\documentclass[11pt, letterpaper]{amsart}

\usepackage{amsthm, amssymb}
\usepackage{graphicx,tikz}
\usepackage[all]{xy}

\usetikzlibrary{decorations.pathreplacing}
\usetikzlibrary{arrows}

\setlength{\parindent}{0pt}
\setlength{\parskip}{\baselineskip}

 \theoremstyle{plain}
 %\newtheorem*{theorem}{Theorem}
 \newtheorem{theorem}{Theorem}[section]
 \newtheorem{thm}{Theorem}[section]
 \newtheorem*{claim}{Claim}
 \newtheorem*{lemma*}{Lemma}
 \newtheorem{lemma}[theorem]{Lemma}
 %\newtheorem*{corollary}{Corollary}
 \newtheorem{cor}[theorem]{Corollary}
 %\newtheorem*{proposition}{Proposition}
 \newtheorem{proposition}[theorem]{Proposition}

 \theoremstyle{definition}
 \newtheorem*{definition}{Definition}
  \newtheorem*{defn}{Definition}
 \newtheorem*{example}{Example}
 \newtheorem*{cexample}{Counter-example}
 \newtheorem*{remark}{Remark}
\newtheorem*{homework}{Homework}
\newtheorem*{question}{Question}
\newtheorem*{answer}{Answer}
\newtheorem*{idea}{Idea}
\newtheorem*{recall}{Recall}
\newtheorem*{note}{Note}
\newtheorem*{fact}{Fact}

\newcommand{\A}{\mathcal{A}}
\newcommand{\B}{\mathcal{B}}
\newcommand{\cC}{\mathcal{C}}
\newcommand{\hilb}{\mathcal{H}}
\newcommand{\cQ}{\mathcal{Q}}

\newcommand{\g}{\mathfrak{g}}
\newcommand{\h}{\mathfrak{h}}

\newcommand{\C}{\mathbb{C}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}

\newcommand{\id}{\mathbf{1}}


\newcommand{\directSum}{\oplus}
\newcommand{\DirectSum}{\bigoplus}
\newcommand{\tensor}{\otimes}
\newcommand{\Tensor}{\bigotimes}
\newcommand{\acton}{\circlearrowright}

\newcommand{\join}{\vee}
\newcommand{\meet}{\wedge}


%\newcommand{}{}
%\newcommand{}{}
%\newcommand{}{}

\begin{document}

\title{Representation theory of $SU(N)$, the Pieri rule, fusion rules for representations of $LSU(N)$ (Monday 10:15am)}
 \author{Speaker:  Hiro Tanaka \\ Typist: Emily Peters}
 \date{\today}
 \thanks{please email texttt{eep@math.mit.edu} with corrections and improvements!}

\maketitle

\begin{abstract}
Notes from the ``Conformal Field Theory and Operator Algebras workshop," August 2010, Oregon.
\end{abstract}

Overview:
$V_f$, $V_g$ are irreps of $G$ (in this talk, we use $G=SU(N)$).  We have decompositions $V_f \tensor V_g = \DirectSum_h N_{fg}^h V_h$.

Meanwhile, if $H_f$ and $H_g$ are irreps of $\tilde{LG}$, we have decompositions $H_f \boxtimes H_g = \DirectSum_h N_{fg}^h \cdot sign(\sigma) H_h$ (for soe $\sigma$ in the affine Weyl group $\Lambda_0 \rtimes S_N$.

\begin{note}
Impliciti in this formula is semisimplicity of the Connes fusion category.
\end{note}

Okay, let's get down to details.

Representations of $SU(N)$:

Two principals to take for granted (cause I don't want to explain them):

1.  studying complex reps of some simply connected Lie group G is the same as studying complex reps of a Lie algebra $\mathfrak{g}$.

2.  complex reps of $\mathfrak{g}$ are in 1-1 correspondence with complex reps of $\mathfrak{g} \tensor_\R \C$.

Goal:  get to define the works signature, highest weight vector.

Down to business.

\begin{example}
($su = $Skew hermetian 3-by-3 matrices)
$su(3) \tensor \C = sl_3(\C)$.

$sl_3$ acts on $sl_3$, the adjoint rep, by:  Given $X$ in first, $v$ in second, $X(v)=[X,v]=Xv-vX$.

This rep splits into a sum of eigenspaces;  $\mathfrak{h} \subset sl_3$ is a Cartan subalgebra; $\mathfrak{h}=$ diagonal matrices in $sl_3$.

How does $\mathfrak{h} \acton sl_3$?  

$\mathfrak{h}$ acting on $\mathfrak{h}$ kills $\h$ (they commute).

Define $E_{ij}$ for $i \neq j$ to be the matrix with one 1, in position $i,j$, and the rest of the entries are 0.  If $X=diag(a_1, a_2, a_3)$ then $X(E_{ij})=(a_i-a_j)E_{ij}$.
\end{example}

\begin{defn}
Let $\L_i \in \h^\vee = hom(\h, \C);$  $L_i:  X \mapsto (X)_{ii}$.  Then $X(E_{ij})=(L_i - L_j)(X) E_{ij}$.
\end{defn}

Picture 1:  how the Cartan subalgebra acts on $sl_3$.

okay, now how do the off-diagonal elements of $sl_3$ act?  

\begin{claim}
The off-diagonal matrices, say $v_\beta$, will take some $v \in \g_{\alpha}$ (here, $\g=sl_3$ and $\alpha \in \h^\vee$) and take it to $v_{\alpha + \beta} \in \g_{\alpha+\beta}$.  elements $\alpha \in \h^\vee$ are called {\em weights} of the adjoint rep.
\end{claim}

\begin{proof}
Take $X \in \h $.  Need to show $X(v_\beta(v_\alpha))=(\alpha+\beta)(X)v_\beta(v_\alpha)$.  By Liebniz, 
\begin{align*}
X(v_\beta(v_\alpha))& = v_\beta(X(v_\alpha)) + [X,v_\beta] (v_\alpha)\\
 & =v_\beta (\alpha(X) v_\alpha) + \beta(X) v_\beta(v_\alpha)\\
 &  = (\alpha(X) + \beta(X))v_\beta(v_\alpha)
\end{align*}
\end{proof}

More picture 1.  
 
 The upshot:  $E_{ij}, i \leq j$ -- the upper triangular matrices -- ``raise vectors''.  Lower triangular matrices ``lower vectors''.  Ie, we've defined a partial order relation on these vector spaces, based on distance (and direction) from line.  

Observation:  there's some $\alpha \in \h^\vee$ such that $\g_\alpha$ is in the kernel of all raising operators.

\begin{defn}
Such an $\alpha$ is called the {\em highest weight} of a rep.
\end{defn}

\begin{example}
$\alpha=L_1-L_3$ is the highest weight of the adjoint representation.
\end{example}

\begin{example}
$sl_3 \acton \C^3=V$ by $V= \C e_1 \directSum \C e_2 \directSum \C e_3$
and $ \C e_1=V_{L_1}$, $ \C e_2=V_{L_2}$, $ \C e_3=V_{L_3}$.

picture 2

Observation:  The orbit of $e_1$ under lowering operators recovers the entire representation $V$.  This is true in general.
\end{example}

\begin{fact}  $V$ is an irrep of $sl_n(\C)$ and $v_\alpha$ is the unique highest weight vector, then $V$ is recovered by applying lowering operators of $v_\alpha$.
\end{fact}

Now, on to the idea of signature:  how to find highest weight vectors.

\begin{defn}
A {\em signature} (called "positive weight" by people other than Wasserman) is $g \in \Z^N$ such that $f_1 \geq f_2 \geq \cdots \geq f_N \geq 0$.
\end{defn}

\begin{question}
Is there a rep of $sl_N(\C)$ such that the highest weight vector has weight $\Sigma f_i L_i$?
\end{question}

\begin{answer}
Yes!  Take $e_f=(e_1)^{\tensor(f_1 - f_2)} \tensor (e_1 \wedge e_2)^{\tensor(f_2 - f_3)} \tensor \cdots (e_1  \wedge \cdots \wedge e_n)^{\tensor(f_n)} \in V^{\tensor(\Sigma f_i)}$.
\end{answer}

\begin{example} In the adjoint action $sl_3 \acton sl_3$, the highest weight vector is $V_\alpha, \alpha=L_1-L_3$.  What's the signature?  $(1,0,-1)$ is not a valid signature; fortunately it's equal to $(2,1,0)$.   This is because:
\end{example}

\begin{fact}
$f = (a,a,\ldots, a)+f$ as the vector sum  from $(a,a,\ldots, a)$ is zero. 
\end{fact}

\begin{defn}
{\em Young diagrams}:  
picture 3

Given a signature $f$, the associated Young diagram to $f$ has $f_i$ boxes in row 2.
\end{defn}

\begin{thm} {\em Pieri rule}:  Notation: $[k] = (1,1,\ldots, 1)$ -- young diagram having $k$ vertical boxes.  $V_f \tensor V_{[k]} = \DirectSum_{g \geq_k f} V_g $ where $ g \geq_k f$ is obtained by adding $k$ boxes to $f$, without adding two boxes in any row.
\end{thm}

\begin{example}
pic 4
\end{example}

\begin{question}
What is trivial rep?
\end{question}

%\begin{answer}
%$V_\empty$, also $V_{[n]}$ for $sl_n$.
%\end{answer}

\begin{remark} When looking at reps of $\tilde{LSU(N)}$, signatures need to be permissible, ie, $f_1-f_N \leq \ell$.
\end{remark}

\begin{theorem} Verlinde Formula:
If $$V_f \tensor V_g= \DirectSum N_{fg}^h V_h$$ then $$H+f \boxtimes H_g = \DirectSum_h N_gh^h sign(\sigma_h) H_{h'}.$$
(Go back and forth between $SU(N)$ and loops by looking at the rep where the auxilery action on the circle, acts trivially).  Here $h'$ is obtained from $h$ but must be permissible.  Details in Wasserman
\end{theorem}

Action of affine Weyl group:

\begin{defn}
$\Lambda_0 = \{ (N+\ell)(m_i) | (m_i) \in \Z^n, \Sigma m_i = 0\}$; $S_N$ =symm group on $N$ letters.

The {\em affine Weyl group} is $\Lambda_0 \rtimes S_n$  (translations and reflections).
\end{defn}

\begin{tikzpicture}
\node (h) at (0,0) {$\h$};
\node (L1) at (0:2cm) {$L_1$};
\node (L2) at (120:2cm) {$L_2$};

\draw [->] (h) -- (L1)+(L2);
\end{tikzpicture}


\begin{example}
\end{example}


\begin{question}
\end{question}



\begin{answer}
\end{answer}



\begin{defn}
\end{defn}  



\end{document}
































