrak h\) with + \([\mathfrak h, \mathfrak g] \subset \mathfrak h\) whose derived series + \[ + \mathfrak h + \supseteq [\mathfrak h, \mathfrak h] + \supseteq [[\mathfrak h, \mathfrak h], [\mathfrak h, \mathfrak h]] + \supseteq + [ + [[\mathfrak h, \mathfrak h], [\mathfrak h, \mathfrak h]], + [[\mathfrak h, \mathfrak h], [\mathfrak h, \mathfrak h]] + ] + \supseteq \cdots + \] + converges to \(0\) in finite time. +\end{definition} + +\begin{example} + The Lie algebras \(\mathfrak{sl}_n(K)\) and \(\mathfrak{sp}_{2 n}(K)\) are both + semisimple -- see the section of \cite{kirillov} on invariant bilinear forms + and the semisimplicity of classical Lie algebras. +\end{example} + +A popular alternative to definition~\ref{thm:sesimple-algebra} is\dots + +\begin{definition}\label{def:semisimple-is-direct-sum} + A Lie algebra \(\mathfrak g\) is called semisimple if it is the direct sum of + simple Lie algebras -- i.e. non-Abelian Lie algebras \(\mathfrak s\) whose + only ideals are \(0\) and \(\mathfrak s\). +\end{definition} + +% TODO: Remove the reference to compact algebras +I suppose this last definition explains the nomenclature, but what does any of +this have to do with complete reducibility? Well, the special thing about +semisimple Lie algebras is that they are \emph{compact algebras}. +Compact Lie algebras are, as you might have guessed, \emph{algebras that come +from compact groups}. In other words\dots + +\begin{theorem} + Every representation of a semisimple Lie algebra is completely reducible. +\end{theorem} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +% TODO: Move this to the introduction +%By the same token, most of other aspects of representation +%theory of compact groups must hold in the context of semisimple algebras. For +%instance, we have\dots +% +%\begin{lemma}[Schur] +% Let \(V\) and \(W\) be two irreducible representations of a complex +% semisimple Lie algebra \(\mathfrak{g}\) and \(T : V \to W\) be an +% intertwining operator. Then either \(T = 0\) or \(T\) is an isomorphism. +% Furthermore, if \(V = W\) then \(T\) is scalar multiple of the identity. +%\end{lemma} +% +%\begin{corollary} +% Every irreducible representation of an Abelian Lie group is 1-dimensional. +%\end{corollary} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +% TODO: Turn this into a proper proof +Alternatively, one could prove the same statement in a purely algebraic manner +by showing the first Lie algebra cohomology group \(H^1(\mathfrak{g}, V) = +\operatorname{Ext}^1(K, V)\) vanishes for all \(V\), as do \cite{kirillov} and +\cite{lie-groups-serganova-student} in their proofs. More precisely, one can +show that there is a natural bijection between \(H^1(\mathfrak{g}, \operatorname{Hom}(V, +W))\) and isomorphism classes of the representations \(U\) of \(\mathfrak{g}\) +such that there is an exact sequence +\begin{center} + \begin{tikzcd} + 0 \arrow{r} & V \arrow{r} & U \arrow{r} & W \arrow{r} & 0 + \end{tikzcd} +\end{center} + +This implies every exact sequence of \(\mathfrak{g}\)-representations splits -- +which, if you recall theorem~\ref{thm:complete-reducibility-equiv}, is +equivalent to complete reducibility -- if, and only if \(H^1(\mathfrak{g}, +\operatorname{Hom}(V, W)) = 0\) for all \(V\) and \(W\). + +% TODO: Comment on the geometric proof by Weyl +%The algebraic approach has the +%advantage of working for Lie algebras over arbitrary fields, but in keeping +%with our principle of preferring geometric arguments over purely algebraic one +%we'll instead focus in the unitarization trick. What follows is a sketch of its +%proof, whose main ingredient is\dots + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +% TODO: Move this to somewhere else: the Killing form is not needed for this +% proof +%\section{The Killing Form} +% +%\begin{definition} +% Given a -- either real or complex -- Lie algebra, its Killing form is the +% symmetric bilinear form +% \[ +% K(X, Y) = \Tr(\ad(X) \ad(Y)) +% \] +%\end{definition} +% +%The Killing form certainly deserves much more attention than what we can +%afford at the present moment, but what's relevant to us is the fact that +%theorem~\ref{thm:compact-form} can be deduced from an algebraic condition +%satisfied by the Killing forms of complex semisimple algebras. Explicitly\dots +% +%\begin{theorem}\label{thm:killing-form-is-negative} +% If \(\mathfrak g\) is semisimple then there exists a semisimple real Lie +% algebra \(\mathfrak{g}_\RR\) whose complexification is precisely \(\mathfrak +% g\) and whose Killing form is negative-definite. +%\end{theorem} +% +%The proof of theorem~\ref{thm:killing-form-is-negative} is combinatorial in +%nature and it can be found in chapter 26 of \cite{fulton-harris}. What we're +%interested in at the moment is showing it implies +%theorem~\ref{thm:compact-form}. We'll start out by showing\dots +% +%\begin{lemma} +% If \(\mathfrak{g}_\RR\) is a real Lie algebra with negative-definite Killing +% form and \(G\) is its simply connected form then \(\mfrac{G}{Z(G)}\) is +% compact. +%\end{lemma} +% +%\begin{proof} +% Let \(G\) be the simply connected form of \(\mathfrak{g}_\RR\). Consider the +% the adjoint action \(\Ad : G \to \Aut(\mathfrak{g}_\RR)\). +% +% We'll start by point out that given \(g \in G\), +% \[ +% \begin{split} +% K(X, Y) +% & = \Tr(\ad(X) \ad(Y)) \\ +% & = \Tr(\Ad(g) (\ad(X) \ad(Y)) \Ad(g)^{-1}) \\ +% & = \Tr((\Ad(g) \ad(X) \Ad(g)^{-1}) (\Ad(g) \ad(Y) \Ad(g)^{-1})) \\ +% \text{(because \(\Ad(g)\) is a homomorphism)} +% & = \Tr(\ad(\Ad(g) X) \ad(\Ad(g) Y)) \\ +% & = K(\Ad(g) X, \Ad(g) Y)) +% \end{split} +% \] +% +% Now since \(K\) is negative-definite, \(\Ad(g)\) is an orthogonal operator. +% Hence \(\Ad(G)\) is a closed subgroup of \(\operatorname{O}(n)\) -- where \(n +% = \dim \mathfrak{g}_\RR\). Notice \(Z(G) = \ker \Ad\). Indeed, if \(\Ad(g) = +% \Id\) by corollary~\ref{thm:lie-group-morphism-at-identity} +% \(h \mapsto g h g^{-1}\) is the identity map -- i.e. \(g \in Z(G)\). It then +% follows from the fact that \(\operatorname{O}(n)\) is compact that +% \[ +% \mfrac{G}{Z(G)} +% = \mfrac{G}{\ker \Ad} +% \cong \Ad(G) +% \] +% is compact. +%\end{proof} +% +%We should point out that this last trick can also be used to prove that +%\(\mathfrak{g}_\RR\) is the direct sum of simple algebras. Indeed, if +%\(\mathfrak{g}_\RR\) is not simple then, by definition, it has a proper +%subalgebra \(\mathfrak h\). We can then consider its orthogonal complement +%\(\mathfrak{h}^\perp\) under the Killing form, so that \(\mathfrak{h}^\perp\) +%is a subalgebra and \(\mathfrak{g}_\RR = \mathfrak{h} \oplus +%\mathfrak{h}^\perp\). Now by induction on the dimension of \(\mathfrak{g}_\RR\) +%we see that theorem~\ref{thm:killing-form-is-negative} implies the +%characterization of definition~\ref{def:semisimple-is-direct-sum}. +% +%To conclude this dubious attempt at a proof, we refer to a theorem by Hermann +%Weyl, whose proof is beyond the scope of this notes as it requires calculating +%the Ricci curvature of \(G\) \footnote{The Ricci curvature is a tensor related +%to any given connection in a manifold. In this proof we're interested in the +%Ricci curvature of the Riemannian connection of \(\widetilde H\) under the +%metric given by the pullback of the unique bi-invariant metric of \(H\) along +%the covering map \(\widetilde H \to H\).} -- for a proof please refer to +%theorem 3.2.15 of \cite{gorodski}. What's interesting about this theorem is it +%implies\dots +% +%\begin{theorem}[Weyl] +% If \(H\) is a compact connected Lie group with discrete center then its +% universal cover \(\widetilde H\) is also compact. +%\end{theorem} +% +%\begin{proof}[Proof of theorem~\ref{thm:compact-form}] +% Let \(\mathfrak{g}_\RR\) be a semisimple real form of \(\mathfrak g\) with +% negative-definite Killing form. Because of the previous lemma, we already +% know \(\mfrac{G}{Z(G)}\) is compact and centerless. Hence by Weyl's theorem +% it suffices to show \(Z(G) = \ker \Ad\) is discrete -- so that the universal +% cover of \(\mfrac{G}{Z(G)}\) is \(G\). +% +% To do so, we consider its Lie algebra \(\mathfrak z = \ker \ad\) -- also +% known as the center of \(\mathfrak{g}_\RR\). Notice \(\mathfrak z\) is an +% ideal. In fact, \(\mathfrak z\) is a solvable ideal of \(\mathfrak{g}_\RR\) +% -- indeed, \([\mathfrak z, \mathfrak z] = 0\). This implies \(\mathfrak z = +% 0\) and therefore \(Z(G)\) is a 0-dimensional Lie group -- i.e. a discrete +% group. We are done. +%\end{proof} +% +%This results can be generalized to a certain extent by considering the exact +%sequence +%\begin{center} +% \begin{tikzcd} +% 0 \arrow{r} & +% \Rad(\mathfrak g) \arrow{r} & +% \mathfrak g \arrow{r} & +% \mfrac{\mathfrak g}{\Rad(\mathfrak g)} \arrow{r} & +% 0 +% \end{tikzcd} +%\end{center} +%where \(\Rad(\mathfrak g)\) is the sum of all solvable ideals of \(\mathfrak +%g\) -- i.e. a maximal solvable ideal -- for arbitrary complex \(\mathfrak g\). +%This implies we can deduce information about the representations of \(\mathfrak +%g\) by studying those of its semisimple part \(\mfrac{\mathfrak +%g}{\Rad(\mathfrak g)}\). In practice though, this isn't quite satisfactory +%because the exactness of this last sequence translates to the +%underwhelming\dots +% +%\begin{theorem}\label{thm:semi-simple-part-decomposition} +% Every irreducible representation of \(\mathfrak g\) is the tensor product of +% an irreducible representation of its semisimple part \(\mfrac{\mathfrak +% g}{\Rad(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak +% g\). +%\end{theorem} +% +%We say that this isn't satisfactory because +%theorem~\ref{thm:semi-simple-part-decomposition} is a statement about +%\emph{irreducible} representations of \(\mathfrak g\). This may sound a bit +%unfair, as theorem~\ref{thm:semi-simple-part-decomposition} does lead to a +%complete classification of a large class of representations of \(\mathfrak g\) +%-- those that are the direct sum of irreducible representations -- but the +%point is that these may not be all possible representations if \(\mathfrak g\) +%is not semisimple. That said, we can finally get to the classification itself. +%Without further ado, we'll start out by highlighting a concrete example of the +%general paradigm we'll later adopt: that of \(\sl_2(\CC)\). +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +% TODO: This shouldn't be considered underwelming! The primary results of this +% notes are concerned with irreducible representations of reducible Lie +% algebras +This results can be generalized to a certain extent by considering the exact +sequence +\begin{center} + \begin{tikzcd} + 0 \arrow{r} & + \operatorname{Rad}(\mathfrak g) \arrow{r} & + \mathfrak g \arrow{r} & + \mfrac{\mathfrak g}{\operatorname{Rad}(\mathfrak g)} \arrow{r} & + 0 + \end{tikzcd} +\end{center} +where \(\operatorname{Rad}(\mathfrak g)\) is the sum of all solvable ideals of \(\mathfrak +g\) -- i.e. a maximal solvable ideal -- for arbitrary \(\mathfrak g\). +This implies we can deduce information about the representations of \(\mathfrak +g\) by studying those of its semisimple part \(\mfrac{\mathfrak +g}{\operatorname{Rad}(\mathfrak g)}\). In practice though, this isn't quite satisfactory +because the exactness of this last sequence translates to the +underwhelming\dots + +\begin{theorem}\label{thm:semi-simple-part-decomposition} + Every irreducible representation of \(\mathfrak g\) is the tensor product of + an irreducible representation of its semisimple part \(\mfrac{\mathfrak + g}{\operatorname{Rad}(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak + g\). +\end{theorem} + +\section{Representations of \(\mathfrak{sl}_2(K)\)} + +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 pointing out 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 \(\lambda\). + +% TODO: We should clarify what right-most means in the context of an arbitrary +% field +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. + +% TODO: This doesn't make any sence in field other than C +In practice this means we'll choose a linear functional \(f : \mathfrak{h}^* +\to \RR\) 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 projection \(Q \to \RR \langle \alpha_1 - +\alpha_3 \rangle \cong \RR\) 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. + +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. + +\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} + +% TODO: Rewrite this: we haven't chosen any line +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 \(\langle \alpha_1 - \alpha_2, \alpha \rangle = 0\). 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\(\langle \alpha_1 - \alpha_2, \alpha \rangle=0\)}; + \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 \(\langle \alpha_3 - +\alpha_2, \alpha \rangle = 0\). 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\(\langle \alpha_1 - \alpha_2, \alpha \rangle=0\)}; + \draw[very thick] \weight{-4}{0} -- \weight{4}{0} + node[right]{\small\(\langle \alpha_3 - \alpha_2, \alpha \rangle=0\)}; + \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 \(\langle \alpha_i - \alpha_j, \alpha \rangle = + 0\). +\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: Rewrite this: we haven't dealt with finite groups at all +The rational behind fixing an Abelian subalgebra is one we have already +encountered when dealing with finite groups: representations of Abelian groups +and 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. + +% TODO: Remove or adjust the comment on maximal tori +% TODO: Turn this into a proper definition +% TODO: Also define the associated Borel subalgebra +Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h} +\subset \mathfrak{g}\). When \(\mathfrak{g}\) is semisimple, these coincide +with the so called \emph{Cartan subalgebras} of \(\mathfrak{g}\) -- i.e. +self-normalizing nilpotent subalgebras. A simple argument via Zorn's lemma is +enough to establish the existence of Cartan subalgebras for semisimple +\(\mathfrak{g}\): it suffices to note that if +\[ + \mathfrak{h}_1 + \subset \mathfrak{h}_2 + \subset \cdots + \subset \mathfrak{h}_n + \subset \cdots +\] +is a chain of Abelian subalgebras, then their sum is again an Abelian +subalgebra. Alternatively, one can show that every compact Lie group \(G\) +contains a maximal tori, whose Lie algebra is therefore a maximal Abelian +subalgebra of \(\mathfrak{g}\). + +That said, we can easily compute 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 subalgebra of \(\mathfrak{gl}_n(K)\). A simple calculation then shows +that if \(X \in \mathfrak{gl}_n(K)\) commutes with every diagonal matrix \(H \in +\mathfrak{h}\) then \(X\) is a diagonal matrix, so that \(\mathfrak{h}\) 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{proposition} + Let \(\mathfrak{g}\) be a Lie algebra, \(\mathfrak{h} \subset \mathfrak{g}\) + be an Abelian 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 is some linear functional + \(\lambda \in \mathfrak{h}^*\) so that + \[ + H v_i = \lambda(H) \cdot v_i + \] + for all \(H \in \mathfrak{h}\) and all \(i\). +\end{proposition} + +% TODO: h is not semisimple. Fix this proof +\begin{proof} + We claim \(\mathfrak{h}\) is semisimple. Indeed, if \(\{H_1, \ldots, H_m\}\) + is basis of \(\mathfrak{h}\) then + \[ + \mathfrak{h} \cong \bigoplus_i K H_i + \] + as vector spaces. Usually this is simply a linear isomorphism, but since + \(\mathfrak{h}\) is Abelian this is an isomorphism of Lie algebras -- here + \(K H_i\) represents the 1-dimensional subalgebra spanned by \(H_i\), which + is isomorphic to the trivial Lie algebra \(K\). Each \(K H_i\) is simple, + so \(\mathfrak{h}\) is isomorphic to a direct sum of simple algebras -- i.e. + \(\mathfrak{h}\) is semisimple. + + Hence + \[ + V + = \operatorname{Res}_{\mathfrak h}^{\mathfrak g} V + \cong \bigoplus V_i, + \] + as representations of \(\mathfrak{h}\), where each \(V_i\) is an irreducible + representation of \(\mathfrak{h}\). Since \(\mathfrak{h}\) is Abelian, it + follows from Schur's lemma that each \(V_i\) is 1-dimensional. Say \(V_i = + k v_i\) and consider the basis \(\{v_1, \ldots, v_n\}\) of \(V\). Now the + assertion that each \(v_i\) is an eigenvector of all elements of + \(\mathfrak{h}\) is equivalent to the statement that each \(K v_i\) is + stable under the action of \(\mathfrak{h}\). +\end{proof} + +As promised, this implies\dots + +\begin{corollary} + Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra + and \(\mathfrak{h}\) be a Cartan subalgebra of \(\mathfrak{g}\). Given a + finite-dimensional representation \(V\) of \(\mathfrak{g}\), + \[ + V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda + \] +\end{corollary} + +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 following theorem. + +% TODO: Add a refenrence to a proof (probably Humphreys) +% TODO: Changed the notation for the Killing form so that there is no conflict +% with the notation for the base field +% TODO: Clarify the meaning of "non-degenerate" +% TODO: Move this to before the analysis of sl3 +\begin{theorem} + If \(\mathfrak g\) is semisimple then its Killing form \(K\) is + non-degenerate. Furthermore, the restriction of \(K\) to \(\mathfrak{h}\) is + non-degenerate. +\end{theorem} + +\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, \(K(X, Y) = \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). + Now if \(- \alpha\) is not an eigenvalue we find \(K(X, \mathfrak{g}_\beta) = + 0\) for all eigenvalues \(\beta\), which contradicts the non-degeneracy of + \(K\). 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} + +% TODO: Rewrite this: the concept of direct has no sence in the general setting +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 \(\mathfrak{h}^* \to +\RR\) 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{theorem} + There is a weight vector \(v \in V\) that is killed by all positive root + spaces of \(\mathfrak{g}\). +\end{theorem} + +\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 \(W = \langle T_\alpha : \alpha \in \Delta^+ \rangle + \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl group of + \(\mathfrak{g}\)}. +\end{definition} + +% TODO: Note that this is the line orthogonal to alpha_i - alpha_j with respect +% to the Killing form +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 \(\langle \alpha_i - \alpha_j, \alpha \rangle = 0\). 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 + \(W\). +\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{theorem}\label{thm:dominant-weight-theo} + For each \(\lambda \in P\) such that \(\lambda(H_\alpha) \ge 0\) for + all positive roots \(\alpha\) there exists precisely one irreducible + representation \(V\) of \(\mathfrak{g}\) whose highest weight is \(\lambda\). +\end{theorem} + +\begin{note} + An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all + \(\alpha \in \Delta^+\) is usually referred to as an \emph{integral + dominant weight of \(\mathfrak{g}\)}. +\end{note} + +Unsurprisingly, our strategy is to copy what we did in the previous section. +The ``uniqueness'' part of the theorem follows at once from the argument used +for \(\mathfrak{sl}_3(K)\), and the proof of existence of can once again be reduced +to the proof of\dots + +\begin{theorem}\label{thm:weak-dominant-weight} + There exists \emph{some} -- not necessarily irreducible -- finite-dimensional + representation of \(\mathfrak{g}\) whose highest weight is \(\lambda\). +\end{theorem} + +The trouble comes when we try to generalize the proof of +theorem~\ref{thm:weak-dominant-weight} we used for the case when \(\mathfrak{g} += \mathfrak{sl}_3(K)\). 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. + +% TODO: Add further details. turn this into a proper proof? +Alternatively, one could construct a potentially infinite-dimensional +representation of \(\mathfrak{g}\) whose highest weight is some fixed dominant +integral weight \(\lambda\) by taking the induced representation +\(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} V_\lambda = \mathcal{U}(\mathfrak{g}) +\otimes_{\mathcal{U}(\mathfrak{b})} V_\lambda\), where \(\mathfrak{b} = +\mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha \subset +\mathfrak{g}\) is the so called \emph{Borel subalgebra of \(\mathfrak{g}\)}, +\(\mathcal{U}(\mathfrak{g})\) denotes the \emph{universal enveloping algebra +of \(\mathfrak{g}\)} and \(\mathfrak{b}\) acts on \(V_\lambda = K v\) via \(H +v = \lambda(H) \cdot v\) and \(X v = 0\) for \(X \in \mathfrak{g}_\alpha\), as +does \cite{humphreys} in his proof. The fact that \(v\) is annihilated by all +positive root spaces guarantees that the maximal weight of \(V\) is at most +\(\lambda\), while the Poincare-Birkhoff-Witt \cite{humphreys} theorem +guarantees that \(v = 1 \otimes v \in V\) is a non-zero weight vector of +\(\lambda\) -- so that \(\lambda\) is the highest weight of \(V\). +The challenge then is to show that the irreducible component of \(v\) in \(V\) +is finite-dimensional -- see chapter 20 of \cite{humphreys} for a proof. +
diff --git a/tcc.tex b/tcc.tex @@ -22,7 +22,7 @@ \pagenumbering{arabic} \setcounter{page}{1} -\input{sections/lie-algebras} +\input{sections/semisimple-algebras} \printbibliography \end{document}