Roots, weights, and Dynkin labels

A highest weight specifies a multiplet once the Lie algebra and node conventions are fixed. Its Dynkin labels are the integer pairings with simple coroots. They are not the components in a simple-root basis, and a representation’s dimension does not uniquely identify its highest weight. The distinction becomes practical when computing dimensions or deciding whether the multiplet belongs to a quotient group.

Prerequisites: representations and irreducibility and Lie algebras versus global groups. This page develops the coordinates behind the highest-weight workbench; highest weights in a physics calculation applies its Casimir, index, and reality outputs.

The category and the coordinates

We consider finite-dimensional complex representations of a complex semisimple Lie algebra $\mathfrak g$, in characteristic zero. Equivalently, these integrate to continuous representations of its compact simply connected group. The examples below have a simple algebra. A noncompact real form’s unitary representations and representations in positive characteristic are different problems.

Choose a Cartan subalgebra $\mathfrak h\subset\mathfrak g$ and positive roots. A weight $\mu$ of a representation $V$ records simultaneous eigenvalues:

$$V_\mu=\{v\in V:H v=\mu(H)v\text{ for every }H\in\mathfrak h\}.$$

A root is a nonzero weight in the adjoint representation: $[H,E_\alpha]=\alpha(H)E_\alpha$. Thus roots describe the algebra’s raising and lowering directions, whereas the weights and their multiplicities depend on the representation. Indeed,

$$H(E_\alpha v)=[H,E_\alpha]v+E_\alpha Hv =(\alpha+\mu)(H)E_\alpha v,$$

so $E_\alpha$ takes a weight vector at $\mu$ to one at $\mu+\alpha$, or to zero. A highest-weight vector is annihilated by all positive-root operators.

The rank $r=\dim\mathfrak h$ counts independent weight coordinates. It is neither $\dim\mathfrak g$ nor $\dim V$. For $\mathfrak{sl}_3(\mathbb C)$ these are $2$, $8$, and the chosen multiplet’s dimension, respectively. See Etingof, §17.2, Proposition 17.4 for the root decomposition.

Fundamental weights and the four lattices

In the real span of the roots, use a Weyl-invariant inner product, with long roots of squared length $2$. For the ordered simple roots $\alpha_1,\ldots,\alpha_r$, define

$$\alpha_i^\vee=\frac{2\alpha_i}{(\alpha_i,\alpha_i)}, \qquad \langle\mu,\alpha_i^\vee\rangle =\frac{2(\mu,\alpha_i)}{(\alpha_i,\alpha_i)}.$$

The fundamental weights are the dual basis to the simple coroots:

$$\langle\omega_j,\alpha_i^\vee\rangle=\delta_{ij}.$$

For $\lambda=\sum_j a_j\omega_j$, the Dynkin label $a_i$ is therefore $\langle\lambda,\alpha_i^\vee\rangle$. Dominant integral weights have $a_i\in\mathbb Z_{\geq0}$; they classify the finite-dimensional complex irreducibles of $\mathfrak g$. This is the highest-weight theorem, Etingof §24.3, Theorem 24.17.

The relevant lattices are

$$\begin{aligned} Q&=\sum_i\mathbb Z\alpha_i,& Q^\vee&=\sum_i\mathbb Z\alpha_i^\vee,\\ P&=\{\mu:\langle\mu,Q^\vee\rangle\subset\mathbb Z\},& P^\vee&=\{x:\langle Q,x\rangle\subset\mathbb Z\}. \end{aligned}$$

Here $Q$ is the root lattice and $P=\sum_i\mathbb Z\omega_i$ the weight lattice. The superscript $\vee$ denotes the coroot or coweight counterpart. The weight lattice is dual to the coroot lattice, while the coweight lattice is dual to the root lattice. These distinctions matter whenever root lengths differ. Etingof, §20.2, printed p. 93.

The site’s Cartan matrix places coroots on rows:

$$A_{ij}=\langle\alpha_j,\alpha_i^\vee\rangle, \qquad \alpha_j=\sum_i A_{ij}\omega_i.$$

Consequently, if $\lambda=\sum_j c_j\alpha_j$, its column vectors of coordinates obey

$$a=A c,\qquad c=A^{-1}a.$$

This gives a direct conversion between root coordinates and Dynkin labels. A source using the transposed definition of $A$ requires the transposed conversion. Check the definition, not just the name “Cartan matrix.” Our convention agrees with Etingof §21.2, Example 21.12 and the SageMath Cartan-matrix documentation.

Work through A₂: the SU(3) labels (p,q)

Take orthonormal vectors $e_1,e_2,e_3$ and the plane with coordinate sum zero. Choose

$$\alpha_1=e_1-e_2,\qquad \alpha_2=e_2-e_3, \qquad A=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.$$

Solving the fundamental-weight pairings gives

$$\omega_1=\frac{2e_1-e_2-e_3}{3}, \qquad \omega_2=\frac{e_1+e_2-2e_3}{3}.$$

For example, $(\omega_1,\alpha_1)=1$ and $(\omega_1,\alpha_2)=0$. Both simple roots have squared length $2$, so roots and coroots coincide in this realization. Node $1$ labels the defining $\mathbf3$ of SU(3); node $2$ labels its dual $\overline{\mathbf3}$.

The three weights of the defining representation are

$$\omega_1,\qquad \omega_1-\alpha_1,\qquad \omega_1-\alpha_1-\alpha_2.$$

They are the restrictions of the three diagonal matrix eigenvalues to the traceless Cartan subalgebra. Their sum is zero, but none is a root. In contrast, the adjoint has the six roots and the zero weight with multiplicity two. A zero weight only means invariance under the Cartan subalgebra: it is not a group singlet. A nonzero traceless diagonal matrix generally fails to commute with a root matrix $E_{ij}$.

Count components with coroots

Write $\rho=\tfrac12\sum_{\alpha>0}\alpha=\sum_i\omega_i$. Weyl’s dimension formula is

$$\dim V_\lambda =\prod_{\beta^\vee>0} \frac{\langle\lambda+\rho,\beta^\vee\rangle} {\langle\rho,\beta^\vee\rangle}.$$

It follows from the identity limit of the character formula; root-length factors cancel between each numerator and denominator. Etingof, §28.1, Proposition 28.1, printed p. 125.

For $A_2$, the positive coroots are $\alpha_1^\vee$, $\alpha_2^\vee$, and $\alpha_1^\vee+\alpha_2^\vee$. At $\lambda=p\omega_1+q\omega_2$ their factors are

$$p+1,\qquad q+1,\qquad\frac{p+q+2}{2},$$

so

$$\dim(p,q)=\frac{(p+1)(q+1)(p+q+2)}{2}.$$

For $(p,q)=(2,1)$ the factors are $3$, $2$, and $5/2$, giving $15$. A fractional individual factor is expected; the product is an integer. The dual has highest weight $(1,2)$ and also dimension $15$. Conjugation exchanges the two fundamental weights because $\mathbf3^*\cong\Lambda^2\mathbf3$. For $(1,1)$ the result is $2\cdot2\cdot2=8$, agreeing with the eight traceless $3\times3$ matrices of the adjoint.

Calculate $(2,1)$ and compare its ordered labels with the SU(3) tensor-product calculator.

Retain the central character

With $z=e^{2\pi i/3}I$, the defining $\mathbf3$ has central phase $e^{2\pi i/3}$ and its exterior square has phase $e^{4\pi i/3}$. The highest-weight constituent of $\mathbf3^{\otimes p}\otimes\overline{\mathbf3}^{\otimes q}$ therefore has phase

$$z\big|_{V_{(p,q)}}=e^{2\pi i(p+2q)/3}\,I.$$

Every constituent has the same phase since the center acts as this scalar on the entire tensor product. This also follows from the lattice calculation:

$$A^{-1}\begin{pmatrix}p\\q\end{pmatrix} =\frac13\begin{pmatrix}2p+q\\p+2q\end{pmatrix}.$$

Thus $\lambda\in Q$ exactly when $p+2q=0\pmod3$. The $15$ with labels $(2,1)$ has charge $1$ and does not descend to PSU(3); the adjoint $(1,1)$ has charge zero and does.

More generally, for a compact simply connected semisimple $\widetilde G$, $P/Q$ is the character group of $Z(\widetilde G)$, and $[\lambda]$ records the central action. Descent to $\widetilde G/K$ requires that character to be trivial on the specified subgroup $K$. The full-center quotient requires $\lambda\in Q$. Conrad, Centers of Simply Connected Semisimple Groups, pp. 1–2. Continue with covers and global forms for the general descent problem.

Why B₂ and C₂ exchange their labels

The compact groups $\operatorname{Spin}(5)$ and $Sp(2)=USp(4)$ are isomorphic; their conventional $B_2$ and $C_2$ node orders differ. Here $Sp(2)$ means the compact quaternionic group, not $Sp(4,\mathbb R)$. The atlas records the group and its double cover of SO(5).

For $B_2$, take $(e_i,e_j)=\delta_{ij}$ and

$$\alpha_1=e_1-e_2,\qquad \alpha_2=e_2.$$

Node $1$ is long and node $2$ short. Their coroots are $e_1-e_2$ and $2e_2$, so

$$A_{B_2}=\begin{pmatrix}2&-1\\-2&2\end{pmatrix}, \qquad \omega_1=e_1,\quad \omega_2=\frac{e_1+e_2}{2}.$$

The positive roots are $e_1-e_2$, $e_2$, $e_1$, and $e_1+e_2$. Applying $\beta^\vee=2\beta/(\beta,\beta)$ gives positive-coroot coefficients

$$(1,0),\ (0,1),\ (2,1),\ (1,1)$$

in the simple-coroot basis. In particular, the root $e_1$ has simple-root coefficients $(1,1)$ but its coroot $2e_1$ has simple-coroot coefficients $(2,1)$. The official Sage type-B reference, simple_root and fundamental_weight, fixes this node order.

For $C_2$, take orthogonal $e_i$ with $(e_i,e_i)=1/2$ so that long roots still have squared length $2$. Choose $\alpha_1=e_1-e_2$ and $\alpha_2=2e_2$. Then

$$A_{C_2}=\begin{pmatrix}2&-2\\-1&2\end{pmatrix}, \qquad \omega_1=e_1,\quad \omega_2=e_1+e_2.$$

Node $2$ is now long. The positive-coroot coefficients are $(1,0)$, $(0,1)$, $(1,1)$, $(1,2)$. This uses the Sage type-C node order, with a uniform inner-product rescaling to the site’s long-root normalization; the root coordinates and Cartan matrix are unchanged.

Substituting these two coroot lists into Weyl’s formula yields

$$\begin{aligned} \dim_{B_2}(a,b)&=\frac{(a+1)(b+1)(a+b+2)(2a+b+3)}{6},\\ \dim_{C_2}(a,b)&=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}{6}. \end{aligned}$$

The check is immediate:

Labels	$B_2$ dimension	$C_2$ dimension
$(1,0)$	$5$	$4$
$(0,1)$	$4$	$5$
$(1,1)$	$16$	$16$

The $B_2$ spinor corresponds to the defining $C_2$ representation, with complex dimension $4$. Reversing the labels gives $\dim_{B_2}(a,b)=\dim_{C_2}(b,a)$, as the group isomorphism requires. At rank $r\geq3$, $B_r$ and $C_r$ are dual root systems but do not define isomorphic complex simple Lie algebras; the rank-two coincidence does not extend to that range.

Compute $B_2(1,0)$ and compute $C_2(1,0)$, retaining the type and node order with each result.

From labels to a reproducible result

Record the algebra type and rank, ordered Dynkin labels, global group, and normalization alongside any dimension or Casimir. The Weyl-formula reference specifies the remaining node orders and quadratic normalizations. A dominant weight supplies an irrep label; a Weyl dimension supplies its component count. Neither operation constructs normalized representation matrices, a tensor-product decomposition, or a basis of invariant contractions.

The current workbench handles its listed finite root systems through rank eight and labels from $0$ to $30$, using exact arithmetic. These are implementation bounds, not bounds on highest-weight theory. For an entire matter sector, test common central kernels; for a restriction, specify the branching map.

Check your understanding: for $SU(3)$, compare $(3,0)$ and $(0,3)$. Both have dimension $10$ and trivial center action, so both descend to $PSU(3)$. They remain distinct conjugate irreps. Quotient descent does not make a complex representation self-dual.

Continue: the Lie groups and topology path.