\section{Algebraic Groups (Adam Brown)}
Aim: Classify algebraic representations of a connected reductive algebraic group in characteristic 0. 
\begin{defn}
Let $k$ be a field. An algebraic group $G$ over $k$ is a group object in the category of algebraic varieties over $k$. Recall that a group object has the following maps \[m: G \times G \rightarrow G, \ \mathrm{inv}: G \rightarrow G,
\ \mathrm{id}: {\text{Spec}(k)} \rightarrow G\] satisfying the expected commutative diagrams.
\end{defn}
\subsection{Functorial Perspective}
We can think of $G$ as a group valued functor 
\begin{align*}
 k\text{-alg}&\rightarrow  \text{Grps}\\
 R&\mapsto G(R):=\text{Hom}(\text{Spec}R,G)
 \end{align*}
 This is a sufficient perspective by Yoneda's lemma. 
 \begin{lem}
 If $\mathcal{C}$ is any (locally small) category then 
 \begin{align*}
 \mathcal{C}&\rightarrow \text{Fun}(\mathcal{C}^{\text{op}},\text{Set})\\
 X&\mapsto h_X:Y\mapsto\text{hom}_\mathcal{C}(Y,X)
 \end{align*}
is a fully faithful embedding. 
\end{lem}
Note: A category is locally small if each class of morphisms hom$(X,Y)$ is a set, rather than a proper class. 
\begin{exmpl}
We can view $G=\text{SL}_n$ as the functor \[ R\mapsto \text{SL}_n(R)=\text{$R$ linear automorphisms of $R^n$ with determinant equal to 1}\]. The coordinate ring for $G$ is \[ k[G]=\text{Spec }k[X_{ij}]/(\text{det}X_{ij}-1)\]

\end{exmpl}
\begin{exmpl}
The coordinate ring of $G=\text{GL}_n$ is \[ \text{Spec } k[X_{ij},d]/(d\cdot \text{det}X_{ij} -1)\]
\end{exmpl}
\begin{exmpl}
Let $E/k$ be an elliptic curve. (Note that this example is not affine)
\end{exmpl}
\begin{exmpl}
Upper triangular matrices with 1's on the diagonal:
\[\left(\begin{matrix}
  1 & &\ast \\
   & \ddots& \\
  0& &1
\end{matrix}\right). \]
Upper triangular matrices:
\[\left(\begin{matrix}
  \ast & &\ast \\
   & \ddots& \\
  0& &\ast
\end{matrix}\right). \]
\end{exmpl}
\begin{prop}
If $G$ is an affine algebraic group over $k$ then $G$ is a closed subgroup of some GL$_n$.
\end{prop}

\subsection{Representations of Algebraic Groups}
\begin{defn}
Let $G$ be an algebraic group over $k$, $V$ be a vector space over $k$ (possibly infinite dimensional). A representation of $G$ on $V$ is a morphism of group valued functors $G\rightarrow GL_V$. Precisely, for all $\phi:R\rightarrow S$ we have the following commutative diagram
\begin{center}
\begin{tikzpicture}
    \node (GR) at (0,0) {$G(R)$};
    \node (GLR) at (3,0) {$\text{GL}_V(R)$};
    \node (GLS) at (3,-2.5){$\text{GL}_V(S)$};
    \node (GS) at (0,-2.5) {$G(S)$};
    
    \draw[->] (GR)--(GLR) node [midway,above] {$f_R$};
    \draw[->] (GR)--(GS);
    \draw[->] (GS)--(GLS) node [midway,above] {$f_S$};
    \draw[->] (GLR)--(GLS) ;
\end{tikzpicture}
\end{center}
Note that if $\dim V<\infty$ then GL$_V$ is an algebraic group.
\end{defn}
\begin{exmpl}
Let us consider an infinite dimensional representation. Let $k[G]$ be the coordinate ring of $G$. Then $k[G]$ is a $G\times G$ representation where
\[ (g_1,g_2)\cdot f(x)=f(g_1^{-1}xg_2)\] where $g_1,g_2,x\in G$ and $f\in k[G]$. To frame this group action in our functorial definition of a representation, we notice that
\[f\in\text{Hom}_{k-\text{alg}}(k[t],k[G])=\text{Hom}(G,\mathbb{A}^1)\]
\end{exmpl}
The groups of primary interest are the following:
\begin{lem}
Let $G$ be an algebraic group over $k$. The following are equivalent:
\begin{enumerate}[1.]
\item Finite dimensional representations of $G$ are semisimple\\
\item Every representation of $G$ is semisimple\\
\item $k[G]$ is semisimple under either the right or left regular representation\\
\item There exists a faithful finite dimensional representation $G\rightarrow \text{GL}_V$ so that for all $n\ge 1$, $V^{\otimes n}$ is semisimple. 
\end{enumerate}
Such a group is called linearly reductive. 
\end{lem}
Some ideas of proof:\\
\indent $1\rightarrow 2$: Every representation is a filtered union of finite dimensional subrepresentations \\
\indent $2\rightarrow 3$: This is clear. \\
\indent $3\rightarrow 1$: 
\begin{lem}
Every finite dimensional $G$ representation $W$ is contained in $k[G]^{\oplus N}$ for some $N\in \mathbb{Z}$. 
\end{lem}
\begin{proof}
Let $W$ be a finite dimensional $G$ representation. Let $\lambda_1,\cdots,\lambda_n$ be a basis of $W^\ast$. Define 
\begin{align*}
W&\rightarrow k[G]^n\\
w&\mapsto(\langle \lambda_i,g_iw\rangle)_{i=1,\cdots,n}
\end{align*}
This map is $G$ equivariant and injective. 
%This agrees with our functorial definition because the element wise formula describes analogous maps for all $k$ algebras. 
\end{proof}

\indent $4\rightarrow 1$: This follows from the following proposition (given without proof). 
\begin{prop}
Let $G\hookrightarrow GL_V$ be a faithful finite dimensional representation. Then every finite dimensional representation of $G$ is isomorphic to a subquotient of some $\bigoplus (V\oplus V^\ast)^{\otimes m}$. 
\end{prop}
Back to general $G$:\\
If $W$ is any irreducible $G$ representation, consider
\begin{align*}
\iota_W:W^\ast\otimes W&\rightarrow k[G]\\
\lambda\otimes w&\mapsto g\mapsto \lambda(gw)
\end{align*}
\begin{lem}
The image of $\iota_W$ is the $W$ isotypic component in $k[G]$ under the right regular action. 


\end{lem}
\begin{proof}
Let $j:W\rightarrow k[G]$ be any $G$ equivariant map. Claim: $j(W)\subset \iota_W(W^\ast\otimes W)$. Let $\lambda\in W^\ast$ be defined by $\lambda(w)=j(w)(1)$. Then $j(w)(g)=j(gw)(1)=\lambda(gw)=\iota_W(\lambda\otimes w)(g)$. 
\end{proof}

\begin{cor}
If $G$ is linearly reductive, then \[k[G]\cong\bigoplus_{W_i\text{ fd irr rep}}W_i^\ast\otimes W_i\] where the sum is countable (finite dimensional irreducible representations form a countable set). 
\end{cor}
\begin{proof}
$G$ linearly reductive implies that $k[G]$ is semisimple. 
\end{proof}
\begin{cor}
If $G$ and $H$ are linearly reductive, then $G\times H$ is linearly reductive.
\end{cor}
\begin{proof}
\[k[G\times H]\cong k[G]\otimes k[H]\cong (\bigoplus_{\text{fd irr of } G}W_i^\ast\otimes W_i)\otimes (\bigoplus_{\text{fd irr of } H}V_j^\ast\otimes V_j)\cong \bigoplus(W_i^\ast\otimes V_j^\ast)\otimes (W_i\otimes V_j)\] Since irreducible $G\times H$ representations are irreducible $G$ representations tensored with irreducible $H$ representations, we have that $k[G\times H]$ is a semisimple $G\times H$ representation. So $G\times H$ is linearly reductive. 
\end{proof}
\subsection{Maximal Tori}
We will begin this section with a short review and some supplemental material from previous lectures: \\
Fact: $W$ is a finite group. 
\begin{proof}
$N_G(T)$ is closed, so $N_G(T)$ is compact. \[\text{exp}_T:\mathfrak{t}/\text{lattice }\Lambda\rightarrow T\] is an isomorphism and $W$ acts faithfully on $\Lambda$. So \[W\hookrightarrow \text{Aut}_\mathbb{Z}\Lambda \] So $W$ is compact and discrete, therefore finite. Example: If $G=SU(n)$ then $W=S_n$. 
\end{proof}
\begin{defn}
a $k$ group scheme is a group object in the category of $k$-schemes. We say it is algebraic if it is finite type over $k$. 

\end{defn}
Our groups of interest will be smooth affine algebraic groups (which we will further refer to as groups over $k$). Fact: If char$k=0$, then an affine algebraic group is smooth. 
\begin{defn}
A linearly reductive group $G$ over $k$ is a smooth affine algebraic group such that all representations are semisimple. 
\end{defn}
\begin{defn}
A group $T$ over $k$ is a torus if \[T_{\bar{k}}\cong \mathbb{G}_m^n\] for some $n\in\mathbb{Z}$, where $\mathbb{G}_m=GL_1$. We say $T$ is split if $T\cong \mathbb{G}_m^n$ over $k$. 
\end{defn}
\begin{exmpl}
Let $k=\mathbb{R}$. $\mathbb{R}^\times $ is a split torus. $S^1=\mathbb{S}^1(\mathbb{R})$ is a non split torus. $\mathbb{C}^\times=\mathbb{S}(\mathbb{R})$ is a non split torus.
 \begin{align*}
\mathbb{S}:\mathbb{R}\text{-alg}&\rightarrow \text{Grp}\\
R&\mapsto \mathbb{G}_m(\mathbb{C}\otimes_\mathbb{R}R )
\end{align*}
Exercise: Show $\mathbb{S}_\mathbb{C}\cong \mathbb{G}_m\times \mathbb{G}_m$. 
\end{exmpl}
\subsection{Representations of Tori}
\begin{lem}
\begin{enumerate}
\item Tori are linearly reductive. 
\item The simple representations of split torus $\mathbb{G}_m^n$ are given by characters \[\chi_{m_1,\cdots,m_n}:(t_1,\cdots,t_n)\mapsto\prod t_i^{m_i}\]
\end{enumerate}\end{lem}
\begin{proof}
Assume $T$ is split, so $T=\mathbb{G}_m^n$. Then \[k[T]=k[X_1^{\pm1},\cdots,X_n^{\pm1}]=\bigoplus_{\mathbb{Z}^n}kX^{m_1}_1\cdots X^{m_n}_n\]
which has summands that are stable under the right regular representation, and have character $\chi_{m_1,\cdots,m_n}$. Since $k[T]$ is semisimple, $T$ is linearly reductive. Since all simples appear in $k[T]$, $\chi_{m_1,\cdots,m_n}$ exhaust characters of simple representations. 
\end{proof}

Exercise: In characteristic 0, SL$_n$ is linearly reductive. In characteristic $p>0$, SL$_n$ is not linearly reductive. \\
There is a different notion of a reductive group:
\begin{defn}
A group $G$ over $k$ is
\begin{enumerate}
\item semisimple if the radical $R(G)$ of $G_{\bar{k}}$ is trivial\\
\item reductive if the unipotent radical $R_u(G_{\bar{k}})$ is trivial
\end{enumerate}
where $R(G)$ is the maximal connected solvable smooth normal subgroup, and $R_u(G)$ is the maximal connected unipotent smooth normal subgroup. 
\end{defn}


\begin{exmpl}
\begin{enumerate}
\item \[\begin{bmatrix}
\ast&\cdots&&\ast\\
0&\ddots&&\vdots\\
\vdots&\ddots&\ddots&\\
0&\cdots&0&\ast
\end{bmatrix}
\subset GL_n\quad\text{ is a solvable group}\]
\item \[\begin{bmatrix}
1&\ast&\cdots&\ast\\
0&\ddots&\ddots&\vdots\\
\vdots&\ddots&\ddots&\ast\\
0&\cdots&0&1
\end{bmatrix}
=\text{ is a unipotent group}\]
\item (Exercise) \[R(GL_n)=\begin{bmatrix}
z&0&\cdots&0\\
0&\ddots&\ddots&\vdots\\
\vdots&\ddots&\ddots&0\\
0&\cdots&0&z
\end{bmatrix}
=\text{ the center of $GL_n$}\]
\item (Exercise) \[R_u(GL_n)=\{1\}\Rightarrow\text{ $GL_n$ is reductive not semisimple}\]
\item $R(SL_n)=\{1\}$ because $(\mu_n)^\circ =\{1\}$ if char $k$ does not divide $n$, and is not smooth if char $k$ divides $n$. \[ Z(SL_n)=\begin{bmatrix}
z&0\\
0&z
\end{bmatrix}\quad\text{ such that $z\in \mu_n$}\]
\end{enumerate}
\end{exmpl}
Fact: in characteristic 0, linearly reductive is equivalent to reductive. \\
When studying representation theory we will focus on connected linearly reductive groups over $k=\bar{k}$ with characteristic 0. 
\subsection{Structure Theory of Reductive Groups over $k=\bar{k}$}

\begin{defn}
The Lie algebra of $G$ is defined as
\begin{align*}
 \mathfrak{g}&=\text{tangent space at $1\in G$}\\
 &= \text{ker}(G(k[\epsilon])\rightarrow G(k))
\end{align*}
where $\epsilon^2=0$. 
\end{defn}
\begin{exmpl}
\begin{enumerate}
\item If $G=GL_n$ then $\mathfrak{g}=\{1+\epsilon A: A\in M_n(k)\}$
\item If $G=SL_n$ then $\mathfrak{g}=\{1+\epsilon A: \text{det}(1+\epsilon A)=1\}=\{A\in M_n(k):\text{trace}(A)=0\}$
\item If $G=Sp_{2n}=\{g\in GL_{2n}: g^tJg=J\}$ where $J$ is an alternating pairing, then $\mathfrak{g}=\{1+\epsilon A\in GL_{2n}(k[\epsilon]):(1+\epsilon A)^t J(1+\epsilon A)=J\quad(\text{equivalently  } A^tJ+JA=0)\}$
\end{enumerate}

\end{exmpl}
$G$ acts on $\mathfrak{g}$ by conjugation:
\begin{align*}
 \text{Ad}:G&\rightarrow \text{Aut}(\mathfrak{g})\\
 \text{ad}:\mathfrak{g}&\rightarrow \text{End}(\mathfrak{g})\\
 \text{ad}(x)(y)&:=[x,y]
\end{align*}
Now we will decompose $\mathfrak{g}$ under $\text{Ad}_T$. Take $G$ to be a connected reductive group with maximal (split) torus $T$. We know $T$ is linearly reductive so there is a semisimple decomposition of any representation of $T$. \[ V\cong \bigoplus_{\lambda\in X^\bullet(T)}V_\lambda\] For $V=\mathfrak{g}$, \[\mathfrak{g}=\mathfrak{g}_0\oplus\bigoplus_{\alpha\in\Phi(G,T)}
\mathfrak{g}_\alpha\]
where $\Phi(G,T)=\{0\neq \lambda\in X^\bullet(T): \mathfrak{g}_\lambda\neq 0\}$ denotes the set of roots of $G$ with respect to $T$. 
\begin{exmpl}
Let $GL_3\supset T=\text{ diagonal matrices}$
\[\mathfrak{g}=\mathfrak{t}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{g}_\alpha\]
where $\alpha\in\{e_1-e_2,e_2-e_3,e_1-e_3,-(e_1-e_2),\cdots\}$. 
\[ X^\bullet(T)=\mathbb{Z}e_1\oplus\mathbb{Z}e_2\oplus\mathbb{Z}e_3\]
where \[ e_i\begin{bmatrix}
t_1&&\\
&t_2&\\
&&t_3
\end{bmatrix}=t_i\] As in $GL_3$, it is always the case that \[\mathfrak{g}_0=\{x\in\mathfrak{g}: \text{Ad}(t)(X)=X \text{ for all }t\in T\}=\mathfrak{t}\]
because the centralizer in $G$ of $T$ is $T$. 
\end{exmpl}
Key facts about maximal tori that generalize from the compact setting:\\
Let $G$ be a connected reductive algebraic group over $k=\bar{k}$, with $T$ a maximal torus. \begin{enumerate}
\item $C_G(T)=T$, $W:=N(T)/C(T)$\\
\item All maximal tori are conjugate (but not all $g$ are conjugate to $t\in T$. Example: $\begin{bmatrix}
1&1\\
0&1
\end{bmatrix}\in SL_2$)
\item $\bigcup_{g\in G(k)}gTg^{-1}$ is Zariski dense in $G$. \\
Note: (2) is not true for $k\neq \bar{k}$. Example: Consider $G=SL_2(\mathbb{R})$\[T_1=\begin{bmatrix}
\text{sin}(\theta)&\text{cos}(\theta)\\
\text{cos}(\theta)&\text{sin}(\theta)
\end{bmatrix}\subset SL_2(\mathbb{R})\quad T_2=\begin{bmatrix}
t&0\\
0&t^{-1}
\end{bmatrix}\subset SL_2(\mathbb{R})
\]
$T_1$ and $T_2$ are two non-conjugate maximal tori.
\end{enumerate}
As before we have a map:\[ R(G)\hookrightarrow R(T)^W\]
Goal: We want to show that this is an isomorphism. First we need more structure theory (this will give us a definition of the Langlands dual group $G^\vee$). Immediate aim: classification of connected reductive algebraic groups over $k=\bar{k}$ in terms of linear algebraic data. \begin{defn}
A Borel subgroup in $G$ is a maximal connected solvable subgroup.
\end{defn}
Fact: All Borel subgroups in $G$ are conjugate (if $k=\bar{k}$). \\
A choice of Borel $T\subset B \subset G$ gives us a decomposition of roots $\Phi=\Phi(G,T)$ into "positive" and "negative" roots $\Phi^+=\{\alpha\in\Phi: \mathfrak{g}_\alpha\subset\mathfrak{b}\}$ and $\Phi^-=-\Phi^+$. \begin{exmpl}
\[\begin{bmatrix}
\ast&\ast&\ast\\
0&\ast&\ast\\
0&0&\ast
\end{bmatrix}\subset GL_3\]
\[\mathfrak{g}=\mathfrak{t}\oplus\bigoplus_\Phi\mathfrak{g}_\alpha\quad \Phi^+=\{e_1-e_2,e_2-e_3,e_1-e_3\}\]
\end{exmpl}
Fact: There is a set $\Delta=\{\alpha_1,\cdots,\alpha_r\}=\Delta(G,B,T)\subset\Phi^+(G,B,T)$ of simple roots such that any positive root $\alpha\in\Phi^+$ is uniquely expressible as $\alpha=\sum c_i\alpha_i$ with $c_i\in\mathbb{Z}_{\ge 0}$.\\
Our classification theorem for connected reductive groups over $k=\bar{k}$ rests on the following refinement of root systems. \begin{defn}
A root datum is a 4-tuple $(X,\Phi,X^\vee,\Phi^\vee)$ where
\begin{enumerate}
\item $X$ and $X^\vee$ are finite free $\mathbb{Z}$-modules with perfect duality \[\langle , \rangle : X\times X^\vee\rightarrow \mathbb{Z}\]
\item $\Phi\subset X$, $\Phi^\vee\subset X^\vee$ are finite subsets with fixed bijection $\alpha\mapsto\alpha^\vee$ such that $\langle \alpha,\alpha^\vee\rangle =2$
\item $s_\alpha(\Phi)=\Phi$ and $s_{\alpha^\vee}(\Phi^\vee)=\Phi^\vee$ 
\end{enumerate}
Where $s_\alpha$ ($s_{\alpha^\vee}$) is a linear endomorphism of $X$ ($X^\vee$) defined by 
\begin{align*}
s_\alpha(x)&=x-\langle x, \alpha^\vee\rangle \alpha\\
s_{\alpha^\vee}(x^\vee)&=x^\vee-\langle \alpha, x^\vee\rangle \alpha^\vee
\end{align*}
\end{defn}
Structure Theorem:
\begin{theorem}
Connected reductive algebraic groups (up to isomorphism) over $k=\bar{k}$ are in bijection with root data (up to isomorphism). 
\end{theorem}
We now show how to associate a root datum with a connected reductive algebraic group. Let $G$ be a connected reductive algebraic group with maximal torus $T$. 
\begin{enumerate}[(a)]
\item $X^\bullet(T)=\text{Hom}(T,\mathbb{G}_m)$\\
\item $\Phi=\Phi(G,T)=\{\alpha\in X^\bullet(T)\setminus 0: \mathfrak{g}_\alpha\neq 0\}$
\item $X_\bullet(T)=\text{Hom}(\mathbb{G}_m,T)$
\item $\Phi^\vee=\Phi^\vee(G,T)$ finite subset of $X_\bullet(T)$ will be defined later
\end{enumerate}
For (a)-(d) to be a root datum, we need 
\begin{enumerate}
\item $\langle, \rangle: X^\bullet\times X_\bullet\rightarrow \mathbb{Z}$\\
\item $\Phi\leftrightarrow \Phi^\vee$, $\alpha\leftrightarrow  \alpha^\vee$  such that $\langle \alpha,\alpha^\vee\rangle =2$
\item $s_\alpha\circlearrowright X^\bullet $ and $s_{\alpha^\vee}\circlearrowright X_\bullet$ preserve $\Phi$ and $\Phi^\vee$. 
\end{enumerate}
Source of (1):
 \begin{align*}
X^\bullet\times X_\bullet &\rightarrow \mathbb{Z}\\
\chi,\lambda&\mapsto \chi\circ\lambda\in \text{Hom}(\mathbb{G}_m,\mathbb{G}_m)\cong\mathbb{Z}
\end{align*} 
Examples: $SL_2$, $PGL_2$. First, we'll typically use the following notation for (co)characters of the diagonal torus:
\begin{align*}
e_i:&\begin{bmatrix}
t_1&&\\
&\ddots&\\
&&t_n
\end{bmatrix}\mapsto t_i\\
e_i^\ast:&t\mapsto\begin{bmatrix}
1&&\\
&t_i&\\
&&1
\end{bmatrix}\text{in the i,i$^\text{th}$ spot}
\end{align*}
\begin{enumerate}
\item \begin{align*}
SL_2&\supset \begin{bmatrix}
t&\\
&t^{-1}
\end{bmatrix}=T\\
X^\bullet&=\frac{\mathbb{Z}e_1\oplus\mathbb{Z}e_2}{\mathbb{Z}(e_1+e_2)}\\
\Phi&=\{e_1-e_2,e_2-e_1\}\\
X_\bullet&=\{a_1e_1^\ast+a_2e_2^\ast: a_1+a_2=0\}\\
\Phi^\vee&=\{\pm(e_1^\ast-e_2^\ast)\}
\end{align*}
\item \begin{align*}
PGL_2&\supset\begin{bmatrix}
t_1&\\
&t_2
\end{bmatrix}/\mathbb{G}_m=T^\vee\\
X^\bullet(T^\vee)&=\{a_1e_1+a_2e_2|a_1+a_2=0\}\\
\Phi&=\{\pm(e_1-e_2)\}\\
X_\bullet(T^\vee)&=\frac{\mathbb{Z}e_1^\ast\oplus\mathbb{Z}e_2^\ast}{\mathbb{Z}(e_1^\ast+e_2^\ast}\\
\Phi^\vee&=\{\pm(e_1^\ast-e_2^\ast)\}
\end{align*}
\end{enumerate}
Note: the root data of SL$_2$ and PGL$_2$ are not isomorphic. But the root datum of SL$_2$( respectively PGL$_2$) is obtained by taking the root datum $(X^\bullet, \Phi,X_\bullet, \Phi^\vee)$ of PGL$_2$ (respectively SL$_2$) and froming the dual root datum $(X_\bullet, \Phi^\vee, X^\bullet, \Phi)$. 
\begin{defn}
If $G$ is a connected reductive group with root datum $(X^\bullet, \Phi,X_\bullet, \Phi^\vee)$, then $G^\vee$ (Langlands dual group of $G$) is the connected reductive group with root datum $(X_\bullet, \Phi^\vee, X^\bullet, \Phi)$. Note that we are only able to define $G^\vee$ in this way because of the previous structure theorem. 
\end{defn}
To complete the description of the root datum associated to $(G,T)$ we need to define coroots $\Phi^\vee\subset X_\bullet(T)$. For $\alpha\in \Phi$, $\alpha^\vee\in\Phi^\vee$ is defined as follows:\\
Basic structure theory of semisimple Lie algebras: For all $\alpha\in\Phi$ there exists $\mathfrak{sl}_2\hookrightarrow \mathfrak{g}$ identified with $\mathfrak{g}_\alpha\oplus\mathfrak{g}_{-\alpha}\oplus[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]$. This lifts to a group homomorphism \[\Psi_\alpha:SL_2\rightarrow G\]
Then $\alpha^\vee$ is by definition the composition \[\mathbb{G}_m\rightarrow SL_2\rightarrow G\] where \[t\mapsto \begin{bmatrix}
t&\\
&t^{-1}
\end{bmatrix}\mapsto \Psi_\alpha\begin{pmatrix}
t&\\
&t^{-1}
\end{pmatrix}\]
Summary: To $G$ and a choice $T$ of maximal torus we associate $(X^\bullet, \Phi,X_\bullet, \Phi^\vee)$. \begin{prop}
This is a root datum. 
\end{prop}
\begin{theorem}
\begin{enumerate}
\item Two reductive groups are isomorphic if they have isomorphic root data\\
\item Given any root datum there exists a reductive group $G$ over $k=\bar{k}$ with given root datum (up to isomorphism)
\end{enumerate}
\end{theorem}
Exercise: Identify Langlads dual group of GL$_n$ and Sp$_{2n}$. (Hint: GL$_n\cong GL_n^\vee$ and $Sp_{2n}^\vee\cong SO_{2n+1}$).\\
Statement of Theorem of highest weight: 
Let $G$ be a connected reductive group over $k=\bar{k}$ with characteristic zero. Fix $B\supset T$, and the corresponding $\Phi=\Phi^+\sqcup\Phi^-$. 
\begin{defn}
Given $B,T$, the dominant weights $X^\bullet(T)_+=\{\lambda\in X^\bullet(T)|\langle \lambda, \alpha^\vee\rangle \ge 0\text{ for all }\alpha\in \Phi^+\}$
\end{defn}
Define a partial order on $X^\bullet(T)$ by $\lambda\ge\mu$ if and only if $\lambda-\mu$ is a non negative integer linear combination of $\alpha\in\Phi^+$. 
\begin{exmpl}
SL$_2$ and weights $e_1$, and $e_1-e_2\equiv 2e_1$ have no order relation
\end{exmpl}
\begin{defn}
A representation $V$ of $G$ has highest weight $\lambda$ (necessarily dominant) if $V_{\lambda} \neq 0$ and $\lambda\ge \mu$ for all $\mu\in X^\bullet(T)$ such that $V_\mu\neq 0$. 
\end{defn}
\begin{theorem}
Every irreducible representation $V$ of $G$ has a highest weight $\lambda$ (necessarily dominant) and the map
\begin{align*}
\{\text{ irr rep of $G$}\}&\rightarrow X^\bullet(T)_+\quad\text{is a bijection}\\
V&\mapsto \lambda=hw(V)
\end{align*}
Moreover, dim$V_\lambda=1$.
\end{theorem}
\begin{cor}
The map \[R(G)\rightarrow R(T)^W\] is an isomorphism 
\end{cor}
\begin{proof}
In every $W$ orbit there is a unique $B$ dominant weight. So a basis of right hand side is given by \[ s_\lambda=\sum_W[w\lambda]\text{ for }\lambda\in X^\bullet(T)_+\]
We can complete the proof by using an upper triangular argument to show that $\{\text{ch}(V)\}_{V\in\text{Irr}}$ have a span containing all the $s_\lambda$ for $\lambda\in X^\bullet(T)_+$. 
\end{proof}
\begin{exmpl}
\[B=\text{upper triangular matrices}\subset SL_3\quad X^\bullet(T)=\frac{\mathbb{Z}e_1\oplus\mathbb{Z}e_2\oplus\mathbb{Z}e_3}{\mathbb{Z}(e_1+e_2+e_2)}\]
$\alpha_1+\alpha_2$ is the highest weight of the Adjoint representation. $\frac{2}{3}\alpha_1+\frac{1}{3}\alpha_2$ is the highest weight of the standard representation. Exercise: $\frac{1}{3}\alpha_1+\frac{2}{3}\alpha_2$ is a highest weight of what representation?
\end{exmpl}


