lie-algebras-and-their-representations

Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules

git clone: git://git.pablopie.xyz/lie-algebras-and-their-representations
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Commit: 8a163c1759003a8381dca2cfe35b6780280dafbe
Parent: 44628b7f88a599c910e33a7a169ef30d986fe91a
Author: Pablo <pablo-escobar@riseup.net>
Date: Fri, 25 Nov 2022 18:04:58 +0000

Standardized the word "nonzero"/"non-zero"

Diffstat

5 files changed, 22 insertions, 22 deletions

Status	File Name	N° Changes	Insertions	Deletions
Modified	sections/complete-reducibility.tex	24	12	12
Modified	sections/introduction.tex	2	1	1
Modified	sections/mathieu.tex	4	2	2
Modified	sections/semisimple-algebras.tex	2	1	1
Modified	sections/sl2-sl3.tex	12	6	6

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -27,18 +27,18 @@ definitions.
 
 \begin{definition}
   A representation of \(\mathfrak{g}\) is called \emph{indecomposable} if it is
-  not isomorphic to the direct sum of two non-zero representations.
+  not isomorphic to the direct sum of two nonzero representations.
 \end{definition}
 
 \begin{definition}
   A representation of \(\mathfrak{g}\) is called \emph{irreducible} if it has
-  no non-zero subrepresentations.
+  no nonzero subrepresentations.
 \end{definition}
 
 \begin{example}
   The trivial representation \(K\) is an example of an irreducible
   representations. In fact, every \(1\)-dimensional representation \(V\) of a
-  Lie algebra \(\mathfrak{g}\) is irreducible: \(V\) has no non-zero proper
+  Lie algebra \(\mathfrak{g}\) is irreducible: \(V\) has no nonzero proper
   subspaces, let alone \(\mathfrak{g}\)-invariant subspaces.
 \end{example}
 
@@ -57,7 +57,7 @@ representations. The existence of the decomposition should be clear from the
 definitions. Indeed, if \(V\) is representation of \(\mathfrak{g}\) a simple
 argument via induction in \(\dim V\) suffices to prove the existence: if \(V\)
 is indecomposable then there is nothing to prove, and if \(V\) is not
-indecomposable then \(V = W \oplus U\) for some \(W, U \subsetneq V\) non-zero
+indecomposable then \(V = W \oplus U\) for some \(W, U \subsetneq V\) nonzero
 subrepresentations, so that their dimensions are both strictly smaller than
 \(\dim V\) and the existence follows from the induction hypothesis. For a proof
 of uniqueness please refer to \cite{etingof}.
@@ -78,7 +78,7 @@ this is not always the case. For instance\dots
               x & \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
   \end{align*}
   is a representation of the Lie algebra \(K[x]\). Notice \(V\) has a single
-  non-zero proper subrepresentation, which is spanned by the vector \((1, 0)\).
+  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
@@ -194,11 +194,11 @@ clear things up.
 
   Suppose without any loss in generality that \(i_k = k\) for all \(k\) and
   let \(j > n\). By the maximality of our set of indexes, there is some
-  non-zero \(w \in (V_j \oplus U) \cap W\). Say \(w = v_j + v_1 + \cdots +
+  nonzero \(w \in (V_j \oplus U) \cap W\). Say \(w = v_j + v_1 + \cdots +
   v_n\) with each \(v_i \in V_i\). Then \(v_j = w - v_1 - \cdots - v_n \in V_j
-  \cap (W \oplus U)\) is non-zero. Indeed, if this is not the case we find \(0
+  \cap (W \oplus U)\) is nonzero. Indeed, if this is not the case we find \(0
   \ne w = v_1 + \cdots + v_n \in \left( \bigoplus_{i = 1}^n V_i \right) \cap
-  W\), a contradiction. This implies \(V_j \cap (W \oplus U)\) is a non-zero
+  W\), a contradiction. This implies \(V_j \cap (W \oplus U)\) is a nonzero
   subrepresentation of \(V_j\). Since \(V_j\) is irreducible, \(V_j = V_j \cap
   (W \oplus U)\) and therefore \(V_j \subset W \oplus U\). Given the arbitrary
   choice of \(j\), it then follows \(V = W \oplus U\).
@@ -609,7 +609,7 @@ a representation}.
 \begin{proposition}
   The Casimir element \(C_V \in \mathcal{U}(\mathfrak{g})\) is central, so that
   \(C_V : W \to W\) is an intertwining operator for any \(\mathfrak{g}\)-module
-  \(W\). Furthermore, \(C_V\) acts in \(V\) as a non-zero scalar operator
+  \(W\). Furthermore, \(C_V\) acts in \(V\) as a nonzero scalar operator
   whenever \(V\) is a non-trivial finite-dimensional irreducible representation
   of \(\mathfrak{g}\).
 \end{proposition}
@@ -683,7 +683,7 @@ establish\dots
       0 \arrow{r} & K \arrow{r} & W \arrow{r}{\pi} & K \arrow{r} & 0
     \end{tikzcd}
   \end{equation}
-  implies \(W\) is 2-dimensional. Take any non-zero \(w \in W\) outside of the
+  implies \(W\) is 2-dimensional. Take any nonzero \(w \in W\) outside of the
   image of the inclusion \(K \to W\).
 
   % TODOOOOOOOOO: Fix this
@@ -696,7 +696,7 @@ 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 non-zero 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
@@ -739,7 +739,7 @@ establish\dots
 
   Finally, we consider the case where \(V\) is not irreducible. Suppose
   \(H^1(\mathfrak{g}, W) = 0\) for all \(\mathfrak{g}\)-modules with \(\dim W <
-  \dim V\) and let \(W \subset V\) be a proper non-zero subrepresentation. Then
+  \dim V\) and let \(W \subset V\) be a proper nonzero subrepresentation. Then
   the exact sequence
   \begin{center}
     \begin{tikzcd}

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -443,7 +443,7 @@ Other interesting classes of Lie algebras are the so called \emph{simple} and
 \begin{definition}\label{thm:sesimple-algebra}
   A Lie algebra \(\mathfrak{g}\) is called \emph{semisimple} if it is the
   direct sum of simple Lie algebras. Equivalently, a Lie algebra
-  \(\mathfrak{g}\) is called \emph{semisimple} if it has no non-zero solvable
+  \(\mathfrak{g}\) is called \emph{semisimple} if it has no nonzero solvable
   ideals.
 \end{definition}

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -128,7 +128,7 @@ A particularly well behaved class of examples are the so called
   \notin 2 \mathbb{Z}\), so that \(K[x, x^{-1}] = \bigoplus_{k \in \mathbb{Z}}
   K x^k\) is a degree \(1\) admissible weight \(\mathfrak{sl}_2(K)\)-module. It
   follows from the remark at the end of example~\ref{ex:submod-is-weight-mod}
-  that any non-zero subrepresentation \(W \subset K[x, x^{-1}]\) must contain a
+  that any nonzero subrepresentation \(W \subset K[x, x^{-1}]\) must contain a
   monomial \(x^k\). But since the operators \(-\frac{\mathrm{d}}{\mathrm{d}x} +
   \frac{x^{-1}}{2}, x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} : K[x,
   x^{-1}] \to K[x, x^{-1}]\) are both injective, this implies all other
@@ -882,7 +882,7 @@ 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,
 this implies the action of \(f\) is actually bijective, so we can obtain a
-non-zero vector in \(K[x, x^{-1}]_{2 k} = K x^k\) for any \(k \in \mathbb{Z}\)
+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
 \(f^{-1}\) denote the differential operator \((-
 \sfrac{\mathrm{d}}{\mathrm{d}x} + \sfrac{x^{-1}}{2})^{-1}\), which is the

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -335,7 +335,7 @@ restriction of the Killing form to the Cartan subalgebra.
 
   For the second statement, note that if the eigenvalues of \(\mathfrak{h}\) do
   not span all of \(\mathfrak{h}^*\) then there is some \(H \in \mathfrak{h}\)
-  non-zero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is
+  nonzero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is
   to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in
   \mathfrak{g}\). Another way of putting it is to say \(H\) is an element of
   the center \(\mathfrak{z}\) of \(\mathfrak{g}\), which is zero by the

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -77,7 +77,7 @@ around \(\lambda\).
 
 Our main objective is to show \(V\) is determined by this string of
 eigenvalues. To do so, we suppose without any loss in generality that
-\(\lambda\) is the right-most eigenvalue of \(h\), fix some non-zero \(v \in
+\(\lambda\) is the right-most eigenvalue of \(h\), fix some nonzero \(v \in
 V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
 
 \begin{theorem}\label{thm:basis-of-irr-rep}
@@ -141,7 +141,7 @@ words\dots
 \begin{proof}
   If \(W\) is an irreducible representation of \(\mathfrak{sl}_2(K)\) whose
   right-most eigenvalue of \(h\) is \(\lambda\) and \(w \in W_\lambda\) is
-  non-zero, consider the linear isomorphism
+  nonzero, consider the linear isomorphism
   \begin{align*}
     T : V     & \to     W      \\
         f^k v & \mapsto f^k w
@@ -323,7 +323,7 @@ half of the remaining elements of \(\mathfrak{sl}_3(K)\). This is exactly
 analogous to the situation we found in \(\mathfrak{sl}_2(K)\): \(h\)
 corresponds to the subalgebra \(\mathfrak{h}\), and the eigenvalues of \(h\) in
 turn correspond to linear functions \(\lambda : \mathfrak{h} \to k\) such that
-\(H v = \lambda(H) \cdot v\) for each \(H \in \mathfrak{h}\) and some non-zero
+\(H v = \lambda(H) \cdot v\) for each \(H \in \mathfrak{h}\) and some nonzero
 \(v \in V\). We call such functionals \(\lambda\) \emph{eigenvalues of
 \(\mathfrak{h}\)}, and we say \emph{\(v\) is an eigenvector of
 \(\mathfrak{h}\)}.
@@ -464,7 +464,7 @@ adjoint action of \(\mathfrak{h}\).
 
 \begin{definition}
   Given a representation \(V\) of \(\mathfrak{sl}_3(K)\), we'll call the
-  non-zero eigenvalues of the action of \(\mathfrak{h}\) in \(V\) \emph{weights
+  nonzero eigenvalues of the action of \(\mathfrak{h}\) in \(V\) \emph{weights
   of \(V\)}. As you might have guessed, we'll correspondingly refer to
   eigenvectors and eigenspaces of a given weight by \emph{weight vectors} and
   \emph{weight spaces}.
@@ -738,7 +738,7 @@ root spaces of \(\mathfrak{sl}_3(K)\) act on the weight spaces of \(V\) via
 translation, this implies that \(E_{1 2}\), \(E_{1 3}\) and \(E_{2 3}\) all
 annihilate \(V_\lambda\), or otherwise one of \(V_{\lambda + \alpha_1 -
 \alpha_2}\), \(V_{\lambda + \alpha_1 - \alpha_3}\) and \(V_{\lambda + \alpha_2
-- \alpha_3}\) would be non-zero -- which contradicts the hypothesis that
+- \alpha_3}\) would be nonzero -- which contradicts the hypothesis that
 \(\lambda\) lies the furthest along the direction we chose. In other words\dots
 
 \begin{theorem}
@@ -1104,7 +1104,7 @@ simpler than that.
   \oplus W\) generated by \(v + w\). Since \(v + w\) is a highest weight of \(V
   \oplus W\), it follows from corollary~\ref{thm:irr-component-of-high-vec}
   that \(U\) is irreducible. The projection maps \(\pi_1 : U \to V\), \(\pi_2 :
-  U \to W\), being non-zero homomorphism between irreducible representations of
+  U \to W\), being nonzero homomorphism between irreducible representations of
   \(\mathfrak{sl}_3(K)\) must be isomorphism. Finally,
   \[
     V \cong U \cong W