lie-algebras-and-their-representations

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -15,9 +15,9 @@ For instance, one can readily check that a representation \(V\) of the
 \(n\)-dimensional Abelian Lie algebra \(K^n\) is nothing more than a choice of
 \(n\) commuting operators \(V \to V\) -- corresponding to the action of the
 canonical basis elements \(e_1, \ldots, e_n \in K^n\). In particular, a
-1-dimensional representation of \(K^n\) is just a choice of \(n\) scalars
+\(1\)-dimensional representation of \(K^n\) is just a choice of \(n\) scalars
 \(\lambda_1, \ldots, \lambda_n\). Different choices of scalars yield
-non-isomorphic representations, so that the 1-dimensional representations of
+non-isomorphic representations, so that the \(1\)-dimensional representations of
 \(K^n\) are parameterized by points in \(K^n\).
 
 This goes to show that classifying the representations of Abelian algebras is
@@ -43,7 +43,7 @@ smaller pieces. This leads us to the following 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}
@@ -87,7 +87,7 @@ 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 \(\lambda \in K\) 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}
 
@@ -706,7 +706,7 @@ establish\dots
   Since \(\dim W = 2\), the irreducible component \(\mathcal{U}(\mathfrak{g})
   \cdot w\) of \(w\) in \(W\) is either \(K w\) or \(W\) itself. But this
   component cannot be \(W\), since the image the inclusion \(K \to W\) is a
-  1-dimensional representation -- i.e. a proper nonzero subrepresentation.
+  \(1\)-dimensional representation -- i.e. a proper nonzero subrepresentation.
   Hence \(K w\) is invariant under the action of \(\mathfrak{g}\). In
   particular, \(X w = 0\) for all \(X \in \mathfrak{g}\). Since \(w\) lies
   outside the image of the inclusion \(K \to W\), \(\pi(w) \ne 0\) -- which is
@@ -910,7 +910,7 @@ 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
-  1-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
@@ -633,7 +633,7 @@ 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
@@ -894,7 +894,7 @@ 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 \(f \in \mathfrak{sl}_2(K)\) on \(K[x, x^{-1}]\) 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
@@ -1427,7 +1427,7 @@ results.
   \[
     V \cong Z \boxtimes V_1 \boxtimes \cdots \boxtimes V_n
   \]
-  where \(Z\) is a 1-dimensional representation of \(\mathfrak{z}\) and \(V_i\)
+  where \(Z\) is a \(1\)-dimensional representation of \(\mathfrak{z}\) and \(V_i\)
   is an irreducible weight \(\mathfrak{s}_i\)-module.
 \end{proposition}
 
@@ -1439,7 +1439,7 @@ results.
 \end{proposition}
 
 We've previously seen that the representations of Abelian Lie algebras,
-particularly the 1-dimensional ones, are well understood. Hence to classify the
+particularly the \(1\)-dimensional ones, are well understood. Hence to classify the
 irreducible representations of an arbitrary reductive algebra it suffices to
 classify those of its simple components. To classify these representations we
 can apply Fernando's results and reduce the problem to constructing the

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -340,7 +340,7 @@ Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and
 \(\mathfrak{sl}_3(K)\) one can show\dots
 
 \begin{proposition}\label{thm:root-space-dim-1}
-  The eigenspaces \(\mathfrak{g}_\alpha\) are all 1-dimensional.
+  The eigenspaces \(\mathfrak{g}_\alpha\) are all \(1\)-dimensional.
 \end{proposition}
 
 The proof of the first statement of
@@ -388,7 +388,7 @@ in the case of \(\mathfrak{sl}_3(K)\) we show\dots
 The proof of proposition~\ref{thm:distinguished-subalgebra} is very technical
 in nature and we won't include it here, but the idea behind it is simple:
 recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both
-1-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\)
+\(1\)-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\)
 is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]
 \ne 0\) and that no generator of \([\mathfrak{g}_\alpha, \mathfrak{g}_{-
 \alpha}] \ne 0\) is annihilated by \(\alpha\), so that by adjusting scalars we

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -139,7 +139,7 @@ self-evident: we have just provided a complete description of the action of
 \(\mathfrak{sl}_2(K)\) on \(V\). In particular, this goes to show\dots
 
 \begin{corollary}
-  Every eigenspace of the action of \(h\) on \(V\) is 1-dimensional.
+  Every eigenspace of the action of \(h\) on \(V\) is \(1\)-dimensional.
 \end{corollary}
 
 \begin{proof}
@@ -217,7 +217,7 @@ Surprisingly, we have already encountered such a \(V\).
 
   Either way, by the previous observation that a finite-dimensional
   representation whose eigenvalues all have the same parity and whose
-  corresponding eigenspace are all 1-dimensional must be irreducible, \(V\) is
+  corresponding eigenspace are all \(1\)-dimensional must be irreducible, \(V\) is
   irreducible. As for the uniqueness of \(V\), it suffices to notice that if
   \(W\) is a finite-dimensional irreducible representation of
   \(\mathfrak{sl}_2(K)\) with right-most eigenvector \(w\) then relations