lie-algebras-and-their-representations

diff --git a/sections/lie-algebras.tex b/sections/lie-algebras.tex
@@ -303,11 +303,9 @@ The primary goal of this section is proving\dots
   \(V\) of \(\sl_2(k)\) with \(\dim V = n\).
 \end{theorem}
 
-It's important to note, however, that -- as promised -- we will end up with an
-explicit construction of \(V\). The general approach we'll take is supposing
-\(V\) is an irreducible representation of \(\sl_2(k)\) and then derive some
-information about its structure. We begin our analysis by pointing out that
-the elements
+The general approach we'll take is supposing \(V\) is an irreducible
+representation of \(\sl_2(k)\) and then derive some information about its
+structure. We begin our analysis by pointing out that the elements
 \begin{align*}
   e & = \begin{pmatrix} 0 & 1 \\ 0 &  0 \end{pmatrix} &
   f & = \begin{pmatrix} 0 & 0 \\ 1 &  0 \end{pmatrix} &
@@ -383,12 +381,12 @@ 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 \(V = \langle v, f v, f^2 v, \ldots \rangle\), which in
-  light of the fact that \(V\) is irreducible is the same as showing \(\langle
-  v, f v, f^2 v, \ldots \rangle\) is invariant under the action of
+  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
   \(\sl_2(k)\).
 
-  The fact that \(h f^k v \in \langle v, f v, f^2 v, \ldots \rangle\) 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,
@@ -547,58 +545,6 @@ To conclude our analysis all it's left is to show that for each \(n\) such
   1, \ldots, n\). In other words, \(W = V\). We are done.
 \end{proof}
 
-A perhaps more elegant way of proving theorem~\ref{thm:irr-rep-of-sl2-exists}
-is to work our way backwards to the corresponding representation \(V\) of
-\(\SU_2\) and then show that its character is irreducible. Computing the
-action of \(\mathfrak{su}_2\) in \(V\) is fairly easy: if we consider the
-standard basis
-\begin{align*}
-  I & = \begin{pmatrix} i &   0 \\   0 & - i \end{pmatrix} &
-  J & = \begin{pmatrix} 0 &   1 \\ - 1 &   0 \end{pmatrix} &
-  K & = \begin{pmatrix} 0 & - i \\ - i &   0 \end{pmatrix}
-\end{align*}
-of \(\mathfrak{su}_2\) one can readily check
-\begin{align*}
-  f^k v
-  & \overset{I}{\mapsto}
-  i (n - 2 k) f^k v \\
-  f^k v
-  & \overset{J}{\mapsto}
-  k (n + 1 - k) f^{k - 1} v - f^{k + 1} v \\
-  f^k v
-  & \overset{K}{\mapsto}
-  - i k (n + 1 - k) f^{k - 1} v - i f^{k + 1} v
-\end{align*}
-
-If we denote by \(\rho_*(X)\) the matrix corresponding to the action of \(X \in
-\mathfrak{su}_2\) in the basis \(\{v, f v, \ldots, f^n v\}\), and given that
-\(\exp : \mathfrak{su}_2 \to \SU_2\) is surjective, to arrive at the action of
-\(\SU_2\) all we have to do is compute \(\exp(\rho_*(X))\) for arbitrary \(X\).
-For instance, if \(n = 1\) we can very quickly check that \(V\) corresponds to
-the natural representation of \(\SU_2\), which can be shown to be irreducible.
-
-In general, however, computing \(\exp(\rho_*(X))\) is \emph{quite hard}.
-Nevertheless, the exceptional isomorphism \(\mathbb{S}^3 \cong \SU_2\) allows
-us to compute its trace: if \(a, b, c \in \RR\) are such that \(a^2 + b^2 + c^2
-= 1\), then the eigenvalues of \(\rho_*(a I + b J + c K)\) are \(\lambda i\),
-where \(\lambda\) ranges over the eigenvalues of \(h\) in \(V\). Hence the
-eigenvalues of \(\exp(\rho_*(t X))\) are \(e^{\lambda i t}\) for \(X = a I +
-b J + c K\). Now if \(p = a i + b j + c k \in \mathbb{S}^3\) then \(\cos t + p
-\sin t = \exp(t X)\), so that
-\[
-  \chi_V(\cos t + p \sin t)
-  = \sum_\lambda e^{\lambda i t}
-  = \sum_\lambda \cos(\lambda t)
-\]
-
-A simple calculation then shows
-\[
-  \norm{\chi_V}^2
-  = \frac{1}{\mu(\mathbb{S}^3)} \int_{\mathbb{S}^3} \abs{\chi_V(q)}^2 \; \dd q
-  = 1,
-\]
-which establishes that \(V\) is irreducible.
-
 Our initial gamble of studying the eigenvalues of \(h\) may have seemed
 arbitrary at first, but it payed off: we've \emph{completely} described
 \emph{all} irreducible representations of \(\sl_2(k)\). It is not yet clear,
@@ -676,8 +622,8 @@ Once again, we'll pay special attention to the eigenvalue decomposition
 \end{equation}
 where \(\lambda\) ranges over all eigenvalues of \(\mathfrak{h}\) and
 \(V_\lambda = \{ v \in V : H v = \lambda(H) \cdot v, \forall H \in \mathfrak{h}
-\}\). We should note that the fact that (\ref{eq:weight-module}) is not at all
-obvious. This is because in general \(V_\lambda\) is not the eigenspace
+\}\). We should note that the fact that (\ref{eq:weight-module}) holds is not
+at all obvious. This is because in general \(V_\lambda\) is not the eigenspace
 associated with an eigenvalue of any particular operator \(H \in
 \mathfrak{h}\), but instead the eigenspace of the action of the entire algebra
 \(\mathfrak{h}\). Fortunately for us, (\ref{eq:weight-module}) always holds,
@@ -833,6 +779,7 @@ Theorem~\ref{thm:sl3-weights-congruent-mod-root} can thus be restated as\dots
   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
@@ -1009,8 +956,8 @@ 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 \(\sl_2\), so that the weights in this
-subspace must be symmetric with respect to the line \(\langle \alpha_3 -
+E_{3 2}]\), 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}
   \begin{tikzpicture}
@@ -1394,6 +1341,7 @@ between these two cases is thus the fact that \(\dim \mathfrak{h} = 1\) 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
 encountered when dealing with finite groups: representations of Abelian groups
 and algebras are generally much simpler to understand than the general case.
@@ -1403,6 +1351,7 @@ 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
 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.
@@ -1465,6 +1414,7 @@ What is simultaneous diagonalization all about then?
   for all \(H \in \mathfrak{h}\) and all \(i\).
 \end{proposition}
 
+% TODO: h is not semisimple. Fix this proof
 \begin{proof}
   We claim \(\mathfrak{h}\) is semisimple. Indeed, if \(\{H_1, \ldots, H_m\}\)
   is basis of \(\mathfrak{h}\) then
@@ -1734,73 +1684,10 @@ 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 \(\sl_3(k)\). Specifically, we used the fact every dominant
-integral weight of \(\sl_3(k)\) can be written as \(n \alpha_1 - m \alpha_3\)
-for unique non-negative integers \(n\) and \(m\). When then constructed
-finite-dimensional representations \(V\) and \(W\) of \(\sl_3(k)\) whose
-highest weights are \(\alpha_1\) and \(- \alpha_3\), so that the highest weight
-of \(\Sym^n V \otimes \Sym^m W\) is \(n \alpha_1 - m \alpha_3\).
-
-A similar construction can be implemented for \(\sl_n(k)\): if \(\mathfrak{h}
-\subset \sl_n(k)\) is the subalgebra of diagonal matrices -- which, as you
-may recall, is a Cartan subalgebra -- and \(\alpha : \mathfrak{h} \to k\) is
-given by \(\alpha(E_{j j}) = \delta_{i j}\), one can show that any dominant
-integral weight of \(\sl_n(k)\) can be uniquely expressed in the form \(k_1
-\alpha_1 + k_2 (\alpha_1 + \alpha_2) + \cdots + k_{n - 1} (\alpha_1 + \cdots +
-\alpha_{n - 1})\) for non-negative integers \(k_1, k_2, \ldots k_{n - 1}\). For
-instance, one may visually represent the roots of \(\sl_4(k)\) by
-\begin{center}
-  \begin{tikzpicture}[scale=3]
-    \draw (0, 0) -- (1, 0) -- (1, 1) -- (0, 1) -- cycle;
-    \draw (0, 1) -- (.4, 1.4) -- (1.4, 1.4) -- (1, 1);
-    \draw (1, 0) -- (1.4, .4) -- (1.4, 1.4);
-    \draw[dotted] (0, 0) -- (.4, .4) -- (1.4, .4);
-    \draw[dotted] (.4, 1.4) -- (.4, .4);
-
-    \filldraw (.5, 0) circle (.7pt);
-    \filldraw (.5, 1) circle (.7pt);
-    \node[below] at (.5, 0) {$\alpha_2 - \alpha_1$};
-
-    \filldraw ( .4, .9) circle (.7pt);
-    \filldraw (1.4, .9) circle (.7pt);
-    \node[right] at (1.4, .9) {$\alpha_1 - \alpha_4$};
-
-    \filldraw (.9,  .4) circle (.7pt);
-    \filldraw (.9, 1.4) circle (.7pt);
-    \node[above] at (.9, 1.4) {$\alpha_1 - \alpha_2$};
-
-    \filldraw (0, .5) circle (.7pt);
-    \filldraw (1, .5) circle (.7pt);
-    \node[left] at (0, .5) {$\alpha_4 - \alpha_1$};
-
-    \filldraw (.2,  .2) circle (.7pt);
-    \filldraw (.2, 1.2) circle (.7pt);
-    \node[above left] at (.2, 1.2) {$\alpha_4 - \alpha_2$};
-
-    \filldraw (1.2,  .2) circle (.7pt);
-    \filldraw (1.2, 1.2) circle (.7pt);
-    \node[below right] at (1.2, .2) {$\alpha_2 - \alpha_4$};
-  \end{tikzpicture}
-\end{center}
-
-% TODO: Historical citation needed!
-% TODO: Mention at the start of the chapter that we are following Weyl's
-% footsteps in here
-One can then construct representations \(V_i\) of \(\sl_n(k)\) whose highest
-weights are \(\alpha_1 + \cdots + \alpha_i\). In fact, whenever we can find
-finitely many generators of \(\beta_i\) of the set of dominant integral weights
-and finite-dimensional representations \(V_i\) of \(\mathfrak{g}\) whose
-highest weights are \(\beta_i\) we can construct a finite-dimensional
-representation of \(\mathfrak{g}\) whose highest weight is some dominant
-integral \(\lambda \in P\) by tensoring symmetric powers of the \(V_i\)'s. This
-is the approach we'll take to prove theorem~\ref{thm:weak-dominant-weight}, as
-historically this was Weyl's first proof of the theorem. As of now, however, we
-don't have the necessary tools to construct a standard set of generators of the
-dominant integral weights of some arbitrary semisimple \(\mathfrak{g}\), let
-alone the representations \(V_i\). Indeed, Weyl's work was based on Cartan's
-classification of finite-dimensional complex simple Lie algebras, which we so
-far have neglected to mention.
+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?
 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
@@ -1819,10 +1706,3 @@ guarantees that \(v = 1 \otimes v \in V\) is a non-zero weight vector of
 The challenge then is to show that the irreducible component of \(v\) in \(V\)
 is finite-dimensional -- see chapter 20 of \cite{humphreys} for a proof.
 
-This approach has the advantage of working over fields other than \(k\), but
-in keeping with our general theme of preferring geometric proofs over purely
-algebraic ones we will instead take this as an opportunity to dive into
-Cartan's classification. In the next chapter we will explore the structure of
-complex semisimple Lie algebras, and in the process of doing so we will reduce
-the proof of theorem~\ref{thm:weak-dominant-weight} to a proof by exhaustion.
-