lie-algebras-and-their-representations

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -80,15 +80,15 @@ this is not always the case. For instance\dots
 \begin{example}\label{ex:indecomposable-not-irr}
   The space \(V = K^2\) endowed with the homomorphism of Lie algebras
   \begin{align*}
-    \rho : K[x] & \to     \mathfrak{gl}(V) \\
-              x & \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
+    x \cdot e_1 & = e_1 & x \cdot e_2 = e_1 + e_2
   \end{align*}
   is a representation of the Lie algebra \(K[x]\). Notice \(V\) has a single
-  nonzero proper subrepresentation, which is spanned by the vector \((1, 0)\).
-  This is because if \((a + b, b) = \rho(x) \ (a, b) = \lambda (a, b)\) for
-  some \(\lambda \in K\) then \(\lambda = 1\) and \(b = 0\). Hence \(V\) is
-  indecomposable -- it cannot be broken into a direct sum of \(1\)-dimensional
-  subrepresentations -- but it is evidently not irreducible.
+  nonzero proper subrepresentation, which is spanned by the vector \(e_1\).
+  This is because if \((a + b) e_1 + b e_2 = x \cdot (a e_1 + b e_2) = \lambda
+  \cdot (a e_1 + b e_2)\) for some \(\lambda \in K\) then \(\lambda = 1\) and
+  \(b = 0\). Hence \(V\) is indecomposable -- it cannot be broken into a direct
+  sum of \(1\)-dimensional subrepresentations -- but it is evidently not
+  irreducible.
 \end{example}
 
 This counterexample poses an interesting question: are there conditions one can