lie-algebras-and-their-representations

diff --git a/preamble.tex b/preamble.tex
@@ -172,3 +172,6 @@
 
 % Quotient object
 \newcommand{\mfrac}[2]{\mathlarger{\sfrac{#1}{#2}}}
+
+% A normal subobject in a pointed cathegory
+\newcommand{\normal}{\triangleleft}

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -55,14 +55,44 @@ A popular alternative to definition~\ref{thm:sesimple-algebra} is\dots
   and an Abelian Lie algebra.
 \end{definition}
 
-% TODO: Give a small proof? (At least for n = 2)
+% TODO: Comment on the standard basis of sl2 beforehand
 \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 algebra \(\mathfrak{sl}_2(K)\). To see this, notice that any ideal
+  \(\mathfrak{a} \normal \mathfrak{sl}_2(K)\) must be stable under the adjoint
+  action of \(h\). But the operator \(\operatorname{ad}(h)\) is diagonalizable,
+  with eigenvalues \(0\) and \(\pm 2\). Hence \(\mathfrak{a}\) must be spanned
+  by some of the eigenvectors \(e, f, h\) of \(\operatorname{ad}(h)\). If \(h
+  \in \mathfrak{a}\), then \([e, h] = - 2 e \in \mathfrak{a}\) and \([f, h] = 2
+  f \in \mathfrak{a}\), so \(\mathfrak{a} = \mathfrak{sl}_2(K)\). If \(e \in
+  \mathfrak{a}\) then \([f, e] = - h \in \mathfrak{a}\), so again
+  \(\mathfrak{a} = \mathfrak{sl}_2(K)\). Similarly, if \(f \in \mathfrak{a}\)
+  then \([e, f] = h \in \mathfrak{a}\) and \(\mathfrak{a} =
+  \mathfrak{sl}_2(K)\). More generaly, the Lie algebra \(\mathfrak{sl}_n(K)\)
+  is simple for each \(n > 0\) -- see the section of \cite[ch. 6]{kirillov} on
+  invariant bilinear forms and the semisimplicity of classical Lie algebras.
 \end{example}
 
-% TODO: Add gl_n(K) as an example of a reductive algebra
+\begin{example}
+  The Lie algebra \(\mathfrak{gl}_n(K)\) is reducible. Indeed,
+  \[
+    X
+    =
+    \begin{pmatrix}
+      a_{1 1} - \frac{\operatorname{Tr}(X)}{n} & \cdots & a_{1 n} \\
+       \vdots & \ddots &  \vdots \\
+      a_{n 1} & \cdots & a_{n n} - \frac{\operatorname{Tr}(X)}{n}
+    \end{pmatrix}
+    +
+    \begin{pmatrix}
+      \frac{\operatorname{Tr}(X)}{n} & \cdots & 0 \\
+      \vdots & \ddots & \vdots \\
+      0 & \cdots & \frac{\operatorname{Tr}(X)}{n}
+    \end{pmatrix}
+  \]
+  for each matrix \(X = (a_{i j})_{i j}\). In other words,
+  \(\mathfrak{gl}_n(K) = \mathfrak{sl}_n(K) \oplus K \operatorname{Id} \cong
+  \mathfrak{sl}_n(K) \oplus K\).
+\end{example}
 
 I suppose this last definition explains the nomenclature, but the reason why
 semisimple Lie algebras are interesting at all is still unclear. In particular,
@@ -400,7 +430,7 @@ we introduce a distinguished element of \(\mathcal{U}(\mathfrak{g})\), known as
   all representations \(V\) of \(\mathfrak{g}\): its action commutes with the
   action of any other element of \(\mathfrak{g}\).
 
-  % TODO: Prove that the action is not zero when V is non-trivial
+  % TODOOO: Prove that the action is not zero when V is non-trivial
   In particular, it follows from Schur's lemma that if \(V\) is
   finite-dimensional and irreducible then \(C\) acts in \(V\) as a scalar
   operator.
@@ -658,9 +688,10 @@ The primary goal of this section is proving\dots
   \(V\) of \(\mathfrak{sl}_2(K)\) with \(\dim V = n\).
 \end{theorem}
 
+% TODO: Point out the standard basis beforehand
 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
+its structure. We begin our analysis by recalling that the elements
 \begin{align*}
   e & = \begin{pmatrix} 0 & 1 \\ 0 &  0 \end{pmatrix} &
   f & = \begin{pmatrix} 0 & 0 \\ 1 &  0 \end{pmatrix} &
@@ -1705,10 +1736,11 @@ The fundamental difference between these two cases is thus the fact that \(\dim
 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
-encountered when dealing with finite groups: representations of Abelian groups
-and algebras are generally much simpler to understand than the general case.
+% TODO: Add a note on how irreducible representations of Abelian algebras are
+% all one dimensional to the previous chapter
+The rational behind fixing an Abelian subalgebra is a simple one: we have seen
+in the previous chapter that representations of Abelian
+algebras are generally much simpler to understand than the general case.
 Thus it make sense to decompose a given representation \(V\) of
 \(\mathfrak{g}\) into subspaces invariant under the action of \(\mathfrak{h}\),
 and then analyze how the remaining elements of \(\mathfrak{g}\) act on this
@@ -1718,17 +1750,15 @@ there are fewer elements outside of \(\mathfrak{h}\) left to analyze.
 Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
 \subset \mathfrak{g}\), which leads us to the following definition.
 
-% TODO: Define the associated Borel subalgebra as soon as possible (we need to
-% fix an ordering of the roots beforehand)
 \begin{definition}
-  An subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a
-  Cartan subalgebra of \(\mathfrak{g}\)} if it is Abelian,
-  \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in
+  An subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a Cartan
+  subalgebra of \(\mathfrak{g}\)} if is self-normalizing -- i.e. \([X, H] \in
+  \mathfrak{h}\) for all \(H \in \mathfrak{h}\) if, and only if \(X \in
+  \mathfrak{h}\) -- and nilpotent. Equivalently for reductive \(\mathfrak{g}\),
+  \(\mathfrak{h}\) is called \emph{a Cartan subalgebra of \(\mathfrak{g}\)} if
+  it is Abelian, \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in
   \mathfrak{h}\) and if \(\mathfrak{h}\) is maximal with respect to the former
-  two properties\footnote{More generally, a Cartan subalgebra of an arbitrary
-  Lie algebra \(\mathfrak{g}\) -- not necessarily semisimple -- is defined as
-  a self-normalizing nilpotent subalgebra. This definition turns out to be
-  equivalent to our characterization whenever \(\mathfrak{g}\) is reductive.}.
+  two properties.
 \end{definition}
 
 \begin{proposition}
@@ -1752,9 +1782,8 @@ Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
   properties -- i.e. a Cartan subalgebra.
 \end{proof}
 
-% TODO: Add futher details to this: why are both this subalgebras ad-diagonal?
-That said, we can easily compute concrete examples. For instance, one can
-readily check that every pair of diagonal matrices commutes, so that
+We have already seen some concrete examples. For instance, one can readily
+check that every pair of diagonal matrices commutes, so that
 \[
   \mathfrak{h} =
   \begin{pmatrix}
@@ -1764,14 +1793,16 @@ readily check that every pair of diagonal matrices commutes, so that
          0 &      0 & \cdots &      K
   \end{pmatrix}
 \]
-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 -- and hence nilpotent -- subalgebra of \(\mathfrak{gl}_n(K)\). A
+simple calculation also shows that if \(i \ne j\) then the coefficient of
+\(E_{i j}\) in \([E_{i i}, X]\) is the same as the coefficient of \(E_{i j}\)
+in \(X\), for all \(X \in \mathfrak{gl}_n(K)\). In particular, if \([E_{i i},
+X]\) is diagonal for all \(i\), then so is \(X\) -- i.e. \(\mathfrak{h}\) is
+self-normalizing. Hence \(\mathfrak{h}\) 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
@@ -1798,7 +1829,7 @@ 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
+% TODOOO: 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
@@ -1838,6 +1869,53 @@ As promised, this implies\dots
   \]
 \end{corollary}
 
+\begin{corollary}
+  The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate.
+\end{corollary}
+
+\begin{proof}
+  Consider the eigenspace decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus
+  \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint representation, where
+  \(\alpha\) ranges over all nonzero eigenvalues of the adjoint action of
+  \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = \mathfrak{h}\).
+
+  Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h})
+  \mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the
+  other hand, since \(\mathfrak{h}\) is self-normalizing, if \([X, H] = 0 \in
+  \mathfrak{h}\) for all \(H \in \mathfrak{h}\) then \(X \in \mathfrak{h}\) --
+  i.e. \(\mathfrak{g}_0 \subset \mathfrak{h}\). So the eigespace decomposition
+  becomes
+  \[
+    \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha
+  \]
+
+  We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to
+  \(\mathfrak{g}_\alpha\) with respect to \(B\) for any \(\alpha \ne 0\).
+  Indeed, given \(X \in \mathfrak{g}_\alpha\) and \(H_1, H_2 \in \mathfrak{h}\)
+  with \(\alpha(H_1) \ne 0\) we have
+  \[
+    \alpha(H_1) \cdot B(X, H_2)
+    = B([H_1, X], H_2)
+    = - B([X, H_1], H_2)
+    = - B(X, [H_1, H_2])
+    = 0
+  \]
+
+  Hence the non-degeneracy of \(B\) implies the non-degenaracy of its
+  restriction.
+\end{proof}
+
+We should point out that the restriction of \(B\) to \(\mathfrak{h}\) is
+\emph{not} the Killing form of \(\mathfrak{h}\). In fact, since
+\(\mathfrak{h}\) is Abelian, its Killing form is identically zero -- which is
+hardly ever a non-degenerate form.
+
+\begin{note}
+  Since \(B\) is induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\),
+  it induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in
+  \(\mathfrak{h}^*\). We denote this form by \(B\).
+\end{note}
+
 We now have most of the necessary tools to reproduce the results of the
 previous chapter in a general setting. Let \(\mathfrak{g}\) be a
 finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\)
@@ -1851,22 +1929,8 @@ 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.
-
-% TODO: Add a refenrence to a proof (probably Humphreys)
-% TODO: Move this to just after the definition of a Cartan subalgebra
-\begin{theorem}
-  The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate.
-\end{theorem}
-
-% TODO: Note that the restriction of the Killing form to h is NOT the Killing
-% form of h! Otherwise h would be semisimple, but h is Abelian
-
-\begin{note}
-  Since \(B\) is induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\),
-  it induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in
-  \(\mathfrak{h}^*\). We denote this form by \(B\).
-\end{note}
+consequence of the nondegeneracy of the restriction of the Killing form to the
+Cartan subalgebra.
 
 \begin{proposition}\label{thm:weights-symmetric-span}
   The eigenvalues \(\alpha\) of the adjoint action of \(\mathfrak{h}\) in
@@ -1940,23 +2004,31 @@ then\dots
   all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\).
 \end{theorem}
 
-% TODO: Rewrite this: the concept of direction has no sence in the general
-% setting
-% TODO: Replace the "fix a direction" shenanigan with a proper discussion of
-% basis
-% TODO: The proof that a basis exists is actually very similiar to the idea of
-% fixing a direction in spirit (see page 48 of Humphreys)
+% TODO: Turn this into a proper discussion of basis and give the idea of the
+% proof of existance of basis?
 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
+\(\mathfrak{h}^* \to \QQ\) such that \(Q\) lies outside of its kernel. This
 choice induces a partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of
 roots of \(\mathfrak{g}\) and once more we find\dots
 
+\begin{definition}
+  The elements of \(\Delta^+\) and \(\Delta^-\) are called \emph{positive} and
+  \emph{negative roots}, respectively. The subalgebra \(\mathfrak{b} =
+  \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is
+  called \emph{the Borel subalgebra associated with \(\mathfrak{h}\)}.
+\end{definition}
+
 \begin{theorem}
   There is a weight vector \(v \in V\) that is killed by all positive root
   spaces of \(\mathfrak{g}\).
 \end{theorem}
 
+% TODO: Here we may take a weight of maximal height, but why is it unique?
+% TODO: We don't really need to talk about height tho, we may simply take a
+% weight that maximizes B(gamma, lambda) in QQ
+% TODO: Either way, we need to move this to after the discussion on the
+% integrality of weights
 \begin{proof}
   It suffices to note that if \(\lambda\) is the weight of \(V\) lying the
   furthest along the direction we chose and \(V_{\lambda + \alpha} \ne 0\) for
@@ -2087,7 +2159,7 @@ theorem~\ref{thm:weak-dominant-weight} we used for the case when \(\mathfrak{g}
 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?
+% TODOOO: 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