lie-algebras-and-their-representations

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -53,8 +53,8 @@ algebra is to classify the indecomposable representations. This is because\dots
 
 \begin{theorem}[Krull-Schmidt]\label{thm:krull-schmidt}
   Every finite-dimensional representation of a Lie algebra can be uniquely --
-  up the isomorphisms and reordering of the summands -- decomposed into a
-  direct sum of indecomposable representations.
+  up to isomorphism and reordering of the summands -- decomposed into a direct
+  sum of indecomposable representations.
 \end{theorem}
 
 Hence finding the indecomposable representations suffices to find \emph{all}
@@ -136,8 +136,8 @@ clear things up.
     \end{tikzcd}
   \end{center}
   be an exact sequence of representations of \(\mathfrak{g}\). We can suppose
-  without any loss of generality that \(W \subset V\) is a subrepresentation
-  and \(i\) is its inclusion in \(V\), for if this is not the case there is an
+  without loss of generality that \(W \subset V\) is a subrepresentation and
+  \(i\) is its inclusion in \(V\), for if this is not the case there is an
   isomorphism of sequences
   \begin{center}
     \begin{tikzcd}

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -66,7 +66,7 @@ and the eigenvalues of \(h\) all have the form \(\lambda - 2 k\) for some
 Even more so, if \(a = \max \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne 0 \}\) and
 \(b = \min \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne 0 \}\) we can see that
 \[
-  \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le n \le b}} V_{\lambda - 2 k}
+  \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le k \le b}} V_{\lambda - 2 k}
 \]
 is also an \(\mathfrak{sl}_2(K)\)-invariant subspace, so that the eigenvalues
 of \(h\) form an unbroken string
@@ -85,9 +85,9 @@ V_\lambda\) and consider the set \(\{v, f v, f^2 v, \ldots\}\).
   action of \(\mathfrak{sl}_2(K)\) on \(V\) is given by the formulas
   \begin{equation}\label{eq:irr-rep-of-sl2}
     \begin{aligned}
-        f^k v & \overset{e}{\mapsto} k(n + 1 - k) f^{k - 1} v
+        f^k v & \overset{e}{\mapsto} k(\lambda + 1 - k) f^{k - 1} v
       & f^k v & \overset{f}{\mapsto} f^{k + 1} v
-      & f^k v & \overset{h}{\mapsto} (n - 2 k) f^k v
+      & f^k v & \overset{h}{\mapsto} (\lambda - 2 k) f^k v
     \end{aligned}
   \end{equation}
 \end{proposition}
@@ -203,10 +203,10 @@ In particular, if the eigenvalues of \(V\) all have the same parity -- i.e.
 they are either all even integers or all odd integers -- and the dimension of
 each eigenspace is no greater than \(1\) then \(V\) must be irreducible, for if
 \(U, W \subset V\) are subrepresentations with \(V = W \oplus U\) then either
-\(W_\lambda = 0\) for all \(\lambda\) or \(U_\lambda = 0\) for all \(\lambda\).
-To conclude our analysis all it is left is to show that for each \(n\) there is
-some finite-dimensional irreducible \(V\) whose highest weight is \(\lambda\).
-Surprisingly, we have already encountered such a \(V\).
+\(W_\lambda = 0\) for all \(\lambda\) or \(U_\lambda = 0\) for all \(\lambda
+\in \mathfrak{h}^*\). To conclude our analysis all it is left is to show that
+for each \(n\) there is some finite-dimensional irreducible \(V\) whose highest
+weight is \(\lambda\). Surprisingly, we have already encountered such a \(V\).
 
 \begin{theorem}\label{thm:sl2-exist-unique}
   For each \(n \ge 0\) there exists a unique irreducible representation of
@@ -536,9 +536,9 @@ Visually,
 In general, we find\dots
 
 \begin{proposition}
-  Given \(i < j\), the subalgebra \(\mathfrak{s}_{\alpha_i - \alpha_j} = K
-  \langle E_{i j}, E_{j i}, [E_{i j}, E_{j i}] \rangle\) is isomorphic to
-  \(\mathfrak{sl}_2(K)\). In addition, given weight \(\lambda \in
+  Given \(i < j\), the subalgebra \(\mathfrak{s}_{i j} = K \langle E_{i j},
+  E_{j i}, [E_{i j}, E_{j i}] \rangle\) is isomorphic to
+  \(\mathfrak{sl}_2(K)\). In addition, given a weight \(\lambda \in
   \mathfrak{h}^*\) of \(V\), the space
   \[
     W = \bigoplus_{k \in \mathbb{Z}} V_{\lambda - k (\alpha_i - \alpha_j)}