diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -35,9 +35,9 @@ most common definition is\dots
\end{definition}
\begin{example}
- The Lie algebras \(\mathfrak{sl}_n(K)\) and \(\mathfrak{sp}_{2 n}(K)\) are both
- semisimple -- see the section of \cite{kirillov} on invariant bilinear forms
- and the semisimplicity of classical Lie algebras.
+ The Lie algebras \(\mathfrak{sl}_n(K)\) and \(\mathfrak{sp}_{2 n}(K)\) are
+ both semisimple -- see the section of \cite{kirillov} on invariant bilinear
+ forms and the semisimplicity of classical Lie algebras.
\end{example}
A popular alternative to definition~\ref{thm:sesimple-algebra} is\dots
@@ -51,9 +51,9 @@ A popular alternative to definition~\ref{thm:sesimple-algebra} is\dots
% TODO: Remove the reference to compact algebras
I suppose this last definition explains the nomenclature, but what does any of
this have to do with complete reducibility? Well, the special thing about
-semisimple Lie algebras is that they are \emph{compact algebras}.
-Compact Lie algebras are, as you might have guessed, \emph{algebras that come
-from compact groups}. In other words\dots
+semisimple Lie algebras is that they are \emph{compact algebras}. Compact Lie
+algebras are, as you might have guessed, \emph{algebras that come from compact
+groups}. In other words\dots
\begin{theorem}
Every representation of a semisimple Lie algebra is completely reducible.
@@ -82,9 +82,9 @@ Alternatively, one could prove the same statement in a purely algebraic manner
by showing the first Lie algebra cohomology group \(H^1(\mathfrak{g}, V) =
\operatorname{Ext}^1(K, V)\) vanishes for all \(V\), as do \cite{kirillov} and
\cite{lie-groups-serganova-student} in their proofs. More precisely, one can
-show that there is a natural bijection between \(H^1(\mathfrak{g}, \operatorname{Hom}(V,
-W))\) and isomorphism classes of the representations \(U\) of \(\mathfrak{g}\)
-such that there is an exact sequence
+show that there is a natural bijection between \(H^1(\mathfrak{g},
+\operatorname{Hom}(V, W))\) and isomorphism classes of the representations
+\(U\) of \(\mathfrak{g}\) such that there is an exact sequence
\begin{center}
\begin{tikzcd}
0 \arrow{r} & V \arrow{r} & U \arrow{r} & W \arrow{r} & 0
@@ -261,19 +261,19 @@ sequence
0
\end{tikzcd}
\end{center}
-where \(\operatorname{Rad}(\mathfrak g)\) is the sum of all solvable ideals of \(\mathfrak
-g\) -- i.e. a maximal solvable ideal -- for arbitrary \(\mathfrak g\).
-This implies we can deduce information about the representations of \(\mathfrak
-g\) by studying those of its semisimple part \(\mfrac{\mathfrak
-g}{\operatorname{Rad}(\mathfrak g)}\). In practice though, this isn't quite satisfactory
-because the exactness of this last sequence translates to the
+where \(\operatorname{Rad}(\mathfrak g)\) is the sum of all solvable ideals of
+\(\mathfrak g\) -- i.e. a maximal solvable ideal -- for arbitrary \(\mathfrak
+g\). This implies we can deduce information about the representations of
+\(\mathfrak g\) by studying those of its semisimple part \(\mfrac{\mathfrak
+g}{\operatorname{Rad}(\mathfrak g)}\). In practice though, this isn't quite
+satisfactory because the exactness of this last sequence translates to the
underwhelming\dots
\begin{theorem}\label{thm:semi-simple-part-decomposition}
Every irreducible representation of \(\mathfrak g\) is the tensor product of
an irreducible representation of its semisimple part \(\mfrac{\mathfrak
- g}{\operatorname{Rad}(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak
- g\).
+ g}{\operatorname{Rad}(\mathfrak g)}\) and a one-dimensional representation of
+ \(\mathfrak g\).
\end{theorem}
\section{Representations of \(\mathfrak{sl}_2(K)\)}
@@ -286,8 +286,8 @@ The primary goal of this section is proving\dots
\end{theorem}
The general approach we'll take is supposing \(V\) is an irreducible
-representation of \(\mathfrak{sl}_2(K)\) and then derive some information about its
-structure. We begin our analysis by pointing out that the elements
+representation of \(\mathfrak{sl}_2(K)\) and then derive some information about
+its structure. We begin our analysis by pointing out that the elements
\begin{align*}
e & = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} &
f & = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} &
@@ -331,8 +331,8 @@ Hence
& \cdots \arrow[bend left=60]{l}
\end{tikzcd}
\end{center}
-and \(\bigoplus_{n \in \ZZ} V_{\lambda + 2 n}\) is an \(\mathfrak{sl}_2(K)\)-invariant
-subspace. This implies
+and \(\bigoplus_{n \in \ZZ} V_{\lambda + 2 n}\) is an
+\(\mathfrak{sl}_2(K)\)-invariant subspace. This implies
\[
V = \bigoplus_{n \in \ZZ} V_{\lambda + 2 n},
\]
@@ -344,8 +344,8 @@ 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 \(\mathfrak{sl}_2(K)\)-invariant subspace, so that the eigenvalues of \(h\)
-form an unbroken string
+is also an \(\mathfrak{sl}_2(K)\)-invariant subspace, so that the eigenvalues
+of \(h\) form an unbroken string
\[
\ldots, \lambda - 4, \lambda - 2, \lambda, \lambda + 2, \lambda + 4, \ldots
\]
@@ -408,7 +408,8 @@ 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 its significance lies in the fact that we have just provided a complete
-description of the action of \(\mathfrak{sl}_2(K)\) in \(V\). In other words\dots
+description of the action of \(\mathfrak{sl}_2(K)\) in \(V\). In other
+words\dots
\begin{corollary}
\(V\) is completely determined by the right-most eigenvalue \(\lambda\) of
@@ -469,8 +470,8 @@ 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 \(\mathfrak{sl}_2(K)\) has the
-form
+We now know every irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\)
+has the form
\begin{center}
\begin{tikzcd}
\cdots \arrow[bend left=60]{r}
@@ -504,15 +505,15 @@ To conclude our analysis all it's left is to show that for each \(n\) such
\begin{proof}
The fact the representation \(V\) from the previous discussion exists is
- clear from the commutator relations of \(\mathfrak{sl}_2(K)\) -- just look at \(f^k
- v\) as abstract symbols and impose the action given by
+ clear from the commutator relations of \(\mathfrak{sl}_2(K)\) -- just look at
+ \(f^k v\) as abstract symbols and impose the action given by
(\ref{eq:irr-rep-of-sl2}). Alternatively, one can readily check that if
- \(K^2\) is the natural representation of \(\mathfrak{sl}_2(K)\), then \(V = \operatorname{Sym}^n
- K^2\) satisfies the relations of (\ref{eq:irr-rep-of-sl2}). To see that
- \(V\) is irreducible let \(W\) be a non-zero subrepresentation and take some
- non-zero \(w \in W\). Suppose \(w = \alpha_0 v + \alpha_1 f v + \cdots +
- \alpha_n f^n v\) and let \(k\) be the lowest index such that \(\alpha_k \ne
- 0\), so that
+ \(K^2\) is the natural representation of \(\mathfrak{sl}_2(K)\), then \(V =
+ \operatorname{Sym}^n K^2\) satisfies the relations of
+ (\ref{eq:irr-rep-of-sl2}). To see that \(V\) is irreducible let \(W\) be a
+ non-zero subrepresentation and take some non-zero \(w \in W\). Suppose \(w =
+ \alpha_0 v + \alpha_1 f v + \cdots + \alpha_n f^n v\) and let \(k\) be the
+ lowest index such that \(\alpha_k \ne 0\), so that
\[
w = \alpha_k f^k v + \cdots + \alpha_n f^n v
\]
@@ -531,55 +532,57 @@ 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 \(\mathfrak{sl}_2(K)\). It is not yet clear,
-however, if any of this can be adapted to a general setting. In the following
-section we shall double down on our gamble by trying to reproduce some of the
-results of this section for \(\mathfrak{sl}_3(K)\), hoping this will \emph{somehow}
-lead us to a general solution. In the process of doing so we'll learn a bit
-more why \(h\) was a sure bet and the race was fixed all along.
+\emph{all} irreducible representations of \(\mathfrak{sl}_2(K)\). It is not yet
+clear, however, if any of this can be adapted to a general setting. In the
+following section we shall double down on our gamble by trying to reproduce
+some of the results of this section for \(\mathfrak{sl}_3(K)\), hoping this
+will \emph{somehow} lead us to a general solution. In the process of doing so
+we'll learn a bit more why \(h\) was a sure bet and the race was fixed all
+along.
\section{Representations of \(\mathfrak{sl}_3(K)\)}\label{sec:sl3-reps}
-The study of representations of \(\mathfrak{sl}_2(K)\) reminds me of the difference the
-derivative of a function \(\RR \to \RR\) and that of a smooth map between
-manifolds: it's a simpler case of something greater, but in some sense it's too
-simple of a case, and the intuition we acquire from it can be a bit misleading
-in regards to the general setting. For instance I distinctly remember my
-Calculus I teacher telling the class ``the derivative of the composition of two
-functions is not the composition of their derivatives'' -- which is, of course,
-the \emph{correct} formulation of the chain rule in the context of smooth
-manifolds.
-
-The same applies to \(\mathfrak{sl}_2(K)\). It's a simple and beautiful example, but
-unfortunately the general picture -- representations of arbitrary semisimple
-algebras -- lacks its simplicity, and, of course, much of this complexity is
-hidden in the case of \(\mathfrak{sl}_2(K)\). The general purpose of this section is
-to investigate to which extent the framework used in the previous section to
-classify the representations of \(\mathfrak{sl}_2(K)\) can be generalized to other
-semisimple Lie algebras, and the algebra \(\mathfrak{sl}_3(K)\) stands as a natural
-candidate for potential generalizations: \(3 = 2 + 1\) after all.
-
-Our approach is very straightforward: we'll fix some irreducible
-representation \(V\) of \(\mathfrak{sl}_3(K)\) and proceed step by step, at each point
-asking ourselves how we could possibly adapt the framework we laid out for
+The study of representations of \(\mathfrak{sl}_2(K)\) reminds me of the
+difference the derivative of a function \(\RR \to \RR\) and that of a smooth
+map between manifolds: it's a simpler case of something greater, but in some
+sense it's too simple of a case, and the intuition we acquire from it can be a
+bit misleading in regards to the general setting. For instance I distinctly
+remember my Calculus I teacher telling the class ``the derivative of the
+composition of two functions is not the composition of their derivatives'' --
+which is, of course, the \emph{correct} formulation of the chain rule in the
+context of smooth manifolds.
+
+The same applies to \(\mathfrak{sl}_2(K)\). It's a simple and beautiful
+example, but unfortunately the general picture -- representations of arbitrary
+semisimple algebras -- lacks its simplicity, and, of course, much of this
+complexity is hidden in the case of \(\mathfrak{sl}_2(K)\). The general
+purpose of this section is to investigate to which extent the framework used in
+the previous section to classify the representations of \(\mathfrak{sl}_2(K)\)
+can be generalized to other semisimple Lie algebras, and the algebra
+\(\mathfrak{sl}_3(K)\) stands as a natural candidate for potential
+generalizations: \(3 = 2 + 1\) after all.
+
+Our approach is very straightforward: we'll fix some irreducible representation
+\(V\) of \(\mathfrak{sl}_3(K)\) and proceed step by step, at each point asking
+ourselves how we could possibly adapt the framework we laid out for
\(\mathfrak{sl}_2(K)\). The first obvious question is one we have already asked
ourselves: why \(h\)? More specifically, why did we choose to study its
eigenvalues and is there an analogue of \(h\) in \(\mathfrak{sl}_3(K)\)?
-The answer to the former question is one we'll discuss at length in the
-next chapter, but for now we note that perhaps the most fundamental
-property of \(h\) is that \emph{there exists an eigenvector \(v\) of
-\(h\) that is annihilated by \(e\)} -- that being the generator of the
-right-most eigenspace of \(h\). This was instrumental to our explicit
-description of the irreducible representations of \(\mathfrak{sl}_2(K)\) culminating in
+The answer to the former question is one we'll discuss at length in the next
+chapter, but for now we note that perhaps the most fundamental property of
+\(h\) is that \emph{there exists an eigenvector \(v\) of \(h\) that is
+annihilated by \(e\)} -- that being the generator of the right-most eigenspace
+of \(h\). This was instrumental to our explicit description of the irreducible
+representations of \(\mathfrak{sl}_2(K)\) culminating in
theorem~\ref{thm:irr-rep-of-sl2-exists}.
-Our fist task is to find some analogue of \(h\) in \(\mathfrak{sl}_3(K)\), but it's
-still unclear what exactly we are looking for. We could say we're looking for
-an element of \(V\) that is annihilated by some analogue of \(e\), but the
+Our fist task is to find some analogue of \(h\) in \(\mathfrak{sl}_3(K)\), but
+it's still unclear what exactly we are looking for. We could say we're looking
+for an element of \(V\) that is annihilated by some analogue of \(e\), but the
meaning of \emph{some analogue of \(e\)} is again unclear. In fact, as we shall
-see, no such analogue exists and neither does such element. Instead, the
-actual way to proceed is to consider the subalgebra
+see, no such analogue exists and neither does such element. Instead, the actual
+way to proceed is to consider the subalgebra
\[
\mathfrak h
= \left\{
@@ -592,13 +595,14 @@ 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 \(\mathfrak{sl}_3(K)\). This is exactly analogous to
-the situation we found in \(\mathfrak{sl}_2(K)\): \(h\) corresponds to the subalgebra
-\(\mathfrak{h}\), and the eigenvalues of \(h\) in turn correspond to linear
-functions \(\lambda : \mathfrak{h} \to k\) such that \(H v = \lambda(H) \cdot
-v\) for each \(H \in \mathfrak{h}\) and some non-zero \(v \in V\). We call such
-functionals \(\lambda\) \emph{eigenvalues of \(\mathfrak{h}\)}, and we say
-\emph{\(v\) is an eigenvector of \(\mathfrak h\)}.
+half of the remaining elements of \(\mathfrak{sl}_3(K)\). This is exactly
+analogous to the situation we found in \(\mathfrak{sl}_2(K)\): \(h\)
+corresponds to the subalgebra \(\mathfrak{h}\), and the eigenvalues of \(h\) in
+turn correspond to linear functions \(\lambda : \mathfrak{h} \to k\) such that
+\(H v = \lambda(H) \cdot v\) for each \(H \in \mathfrak{h}\) and some non-zero
+\(v \in V\). We call such functionals \(\lambda\) \emph{eigenvalues of
+\(\mathfrak{h}\)}, and we say \emph{\(v\) is an eigenvector of \(\mathfrak
+h\)}.
Once again, we'll pay special attention to the eigenvalue decomposition
\begin{equation}\label{eq:weight-module}
@@ -613,22 +617,24 @@ 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 section.
-Next we turn our attention to the remaining elements of \(\mathfrak{sl}_3(K)\). In our
-analysis of \(\mathfrak{sl}_2(K)\) we saw that the eigenvalues of \(h\) differed from
-one another by multiples of \(2\). A possible way to interpret this is to say
-\emph{the eigenvalues of \(h\) differ from one another by integral linear
-combinations of the eigenvalues of the adjoint action of \(h\)}. In English,
-the eigenvalues of of the adjoint actions of \(h\) are \(\pm 2\) since
+Next we turn our attention to the remaining elements of \(\mathfrak{sl}_3(K)\).
+In our analysis of \(\mathfrak{sl}_2(K)\) we saw that the eigenvalues of \(h\)
+differed from one another by multiples of \(2\). A possible way to interpret
+this is to say \emph{the eigenvalues of \(h\) differ from one another by
+integral linear combinations of the eigenvalues of the adjoint action of
+\(h\)}. In English, the eigenvalues of of the adjoint actions of \(h\) are
+\(\pm 2\) since
\begin{align*}
[h, f] & = -2 f &
[h, e] & = 2 e
\end{align*}
and the eigenvalues of the action of \(h\) in an irreducible
-\(\mathfrak{sl}_2(K)\)-representation differ from one another by multiples of \(\pm 2\).
+\(\mathfrak{sl}_2(K)\)-representation differ from one another by multiples of
+\(\pm 2\).
-In the case of \(\mathfrak{sl}_3(K)\), a simple calculation shows that if \([H, X]\) is
-scalar multiple of \(X\) for all \(H \in \mathfrak{h}\) then all but one entry
-of \(X\) are zero. Hence the eigenvectors of the adjoint action of
+In the case of \(\mathfrak{sl}_3(K)\), a simple calculation shows that if \([H,
+X]\) is scalar multiple of \(X\) for all \(H \in \mathfrak{h}\) then all but
+one entry of \(X\) are zero. Hence the eigenvectors of the adjoint action of
\(\mathfrak{h}\) are \(E_{i j}\) and its eigenvalues are \(\alpha_i -
\alpha_j\), where
\[
@@ -672,9 +678,9 @@ Visually we may draw
\end{figure}
If we denote the eigenspace of the adjoint action of \(\mathfrak{h}\) in
-\(\mathfrak{sl}_3(K)\) associated to \(\alpha\) by \(\mathfrak{sl}_3(K)_\alpha\) and fix some
-\(X \in \mathfrak{sl}_3(K)_\alpha\), \(H \in \mathfrak{h}\) and \(v \in V_\lambda\)
-then
+\(\mathfrak{sl}_3(K)\) associated to \(\alpha\) by
+\(\mathfrak{sl}_3(K)_\alpha\) and fix some \(X \in \mathfrak{sl}_3(K)_\alpha\),
+\(H \in \mathfrak{h}\) and \(v \in V_\lambda\) then
\[
\begin{split}
H (X v)
@@ -684,11 +690,11 @@ then
\end{split}
\]
so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words,
-\(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors between
-eigenspaces}.
+\(\mathfrak{sl}_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors
+between eigenspaces}.
-For instance \(\mathfrak{sl}_3(K)_{\alpha_1 - \alpha_3}\) will act on the adjoint
-representation of \(\mathfrak{sl}_3(K)\) via
+For instance \(\mathfrak{sl}_3(K)_{\alpha_1 - \alpha_3}\) will act on the
+adjoint representation of \(\mathfrak{sl}_3(K)\) via
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=2.5]
@@ -708,13 +714,13 @@ representation of \(\mathfrak{sl}_3(K)\) via
\end{tikzpicture}
\end{figure}
-This is again entirely analogous to the situation we observed in \(\mathfrak{sl}_2(K)\).
-In fact, we may once more conclude\dots
+This is again entirely analogous to the situation we observed in
+\(\mathfrak{sl}_2(K)\). In fact, we may once more conclude\dots
\begin{theorem}\label{thm:sl3-weights-congruent-mod-root}
The eigenvalues of the action of \(\mathfrak{h}\) in an irreducible
- \(\mathfrak{sl}_3(K)\)-representation \(V\) differ from one another by integral
- linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of
+ \(\mathfrak{sl}_3(K)\)-representation \(V\) differ from one another by
+ integral linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of
adjoint action of \(\mathfrak{h}\) in \(\mathfrak{sl}_3(K)\).
\end{theorem}
@@ -733,9 +739,9 @@ eigenvalues of the action of \(\mathfrak{h}\) in \(V\) and eigenvalues of the
adjoint action of \(\mathfrak{h}\).
\begin{definition}
- Given a representation \(V\) of \(\mathfrak{sl}_3(K)\), we'll call the non-zero
- eigenvalues of the action of \(\mathfrak{h}\) in \(V\) \emph{weights of
- \(V\)}. As you might have guessed, we'll correspondingly refer to
+ Given a representation \(V\) of \(\mathfrak{sl}_3(K)\), we'll call the
+ non-zero eigenvalues of the action of \(\mathfrak{h}\) in \(V\) \emph{weights
+ of \(V\)}. As you might have guessed, we'll correspondingly refer to
eigenvectors and eigenspaces of a given weight by \emph{weight vectors} and
\emph{weight spaces}.
\end{definition}
@@ -744,17 +750,17 @@ It's clear from our previous discussion that the weights of the adjoint
representation of \(\mathfrak{sl}_3(K)\) deserve some special attention.
\begin{definition}
- The weights of the adjoint representation of \(\mathfrak{sl}_3(K)\) are called
- \emph{roots of \(\mathfrak{sl}_3(K)\)}. Once again, the expressions \emph{root
- vector} and \emph{root space} are self-explanatory.
+ The weights of the adjoint representation of \(\mathfrak{sl}_3(K)\) are
+ called \emph{roots of \(\mathfrak{sl}_3(K)\)}. Once again, the expressions
+ \emph{root vector} and \emph{root space} are self-explanatory.
\end{definition}
Theorem~\ref{thm:sl3-weights-congruent-mod-root} can thus be restated as\dots
\begin{corollary}
- The weights of an irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) are all
- congruent module the lattice \(Q\) generated by the roots \(\alpha_i -
- \alpha_j\) of \(\mathfrak{sl}_3(K)\).
+ The weights of an irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\)
+ are all congruent module the lattice \(Q\) generated by the roots \(\alpha_i
+ - \alpha_j\) of \(\mathfrak{sl}_3(K)\).
\end{corollary}
\begin{definition}
@@ -765,10 +771,11 @@ Theorem~\ref{thm:sl3-weights-congruent-mod-root} can thus be restated as\dots
To proceed we once more refer to the previously established framework: next we
saw that the eigenvalues of \(h\) formed an unbroken string of integers
symmetric around \(0\). To prove this we analyzed the right-most eigenvalue of
-\(h\) and its eigenvector, providing an explicit description of the
-irreducible representation of \(\mathfrak{sl}_2(K)\) in terms of this vector. We may
-reproduce these steps in the context of \(\mathfrak{sl}_3(K)\) by fixing a direction in
-the place an considering the weight lying the furthest in that direction.
+\(h\) and its eigenvector, providing an explicit description of the irreducible
+representation of \(\mathfrak{sl}_2(K)\) in terms of this vector. We may
+reproduce these steps in the context of \(\mathfrak{sl}_3(K)\) by fixing a
+direction in the place an considering the weight lying the furthest in that
+direction.
% TODO: This doesn't make any sence in a field other than C
In practice this means we'll choose a linear functional \(f : \mathfrak{h}^*
@@ -827,12 +834,12 @@ sort of \(\frac{1}{3}\)-plane with corners at \(\lambda\), as shown in
% TODO: Rewrite this: we haven't chosen any line
Indeed, if this is not the case then, by definition, \(\lambda\) is not the
furthest weight along the line we chose. Given our previous assertion that the
-root spaces of \(\mathfrak{sl}_3(K)\) act on the weight spaces of \(V\) via translation,
-this implies that \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\) all annihilate
-\(V_\lambda\), or otherwise one of \(V_{\lambda + \alpha_1 - \alpha_2}\),
-\(V_{\lambda + \alpha_1 - \alpha_3}\) and \(V_{\lambda + \alpha_2 - \alpha_3}\)
-would be non-zero -- which contradicts the hypothesis that \(\lambda\) lies the
-furthest along the direction we chose. In other words\dots
+root spaces of \(\mathfrak{sl}_3(K)\) act on the weight spaces of \(V\) via
+translation, this implies that \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\) all
+annihilate \(V_\lambda\), or otherwise one of \(V_{\lambda + \alpha_1 -
+\alpha_2}\), \(V_{\lambda + \alpha_1 - \alpha_3}\) and \(V_{\lambda + \alpha_2
+- \alpha_3}\) would be non-zero -- which contradicts the hypothesis that
+\(\lambda\) lies the furthest along the direction we chose. In other words\dots
\begin{theorem}
There is a weight vector \(v \in V\) that is killed by all positive root
@@ -840,20 +847,21 @@ furthest along the direction we chose. In other words\dots
\end{theorem}
\begin{proof}
- It suffices to note that the positive roots of \(\mathfrak{sl}_3(K)\) are precisely
- \(\alpha_1 - \alpha_2\), \(\alpha_1 - \alpha_3\) and \(\alpha_2 - \alpha_3\).
+ It suffices to note that the positive roots of \(\mathfrak{sl}_3(K)\) are
+ precisely \(\alpha_1 - \alpha_2\), \(\alpha_1 - \alpha_3\) and \(\alpha_2 -
+ \alpha_3\).
\end{proof}
We call \(\lambda\) \emph{the highest weight of \(V\)}, and we call any \(v \in
V_\lambda\) \emph{a highest weight vector}. Going back to the case of
-\(\mathfrak{sl}_2(K)\), we then constructed an explicit basis of our irreducible
-representations in terms of a highest weight vector, which allowed us to
-provide an explicit description of the action of \(\mathfrak{sl}_2(K)\) in terms of
-its standard basis and finally we concluded that the eigenvalues of \(h\) must
-be symmetrical around \(0\). An analogous procedure could be implemented for
-\(\mathfrak{sl}_3(K)\) -- and indeed that's what we'll do later down the line -- but
-instead we would like to focus on the problem of finding the weights of \(V\)
-for the moment.
+\(\mathfrak{sl}_2(K)\), we then constructed an explicit basis of our
+irreducible representations in terms of a highest weight vector, which allowed
+us to provide an explicit description of the action of \(\mathfrak{sl}_2(K)\)
+in terms of its standard basis and finally we concluded that the eigenvalues of
+\(h\) must be symmetrical around \(0\). An analogous procedure could be
+implemented for \(\mathfrak{sl}_3(K)\) -- and indeed that's what we'll do later
+down the line -- but instead we would like to focus on the problem of finding
+the weights of \(V\) for the moment.
We'll start out by trying to understand the weights in the boundary of
\(\frac{1}{3}\)-plane previously drawn. Since the root spaces act by
@@ -897,8 +905,8 @@ What's remarkable about all this is the fact that the subalgebra spanned by
H & \mapsto h
\end{align*}
-In other words, \(W\) is a representation of \(\mathfrak{sl}_2(K)\). Even more so, we
-claim
+In other words, \(W\) is a representation of \(\mathfrak{sl}_2(K)\). Even more
+so, we claim
\[
V_{\lambda + k (\alpha_2 - \alpha_1)} = W_{\lambda(H) - 2k}
\]
@@ -1005,9 +1013,9 @@ 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 \(\mathfrak{sl}_2(K)\) whose weights correspond to weights of \(V\) lying in a
-string inside the hexagon, and whose right-most weight is precisely the weight
-of \(V\) we started with.
+of \(\mathfrak{sl}_2(K)\) whose weights correspond to weights of \(V\) lying in
+a string inside the hexagon, and whose right-most weight is precisely the
+weight of \(V\) we started with.
\begin{center}
\begin{tikzpicture}
\AutoSizeWeightLatticefalse
@@ -1044,9 +1052,9 @@ of \(V\) we started with.
\end{center}
By construction, \(\nu\) corresponds to the right-most weight of the
-representation of \(\mathfrak{sl}_2(K)\), so that all dots lying on the gray string
-must occur in the representation of \(\mathfrak{sl}_2(K)\). Hence they must also be
-weights of \(V\). The final picture is thus
+representation of \(\mathfrak{sl}_2(K)\), so that all dots lying on the gray
+string must occur in the representation of \(\mathfrak{sl}_2(K)\). Hence they
+must also be weights of \(V\). The final picture is thus
\begin{center}
\begin{tikzpicture}
\AutoSizeWeightLatticefalse
@@ -1128,9 +1136,9 @@ Finally\dots
reflections across the lines \(B(\alpha_i - \alpha_j, \alpha) = 0\).
\end{theorem}
-Once more there's a clear parallel between the case of \(\mathfrak{sl}_3(K)\) and that
-of \(\mathfrak{sl}_2(K)\), where we observed that the weights all lied in the lattice
-\(P = \ZZ\) and were congruent modulo the lattice \(Q = 2 \ZZ\).
+Once more there's a clear parallel between the case of \(\mathfrak{sl}_3(K)\)
+and that of \(\mathfrak{sl}_2(K)\), where we observed that the weights all lied
+in the lattice \(P = \ZZ\) and were congruent modulo the lattice \(Q = 2 \ZZ\).
Having found all of the weights of \(V\), the only thing we're missing is an
existence and uniqueness theorem analogous to
theorem~\ref{thm:sl2-exist-unique}. In other words, our next goal is
@@ -1138,33 +1146,33 @@ establishing\dots
\begin{theorem}\label{thm:sl3-existence-uniqueness}
For each pair of positive integers \(n\) and \(m\), there exists precisely
- one irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) whose highest weight
- is \(n \alpha_1 - m \alpha_3\).
+ one irreducible representation \(V\) of \(\mathfrak{sl}_3(K)\) whose highest
+ weight is \(n \alpha_1 - m \alpha_3\).
\end{theorem}
To proceed further we once again refer to the approach we employed in the case
-of \(\mathfrak{sl}_2(K)\): next we showed in theorem~\ref{thm:basis-of-irr-rep} that
-any irreducible representation of \(\mathfrak{sl}_2(K)\) is spanned by the images of
-its highest weight vector under \(f\). A more abstract way of putting it is to
-say that an irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\) is spanned by
-the images of its highest weight vector under successive applications by half
-of the root spaces of \(\mathfrak{sl}_2(K)\). The advantage of this alternative
-formulation is, of course, that the same holds for \(\mathfrak{sl}_3(K)\).
-Specifically\dots
+of \(\mathfrak{sl}_2(K)\): next we showed in theorem~\ref{thm:basis-of-irr-rep}
+that any irreducible representation of \(\mathfrak{sl}_2(K)\) is spanned by the
+images of its highest weight vector under \(f\). A more abstract way of putting
+it is to say that an irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\)
+is spanned by the images of its highest weight vector under successive
+applications by half of the root spaces of \(\mathfrak{sl}_2(K)\). The
+advantage of this alternative formulation is, of course, that the same holds
+for \(\mathfrak{sl}_3(K)\). Specifically\dots
\begin{theorem}\label{thm:irr-sl3-span}
- Given an irreducible \(\mathfrak{sl}_3(K)\)-representation \(V\) and a highest
- weight vector \(v \in V\), \(V\) is spanned by the images of \(v\) under
- successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\).
+ Given an irreducible \(\mathfrak{sl}_3(K)\)-representation \(V\) and a
+ highest weight vector \(v \in V\), \(V\) is spanned by the images of \(v\)
+ under successive applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\).
\end{theorem}
The proof of theorem~\ref{thm:irr-sl3-span} is very similar to that of
theorem~\ref{thm:basis-of-irr-rep}: we use the commutator relations of
-\(\mathfrak{sl}_3(K)\) to inductively show that the subspace spanned by the images of a
-highest weight vector under successive applications of \(E_{2 1}\), \(E_{3 1}\)
-and \(E_{3 2}\) is invariant under the action of \(\mathfrak{sl}_3(K)\) -- please refer
-to \cite{fulton-harris} for further details. The same argument also goes to
-show\dots
+\(\mathfrak{sl}_3(K)\) to inductively show that the subspace spanned by the
+images of a highest weight vector under successive applications of \(E_{2 1}\),
+\(E_{3 1}\) and \(E_{3 2}\) is invariant under the action of
+\(\mathfrak{sl}_3(K)\) -- please refer to \cite{fulton-harris} for further
+details. The same argument also goes to show\dots
\begin{corollary}
Given a representation \(V\) of \(\mathfrak{sl}_3(K)\) with highest weight
@@ -1181,8 +1189,8 @@ representation turns out to be quite simple.
\begin{proof}[Proof of existence]
Consider the natural representation \(V = K^3\) of \(\mathfrak{sl}_3(K)\). We
- claim that the highest weight of \(\operatorname{Sym}^n V \otimes \operatorname{Sym}^m V^*\)
- is \(n \alpha_1 - m \alpha_3\).
+ claim that the highest weight of \(\operatorname{Sym}^n V \otimes
+ \operatorname{Sym}^m V^*\) is \(n \alpha_1 - m \alpha_3\).
First of all, notice that the eigenvectors of \(V\) are the canonical basis
vectors \(e_1\), \(e_2\) and \(e_3\), whose eigenvalues are \(\alpha_1\),
@@ -1225,8 +1233,8 @@ 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 \(\mathfrak{sl}_3(K)\)-representations \(U\) and
- \(W\), by computing
+ On the other hand if we fix two \(\mathfrak{sl}_3(K)\)-representations \(U\)
+ and \(W\), by computing
\[
\begin{split}
H (u \otimes w)
@@ -1239,20 +1247,22 @@ representation turns out to be quite simple.
we can see that the weights of \(U \otimes W\) are precisely the sums of the
weights of \(U\) with the weights of \(W\).
- This implies that the maximal weights of \(\operatorname{Sym}^n V\) and \(\operatorname{Sym}^m V^*\) are
- \(n \alpha_1\) and \(- m \alpha_3\) respectively -- with maximal weight
- vectors \(e_1^n\) and \(f_3^m\). Furthermore, by the same token the highest
- weight of \(\operatorname{Sym}^n V \otimes \operatorname{Sym}^m V^*\) must be \(n e_1 - m e_3\) -- with
- highest weight vector \(e_1^n \otimes f_3^m\).
+ This implies that the maximal weights of \(\operatorname{Sym}^n V\) and
+ \(\operatorname{Sym}^m V^*\) are \(n \alpha_1\) and \(- m \alpha_3\)
+ respectively -- with maximal weight vectors \(e_1^n\) and \(f_3^m\).
+ Furthermore, by the same token the highest weight of \(\operatorname{Sym}^n V
+ \otimes \operatorname{Sym}^m V^*\) must be \(n e_1 - m e_3\) -- with highest
+ weight vector \(e_1^n \otimes f_3^m\).
\end{proof}
The ``uniqueness'' part of theorem~\ref{thm:sl3-existence-uniqueness} is even
simpler than that.
\begin{proof}[Proof of uniqueness]
- Let \(V\) and \(W\) be two irreducible representations of \(\mathfrak{sl}_3(K)\) with
- highest weight \(\lambda\). By theorem~\ref{thm:sl3-irr-weights-class}, the
- weights of \(V\) are precisely the same as those of \(W\).
+ Let \(V\) and \(W\) be two irreducible representations of
+ \(\mathfrak{sl}_3(K)\) with highest weight \(\lambda\). By
+ theorem~\ref{thm:sl3-irr-weights-class}, the weights of \(V\) are precisely
+ the same as those of \(W\).
Now by computing
\[
@@ -1267,62 +1277,64 @@ 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 \(U = \mathfrak{sl}_3(K) \cdot v + w\) generated by \(v +
- w\). The projection maps \(\pi_1 : U \to V\), \(\pi_2 : U \to W\), being
- non-zero homomorphism between irreducible representations of \(\mathfrak{sl}_3(K)\)
- must be isomorphism. Finally,
+ irreducible representation \(U = \mathfrak{sl}_3(K) \cdot v + w\) generated
+ by \(v + w\). The projection maps \(\pi_1 : U \to V\), \(\pi_2 : U \to W\),
+ being non-zero homomorphism between irreducible representations of
+ \(\mathfrak{sl}_3(K)\) must be isomorphism. Finally,
\[
V \cong U \cong W
\]
\end{proof}
The situation here is analogous to that of the previous section, where we saw
-that the irreducible representations of \(\mathfrak{sl}_2(K)\) are given by symmetric
-powers of the natural representation.
+that the irreducible representations of \(\mathfrak{sl}_2(K)\) are given by
+symmetric powers of the natural representation.
We've been very successful in our pursue for a classification of the
-irreducible representations of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\), but so far
-we've mostly postponed the discussion on the motivation behind our methods. In
-particular, we did not explain why we chose \(h\) and \(\mathfrak{h}\), and
-neither why we chose to look at their eigenvalues. Apart from the obvious fact
-we already knew it would work a priory, why did we do all that? In the
-following section we will attempt to answer this question by looking at what we
-did in the last chapter through more abstract lenses and studying the
-representations of an arbitrary finite-dimensional semisimple Lie
-algebra \(\mathfrak{g}\).
+irreducible representations of \(\mathfrak{sl}_2(K)\) and
+\(\mathfrak{sl}_3(K)\), but so far we've mostly postponed the discussion on the
+motivation behind our methods. In particular, we did not explain why we chose
+\(h\) and \(\mathfrak{h}\), and neither why we chose to look at their
+eigenvalues. Apart from the obvious fact we already knew it would work a
+priory, why did we do all that? In the following section we will attempt to
+answer this question by looking at what we did in the last chapter through more
+abstract lenses and studying the representations of an arbitrary
+finite-dimensional semisimple Lie algebra \(\mathfrak{g}\).
\section{Simultaneous Diagonalization \& the General Case}
-At the heart of our analysis of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) was the
-decision to consider the eigenspace decomposition
+At the heart of our analysis of \(\mathfrak{sl}_2(K)\) and
+\(\mathfrak{sl}_3(K)\) was the decision to consider the eigenspace
+decomposition
\begin{equation}\label{sym-diag}
V = \bigoplus_\lambda V_\lambda
\end{equation}
-This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the reasoning
-behind it, as well as the mere fact equation (\ref{sym-diag}) holds, are harder
-to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace decomposition
-associated with an operator \(V \to V\) is a very well-known tool, and this
-type of argument should be familiar to anyone familiar with basic concepts of
-linear algebra. On the other hand, the eigenspace decomposition of \(V\) with
-respect to the action of an arbitrary subalgebra \(\mathfrak{h} \subset
-\mathfrak{gl}(V)\) is neither well-known nor does it hold in general: as previously
-stated, it may very well be that
+This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the
+reasoning behind it, as well as the mere fact equation (\ref{sym-diag}) holds,
+are harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace
+decomposition associated with an operator \(V \to V\) is a very well-known
+tool, and this type of argument should be familiar to anyone familiar with
+basic concepts of linear algebra. On the other hand, the eigenspace
+decomposition of \(V\) with respect to the action of an arbitrary subalgebra
+\(\mathfrak{h} \subset \mathfrak{gl}(V)\) is neither well-known nor does it
+hold in general: as previously stated, it may very well be that
\[
\bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \subsetneq V
\]
We should note, however, that this two cases are not as different as they may
sound at first glance. Specifically, we can regard the eigenspace decomposition
-of a representation \(V\) of \(\mathfrak{sl}_2(K)\) with respect to the eigenvalues of
-the action of \(h\) as the eigenvalue decomposition of \(V\) with respect to
-the action of the subalgebra \(\mathfrak{h} = K h \subset \mathfrak{sl}_2(K)\).
-Furthermore, in both cases \(\mathfrak{h} \subset \mathfrak{sl}_n(K)\) is the
-subalgebra of diagonal matrices, which is Abelian. The fundamental difference
-between these two cases is thus the fact that \(\dim \mathfrak{h} = 1\) for
-\(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim \mathfrak{h} > 1\) for
-\(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The question then is: why did we choose
-\(\mathfrak{h}\) with \(\dim \mathfrak{h} > 1\) for \(\mathfrak{sl}_3(K)\)?
+of a representation \(V\) of \(\mathfrak{sl}_2(K)\) with respect to the
+eigenvalues of the action of \(h\) as the eigenvalue decomposition of \(V\)
+with respect to the action of the subalgebra \(\mathfrak{h} = K h \subset
+\mathfrak{sl}_2(K)\). Furthermore, in both cases \(\mathfrak{h} \subset
+\mathfrak{sl}_n(K)\) is the subalgebra of diagonal matrices, which is Abelian.
+The fundamental difference between these two cases is thus the fact that \(\dim
+\mathfrak{h} = 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim
+\mathfrak{h} > 1\) for \(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The
+question then is: why did we choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} >
+1\) for \(\mathfrak{sl}_3(K)\)?
% TODO: Rewrite this: we haven't dealt with finite groups at all
The rational behind fixing an Abelian subalgebra is one we have already
@@ -1366,13 +1378,14 @@ readily check that every pair of diagonal matrices commutes, so that
0 & 0 & \cdots & K
\end{pmatrix}
\]
-is an Abelian subalgebra of \(\mathfrak{gl}_n(K)\). A simple calculation then shows
-that if \(X \in \mathfrak{gl}_n(K)\) commutes with every diagonal matrix \(H \in
-\mathfrak{h}\) then \(X\) is a diagonal matrix, so that \(\mathfrak{h}\) is a
-Cartan subalgebra of \(\mathfrak{gl}_n(K)\). The intersection of such subalgebra with
-\(\mathfrak{sl}_n(K)\) -- i.e. the subalgebra of traceless diagonal matrices -- is a
-Cartan subalgebra of \(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\)
-we get to the subalgebras described the previous two sections.
+is an Abelian subalgebra of \(\mathfrak{gl}_n(K)\). A simple calculation then
+shows that if \(X \in \mathfrak{gl}_n(K)\) commutes with every diagonal matrix
+\(H \in \mathfrak{h}\) then \(X\) is a diagonal matrix, so that
+\(\mathfrak{h}\) is a Cartan subalgebra of \(\mathfrak{gl}_n(K)\). The
+intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the
+subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of
+\(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the
+subalgebras described the previous two sections.
The remaining question then is: if \(\mathfrak{h} \subset \mathfrak{g}\) is a
Cartan subalgebra and \(V\) is a representation of \(\mathfrak{g}\), does the
@@ -1450,9 +1463,9 @@ analogous to the ones on the previous sections. Further details can be found in
appendix D of \cite{fulton-harris} and in \cite{humphreys}.
We begin our analysis by remarking that in both \(\mathfrak{sl}_2(K)\) and
-\(\mathfrak{sl}_3(K)\), the roots were symmetric about the origin and spanned all of
-\(\mathfrak{h}^*\). This turns out to be a general fact, which is a consequence
-of the following theorem.
+\(\mathfrak{sl}_3(K)\), the roots were symmetric about the origin and spanned
+all of \(\mathfrak{h}^*\). This turns out to be a general fact, which is a
+consequence of the following theorem.
% TODO: Add a refenrence to a proof (probably Humphreys)
% TODO: Clarify the meaning of "non-degenerate"
@@ -1565,7 +1578,8 @@ Accordingly, we call \(\lambda\) \emph{the highest weight of \(V\)}, and we
call any \(v \in V_\lambda\) \emph{a highest weight vector}. The strategy then
is to describe all weight spaces of \(V\) in terms of \(\lambda\) and \(v\), as
in theorem~\ref{thm:sl3-irr-weights-class}, and unsurprisingly we do so by
-reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). Namely, we show\dots
+reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). Namely, we
+show\dots
\begin{proposition}\label{thm:distinguished-subalgebra}
Given a root \(\alpha\) of \(\mathfrak{g}\) the subspace
@@ -1602,7 +1616,8 @@ The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely
determined by this condition, but \(H_\alpha\) is. The second statement of
corollary~\ref{thm:distinguished-subalg-rep} imposes a restriction on the
weights of \(V\). Namely, if \(\mu\) is a weight, \(\mu(H_\alpha)\) is an
-eigenvalue of \(h\) in some representation of \(\mathfrak{sl}_2(K)\), so that\dots
+eigenvalue of \(h\) in some representation of \(\mathfrak{sl}_2(K)\), so
+that\dots
\begin{proposition}
The weights \(\mu\) of an irreducible representation \(V\) of
@@ -1667,8 +1682,8 @@ 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 \(\mathfrak{sl}_3(K)\), and the proof of existence of can once again be reduced
-to the proof of\dots
+for \(\mathfrak{sl}_3(K)\), and the proof of existence of can once again be
+reduced to the proof of\dots
\begin{theorem}\label{thm:weak-dominant-weight}
There exists \emph{some} -- not necessarily irreducible -- finite-dimensional
@@ -1677,26 +1692,27 @@ 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 \(\mathfrak{g}
-= \mathfrak{sl}_3(K)\). The issue is that our proof relied heavily on our knowledge of
-the roots of \(\mathfrak{sl}_3(K)\). Instead, we need a new strategy for the general
-setting.
+= \mathfrak{sl}_3(K)\). The issue is that our proof relied heavily on our
+knowledge of the roots of \(\mathfrak{sl}_3(K)\). Instead, we need a new
+strategy for the general setting.
% TODO: Add further details. turn this into a proper proof?
Alternatively, one could construct a potentially infinite-dimensional
representation of \(\mathfrak{g}\) whose highest weight is some fixed dominant
integral weight \(\lambda\) by taking the induced representation
-\(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} V_\lambda = \mathcal{U}(\mathfrak{g})
-\otimes_{\mathcal{U}(\mathfrak{b})} V_\lambda\), where \(\mathfrak{b} =
-\mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha \subset
-\mathfrak{g}\) is the so called \emph{Borel subalgebra of \(\mathfrak{g}\)},
-\(\mathcal{U}(\mathfrak{g})\) denotes the \emph{universal enveloping algebra
-of \(\mathfrak{g}\)} and \(\mathfrak{b}\) acts on \(V_\lambda = K v\) via \(H
-v = \lambda(H) \cdot v\) and \(X v = 0\) for \(X \in \mathfrak{g}_\alpha\), as
-does \cite{humphreys} in his proof. The fact that \(v\) is annihilated by all
-positive root spaces guarantees that the maximal weight of \(V\) is at most
-\(\lambda\), while the Poincare-Birkhoff-Witt \cite{humphreys} theorem
-guarantees that \(v = 1 \otimes v \in V\) is a non-zero weight vector of
-\(\lambda\) -- so that \(\lambda\) is the highest weight of \(V\).
-The challenge then is to show that the irreducible component of \(v\) in \(V\)
-is finite-dimensional -- see chapter 20 of \cite{humphreys} for a proof.
+\(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} V_\lambda =
+\mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{b})} V_\lambda\),
+where \(\mathfrak{b} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+}
+\mathfrak{g}_\alpha \subset \mathfrak{g}\) is the so called \emph{Borel
+subalgebra of \(\mathfrak{g}\)}, \(\mathcal{U}(\mathfrak{g})\) denotes the
+\emph{universal enveloping algebra of \(\mathfrak{g}\)} and \(\mathfrak{b}\)
+acts on \(V_\lambda = K v\) via \(H v = \lambda(H) \cdot v\) and \(X v = 0\)
+for \(X \in \mathfrak{g}_\alpha\), as does \cite{humphreys} in his proof. The
+fact that \(v\) is annihilated by all positive root spaces guarantees that the
+maximal weight of \(V\) is at most \(\lambda\), while the
+Poincare-Birkhoff-Witt \cite{humphreys} theorem guarantees that \(v = 1 \otimes
+v \in V\) is a non-zero weight vector of \(\lambda\) -- so that \(\lambda\) is
+the highest weight of \(V\). The challenge then is to show that the irreducible
+component of \(v\) in \(V\) is finite-dimensional -- see chapter 20 of
+\cite{humphreys} for a proof.