# Prologue

## 0.1 References

• Carter’s “Finite Groups of Lie Type”[1]

• Humphreys’ “Linear Algebraic Groups”[2]

# 1 Friday, August 21

## 1.1 Intro and Definitions

Let $$k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ be algebraically closed (e.g. $$k = {\mathbb{C}}, \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}_p\mkern-1.5mu}\mkern 1.5mu$$). A variety $$V\subseteq k^n$$ is an affine $$k{\hbox{-}}$$variety iff $$V$$ is the zero set of a collection of polynomials in $$k[x_1, \cdots, x_n]$$.

Here $${\mathbb{A}}^n\mathrel{\vcenter{:}}= k^n$$ with the Zariski topology, so the closed sets are varieties.

An affine algebraic $$k{\hbox{-}}$$group is an affine variety with the structure of a group, where the multiplication and inversion maps \begin{align*} \mu: G\times G &\to G \\ \iota: G&\to G \end{align*} are continuous.

$$G = {\mathbb{G}}_a \subseteq k$$ the additive group of $$k$$ is defined as $${\mathbb{G}}_a \mathrel{\vcenter{:}}=(k, +)$$. We then have a coordinate ring $$k[{\mathbb{G}}_a] = k[x] / I = k[x]$$.

$$G = \operatorname{GL}(n, k)$$, which has coordinate ring $$k[x_{ij}, T] / \left\langle{\det(x_{ij})\cdot T = 1}\right\rangle$$.

Setting $$n=1$$ above, we have $${\mathbb{G}}_m \mathrel{\vcenter{:}}=\operatorname{GL}(1, k) = (k^{\times}, \cdot)$$. Here the coordinate ring is $$k[x, T] / \left\langle{xT = 1}\right\rangle$$.

$$G = {\operatorname{SL}}(n, k) \leq \operatorname{GL}(n, k)$$, which has coordinate ring $$k[G] = k[x_{ij}] / \left\langle{\det(x_{ij}) = 1}\right\rangle$$.

A variety $$V$$ is irreducible iff $$V$$ can not be written as $$V = \cup_{i=1}^n V_i$$ with each $$V_i \subseteq V$$ a proper subvariety.

There exists a unique irreducible component of $$G$$ containing the identity $$e$$. Notation: $$G^0$$.

$$G$$ is the union of translates of $$G^0$$, i.e. there is a decomposition \begin{align*} G = {\coprod}_{g\in \Gamma} \, g\cdot G^0 ,\end{align*} where we let $$G$$ act on itself by left-translation and define $$\Gamma$$ to be a set of representatives of distinct orbits.

One can define solvable and nilpotent algebraic groups in the same way as they are defined for finite groups, i.e. as having a terminating derived or lower central series respectively.

## 1.2 Jordan-Chevalley Decomposition

There is a maximal connected normal solvable subgroup $$R(G)$$, denoted the radical of $$G$$.

• $$\left\{{e}\right\} \subseteq R(G)$$, so the radical exists.
• If $$A, B \leq G$$ are solvable then $$AB$$ is again a solvable subgroup.

An element $$u$$ is unipotent $$\iff$$ $$u = 1+n$$ where $$n$$ is nilpotent $$\iff$$ its the only eigenvalue is $$\lambda = 1$$.

For any $$G$$, there exists a closed embedding $$G\hookrightarrow\operatorname{GL}(V) = \operatorname{GL}(n , k)$$ and for each $$x\in G$$ a unique decomposition $$x=su$$ where $$s$$ is semisimple (diagonalizable) and $$u$$ is unipotent.

Define $$R_u(G)$$ to be the subgroup of unipotent elements in $$R(G)$$.

Suppose $$G$$ is connected, so $$G = G^0$$, and nontrivial, so $$G\neq \left\{{e}\right\}$$. Then

• $$G$$ is semisimple iff $$R(G) = \left\{{e}\right\}$$.
• $$G$$ is reductive iff $$R_u(G) = \left\{{e}\right\}$$.

$$G = \operatorname{GL}(n, k)$$, then $$R(G) = Z(G) = kI$$ the scalar matrices, and $$R_u(G) = \left\{{e}\right\}$$. So $$G$$ is reductive and semisimple.

$$G = {\operatorname{SL}}(n , k)$$, then $$R(G) = \left\{{I}\right\}$$.

Is this semisimple? Reductive? What is $$R_u(G)$$?

A torus $$T\subseteq G$$ in $$G$$ an algebraic group is a commutative algebraic subgroup consisting of semisimple elements.

Let \begin{align*} T \mathrel{\vcenter{:}}= \left\langle{ \begin{bmatrix} a_1 & & \mathbf 0\\ & \ddots & \\ \mathbf 0 & & a_n \end{bmatrix} \subseteq \operatorname{GL}(n ,k) }\right\rangle .\end{align*}

Why are torii useful? For $$g = \mathrm{Lie}(G)$$, we obtain a root space decomposition \begin{align*} g = \qty{\bigoplus_{\alpha \in \Phi_- }g_\alpha} \oplus t \oplus \qty{\bigoplus_{\alpha \in \Phi_+ }g_\alpha} .\end{align*}

When $$G$$ is a simple algebraic group, there is a classification/correspondence: \begin{align*} (G, T) \iff (\Phi, W) .\end{align*} where $$\Phi$$ is an irreducible root system and $$W$$ is a Weyl group.

# 2 Monday, August 24

## 2.1 Review and General Setup

• $$k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ is algebraically closed
• $$G$$ is a reductive algebraic group
• $$T\subseteq G$$ is a maximal split torus

Split: $$T\cong \bigoplus {\mathbb{G}}_m$$.

We’ll associate to this a root system, not necessarily irreducible, yielding a correspondence \begin{align*} (G, T) \iff (\Phi, W) \end{align*} with $$W$$ a Weyl group.

This will be accomplished by looking at $${\mathfrak{g}}= \mathrm{Lie}(G)$$. If $$G$$ is simple, then $${\mathfrak{g}}$$ is “simple,” and $$\Phi$$ irreducible will correspond to a Dynkin diagram.

There is this a 1-to-1 correspondence \begin{align*} G \text{ simple}/\sim \iff A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4, G_2 \end{align*} where $$\sim$$ denotes isogeny.

Taking the Zariski tangent space at the identity “linearizes” an algebraic group, yielding a Lie algebra.

We have the coordinate ring $$k[G] = k[x_1, \cdots, x_n] / \mathcal{I}(G)$$ where $$\mathcal{I}(G)$$ is the zero set. This is equal to $$\left\{{f:G\to k}\right\}$$,

## 2.2 The Associated Lie Algebra

Define left translation is \begin{align*} \lambda_x: k[G] &\to k[G] \\ y &\mapsto f(x^{-1} y) .\end{align*}

Define derivations as \begin{align*} \mathrm{Der} ~k[G] = \left\{{D: k[G] \to k[G] {~\mathrel{\Big|}~}D(fg) = D(f) g + f D(g) }\right\} .\end{align*}

We can then realize the Lie algebra as \begin{align*} {\mathfrak{g}}= \mathrm{Lie}(G) = \left\{{D\in \mathrm{Der} k[G] {~\mathrel{\Big|}~}\lambda_x \circ D = D\circ \lambda_x}\right\} ,\end{align*} the left-invariant derivations.

• $$G = \operatorname{GL}(n, k) \implies{\mathfrak{g}}= {\mathfrak{gl}}(n, k)$$
• $$G = {\operatorname{SL}}(n, k) \implies{\mathfrak{g}}= {\mathfrak{sl}}(n, k)$$

Let $$G$$ be reductive and $$T$$ be a split torus. Then $$T$$ acts on $${\mathfrak{g}}$$ via an adjoint action. (For $$\operatorname{GL}_n, {\operatorname{SL}}_n$$, this is conjugation.)

There is a decomposition into eigenspaces for the action of $$T$$, \begin{align*} {\mathfrak{g}}= \qty{\bigoplus_{\alpha\in \Phi} g_\alpha} \oplus t \end{align*} where $$t = \mathrm{Lie}(T)$$ and $$g_\alpha \mathrel{\vcenter{:}}=\left\{{x\in {\mathfrak{g}}{~\mathrel{\Big|}~}t.x = \alpha(t) x\,\, \forall t\in T}\right\}$$ with $$\alpha: T\to K^{\times}$$ a rational function (a root).

In general, take $$\alpha\in\hom_{\text{AlgGrp}}(T, {\mathbb{G}}_m)$$.

Let $$G = \operatorname{GL}(n, k)$$ and \begin{align*} T = \left\{{ \begin{bmatrix} a_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & a_n \end{bmatrix} {~\mathrel{\Big|}~}a_j\in k^{\times} }\right\} .\end{align*}

Then check the following action:

which indeed acts by a rational function.

Then \begin{align*} g_\alpha = {\operatorname{span}}\left\{{ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} }\right\} = g_{(1, -1, 0)} .\end{align*}

For $${\mathfrak{g}}= {\mathfrak{gl}}(3, k)$$, we have \begin{align*} {\mathfrak{g}}= t &\oplus g_{(1, -1, 0)} \oplus g_{(-1, 1, 0)} \\ &\oplus g_{(0, 1, -1)} \oplus g_{(0, -1, 1)} \\ &\oplus g_{(1, 0, -1)} \oplus g_{(-1, 0, 1)} .\end{align*}

## 2.3 Representations

Let $$\rho: G\to \operatorname{GL}(V)$$ be a group homomorphisms, then equivalently $$V$$ is a (rational) $$G{\hbox{-}}$$module.

For $$T\subseteq G$$, $$T\curvearrowright G$$ semisimply, so we can simultaneously diagonalize these operators to obtain a weight space decomposition $$V = \bigoplus_{\lambda \in X(T)} V_\lambda$$, where \begin{align*} V_\lambda &\mathrel{\vcenter{:}}=\left\{{v\in V{~\mathrel{\Big|}~}t.v = \lambda(t)v\,\, \forall t\in T}\right\} \\\ X(T) &\mathrel{\vcenter{:}}=\hom(T, {\mathbb{G}}_m) .\end{align*}

Let $$G = \operatorname{GL}(n, k)$$ and $$V$$ the $$n{\hbox{-}}$$dimensional natural representation as column vectors, \begin{align*} V = \left\{{{\left[ {v_1, \cdots, v_n} \right]} {~\mathrel{\Big|}~}v_j \in k}\right\} .\end{align*}

Then \begin{align*} T = \left\{{ \begin{bmatrix} a_1 & 0 & 0 \\ 0 & \ddots & 0\\ 0 & 0 & a_n \end{bmatrix} {~\mathrel{\Big|}~}a_j \in k^{\times} }\right\} .\end{align*}

Consider the basis vectors $$\mathbf{e}_j$$, then \begin{align*} \begin{bmatrix} a_1 & 0 & 0 \\ 0 & \ddots & 0\\ 0 & 0 & a_n \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = a_j \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = a_1^0 a_2^0 \cdots a_j^0 \cdots a_n^0 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} .\end{align*}

Here the weights are of the form $$\varepsilon_j\mathrel{\vcenter{:}}={\left[ {0, 0, \cdots, 1, \cdots, 0} \right]}$$ with a $$1$$ in the $$j$$th spot, so we have \begin{align*} V = V_{\varepsilon_1} \oplus V_{\varepsilon_2} \oplus \cdots \oplus V_{{\varepsilon_n}} .\end{align*}

For $$V = {\mathbb{C}}$$, we have $$t.v = (a_1^0 \cdots a_n^0)v$$ and $$V = V_{(0, 0, \cdots, 0)}$$.

## 2.4 Classification

Let $$G$$ be a simple algebraic group (ano closed, connected, normal subgroups other than $$\left\{{e}\right\}, G$$) that is nonabelian that is nonabelian.

Let $$G = {\operatorname{SL}}(3, k)$$. Then \begin{align*} T = \left\{{ t = \begin{bmatrix} a_1 & 0 & 0 \\ 0 & a_1 a_2^{-1} & 0\\ 0 & 0 & a_2^{-1} \end{bmatrix} {~\mathrel{\Big|}~} a_1, a_2\in k^{\times} }\right\} \end{align*} and \begin{align*} t. \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} = a_1^2 a_2^{-1} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} .\end{align*} and $$\alpha_1 = (2, -1)$$.

Then \begin{align*} {\mathfrak{g}}= {\mathfrak{g}}_{(2, -1)} \oplus {\mathfrak{g}}_{(-2, 1)} \oplus {\mathfrak{g}}_{(-1, 2)} \oplus {\mathfrak{g}}_{(1, -2)} \oplus {\mathfrak{g}}_{(1, 1)} \oplus {\mathfrak{g}}_{(-1, -1)} .\end{align*}

Then $$\alpha_2 = (-1, 2)$$ and $$\alpha_1 + \alpha_2 = ( 1, 1)$$.

This gives the root space decomposition for $${\mathfrak{sl}}_3$$:

Then the Weyl group will be generated by reflections through these hyperplanes.

# 3 Wednesday, August 26

## 3.1 Review

• $$G$$ a reductive algebraic group over $$k$$
• $$T = \prod_{i=1}^n {\mathbb{G}}_m$$ a maximal split torus
• $${\mathfrak{g}}= \mathrm{Lie}(G)$$
• There’s an induced root space decomposition $${\mathfrak{g}}= t\oplus \bigoplus_{\alpha\in \Phi}{\mathfrak{g}}_\alpha$$
• When $$G$$ is simple, $$\Phi$$ is an irreducible root system
• There is a classification of these by Dynkin diagrams

$$A_n$$ corresponds to $${\mathfrak{sl}}(n+1, k)$$ (mnemonic: $$A_1$$ corresponds to $${\mathfrak{sl}}(2)$$)

• We have representations $$\rho: G\to \operatorname{GL}(V)$$, i.e. $$V$$ is a $$G{\hbox{-}}$$module

• For $$T\subseteq G$$, we have a weight space decomposition: $$V = \bigoplus_{\lambda \in X(T)} V_\lambda$$ where $$X(T) = \hom(T, {\mathbb{G}}_m)$$.

Note that $$X(T) \cong {\mathbb{Z}}^n$$, the number of copies of $${\mathbb{G}}_m$$ in $$T$$.

## 3.2 Root Systems and Weights

Let $$\Phi = A_2$$, then we have the following root system:

In general, we’ll have $$\Delta = \left\{{\alpha_1, \cdots, \alpha_n}\right\}$$ a basis of simple roots.

Every root $$\alpha\in I$$ can be expressed as either positive integer linear combination (or negative) of simple roots.

For any $$\alpha\in \Phi$$, let $$s_\alpha$$ be the reflection across $$H_\alpha$$, the hyperplane orthogonal to $$\alpha$$. Then define the Weyl group $$W = \left\{{s_\alpha {~\mathrel{\Big|}~}\alpha\in \Phi}\right\}$$.

Here the Weyl group is $$S_3$$:

$$W$$ acts transitively on bases.

$$X(T) \subseteq {\mathbb{Z}}\Phi$$, recalling that $$X(T) = \hom(T, {\mathbb{G}}_m) = {\mathbb{Z}}^n$$ for some $$n$$. Denote $${\mathbb{Z}}\Phi$$ the root lattice and $$X(T)$$ the weight lattice.

Let $$G = {\mathfrak{sl}}(2, {\mathbb{C}})$$ then $$X(T) = {\mathbb{Z}}\omega$$ where $$\omega = 1$$, $${\mathbb{Z}}\Phi = {\mathbb{Z}}\left\{{\alpha}\right\}$$ Then there is one weight $$\alpha$$, and the root lattice $${\mathbb{Z}}\Phi$$ is just $$2{\mathbb{Z}}$$. However, the weight lattice is $${\mathbb{Z}}\omega = {\mathbb{Z}}$$, and these are not equal in general.

There is partial ordering on $$X(T)$$ given by $$\lambda \geq \mu \iff \lambda - \mu = \sum_{\alpha\in \Delta} n_\alpha \alpha$$ where $$n_\alpha \geq 0$$. (We say $$\lambda$$ dominates $$\mu$$.)

We extend scalars for the weight lattice to obtain $$E \mathrel{\vcenter{:}}= X(T) \otimes_{\mathbb{Z}}{\mathbb{R}}\cong {\mathbb{R}}^n$$, a Euclidean space with an inner product $${\left\langle {{\,\cdot\,}},~{{\,\cdot\,}} \right\rangle}$$.

For $$\alpha\in \Phi$$, define its coroot $$\alpha^\vee\mathrel{\vcenter{:}}={2\alpha \over {\left\langle {\alpha},~{\alpha} \right\rangle}}$$. Define the simple coroots as $$\Delta^\vee\mathrel{\vcenter{:}}=\left\{{\alpha_i^\vee}\right\}_{i=1}^n$$, which has a dual basis $$\Omega \mathrel{\vcenter{:}}=\left\{{\omega_i}\right\}_{i=1}^n$$ the fundamental weights. These satisfy $${\left\langle {\omega_i},~{\alpha_j^\vee} \right\rangle} = \delta_{ij}$$.

Important because we can index irreducible representations by fundamental weights.

A weight $$\lambda\in X(T)$$ is dominant iff $$\lambda \in {\mathbb{Z}}^{\geq 0} \Omega$$, i.e. $$\lambda = \sum n_i \omega_i$$ with $$n_i \in {\mathbb{Z}}^{\geq 0}$$.

If $$G$$ is simply connected, then $$X(T) = \bigoplus {\mathbb{Z}}\omega_i$$.

See Jantzen for definition of simply-connected, $${\operatorname{SL}}(n+1)$$ is simply connected but its adjoint $$PGL(n+1)$$ is not simply connected.

## 3.3 Complex Semisimple Lie Algebras

When doing representation theory, we look at the Verma modules $$Z(\lambda) = U({\mathfrak{g}}) \otimes_{U({\mathfrak{b}}^+)} \lambda \twoheadrightarrow L(\lambda)$$.

$$L(\lambda)$$ as a finite-dimensional $$U({\mathfrak{g}}){\hbox{-}}$$module $$\iff$$ $$\lambda$$ is dominant, i.e. $$\lambda \in X(T)_+$$.

Thus the representations are indexed by lattice points in a particular region: