Examples of root, parabolic, and borel subgroups corresponding to roots

Question

I'm interested in seeing a few examples of root, parabolic, and Borel subgroups given a specific reductive group $G$. Here is what I know.

Let $G$ be a reductive algebraic group over an algebraically closed field $k$ of characteristic $p$. Fix a maximal torus $T\subset G$, and let $T$ act on Lie$(G)=\mathfrak{g}$ via the adjoint action. Then $\mathfrak{g}$ decomposes as a direct sum of eigenspaces corresponding to this action. The eigenvalues are the roots in the group of characters of the torus $T$. Let the roots of $G$ with respect to $T$ be denoted by $\Phi$. Choose a set of simple roots $\Pi\subset\Phi$.

For each root $\alpha$, we can uniquely define a root subgroup $U_{\alpha}$ of $G$ as the image of a certain one-parameter subgroup $\mathbb{G}_a\to G$. Furthermore, given a subset $J\subset\Pi$ of simple roots, we can define $P_J=\langle T,U_{\alpha}|\alpha\in J\rangle$. Finally, there is exactly one Borel subgroup which contains the root subgroups corresponding to all positive roots (which depend on the choice of $\Pi$).

Is all of this correct? If so, it doesn't mean much to me without seeing an example or two worked through. What are examples of these subgroups in the case that $G=GL_n$ and $T$ is the diagonal subgroup? If this is too much to type out here, I would appreciate a link to any online notes that work through this example thoroughly. It would be nice to see how $T$ determines the set of roots $\Phi$, what the root subgroups are for an arbitrary root, what the parabolic subgroups are for subsets of a choice of simple roots, and what the corresponding Borel subgroup is.

Answer 1

I am also a beginner in this field. The following is an example.

For $G=GL_n$, $T=\{\mathrm{diag}(c_1,\cdots c_n),c_i\in G_m\}$ the set of diagonal matrices in $G$. $T\simeq G_m^n$ is a torus of $G$.

1. Consider the character group $X^*(T)$. It consists a natural basis $$ \epsilon_j:T\rightarrow G_m,\quad \epsilon_j(\mathrm{diag}(c_1,\cdots c_n))=c_j $$ Every characters of $T$ is a $\mathbb{Z}$-linear combinatinon of $\epsilon_j$.

2. $Lie(G)=\mathfrak{gl}_n$, the set of $n\times n$ matices. It has a natrual basis $E_{ij}$, the matrix with $1$ at $(i,j)$ and $0$ otherwise.

3. For $t=\mathrm{diag}(c_1,\cdots c_n)\in T$, consider the adjoint action on $E_{ij}$. \begin{eqnarray*} \mathrm{Ad}t.E_{ij}=\begin{cases}E_{ij},&\quad i=j\\ (c_ic_j^{-1})E_{ij}=(\epsilon_i-\epsilon_j)(t)E_{ij}&\quad i\neq j\end{cases} \end{eqnarray*} It shows that $E_{i,i}$ are in weiht zero space, and $E_{ij},i\neq j$ are non-zero weight space of weight $\alpha_{ij}=\epsilon_i-\epsilon_j$.

4. Thus the root system of $G=GL_n$ is $$\Phi=\{\pm\alpha_{ij}=\pm(\epsilon_i-\epsilon_j),1\leq i< j\leq n\}.$$ One can choose a set of positive roots $$\Phi^+=\{\alpha_{ij}=(\epsilon_i-\epsilon_j),1\leq i< j\leq n\}.$$ and the set of simple roots $$\Delta=\{\alpha_{i,i+1}=(\epsilon_i-\epsilon_{i+1}),1\leq i< j\leq n\}.$$ It is easy to see that elements in $\Phi$ are $\mathrm{Z}$-linear combinition of $\Delta$, and elements in $\Phi^+$ are with non-negative coefficients.

Answer 2

We cosider $X_*(T)=\{\lambda^\vee:G_m\rightarrow T\}$, the set of $1$-parameters subgroups.

1. It is easy to see that $X_*(T)$ is a free $\mathbb{Z}$-module with generators \begin{eqnarray} \epsilon_j^\vee:G_m\rightarrow T,\quad c\mapsto \mathrm{diag}(1,\cdots c,\cdots 1) \end{eqnarray}

2. Consider the composition: \begin{eqnarray} \epsilon_j\circ\epsilon_i^\vee:G_m\rightarrow T\rightarrow G_m \end{eqnarray} It is easy to see that $\epsilon_j\circ\epsilon_i^\vee=0$ if $i=j$, and $\epsilon_j\circ\epsilon^\vee_j=1$ is the identity map on $G_m$. Thus we have a $\mathbb{Z}$-binlinear map \begin{eqnarray} X^*(T)\times X_*(T)\rightarrow \mathbb{Z},\quad (\lambda,\mu^\vee)\mapsto\lambda\circ\mu^\vee \end{eqnarray} and $\{\epsilon_i\}$ and $\{\epsilon_i^\vee\}$ are the dual $\mathrm{Z}$-basis.

3. Given $\alpha_{ij}=\epsilon_i-\epsilon_j$. Consider $\alpha_{ij}^\vee=\epsilon_i-\epsilon_j$. It is an element in $X_*(T)$ and one has $$\alpha_{ij}\circ\alpha_{ij}^\vee= \epsilon_i\circ\epsilon_i^\vee+(-\epsilon_j)\circ(-\epsilon_j^\vee)=2.$$ Thus $\alpha_{ij}^\vee$ are the coroots of $\alpha_{ij}$. The set of corroots are $$ \Phi^\vee=\{\pm\alpha_{ij}^\vee=\pm(\epsilon_i-\epsilon_j),1\leq i<j\leq n\} $$

Answer 3

Now we have the root datum of $G=GL_n$. We also have chosen a set of positive roots $\Phi^+$ as above. For each $\alpha_{ij}\in\Phi^+$, as the above argument, the corresponding root space are $$\mathfrak{gl_n}_{\alpha_{ij}}=\mathbb{C}E_{ij}.$$ Roughly speaking, the torus $T$ and all positive root spaces corresponds to upper triangular matrix, which is just the Borel subgroup.