= Solution
The PDF's $J$ is the anti-diagonal identity matrix. Write $\bar i=2n+2-i$ and $o=n+1$. Its diagonal <Cartan subalgebra> consists of
$$
H=\operatorname{diag}(h_1,\ldots,h_n,0,-h_n,\ldots,-h_1),\qquad
\varepsilon_i(H)=h_i.
$$
The defining matrix identity says $A_{ij}=-A_{\bar j,\bar i}$. The <root spaces> have the following <root vectors>, together with their opposites:
$$
\begin{array}{c|c}
\text{root}&\text{root vector}\\\hline
\varepsilon_i-\varepsilon_j&E_{ij}-E_{\bar j,\bar i}\quad(i\ne j)\\
\varepsilon_i+\varepsilon_j&E_{i,\bar j}-E_{j,\bar i}\quad(i<j)\\
\varepsilon_i&E_{i,o}-E_{o,\bar i}
\end{array}
$$
The negative short root has vector $E_{o,i}-E_{\bar i,o}$. Hence the <root-space decomposition> is $\mathfrak g=\mathfrak t\oplus\bigoplus_{\alpha\in R}\mathfrak g_\alpha$, each root space is one-dimensional, and the <Bn root system> is
$$
\boxed{R=\{\pm\varepsilon_i\pm\varepsilon_j:i<j\}\cup\{\pm\varepsilon_i\}.}
$$
There are $2n^2$ roots and $n$ zero-weight dimensions, giving $\dim\mathfrak g=n(2n+1)$ as a check. Upper triangular matrices give
$$
\boxed{R^+=\{\varepsilon_i-\varepsilon_j,\varepsilon_i+\varepsilon_j:i<j\}\cup\{\varepsilon_i\},\quad
\alpha_i=\varepsilon_i-\varepsilon_{i+1}\ (i<n),\quad\alpha_n=\varepsilon_n.}
$$
For $n\geq2$, the <highest root> is $\theta=\varepsilon_1+\varepsilon_2=\alpha_1+2\alpha_2+\cdots+2\alpha_n$. The <fundamental weights> and <Weyl vector> are
$$
\boxed{\omega_i=\varepsilon_1+\cdots+\varepsilon_i\ (i<n),\quad
\omega_n=\tfrac12(\varepsilon_1+\cdots+\varepsilon_n),\quad
\rho=\sum_{i=1}^n\left(n-i+\tfrac12\right)\varepsilon_i.}
$$
These weights pair to $\delta_{ij}$ with the <simple coroots>; summing the positive roots gives the displayed $\rho$.
The <Dynkin diagram> is the chain $1,\ldots,n$, with single edges except a double edge between $n-1$ and $n$, whose arrow points to the short root $\alpha_n$. In the <Extended Dynkin diagram>, $\alpha_0=-\theta$ attaches by a single edge to $\alpha_2$ when $n\geq3$. For $n=2$, both long nodes $\alpha_0,\alpha_1$ have a double edge to the short node $\alpha_2$. The <Bn Dynkin diagram and affine extension> below distinguishes these small-rank cases.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-102-bn-dynkin.png]
{title=Ordinary and affine Dynkin diagrams of type B}
For $n=1$, the system is $A_1$: $R^+=\{\varepsilon_1\}$, $\theta=\varepsilon_1$, $\omega_1=\rho=\varepsilon_1/2$. Its ordinary diagram is one node; its affine diagram has two equal-length nodes with the usual affine $A_1$ double bond.
Back to article page
