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Write an element of the diagonal torus asand let ${\epsilon }_i$ extract $t_i$. The root-space decomposition iswith one-dimensional root spaces. Thus the $D_n$ root system is

The upper-triangular choice givesA compatible simple system isThe highest root and Weyl vector areThe fundamental weights are

Using the paper's letters, the root lattice and weight lattice are respectivelyTheir quotient is

The Dynkin diagram is the $D_n$ diagram: a chain ${\alpha }_1-\cdots -{\alpha }_{n-2}$ whose last node is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. The Extended Dynkin diagram adds ${\alpha }_0=-\theta$ joined to ${\alpha }_2$. For $D_4$, the central node ${\alpha }_2$ consequently has the four leaves ${\alpha }_0,{\alpha }_1,{\alpha }_3,{\alpha }_4$.

A Cartan subalgebra $\mathfrak{h}$ of a complex semisimple Lie algebra is a maximal commuting subalgebra consisting of semisimple elements. A nonzero functional $\alpha \in {\mathfrak{h}}^{\ast }$ is a root when its space in the root-space decompositionis nonzero. A Cartan-Weyl basis combines a basis of $\mathfrak{h}$ with root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$.

A choice of regular hyperplane divides roots into positive and negative roots. The simple roots are the positive roots that are not sums of two positive roots; they form a basis of the real root span. The Cartan matrix isup to the equivalent transposed indexing convention.

Use the nondegenerate restriction of the Killing form to $\mathfrak{t}$ to define $t_{\alpha }\in \mathfrak{t}$ byKilling-form invariance and the root-space decomposition show that ${\mathfrak{g}}_{\alpha }$ pairs nondegenerately with ${\mathfrak{g}}_{-\alpha }$ and orthogonally with every other root space. Choose nonzero $e\in {\mathfrak{g}}_{\alpha }$ and $f\in {\mathfrak{g}}_{-\alpha }$ with $\kappa (e,f)\ne 0$. For $t\in \mathfrak{t}$,so

We need $\alpha (t_{\alpha })\ne 0$. If it were zero, the span of $e,f,t_{\alpha }$ would be a solvable Heisenberg-type Lie algebra with central commutator $[e,f]$. By Lie theorem, its adjoint action on $\mathfrak{g}$ can be upper triangularized, so $\mathrm{ad}[e,f]$ is nilpotent. But $[e,f]\in \mathfrak{t}$, and elements of the Cartan subalgebra act semisimply; hence $\mathrm{ad}[e,f]=0$. A semisimple Lie algebra has zero center, contradicting $[e,f]\ne 0$.

SetRescale $f$ so that $[e_{\alpha },f_{\alpha }]=h_{\alpha }$. Since $e_{\alpha }$ and $f_{\alpha }$ lie in the $\alpha$ and $-\alpha$ root spaces,Thus ${\mathfrak{m}}_{\alpha }=⟨e_{\alpha },h_{\alpha },f_{\alpha }⟩$ is the sl2 subalgebra associated with a root.

After ordering a symplectic basis in two blocks, writeMatrices in the symplectic Lie algebra have block formThe root-space decomposition isFor example, these one-dimensional spaces are spanned respectively byThus this is the Cn root systemThe upper-triangular choice givesIts simple roots, highest root, fundamental weights, and half-sum of positive roots are

Using the notation requested in the paper, the root lattice $P$ and weight lattice $Q$ areso $Q/P\cong \mathbb{Z}/2\mathbb{Z}$. This reverses the common notation in which the root lattice is called $Q$ and the weight lattice is called $P$.

Since a multiple-edge arrow in a Dynkin diagram points toward the shorter root, the finite and extended diagrams areand

For $t$ in a Cartan subalgebra $\mathfrak{t}$, the root-space decomposition givesThus $t$ is regular exactly when $\alpha (t)\ne 0$ for every root $\alpha$.

A root vector is a nonzero element of a root space ${\mathfrak{g}}_{\alpha }$ in the root-space decomposition of a semisimple Lie algebra.

A semisimple Lie algebra over a field of characteristic zero is split if it has a Cartan subalgebra whose adjoint action is simultaneously diagonalizable over that field. Its root-space decomposition then needs no extension of scalars. The sl2R Lie algebra is split, with the real diagonal Cartan subalgebra. The real special unitary Lie algebra $\mathfrak{su}(2)$ is not: its adjoint operators are skew-adjoint for the positive definite inner product obtained by negating its Killing form. Thus a split adjoint operator would have to be zero.