Compact Lie representation formulas

A highest weight determines the dimension and quadratic invariants of a finite-dimensional complex irrep. A global group imposes a separate condition on its central character. The formulas below keep these two calculations distinct and fix the normalization needed to use their numerical outputs in physics.

The domain is a compact simply connected group $\widetilde G$ with simple Lie algebra, or equivalently finite-dimensional modules of its complexified algebra $\mathfrak g$, over $\mathbb C$ in characteristic zero. A quotient $\widetilde G/K$ has the same local invariants for every representation that descends. Infinite-dimensional noncompact unitary representations are outside this formula sheet.

Open the highest-weight calculator · Learn the coordinates · Full method and node order

Fixed conventions

Choose ordered simple roots $\alpha_i$, fundamental weights $\omega_i$, and positive roots $\Phi_+$. Write

$$\lambda=\sum_{i=1}^r a_i\omega_i,\qquad \langle\omega_j,\alpha_i^\vee\rangle=\delta_{ij},\qquad a_i\in\mathbb Z_{\geq0},\qquad \rho=\frac12\sum_{\alpha\in\Phi_+}\alpha=\sum_i\omega_i.$$

Long roots have squared length $2$. The Cartan matrix has coroots on rows:

$$A_{ij}=\langle\alpha_j,\alpha_i^\vee\rangle, \qquad D_{ii}=\frac{(\alpha_i,\alpha_i)}2, \qquad M_{ij}=(\omega_i,\omega_j)=(D A^{-1})_{ij}.$$

The matrix $M$ is symmetric. To verify the order of the matrices, use $\alpha_j=\sum_i A_{ij}\omega_i$ and $(\omega_i,\alpha_j)=\delta_{ij}D_{jj}$, giving $M A=D$. Transposing the Cartan convention changes the coordinate formulas. Etingof, §§20.2 and 21.2, Example 21.12, uses this row-coroot convention.

For the generators, use Hermitian matrices $T_R^a$ in a unitary realization and define

$$[T_R^a,T_R^b]=i f^{abc}T_R^c, \qquad \sum_a T_R^aT_R^a=C_2(R)I_R, \qquad \operatorname{tr}_R(T_R^aT_R^b)=T(R)\delta^{ab}.$$

The same basis of the algebra is used in every representation. Its scale is fixed here by $C_2(\mathrm{adj})=h^\vee$, the dual Coxeter number; equivalently $f^{acd}f^{bcd}=h^\vee\delta^{ab}$. For $SU(N)$ this is the familiar normalization $T(\mathbf N)=1/2$, or $\operatorname{tr}_{\mathbf N}(T^aT^b)=\delta^{ab}/2$. See Haber, equations (1), (3)–(5), and (45). The factors of $1/2$ below are part of this convention.

Dimensions, Casimirs, and trace indices

The three principal formulas are

$$\boxed{\begin{aligned} \dim V_\lambda &=\prod_{\beta^\vee\in\Phi_+^\vee} \frac{\langle\lambda+\rho,\beta^\vee\rangle} {\langle\rho,\beta^\vee\rangle},\\ C_2(\lambda) &=\frac12(\lambda,\lambda+2\rho) =\frac12 a^{\mathsf T}M(a+2\mathbf1),\\ T(V_\lambda) &=\frac{\dim V_\lambda}{\dim\mathfrak g}\,C_2(\lambda). \end{aligned}}$$

Here $\mathbf1$ is the column of $r$ ones; $\dim\mathfrak g$ is the complex dimension of the algebra, also the real dimension of its compact form. The dimension formula is Etingof, §28.1, Proposition 28.1. If $\beta^\vee=\sum_i m_i\alpha_i^\vee$, its Weyl factor is simply

$$\frac{\sum_i m_i(a_i+1)}{\sum_i m_i}.$$

Use coefficients of positive coroots in this expression. Replacing them with coefficients of positive roots produces wrong dimensions in non-simply-laced examples.

The Casimir follows by applying the central quadratic element to a highest-weight vector: positive-root operators annihilate it, leaving the Cartan term $(\lambda,\lambda)$ and the commutator contribution $2(\lambda,\rho)$. Etingof’s §25.4, Lemma 25.6, gives that eigenvalue without the factor $1/2$; the site uses half that quadratic operator to match the generator normalization above.

Finally, tracing $\sum_a T_R^aT_R^a=C_2(R)I_R$ gives $T(R)\dim\mathfrak g=C_2(R)\dim R$. This derives the trace-index formula without an independent table convention. Rescaling every generator by $s$ rescales $C_2$ and $T$ by $s^2$. In particular, a table whose defining $SU(N)$ index is $1$ has quadratic values twice those used here. The word “index” alone does not fix this scale.

Useful checks are

$$\begin{aligned} T(R\oplus S)&=T(R)+T(S),\\ T(R\otimes S)&=\dim S\,T(R)+\dim R\,T(S),\\ C_2(R^*)&=C_2(R),\qquad T(R^*)=T(R). \end{aligned}$$

For the tensor identity, $T^a_{R\otimes S}=T_R^a\otimes I+I\otimes T_S^a$; the mixed traces vanish because generators of a simple algebra have zero trace. For a reducible representation, the quadratic operator need not be one scalar on all summands, although its trace index is still defined.

SU(N): fundamental and adjoint checks

For type $A_{N-1}$, $N\ge2$, order the nodes so $\omega_i$ labels $\Lambda^i\mathbb C^N$. In the usual coordinate plane with $\sum x_i=0$,

$$(\omega_1,\omega_1)=\frac{N-1}{N},\qquad (\omega_1,\rho)=\frac{N-1}{2}.$$

Substitution gives $C_2(\mathbf N)=(N^2-1)/(2N)$. The Weyl factors for the roots involving the first coordinate telescope to $\prod_{j=2}^N j/(j-1)=N$; the other factors are one. For the adjoint, the highest root is $\theta=e_1-e_N$, with $(\theta,\theta)=2$ and $(\theta,\rho)=N-1$, giving $C_2(\mathrm{adj})=N$. Its dimension counts traceless $N\times N$ matrices.

Representation	Highest weight	Dimension	$C_2$	$T$
Defining	$\omega_1$	$N$	$(N^2-1)/(2N)$	$1/2$
Adjoint	$\omega_1+\omega_{N-1}$	$N^2-1$	$N$	$N$

For $N=2$, the two weights in the adjoint expression coincide, so its label is $2\omega_1$. The matrix normalization check for $N=2$ is immediate from $T^a=\sigma^a/2$: each square is $I/4$, and their sum is $3I/4$. The general defining and adjoint values agree with Haber, equations (5) and (45).

For $SU(3)$, $a=(p,q)$ and

$$M=\frac13\begin{pmatrix}2&1\\1&2\end{pmatrix}, \qquad \dim(p,q)=\frac{(p+1)(q+1)(p+q+2)}2, \qquad C_2(p,q)=\frac{p^2+pq+q^2+3p+3q}{3}.$$

Labels	Dimension	$C_2$	$T$
$(1,0)$	$3$	$4/3$	$1/2$
$(1,1)$	$8$	$3$	$3$
$(3,0)$	$10$	$6$	$15/2$

As an independent trace check, $\mathbf3\otimes\overline{\mathbf3}=\mathbf1\oplus\mathbf8$ gives $T=3(1/2)+3(1/2)=3$, precisely the adjoint index. Similarly, for $SU(2)$, $\mathbf2\otimes\mathbf2=\mathbf1\oplus\mathbf3$ gives $T=2$. These decompositions follow from endomorphisms splitting into scalar and traceless parts, with self-duality of the $SU(2)$ doublet.

Compute the SU(3) adjoint or check its tensor origin.

A non-simply-laced check: compact Sp(2)

Use $C_2$ node order, with the second simple root long. The compact group is $Sp(2)=USp(4)\cong\operatorname{Spin}(5)$; it is not the split group $Sp(4,\mathbb R)$. Choose orthogonal $e_i$ with $(e_i,e_i)=1/2$, simple roots $e_1-e_2,2e_2$, and fundamental weights $\omega_1=e_1$, $\omega_2=e_1+e_2$. This gives

$$A=\begin{pmatrix}2&-2\\-1&2\end{pmatrix},\quad D=\begin{pmatrix}1/2&0\\0&1\end{pmatrix},\quad M=\begin{pmatrix}1/2&1/2\\1/2&1\end{pmatrix}.$$

The node order and coordinate realization are those of the SageMath type-C reference, simple_root and fundamental_weight, with the displayed metric imposing squared long-root length two. The positive-coroot coefficients are $(1,0),(0,1),(1,1),(1,2)$. Therefore

$$\begin{aligned} \dim(a,b)&=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6,\\ C_2(a,b)&=\frac{a^2+2ab+2b^2+4a+6b}{4},\\ T(a,b)&=\frac{\dim(a,b)\,C_2(a,b)}{10}. \end{aligned}$$

For the defining $\mathbf4=(1,0)$, the Weyl factors are $2,1,3/2,4/3$, giving $4$. The metric gives $C_2=5/4$ and hence $T=1/2$. For the adjoint $(2,0)$, the factors $3,1,2,5/3$ give dimension $10$, while $C_2=T=3=h^\vee$. A third check is $(0,1)$, the primitive part of $\Lambda^2\mathbf4$: its dimension is $\binom42-1=5$, its Casimir is $2$, and its index is $1$.

The same compact group written in $B_2$ conventions exchanges the labels: $(1,0)_{C_2}$ is $(0,1)_{B_2}$. Its Casimir and index agree after this relabeling because both metrics use squared long-root length two. The roots lesson shows the coroot calculation and the Sp(2) record fixes the global group.

Calculate the defining four.

Dual weights and global descent

If $w_0$ is the longest Weyl-group element, the dual highest weight is

$$\lambda^*=-w_0\lambda.$$

Indeed, dualizing negates all weights, and $w_0\lambda$ is the lowest weight. For $SU(N)$ this reverses the Dynkin labels; for $SU(3)$, $(p,q)^*=(q,p)$. Equality with the dual is a separate question from its reality type: the self-dual $SU(2)$ doublet is quaternionic, whereas its adjoint is real. Neither equality of dimensions nor equality of quadratic invariants proves equivalence of irreps.

Let $P$ and $Q$ be the weight and root lattices. The center of $\widetilde G$ is dual to $P/Q$, and the central character of $V_\lambda$ is the class $[\lambda]$. Conrad, Centers of Simply Connected Semisimple Groups, pp. 1–4. In the chosen matrix convention,

$$\lambda\in Q\iff A^{-1}a\in\mathbb Z^r \iff V_\lambda\text{ descends to }\widetilde G/Z(\widetilde G).$$

The order of this central character is the least common multiple of the reduced denominators of the entries of $A^{-1}a$. For an intermediate quotient by $K$, test the actual character on every generator of $K$; knowing only its kernel’s order does not identify its embedding.

For $SU(N)$, choose $z=e^{2\pi i/N}I$ and set

$$t=\sum_{i=1}^{N-1}i a_i\pmod N.$$

Then $\rho_\lambda(z)=e^{2\pi i t/N}I$. For $k\mid N$, the embedded subgroup $K=\langle z^{N/k}\rangle$ acts trivially exactly when $t=0\pmod k$. The defining multiplet has $t=1$; the adjoint has $t=0$. For $SU(3)$, $(3,0)$ also has $t=0$, so this complex ten descends to $PSU(3)$ despite being inequivalent to its dual $(0,3)$.

For compact $Sp(2)$ in the $C_2$ labels above, $-I$ acts by $(-1)^a$. The defining four has odd $a$ and fails to descend to $SO(5)$; both the five $(0,1)$ and adjoint $(2,0)$ descend. This follows by realizing $\omega_1$ in the defining representation and $\omega_2$ in its exterior square. Compare the covering map and the global-form lesson.

Scope of a numerical result

The current calculator accepts $A_1$–$A_8$, $B_2$–$B_8$, $C_2$–$C_8$, $D_4$–$D_8$, $E_6,E_7,E_8,F_4,G_2$, and Dynkin labels $0$–$30$. These are implementation bounds. Low-rank coincidences must be translated into the stated node order; the simple-type classification identifies them.

A product of simple groups has a separate quadratic index for each factor, with spectator dimensions multiplying the trace. An abelian factor needs a chosen charge normalization; it has no simple-root Weyl formula of this kind. Quadratic indices also do not test cubic gauge anomalies or count independent interaction operators. Use the relevant matter and operator workflows for those additional questions.

Keep the highest weight, global group, and convention profile with the exported exact values. A dimension, Casimir, and center character do not supply representation matrices or normalized contraction tensors.