lie-algebras-and-their-representations

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -37,7 +37,7 @@ definitions.
 
 \begin{example}
   The trivial representation \(K\) is an example of an irreducible
-  representations. In fact, every \(1\)-dimensional representation \(V\) of a
+  representations. In fact, every 1-dimensional representation \(V\) of a
   Lie algebra \(\mathfrak{g}\) is irreducible: \(V\) has no nonzero proper
   subspaces, let alone \(\mathfrak{g}\)-invariant subspaces.
 \end{example}
@@ -81,7 +81,7 @@ this is not always the case. For instance\dots
   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 \(\lambda \in \CC\) then \(\lambda = 1\) and \(b = 0\). Hence \(V\) is
-  indecomposable -- it cannot be broken into a direct sum of \(1\)-dimensional
+  indecomposable -- it cannot be broken into a direct sum of 1-dimensional
   subrepresentations -- but it is evidently not irreducible.
 \end{example}
 
@@ -900,7 +900,7 @@ proposition~\ref{thm:quotients-by-rads}. In practice this translates to\dots
   Every irreducible representation of \(\mathfrak{g}\) is the tensor product of
   an irreducible representation of its semisimple part
   \(\mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})}\) and a
-  one-dimensional representation of \(\mathfrak{g}\).
+  1-dimensional representation of \(\mathfrak{g}\).
 \end{theorem}
 
 Having finally reduced our initial classification problem to that of

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -625,7 +625,7 @@ algebras \(\mathcal{U}(f) : \mathcal{U}(\mathfrak{g}) \to
 It is important to note, however, that \(\mathcal{U} : K\text{-}\mathbf{LieAlg}
 \to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of our functor
 \(K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). For instance, if
-\(\mathfrak{g} = K\) is the \(1\)-dimensional Abelian Lie algebra then
+\(\mathfrak{g} = K\) is the 1-dimensional Abelian Lie algebra then
 \(\mathcal{U}(\mathfrak{g}) \cong K[x]\), which is infinite-dimensional.
 Nevertheless, proposition~\ref{thm:universal-env-uni-prop} may be restated
 as\dots

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -880,7 +880,7 @@ find \(V \subsetneq \mathcal{M}[\lambda]\). In fact, we may find
 
 This wasn't an issue an example~\ref{ex:laurent-polynomial-mod} because we
 verified that the action of \(f \in \mathfrak{sl}_2(K)\) in \(K[x, x^{-1}]\) is
-injective. Since all weight spaces of \(K[x, x^{-1}]\) are \(1\)-dimensional,
+injective. Since all weight spaces of \(K[x, x^{-1}]\) are 1-dimensional,
 this implies the action of \(f\) is actually bijective, so we can obtain a
 nonzero vector in \(K[x, x^{-1}]_{2 k} = K x^k\) for any \(k \in \mathbb{Z}\)
 by translating between weight spaced using \(f\) and \(f^{-1}\) -- here

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -157,7 +157,7 @@ words\dots
 Other important consequences of theorem~\ref{thm:basis-of-irr-rep} are\dots
 
 \begin{corollary}
-  Every \(h\) eigenspace is one-dimensional.
+  Every \(h\) eigenspace is 1-dimensional.
 \end{corollary}
 
 \begin{proof}
@@ -203,7 +203,7 @@ has the form
     & V_n \arrow[bend left=60]{l}{f}
   \end{tikzcd}
 \end{center}
-where \(V_{n - 2 k}\) is the one-dimensional eigenspace of \(h\) associated to
+where \(V_{n - 2 k}\) is the 1-dimensional eigenspace of \(h\) associated to
 \(n - 2 k\) and \(n = \dim V - 1\). Even more so, we explicitly know
 \[
   V = \bigoplus_{k = 0}^n K f^k v