\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{ifthen}
\usepackage{ulem} % for strikethrough \sout{...} command
\usepackage{color}

\usepackage{wasysym}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{-0.25in}
\setlength{\evensidemargin}{-0.25in}
\setlength{\topmargin}{-0.25in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{9.75in}
\setlength{\textwidth}{7in}

\newcommand{\bdot}{\, {\scriptstyle{\bullet}} \,}
\newcommand{\B}[1]{{\bf {#1}}}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}

\newcommand{\myspace}{\parbox[c][0.25in]{0.25in}{\hfill}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}
\newcommand{\force}{{\noindent $\mbox{\;}$}}

\newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}}

\begin{document}

\pagestyle{empty}

\noindent
\parbox{2.25in}{\bf Math 4010/5530}
\hfill {\Large \bf Homework \#6} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Apr. $5^\mathrm{th}$, 2023}

\vspace{0.2in}

Let $\mathfrak{g}$ be a Lie algebra over a field $\mathbb{F}$ and let $V$ be a $\mathfrak{g}$-module.

\vspace{0.1in}

Also, let $\mathfrak{sl}_2(\mathbb{C}) = \mathrm{span} \{ f,h,e \}$ where as usual: $[h,f]=-2f$, $[h,e]=2e$, and $[e,f]=h$.

\vspace{0.1in}

\begin{description}
\item[\#1 Restriction] %(EW page 74 Exercise 8.3) 
Given a $\mathfrak{g}$-module $V$ and a subalgebra $\mathfrak{h}$ of $\mathfrak{g}$, we can think of $V$ as a $\mathfrak{h}$-module by restricting $\mathfrak{g}$'s action to $\mathfrak{h}$: \quad $h \bdot v$ where $h \in \mathfrak{h}$ and $v \in V$.

Notice that the matrices of the form $\displaystyle \begin{bmatrix} \star & \star & 0 \\ \star & \star & 0 \\ 0 & 0 & 0 \end{bmatrix}$ sitting inside $\mathfrak{sl}_3(\mathbb{C})$ give us an isomorphic copy of $\mathfrak{sl}_2(\mathbb{C})$ (similarly we can treat $\mathfrak{sl}_k(\mathbb{C})$ as a subalgebra of $\mathfrak{sl}_n(\mathbb{C})$ for any $k \leq n$). This allows us to treat $V = \mathfrak{sl}_3(\mathbb{C})$ as an $\mathfrak{sl}_2(\mathbb{C})$-module with action: $\B x \bdot \B y = [\B x, \B y]$ (we are restricting $\mathfrak{sl}_3$'s adjoint action to this subalgebra isomorphic to $\mathfrak{sl}_2(\mathbb{C})$). By Weyl's theorem, we know that $V$ is completely reducible. 

Show that $V \cong V(2) \oplus V(1) \oplus V(1) \oplus V(0)$ by finding a basis of $h$-eigenvectors ($h = \mathrm{diag}(1,-1,0)$).

{}[{\it Note:} $V(m)$ denotes the irreducible $\mathfrak{sl}_2(\mathbb{C})$-module of highest weight $m$ whose dimension is $m+1$.]

\item[\#2 Eigen-Determinism] %(EW page 75 Exercise 8.4) 
Suppose that $V$ is a finite dimensional $\mathfrak{sl}_2(\mathbb{C})$-module. Using Weyl's theorem and the classification of irreducible $\mathfrak{sl}_2(\mathbb{C})$-modules, show that $V$ is the direct sum of $k$ irreducible submodules where $k = \dim(V_0)+\dim(V_1)$ and $V_\lambda = \{ \B v \in V \;|\; h \bdot \B v = \lambda \B v \}$. 

\vspace{0.05in}

Explain why $V$ is {\bf not} determined by the set of $h$'s eigenvalues alone. However, it is determined (up to module isomorphism) by $h$'s eigenvalues {\it counting multiplicity}.

\item[\#3 Not Quite Trivial] %(EW page 65 Exercise 7.11) 
We explore $1$-dimensional representations.

\begin{enumerate}[(a)]
\item Suppose that $\varphi : \mathfrak{g} \to \mathfrak{gl}(1,\mathbb{F})$ is a 1-dimensional representation of $\mathfrak{g}$. Show that $\varphi(\mathfrak{g}') = \{\B 0\}$. \quad 

{}[Recall that $\mathfrak{g}' = [\mathfrak{g},\mathfrak{g}]$.]

\item {\bf (Grad Students)} Show that any representation of $\mathfrak{g}/\mathfrak{g}'$ can be viewed as a representation of $\mathfrak{g}$ on which $\mathfrak{g}'$ acts trivially. 

\item {\bf (Grad Students)} When $\mathbb{F}=\mathbb{C}$, show that if $\mathfrak{g}' \not= \mathfrak{g}$, then $\mathfrak{g}$ has infinitely many non-isomorphic 1-dimensional modules, but if $\mathfrak{g}'=\mathfrak{g}$, then the only 1-dimensional representation of $\mathfrak{g}$ is the trivial representation.

\end{enumerate}

{\it Note:} For 3(a) it may be worth noting that we can identify $\mathfrak{gl}(1,\mathbb{F})=\mathbb{F}$ (these are naturally isomorphic). Having done so, in part (a) we are showing that if $\varphi$ is a 1-dimensional representation (so $\varphi \in \mathfrak{g}^*$ and it is a homomorphism), then $\varphi(\mathfrak{g}')= \B 0$. It might be helpful to note that the converse is also true: if $\varphi$ is a dual vector vanishing on $\mathfrak{g}'$, then $\varphi$ is a 1-dimensional representation (if we identify $\mathfrak{gl}(1,\mathbb{F})$ and $\mathbb{F}$).

\end{description}

\end{document}

