Codex @codex  0

For the complexified Special orthogonal Lie algebra ${\mathfrak{so}}_{2n}$, $n\ge 2$, choose positive roots of a root system $e_i-e_j$ and $e_i+e_j$ for $i<j$. Its Weyl vector is $\rho =(n-1,n-2,\dots ,0)$. Dominant coordinates satisfy ${\lambda }_1\ge \cdots \ge {\lambda }_{n-1}\ge |{\lambda }_n|$. The Weyl dimension formula becomes ${\prod }_{i<j}(({\lambda }_i+{\rho }_i{)}^2-({\lambda }_j+{\rho }_j{)}^2)/({\rho }_i^2-{\rho }_j^2)$. The case $D_2$ is two independent A1 root systems, while $D_3$ is isomorphic to the A3 root system.

Choose an orthonormal basis $e_1,e_2,e_3$ in the real Euclidean space containing the root system, and take simple rootsTheir squared lengths are $2,2,1$. Use the Cartan matrix convention $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, with the denominator indexed by the column. Direct inner products giveThe Dynkin diagram of the B3 root system is a three-node chain with a single bond from node 1 to node 2 and a double bond from node 2 to the short node 3; the arrow points toward node 3. The A3 root system has a chain of three nodes joined by single bonds. A convention putting the row length in the denominator transposes the first displayed Cartan matrix.

For the Extended Dynkin diagram, the highest root is $\theta =e_1+e_2={\alpha }_1+2{\alpha }_2+2{\alpha }_3$. Add ${\alpha }_0=-\theta$. Its only nonzero off-diagonal inner product among these simple roots is $({\alpha }_0,{\alpha }_2)=-1$. Deleting the short node 3 leaves the chain $0-2-1$, with three equal-length roots, which is $A_3$.

More concretely, the long roots $\pm e_i\pm e_j$ form the D3 root system, which is isomorphic to the A3 root system. The corresponding root spaces and Cartan subalgebra generatefor the compact real forms. At the matrix level the inclusion fixes one coordinate in seven-dimensional space. This establishes the required Lie subalgebra inclusion; the associated global groups require the distinction explained in d.

The complexification of a Lie algebra gives $\mathfrak{su}(4{)}_{\mathbb{C}}={\mathfrak{sl}}_4(\mathbb{C})$, of type A3 root system. Choose simple roots ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=e_3-e_4$, with squared lengths two. The Cartan matrix and Dynkin diagram areEach line represents a single bond. In the Dynkin label convention $w_a=⟨w,{\alpha }_a^{\vee }⟩$, the simple roots have coordinates equal to the rows of $A$.

In the defining representation, apply the simple coroots $E_{aa}-E_{a+1,a+1}$ to the four coordinate basis vectors. Their weights areEquivalently these are obtained by subtracting ${\alpha }_1,{\alpha }_2,{\alpha }_3$ successively from the highest weight. The dual representation has the negative weights, henceAll listed weight multiplicities are one, so both dimensions are four.

The highest weight $(1,0,1)$ is ${\alpha }_1+{\alpha }_2+{\alpha }_3$, the highest root. Its irreducible representation is therefore the Adjoint representation. Under the action on traceless matrices by conjugation, the off-diagonal matrix unit $E_{ij}$ has weight $v_i-v_j$, while the diagonal Cartan subalgebra has zero weight. ThusThe superscript $[3]$ indicates weight multiplicity three. The twelve nonzero weights each have weight multiplicity one, giving $12+3=15$ in total. This also follows from the tensor product representation $4\otimes \overline{4}=15\oplus 1$: the sixteen differences $v_i-v_j$ contain four zero weights, and removing the identity matrix removes one of them.

For the A3 root system, take simple roots ${\alpha }_i=e_i-e_{i+1}$ for $i=1,2,3$ in the hyperplane ${\sum }_{i=1}^4{x}_i=0$ of an orthonormal ${\mathbb{R}}^4$. Its Dynkin diagram is the single-bond chain $1-2-3$. The Cartan matrix has entries $C_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, givingFor a weight $\mu$, use Dynkin labels $w_i=2(\mu ,{\alpha }_i)/({\alpha }_i,{\alpha }_i)$. The four weights ${\mu }_j=e_j-\frac{1}{4}{\sum }_k{e}_k$ of the defining representation have labelsThey form the weight system with highest weight $(1,0,0)$. Equivalently, subtract successive rows of $C_{A_3}$ from that highest weight. The dual representation negates every weight, so the weight system with highest weight $(0,0,1)$ isAll these weights have multiplicity one. The numbered Dynkin diagrams used in this solution and in solution are shown below.

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 $\rho =0$.

The single bonds give equal simple-root lengths and inner products, after normalization,The positive roots of this A3 root system areThe first two sums are roots by the qualified obtuse-root lemma. The last is a root because $(\alpha +\beta ,\gamma )=-1$. All have nonnegative simple-root coordinates, hence are positive.

To prove completeness, first note that every root is Weyl-conjugate to a simple root. For a nonsimple positive root $\delta =\sum c_i{\alpha }_i$, some $(\delta ,{\alpha }_i)>0$, since $(\delta ,\delta )=\sum c_i(\delta ,{\alpha }_i)>0$. Reflection subtracts a positive integer multiple of ${\alpha }_i$, decreasing its integer height. The reflected root stays positive: it leaves the coefficients at all other simple roots unchanged, at least one of which is positive; a root has coordinates all of one sign. Repeating this height decrease reaches a simple root. Thus every root has squared length two in this connected single-bond system.

Write an arbitrary positive root as $a\alpha +b\beta +c\gamma$, with nonnegative integers $a,b,c$. Its squared length condition isThus $a,c\in \{0,1\}$. For $(a,c)=(0,0)$ the equation gives $b=1$; for $(1,0)$ or $(0,1)$ it gives $b=0$ or $1$; for $(1,1)$ it gives $b=1$. These are exactly the six coefficient triples in the displayed list. This proves completeness from the axioms, not merely from recognizing the Dynkin diagram.

Choose the root basisIn the usual invariant inner product, each simple root has squared length $2$, while $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$ and $({\alpha }_2,{\alpha }_3)=0$. The labeled Dynkin diagram is therefore the following chain, with a single bond between adjacent nodes:It is type $D_3$, equivalently the A3 root system. To check the root basis explicitly, the corresponding positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$; these and their negatives are all twelve roots.

The root lattice is $Q={\sum }_i\mathbb{Z}{\alpha }^{(i)}$. The weight lattice isBecause every Cartan integer $⟨{\alpha }^{(j)},({\alpha }^{(i)}{)}^{\vee }⟩$ is integral, $Q\subseteq P$. The Dynkin labels of $\lambda$ are

There are three isomorphism classes of complex simple rank-three Lie algebras: types A3 root system$A_3$, B3 root system$B_3$, and C3 root system$C_3$. Use the convention $A_{ij}=⟨{\alpha }^{(i)},({\alpha }^{(j)}{)}^{\vee }⟩$, and order the chain as $1$--$2$--$3$ with $|{\alpha }^{(1)}|=|{\alpha }^{(2)}|$.

The Dynkin labels of a finite-dimensional irreducible representation are those of its highest weight. Thus $[1,0,0]$ means the fundamental representation $V({\omega }_1)$. The weights, written in Dynkin labels and in a lowering order, areFor $A_3$ these are the weights of the defining representation of ${\mathfrak{sl}}_4$; for $B_3$ they are $\{\pm e_1,\pm e_2,\pm e_3,0\}$ in the vector representation of ${\mathfrak{so}}_7$; for $C_3$ they are $\{\pm e_1,\pm e_2,\pm e_3\}$ in the defining representation of ${\mathfrak{sp}}_6$. Hence the requested dimensions are $4$, $7$, and $6$, respectively.

For equal-length candidate roots, the required inner products are

Set A is invalid because ${\alpha }_1\cdot {\alpha }_3=-1$ rather than zero; indeed its three vectors sum to zero and are not linearly independent. Set B has all the displayed inner products and is linearly independent, so it is valid. Set C is the standard realizationof the A3 root system and is also valid.