diff --git a/sections/lie-algebras.tex b/sections/lie-algebras.tex
@@ -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 \(\mathfrak{sl}_n(K)\) 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}
@@ -80,9 +80,9 @@ 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
+\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}, \Hom(V,
+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}
@@ -94,7 +94,7 @@ 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 \(W\).
+\operatorname{Hom}(V, W)) = 0\) for all \(V\) and \(W\).
% TODO: Comment on the geometric proof by Weyl
%The algebraic approach has the
@@ -255,45 +255,45 @@ sequence
\begin{center}
\begin{tikzcd}
0 \arrow{r} &
- \Rad(\mathfrak g) \arrow{r} &
+ \operatorname{Rad}(\mathfrak g) \arrow{r} &
\mathfrak g \arrow{r} &
- \mfrac{\mathfrak g}{\Rad(\mathfrak g)} \arrow{r} &
+ \mfrac{\mathfrak g}{\operatorname{Rad}(\mathfrak g)} \arrow{r} &
0
\end{tikzcd}
\end{center}
-where \(\Rad(\mathfrak g)\) is the sum of all solvable ideals of \(\mathfrak
+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 \(\mfrac{\mathfrak
-g}{\Rad(\mathfrak g)}\). In practice though, this isn't quite satisfactory
+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 \(\mfrac{\mathfrak
- g}{\Rad(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak
+ g}{\operatorname{Rad}(\mathfrak g)}\) and a one-dimensional representation of \(\mathfrak
g\).
\end{theorem}
-\section{Representations of \(\sl_2(K)\)}
+\section{Representations of \(\mathfrak{sl}_2(K)\)}
The primary goal of this section is proving\dots
\begin{theorem}\label{thm:sl2-exist-unique}
For each \(n > 0\), there exists precisely one irreducible representation
- \(V\) of \(\sl_2(K)\) with \(\dim V = n\).
+ \(V\) of \(\mathfrak{sl}_2(K)\) with \(\dim V = n\).
\end{theorem}
The general approach we'll take is supposing \(V\) is an irreducible
-representation of \(\sl_2(K)\) and then derive some information about its
+representation of \(\mathfrak{sl}_2(K)\) and then derive some information about its
structure. We begin our analysis by pointing out that the elements
\begin{align*}
e & = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} &
f & = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} &
h & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{align*}
-form a basis of \(\sl_2(K)\) and satisfy
+form a basis of \(\mathfrak{sl}_2(K)\) and satisfy
\begin{align*}
[e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e
\end{align*}
@@ -331,7 +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 \(\mathfrak{sl}_2(K)\)-invariant
subspace. This implies
\[
V = \bigoplus_{n \in \ZZ} V_{\lambda + 2 n},
@@ -344,7 +344,7 @@ 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 \(\sl_2(K)\)-invariant subspace, so that the eigenvalues of \(h\)
+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
@@ -368,7 +368,7 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
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
- \(\sl_2(K)\).
+ \(\mathfrak{sl}_2(K)\).
The fact that \(h f^k v \in K \langle v, f v, f^2 v, \ldots \rangle\) follows
immediately from our previous assertion that \(f^k v \in V_{\lambda - 2 k}\)
@@ -408,7 +408,7 @@ 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 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
+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
@@ -416,7 +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 \(\mathfrak{sl}_2(K)\) whose
right-most eigenvalue of \(h\) is \(\lambda\) and \(w \in W_\lambda\) is
non-zero, consider the linear isomorphism
\begin{align*}
@@ -469,7 +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 \(\mathfrak{sl}_2(K)\) has the
form
\begin{center}
\begin{tikzcd}
@@ -499,15 +499,15 @@ 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 \(n\).
+ \(\mathfrak{sl}_2(K)\) whose left-most eigenvalue of \(h\) is \(n\).
\end{theorem}
\begin{proof}
The fact the representation \(V\) from the previous discussion exists is
- clear from the commutator relations of \(\sl_2(K)\) -- just look at \(f^k
+ 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 \(\sl_2(K)\), then \(V = \Sym^n
+ \(K^2\) 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 +
@@ -531,16 +531,16 @@ 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 \(\mathfrak{sl}_2(K)\). 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 \(\sl_3(K)\), hoping this will \emph{somehow}
+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 \(\sl_3(K)\)}\label{sec:sl3-reps}
+\section{Representations of \(\mathfrak{sl}_3(K)\)}\label{sec:sl3-reps}
-The study of representations of \(\sl_2(K)\) reminds me of the difference the
+The study of representations of \(\mathfrak{sl}_2(K)\) 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
@@ -550,31 +550,31 @@ 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 \(\mathfrak{sl}_2(K)\). 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 \(\sl_2(K)\). The general purpose of this section 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 \(\sl_2(K)\) can be generalized to other
-semisimple Lie algebras, and the algebra \(\sl_3(K)\) stands as a natural
+classify the representations of \(\mathfrak{sl}_2(K)\) 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 \(\sl_3(K)\) and proceed step by step, at each point
+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
-\(\sl_2(K)\). The first obvious question is one we have already asked
+\(\mathfrak{sl}_2(K)\). 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 \(\sl_3(K)\)?
+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 \(\sl_2(K)\) culminating in
+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 \(\sl_3(K)\), but it's
+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
@@ -585,15 +585,15 @@ actual way to proceed is to consider the subalgebra
= \left\{
X \in
\begin{pmatrix} K & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K \end{pmatrix}
- : \Tr(X) = 0
+ : \operatorname{Tr}(X) = 0
\right\}
\]
The choice of \(\mathfrak{h}\) may seem like an odd choice at the moment, but
the point is we'll later show that there exists some \(v \in V\) that is
simultaneously an eigenvector of each \(H \in \mathfrak{h}\) and annihilated by
-half of the remaining elements of \(\sl_3(K)\). This is exactly analogous to
-the situation we found in \(\sl_2(K)\): \(h\) corresponds to the subalgebra
+half of the remaining elements of \(\mathfrak{sl}_3(K)\). 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
@@ -613,8 +613,8 @@ 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 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 \(\mathfrak{sl}_3(K)\). 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,
@@ -624,9 +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 \(\pm 2\).
+\(\mathfrak{sl}_2(K)\)-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 \(\mathfrak{sl}_3(K)\), 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 -
@@ -672,8 +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 \(v \in V_\lambda\)
+\(\mathfrak{sl}_3(K)\) 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
\[
\begin{split}
@@ -684,11 +684,11 @@ then
\end{split}
\]
so that \(X\) carries \(v\) to \(V_{\alpha + \lambda}\). In other words,
-\(\sl_3(k)_\alpha\) \emph{acts on \(V\) by translating vectors between
+\(\mathfrak{sl}_3(k)_\alpha\) \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 \(\mathfrak{sl}_3(K)_{\alpha_1 - \alpha_3}\) will act on the adjoint
+representation of \(\mathfrak{sl}_3(K)\) via
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=2.5]
@@ -708,19 +708,19 @@ representation of \(\sl_3(K)\) via
\end{tikzpicture}
\end{figure}
-This is again entirely analogous to the situation we observed in \(\sl_2(K)\).
+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
- \(\sl_3(K)\)-representation \(V\) differ from one another by integral
+ \(\mathfrak{sl}_3(K)\)-representation \(V\) differ from one another by integral
linear combinations of the eigenvalues \(\alpha_i - \alpha_j\) of
- adjoint action of \(\mathfrak{h}\) in \(\sl_3(K)\).
+ adjoint action of \(\mathfrak{h}\) in \(\mathfrak{sl}_3(K)\).
\end{theorem}
\begin{proof}
This proof goes exactly as that of the analogous statement for
- \(\sl_2(K)\): it suffices to note that if we fix some eigenvalue
+ \(\mathfrak{sl}_2(K)\): it suffices to note that if we fix some eigenvalue
\(\lambda\) of \(\mathfrak{h}\) and let \(i\) and \(j\) vary then
\[
\bigoplus_{i j} V_{\lambda + \alpha_i - \alpha_j}
@@ -733,7 +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 \(\mathfrak{sl}_3(K)\), 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
@@ -741,33 +741,33 @@ adjoint action of \(\mathfrak{h}\).
\end{definition}
It's clear from our previous discussion that the weights of the adjoint
-representation of \(\sl_3(K)\) deserve some special attention.
+representation of \(\mathfrak{sl}_3(K)\) 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 \(\mathfrak{sl}_3(K)\) 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 \(\sl_3(K)\) are all
+ 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 \(\sl_3(K)\).
+ \alpha_j\) of \(\mathfrak{sl}_3(K)\).
\end{corollary}
\begin{definition}
The lattice \(Q = \ZZ \langle \alpha_i - \alpha_j : i, j = 1, 2, 3 \rangle\)
- is called \emph{the root lattice of \(\sl_3(K)\)}.
+ is called \emph{the root lattice of \(\mathfrak{sl}_3(K)\)}.
\end{definition}
To proceed we once more refer to the previously established framework: next we
saw that the eigenvalues of \(h\) formed an unbroken string of integers
symmetric around \(0\). To prove this we analyzed the right-most eigenvalue of
\(h\) and its eigenvector, providing an explicit description of the
-irreducible representation of \(\sl_2(K)\) 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 \(\mathfrak{sl}_2(K)\) in terms of this vector. We may
+reproduce these steps in the context of \(\mathfrak{sl}_3(K)\) by fixing a direction in
the place an considering the weight lying the furthest in that direction.
% TODO: This doesn't make any sence in field other than C
@@ -827,7 +827,7 @@ 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 \(\sl_3(K)\) act on the weight spaces of \(V\) via translation,
+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 \(V_{\lambda + \alpha_2 - \alpha_3}\)
@@ -836,22 +836,22 @@ 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 \(\mathfrak{sl}_3(K)\).
\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 \(\mathfrak{sl}_3(K)\) 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
-\(\sl_2(K)\), we then constructed an explicit basis of our irreducible
+\(\mathfrak{sl}_2(K)\), 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 \(\sl_2(K)\) in terms of
+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
-\(\sl_3(K)\) -- and indeed that's what we'll do later down the line -- but
+\(\mathfrak{sl}_3(K)\) -- 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.
@@ -890,14 +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
+\(\mathfrak{sl}_2(K)\) via
\begin{align*}
E_{2 1} & \mapsto e &
E_{1 2} & \mapsto f &
H & \mapsto h
\end{align*}
-In other words, \(W\) is a representation of \(\sl_2(K)\). Even more so, we
+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}
@@ -940,7 +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 \(\mathfrak{sl}_2(K)\), so that the weights in
this subspace must be symmetric with respect to the line \(\langle \alpha_3 -
\alpha_2, \alpha \rangle = 0\). The picture is now
\begin{center}
@@ -970,7 +970,7 @@ In general, given a weight \(\mu\), the space
\]
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
+is once more isomorphic to \(\mathfrak{sl}_2(K)\), and again the weight spaces in this
string match precisely the eigenvalues of \(h\). Needless to say, we could keep
applying this method to the weights at the ends of our string, arriving at
\begin{center}
@@ -1005,7 +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 \(\mathfrak{sl}_2(K)\) 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}
@@ -1044,8 +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 \(\mathfrak{sl}_2(K)\), 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}
@@ -1085,7 +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 \(\mathfrak{sl}_2(K)\), so it must be an integer. Now since
\[
\lambda
\begin{pmatrix}
@@ -1116,7 +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 \(\mathfrak{sl}_3(K)\)}.
\end{definition}
Finally\dots
@@ -1129,8 +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 \(\mathfrak{sl}_3(K)\) 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
@@ -1139,36 +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 \(\mathfrak{sl}_3(K)\) 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 \(\sl_2(K)\): 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 \(\mathfrak{sl}_2(K)\): 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 \(\sl_2(K)\) is spanned by
+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 \(\sl_2(K)\). The advantage of this alternative
-formulation is, of course, that the same holds for \(\sl_3(K)\).
+of the root spaces of \(\mathfrak{sl}_2(K)\). 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 \(\sl_3(K)\)-representation \(V\) and a highest
+ 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
-\(\sl_3(K)\) to inductively show that the subspace spanned by the images of a
+\(\mathfrak{sl}_3(K)\) 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 \(\sl_3(K)\) -- please refer
+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 \(\sl_3(K)\) with highest weight
+ Given a representation \(V\) of \(\mathfrak{sl}_3(K)\) with highest weight
\(\lambda\) and \(v \in V_\lambda\), the subspace spanned by successive
applications of \(E_{2 1}\), \(E_{3 1}\) and \(E_{3 2}\) to \(v\) is an
irreducible subrepresentation whose highest weight is \(\lambda\).
@@ -1181,8 +1181,8 @@ 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
- claim that the highest weight of \(\Sym^n V \otimes \Sym^m V^*\)
+ Consider the natural representation \(V = K^3\) of \(\mathfrak{sl}_3(K)\). We
+ claim that the highest weight of \(\operatorname{Sym}^n V \otimes \operatorname{Sym}^m V^*\)
is \(n \alpha_1 - m \alpha_3\).
First of all, notice that the eigenvectors of \(V\) are the canonical basis
@@ -1226,7 +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 \(\mathfrak{sl}_3(K)\)-representations \(U\) and
\(W\), by computing
\[
\begin{split}
@@ -1240,10 +1240,10 @@ 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 \(\Sym^n V\) and \(\Sym^m V^*\) are
+ 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 \(\Sym^n V \otimes \Sym^m V^*\) must be \(n e_1 - m e_3\) -- with
+ 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}
@@ -1251,7 +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 \(\mathfrak{sl}_3(K)\) with
highest weight \(\lambda\). By theorem~\ref{thm:sl3-irr-weights-class}, the
weights of \(V\) are precisely the same as those of \(W\).
@@ -1268,9 +1268,9 @@ 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 \(U = \sl_3(K) \cdot v + w\) generated by \(v +
+ 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 \(\sl_3(K)\)
+ non-zero homomorphism between irreducible representations of \(\mathfrak{sl}_3(K)\)
must be isomorphism. Finally,
\[
V \cong U \cong W
@@ -1278,11 +1278,11 @@ simpler than that.
\end{proof}
The situation here is analogous to that of the previous section, where we saw
-that the irreducible representations of \(\sl_2(K)\) are given by symmetric
+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 \(\sl_2(K)\) and \(\sl_3(K)\), but so far
+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
@@ -1294,20 +1294,20 @@ 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 \(\mathfrak{sl}_2(K)\) 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 \(\sl_2(K)\), but the reasoning
+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 \(\sl_3(K)\). The eigenspace decomposition
+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 \(\mathfrak{h} \subset
-\gl(V)\) is neither well-known nor does it hold in general: as previously
+\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
@@ -1315,15 +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 \(\mathfrak{sl}_2(K)\) with respect to the eigenvalues of
the action of \(h\) as the eigenvalue decomposition of \(V\) with respect to
-the action of the subalgebra \(\mathfrak{h} = K h \subset \sl_2(K)\).
-Furthermore, in both cases \(\mathfrak{h} \subset \sl_n(K)\) is the
+the action of the subalgebra \(\mathfrak{h} = K h \subset \mathfrak{sl}_2(K)\).
+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
-\(\mathfrak{h} \subset \sl_2(K)\) 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 \(\sl_3(K)\)?
+\(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) 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
@@ -1367,12 +1367,12 @@ readily check that every pair of diagonal matrices commutes, so that
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 \(\mathfrak{gl}_n(K)\). 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 \(\gl_n(K)\). 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 \(\mathfrak{gl}_n(K)\). 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
@@ -1417,7 +1417,7 @@ What is simultaneous diagonalization all about then?
Hence
\[
V
- = \Res{\mathfrak g}{\mathfrak h} V
+ = \operatorname{Res}_{\mathfrak h}^{\mathfrak g} V
\cong \bigoplus V_i,
\]
as representations of \(\mathfrak{h}\), where each \(V_i\) is an irreducible
@@ -1450,8 +1450,8 @@ 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 \(\sl_2(K)\) and
-\(\sl_3(K)\), the roots were symmetric about the origin and spanned all of
+We begin our analysis by remarking that in both \(\mathfrak{sl}_2(K)\) 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.
@@ -1485,15 +1485,15 @@ of the following theorem.
\]
for all \(H \in \mathfrak{h}\).
- This implies that if \(\alpha + \beta \ne 0\) then \(\ad(X) \ad(Y)\) is
+ This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X) \operatorname{ad}(Y)\) is
nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then
\[
- (\ad(X) \ad(Y))^n Z
+ (\operatorname{ad}(X) \operatorname{ad}(Y))^n Z
= [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ]
\in \mathfrak{g}_{n \alpha + n \beta + \gamma}
= 0
\]
- for \(n\) large enough. In particular, \(K(X, Y) = \Tr(\ad(X) \ad(Y)) = 0\).
+ for \(n\) large enough. In particular, \(K(X, Y) = \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\).
Now if \(- \alpha\) is not an eigenvalue we find \(K(X, \mathfrak{g}_\beta) =
0\) for all eigenvalues \(\beta\), which contradicts the non-degeneracy of
\(K\). Hence \(- \alpha\) must be an eigenvalue of the adjoint action of
@@ -1502,12 +1502,12 @@ of the following theorem.
For the second statement, note that if the eigenvalues of \(\mathfrak{h}\) do
not span all of \(\mathfrak{h}^*\) then there is some \(H \in \mathfrak{h}\)
non-zero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is
- to say, \(\ad(H) X = [H, X] = 0\) for all \(X \in \mathfrak{g}\). Another way
+ to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in \mathfrak{g}\). Another way
of putting it is to say \(H\) is an element of the center \(\mathfrak{z}\) of
\(\mathfrak{g}\), which is zero by the semisimplicity -- a contradiction.
\end{proof}
-Furthermore, as in the case of \(\sl_2(K)\) and \(\sl_3(K)\) one can show\dots
+Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) one can show\dots
\begin{proposition}\label{thm:root-space-dim-1}
The eigenspaces \(\mathfrak{g}_\alpha\) are all 1-dimensional.
@@ -1536,7 +1536,7 @@ then\dots
\end{theorem}
% 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
+To proceed further, as in the case of \(\mathfrak{sl}_3(K)\) we have to fix a direction
in \(\mathfrak{h}^*\) -- i.e. we fix a linear function \(\mathfrak{h}^* \to
\RR\) such that \(Q\) lies outside of its kernel. This choice induces a
partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of roots of
@@ -1559,13 +1559,13 @@ 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 \(\sl_3(K)\). 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
\(\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{- \alpha}
\oplus [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) is a subalgebra
- isomorphic to \(\sl_2(k)\).
+ isomorphic to \(\mathfrak{sl}_2(k)\).
\end{proposition}
\begin{corollary}\label{thm:distinguished-subalg-rep}
@@ -1596,7 +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 \(\mathfrak{sl}_2(K)\), so that\dots
\begin{proposition}
The weights \(\mu\) of an irreducible representation \(V\) of
@@ -1632,7 +1632,7 @@ is\dots
% 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
+This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we found
that the weights of the irreducible representations were symmetric with respect
to the lines \(\langle \alpha_i - \alpha_j, \alpha \rangle = 0\). Indeed, the
same argument leads us to the conclusion\dots
@@ -1663,7 +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 \(\mathfrak{sl}_3(K)\), and the proof of existence of can once again be reduced
to the proof of\dots
\begin{theorem}\label{thm:weak-dominant-weight}
@@ -1673,15 +1673,15 @@ 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
+= \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
-\(\Ind{\mathfrak{g}}{\mathfrak{b}} V_\lambda = \mathcal{U}(\mathfrak{g})
+\(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} V_\lambda = \mathcal{U}(\mathfrak{g})
\otimes_{\mathcal{U}(\mathfrak{b})} V_\lambda\), where \(\mathfrak{b} =
\mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha \subset
\mathfrak{g}\) is the so called \emph{Borel subalgebra of \(\mathfrak{g}\)},