lie-algebras-and-their-representations

diff --git a/sections/coherent-families.tex b/sections/coherent-families.tex
@@ -104,7 +104,8 @@ combinatorial counterpart.
 \section{Coherent \(\mathfrak{sp}_{2n}(K)\)-families}
 
 % TODO: Fix n >= 2
-% TODO: Does the analysis in here work for n = 1?
+% TODO: Does the analysis in here work for n = 1? We certainly need to adapt
+% the contitions of the following lemma in this situation
 
 Consider the Cartan subalgebra \(\mathfrak{h} \subset \mathfrak{sp}_{2n}(K)\)
 of diagonal matrices, as in Example~\ref{ex:cartan-of-sp}, and the basis
@@ -210,8 +211,13 @@ Example~\ref{ex:sl-canonical-basis}. Also fix \(\rho = \sfrac{1}{2} \beta_1 +
 \end{lemma}
 
 \begin{definition}
-  A \emph{\(\mathfrak{sl}_n\)-sequence} \(m\) is a \(n\)-tuple \(m \in K^n\)
-  such that \(m_1 + \cdots + m_n = 0\).
+  A \emph{\(\mathfrak{sl}_n\)-sequence} \(m\) is a \(n\)-tuple \(m = (m_1,
+  \ldots, m_n) \in K^n\) such that \(m_1 + \cdots + m_n = 0\).
+\end{definition}
+
+\begin{definition}
+  A \(k\)-tuple \(m = (m_1, \ldots, m_k) \in K^k\) is called \emph{ordered} if
+  \(m_i - m_{i + 1}\) is a positive integer for all \(i < k\).
 \end{definition}
 
 % TODO: Revise the notation for this? I don't really like calling this
@@ -235,31 +241,32 @@ Example~\ref{ex:sl-canonical-basis}. Also fix \(\rho = \sfrac{1}{2} \beta_1 +
   \(\mathfrak{h}^*\) is given by the dot action and the action of \(W\) on the
   space of \(\mathfrak{sl}_n\)-sequences is given my permuting coordinates. A
   weight \(\lambda \in \mathfrak{h}^*\) satisfies the conditions of
-  Lemma~\ref{thm:sl-bounded-weights} if, and only if the diferences between all
-  but one consecutive coordinates of \(m(\lambda)\) are positive integers --
-  i.e. there is some unique \(i < n\) such that \(m(\lambda)_i - m(\lambda)_{i
-  + 1}\) is \emph{not} a positive integer.
+  Lemma~\ref{thm:sl-bounded-weights} if, and only if \(m(\lambda)\) is
+  \emph{not} ordered, but becomes ordered after removing one term.
 \end{proposition}
 
 % TODO: Explain the significance of 𝓑 + and 𝓑 -: these are the subsets whose
 % union corresponds to condition (i)
 \begin{definition}
   We denote by \(\mathscr{B}\) the set of \(\mathfrak{sl}_n\)-sequences \(m\)
-  such that \(m_i - m_{i + 1}\) is a positive integer for all but a single \(i
-  < n\). We also consider the subsets \(\mathscr{B}^+ = \{m \in \mathscr{B} :
-  m_1 - m_2 \ \text{is \emph{not} a positive integer}\}\) and \(\mathscr{B}^- =
-  \{m \in \mathscr{B} : m_{n-1} - m_n \ \text{is \emph{not} a positive
-  integer}\}\).
+  which are \emph{not} ordered, but becomes ordered after removing one term. We
+  also consider the \emph{extremal} subsets \(\mathscr{B}^+ = \{m \in
+  \mathscr{B} : (m_2, m_3, \ldots, m_n) \ \text{is ordered}\}\) and
+  \(\mathscr{B}^- = \{m \in \mathscr{B} : (m_1, m_2, \ldots, m_{n - 1}) \
+  \text{is ordered}\}\).
 \end{definition}
 
+% TODO: Explain that for each m ∈ 𝓑  there is a unique i such that so that
+% m_i - m_i+1 is not a positive integer. For m ∈ 𝓑 + this is i = 1, while for
+% m ∈ 𝓑 - this is i = n-1
 % TODO: Explain the intuition behind defining the arrows like so: the point is
 % that if there is an arrow m(λ) → m(μ) then μ = σ_i ∙ λ for some i, which
 % implies L(μ) is contained in 𝓔𝔁𝓽(L(λ)) - so that L(μ) is also bounded and
 % 𝓔𝔁𝓽(L(λ)) ≅ 𝓔𝔁𝓽(L(μ))
 \begin{definition}
-  Given \(m, m' \in \mathscr{B}\), say there is an arrow \(m \to m'\) if the
-  unique \(i\) such that \(m_i - m_{i + 1}\) is not a positive integer is such
-  that \(m' = \sigma_i \cdot m\).
+  Given \(m, m' \in \mathscr{B}\), say there is an arrow \(m \to m'\) if there
+  some \(i\) such that \(m_i - m_{i + 1}\) is \emph{not} a positive integer and
+  \(m' = \sigma_i \cdot m\).
 \end{definition}
 
 It should then be obvious that\dots
@@ -296,9 +303,9 @@ all \(i\) and \(j\).
   following.
   \begin{enumerate}
     \item If \(m\) is regular and integral then there exists\footnote{Notice
-      that in this case $m' \notin \mathscr{B}$, however.} a unique \(m' \in W
-      \cdot m\) such that \(m_1' > m_2' > \cdots > m_n'\), in which case the
-      connected component of \(m\) is given by
+      that in this case $m' \notin \mathscr{B}$, however.} a unique ordered
+      \(m' \in W \cdot m\), in which case the connected component of \(m\) is
+      given by
       \[
         \begin{tikzcd}[cramped, sep=small]
           \sigma_1 \sigma_2 \cdots \sigma_i \cdot m'           \rar &
@@ -320,9 +327,12 @@ all \(i\) and \(j\).
       \mathscr{B}^+\) and \(\sigma_{n-1} \cdots \sigma_i \cdot m' \in
       \mathscr{B}^-\).
 
+    % TODOO: What happens when i = 1?? Do we need to suppose i > 1?
+    % TODO: For instance, consider m = (1, 1, -2)
     \item If \(m\) is singular then there exists unique \(m' \in W \cdot m\)
-      and \(i\) such that \(m_1' > m_2' > \cdots > m_i' = m_{i + 1}' > \cdots >
-      m_n'\), in which case the connected component of \(m\) is given by
+      and \(i\) such that \(m'_i = m'_{i + 1}\) and \((m_1', \cdots, m_i')\)
+      and \((m_{i + 1}', \ldots, m_n')\) are ordered, in which case the
+      connected component of \(m\) is given by
       \[
         \begin{tikzcd}[cramped, sep=small]
           \sigma_1 \sigma_2 \cdots \sigma_{i-1} \cdot m'       \rar &
@@ -344,8 +354,8 @@ all \(i\) and \(j\).
       \(\sigma_{n-1} \cdots \sigma_{i+1} \cdot m' \in \mathscr{B}^-\).
 
     \item If \(m\) is non-integral then there exists unique \(m' \in W \cdot
-      m\) such that \(m_2' > m_3' > \cdots > m_n'\), in which case the
-      connected component of \(m\) is given by
+      m\) with \(m' \in \mathscr{B}^+\), in which case the connected component
+      of \(m\) is given by
       \[
         \begin{tikzcd}[cramped]
           m'                                    \rar      &
@@ -355,8 +365,7 @@ all \(i\) and \(j\).
           \sigma_{n-1} \cdots \sigma_1 \cdot m'      \lar &
         \end{tikzcd}
       \]
-      with \(m' \in \mathscr{B}^+\) and \(\sigma_{n-1} \cdots \sigma_1 \cdot m'
-      \in \mathscr{B}^-\).
+      with \(\sigma_{n-1} \cdots \sigma_1 \cdot m' \in \mathscr{B}^-\).
   \end{enumerate}
 \end{lemma}