This week Sükrü led the discussion which focussed on root systems, algebraic groups and the Curtis-Tits Theorem.

Recall from an earlier week that given ${\alpha\in V=\mathbb{R}^n}$ and a positive definite bilinear form ${(.,.)}$ on ${V}$, we can define the reflection

$\displaystyle \sigma_{\alpha}:\beta\mapsto \beta-\frac{2(\alpha,\beta)}{(\alpha,\alpha)}\alpha$

Then a root system is a subset ${\Phi}$ of ${V}$ such that

1. ${\langle \Phi\rangle=V}$ and ${0\notin \Phi}$,
2. If ${\alpha\in\Phi}$ then the only scalar multiples of ${\alpha}$ in ${\Phi}$ are ${\pm \alpha}$,
3. ${\sigma_{\alpha}(\Phi)=\Phi}$ for all ${\alpha\in\Phi}$,
4. ${2(\alpha,\beta)/(\alpha,\alpha)\in\mathbb{Z}}$ for all ${\alpha,\beta\in\Phi}$.

A root system is called irreducible if it cannot be written as a disjoint union of ${\Phi_1}$ and ${\Phi_2}$ such that ${\langle \Phi_1\rangle \perp \langle \Phi_2\rangle}$.

A fundamental system is then a subset ${\Pi}$ of ${\Phi}$ which is a basis for ${V}$ and such that each element of ${\Phi}$ can be written as a linear combination of the elements of ${\Pi}$ such that the coefficients are either all nonnegative integers or all nonpositive integers.

The Dynkin diagram is the graph with ${n}$ vertices such that the ${i^{\mathrm{th}}}$ vertex is joined to the ${j^{\mathrm{th}}}$ vertex with strength

$\displaystyle n_{ij}=\frac{4(\alpha_i,\alpha_j)^2}{\alpha_i,\alpha_i)(\alpha_j,\alpha_j)}\in\mathbb{Z}$

Note that ${n_{ij}\in\{0,1,2,3\}}$. The table below shows what happens for each value of ${n_{ij}}$.

 ${n_{ij}}$ angle between ${\alpha_i}$ and ${\alpha_j}$ name of restriction of diagram to ${\{\alpha_i,\alpha_j\}}$ ${0}$ ${\pi/2}$ ${A_1\times A_1}$ ${1}$ ${\pi/3}$ ${A_2}$ ${2}$ ${\pi/4}$ ${B_2}$ ${3}$ ${\pi/6}$ ${G_2}$

Now let ${G=\mathrm{SL}_n(\overline{K})}$, where ${\overline{K}}$ is an algebraically closed field. A maximal torus ${T}$ is a subgroup maximal subject to being isomorphic to a subgroup of ${\overline{K}^{\times}\times\cdots \times \overline{K}^{\times}}$ and hence can be represented by all diagonal matrices. For each ${i,j\in\{1,\ldots,n\}}$ we can define

$\displaystyle \begin{array}{lcll} \alpha_{ij}:&T&\rightarrow &\overline{K}^{\times}\\ &\left(\begin{array}{ccc} \lambda_1 & &0\\ & \ddots &\\ 0& & \lambda_n\end{array}\right) & \mapsto & \lambda_i\lambda_j^{-1} \end{array}$

Let ${\Phi=\{\alpha_{ij}\mid i\neq j\}}$. The elements ${\alpha_{ij}}$ lie in the dual space ${E}$ of ${\overline{K}^n}$ and we can think of ${\Phi}$ as a subset of ${\mathbb{R}\otimes E}$, a vector space over ${\mathbb{R}}$. In this setting ${\Phi}$ is a root system with fundamental system ${\Pi=\{\alpha_{12},\alpha_{23},\ldots,\alpha_{n-1,n}\}}$.

An alternative way to see these roots is as follows. Let ${B}$ be a Borel subgroup of ${G}$, that is, a maximal connected solvable subgroup. Then ${B=UT}$ where ${U}$ is unipotent and normal in ${B}$, while ${T}$ is a torus. We can then choose a minimal ${T}$-invariant subgroup ${U_{\alpha}}$ of ${U}$. Then ${U_{\alpha}\cong \overline{K}^{+}}$. Since ${T}$ normalises ${U_{\alpha}}$ we obtain a map ${\alpha:T\rightarrow \mathrm{Aut}(\overline{K}^+)=\overline{K}^\times}$.

We can take ${B}$ to be the group of all upper triangular matrices and then ${U}$ is the subgroup of all upper triangular matrices such that the entries on the diagonal are all equal to 1, and ${T}$ is the subgroup of all diagonal matrices. Then there exists ${i,j}$ such that ${U_{\alpha}=\{X_{\alpha}(\lambda)=I+\lambda E_{ij}\mid \lambda\in \overline{K}\}}$. Now ${B}$ has a unique conjugate ${B^-}$ such that ${B\cap B^-=T}$ and then ${B^-=U^-T}$. The group ${B^-}$ is the group of all lower triangular matrices.

Note that for ${U_{\alpha_1}=\{I+\lambda E_{12}\mid \lambda\in \overline{K}\}}$ and ${U_{-\alpha_1}=\{I+\lambda E_{21}\mid \lambda\in \overline{K}\}}$ we have ${\langle U_{\alpha_1},U_{\alpha_2}\rangle=\mathrm{SL}_2(\overline{K})}$.

Now ${G=\mathrm{SL}_n(\overline{K})=\langle X_{\alpha}(\lambda)\mid \lambda\in\overline{K},\alpha\in\Phi\rangle}$. These generators are known as the Steinberg generators.

We can now state the Curtis-Tits Theorem.

Theorem Let ${\Phi}$ be an irreducible root system of rank at least 3 with fundamental system ${\Pi}$ and Dynkin diagram ${\Delta}$. Assume that ${G}$ is a group which satisfies the following conditions.

1. There exist subgroups ${K_{\alpha}=\langle U_{\alpha},U_{-\alpha}\rangle}$ for ${\alpha\in \Pi}$ with ${U_{\alpha}\cong U_{-\alpha}\cong K^+}$ for some field ${K}$ and ${K_{\alpha}/Z(K_{\alpha})\cong \mathrm{PSL}_2(K)}$ such that ${G=\langle K_{\alpha}\mid \alpha\in \Pi\rangle}$.
2. ${[K_{\alpha},K_{\alpha}]=1}$ if ${\alpha}$ and ${\beta}$ are not adjacent in the Dynkin diagram.
3. ${\langle K_{\alpha},K_\beta\rangle/Z\cong \mathrm{PSL}_3(K)}$ if ${\alpha}$ and ${\beta}$ are joined by a single bond.
4. ${\langle K_{\alpha},K_\beta\rangle/Z\cong\mathrm{PSp}_4(K)}$ if ${\alpha}$ and ${\beta}$ are joined by a double bond.
5. ${N_{K_{\alpha}}(U_{\alpha})\cap N_{K_{\alpha}}(U_{-\alpha})\leqslant N_G(U_{\beta})}$ for all ${\alpha,\beta\in \Pi}$.

Then there exists a group of Lie type ${\tilde{G}}$ with corresponding root system ${\Phi}$, fundamental system ${\Pi}$ and Dynkin diagram ${\Delta}$ and a surjective homomorphism from ${G}$ onto ${\tilde{G}}$.

We note it is possible to have ${K_{\alpha}}$ isomorphic to ${\mathrm{SL}_2(K)}$ or ${\mathrm{PSL}_2(K)}$. For example, for groups of Lie type with root systems of type ${B_n}$, the long roots all have ${K_{\alpha}\cong \mathrm{SL}_2(K)}$ while the short root has ${K_{\alpha}\cong \mathrm{PSL}_2(K)}$.

Sükrü will continue next week. He is also giving a Groups and Combinatorics seminar on Tuesday on related material.