\section{Overview (notes by Michael Zhao)}
\subsection{Next Few Weeks}
Over the next few weeks, we will cover

\begin{enumerate}[1.]
\item compact groups,
\item basics of algebraic groups,
\item Flag varieties and Bruhat decomposition,
\item Highest weight theory and Borel-Weil theorem,
\item Tannakian categories.
\end{enumerate}

The next unit will start with the smooth representations of reductive
groups over local fields.

\subsection{Course Overview}

Our goal is to get to the geometric Satake equivalence. Let $k$ be an
algebraically closed field. Let $G$ be a connected reductive algebraic group
over $k$ (e.g. $\gl n k$, $\sl n k$, $\sp {2n} k$, $G_2$). The geometric Satake
equivalence says there is the equivalence of abelian tensor (!) categories
\[ \perv_{``G(k[[t]])"}\left(``\quotient{G(k((t)))}{G(k[[t]])}"\right)
\isomorphism \rep(\langdual G). \]
The left-hand side is the category of perverse sheaves on $\gr G$,
the affine Grassmannian of $G$, and the right-hand the category of finite
dimensional algebraic representations of the Langlands dual group of $G$
(e.g if $G = \gl n k$, $\langdual G = G$, and if $G = \sp {2n} k$,
$\langdual G = \so {2n + 1} k$).

\subsubsection{Part I of the Class} Understand $\rep(\langdual G)$.

\subsubsection{Part II of the Class} The classical Satake isomorphism and
smooth representations of reductive groups over local fields. It is the
isomorphism
\[ \mathcal{C}^\infty_{c, ``G(k[[t]])"}\left(``\quotient{G(k((t)))}{G(k[[t]])}"
  \right) \longiso R(\langdual G), \]
obtained by taking the Grothendieck group of both sides (applying $\ggroup$)
of the geometric Satake equivalence. The left-hand side is called the
spherical Hecke algebra of $G$, which arises from studying the smooth
representations of $G(k((t)))$ or $G(\Qp)$, and is made into an algebra by
convolution of functions. The right-hand side is the representation ring of
the Langlands dual of $G$.

\subsubsection{Part III of the Class} Understand the $\perv$ of the geometric
Satake equivalence. This entails understanding the six operations (e.g. $f_*,
f^*$) on derived categories of (constructible) sheaves on algebraic varieties,
and the full subcategory of perverse sheaves.

\subsubsection{Part IV of the Class} This will be distributed throughout the
course, but the goal is to understand what $\gr G$ is.

\subsubsection{Part V of the Class} The geometric Satake equivalence.

\section{Representation and Structure Theory of Compact Lie Groups (notes by Michael Zhao)}

\begin{defn}
A \textbf{Lie group} is a group object in the category of smooth manifolds, i.e.
$m: G \times G \rightarrow G, \ \mathrm{inv}: G \rightarrow G,
\ \mathrm{id}: {1} \rightarrow G$, satisfying the expected commutative diagrams.
\end{defn}

\begin{exmpl}
The following are Lie groups.

\begin{itemize}
\item $\gl n \reals \subset M_n(\reals) \iso \reals^{n^2}$
\item $\sp {2n} \reals = \{ g \in \sl {2n} \reals \ | \
  g \ \textrm{preserves a non-degenerate alternating form} \}$
\item More generally, closed subgroups $G \xhookrightarrow{} \gl n \reals$.
\item Unipotent matrices, invertible upper triangular matrices (Borel subgroup).
\item $\sl 2 \reals$. $\pi_1(\sl 2 \reals) \iso \ints$. This can be
established by looking at the Iwasawa decomposition of $\sl 2 \reals$.
The covering space corresponding to $2 \ints \subset \pi_1(\sl 2 \reals)$ is
denoted by $\widetilde{\sl 2 \reals}$, and called the metaplectic cover of
$\sl 2 \reals$. It is a Lie group but can't be embedded into $\gl n \reals$.
\end{itemize}
\end{exmpl}

The metaplectic group is not relevant for us.

\subsection{Lie Algebras}
The Lie algebra $\mathfrak{g}$ of $G$ is given by the tangent space at the 
identity. It is a priori a vector space, but the Lie bracket gives it an algebra
structure, which we will describe briefly.

There is a map $\Psi: G \rightarrow \aut(G)$ since $G$ acts on $G$ by 
conjugation. Since $\Psi(g)$ preserves the identity, we can look at the map
\[ \operatorname{Ad}: G \rightarrow \aut_\reals(\mathfrak{g}), \]
given by \[ g \rightarrow \restr{d\Psi(g)}{1}.\] By differentiating at the 
identity, we obtain the map
\[ \operatorname{ad}: \mathfrak{g} \rightarrow \into_\reals(\mathfrak{g}). \]
Then we can define the Lie bracket as $[X, Y] := \operatorname{ad}(X)(Y)$.

\begin{exmpl}
For $G = \gl n \reals$, $\mathfrak{g} = M_n(\reals)$, and $[X, Y] = XY - YX$.
\end{exmpl}

\subsection{Exponential Map}

Here we will explain the role of Lie algebras, by constructing a map
$\op{exp} : \mathfrak{g} \rightarrow G$. For $G = \gl n k$,
$\op{exp}(X) = \sum_n X^n / n!$.

\subsubsection{$\op{exp}$ Facts} Before giving the construction, note several
facts.

\begin{enumerate}[1.]
\item $\op{exp}$ is a local diffeomorphism near 0, by the implicit function
theorem.
\item $\op{exp}$ is not a homomorphism once $G$ is non-abelian.
\end{enumerate}

\subsubsection{$\op{exp}$ Construction}
Take $X \in \mathfrak{g}$. Use left-multiplication to propagate $X$ to a
left-invariant vector field $\xi_X$ on $G$, i.e. $\xi_X(g) = (m_g)_*(X).$

Construct an integral curve  $\varphi_X$ for this vector field by the following 
procedure. Solve the  differential equation
$\psi(0) = 1, \psi_X'(t)
= \xi_X(\psi_X(t))$. This gives a 
function $\psi_X : (-\varepsilon, \varepsilon) \rightarrow G$.

Note that $\psi_X$ is a homomorphism. Fix $s$ and let $t$ vary. Let
$\alpha(t) = \psi_X(s) \psi_X(t)$ and let $\beta(t) = \psi_X(s + t)$. Then
$\alpha(0) = \beta(0)$, and $\alpha'(t) = \xi_X(\alpha(t))$ and
$\beta'(t) = \xi_X(\beta(t))$ by left-invariance of $\xi_X$. Thus,
$\alpha = \beta$.

This can be extended to a homomorphism $\varphi_X : \reals \rightarrow G$,
by setting $\varphi_X(t) = \phi_X(t/N)^N$, where $N$ is an integer such that
$t/N < \varepsilon/2$. This does not depend on the choice of $N$; if $M$ was
another such integer, we would have
\[ \psi_X\left(\dfrac{t}{MN}\right)^N = \psi_X\left(\dfrac{t}{M}\right), 
  \psi_X\left(\dfrac{t}{MN}\right)^M = \psi_X\left(\dfrac{t}{N}\right). \]
Hence \[ \psi_X\left(\dfrac{t}{M}\right)^M
 = \phi_X\left(\dfrac{t}{MN}\right)^{MN}
 = \phi_X\left(\dfrac{t}{N}\right)^N. \]
Using this definition and the fact that $\varphi_X$ is a homomorphism in
$(-\varepsilon, \varepsilon)$, it can be shown $\varphi_X$ is a homomorphism
$\reals \rightarrow G$.

\begin{defn}
$\op{exp}_G(X) := \varphi_X(1)$.
\end{defn}

\begin{rmk}
For $G$ a compact connected Lie group, $\mathfrak{g}$ its Lie algebra,
$\op{exp}_G: \mathfrak{g} \rightarrow G$ is surjective. The basic idea of the
proof is to use the Killing form on a Lie algebra. Use multiplication to 
transport it to a Riemannian metric. The integral curves are geodesics, and any
two points are connected by a geodesic.
\end{rmk}

\begin{ex}
Find a $G$ where $\op{exp}_G$ is not surjective.

We do not need to look far for such an example. If $G = \gl 2 \reals$, and
$\lambda_1, \lambda_2$ are two \emph{distinct} negative real numbers, consider
\[B = \left(\begin{matrix}
  \lambda_1 & 0 \\
  0 & \lambda_2
\end{matrix}\right). \]
If there is an $A \in M_n(\reals)$ with $\op{exp}(A) = B$, and $A$ has
eigenvalues $\alpha_1, \alpha_2$, then $e^{\alpha_i} = \lambda_i$. Since
$\lambda_i$ are negative, $\alpha_1$ and $\alpha_2$ are complex conjugates. Thus
$\lambda_1, \lambda_2$ have the same absolute value, a contradiction.
\end{ex}

\subsection{Representation of Compact Lie Groups}

\begin{defn}
A representation of $G$ a compact Lie group is a continuous (hence smooth)
homomorphism $G \rightarrow \gl n \cmplx$.
\end{defn}

The motto here is to study representations of $G$ by restricting to a maximal
tori.

\begin{defn}
A (compact) torus is a (compact) connected abelian Lie group.
\end{defn}

\begin{exmpl}
$S^1$ is a torus, and all compact examples are isomorphic to $(S^1)^n$.
\end{exmpl}

\begin{defn}
A \textbf{character} (or \textbf{weight}) of $T$ is a continuous homomorphism
$T \rightarrow S^1$. Write $\cgroup T$ for the abelian group of characters of
$T$.
\end{defn}

\begin{ex}
Prove the following lemma.
\begin{lem}
If $T \iso \reals^n/\ints^n$, then all characters of $T$ have the form
\[ (x_1, ..., x_n) \rightarrow \prod_{r = 1}^n e^{2 \pi i a_r x_r}, \]
where all $a_r \in \ints$.
\end{lem}
\begin{proof}
We just need to know what homomorphisms there are from $\reals/\ints$ to $S^1$ 
(viewed as the unit circle in $\cmplx$). This is the same as knowing the 
homomorphisms from $\reals/\ints \rightarrow \reals/\ints$, and any such
$\varphi$ lifts to a homomorphism $\phi$ from $\reals$ to $\reals$, by using 
twice the fact that a universal covering space covers any connected cover.

This construction shows $\phi(1) \equiv 0 \pmod{1}$, so
for some integer $n$, $\phi(1) = n$, $\phi(a) = na$,
$b \phi(a/b) = \phi(a) = na$. By continuity,
$\phi(x) = nx$ for any $x$. Then $\varphi(x) = \phi(x) = nx \pmod{1}$. As
$x \rightarrow e^{2 \pi i x}$ is a homeomorphism $\reals/\ints \rightarrow S^1$,
any homomorphism $\reals/\ints \rightarrow S^1$ is given by
$x \rightarrow e^{2\pi i n x}$, for some $n$.
\end{proof}
\end{ex}

\begin{lem}
Any compact torus has a (dense set of) topological generators, i.e. there is
$t \in T$ with $T = \overline{\langle t \rangle}.$
\end{lem}

\begin{proof}
Assume $T \iso \reals^n/\ints^n$, and take $T \ni t = (t_1, ..., t_n)$. Let
$H := \overline{\langle t \rangle}$. Then use the

\begin{lem}
$H = T$ if and only if $1, t_1, ..., t_n$ are $\rats$-linearly independent.
\end{lem}

If $H \neq T$, then $T/H$, a compact torus, has a non-trivial character $\chi$,
and \[ \chi(x_1, ..., x_n) = \prod_r e^{2 \pi i a_r x_r}, \]  for some integers
$a_r$ not all zero. Then $\restr{\chi}{H} = 1$ if and only if $\chi(t) = 1$
if and only if $\sum a_r t_r \in \ints$, which implies $1, t_1, ..., t_n$ are
$\rats$-linearly dependent.
\end{proof}

\begin{thm}
Let $G$ be a compact connected Lie group. Fix a maximal torus $T \subset G$.
\begin{enumerate}[(1)]
\item\label{conju} Every $g \in G$ is conjugate to an element of $T$.
\item Any other maximal torus $T' \subset G$ is $G$-conjugate to $T$.
\item $C_G(T) = T$.
\end{enumerate}
\end{thm}

\begin{ex}
Find a counter-example for non-compact groups. We already know \ref{conju}
for $U(n)$. Why?

An example of a non-compact Lie group without any tori is $\reals^n$. A maximal 
torus of $U(n)$ is the subgroup of all diagonal matrices. The spectral theorem
tells us that any unitary matrix can be diagonalized by another unitary matrix 
gives \ref{conju} for $U(n)$.
\end{ex}

\begin{proof} Idea of the proof:

\begin{enumerate}[(1)]
\item\label{surj} follows from surjectivity of $\op{exp}_G$.
\item is immediate from \ref{surj}.
\item uses surjectivity of $\op{exp}_G$.
\end{enumerate}

\end{proof}

\subsection{Onto Representation Theory}
Let $G$ be a compact connected group as before.

\begin{prop}
Any finite dimensional representation of $G$ is isomorphic to a direct sum of
irreducible representations.
\end{prop}

\begin{proof}
The proof is the same as in the finite case. Let $V \in \rep(G)$. Fix a 
hermitian inner product $( \ , \ )$ on $V$. Produce a $G$-invariant inner product $\langle \ , \ \rangle$ by using the Haar measure that exists on $G$:
\[ \langle v, w \rangle = \int_G (g \cdot v, g \cdot w) \ dg. \]
If $W$ is a $G$-subrepresentation of $V$, then $V$ decomposes as
$V = W \oplus W^\perp$ ($W^\perp$ is $G$-stable since $\langle \ , \ \rangle$ is
$G$-invariant.
\end{proof}

The following is due to Schur.

\begin{cor}
If $V$ is irreducible, then $\into_{\cmplx[G]}(V) = \cmplx$.
\end{cor}

\begin{proof}
Let $T \in \into_{\cmplx[G]}(V)$. Since $V$ is finite-dimensional, $T$ has an
eigenvalue $\lambda$. Then if $T$ is non-constant, $T - \lambda \cdot 1$ acts 
on $V$ and satisfies $0 \neq \ker(T - \lambda \cdot 1) \subset V$. Since $V$ is
irreducible, $\ker(T -\lambda \cdot 1) = V$, so $T = \lambda$.
\end{proof}

\begin{defn}
Let $V \in \rep(G)$. The \textbf{character} $\chi_V$ of $V$ is a homomorphism
$G \rightarrow \cmplx$ given by $g \rightarrow \tr(\restr{g}{V})$.
\end{defn}

Some basic facts about characters.

\begin{enumerate}[1.]
\item $\chi_V$ only depends on the isomorphism class of $V$.
\item $\chi_V$ only depends on the conjugacy class of its argument.
\item $\chi_{V \oplus W} = \chi_V + \chi_W$.
\item $\chi_{V \otimes W} = \chi_V \chi_W$.
\item $\chi_{V^*} = \conj{\chi_V}$, where $V = \operatorname{Hom}(V, \cmplx)$.
\end{enumerate}

The first four properties give us a homomorphism
$K_0(\rep(G)) = R(G) \rightarrow C(G)$. Viewing $\chi_V$ in $L^2(G)$ with an
inner product
\[ \langle f_1, f_2 \rangle = \int_G \conj{f_1(g)} f_2(g) \ dg, \]
we have the following basic calculation.

\begin{prop}
\begin{enumerate}[(1)]
\item\label{dim-formula} For $V, W \in \rep(G)$,
$\langle \chi_V, \chi_W \rangle = \dim \operatorname{Hom}_G (V, W).$
\item\label{schur} Characters of irreducible representations are orthnormal 
(Schur).
\item\label{char-iso}$\chi_V = \chi_W$ if and only if $V \iso W$.
\end{enumerate}
\end{prop}

\begin{proof}
Prior to describing a sketch of the proof, we will need the following two
lemmas (\textbf{left as exercises}).

\begin{lem}\label{dims}
For all $V \in \rep(G)$, $\dim V^G = \displaystyle\int_G \chi_V(g) \ dg$.
\end{lem}

\begin{proof}
Let $\rho$ be the map $G \rightarrow \op{GL}_n(V)$ associated to $V$. Recall
that there exists a unique linear transformation $A$ such that for any
$f \in V^*$, \[ \int_G f(\rho(g) v) \ dg = f(Av), \] and that the action of
$\int_G \rho(g) \ dg$ is defined to be the action of $A$. We will show that
$\op{im} A = V^G$ and that $A$ is a projection operator. Since the trace of
a projection operator and its rank coincide, this will prove the lemma.

Due to left-invariance of Haar measure and this fact, we can prove
\[\rho(h) \int_G \rho(g) \ dg = \int_G \rho(h) \rho(g) \ dg
    = \int_G \rho(hg) \ dg = \int_G \rho(g) \ dg,\]
when viewed as operators on $V$. For instance, the first equality follows from
the fact that for any linear form $f$, $f \circ \rho(h)$ is also a linear form.
This lets us prove that $\rho(h) Av = Av$ for $A = \int_G \rho(g) \ dg$. Hence
$\op{im} A \subset V^G$. Now suppose we have $v$ such that $\rho(g) \cdot v = v$
for every $g \in G$. Then for any $f \in V^*$, $f(\rho(g) \cdot v) = f(v)$.
Taking the integral over $G$, $f(Av) = f(v)$. On $V^G$, then, $A$ acts as
the identity. From this, surjectivity and idempotency follow.
\end{proof}

\begin{lem}\label{b}
For all $V$, the natural map
\[ \bigoplus_{W} W \otimes \op{Hom}_G(W, V) \rightarrow V, \]
where the sum is taken over all irreducible representations $W$, given by
\[ w \otimes \varphi \rightarrow \varphi(w), \]
is an isomorphism of $G$-representations (the action of $G$ on
$\op{Hom}_G(W, V)$ is trivial).
\end{lem}

\begin{proof}
Since $W$ is irreducible, any element of $\op{Hom}_G(W, V)$ is an embedding.
Since $\rep(G)$ is semi-simple, the dimension of $\op{Hom}_G(W, V)$ counts
the multiplicitly of $W$ in $V$. Then it suffices to show that for every
irreducible $W$,
\[ W \otimes \op{Hom}_G(W, V) \iso W^d, \]
where $d := \dim \op{Hom}_G(W, V)$. Taking a basis of
$\op{Hom}_G(W, V)$ gives an identification with $\mathbb{C}^d$, and provides the
isomorphism.
\end{proof}

Then we can calculate
\begin{align*}
\dim \op{Hom}_G(V, W) & = \dim (V^* \otimes W)^G \\
& \stackrel{\ref{dims}}{=} \int_G \chi_{V^* \otimes W} (g) \ dg\\
& = \int_G \conj{\chi_V(g)} \cdot \chi_W(g) \ dg \\
& = \langle \chi_V, \chi_W \rangle,
\end{align*}
which proves \ref{dim-formula}. Part~\ref{schur} is proven by using
\ref{dim-formula} and Schur's Lemma. For part~\ref{char-iso}, use lemma~\ref{b},
with part~\ref{dim-formula}. If $\chi_{V_1} = \chi_{V_2}$, then for all
irreducible $W$, $\dim \op{Hom}_G(W, V_i)$ is the same for $i = 1, 2$. Then 
apply the lemma.
\end{proof}

\begin{prop}\label{motto-justification}
Let $T < G$ be a maximal torus. Then $V, W \in \rep(G)$ are isomorphic if and
only if $\restr{V}{T} \iso \restr{W}{T}$.
\end{prop}

\begin{proof}
If $\restr{V}{T} \iso \restr{W}{T}$, then
$\restr{\chi_V}{T} = \restr{\chi_W}{T}$. Since characters are invariant under
conjugation, and the union of the conjugates of $T$ is $G$, $\chi_V = \chi_W$ on
all of $G$, hence $V \iso W$.
\end{proof}

This justifies the motto we had earlier, that to study representations of $G$ 
we can restrict our attention to a maximal torus.

\subsection{The Representation Ring}

We can rephrase proposition~\ref{motto-justification}; recall that
\[ R(G) = \ggroup(\rep(G)). \]

\begin{prop}
The restriction map $\op{res}: R(G) \rightarrow R(T)$ given by 
\[ R(G) \ni V \rightarrow \restr{V}{T}
    = \bigoplus_{\lambda \in \cgroup T} V_\lambda
    \ (= \{ v \in V | t \cdot v = \lambda(t) v, \ \forall t \in T \}) \]
is injective.
\end{prop}

Thus to classify representations of $G$, we can just identify the image.
Moreover, there is a symmetry present in the image of this map.

\subsubsection{Weyl Group} That symmetry is related to the Weyl group.

\begin{defn}
Let $N_G(T)$ be the normalizer of $T$ in $G$. The \textbf{Weyl group}
$W(G, T) := N_G(T)/T$. Often we will just write $W$, as this group doesn't
depend on the choice of $T$, up to isomorphism.
\end{defn}

\begin{rmk}
Note that $W$ is a finite group. Note that $N_G(T)$ is closed, hence a compact
subgroup. Also, $\op{exp}_T$ induces an isomorphism of $T$ with $\op{Lie}(T)$
modulo some lattice $\Lambda$. $W$ acts faithfully on this lattice, i.e. we 
have an embedding \[ W \xhookrightarrow{} \op{Aut}_\ints(\Lambda), \]
so $W$ is both compact and discrete, hence finite.
\end{rmk}

\begin{exmpl}
If $G = \op{SU}(n)$, then $W \iso S_n$ (think permutation matrices).
\end{exmpl}


The action of $W$ on $T$ induces an action of $W$ on $\cgroup T$, given by
\[ ^\omega \chi(t) = \chi(\omega^{-1} \cdot t). \] Hence $\op{res}(R(G))$ lands
in $\ints[\cgroup T]^W \subset \ints[\cgroup T],$ because characters are
conjugacy invariant.

\begin{theorem}[of the Highest Weight]
\[ \op{res}: R(G) \iso \ints[\cgroup T]^W \]
\end{theorem}

Later, we'll prove this in the setting of reductive algebraic groups.

\begin{exmpl}
Let $G = SU(2) = \{g \in \sl 2 \cmplx \ | \ \ ^t \conj{g} \cdot g = 1 \}$ and
take \[ T := \left\{ \left(\begin{matrix}
t & \\
  & t^{-1} \\
\end{matrix}\right) \ | \ t \in S^1 \right\}. \]

To prove surjectivity of $\op{res}$ (onto $\ints[\cgroup T]^W$), we just need to
write down enough representations of $G$. Let $V$ be the standard 
two-dimensional representation. Then for all $k \ge 0$, $\op{Sym}^k(V)$ is an
irreducible representation of $G$ of dimension $k + 1$.

Let $e_1, e_2$ be a basis of $V$. Then a basis of $\op{Sym}^k(V)$ is given by
$e_1^k, e_1^{k - 1} e_2, ..., e_2^k$, and the action on these basis elements
is $t^k, t^{k - 1}t^{-1} = t^{k - 2}, ..., t^{-k}$. Let $\chi_n$ be the
irreducible character
\[ \left(\begin{matrix}
t & \\
  & t^{-1} \\
\end{matrix}\right) \rightarrow t^n \]
of $T$. Then 
\[ \restr{\op{Sym}^k(V)}{T}
    = \chi_k \oplus \chi_{k - 2} \cdots \oplus \chi_{-k}.\]

Note that if $\omega \in W \setminus \{ 1 \},$ then
$^\omega \chi_k = \chi_{-k}$, because
\[ \left(\begin{matrix}
& 1 \\
-1 & \\
\end{matrix}\right) \left(\begin{matrix}
t & \\
& t^{-1} \\
\end{matrix}\right) \left(\begin{matrix}
& -1 \\
1 & \\
\end{matrix}\right) = \left(\begin{matrix}
t^{-1} & \\
& t \\
\end{matrix}\right). \]

These $\restr{\op{Sym}^k(V)}{T}$ for all $k \ge 0$ visibly span
$\ints[\cgroup T]^W$.
\end{exmpl}

