diff --git a/sections/lie-algebras.tex b/sections/lie-algebras.tex
@@ -11,7 +11,7 @@ holds for them?
\section{Semisimplicity \& Complete Reducibility}
-Let \(k\) be an algebraicly closed field of characteristic \(0\).
+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
@@ -35,7 +35,7 @@ most common definition is\dots
\end{definition}
\begin{example}
- The Lie algebras \(\sl_n(k)\) and \(\mathfrak{sp}_{2 n}(k)\) are both
+ 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}
@@ -48,21 +48,7 @@ A popular alternative to definition~\ref{thm:sesimple-algebra} is\dots
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
- \[
- \mathfrak{g} = \bigoplus_i k X_i
- \]
- 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
- \(\mathfrak{g} \cong \bigoplus_i k X_i\) as a Lie algebra.
-\end{example}
-
+% 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}.
@@ -94,18 +80,14 @@ from compact groups}. In other words\dots
% 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 \(H^1(\mathfrak{g}, V) =
-\Ext^1(k, V)\) 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 \(\mathfrak{g}\)
such that there is an exact sequence
\begin{center}
\begin{tikzcd}
- 0 \arrow{r} &
- V \arrow{r} &
- U \arrow{r} &
- W \arrow{r} &
- 0
+ 0 \arrow{r} & V \arrow{r} & U \arrow{r} & W \arrow{r} & 0
\end{tikzcd}
\end{center}
@@ -294,24 +276,24 @@ underwhelming\dots
g\).
\end{theorem}
-\section{Representations of \(\sl_2(k)\)}
+\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 \(\sl_2(k)\) with \(\dim V = n\).
+ \(V\) of \(\sl_2(K)\) with \(\dim V = n\).
\end{theorem}
The general approach we'll take is supposing \(V\) is an irreducible
-representation of \(\sl_2(k)\) and then derive some information about its
+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*}
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 \(\sl_2(k)\) 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*}
@@ -349,7 +331,7 @@ Hence
& \cdots \arrow[bend left=60]{l}
\end{tikzcd}
\end{center}
-and \(\bigoplus_{n \in \ZZ} V_{\lambda + 2 n}\) is an \(\sl_2(k)\)-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},
@@ -362,13 +344,15 @@ 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 \(\sl_2(k)\)-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
\[
\ldots, \lambda - 4, \lambda - 2, \lambda, \lambda + 2, \lambda + 4, \ldots
\]
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 \(v \in
@@ -381,15 +365,15 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
\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
+ 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
- \(\sl_2(k)\).
+ \(\sl_2(K)\).
- The fact that \(h f^k v \in k \langle v, f v, f^2 v, \ldots \rangle\) follows
+ 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 \(e f^k v \in \langle
- v, f v, f^2 v, \ldots \rangle\) is a bit more complex. Clearly,
+ -- 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
@@ -423,8 +407,8 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
\end{note}
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 \(\sl_2(k)\) in \(V\). In other words\dots
+but its significance lies in the fact that we have just provided a complete
+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
@@ -432,7 +416,7 @@ description of the action of \(\sl_2(k)\) in \(V\). In other words\dots
\end{corollary}
\begin{proof}
- If \(W\) is an irreducible representation of \(\sl_2(k)\) 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*}
@@ -485,7 +469,7 @@ 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 \(\sl_2(k)\) has the
+We now know every irreducible representation \(V\) of \(\sl_2(K)\) has the
form
\begin{center}
\begin{tikzcd}
@@ -499,7 +483,7 @@ 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
\[
- V = \bigoplus_{k = 0}^n k f^k v
+ V = \bigoplus_{k = 0}^n K f^k v
\]
and
\begin{equation}\label{eq:irr-rep-of-sl2}
@@ -515,16 +499,16 @@ 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
- \(\sl_2(k)\) 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 \(\sl_2(k)\) -- 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
- \(k^2\) 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
+ \(K^2\) 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
@@ -547,50 +531,50 @@ To conclude our analysis all it's left is to show that for each \(n\) such
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 \(\sl_2(k)\). 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 \(\sl_3(k)\), 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 \(\sl_3(k)\)}\label{sec:sl3-reps}
+\section{Representations of \(\sl_3(K)\)}\label{sec:sl3-reps}
-The study of representations of \(\sl_2(k)\) 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
-in regards to the general one. 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
+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 \(\sl_2(k)\). 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 \(\sl_2(k)\). 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 \(\sl_2(k)\) can be generalized to other
-semisimple Lie algebras, and the algebra \(\sl_3(k)\) 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 \(\sl_3(k)\) 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
-\(\sl_2(k)\). 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 \(\sl_3(k)\)?
+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 \(\sl_2(k)\) 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 \(\sl_3(k)\), 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
@@ -600,7 +584,7 @@ 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}
+ \begin{pmatrix} K & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K \end{pmatrix}
: \Tr(X) = 0
\right\}
\]
@@ -608,8 +592,8 @@ 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 \(\sl_3(k)\). This is exactly analogous to
-the situation we found in \(\sl_2(k)\): \(h\) 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 \(\lambda : \mathfrak{h} \to k\) such that \(H v = \lambda(H) \cdot
v\) for each \(H \in \mathfrak{h}\) and some non-zero \(v \in V\). We call such
@@ -627,10 +611,10 @@ 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 chapter.
+but we will postpone its proof to the next section.
-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
+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,
@@ -640,10 +624,9 @@ 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
-\(\sl_2(k)\)-representation differ from one another by multiples of \(\pm
-2\).
+\(\sl_2(K)\)-representation differ from one another by multiples of \(\pm 2\).
-In the case of \(\sl_3(k)\), 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 -
@@ -689,8 +672,8 @@ Visually we may draw
\end{figure}
If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) in
-\(\sl_3(k)\) associated to \(\alpha\) by \(\sl_3(k)_\alpha\) and fix some
-\(X \in \sl_3(k)_\alpha\), \(H \in \mathfrak{h}\) and \(v \in V_\lambda\)
+\(\sl_3(K)\) 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}
@@ -704,8 +687,8 @@ so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words,
\(\sl_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors between
eigenspaces}.
-For instance \(\sl_3(k)_{\alpha_1 - \alpha_3}\) will act on the adjoint
-representation of \(\sl_3(k)\) 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]
@@ -725,19 +708,19 @@ representation of \(\sl_3(k)\) via
\end{tikzpicture}
\end{figure}
-This is again entirely analogous to the situation we observed in
-\(\sl_2(k)\). In fact, we may once more conclude\dots
+This is again entirely analogous to the situation we observed in \(\sl_2(K)\).
+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
- \(\sl_3(k)\)-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 \(\sl_3(k)\).
+ adjoint action of \(\mathfrak{h}\) in \(\sl_3(K)\).
\end{theorem}
\begin{proof}
This proof goes exactly as that of the analogous statement for
- \(\sl_2(k)\): 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}
@@ -750,7 +733,7 @@ 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 \(\sl_3(k)\), 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
@@ -758,36 +741,36 @@ adjoint action of \(\mathfrak{h}\).
\end{definition}
It's clear from our previous discussion that the weights of the adjoint
-representation of \(\sl_3(k)\) deserve some special attention.
+representation of \(\sl_3(K)\) deserve some special attention.
\begin{definition}
- The weights of the adjoint representation of \(\sl_3(k)\) are called
- \emph{roots of \(\sl_3(k)\)}. 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 \(\sl_3(k)\) 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 \(\sl_3(k)\).
+ \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 \(\sl_3(k)\)}.
+ is called \emph{the root lattice of \(\sl_3(K)\)}.
\end{definition}
-% TODO: This doesn't make any sence in an arbitrary field
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 \(\sl_2(k)\) in terms of this vector. We may
-reproduce these steps in the context of \(\sl_3(k)\) 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.
+% 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
@@ -841,33 +824,34 @@ sort of \(\frac{1}{3}\)-plane with corners at \(\lambda\), as shown in
\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 \(\sl_3(k)\) 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
+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 \(V_{\lambda + \alpha_2 - \alpha_3}\)
+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 \(\sl_3(k)\).
+ spaces of \(\sl_3(K)\).
\end{theorem}
\begin{proof}
- It suffices to note that the positive roots of \(\sl_3(k)\) 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
-\(\sl_2(k)\), 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 \(\sl_2(k)\) 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
-\(\sl_3(k)\) -- 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.
@@ -906,14 +890,14 @@ 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
-\(\sl_2(k)\) 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 \(\sl_2(k)\). 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}
@@ -956,7 +940,7 @@ thus
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 \(\sl_2(k)\), so that the weights in
+E_{3 2}]\), which is again isomorphic to \(\sl_2(K)\), 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}
@@ -985,8 +969,8 @@ 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 \(\mathfrak{s}_{\alpha_i -
-\alpha_j} = k \langle E_{i j}, E_{j i}, [E_{i j}, E_{j i}] \rangle\), which
-is once more isomorphic to \(\sl_2(k)\), and again the weight spaces in this
+\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}
@@ -1021,7 +1005,7 @@ 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 \(\sl_2(k)\) 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}
@@ -1060,8 +1044,8 @@ of \(V\) we started with.
\end{center}
By construction, \(\nu\) corresponds to the right-most weight of the
-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
+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}
@@ -1101,7 +1085,7 @@ 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 \(\sl_2(k)\), so it must be an integer. Now since
+representation of \(\sl_2(K)\), so it must be an integer. Now since
\[
\lambda
\begin{pmatrix}
@@ -1132,7 +1116,7 @@ P\).
\begin{definition}
The lattice \(P = \ZZ \alpha_1 \oplus \ZZ \alpha_2 \oplus \ZZ \alpha_3\) is
- called \emph{the weight lattice of \(\sl_3(k)\)}.
+ called \emph{the weight lattice of \(\sl_3(K)\)}.
\end{definition}
Finally\dots
@@ -1145,8 +1129,8 @@ Finally\dots
0\).
\end{theorem}
-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
+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
@@ -1155,36 +1139,36 @@ 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 \(\sl_3(k)\) 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 \(\sl_2(k)\): 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
+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 \(\sl_2(k)\) 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 \(\sl_2(k)\). The advantage of this alternative
-formulation is, of course, that the same holds for \(\sl_3(k)\).
+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 \(\sl_3(k)\)-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
-\(\sl_3(k)\) 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 \(\sl_3(k)\) -- 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 \(\sl_3(k)\) 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 \(\lambda\).
@@ -1197,7 +1181,7 @@ 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 \(V = k^3\) of \(\sl_3(k)\). 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 \(n \alpha_1 - m \alpha_3\).
@@ -1242,7 +1226,7 @@ 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 \(\sl_3(k)\)-representations \(U\) and
+ On the other hand if we fix two \(\sl_3(K)\)-representations \(U\) and
\(W\), by computing
\[
\begin{split}
@@ -1267,7 +1251,7 @@ 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 \(\sl_3(k)\) 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 \(W\).
@@ -1284,21 +1268,21 @@ 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 \(U = \sl_3(k) \ v + w\) 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 \(\sl_3(k)\) must be
- isomorphism. Finally,
+ irreducible representation \(U = \sl_3(K) \cdot v + w\) 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 \(\sl_3(K)\)
+ 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 \(\sl_2(k)\) 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 \(\sl_2(k)\) and \(\sl_3(k)\), 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
@@ -1310,15 +1294,15 @@ algebra \(\mathfrak{g}\).
\section{Simultaneous Diagonalization \& the General Case}
-At the heart of our analysis of \(\sl_2(k)\) and \(\sl_3(k)\) 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 \(\sl_2(k)\), 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 \(\sl_3(k)\). 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
@@ -1331,15 +1315,15 @@ 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 \(\sl_2(k)\) 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 \(\mathfrak{h} = k h \subset \sl_2(k)\).
-Furthermore, in both cases \(\mathfrak{h} \subset \sl_n(k)\) 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
-\(\mathfrak{h} \subset \sl_2(k)\) 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 \(\sl_3(k)\)?
+\(\mathfrak{h} \subset \sl_2(K)\) 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 \(\sl_3(K)\)?
% TODO: Rewrite this: we haven't dealt with finite groups at all
The rational behind fixing an Abelian subalgebra is one we have already
@@ -1351,7 +1335,9 @@ 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 adjunt the comment on maximal tori
+% 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.
@@ -1375,18 +1361,18 @@ readily check that every pair of diagonal matrices commutes, so that
\[
\mathfrak{h} =
\begin{pmatrix}
- k & 0 & \cdots & 0 \\
- 0 & k & \cdots & 0 \\
+ K & 0 & \cdots & 0 \\
+ 0 & K & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
- 0 & 0 & \cdots & k
+ 0 & 0 & \cdots & K
\end{pmatrix}
\]
-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
+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 \(\gl_n(k)\). 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 \(n = 3\)
+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
@@ -1419,12 +1405,12 @@ 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
\[
- \mathfrak{h} \cong \bigoplus_i k H_i
+ \mathfrak{h} \cong \bigoplus_i K H_i
\]
as vector spaces. Usually this is simply a linear isomorphism, but since
\(\mathfrak{h}\) is Abelian this is an isomorphism of Lie algebras -- here
- \(k H_i\) represents the 1-dimensional subalgebra spanned by \(H_i\), which
- is isomorphic to the trivial Lie algebra \(k\). Each \(k H_i\) is simple,
+ \(K H_i\) 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.
@@ -1439,7 +1425,7 @@ What is simultaneous diagonalization all about then?
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
+ \(\mathfrak{h}\) is equivalent to the statement that each \(K v_i\) is
stable under the action of \(\mathfrak{h}\).
\end{proof}
@@ -1464,14 +1450,16 @@ 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 \(\sl_2(k)\) and
-\(\sl_3(k)\), 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.
-% TODO: Add a proof? The proof of FH turns out to be recursive!!!!
-% TODO: Note that this is where the maximality of the Cartan subalgebra comes
-% into play
+% 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
@@ -1519,8 +1507,7 @@ of the following theorem.
\(\mathfrak{g}\), which is zero by the semisimplicity -- a contradiction.
\end{proof}
-Furthermore, as in the case of \(\sl_2(k)\) and \(\sl_3(k)\) one can
-show\dots
+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}
The eigenspaces \(\mathfrak{g}_\alpha\) are all 1-dimensional.
@@ -1548,7 +1535,8 @@ then\dots
all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\).
\end{theorem}
-To proceed further, as in the case of \(\sl_3(k)\) we have to fix a direction
+% TODO: Rewrite this: the concept of direct has no sence in the general setting
+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
@@ -1571,7 +1559,7 @@ 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 \(\sl_3(k)\). 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
@@ -1608,7 +1596,7 @@ 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 \(\sl_2(k)\), 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
@@ -1642,7 +1630,9 @@ is\dots
\(\mathfrak{g}\)}.
\end{definition}
-This is entirely analogous to the situation of \(\sl_3(k)\), where we found
+% 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 \(\sl_3(K)\), 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
@@ -1673,7 +1663,7 @@ 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 \(\sl_3(k)\), 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}
@@ -1683,8 +1673,8 @@ 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 \(\mathfrak{g}
-= \sl_3(k)\). The issue is that our proof relied heavily on our knowledge of
-the roots of \(\sl_3(k)\). Instead, we need a new strategy for the general
+= \sl_3(K)\). The issue is that our proof relied heavily on our knowledge of
+the roots of \(\sl_3(K)\). Instead, we need a new strategy for the general
setting.
% TODO: Add further details. turn this into a proper proof?
@@ -1696,7 +1686,7 @@ 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 \(V_\lambda = k v\) 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