= \

= \sum_{\beta, \beta'} p_\beta n_{\beta'} \<\beta, \beta'\> \leq 0.$$ So $\

=0$, which implies $0=p=n$. Thus $c_\beta =0$ for all $\beta$.
\end{proof}
It is additionally straightforward to show that every base $B$ of $R$ is of the form $B=B(\gamma)$ for any regular $\gamma$ satisfying $\<\beta, \gamma\> > 0$ for all $\beta \in B$ (see, for example \cite[Theorem 2, \S 10.1]{Hum}). So there is a base corresponding to every \dfn{Weyl chamber} (connected subset of $\fh^*_\RR - \bigcup_{\alpha \in R}$). In particular, the regular value determining the base sits in the fundamental chamber, whose walls are exactly $\{\fh_\beta ~|~ \beta \in B\}$ (see, for example, \cite[Ch VI, \S1, no. 9, Thm 2]{Bou}). In short, choosing a base, and therefore the positive roots, is the same thing as choosing a fundamental chamber.
\begin{example}For example, there are six different bases of the root system for $\fsl_3$, one for each choice of Weyl chamber:
$$\begin{tikzpicture}[scale=.2]
\begin{scope}[thick, <->]
\draw (0,-12) to (0,12) node[above]{\small$\fh_{\vep_1- \vep_2}$};
\draw (30:-12) to (30:12) node[above right]{\small$\fh_{\vep_2-\vep_3}$};
\draw (-30:12) to (-30:-12) node[above left]{\small$\fh_{\vep_1-\vep_3}$};
\end{scope}
\begin{scope}[thick, ->]
\draw (0,0) to (3,0) node[right] {\tiny $\vep_1\! - \!\vep_2$};
\draw (0,0) to (120:3) node[anchor=-60] {\tiny $\vep_2\! - \!\vep_3$};
\draw (0,0) to (60:3) node[anchor=-137] {\tiny $\vep_1\!-\!\vep_3$};
\end{scope}
\node[right] at (16,5) {$C_1: B= \{\vep_1- \vep_2, \vep_2-\vep_3\}$};
\node at (60:9){\NOTE{$C_1$}};
\node[right] at (16,3) {$C_2: B= \{\vep_2-\vep_1, \vep_1-\vep_3\}$};
\node at (120:9){\NOTE{{$C_2$}}};
\node[right] at (16,1) {$C_3: B= \{\vep_2-\vep_3, \vep_3-\vep_1\}$};
\node at (180:9){\NOTE{$C_3$}};
\node[right] at (16,-1) {$C_4: B= \{\vep_2-\vep_1, \vep_3-\vep_2\}$};
\node at (240:9){\NOTE{$C_4$}};
\node[right] at (16,-3) {$C_5: B= \{\vep_1-\vep_2, \vep_3-\vep_1\}$};
\node at (300:9){\NOTE{$C_5$}};
\node[right] at (16,-5) {$C_6: B= \{\vep_1-\vep_3, \vep_3-\vep_2\}$};
\node at (0:9){\NOTE{$C_6$}};
\end{tikzpicture}$$
\end{example}
Further, with $R^\vee = \{\alpha^\vee ~|~ \alpha \in R\}$, the elements of $R$ are all scalars of elements of $R^\vee$, and so the following corollary is immediate.
\begin{cor}\label{cor:base-R-check}
Let $R^\vee = \{\alpha^\vee ~|~ \alpha \in R\}$. If $\gamma \in \fh^*$ is a regular weight, then $B^\vee(\gamma) = \{ \beta^\vee ~|~ \beta \in B(\gamma)\}$ is a base of $R^\vee$ (is a basis of $\fh$ contained in $R^\vee$ satisfying \eqref{eq:base-dfn}).
\end{cor}
A choice of fundamental chamber $C$, and therefore base and positive roots, also determines a \subdfn{weights}{partial order} on $\fh_\RR^*$. Namely, with $\mu, \lambda \in \fh_\RR^*$, let
\begin{equation}\label{eq:partial-order-on-weights}
\lambda > \mu \quad \text{ if } \lambda - \mu \text{ is the sum of simple roots.}
\end{equation}
Note that this means $\alpha \in R$ is a positive root if and only if $\alpha > 0$.
Finally, a couple of loose ends.
\begin{lemma} Let $B$ be a base of $R$.
\begin{enumerate}
\item For every $\beta, \beta' \in B$, $\<\beta, \beta'\> \leq 0$ and $\beta - \beta' \notin B$.
\item Each $\alpha \in R^+$ can be written as $\alpha = \gamma_1 + \gamma_2 + \cdots + \gamma_m$ with $\gamma_i \in B$ (not necessarily distinct) in such a way that $\gamma_1 + \cdots + \gamma_j \in R^+$ for each $1 \leq j \leq m$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) is a direct consequence of Lemma \ref{lem:acute-obtuse-roots}.
For part (2), if $\alpha \in B$, then this is trivial. If not,
in the proof of linear independence in Proposition \ref{prop:Base}, we actually showed that any set of pairwise obtuse weights sitting to one side of a hyperplane forms a linearly independent set.
But since $B$ is a basis, $\{\alpha\} \cup B$ is not linearly independent, and so $\< \alpha, \beta\> > 0$ for some $\beta \in B$, and so $\alpha - \beta \in R$.
So inductively, since $R$ is finite, $\alpha$ can be written as $\alpha = \gamma_1 + \gamma_2 + \cdots + \gamma_m$ with $\gamma_i \in B$ (not necessarily distinct) in such a way that $\alpha_{(j)} = \gamma_1 + \cdots + \gamma_j \in R^+$ for each $1 \leq j \leq m$. Further, since $\alpha_{(j)}$ is the positive sum of simple roots, $\beta \in R^+$.
\end{proof}
\subsection{Abstract root systems, Coxeter diagrams, and Dynkin diagrams}
In light of the table in \eqref{table:root-angles}, we can now classify all possible sets of roots. Often in the literature, roots are handles completely abstractly, and then root systems associated to Lie algebras are presented as examples. Here, we have mostly avoided this abstraction until now. However some abstraction will help us with classification, so we start with some axiomatics about roots in general.
Let $E$ be a euclidean space over $\RR$ with inner product $\<,\>$. A \dfn{symmetry} $s_\lambda$ associated to $\lambda \in E$ is an automorphism of $E$ satisfying
$$s_\lambda(\lambda) = -\lambda \qquad \text{ and } \qquad E_\lambda = \{ v \in E ~|~ s_\lambda(v) = v\} \text{ is a hyperplane in }E.$$
It is immediate that $E_\lambda^\perp = \RR \lambda$, and $s_\lambda$ has order 2 and is determined by $\RR \lambda$. With $\lambda^\vee$ the element which is uniquely determined by
$$\<\lambda^\vee, E_\lambda\> = 0 \qquad \text{and} \qquad \<\lambda^\vee, \lambda = 2\>,$$
we have $s_\lambda(\mu) = \mu - \<\lambda^\vee, \mu\> \lambda.$
A subset $R \subset E$ is called a \dfn{root system} in $E$ if
\begin{enumerate}[(R1)]
\item $R$ is finite, spans $E$, and does not contain 0;
\item if $\alpha \in R$, then there is a symmetry $s_\alpha$ acting on $E$ leaving $R$ invariant; and
\item for each $\alpha, \beta \in R$, $s_\alpha(\beta) - \beta$ is an integer multiple of $\alpha$.
\end{enumerate}
Note that $s_\alpha(\alpha) = \alpha - 2 \alpha = -\alpha \in R$ necessarily. Also,
for any finite spanning set $R$for $E$, there is at most on symmetry associated to any vector $\lambda$ which leaves $R$ invariant (this follows from an analysis of eigenvalues associated to the product of any two such symmetries).
A root system is said to be \subdfn{root system}{reduced} if for all $\alpha \in R$, the roots proportional to $\alpha$ are $\pm \alpha$. If a root system is not reduced, then it $R$ contains a pair $\alpha, t \alpha \in R$ with $0 < t < 1$. Then (R3) forces $t = \half$. So for any $\alpha \in R$, the only roots proportional to $\alpha$ are $\pm \alpha$ and either $\pm \half \alpha$ or $\pm 2 \alpha$.
\begin{thm}
The roots associated to finite-dimensional semisimple complex Lie algebras form reduced root systems.
\end{thm}
In particular, everything we've shown about roots associated to Lie algebras apply to reduced root systems. The non-reduced root systems arise for Lie algebras generated over non-algebraically closed fields, such as $\RR$.
A subset of $B$ a root system $R$ is called a \subdfn{root system}{base} if $S$ is a basis for $E$ and for all $\alpha \in R$,
$$\alpha = \pm \sum_{\beta \in B} z_\beta \beta \qquad \text{with } z_\beta \in \ZZ_{\geq 0}.$$
The elements of $B$ are called the simple roots. Our proof of existence of bases in the previous section holds here as well.
Let $E$ be a euclidean space/$\RR$ with inner product $\<,\>$ (of any big dimension). Call a finite subset $A=\{\alpha_1, \dots, \alpha_r\}\subset E$ \emph{admissible} if
\begin{enumerate}[(i)]
\item $A$ is a set of linearly independent unit vectors ($\<\alpha_i, \alpha_i\>=1$),
\item $\<\alpha_i, \alpha_j\> \leq 0$ whenever $i \neq j$, and
\item $4\<\alpha_i,\alpha_j\>^2 \in \{0,1,2,3\}$ whenever $i \neq j$.
\end{enumerate}
Associate to any admissible set $A$ a graph $\Gamma(A)$ (called the Coxeter diagram) with vertices labeled by elements of $A$ (or $i$ short for $\alpha_i$), with $m_{i,j} = 4\<\alpha_i,\alpha_j\>^2$ edges connecting $i$ to $j$:
$$\begin{array}{c@{\qquad}l}
\begin{matrix}\begin{tikzpicture}
\node[C, label=above:{$i$}] (i) at (0,0) {};
\node[C, label=above:{$j$}] (j) at (1,0) {};
\end{tikzpicture}\end{matrix} & \text{if } 4\<\alpha_i,\alpha_j\>^2 = 0,\\
\begin{matrix}\begin{tikzpicture}
\node[C, label=above:{$i$}] (i) at (0,0) {};
\node[C, label=above:{$j$}] (j) at (1,0) {};
\draw[thick] (i) to (j);
\end{tikzpicture}\end{matrix} & \text{if } 4\<\alpha_i,\alpha_j\>^2 = 1, \\
\begin{matrix}\begin{tikzpicture}
\node[C, label=above:{$i$}] (i) at (0,0) {};
\node[C, label=above:{$j$}] (j) at (1,0) {};
\draw[thick, double distance=2pt] (i) to (j);
\end{tikzpicture}\end{matrix} & \text{if } 4\<\alpha_i,\alpha_j\>^2 = 2,\\
\begin{matrix}\begin{tikzpicture}
\node[C, label=above:{$i$}] (i) at (0,0) {};
\node[C, label=above:{$j$}] (j) at (1,0) {};
\draw[thick, double distance=3pt] (i) to (j);
\draw[thick] (i) to (j);
\end{tikzpicture}\end{matrix} & \text{if } 4\<\alpha_i,\alpha_j\>^2 = 3.
\end{array}$$
Note that by normalizing the elements of any base $B$ for a set of roots $R$, you get an admissible set $A$. The \dfn{Coxeter diagram} associated to a root system $R$ is the graph associated to the normalization of a base $B$. In particular, the diagram does not depend on the chosen base (we will see that the Weyl group acts transitively on Weyl chambers, and so all Weyl chambers have the same set of angles between pairs of chamber walls).
\begin{thm}[{\cite[\S 11,12]{Ser}}]
Every root system is a sum of irreducible root systems, and a root system is irreducible if and only if the associated Coxeter diagram is connected.
\end{thm}
This follows because $R_1$ decomposes into $R_1$ and $R_2$ if and only if $R_1$ and $R_2$ are orthogonal to each other, in which case there are no edges connecting to the vertices for $R_1$ to the vertices for $R_2$.
A \dfn{Dynkin diagram} associated to a base $B$ for a root system is a decorated Coxeter graph for the associated normalized admissible set. If $\alpha_i$ is adjacent to $\alpha_j$, and the root $\beta_i$ associated to $\alpha_i$ is longer than the root $\beta_j$ associated to $\alpha_j$, decorate the $m_{i,j}$ edges connecting $\alpha_i$ to $\alpha_j$ with an arrow pointing to $\alpha_i$ (the normalization of the longer root).
Suppose that $E$ is the direct sum of (non-trivial) subspaces $E_i$, $i=1, \dots, \ell$, such that $R \subset \bigcup_{i=1}^\ell E_i$. Then $R_i = E_i \cap R$ is a root system for $E_i$. We say $R$ is the sum of subsystems $R_i$. If the only decompositions trivial ($\ell = 1$), we say $R$ is \subdfn{root system}{irreducible}. For example, the root systems associated to simple Lie algebras are irreducible, while semisimple but not simple Lie algebras have reducible root systems.
The \subdfn{root system}{Cartan matrix} associated with $R$ is the matrix $(\<\alpha, \beta^\vee\>)_{\alpha, \beta \in B}$. In particular, since $\<\alpha, \alpha^\vee\> = 2$, the diagonal entries are all 2, and since $\<\alpha, \beta\> \leq 0$ for $\alpha \neq \beta$, the off-diagonal entries are all $0, -1, -2,$ or $-3$.
\begin{prop}[{\cite[\S11, Prop 8 and \S15, Prop 13]{Ser}}]
A reduced root system is determined (up to isomorphism) by its Cartan matrix and vice versa. A root system is also determined (up to isomorphism) by its dynkin diagram.
\end{prop}
Then due to the triangular decomposition of any finite-dimensional semisimple Lie algebra, the Lie algebra is determined by its root system. So to classify all simple finite-dimensional Lie algebras, one must simply classify all connected Dynkin diagrams (of finite type). On the homework, you're asked to (1) narrow down the possible connected Coxeter diagrams, (2) show existence of admissible sets for the leftover graphs, and (3) classify the connected Dynkin diagrams (of finite type).
\medskip
\paragraph{\bf Exercise 6:} Some things about classification.
\begin{enumerate}
\item Let $A=\{\alpha_1, \dots, \alpha_r\} $ be an admissible set yielding a connected graph $\Gamma(A)$.
\begin{enumerate}
\item Show that the number of pairs of vertices connected by at least one edge strictly less than $r$. \\
{}[What is the condition on vertices being adjacent? Consider $\<\alpha, \alpha\>$ where $\alpha = \sum_{A} \alpha_i$.]
\item Show that $\Gamma(A)$ contains no cycles. [Note that any subset of an admissible set is admissible.
]
\item Show that the degree (counting multiple edges) of any vertex in $\Gamma(A)$ is no more than three.\\
{}[Take a vertex $\alpha \in A$, and let $S$ be the set containing $\alpha$ together with its neighborhood (the vertices adjacent to it). Note that in the span of $S$ is a unit vector $\beta$ which is orthogonal to $S - \{\alpha\}$, so that $\alpha = \sum_{\gamma \in S - \{\alpha\} + \{\beta\}} \<\alpha, \gamma\> \gamma$ and $\<\alpha, \beta\> \neq 0$ (why??).]
\item Show that if $S \subseteq A$ has graph
$$\Gamma(S) =
\begin{matrix}\begin{tikzpicture}
\foreach \x in {0, 1, 2, 4, 5}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (0) to (1) to (2) to (2.5,0) (3.5,0) to (4) to (5);
\node at (3,0) {$\cdots$};
\end{tikzpicture}\end{matrix},$$
then $A' = A - S + \{\sum_S \alpha\}$ is admissible (with graph $\Gamma(A')$ obtained by collapsing the subgraph $\Gamma(S)$ to a single vertex).
\item Show that $\Gamma(A)$ cannot contain any of the following graphs as subgraphs:
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {0, 1, 2, 4, 5,6}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (1) to (2) to (2.5,0) (3.5,0) to (4) to (5);
\draw[thick, double distance=2pt] (0) to (1) (5) to (6);
\node at (3,0) {$\cdots$};
\end{tikzpicture}\end{matrix}$$
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {0, 1, 2, 4, 5}{\node[C] (\x) at (\x,0) {};}
\node[C] (6a) at (6,-.5) {}; \node[C] (6b) at (6,.5) {};
\draw[thick] (1) to (2) to (2.5,0) (3.5,0) to (4) to (5) to (6a) (5) to (6b);
\draw[thick, double distance=2pt] (0) to (1);
\node at (3,0) {$\cdots$};
\end{tikzpicture}\end{matrix}$$
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 4, 5}{\node[C] (\x) at (\x,0) {};}
\node[C] (6a) at (6,-.5) {}; \node[C] (6b) at (6,.5) {};
\node[C] (0a) at (0,-.5) {}; \node[C] (0b) at (0,.5) {};
\draw[thick] (0b) to (1) (0a) to (1) to (2) to (2.5,0) (3.5,0) to (4) to (5) to (6a) (5) to (6b);
\node at (3,0) {$\cdots$};
\end{tikzpicture}\end{matrix}$$
[Use the previous part]
\item Show that the only remaining possible graphs associated to admissible sets are of one of the following four forms:
$$\begin{matrix}\begin{tikzpicture}
\node[C] (i) at (0,0) {};
\node[C] (j) at (1,0) {};
\draw[thick, double distance=3pt] (i) to (j);
\draw[thick] (i) to (j);
\end{tikzpicture}\end{matrix} $$
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {0, 1, 2, 4, 5}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (0) to (1) to (2) to (2.5,0) (3.5,0) to (4) to (5);
\node at (3,0) {$\cdots$};
\end{tikzpicture}\end{matrix}$$
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4,...,8}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (1) to (2) to (2.2,0) (2.8,0) to (3) to (4)
(5) to (6) to (6.2,0) (6.8,0) to (7) to (8);
\draw[thick, double distance=2pt] (4) to (5);
\node at (2.5,0) {$\cdots$}; \node at (6.5,0) {$\cdots$};
\end{tikzpicture}\end{matrix}$$
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4}{\node[C] (\x) at (\x-4,0) {};}
\foreach \x in {1, 2, 3}{
\node[C] (u\x) at (15:\x) {};
\node[C] (d\x) at (-15:\x) {};}
\draw[thick] (1) to (2) to (2.2-4,0) (2.8-4,0) to (3) to (4)
(4) to (u1) to (15:1.2) (15:1.8) to (u2) to (u3)
(4) to (d1) to (-15:1.2) (-15:1.8) to (d2) to (d3);
\node at (2.5-4,0) {$\cdots$};
\node[rotate=15] at (15:1.5) {$\cdots$};
\node[rotate=-15] at (-15:1.5) {$\cdots$};
\end{tikzpicture}\end{matrix}$$
\item Show the only possible graphs of the third type are
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4,5}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (1) to (2) to (2.2,0) (2.8,0) to (3) to (4);
\draw[thick, double distance=2pt] (4) to (5);
\node at (2.5,0) {$\cdots$};
\end{tikzpicture}\end{matrix}
\qquad \text{and} \qquad
\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (1) to (2) (3) to (4);
\draw[thick, double distance=2pt] (2) to (3);
\end{tikzpicture}\end{matrix}$$
[Suppose the vectors corresponding to the vertices to the left of the double bond are $\lambda_1, \dots, \lambda_\ell$ (from left to right) and the vertices to the rights of the double bond are $\mu_1, \dots, \mu_m$ (from right to left). Let $\lambda = \sum_i i \lambda_i$ and $\mu = \sum_i i \mu_i$. Show that
$\<\lambda, \lambda\> = \ell(\ell+1)/2$, $\< \mu, \mu\> = m(m+1)/2$, and $\<\lambda,\mu\>^2 = \ell^2 m^2/2$, and use the Cauchy-Schwarz inequality for inner products.]
\item Bonus: Show the only graphs of the fourth kind are
$$\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4}{\node[C] (\x) at (\x-4,0) {};}
\foreach \x in {1}{
\node[C] (u\x) at (15:\x) {};
\node[C] (d\x) at (-15:\x) {};}
\draw[thick] (1) to (2) to (2.2-4,0) (2.8-4,0) to (3) to (4)
(4) to (u1)
(4) to (d1) ;
\node at (2.5-4,0) {$\cdots$};
\end{tikzpicture}\end{matrix}
\qquad \text{ and } \qquad
\begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4}{\node[C] (\x) at (\x-4,0) {};}
\foreach \x in {1, 2}{\node[C] (u\x) at (15:\x) {};}
\node[C] (d1) at (-15:1) {};
\draw[thick] (1) to (2) to (2.2-4,0) (2.8-4,0) to (3) to (4)
(4) to (u1) to (u2)
(4) to (d1) ;
\node at (2.5-4,0) {$\cdots$};
\draw[|-|] (-3,-.5) to node[midway, below] {$4\leq * \leq 6$} (0, -.5);
\end{tikzpicture}\end{matrix}
$$
[This is like the previous part, only more so]
\end{enumerate}
\item Show that there's an admissible set associated to every remaining graph by displaying existence. Namely, associate most of the remaining possible graphs to a classical root systems (showing existence), and take for granted that the remaining five are associated to the \emph{exceptional simple Lie algebras}, $E_6, E_7, E_8, F_4$, and $G_2$:
$$E_6, E_7, E_8: \begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4}{\node[C] (\x) at (\x-4,0) {};}
\foreach \x in {1, 2}{\node[C] (u\x) at (15:\x) {};}
\node[C] (d1) at (-15:1) {};
\draw[thick] (1) to (2) to (2.2-4,0) (2.8-4,0) to (3) to (4)
(4) to (u1) to (u2)
(4) to (d1) ;
\node at (2.5-4,0) {$\cdots$};
\end{tikzpicture}\end{matrix}
$$
$$F_4: \begin{matrix}\begin{tikzpicture}
\foreach \x in {1, 2, 3, 4}{\node[C] (\x) at (\x,0) {};}
\draw[thick] (1) to (2) (3) to (4);
\draw[thick, double distance=2pt] (2) to (3);
\end{tikzpicture}\end{matrix}$$
$$G_2: \begin{matrix}\begin{tikzpicture}
\node[C] (i) at (0,0) {};
\node[C] (j) at (1,0) {};
\draw[thick, double distance=3pt] (i) to (j);
\draw[thick] (i) to (j);
\end{tikzpicture}\end{matrix} $$
\item Classify all (finite type) connected Dynkin diagrams.
\end{enumerate}
\section{Weyl groups}
Now that we're armed with all this great structure on $R$ and understand a little more about Weyl chambers and roots, let's explore some properties of Weyl groups.
Fix a fundamental chamber $C$, and therefore a base $B$ and positive set of roots $R^+$. With $B = \{\beta_1, \dots, \beta_r\}$, let $s_i = s_{\beta_i}$. Let $W$ be the group generated by $\{s_\alpha ~|~ \alpha \in R\}$.
\begin{lemma}\label{lem:simple-reflections} ~
\begin{enumerate}
\item The Weyl group $W$ is finite.
\item The form $\<,\>$ on $\fh^*_\RR$ is $W$-invariant, i.e.\
$$\ $. Consider the region contained between $\fh_{\beta_i^\vee - a}$ and $\fh_{\beta_i^\vee - (a+1)}$. Highlight parts of the paths in this region as follows.
Let $t_L$ be maximal such that $\ = a$ (the last place where the path hits $\fh_{\beta_i^\vee - a}$). If $\ \geq a+1$, then let $t_R$ be minimal such that $\ \geq a + 1$ for all $[t_R,1]$ (the first point where $p$ crosses $\fh_{\beta_i^\vee - (a+1)}$. Then fix a finite partition of $[0,1]$ given by
$$t_L = t_0 < t_1 < \dots < t_m = t_R$$
such that either
\begin{enumerate}
\item $\ = \ $ and $\ \geq \ $ for $t \in [t_j, t_{j+1}]$ ($p$ starts on the hyperplane intersecting $p(t_j)$, heads to the positive side and doubles back to that same hyperplane), or
\item $\ $ is strictly increasing for $t \in [t_j, t_{j+1}]$ and $\ \geq \ $ for all $t \geq t_{j+1}$ ($p$ heads in the positive direction, and there's nothing later on the path further to the negative).
\end{enumerate}
Highlight all segments of the path $p$ on intervals of the second kind.
For example,
\def\A{1.732}
$$\begin{tikzpicture}[scale=1]
\begin{scope}[scale = \A/2]
\filldraw[blue!10] (-2,-.5) to (-2,7) to (-3,7) to (-3,-.5);
\foreach \x in {-5, ..., 3}{\draw[thin, black!70] (\x, 7) to (\x,-.5);}
\draw[thick] (0,-.5) to (0,7) node[above]{$\fh_{\beta_i}$};
\draw[thick] (-2,-.5) to (-2,7) node[above]{$\fh_{\beta_i}+2$};
\draw[thick] (-3,7) to (-3,-.5) node[below]{$\fh_{\beta_i}+3$};
\draw[very thick, ->, red] (0,0) to (2,0) node[right, inner sep=2pt] {\tiny$\beta_i$};
\node[draw, black, fill =black, circle, inner sep=0pt, minimum size=5pt] (0) at (0,0){};
\draw[green!70!blue!90!black, thick, ->] (0,0) to (1.5,1.5/2) to (-2.5,5.5/2)
to (-1.5,6.5/2) to (-3,8/2) to (-1,10/2) to (-2.5, 11.5/2) to (-1,13/2)
node[above right] {\tiny$p$};
\draw[line width=2pt, blue] (-3,8/2) to (-2.5, 8.5/2) (-2.5,11.5/2) to (-2,12/2);
\node[below left] at (0) {$0$};
\end{scope}
\end{tikzpicture}$$
If $\ < a+1$, the operator $f_i$ acts by $0$. Otherwise, then operator $f_i$ reflects each highlighted segment from the positive to the negative side the hyperplane $\fh_{\beta_i-\ }$, dragging the rest of the path with it:
$$\begin{tikzpicture}[scale=1]
\begin{scope}[scale = \A/2]
\foreach \x in {-5, ..., 3}{\draw[thin, black!70] (\x, 7) to (\x,-.5);}
\draw[thick] (0,-.5) to (0,7) node[above]{$\fh_{\beta_i}$};
\draw[thick] (-2,-.5) to (-2,7) node[above]{$\fh_{\beta_i}+2$};
\draw[thick] (-3,7) to (-3,-.5) node[below]{$\fh_{\beta_i}+3$};
\draw[red, very thick, ->] (0,0) to (2,0) node[right, inner sep=2pt] {\tiny$\beta_i$};
\node[draw, black, fill =black, circle, inner sep=0pt, minimum size=5pt] (0) at (0,0){};
\draw[green!70!blue!90!black, thick, ->] (0,0) to (1.5,1.5/2) to (-2.5,5.5/2)
to (-1.5,6.5/2) to (-3.5,8.5/2) to (-2,10/2) to (-4, 12/2) to (-3,13/2)
node[above right] {\tiny$f_i p$};
\draw[line width=2pt, blue] (-3,8/2) to (-3.5, 8.5/2) (-4,12/2) to (-3.5,11.5/2);
\node[below left] at (0) {$0$};
\end{scope}
\end{tikzpicture}$$
Note that the result move the end of the path by $-\beta_i$. The operators $e_i$ reverses this operation, i.e.\ it is defined by $e_i f_i = 1$ (whenever $f_i$ does not act by $0$) and $f_i e_i = 1$ (whenever $e_i$ does not act by $0$). It still uses the part of the part furthest in the $-\beta_i$-direction, but reflecting from the negative side to the positive side of a hyperplane. So
$$f_i p = 0 \text{ or } (f_i p)(1) = p(1) - \beta_i \qquad \text{and} \qquad
e_i p = 0 \text{ or } (e_i p)(1) = p(1) + \beta_i.$$
A \subdfn{path model}{crystal} $\cB$ is a set of paths which is closed under the action of root operators $\{ e_i, f_i ~|~ \beta_i \in B \}$. Let $\cB(p)$ be the minimal crystal containing $p$.
\begin{example}\label{ex:L(rho)-crystal}
Let $\fg = \fsl_3$ with base $B =\{ \beta_1 = \vep_1 - \vep_2, \beta_2 = \vep_2 - \vep_3\}$. Let $p$ be the straight-line path from $0$ to $\lambda = \vep_1 - \vep_3$. Then $\cB(p)$, together with its actions by $e_1, e_2, f_1, f_2$, is given in Figure \ref{fig:L(rho)-crystal}.
\begin{figure}[b]
\caption{The crystal generated by the straight-line path to $\rho$ for $\fsl_3$.}\label{fig:L(rho)-crystal}
$$
\begin{tikzpicture}[scale=1]
\node (p) at (0,12) {\Path{(0,0) to (w1)}};
\node (1p) at (-3,9) {\Path{(0,0) to (w2)}};
\node (2p) at (3,9) {\Path{(0,0) to (w6)}};
\node (21p) at (-1.5,6) {\Path{(0,0) .. controls (-75:\A*.75) and (-45:\A*.75) .. (0,0)}};
\node (12p) at (1.5,6) {\Path{(0,0) .. controls (180+15:\A*.75) and (180-15:\A*.75) .. (0,0)}};
\node (112p) at (-3,3) {\Path{(0,0) to (w3)}};
\node (221p) at (3,3) {\Path{(0,0) to (w5)}};
\node (1221p) at (0,0) {\Path{(0,0) to (w4)}};
\node (p2) at (3,12) {$0$}; \node (p1) at (-3,12) {$0$};
\node (1p2) at (-6,9) {$0$}; \node (2p1) at (6,9) {$0$};
\node (112p2) at (-6,3) {$0$}; \node (221p1) at (6,3) {$0$};
\node (1221p2) at (3,0) {$0$}; \node (1221p1) at (-3,0) {$0$};
\begin{scope}[->, bend right=10]
\draw (p) to node[above left] {$f_1$} (1p);
\draw (1p) to node[below right] {$e_1$} (p);
\draw (2p) to node[above left] {$f_1$} (12p);
\draw (12p) to node[below right] {$e_1$} (2p);
\draw (12p) to node[above left, pos=.8] {$f_1$} (112p);
\draw (112p) to node[below right, pos=.2] {$e_1$} (12p);
\draw (221p) to node[above left] {$f_1$} (1221p);
\draw (1221p) to node[below right] {$e_1$} (221p);
\end{scope}
\begin{scope}[->, bend left=10]
\draw (p) to node[above right] {$f_2$} (2p);
\draw (2p) to node[below left] {$e_2$} (p);
\draw (1p) to node[above right] {$f_2$} (21p);
\draw (21p) to node[below left] {$e_2$} (1p);
\draw (21p) to node[above right, pos=.8] {$f_2$} (221p);
\draw (221p) to node[below left, pos=.2] {$e_2$} (21p);
\draw (112p) to node[above right] {$f_2$} (1221p);
\draw (1221p) to node[below left] {$e_2$} (112p);
\end{scope}
\begin{scope}[->]
\draw (p) to node[midway, above] {$e_2$} (p2);
\draw (p) to node[midway, above] {$e_1$} (p1);
\draw (1p) to node[midway, above] {$e_2, f_1$} (1p2);
\draw (2p) to node[midway, above] {$e_1, f_2$} (2p1);
\draw (112p) to node[midway, above] {$e_2, f_1$} (112p2);
\draw (221p) to node[midway, above] {$e_1, f_2$} (221p1);
\draw (1221p) to node[midway, below] {$e_2$} (1221p2);
\draw (1221p) to node[midway, below] {$e_1$} (1221p1);
\end{scope}
\end{tikzpicture}
$$
\end{figure}
\end{example}
We say $\cB$ and $\cB'$ are \subdfn{path model}{isomorphic crystals} if there is a bijection $\phi: \cB(p) \to \cB(p')$ with $f_i \phi(q) = \phi(f_i q)$ and $e_i \phi(q) = \phi(e_i q)$ for all $q \in \cB(p)$ and simple root operators $f_i, e_i$. The \emph{path model}{crystal graph is the graph with
$$\text{vertices } p \in \cB \qquad \text{ and } \qquad \text{ labeled edges } p \stackrel{i}{\longrightarrow} f_i p.$$
See, for example, Figure \ref{fig:L(rho)-crystal-tableau-path}. Two crystals are isomorphic if and only if they have isomorphic graphs (with the same labelings of edges).
Note that since $\rho = \sum_i \omega_i$, by pulling the (open) fundamental chamber $C$ back by $\rho$, $C-\rho$ contains the closed chamber $\bar{C}$, but not any of the walls $\fh_{\beta_i + 1}$. In fact,
$$P^{++} \to P^+ \qquad\text{defined by}\qquad \lambda \mapsto \lambda-\rho$$
is a bijection.
A \subdfn{path model}{highest weight path} is a path $p$ satisfying
$$e_i p =0 \qquad \text{for all } i=1, \dots, r.$$
For $e_i$ to act by 0 means that $\ > -1$ for all $t$ and $i$. So a path is highest weight if and only if
$$p(1) \in P^+ \qquad \text{and} \qquad p(t)\in C - \rho \text{ for all $t \in [0,1]$.}$$
The \subdfn{path model}{weight} of any path $p$ is $\wt(p) = p(1)$.
\begin{prop}Let $p$ and $p'$ be highest weight paths of the same weight. Then the crystals generated $p$ and $p'$ are isomorphic.
\end{prop}
So without ambiguity, for $\lambda \in P^+$, define \subdfn{path model}{$\cB(\lambda)$} as the crystal generated any fixed highest weight path $p_\lambda^+$ of weight $\lambda$. In Example \ref{ex:L(rho)-crystal}, we computed $\cB(\rho)$. The same crystal with a different highest weight path is in Figure \ref{fig:L(rho)-crystal-tableau-path}.
The \subdfn{path model}{character of a crystal} is
\begin{equation}\label{eq:crystal-character}
\ch(\cB) = \sum_{p \in \cB} X^{\wt(p)}.
\end{equation}
\begin{thm}For $\lambda \in P^+$,
$$\ch(\cB(\lambda)) = \ch(L(\lambda)).$$
\end{thm}
So, for example, Example \ref{ex:L(rho)-crystal} shows once again that the weight space of weight 0 in $L(\rho)$ has dimension 2.
\begin{remark}
One thing that's more exciting about the path model than methods in the previous two sections is that we don't just have a count of the multiplicities of weights, but we have a set indexing the individual dimensions of the weight spaces of any finite-dimensional $\fg$-module. In other words,
for any finite-dimensional $\fg$-module $V = \sum_{\lambda \in \hat V} L(\lambda)$, there is a a weight basis of $V$ indexed by paths in $p \in \sqcup_{\lambda \in \hat{V}} \cB(\lambda)$, where
if $v_p$ is the basis element of $V$ indexed by path $p$, $v_P$ has weight $\wt(p)$.
The action is a little trickier, though. In an ideal world, we would hope that since $f_i$ changes the weight of $p$ by $-\beta_i$, maybe we would have relations like $y_{\beta_i} v_p = v_{f_i p}$ and $x_{\beta_i} v_p = v_{e_i p}$. However, this action would not satisfy the bracket relation $[x_{\beta_i}, y_{\beta_i}] = h_{\beta_i^\vee}$. There is an initial attempt to deal with this in \cite[\S2.1]{Li95}, where they build operators for each simple root that form an $\fsl_2$-triple. However, the various $\fsl_2$-triples for various $\beta_i$'s do not interact properly with each other (they don't satisfy what are called the Chevalley-Serre relations, like $[h_\alpha, x_\beta] = \<\alpha, \beta\> x_\alpha$ and $[x_\alpha, y_\beta] = \delta_{\alpha, \beta} h_{\alpha^\vee}$). There are proper normalizations from the Lie algebras side, which fall into the study of crystal bases.
\end{remark}
\begin{prop} Let $\cB, \cB'$ be finite crystals.
\begin{enumerate}
\item $\ch(\cB) = \ch(\cB')$ if and only if $\cB \cong \cB'$.
\item The union $\cB\sqcup \cB'$ is a crystal, and
$$\ch(\cB \sqcup \cB') = \ch(\cB) + \ch(\cB').$$
\item
$$\ch(\cB) = \sum_{p \in \cB \atop p \text{ is highest weight}} \ch(\cB(\wt(p))).$$
\end{enumerate}
\end{prop}
It begins to make sense now to intro duct the notation $\cB(V)$ to denote a crystal associated to a finite-dimensional module $V$. Namely, if $V = \sum_{\lambda \in \hat{V}} L(\lambda)$, then
$\cB(V) = \bigsqcup_{\lambda \in \hat{V}} \cB(\lambda)$.
\subsubsection{Tensor product decomposition}
The concatenation of two paths $p,p'$ is defined by
$$pp' = \begin{cases}
p(2t) & 0 \leq t \leq 1/2,\\
p(1) + p'(2(t-1/2)) & 1/2 \leq t \leq 1.
\end{cases}$$
Pictorially, think of sticking $p'$ onto the end of $p$ (like vector addition). Note that $\wt(pp') = \wt(p) + \wt(p')$.
\begin{thm}~
\begin{enumerate}
\item For finite-dimensional $\fg$-modules $V, V'$,
$$\cB(V \otimes V') = \{ pp' ~|~ p \in \cB(V), p' \in \cB(V')\}.$$
\item With $\lambda, \mu \in P^+$, and $p_\lambda^+$ highest weight in $\cB(\lambda)$,
$$\ch(L(\lambda) \otimes L(\mu)) = \sum_{q \in \cB(\mu) \atop p^+_\lambda q \text{ highest weight}} \ch(L(\lambda+ \wt(q))).$$
\end{enumerate}
\end{thm}
\paragraph{\bf Type $A_r$} First, let's return to the example where $\fg=\fsl_3$. Recall the fundamental weights with resect to the base $B = \{ \beta_1 = \vep_1 - \vep_2, \beta_2 = \vep_2 - \vep_3\}$ are
$$\omega_1 = \vep_1 - \frac{1}{3}(\vep_1 + \vep_2 + \vep_3),
\quad \text{ and } \quad
\omega_2 = \vep_1 + \vep_2- \frac{2}{3}(\vep_1 + \vep_2 + \vep_3).$$
The crystal $\cB(\omega_1)$ is generated by highest weight path
$$p_{\omega_1}^+: \qquad \begin{matrix}\Path{(0,0) to (L01)}\end{matrix},$$
and contains the three paths
$$p_1 =p_{\omega_1}^+ =
\begin{matrix}\begin{tikzpicture}[scale=.5]
\draw[thin, black!50] (30:-.5) to (30:.5) (-30:-.5) to (-30:.5) (0,-.5) to (0,.5);
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (30:1){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b);
\end{scope}
\end{tikzpicture}\end{matrix} \qquad
p_2 = f_1 p_{\omega_1}^+ = \begin{matrix}\begin{tikzpicture}[scale=.5]
\draw[thin, black!50] (30:-.5) to (30:.5) (-30:-.5) to (-30:.5) (0,-.5) to (0,.5);
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (-30:-1){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b);
\end{scope}
\end{tikzpicture}\end{matrix} \qquad
p_3 = f_2f_1 p_{\omega_1}^+ = \begin{matrix}\begin{tikzpicture}[scale=.5]
\draw[thin, black!50] (30:-.5) to (30:.5) (-30:-.5) to (-30:.5) (0,-.5) to (0,.5);
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (0,-1){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b);
\end{scope}
\end{tikzpicture}\end{matrix}.$$
The weights of these paths are
\begin{align*}
\wt(p_1) = \vep_1 - \frac{1}{3}(\vep_1 + \vep_2 + \vep_3) &= \omega_1,\\
\wt(p_2) = \vep_2 - \frac{1}{3}(\vep_1 + \vep_2 + \vep_3) &= \omega_2- \omega_1,\\
\wt(p_1) = \vep_3 - \frac{1}{3}(\vep_1 + \vep_2 + \vep_3) &= \omega_3-\omega_2.
\end{align*}
where $\omega_3 = 0 = \vep_1+ \vep_2 + \vep_3 - \frac{3}{3}(\vep_1 + \vep_2 + \vep_3).$
For general $r$, we saw on the homework that the standard representation of $A_r$ is $L(\omega_1)$, and now it's not difficult to compute that weights in $L(\omega_1)$ are exactly analogous to the $r=2$ case, namely,
$$P_{\omega_1} = \{ \omega_i - \omega_{i-1} = \vep_i - \frac{1}{r+1}\sum_{i=1}^r \vep_i ~|~ i=1, \dots, r+1, \omega_0 = \omega_{r+1} = 0\}.$$
\newcommand\TensorTwo[2]{
\begin{matrix}\begin{tikzpicture}[scale=.5]
\draw[thin, black!50] (30:-.75) to (30:.75) (-30:-.75) to (-30:.75) (0,-.75) to (0,.75);
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (#1){}; \node (c) at (#2){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b) to (c);
\end{scope}
\end{tikzpicture}\end{matrix}}
Back in the case where $r=2$, the crystal for $\cB(L(\omega_1) \otimes L(\omega_1))$
is the set containing
$$
\begin{array}{c@{\qquad}c@{\qquad}c}
p_1^2 = \TensorTwo{30:1}{30:2}~,&
p_1 p_2 = \TensorTwo{30:1}{0,1}~,&
p_1 p_3 = \TensorTwo{30:1}{-30:1}~,
\\
p_2 p_1 = \TensorTwo{-30:-1}{0,1}~, &
p_2^2= \TensorTwo{-30:-1}{-30:-2}~, &
p_2 p_3 = \TensorTwo{-30:-1}{30:-1}~,\\
p_3 p_1 = \TensorTwo{0,-1}{-30:1}~, &
p_3 p_2 = \TensorTwo{0,-1}{30:-1}~,&
p_3^2= \TensorTwo{0,-1}{0,-2}~.
\end{array}
$$
The two of these that are highest weight are $p_1^2$, which has weight $2 \omega_1$, and $p_1p_2$, which has weight $\omega_2$. This is reflected in the fact that the crystal graph for $\cB(L(\omega_1) \otimes L(\omega_1))$ has two connected components:
$$\begin{matrix}
\begin{tikzpicture}[yscale=-1]
\node (p) at (0,0) {$\TensorTwo{30:1}{30:2}$};
\node (2p) at (0,2) {$\TensorTwo{-30:-1}{0,1}$};
\node (12p) at (2,3.5) {$\TensorTwo{0,-1}{-30:1}$};
\node (22p) at (-2,3.5) {$\TensorTwo{-30:-1}{-30:-2}$};
\node (122p) at (0,5) {$\TensorTwo{0,-1}{30:-1}$};
\node (1122p) at (0,7) {$\TensorTwo{0,-1}{0,-2}$};
\begin{scope}[->]
\draw (p) to node[midway,right] {$1$} (2p);
\draw (2p) to node[midway,above right] {$2$} (12p);
\draw (2p) to node[midway,above left] {$1$} (22p);
\draw (22p) to node[midway,below left] {$2$} (122p);
\draw (12p) to node[midway,below right] {$1$} (122p);
\draw (122p) to node[midway,right] {$2$} (1122p);
\end{scope}
\end{tikzpicture}\end{matrix} \qquad \text{ and } \qquad
\begin{matrix}\begin{tikzpicture}[yscale=-1]
\node (p) at (0,0) {$ \TensorTwo{30:1}{0,1}$};
\node (1p) at (0,2) {$\TensorTwo{30:1}{-30:1}$};
\node (21p) at (0,4) {$\TensorTwo{-30:-1}{30:-1}$};
\begin{scope}[->]
\draw (p) to node[midway,right] {$2$} (1p);
\draw (1p) to node[midway,right] {$1$} (21p);
\end{scope}
\end{tikzpicture}\end{matrix}
$$
So
$$\ch(L(\omega_1) \otimes L(\omega_1)) = \ch(L(2 \omega_1)) + \ch(L(\omega_2)),$$
implying
$$L(\omega_1) \otimes L(\omega_1) \cong L(2 \omega_1) \oplus L(\omega_2).$$
Stepping back, since $p_1$ is the highest weight path of $\cB(\omega_1)$, the highest weight paths in this tensor product are those for which $p_1 p_i$ are in $C - \rho$. More to the point, since concatenation by $p_i$ walks from one integral weight to one to the six nearest integral weights, $p_1 p_i$ is highest weight exactly when $\wt(p_1p_i) \in P^+$.
In general, for $\lambda \in P^+$, we have that
\begin{align*}
L(\lambda) \otimes L(\omega_1)
&= \bigoplus_{i = 1, \dots , r+1 \atop \lambda + \omega_i - \omega_{i-1} \in P^+}
L(\lambda + \omega_i - \omega_{i-1}),
\end{align*}
which has exactly $r+1$ terms when $\lambda \in P^{++}$.
\def\A{1.732}
\def\B{8}
$$\begin{tikzpicture}[scale=.75]
\filldraw[blue!10] (0,0)--(0,\B)--(30:\B)--(0,0);
\begin{scope}[thick]
\draw (0,-1) to (0,\B) node[above]{$\fh_{\alpha_1}$};
\draw (30:-2) to (30:\B) node[above right]{$\fh_{\alpha_2}$};
\draw (-30:2) to (-30:-2) ;
\end{scope}
%----- INTEGRAL WEIGHTS ---%
\begin{scope}
\clip (0,\B+.2)--(30:\B+.2)--(\B+.2,-.2)--(-.2,-.2) --(-.2,\B+.2)--(0,\B+.2);
\foreach \x in {0,1,...,7}{
\coordinate (c\x) at (0,\x);
\begin{scope}[shift=(c\x)]
\foreach \y in {0,1,..., 8}{
\node[draw, blue, fill =blue, circle, inner sep=0pt, minimum size=2pt] (L\x\y) at (30:\y){};
}
\end{scope}
}
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (L33) to (L34);
\draw (L33) to (L42);
\draw (L33) to (L23);
\end{scope}
\draw[->, very thick] (L00) to (75:2) to (42:5) to (L33);
\end{tikzpicture}$$
In the language of partitions, since adding $\vep_i - \frac{1}{r+1}\sum_{i=1}^{r+1} \vep_i$ to $\lambda$ is the same as adding a box to the partition corresponding to $\lambda$ in the $i$'s row (where adding a box in the $r+1$ row is equivalent to subtracting the first column). The result yields a dominant weight exactly when adding a box yields a partition (rather than a non-partition composition). Finally, $\omega_1$ corresponds to the partition of 1, a single box. So put in this combinatorial language
\begin{equation}
L(\lambda)\otimes L(\PART{1}) = \sum_{\mu \in \lambda^+} L(\mu)
\label{eq:Ar-add-a-box}
\end{equation}
where
$$\lambda^+ = \{ \text{ partitions obtained from $\lambda$ by adding a box }\}.$$
In our example above, we saw that
$$L(\PART{1}) \otimes L(\PART{1}) = L(\PART{2}) \oplus L(\PART{1,1}).$$
since $\omega_2 = \PART{1,1}$ and $2 \omega_1 = \PART{2}$.
\begin{figure}
\caption{The $\fsl_3$ crystal graph generated corresponding to $\rho$.
The left is generated by the path to $\rho$ given by $p = p_1^{\lambda_1} p_2^{\lambda_2} \dots p_4^{\lambda_r}$. The right shows the corresponding fillings of the partition corresponding to $\rho$. Labeled edges indicate $p \stackrel{i}{\to} f_i p$; missing edges indicate $e_i$ or $f_i$ acts by 0.
}\label{fig:L(rho)-crystal-tableau-path}
$$\begin{matrix}
\begin{tikzpicture}[scale=1]
\node (p) at (0,12) {\Path{(0,0) to (L02) to (L11)}};
\node (1p) at (3,9) {\Path{(0,0) to (L02) to (L-12)}};
\node (2p) at (-3,9) {\Path{(0,0) to (L1-1) to (L10) to (L2-1)}};
\node (21p) at (1.5,6) {\Path{(0,0) to (L1-1) to (L10) to (L00)}};
\node (12p) at (-1.5,6) {\Path{(0,0) to (L-10) to (L-11) to (L00)}};
\node (112p) at (3,3) {\Path{(0,0) to (L-10) to (L-11) to (L-21)}};
\node (221p) at (-3,3) {\Path{(0,0) to (L2-2) to (L1-2)}};
\node (1221p) at (0,0) {\Path{(0,0) to (L-10) to (L0-1) to (L-1-1)}};
\begin{scope}[->]
\draw (p) to node[above right] {$2$} (1p);
\draw (2p) to node[above right] {$2$} (12p);
\draw (12p) to node[above right, pos=.8] {$2$} (112p);
\draw (221p) to node[above right] {$2$} (1221p);
\draw (p) to node[above left] {$1$} (2p);
\draw (1p) to node[above left] {$1$} (21p);
\draw (21p) to node[above left, pos=.8] {$1$} (221p);
\draw (112p) to node[above left] {$1$} (1221p);
\end{scope}
\end{tikzpicture}\end{matrix}\qquad\qquad
\begin{matrix}
\begin{tikzpicture}[scale=1]
\node (p) at (0,12) {$\TAB{2,1}{1,1,2}$};
\node (1p) at (-3,9) {$\TAB{2,1}{2,1,2}$};
\node (2p) at (3,9) {$\TAB{2,1}{1,1,3}$};
\node (21p) at (-1.5,6) {$\TAB{2,1}{3,1,2}$};
\node (12p) at (1.5,6) {$\TAB{2,1}{2,1,3}$};
\node (112p) at (-3,3) {$\TAB{2,1}{2,2,3}$};
\node (221p) at (3,3) {$\TAB{2,1}{3,1,3}$};
\node (1221p) at (0,0) {$\TAB{2,1}{3,2,3}$};
\begin{scope}[<->]
\draw (p) to node[above left] {$e_1, f_1$} (1p);
\draw (2p) to node[above left] {$e_1, f_1$} (12p);
\draw (12p) to node[above left, pos=.8] {$e_1, f_1$} (112p);
\draw (221p) to node[above left] {$e_1, f_1$} (1221p);
\draw (p) to node[above right] {$e_2, f_2$} (2p);
\draw (1p) to node[above right] {$e_2, f_2$} (21p);
\draw (21p) to node[above right, pos=.8] {$e_2, f_2$} (221p);
\draw (112p) to node[above right] {$e_2, f_2$} (1221p);
\end{scope}
\end{tikzpicture}
\end{matrix}
$$
\end{figure}
To connect back to the tableaux in Section \ref{sec:typeA-YoungTab}, let's start to use these more discrete paths to generate crystals. Keeping with $p_i$ being the straight-line path to $\vep_i - \frac{1}{r+1}(\vep_1 + \vep_2 + \cdots + \vep_{r+1})$, we have $p_{\omega_1}^+ = p_1$, and $\cB(\omega_1) =\{p_i ~|~ i=1, \dots, r+1\}$. With $\lambda \in P^+$, instead of starting with the straight line path to $\lambda$, rewrite
$$\lambda = \lambda_1 \vep_1 + \lambda_2 \vep_2 + \cdots + \lambda_r \vep_r
- \frac{|\lambda|}{r+1}(\vep_1 + \dots + \vep_{r+1})$$
(where $|\lambda| = \lambda_1 + \cdots + \lambda_r$ is not the same thing as $||\lambda||$).
Then take the path which is the concatenation of paths $p_i$ according to
$$p = \underbrace{p_1p_1 \cdots p_1}_{\lambda_1}
\underbrace{p_2p_2 \cdots p_2}_{\lambda_2} \cdots = p_1^{\lambda_1}p_1^{\lambda_1}\cdots p_r^{\lambda_r}.$$
Since $\wt(p_i) = \omega_i - \omega_{i-1}$ and $\lambda_i \leq \lambda_{i-1}$, we have $p = p_1^{\lambda_1}p_1^{\lambda_1}\cdots p_r^{\lambda_r} \in C - \rho$ (in fact, $p \in \bar{C}$).
In fact, for any path constructed as
$$p = p_{i_1} p_{i_2} \cdots p_{i_n},$$ $p$ is in $C - \rho$ if and only if every initial path $p_{i_1} \cdots p_{i_j}$ has weight
$$\wt(p_{i_1} \cdots p_{i_j}) = \sum_{k = 1}^j (\omega_{i_k} - \omega_{i_k - 1}) \in P^+$$
( if any only if it's the positive sum of $\omega$'s). Another way to put this is that if we define the \emph{reading word} of $p$ to be $i_1 i_2 \cdots i_n$, then $p$ is in $C - \rho$ if any only if
every initial subword $i_1 i_2 \cdots i_j$ of the reading word of $p$ has the property that is contains more 1's than 2's, more 2's than 3's, and so on. This should sound wildly familiar if you are acquainted with the Littlewood-Richardson rule, which we'll get to momentarily.
\begin{example}\label{ex:L(rho)-crystal-tableau-path}
Let's return to the example where $r =2$ and $\lambda = \rho$, and
$$p_1 = \begin{matrix}\begin{tikzpicture}[scale=.5]
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (30:1){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b);
\end{scope}
\end{tikzpicture}\end{matrix} \qquad
p_2 = \begin{matrix}\begin{tikzpicture}[scale=.5]
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (-30:-1){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b);
\end{scope}
\end{tikzpicture}\end{matrix} \qquad \text{and} \qquad
p_3 = \begin{matrix}\begin{tikzpicture}[scale=.5]
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (0,-1){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b);
\end{scope}
\end{tikzpicture}~.\end{matrix}$$
Since
$$\rho = \vep_1 - \vep_3 = 2\vep_1 + \vep_2 - \frac{3}{3}(\vep_1 + \vep_2 + \vep_3) \qquad \text{ corresponds to } \PART{2,1},$$
the highest weight path we want to start with is
\def\A{1.732}
$$p = p_1 p_1 p_2
= \begin{matrix}\begin{tikzpicture}[scale=.5]
\begin{scope}
[every node/.style={draw, blue, fill =blue, circle, inner sep=0pt, minimum size=1.5pt}]
\node (a) at (0,0){}; \node (b) at (30:1){};
\node (c) at (30:2){}; \node (d) at (60:\A){};
\end{scope}
\begin{scope}[->,very thick, red ]
\draw (a) to (b) to (c) to (d);
\end{scope}
\end{tikzpicture}\end{matrix}$$
Then the resulting crystal is in Figure \ref{fig:L(rho)-crystal-tableau-path}.
\end{example}
When we construct a crystal in this way, generated by highest weight path $p = p_1^{\lambda_1} \cdots p_\ell^{\lambda_\ell}$, there is a very straightforward correspondence between the semistandard fillings of the partition $\lambda$ in Section \ref{sec:typeA-YoungTab} and the reading words of the paths in $\cB(p_1^{\lambda_1} \cdots p_\ell^{\lambda_\ell})$. Namely, every path in $\cB(p_1^{\lambda_1} \cdots p_\ell^{\lambda_\ell})$ is of the form $q= p_{i_1} \cdots p_{i_n}$ with $n = |\lambda|$, and
\begin{enumerate}
\item if $q$ has weight $\mu$, the collection of integers $i_1,i_2,\dots,i_n$ is the weight of the composition $\mu$; and
\item filling $\lambda$, reading right to left, top to bottom, with the word $i_1 i_2 \cdots i_n$ yields a semistandard filling of $\lambda$ with weight $\mu$.
\end{enumerate}
Moreover, this correspondence gives a bijection. The example where $\fg = \fsl_3$ and $\lambda = \rho$ is also in Figure \ref{fig:L(rho)-crystal-tableau-path}.
\paragraph{\bf Type $C_r$} In type $C_r$, there's a similar story. Choose the base
$$B= \{\vep_1 - \vep_2, \vep_2- \vep_3, \dots, \vep_{r-1} - \vep_r , 2 \vep_r\}$$
so that the fundamental weights are given by
$$\omega_i = \vep_1 + \dots + \vep_i \qquad \text{ for } i=1, \dots, r.$$
We saw on the homework that for $r=2$, $L(\omega_1)$ is a four-dimensional module, with all one-dimensional weight spaces corresponding to the weights
$$\{\pm \vep_1, \pm \vep_2\}.$$
Similar to the proof on the homework for type $A_r$, you can check that $L(\omega_1)$ is the standard representation for $C_r$ in general. (In fact, that's true across types!) It has one-dimensional weight spaces of weights
$$\{\pm \vep_1, \pm \vep_2, \cdots, \pm\vep_r\}$$
(so that it has dimension $2r$, as expected).
Similar to the type $A_r$ case, there's also a connection to partitions in the $C_r$ case as well.
Namely,
$$P^+ = \left\{\lambda_1 \vep_1 + \cdots + \lambda_r \vep_r ~\left|~
\begin{matrix} \lambda_i \in \ZZ \\
\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_r \geq 0\end{matrix}\right.\right\}$$
is in bijection with integer partitions of length at most $r$ (with less work than in type $A_r$, even).
On the homework, you're asked to give the analogous decomposition to \eqref{eq:Ar-add-a-box} for type $C_r$.
\section{Centralizer algebras}
As I hope I've conveyed, one big goal in representation theory is to know how representations decompose into irreducible components. In general, indecomposable (breaks further into direct summands) doesn't imply irreducible (contains no proper non-trivial invariant subspaces). But by definition, modules for semisimple algebras are indecomposable if and only if they are irreducible. So the trick is to figure out \emph{how} to decompose a given module. A good reference for this section is \cite[\S 4]{GW}.
At the level of individual modules, this can be difficult, since the decomposition might not be unique. But at the level of isotypic components, the decomposition is canonical. Namely, let $\widehat{A}$ be an indexing set for the isomorphism classes of irreducible $A$-modules, and for $\lambda \in \widehat{A}$, let $A^\lambda$ be the $A$-module indexed by $\lambda$. Then for an $A$ module $M$, the \dfn{isotypic component} of $M$ corresponding to $\lambda$ is
$$M^{(\lambda)} = \sum_{U \subseteq M \atop U \cong A^\lambda} U,$$
the subspace of $M$ generated by all submodules isomorphic to $A^\lambda$. Though the multiplicity $m_M(\lambda)$ of $A^\lambda$ in $M$ (and so in $M^{(\lambda)}$) is well-defined, the mechanical decomposition
$$M^{(\lambda)} = \bigoplus_{i=1}^{m_M(\lambda)} A^\lambda = m_M(\lambda) A^\lambda$$
is not unique. However, the decomposition
$$M = \bigoplus_{\lambda \in \widehat{M}} M^{(\lambda)} \qquad \text{ where }
\widehat{M} = \{ \lambda \in \widehat{A} ~|~ M^{(\lambda)} \neq 0 \},$$
is unique.
In the case where $A$ is finite-dimensional, the decomposition of $M$ into styptic components comes straight from Wedderburn's theorem: If $A$ is finite-dimensional, \dfn{Wedderburn's theorem} says
$$A \cong \bigoplus_{\lambda\in \widehat{A}} \End(A^\lambda)$$
where $\End(A^\lambda)$ is the algebra of endomorphisms of the vector space $A^\lambda$ (coming from the action of $A$ on itself).
So on each block $\End(A^\lambda)$, there is an identity operator $I_\lambda$ which looks like $1$ on $\End(A^\lambda)$ and 0 on $\End(A^\mu)$ for $\mu \neq \lambda$.
These operators satisfy
\begin{enumerate}
\item $I_\lambda^2 = I_\lambda$ ($I_\lambda$ is an \emph{idempotent});
\item $I_\lambda I_\mu = I_\mu I_\lambda = 0$ for $\lambda \neq 0$ (they are pairwise orthogonal);
\item $\sum_{\lambda \in \widehat{A}} I_\lambda = 1$;
\item $Z(A) = \CC\{I_\lambda ~|~ \lambda \in \widehat{A}\}$; and
\item the action of $I_\lambda$ on any $A$-module $M$ projects onto $M^{(\lambda)}$.
\end{enumerate}
The $I_\lambda$'s are called the \dfn{centrally primitive idempotents} of $A$.
In the case where $A$ is infinite dimensional, we need a replacement for the projection operation from $I_\lambda$.
To this end, let $\Hom(A^\lambda, M)$ be the set of $A$-module homomorphisms from $A^\lambda$ into $M$. Since $A^\lambda$ is simple, Schur's lemma tells us that every non-zero $\phi \in \Hom(A^\lambda, M)$ gives $\phi(A^\lambda) \cong A^\lambda$, and so $\phi(A^\lambda) \subseteq M^{(\lambda)}$. There's then a canonical map
$$ \Hom(A^\lambda, M) \otimes A^\lambda \to M \quad \text{ defined by } \quad
\phi \otimes u \mapsto \phi(u),$$
which produces an isomorphism
$$\Hom(A^\lambda, M) \otimes A^\lambda \cong M^{(\lambda)}.$$
This shows that $m_M(\lambda) = \dim(\Hom(A^\lambda, M))$. In fact, more is true! There is a bilinear form on the vector space of $A$-modules given by
$\