diff --git a/sections/lie-algebras.tex b/sections/lie-algebras.tex
@@ -3,50 +3,7 @@
\epigraph{Nobody has ever bet enough on a winning horse.}{\textit{Some
gambler}}
-We've just established that \(\Rep(G) \cong \Rep(\mathfrak{g}_\CC)\) for each
-simply connected \(G\), but how do we go about classifying the representations
-of \(\mathfrak{g}_\CC\)? We can very quickly see that many of the aspects of
-the representation theory of compact groups that were essential for the
-solution our classification problem are no longer valid in this context. For
-instance if we take
-\[
- \mathfrak g =
- \begin{pmatrix}
- \CC & \CC \\
- 0 & 0
- \end{pmatrix}
- \subset \gl_2(\CC)
-\]
-and consider its natural representation \(V = \CC^2\) we can see that \(W =
-\langle (1, 0) \rangle\) is a subrepresentation with no \(\mathfrak
-g\)-invariant complement. The issue we now face is, of course, the fact that
-complete reducibility fails for some complex Lie algebras. In other words,
-understanding the irreducible representations of an algebra isn't enough,
-because there may be representations which cannot be expressed as the direct
-sum of irreducible ones.
-
-% TODO: Update the "two chapter!" thing after those chapters are done?
-This is not going to be an easy ride\dots \ In fact, even after some further
-restrictions we will soon impose, this classification
-problem will keep us busy for the next \emph{two} chapters! The primary goal
-of this chapter is to highlight this restrictions and study some particular
-examples in the hopes of getting some insight into the general case. We should
-point that the following discussion is \emph{immensely} inspired by the third
-part of \citetitle{fulton-harris} \cite{fulton-harris}: Fulton \& Harris are
-the real authors of this chapter, I am merely a commentator on their work.
-
-The restriction we'll make is, as you might have guessed, that we'll primarily
-focus on \emph{semisimple Lie algebras} -- those for whom \emph{complete
-reducibility}, also known as \emph{semisimplicity}, holds.
-This is a bit of an admission of defeat from my end, as we won't ever get to
-classify the smooth representations of \emph{all} simply connected Lie groups,
-but keep in mind that the problem of classifying the finite-dimensional
-representations of an arbitrary finite-dimensional Lie algebra is still an open
-one. In other words, the semisimple Lie algebras are the only algebras whose
-representation we can expect to understand in full generality. This goes to
-show the importance of complete reducibility in representation theory.
-
-I guess we could simply define semisimple Lie algebras as the class of complex
+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
@@ -54,13 +11,14 @@ 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 \(\mathfrak g\) is called \emph{semisimple} if it has no
- non-zero solvable ideals -- i.e. subalgebras \(\mathfrak h\) with
+ 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
@@ -77,109 +35,66 @@ most common definition is\dots
\end{definition}
\begin{example}
- The complex Lie algebras \(\sl_n(\CC)\) and \(\mathfrak{sp}_{2 n}(\CC)\) are
- both semisimple -- see the section of \cite{kirillov} on invariant bilinear
- forms and the semisimplicity of classical Lie algebras.
+ The Lie algebras \(\sl_n(k)\) 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. Lie algebras \(\mathfrak s\) whose only ideals
- are \(0\) and \(\mathfrak s\).
+ simple Lie algebras -- i.e. non-Abelian Lie algebras \(\mathfrak s\) whose
+ only ideals are \(0\) and \(\mathfrak s\).
\end{definition}
+% TODO: This is plain wrong, fix this
\begin{example}
Every Abelian Lie algebra \(\mathfrak{g}\) is semisimple. Indeed, if \(\{X_i
: i \in \Lambda\}\) is a basis of \(\mathfrak{g}\) then
\[
- \mathfrak{g} = \bigoplus_i \CC X_i
+ \mathfrak{g} = \bigoplus_i k X_i
\]
- as a vector space. Clearly, each \(\CC X_i\) is a subalgebra of
- \(\mathfrak{g}\). But \(\dim \CC X_i = 1\), so that the only subspaces of
- \(\CC X_i\) are \(0\) and \(\CC X_i\). In particular, the only ideals of
- \(\CC X_i\) are \(0\) and \(\CC X_i\) -- i.e. \(\CC X_i\) is simple.
+ as a vector space. Clearly, each \(k X_i\) is a subalgebra of
+ \(\mathfrak{g}\). But \(\dim k X_i = 1\), so that the only subspaces of
+ \(k X_i\) are \(0\) and \(k X_i\). In particular, the only ideals of
+ \(k X_i\) are \(0\) and \(k X_i\) -- i.e. \(k X_i\) is simple.
Moreover, the fact that \([X_i, X_j] = 0\) for all \(i\) and \(j\) implies
- \(\mathfrak{g} \cong \bigoplus_i \CC X_i\) as a Lie algebra.
-\end{example}
-
-\begin{example}
- The algebra \(\gl_n(\CC) = \sl_n(\CC) \oplus \CC\) is semisimple. In fact,
- the direct sum of any two semisimple Lie algebras is semisimple.
+ \(\mathfrak{g} \cong \bigoplus_i k X_i\) as a Lie algebra.
\end{example}
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 complex Lie algebras is that they are \emph{compact algebras}.
+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}\label{thm:compact-form}
- If \(\mathfrak g\) is a complex semisimple Lie algebra, there exists a
- (unique) semisimple real form of \(\mathfrak g\) whose simply connected form
- is compact.
-\end{theorem}
-
-The proof of theorem~\ref{thm:compact-form} is quite involved and we will only
-provide a rough outline -- mostly based on the proof by \cite{fegan} -- but the
-interesting thing about it is the fact that we have already classified the
-smooth (complex) representations of compact Lie groups: because of
-chapter~\ref{ch:compacts}, we already know every smooth representation of a
-compact group is completely reducible. We can then use this knowledge to lift
-complete reducibility to our semisimple algebra. For instance, given a complex
-representation \(V\) of \(\sl_n(\CC)\) and a subrepresentation \(W \subset V\),
-\begin{enumerate}
- \item Since \(\mathfrak{su}_n \otimes \CC \cong \sl_n(\CC)\), \(V\) and \(W\)
- correspond to complex representations of \(\mathfrak{su}_n\)
- \item Since \(\SU_n\) is simply connected, \(V\) and \(W\) correspond to
- smooth representations of \(\SU_n\)
- \item Since \(\SU_n\) is compact, \(W\) admits a \(\SU_n\)-invariant
- complement \(U \subset V\)
- \item \(U\) is invariant under the action of \(\mathfrak{su}_n\), so\dots
- \item \(U\) is invariant under the action of \(\sl_n(\CC)\) and
- therefore\dots
- \item \(U\) is a \(\sl_n(\CC)\)-invariant complement of \(W\) in \(V\)
-\end{enumerate}
-
-If we assume theorem~\ref{thm:compact-form} and replace \(\sl_n(\CC)\) with
-some arbitrary semisimple \(\mathfrak{g}\) in the previous paragraph we arrive
-at a proof of\dots
-
\begin{theorem}
Every representation of a semisimple Lie algebra is completely reducible.
\end{theorem}
-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}
-
-Indeed, if we take theorem~\ref{thm:compact-form} at face
-value we can easily see that given a semisimple Lie algebra \(\mathfrak g\),
-\[
- \Rep(\mathfrak g)
- \cong \Rep(\mathfrak{g}_\RR)
- \cong \Rep(G),
-\]
-where \(\mathfrak{g}_\RR\) is some real form of \(\mathfrak g\) with compact
-simply connected form \(G\). This is what's known as \emph{Weyl's unitarization
-trick}, and historically it was the first proof of complete reducibility for
-semisimple Lie algebras.
-
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% 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 \(H^1(\mathfrak{g}, V) =
-\Ext^1(\CC, V)\) vanishes for all \(V\), as do \cite{kirillov} and
+\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 \(H^1(\mathfrak{g}, \Hom(V,
W))\) and isomorphism classes of the representations \(U\) of \(\mathfrak{g}\)
@@ -197,115 +112,162 @@ such that there is an exact sequence
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 \(H^1(\mathfrak{g},
-\Hom(V, W)) = 0\) for all \(V\) and \(W\). 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
-
-\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}
-
+\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}
@@ -318,7 +280,7 @@ sequence
\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 \(\mathfrak g\).
+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 \(\mfrac{\mathfrak
g}{\Rad(\mathfrak g)}\). In practice though, this isn't quite satisfactory
@@ -332,29 +294,18 @@ underwhelming\dots
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 \(\sl_2(\CC)\).
-
-\section{Representations of \(\sl_2(\CC)\)}
+\section{Representations of \(\sl_2(k)\)}
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 \(\sl_2(\CC)\) with \(\dim V = n\).
+ \(V\) of \(\sl_2(k)\) with \(\dim V = n\).
\end{theorem}
It's important to note, however, that -- as promised -- we will end up with an
explicit construction of \(V\). The general approach we'll take is supposing
-\(V\) is an irreducible representation of \(\sl_2(\CC)\) and then derive some
+\(V\) is an irreducible representation of \(\sl_2(k)\) and then derive some
information about its structure. We begin our analysis by pointing out that
the elements
\begin{align*}
@@ -362,7 +313,7 @@ the elements
f & = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} &
h & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{align*}
-form a basis of \(\sl_2(\CC)\) and satisfy
+form a basis of \(\sl_2(k)\) and satisfy
\begin{align*}
[e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e
\end{align*}
@@ -400,7 +351,7 @@ Hence
& \cdots \arrow[bend left=60]{l}
\end{tikzcd}
\end{center}
-and \(\bigoplus_{n \in \ZZ} V_{\lambda + 2 n}\) is an \(\sl_2(\CC)\)-invariant
+and \(\bigoplus_{n \in \ZZ} V_{\lambda + 2 n}\) is an \(\sl_2(k)\)-invariant
subspace. This implies
\[
V = \bigoplus_{n \in \ZZ} V_{\lambda + 2 n},
@@ -413,7 +364,7 @@ Even more so, if \(a = \min \{ n \in \ZZ : V_{\lambda + 2 n} \ne 0 \}\) and
\[
\bigoplus_{\substack{n \in \ZZ \\ a \le n \le b}} V_{\lambda + 2 n}
\]
-is also an \(\sl_2(\CC)\)-invariant subspace, so that the eigenvalues of \(h\)
+is also an \(\sl_2(k)\)-invariant subspace, so that the eigenvalues of \(h\)
form an unbroken string
\[
\ldots, \lambda - 4, \lambda - 2, \lambda, \lambda + 2, \lambda + 4, \ldots
@@ -435,7 +386,7 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
suffices to show \(V = \langle v, f v, f^2 v, \ldots \rangle\), which in
light of the fact that \(V\) is irreducible is the same as showing \(\langle
v, f v, f^2 v, \ldots \rangle\) is invariant under the action of
- \(\sl_2(\CC)\).
+ \(\sl_2(k)\).
The fact that \(h f^k v \in \langle v, f v, f^2 v, \ldots \rangle\) follows
immediately from our previous assertion that \(f^k v \in V_{\lambda - 2 k}\)
@@ -475,7 +426,7 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
Theorem~\ref{thm:basis-of-irr-rep} may seem unrelated to our problem at first,
but it's significance lies in the fact that we have just provided a complete
-description of the action of \(\sl_2(\CC)\) in \(V\). In other words\dots
+description of the action of \(\sl_2(k)\) in \(V\). In other words\dots
\begin{corollary}
\(V\) is completely determined by the right-most eigenvalue \(\lambda\) of
@@ -483,7 +434,7 @@ description of the action of \(\sl_2(\CC)\) in \(V\). In other words\dots
\end{corollary}
\begin{proof}
- If \(W\) is an irreducible representation of \(\sl_2(\CC)\) whose
+ If \(W\) is an irreducible representation of \(\sl_2(k)\) whose
right-most eigenvalue of \(h\) is \(\lambda\) and \(w \in W_\lambda\) is
non-zero, consider the linear isomorphism
\begin{align*}
@@ -536,7 +487,7 @@ Other important consequences of theorem~\ref{thm:basis-of-irr-rep} are\dots
left-most eigenvalue of \(h\) is precisely \(n - 2 (m - 1) = -n\).
\end{proof}
-We now know every irreducible representation \(V\) of \(\sl_2(\CC)\) has the
+We now know every irreducible representation \(V\) of \(\sl_2(k)\) has the
form
\begin{center}
\begin{tikzcd}
@@ -550,7 +501,7 @@ form
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
\[
- V = \bigoplus_{k = 0}^n \CC f^k v
+ V = \bigoplus_{k = 0}^n k f^k v
\]
and
\begin{equation}\label{eq:irr-rep-of-sl2}
@@ -566,16 +517,16 @@ To conclude our analysis all it's left is to show that for each \(n\) such
\begin{theorem}\label{thm:irr-rep-of-sl2-exists}
For each \(n \ge 0\) there exists a (unique) irreducible representation of
- \(\sl_2(\CC)\) whose left-most eigenvalue of \(h\) is \(n\).
+ \(\sl_2(k)\) 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 \(\sl_2(\CC)\) -- just look at \(f^k
+ clear from the commutator relations of \(\sl_2(k)\) -- 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
- \(\CC^2\) is the natural representation of \(\sl_2(\CC)\), then \(V = \Sym^n
- \CC^2\) satisfies the relations of (\ref{eq:irr-rep-of-sl2}). To see that
+ \(k^2\) is the natural representation of \(\sl_2(k)\), then \(V = \Sym^n
+ k^2\) 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
@@ -650,16 +601,16 @@ which establishes that \(V\) is irreducible.
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 \(\sl_2(\CC)\). It is not yet clear,
+\emph{all} irreducible representations of \(\sl_2(k)\). 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 \(\sl_3(\CC)\), hoping this will \emph{somehow}
+results of this section for \(\sl_3(k)\), 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 \(\sl_3(\CC)\)}\label{sec:sl3-reps}
+\section{Representations of \(\sl_3(k)\)}\label{sec:sl3-reps}
-The study of representations of \(\sl_2(\CC)\) reminds me of the difference the
+The study of representations of \(\sl_2(k)\) 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
@@ -669,31 +620,31 @@ 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 \(\sl_2(\CC)\). It's a simple and beautiful example, but
+The same applies to \(\sl_2(k)\). 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 \(\sl_2(\CC)\). The general purpose of this section is
+hidden in the case of \(\sl_2(k)\). The general purpose of this section is
to investigate to which extent the framework used in the previous section to
-classify the representations of \(\sl_2(\CC)\) can be generalized to other
-semisimple Lie algebras, and the algebra \(\sl_3(\CC)\) stands as a natural
+classify the representations of \(\sl_2(k)\) can be generalized to other
+semisimple Lie algebras, and the algebra \(\sl_3(k)\) 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 \(\sl_3(\CC)\) and proceed step by step, at each point
+representation \(V\) of \(\sl_3(k)\) and proceed step by step, at each point
asking ourselves how we could possibly adapt the framework we laid out for
-\(\sl_2(\CC)\). The first obvious question is one we have already asked
+\(\sl_2(k)\). 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 \(\sl_3(\CC)\)?
+eigenvalues and is there an analogue of \(h\) in \(\sl_3(k)\)?
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 \(\sl_2(\CC)\) culminating in
+description of the irreducible representations of \(\sl_2(k)\) culminating in
theorem~\ref{thm:irr-rep-of-sl2-exists}.
-Our fist task is to find some analogue of \(h\) in \(\sl_3(\CC)\), but it's
+Our fist task is to find some analogue of \(h\) in \(\sl_3(k)\), 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
@@ -703,7 +654,7 @@ actual way to proceed is to consider the subalgebra
\mathfrak h
= \left\{
X \in
- \begin{pmatrix} \CC & 0 & 0 \\ 0 & \CC & 0 \\ 0 & 0 & \CC \end{pmatrix}
+ \begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix}
: \Tr(X) = 0
\right\}
\]
@@ -711,10 +662,10 @@ actual way to proceed is to consider the subalgebra
The choice of \(\mathfrak{h}\) may seem like an odd choice at the moment, but
the point is we'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 \(\sl_3(\CC)\). This is exactly analogous to
-the situation we found in \(\sl_2(\CC)\): \(h\) corresponds to the subalgebra
+half of the remaining elements of \(\sl_3(k)\). This is exactly analogous to
+the situation we found in \(\sl_2(k)\): \(h\) corresponds to the subalgebra
\(\mathfrak{h}\), and the eigenvalues of \(h\) in turn correspond to linear
-functions \(\lambda : \mathfrak{h} \to \CC\) such that \(H v = \lambda(H) \cdot
+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 \(\mathfrak h\)}.
@@ -732,8 +683,8 @@ associated with an eigenvalue of any particular operator \(H \in
\(\mathfrak{h}\). Fortunately for us, (\ref{eq:weight-module}) always holds,
but we will postpone its proof to the next chapter.
-Next we turn our attention to the remaining elements of \(\sl_3(\CC)\). In our
-analysis of \(\sl_2(\CC)\) we saw that the eigenvalues of \(h\) differed from
+Next we turn our attention to the remaining elements of \(\sl_3(k)\). In our
+analysis of \(\sl_2(k)\) 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,
@@ -743,10 +694,10 @@ the eigenvalues of of the adjoint actions of \(h\) are \(\pm 2\) since
[h, e] & = 2 e
\end{align*}
and the eigenvalues of the action of \(h\) in an irreducible
-\(\sl_2(\CC)\)-representation differ from one another by multiples of \(\pm
+\(\sl_2(k)\)-representation differ from one another by multiples of \(\pm
2\).
-In the case of \(\sl_3(\CC)\), a simple calculation shows that if \([H, X]\) is
+In the case of \(\sl_3(k)\), 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 -
@@ -792,8 +743,8 @@ Visually we may draw
\end{figure}
If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) in
-\(\sl_3(\CC)\) associated to \(\alpha\) by \(\sl_3(\CC)_\alpha\) and fix some
-\(X \in \sl_3(\CC)_\alpha\), \(H \in \mathfrak{h}\) and \(v \in V_\lambda\)
+\(\sl_3(k)\) associated to \(\alpha\) by \(\sl_3(k)_\alpha\) and fix some
+\(X \in \sl_3(k)_\alpha\), \(H \in \mathfrak{h}\) and \(v \in V_\lambda\)
then
\[
\begin{split}
@@ -804,11 +755,11 @@ then
\end{split}
\]
so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words,
-\(\sl_3(\CC)_\alpha\) \emph{acts on \(V\) by translating vectors between
+\(\sl_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors between
eigenspaces}.
-For instance \(\sl_3(\CC)_{\alpha_1 - \alpha_3}\) will act on the adjoint
-representation of \(\sl_3(\CC)\) via
+For instance \(\sl_3(k)_{\alpha_1 - \alpha_3}\) will act on the adjoint
+representation of \(\sl_3(k)\) via
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=2.5]
@@ -829,18 +780,18 @@ representation of \(\sl_3(\CC)\) via
\end{figure}
This is again entirely analogous to the situation we observed in
-\(\sl_2(\CC)\). In fact, we may once more conclude\dots
+\(\sl_2(k)\). 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
- \(\sl_3(\CC)\)-representation \(V\) differ from one another by integral
+ \(\sl_3(k)\)-representation \(V\) differ from one another by integral
linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of
- adjoint action of \(\mathfrak{h}\) in \(\sl_3(\CC)\).
+ adjoint action of \(\mathfrak{h}\) in \(\sl_3(k)\).
\end{theorem}
\begin{proof}
This proof goes exactly as that of the analogous statement for
- \(\sl_2(\CC)\): it suffices to note that if we fix some eigenvalue
+ \(\sl_2(k)\): 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}
@@ -853,7 +804,7 @@ 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 \(\sl_3(\CC)\), we'll call the non-zero
+ Given a representation \(V\) of \(\sl_3(k)\), 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
@@ -861,33 +812,33 @@ adjoint action of \(\mathfrak{h}\).
\end{definition}
It's clear from our previous discussion that the weights of the adjoint
-representation of \(\sl_3(\CC)\) deserve some special attention.
+representation of \(\sl_3(k)\) deserve some special attention.
\begin{definition}
- The weights of the adjoint representation of \(\sl_3(\CC)\) are called
- \emph{roots of \(\sl_3(\CC)\)}. Once again, the expressions \emph{root
+ The weights of the adjoint representation of \(\sl_3(k)\) are called
+ \emph{roots of \(\sl_3(k)\)}. 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 \(\sl_3(\CC)\) are all
+ The weights of an irreducible representation \(V\) of \(\sl_3(k)\) are all
congruent module the lattice \(Q\) generated by the roots \(\alpha_i -
- \alpha_j\) of \(\sl_3(\CC)\).
+ \alpha_j\) of \(\sl_3(k)\).
\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 \(\sl_3(\CC)\)}.
+ is called \emph{the root lattice of \(\sl_3(k)\)}.
\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 \(\sl_2(\CC)\) in terms of this vector. We may
-reproduce these steps in the context of \(\sl_3(\CC)\) by fixing a direction in
+irreducible representation of \(\sl_2(k)\) in terms of this vector. We may
+reproduce these steps in the context of \(\sl_3(k)\) by fixing a direction in
the place an considering the weight lying the furthest in that direction.
In practice this means we'll choose a linear functional \(f : \mathfrak{h}^*
@@ -945,7 +896,7 @@ sort of \(\frac{1}{3}\)-plane with corners at \(\lambda\), as shown in
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 \(\sl_3(\CC)\) act on the weight spaces of \(V\) via
+root spaces of \(\sl_3(k)\) 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 -
@@ -954,22 +905,22 @@ annihilate \(V_\lambda\), or otherwise one of \(V_{\lambda + \alpha_1 -
\begin{theorem}
There is a weight vector \(v \in V\) that is killed by all positive root
- spaces of \(\sl_3(\CC)\).
+ spaces of \(\sl_3(k)\).
\end{theorem}
\begin{proof}
- It suffices to note that the positive roots of \(\sl_3(\CC)\) are precisely
+ It suffices to note that the positive roots of \(\sl_3(k)\) 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
-\(\sl_2(\CC)\), we then constructed an explicit basis of our irreducible
+\(\sl_2(k)\), 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 \(\sl_2(\CC)\) in terms of
+provide an explicit description of the action of \(\sl_2(k)\) 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
-\(\sl_3(\CC)\) -- and indeed that's what we'll do later down the line -- but
+\(\sl_3(k)\) -- 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.
@@ -1008,14 +959,14 @@ To draw a familiar picture
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
-\(\sl_2(\CC)\) via
+\(\sl_2(k)\) 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 \(\sl_2(\CC)\). Even more so, we
+In other words, \(W\) is a representation of \(\sl_2(k)\). Even more so, we
claim
\[
V_{\lambda + k (\alpha_2 - \alpha_1)} = W_{\lambda(H) - 2k}
@@ -1087,8 +1038,8 @@ 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 \(\mathfrak{s}_{\alpha_i -
-\alpha_j} = \CC \langle E_{i j}, E_{j i}, [E_{i j}, E_{j i}] \rangle\), which
-is once more isomorphic to \(\sl_2(\CC)\), and again the weight spaces in this
+\alpha_j} = k \langle E_{i j}, E_{j i}, [E_{i j}, E_{j i}] \rangle\), which
+is once more isomorphic to \(\sl_2(k)\), 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}
@@ -1123,7 +1074,7 @@ applying this method to the weights at the ends of our string, arriving at
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 \(\sl_2(\CC)\) whose weights correspond to weights of \(V\) lying in a
+of \(\sl_2(k)\) 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}
@@ -1162,8 +1113,8 @@ of \(V\) we started with.
\end{center}
By construction, \(\nu\) corresponds to the right-most weight of the
-representation of \(\sl_2(\CC)\), so that all dots lying on the gray string
-must occur in the representation of \(\sl_2(\CC)\). Hence they must also be
+representation of \(\sl_2(k)\), so that all dots lying on the gray string
+must occur in the representation of \(\sl_2(k)\). Hence they must also be
weights of \(V\). The final picture is thus
\begin{center}
\begin{tikzpicture}
@@ -1203,7 +1154,7 @@ weights of \(V\). The final picture is thus
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 \(\sl_2(\CC)\), so it must be an integer. Now since
+representation of \(\sl_2(k)\), so it must be an integer. Now since
\[
\lambda
\begin{pmatrix}
@@ -1234,7 +1185,7 @@ P\).
\begin{definition}
The lattice \(P = \ZZ \alpha_1 \oplus \ZZ \alpha_2 \oplus \ZZ \alpha_3\) is
- called \emph{the weight lattice of \(\sl_3(\CC)\)}.
+ called \emph{the weight lattice of \(\sl_3(k)\)}.
\end{definition}
Finally\dots
@@ -1247,8 +1198,8 @@ Finally\dots
0\).
\end{theorem}
-Once more there's a clear parallel between the case of \(\sl_3(\CC)\) and that
-of \(\sl_2(\CC)\), where we observed that the weights all lied in the lattice
+Once more there's a clear parallel between the case of \(\sl_3(k)\) and that
+of \(\sl_2(k)\), 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
@@ -1257,36 +1208,36 @@ 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 \(\sl_3(\CC)\) whose highest weight
+ one irreducible representation \(V\) of \(\sl_3(k)\) 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 \(\sl_2(\CC)\): next we showed in theorem~\ref{thm:basis-of-irr-rep} that
-any irreducible representation of \(\sl_2(\CC)\) is spanned by the images of
+of \(\sl_2(k)\): next we showed in theorem~\ref{thm:basis-of-irr-rep} that
+any irreducible representation of \(\sl_2(k)\) 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 \(\sl_2(\CC)\) is spanned by
+say that an irreducible representation \(V\) of \(\sl_2(k)\) is spanned by
the images of its highest weight vector under successive applications by half
-of the root spaces of \(\sl_2(\CC)\). The advantage of this alternative
-formulation is, of course, that the same holds for \(\sl_3(\CC)\).
+of the root spaces of \(\sl_2(k)\). The advantage of this alternative
+formulation is, of course, that the same holds for \(\sl_3(k)\).
Specifically\dots
\begin{theorem}\label{thm:irr-sl3-span}
- Given an irreducible \(\sl_3(\CC)\)-representation \(V\) and a highest
+ Given an irreducible \(\sl_3(k)\)-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
-\(\sl_3(\CC)\) to inductively show that the subspace spanned by the images of a
+\(\sl_3(k)\) 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 \(\sl_3(\CC)\) -- please refer
+and \(E_{3 2}\) is invariant under the action of \(\sl_3(k)\) -- please refer
to \cite{fulton-harris} for further details. The same argument also goes to
show\dots
\begin{corollary}
- Given a representation \(V\) of \(\sl_3(\CC)\) with highest weight
+ Given a representation \(V\) of \(\sl_3(k)\) 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 \(\lambda\).
@@ -1299,7 +1250,7 @@ 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 \(V = \CC^3\) of \(\sl_3(\CC)\). We
+ Consider the natural representation \(V = k^3\) of \(\sl_3(k)\). We
claim that the highest weight of \(\Sym^n V \otimes \Sym^m V^*\)
is \(n \alpha_1 - m \alpha_3\).
@@ -1344,7 +1295,7 @@ representation turns out to be quite simple.
is the weight diagram of \(V^*\) and \(\alpha_3\) is the highest weight of
\(V^*\).
- On the other hand if we fix two \(\sl_3(\CC)\)-representations \(U\) and
+ On the other hand if we fix two \(\sl_3(k)\)-representations \(U\) and
\(W\), by computing
\[
\begin{split}
@@ -1369,7 +1320,7 @@ 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 \(\sl_3(\CC)\) with
+ Let \(V\) and \(W\) be two irreducible representations of \(\sl_3(k)\) with
highest weight \(\lambda\). By theorem~\ref{thm:sl3-irr-weights-class}, the
weights of \(V\) are precisely the same as those of \(W\).
@@ -1386,9 +1337,9 @@ simpler than that.
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 \(U = \sl_3(\CC) \ v + w\) generated by \(v + w\).
+ irreducible representation \(U = \sl_3(k) \ v + w\) 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 \(\sl_3(\CC)\) must be
+ homomorphism between irreducible representations of \(\sl_3(k)\) must be
isomorphism. Finally,
\[
V \cong U \cong W
@@ -1396,31 +1347,31 @@ simpler than that.
\end{proof}
The situation here is analogous to that of the previous section, where we saw
-that the irreducible representations of \(\sl_2(\CC)\) are given by symmetric
+that the irreducible representations of \(\sl_2(k)\) 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 \(\sl_2(\CC)\) and \(\sl_3(\CC)\), but so far
+irreducible representations of \(\sl_2(k)\) and \(\sl_3(k)\), 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 complex semisimple Lie
+representations of an arbitrary finite-dimensional semisimple Lie
algebra \(\mathfrak{g}\).
\section{Simultaneous Diagonalization \& the General Case}
-At the heart of our analysis of \(\sl_2(\CC)\) and \(\sl_3(\CC)\) was the
+At the heart of our analysis of \(\sl_2(k)\) and \(\sl_3(k)\) 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 \(\sl_2(\CC)\), but the reasoning
+This was simple enough to do in the case of \(\sl_2(k)\), but the reasoning
behind it, as well as the mere fact equation (\ref{sym-diag}) holds, are harder
-to explain in the case of \(\sl_3(\CC)\). The eigenspace decomposition
+to explain in the case of \(\sl_3(k)\). 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
@@ -1433,15 +1384,15 @@ stated, it may very well be that
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 \(\sl_2(\CC)\) with respect to the eigenvalues of
+of a representation \(V\) of \(\sl_2(k)\) with respect to the eigenvalues of
the action of \(h\) as the eigenvalue decomposition of \(V\) with respect to
-the action of the subalgebra \(\mathfrak{h} = \CC h \subset \sl_2(\CC)\).
-Furthermore, in both cases \(\mathfrak{h} \subset \sl_n(\CC)\) is the
+the action of the subalgebra \(\mathfrak{h} = k h \subset \sl_2(k)\).
+Furthermore, in both cases \(\mathfrak{h} \subset \sl_n(k)\) 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
-\(\mathfrak{h} \subset \sl_2(\CC)\) while \(\dim \mathfrak{h} > 1\) for
-\(\mathfrak{h} \subset \sl_3(\CC)\). The question then is: why did we choose
-\(\mathfrak{h}\) with \(\dim \mathfrak{h} > 1\) for \(\sl_3(\CC)\)?
+\(\mathfrak{h} \subset \sl_2(k)\) while \(\dim \mathfrak{h} > 1\) for
+\(\mathfrak{h} \subset \sl_3(k)\). The question then is: why did we choose
+\(\mathfrak{h}\) with \(\dim \mathfrak{h} > 1\) for \(\sl_3(k)\)?
The rational behind fixing an Abelian subalgebra is one we have already
encountered when dealing with finite groups: representations of Abelian groups
@@ -1475,18 +1426,18 @@ readily check that every pair of diagonal matrices commutes, so that
\[
\mathfrak{h} =
\begin{pmatrix}
- \CC & 0 & \cdots & 0 \\
- 0 & \CC & \cdots & 0 \\
+ k & 0 & \cdots & 0 \\
+ 0 & k & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
- 0 & 0 & \cdots & \CC
+ 0 & 0 & \cdots & k
\end{pmatrix}
\]
-is an Abelian subalgebra of \(\gl_n(\CC)\). A simple calculation then shows
-that if \(X \in \gl_n(\CC)\) commutes with every diagonal matrix \(H \in
+is an Abelian subalgebra of \(\gl_n(k)\). A simple calculation then shows
+that if \(X \in \gl_n(k)\) commutes with every diagonal matrix \(H \in
\mathfrak{h}\) then \(X\) is a diagonal matrix, so that \(\mathfrak{h}\) is a
-Cartan subalgebra of \(\gl_n(\CC)\). The intersection of such subalgebra with
-\(\sl_n(\CC)\) -- i.e. the subalgebra of traceless diagonal matrices -- is a
-Cartan subalgebra of \(\sl_n(\CC)\). In particular, if \(n = 2\) or \(n = 3\)
+Cartan subalgebra of \(\gl_n(k)\). The intersection of such subalgebra with
+\(\sl_n(k)\) -- i.e. the subalgebra of traceless diagonal matrices -- is a
+Cartan subalgebra of \(\sl_n(k)\). 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
@@ -1518,12 +1469,12 @@ What is simultaneous diagonalization all about then?
We claim \(\mathfrak{h}\) is semisimple. Indeed, if \(\{H_1, \ldots, H_m\}\)
is basis of \(\mathfrak{h}\) then
\[
- \mathfrak{h} \cong \bigoplus_i \CC H_i
+ \mathfrak{h} \cong \bigoplus_i k H_i
\]
as vector spaces. Usually this is simply a linear isomorphism, but since
\(\mathfrak{h}\) is Abelian this is an isomorphism of Lie algebras -- here
- \(\CC H_i\) represents the 1-dimensional subalgebra spanned by \(H_i\), which
- is isomorphic to the trivial Lie algebra \(\CC\). Each \(\CC H_i\) is simple,
+ \(k H_i\) 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.
@@ -1536,16 +1487,16 @@ What is simultaneous diagonalization all about then?
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 \(V_i =
- \CC v_i\) and consider the basis \(\{v_1, \ldots, v_n\}\) of \(V\). Now the
+ 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 \(\CC v_i\) is
+ \(\mathfrak{h}\) 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 complex semisimple Lie algebra
+ 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 \(\mathfrak{g}\),
\[
@@ -1563,8 +1514,8 @@ 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 \(\sl_2(\CC)\) and
-\(\sl_3(\CC)\), the roots were symmetric about the origin and spanned all of
+We begin our analysis by remarking that in both \(\sl_2(k)\) and
+\(\sl_3(k)\), 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.
@@ -1618,7 +1569,7 @@ of the following theorem.
\(\mathfrak{g}\), which is zero by the semisimplicity -- a contradiction.
\end{proof}
-Furthermore, as in the case of \(\sl_2(\CC)\) and \(\sl_3(\CC)\) one can
+Furthermore, as in the case of \(\sl_2(k)\) and \(\sl_3(k)\) one can
show\dots
\begin{proposition}\label{thm:root-space-dim-1}
@@ -1647,7 +1598,7 @@ then\dots
all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\).
\end{theorem}
-To proceed further, as in the case of \(\sl_3(\CC)\) we have to fix a direction
+To proceed further, as in the case of \(\sl_3(k)\) 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
@@ -1670,13 +1621,13 @@ 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 \(\sl_3(\CC)\). Namely, we show\dots
+reproducing the proof of the case of \(\sl_3(k)\). 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 \(\sl_2(\CC)\).
+ isomorphic to \(\sl_2(k)\).
\end{proposition}
\begin{corollary}\label{thm:distinguished-subalg-rep}
@@ -1707,7 +1658,7 @@ 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 \(\sl_2(\CC)\), so that\dots
+eigenvalue of \(h\) in some representation of \(\sl_2(k)\), so that\dots
\begin{proposition}
The weights \(\mu\) of an irreducible representation \(V\) of
@@ -1741,7 +1692,7 @@ is\dots
\(\mathfrak{g}\)}.
\end{definition}
-This is entirely analogous to the situation of \(\sl_3(\CC)\), where we found
+This is entirely analogous to the situation of \(\sl_3(k)\), 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
@@ -1772,7 +1723,7 @@ theorem~\ref{thm:sl3-existence-uniqueness}. Lo and behold\dots
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 \(\sl_3(\CC)\), and the proof of existence of can once again be reduced
+for \(\sl_3(k)\), and the proof of existence of can once again be reduced
to the proof of\dots
\begin{theorem}\label{thm:weak-dominant-weight}
@@ -1782,22 +1733,22 @@ to the proof of\dots
The trouble comes when we try to generalize the proof of
theorem~\ref{thm:weak-dominant-weight} we used for the case when \(\mathfrak{g}
-= \sl_3(\CC)\). The issue is that our proof relied heavily on our knowledge of
-the roots of \(\sl_3(\CC)\). Specifically, we used the fact every dominant
-integral weight of \(\sl_3(\CC)\) can be written as \(n \alpha_1 - m \alpha_3\)
+= \sl_3(k)\). The issue is that our proof relied heavily on our knowledge of
+the roots of \(\sl_3(k)\). Specifically, we used the fact every dominant
+integral weight of \(\sl_3(k)\) can be written as \(n \alpha_1 - m \alpha_3\)
for unique non-negative integers \(n\) and \(m\). When then constructed
-finite-dimensional representations \(V\) and \(W\) of \(\sl_3(\CC)\) whose
+finite-dimensional representations \(V\) and \(W\) of \(\sl_3(k)\) whose
highest weights are \(\alpha_1\) and \(- \alpha_3\), so that the highest weight
of \(\Sym^n V \otimes \Sym^m W\) is \(n \alpha_1 - m \alpha_3\).
-A similar construction can be implemented for \(\sl_n(\CC)\): if \(\mathfrak{h}
-\subset \sl_n(\CC)\) is the subalgebra of diagonal matrices -- which, as you
-may recall, is a Cartan subalgebra -- and \(\alpha : \mathfrak{h} \to \CC\) is
+A similar construction can be implemented for \(\sl_n(k)\): if \(\mathfrak{h}
+\subset \sl_n(k)\) is the subalgebra of diagonal matrices -- which, as you
+may recall, is a Cartan subalgebra -- and \(\alpha : \mathfrak{h} \to k\) is
given by \(\alpha(E_{j j}) = \delta_{i j}\), one can show that any dominant
-integral weight of \(\sl_n(\CC)\) can be uniquely expressed in the form \(k_1
+integral weight of \(\sl_n(k)\) can be uniquely expressed in the form \(k_1
\alpha_1 + k_2 (\alpha_1 + \alpha_2) + \cdots + k_{n - 1} (\alpha_1 + \cdots +
\alpha_{n - 1})\) for non-negative integers \(k_1, k_2, \ldots k_{n - 1}\). For
-instance, one may visually represent the roots of \(\sl_4(\CC)\) by
+instance, one may visually represent the roots of \(\sl_4(k)\) by
\begin{center}
\begin{tikzpicture}[scale=3]
\draw (0, 0) -- (1, 0) -- (1, 1) -- (0, 1) -- cycle;
@@ -1835,7 +1786,7 @@ instance, one may visually represent the roots of \(\sl_4(\CC)\) by
% TODO: Historical citation needed!
% TODO: Mention at the start of the chapter that we are following Weyl's
% footsteps in here
-One can then construct representations \(V_i\) of \(\sl_n(\CC)\) whose highest
+One can then construct representations \(V_i\) of \(\sl_n(k)\) whose highest
weights are \(\alpha_1 + \cdots + \alpha_i\). In fact, whenever we can find
finitely many generators of \(\beta_i\) of the set of dominant integral weights
and finite-dimensional representations \(V_i\) of \(\mathfrak{g}\) whose
@@ -1858,7 +1809,7 @@ integral weight \(\lambda\) by taking the induced representation
\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 \(V_\lambda = \CC v\) via \(H
+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
@@ -1868,7 +1819,7 @@ guarantees that \(v = 1 \otimes v \in V\) is a non-zero weight vector of
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.
-This approach has the advantage of working over fields other than \(\CC\), but
+This approach has the advantage of working over fields other than \(k\), but
in keeping with our general theme of preferring geometric proofs over purely
algebraic ones we will instead take this as an opportunity to dive into
Cartan's classification. In the next chapter we will explore the structure of