lie-algebras-and-their-representations

diff --git a/preamble.tex b/preamble.tex
@@ -175,3 +175,22 @@
 
 % A normal subobject in a pointed cathegory
 \newcommand{\normal}{\triangleleft}
+
+% Command for marking a node inside a matrix
+\newcommand{\tm}[2]{%
+  \tikz[overlay,remember picture,baseline] \node [anchor=base] (#1) {$#2$};%
+}
+
+% Command for drawing a vertical line between nodes in a matrix
+\newcommand{\DrawVLine}[3][]{%
+  \begin{tikzpicture}[overlay,remember picture]
+    \draw[shorten <=0.3ex, #1] (#2.north) -- (#3.south);
+  \end{tikzpicture}
+}
+
+% Command for drawing a horizontal line between nodes in a matrix
+\newcommand{\DrawHLine}[3][]{%
+  \begin{tikzpicture}[overlay,remember picture]
+    \draw[shorten <=0.2em, #1] (#2.west) -- (#3.east);
+  \end{tikzpicture}
+}

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -564,17 +564,41 @@ In general, we find\dots
 \end{proposition}
 
 \begin{proof}
-  For the first claim, it suffices to notice the map
+  In effect, if \(i \ne k \ne j\) then \(\mathfrak{s}_{i j}\) is the subalgebra
+  of matrices whose \(k\)th row and \(k\)th collumn are nil. For instance, if
+  \(i = 1\) and \(j = 3\) then
+  \[
+    \mathfrak{s}_{1 3}
+    = \begin{pmatrix} K & 0 & K \\ 0 & 0 & 0 \\ K & 0 & K \end{pmatrix}
+      \cap \mathfrak{sl}_3(K)
+  \]
+
+  In this case, the map
   \begin{align*}
-    \mathfrak{sl}_2(K) & \to \mathfrak{s}_{i j}     \\
-                     e & \mapsto E_{i j}            \\
-                     f & \mapsto E_{j i}            \\
-                     h & \mapsto [E_{i j}, E_{j i}]
+    \mathfrak{s}_{1 3} & \to \mathfrak{sl}_2(K) \\
+    \begin{pmatrix} a & 0 & b \\ 0 & 0 & 0 \\ c & 0 & -a \end{pmatrix}
+    & \mapsto
+    \begin{pmatrix}
+                  a &    \tm{topA}{0} &              b \\
+      \tm{leftA}{0} &               0 & \tm{rightA}{0} \\
+                  c & \tm{bottomA}{0} &             -a
+    \end{pmatrix}
+    = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}
+    \DrawVLine[black, thick, opacity=0.5]{topA}{bottomA}
+    \DrawHLine[black, thick, opacity=0.5]{leftA}{rightA}
   \end{align*}
-  is an isomorphism.
+  is an isomorphism of Lie algebras. In general, the map
+  \begin{align*}
+    \mathfrak{s}_{i j} & \to     \mathfrak{sl}_2(K) \\
+    E_{i j}            & \mapsto e                  \\
+    E_{j i}            & \mapsto f                  \\
+    [E_{i j}, E_{j i}] & \mapsto h
+  \end{align*}
+  which ``erases the \(k\)th row and the \(k\)th collumn'' of a matrix is an
+  isomorphism.
 
   To see that \(W\) is invariant under the action of \(\mathfrak{s}_{i j}\), it
-  suffices to notice \(E_{i j}\) and \(E_{j i}\) map \(v \in V_{\lambda + k
+  suffices to notice \(E_{i j}\) and \(E_{j i}\) map \(v \in V_{\lambda - k
   (\alpha_i - \alpha_j)}\) to \(E_{i j} v \in V_{\lambda - (k - 1) (\alpha_i -
   \alpha_j)}\) and \(E_{j i} v \in V_{\lambda - (k + 1) (\alpha_i -
   \alpha_j)}\). Moreover,
@@ -596,9 +620,6 @@ In general, we find\dots
   a contradiction.
 \end{proof}
 
-% TODO: Note that the subalgebra s_ij is obtained by "excluding" a given row
-% and a given column from the matrices in sl3
-
 As a first consequence of this, we show\dots
 
 \begin{definition}