Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 3 / 2 / a / Solution

Put $\bar{i}=2n+1-i$ and let $E_{ij}$ denote a matrix unit. The antidiagonal matrix satisfies $J^{-1}=J$, so the orthogonal condition is $A=-J{A}^{T}J$, or entrywise $A_{ij}=-A_{\bar{j}\bar{i}}$. Consequently the diagonal Cartan subalgebra isWe first take $n\ge 2$. Since $[H,E_{ab}]=(H_{aa}-H_{bb})E_{ab}$, the matrix root basis of the even orthogonal Lie algebra givesThe zero weight space is $\mathfrak{t}$; its basis is $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. No vector of weight $2{\epsilon }_i$ occurs, because the corresponding matrix unit is forced to be its own negative. These spaces account for $n+2n(n-1)=n(2n-1)$ dimensions, exhausting the Special orthogonal Lie algebra. Thus its decomposition as a $\mathfrak{t}$-module and its Dn root system areEach nonzero root space has dimension one; the zero weight has multiplicity $n$.

The upper triangular root spaces are precisely those of weights ${\epsilon }_i-{\epsilon }_j$ and ${\epsilon }_i+{\epsilon }_j$ with $i<j$. ThereforeThese $n$ roots form the simple roots. To check that they generate the chosen positive system with nonnegative coefficients, write ${\epsilon }_i-{\epsilon }_j={\sum }_{k=i}^{j-1}{\alpha }_k$. For $j<n$,and for $j=n$ use ${\epsilon }_i+{\epsilon }_n={\sum }_{k=i}^{n-2}{\alpha }_k+{\alpha }_n$. Empty sums are zero. Their linear independence and these expansions identify the base of the root system.

Equip the real span of the roots in ${\mathfrak{t}}^{\ast }$ with the inner product making the ${\epsilon }_i$ orthonormal. All roots then have squared length two. The coroots of ${\epsilon }_i\pm {\epsilon }_j$ are $H_i\pm H_j$, so the fundamental weights of Dn, characterized by ${\omega }_i({\alpha }_j^{\vee })={\delta }_{ij}$, areTaking their pairings with $H_j-H_{j+1}$ and $H_{n-1}+H_n$ verifies the defining identities directly. For each pair $i<j$, the sum of its two positive roots is $2{\epsilon }_i$. Hence the half-sum of positive roots is

The Dynkin diagram has a chain ending at ${\alpha }_{n-2}$, which is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. Every edge is a single bond because the root lengths agree; the nonzero off-diagonal simple-root pairings are minus one. The labelled Dn Dynkin diagram and its low-rank versions are shown below.

For $n=2$, ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_1+{\epsilon }_2$ are orthogonal, giving two disconnected A1 root systems. For $n=3$, the diagram is the chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$, of type A3 root system; the displayed weight formulas still apply. If $n=1$ is allowed, ${\mathfrak{so}}_2=\mathfrak{t}$ is one-dimensional and abelian: there are no roots or fundamental weights, and $\rho =0$.