Root Systems and Cartan Classification

Continuous symmetry looks infinitely varied — every rotation, phase, and gauge transformation seems to demand its own bespoke algebra. Cartan's classification is the discovery that it is not varied at all. Take any finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$ , isolate a maximal set of commuting generators — the Cartan subalgebra $\mathfrak{h}$ — and diagonalize the adjoint action. The algebra shatters into $\mathfrak{h}$ plus a family of one-dimensional ladder spaces, each tagged by a vector $\alpha\in\mathfrak{h}^*$ called a root. That finite constellation, floating in a Euclidean space whose metric is the Killing form, obeys axioms so rigid it reconstructs $\mathfrak{g}$ completely. The rigidity is almost unreasonable. Roots come in $\pm$ pairs; reflection perpendicular to any root permutes the whole set; and the only angles two roots may subtend are $90^\circ, 120^\circ, 135^\circ, 150^\circ$ . Split the roots into positive and negative with a hyperplane and a minimal generating set of simple roots appears, whose geometry compresses into a small matrix of integers — the Cartan matrix — drawn as a Dynkin diagram of nodes and bonds. Killing, Cartan, Coxeter, and Dynkin proved that the connected admissible diagrams form exactly four infinite families $A_n, B_n, C_n, D_n$ and precisely five exceptions $G_2, F_4, E_6, E_7, E_8$ — the entire universe of simple symmetry, on one page. For physics this is infrastructure, not decoration: it fixes the compact gauge groups, the $\mathfrak{su}(3)$ of colour and flavour, the grand-unified $\mathfrak{su}(5)$ and $\mathfrak{so}(10)$ , the $E_8\times E_8$ of the heterotic string, and the highest-weight machine that turns a diagram into a particle multiplet.

Key takeaways

A semisimple Lie algebra is reconstructed from its root system $\Delta$ : the nonzero weights of the adjoint action of a maximal abelian (Cartan) subalgebra $\mathfrak{h}$ , with $\dim\mathfrak{g}=\mathrm{rank}+|\Delta|$ .
Choosing positive roots singles out a basis of simple roots, whose angles and length ratios are packed into the integer Cartan matrix $A_{ij}=2\langle\alpha_i,\alpha_j\rangle/\langle\alpha_j,\alpha_j\rangle$ .
The one hard fact, integrality of $\langle\beta,\alpha^\vee\rangle$ , comes from an embedded $\mathfrak{su}(2)$ ladder and forces $A_{ij}A_{ji}=4\cos^2\theta\in\{0,1,2,3\}$ — only four angles survive.
Cartan matrices are drawn as Dynkin diagrams; positive-definiteness forbids loops and multiple branches, cutting the admissible list to $A_n, B_n, C_n, D_n$ plus $G_2, F_4, E_6, E_7, E_8$ .
The same root data run representation theory: highest weights $\lambda=\sum n_i\omega_i$ , the Weyl group of reflections, and the weight diagrams that are physical multiplets.
Key idea: encode a semisimple Lie algebra in the finite geometry of its roots, compress that geometry into a small integer matrix and its diagram, and enumerate every admissible diagram — that finite list is every simple symmetry there is.

The core insight

Encode a semisimple Lie algebra in the finite geometry of its roots; reduce that geometry to a small integer matrix — the Cartan matrix — and its Dynkin diagram, then enumerate all admissible diagrams. The four families plus five exceptionals are the complete list of simple complex Lie algebras.

Where this lives today

The most striking modern appearance of a root system is not a string-theory blackboard but a laboratory magnet. In 2010 Coldea and collaborators cooled the quasi-one-dimensional Ising ferromagnet cobalt niobate ( $\mathrm{CoNb_2O_6}$ ) to its quantum critical point in a transverse field and read out its quasiparticle spectrum with neutron scattering. Zamolodchikov had predicted in 1989 that the critical Ising chain in a longitudinal field is governed by an integrable theory whose eight massive particles carry the hidden symmetry of $E_8$ — and that the two lightest masses must stand in the ratio $m_2/m_1=2\cos(\pi/5)=1.618\ldots$ , the golden ratio, a number built directly from the $E_8$ root geometry. The experiment found the first two peaks in precisely that ratio: a rank-8 root system with 240 roots, discovered by Killing while classifying abstract symmetry, sets the mass spectrum of a real solid. Root systems are also the daily grammar of high-energy physics. Grand-unified model builders sift the Dynkin diagrams for simple groups containing $SU(3)\times SU(2)\times U(1)$ as a sub-diagram — deleting a node of $A_4=\mathfrak{su}(5)$ or $D_5=\mathfrak{so}(10)$ yields the Standard Model, with branching rules read straight off the sub-diagrams. The two anomaly-free heterotic strings carry gauge algebra $E_8\times E_8$ or $\mathfrak{so}(32)$ , chosen for the root-lattice reasons the classification supplies. And the $ADE$ half of the list reappears far from Lie theory — in the McKay correspondence, in simple surface singularities, and in the quiver gauge theories of condensed matter.

The physical picture

Think of the Cartan subalgebra as a rack of mutually commuting "measuring rods" — the analogue of $J_z$ in angular momentum, or the pair $I_3$ (isospin) and $Y$ (hypercharge) in $\mathfrak{su}(3)$ . Being commuting, they diagonalize at once, and every remaining generator becomes a ladder operator that shifts those simultaneous eigenvalues by a fixed vector — the root. The algebra is a lattice of ladder operators hanging off the Cartan torus, the roots the arrows between rungs. The first figure is the smallest interesting case, $\mathfrak{su}(3)=A_2$ : six roots on a regular hexagon in the $(I_3, Y)$ plane, two singled out as simple roots $\alpha_1, \alpha_2$ at $120^\circ$ , every other generator reached by adding them. The hexagonal symmetry is the Weyl group — reflections perpendicular to each root — the same symmetry decorating the meson and baryon multiplets of the eightfold way.

Figure

The su(3) = A₂ root system: six roots on a regular hexagon in the isospin–hypercharge plane, with simple roots α₁ and α₂ at 120° and equal length, all others reached by adding them — so dim g = rank 2 + 6 roots = 8

Each opposite pair $\pm\alpha$ , with the coroot $\alpha^\vee$ , closes into a private copy of $\mathfrak{su}(2)$ . This is the secret engine of the classification: $\mathfrak{su}(2)$ 's rigidly integer-spaced ladders are imported wholesale, forcing the Cartan matrix entries to be integers — and integers admit only finitely many geometric solutions. Everything downstream — the four angles, the ban on loops, the five exceptionals — is $\mathfrak{su}(2)$ ladder theory iterated over every pair of roots. The second figure is the destination: the complete gallery of connected Dynkin diagrams. A node is a simple root; a missing line means two commuting $\mathfrak{su}(2)$ 's; single, double, and triple bonds record ever-tighter angles, arrows pointing from long root to short. Reading this chart is reading the list of every simple symmetry the universe permits.

Figure

The complete Cartan classification as connected Dynkin diagrams: four infinite families A B C D and five exceptionals G₂ F₄ E₆ E₇ E₈, node = simple root, bond thickness = angle, arrow long→short — e.g. E₈ has rank 8 and 240 roots so dim = 8 + 240 = 248

The core equations

The starting move is to diagonalize the algebra by its own commuting core. The operators $\mathrm{ad}_H$ for $H\in\mathfrak{h}$ commute and are simultaneously diagonalizable, splitting $\mathfrak{g}$ into joint eigenspaces whose nonzero eigenvalue-functionals are the roots — replacing the algebra by the finite set $\Delta$ of arrows in $\mathfrak{h}^*$ .

\mathfrak{g}=\mathfrak{h}\ \oplus\ \bigoplus_{\alpha\in\Delta}\mathfrak{g}_\alpha,\qquad [H,E_\alpha]=\alpha(H)\,E_\alpha\ \ \forall H\in\mathfrak{h} \qquad (1)

where: $\mathfrak{h}$ — the Cartan subalgebra, maximal abelian and ad-diagonalizable; $\Delta\subset\mathfrak{h}^*$ — the roots, nonzero weights of the adjoint action; $\mathfrak{g}_\alpha$ — the one-dimensional root space, spanned by $E_\alpha$ ; $\alpha(H)$ — the root evaluated on $H$ . Reading it: $H$ scales $E_\alpha$ by $\alpha(H)$ , so $E_\alpha$ shifts every Cartan eigenvalue by the fixed vector $\alpha$ — a raising operator. The one genuinely hard fact of the subject is that a certain projection of one root onto another is always an integer — it descends from the $\mathfrak{su}(2)$ hidden inside each root pair (Derivation A) and seeds every angle and length constraint.

\langle\beta,\alpha^\vee\rangle=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\in\mathbb{Z},\qquad \alpha^\vee=\frac{2\alpha}{\langle\alpha,\alpha\rangle} \qquad (2)

where: $\langle\cdot,\cdot\rangle$ — the inner product on $\mathfrak{h}^*$ inherited from the Killing form; $\alpha^\vee$ — the coroot dual to $\alpha$ ; $\mathbb{Z}$ — the projection is always an integer. Reading it: the angles between roots may be irrational, but the asymmetric pairing of a root with a coroot is quantized — and quantization is what makes the geometry finite. Restricting equation (2) to the simple roots packages all their geometry into one integer matrix: diagonal fixed at $2$ , off-diagonal non-positive, and — crucially — generally not symmetric, only symmetrizable, since roots may differ in length.

A_{ij}=\langle\alpha_i,\alpha_j^\vee\rangle=\frac{2\langle\alpha_i,\alpha_j\rangle}{\langle\alpha_j,\alpha_j\rangle},\qquad A_{ii}=2,\qquad A_{ij}\le 0\ (i\ne j) \qquad (3)

where: $\alpha_i$ — the simple roots, $i=1,\dots,n$ with $n=\mathrm{rank}$ ; $A_{ij}$ — the Cartan integer, encoding both the angle and the length ratio of $\alpha_i$ and $\alpha_j$ . The whole algebra is recoverable from this $n\times n$ table of integers — the most economical fingerprint of a simple symmetry. Multiplying the two off-diagonal partners cancels the length information, leaving the angle alone; the product of two non-positive integers, bounded by Cauchy–Schwarz, leaves only four values.

A_{ij}A_{ji}=4\cos^2\theta_{ij}\in\{0,1,2,3\} \qquad (4)

where: $\theta_{ij}$ — the angle between the simple roots $\alpha_i$ and $\alpha_j$ . Reading it: $0\Rightarrow90^\circ$ (no bond), $1\Rightarrow120^\circ$ (single, equal length), $2\Rightarrow135^\circ$ (double, ratio $\sqrt2$ ), $3\Rightarrow150^\circ$ (triple, ratio $\sqrt3$ ); $4$ is excluded as it makes the roots parallel. Each root defines a reflection of the ambient space, and these reflections map the root system to itself. The finite group they generate is the Weyl group, the symmetry group of every weight diagram.

s_\alpha(\beta)=\beta-\langle\beta,\alpha^\vee\rangle\,\alpha,\qquad W=\langle s_{\alpha_1},\dots,s_{\alpha_n}\rangle,\qquad s_\alpha:\Delta\to\Delta \qquad (5)

where: $s_\alpha$ — the orthogonal reflection in the hyperplane perpendicular to $\alpha$ ; $W$ — the Weyl group, generated by the simple reflections. Since $\langle\beta,\alpha^\vee\rangle$ is an integer, $s_\alpha$ moves roots by integer multiples of $\alpha$ and never leaves $\Delta$ — the symmetry that hexagons the $\mathfrak{su}(3)$ multiplets. The unbroken chain of roots obtained by stepping $\beta$ along $\alpha$ is symmetric under $s_\alpha$ , and its imbalance is exactly the Cartan integer — the identity behind the integrality proof.

\beta+k\alpha\in\Delta\cup\{0\}\ \text{for}\ -p\le k\le q,\qquad p-q=\langle\beta,\alpha^\vee\rangle \qquad (6)

where: $p, q\ge0$ — how far the string extends below and above $\beta$ in the $\alpha$ -direction. The string is unbroken and $s_\alpha$ reverses it end for end, so its asymmetry $p-q$ equals the integer $\langle\beta,\alpha^\vee\rangle$ . Representations run the same data in reverse: a finite-dimensional irreducible is fixed by one highest weight, a non-negative-integer combination of the fundamental weights dual to the coroots.

\lambda=\sum_i n_i\,\omega_i,\quad n_i=\langle\lambda,\alpha_i^\vee\rangle\in\mathbb{Z}_{\ge0},\qquad \langle\omega_i,\alpha_j^\vee\rangle=\delta_{ij} \qquad (7)

where: $\omega_i$ — the fundamental weights, the dual basis to the coroots; $n_i$ — the Dynkin labels of $\lambda$ . Attach non-negative integers to the diagram's nodes and you have named an irreducible representation; the Cartan matrix then unfolds it into a full weight diagram. Finally, one master count reconciles the sizes: the algebra is its rank plus its roots, and the root total factors through the Coxeter number.

\dim\mathfrak{g}=n+|\Delta|,\qquad |\Delta|=n\,h,\qquad n=\mathrm{rank}=\dim\mathfrak{h} \qquad (8)

where: $|\Delta|$ — the number of roots, equal to the number of ladder operators; $h$ — the Coxeter number; $n$ — the rank. For $E_8$ , $n=8$ and $h=30$ , so $|\Delta|=240$ and $\dim=8+240=248$ — the largest exceptional algebra, on the back of an envelope.

Derivation AQuantization of the Cartan integers from an embedded su(2)

The goal is to prove the single fact on which the whole classification balances: for any two roots $\alpha,\beta$ , the projection $\langle\beta,\alpha^\vee\rangle$ is an integer. Everything else — the four angles, the acyclic diagrams, the finite list — is a corollary. The strategy: each root pair $\pm\alpha$ generates a copy of $\mathfrak{sl}(2,\mathbb{C})$ ; let it act on the chain of root spaces through $\beta$ and import the rigidly integer weight-spectrum of finite-dimensional $\mathfrak{sl}(2)$ modules.

Step 1. Take $\alpha,\beta\in\Delta$ with $\beta\ne\pm\alpha$ and set $e=E_\alpha$ , $f=E_{-\alpha}$ , $h=H_\alpha$ with $[h,e]=2e$ , $[h,f]=-2f$ , $[e,f]=h$ — the standard $\mathfrak{sl}(2)$ , $h$ acting as the coroot $\alpha^\vee$ . Why: the bracket-grading $[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]\subseteq\mathfrak{h}$ makes these close, so $\mathfrak{sl}(2)$ theory becomes available inside any semisimple $\mathfrak{g}$ .

Step 2. Collect the root spaces along the $\alpha$ -string through $\beta$ into $V=\bigoplus_{k=-p}^{q}\mathfrak{g}_{\beta+k\alpha}$ ; $e$ raises $k$ and $f$ lowers it, so $V$ is a finite-dimensional $\mathfrak{sl}(2)$ -module. Why: a raising and a lowering operator on an unbroken ladder is exactly what $\mathfrak{sl}(2)$ theory constrains.

Step 3. The $h$ -eigenvalue on rung $\mathfrak{g}_{\beta+k\alpha}$ is $\langle\beta,\alpha^\vee\rangle+2k$ , since $\langle\alpha,\alpha^\vee\rangle=2$ : consecutive rungs carry $h$ -weights spaced by exactly $2$ . Why: spacing $2$ is the fingerprint of an $\mathfrak{sl}(2)$ ladder, locking the string into one irreducible tower.

Step 4. Finite-dimensional $\mathfrak{sl}(2)$ weights are symmetric about zero, so top $=-$ bottom: $\langle\beta,\alpha^\vee\rangle+2q=-\big(\langle\beta,\alpha^\vee\rangle-2p\big)$ . Why: this is just $J_z$ running from $-j$ to $+j$ in any spin- $j$ multiplet, pinning the two ends of the string together.

Step 5. Solve the symmetry condition to get $\langle\beta,\alpha^\vee\rangle=p-q$ . Since $p$ and $q$ are non-negative integers, the projection is an integer — the boxed result. Why: this integer, differenced from two ladder lengths, is the whole arithmetic backbone of the classification.

\boxed{\ \langle\beta,\alpha^\vee\rangle=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=p-q\in\mathbb{Z}\ }

Note that the string relation $p-q=\langle\beta,\alpha^\vee\rangle$ is exactly the Weyl-reflection reversal $s_\alpha(\beta)=\beta-(p-q)\alpha$ , confirming that the $\mathfrak{sl}(2)$ ladder structure and the integer Cartan matrices are two faces of the same fact.

Go deeper · Derivation B · 1 minEnumerating the four rank-2 root systems

Derivation BEnumerating the four rank-2 root systems

The goal is to show, by direct enumeration, that a rank-2 simple Lie algebra takes only four shapes — $A_1\times A_1$ , $A_2$ , $B_2=C_2$ , $G_2$ — watching the integrality of Derivation A collapse a seemingly continuous family into a finite list. Two simple roots $\alpha_1,\alpha_2$ are constrained only by integrality of the off-diagonal Cartan entries and the bound $\cos^2\theta\le1$ ; we run through the allowed products $A_{12}A_{21}=4\cos^2\theta$ and read off the algebra.

Step 1. If $A_{12}A_{21}=0$ then $\theta=90^\circ$ and the Cartan matrix is diagonal, $A=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ . The two roots are orthogonal, so the algebra splits into $A_1\times A_1=\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ — two disconnected nodes, not simple. Why: orthogonal simple roots mean the two $\mathfrak{su}(2)$ 's commute — a disconnected diagram is a direct sum.

Step 2. If $A_{12}A_{21}=1$ then $\theta=120^\circ$ , the roots have equal length, and $A=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ . This is $A_2=\mathfrak{su}(3)$ : six roots forming the regular hexagon of the first figure, $\dim=2+6=8$ . Why: equal-length roots at $120^\circ$ tile the plane hexagonally — the eightfold-way geometry.

Step 3. If $A_{12}A_{21}=2$ then $\theta=135^\circ$ , the length-squared ratio is $2$ , and $A=\begin{pmatrix}2&-1\\-2&2\end{pmatrix}$ . This is $B_2\cong C_2=\mathfrak{so}(5)\cong\mathfrak{sp}(4)$ : eight roots, $\dim=2+8=10$ . Why: a double bond encodes one root $\sqrt2$ longer — the orthogonal/symplectic asymmetry in rank 2.

Step 4. If $A_{12}A_{21}=3$ then $\theta=150^\circ$ , the length-squared ratio is $3$ , and $A=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$ . This is $G_2$ : twelve roots, $\dim=2+12=14$ , the smallest exceptional algebra. Why: the triple bond is the most lopsided angle integrality allows, and it occurs exactly once.

Step 5. If $A_{12}A_{21}=4$ then $\cos^2\theta=1$ , forcing $\alpha_1\parallel\alpha_2$ , which is impossible for distinct simple roots. The list therefore terminates at four. Why: parallel simple roots would be equal up to sign, contradicting independence — no fifth system exists.

\boxed{\ A_{12}A_{21}=4\cos^2\theta\in\{0,1,2,3\}\ \Rightarrow\ A_1{\times}A_1,\ A_2,\ B_2{=}C_2,\ G_2\ }

The four Cartan matrices and their root counts — $4, 6, 8, 12$ — give dimensions $6+2=8$ , $8$ , $10$ , $14$ for $A_1{\times}A_1$ , $\mathfrak{su}(3)$ , $\mathfrak{so}(5)$ , and $G_2$ respectively, as expected from the master count $\dim\mathfrak{g}=n+|\Delta|$ .

Go deeper · Derivation C · 1 minWhy Dynkin diagrams have no loops and branch at most once

Derivation CWhy Dynkin diagrams have no loops and branch at most once

The goal is to see how positive-definiteness — the mere fact that the simple roots are linearly independent — eliminates all but the finite list. The device: normalize each simple root to a unit vector $\epsilon_i=\alpha_i/|\alpha_i|$ and study the Gram matrix $\langle\epsilon_i,\epsilon_j\rangle$ . Because the simple roots form a basis, this matrix is positive-definite, and positivity is a surprisingly sharp bound on how many bonds a connected diagram may carry.

Step 1. Linear independence of $\{\alpha_i\}$ means $\sum_i c_i\alpha_i=0$ forces all $c_i=0$ , so any nonzero combination has strictly positive norm and the Gram matrix of the $\epsilon_i$ is positive-definite. Why: positive-definiteness is the one global fact surviving from "the roots are a basis," and every inequality is squeezed from it.

Step 2. Two nodes joined by at least a single bond have angle $\ge120^\circ$ , so $\langle\epsilon_i,\epsilon_j\rangle=\cos\theta_{ij}\le-\tfrac12$ ; unlinked nodes are orthogonal. Why: this turns a diagram's bond count into a numerical bound on overlaps — the bridge from combinatorics to geometry.

Step 3. Suppose a cycle $C$ of single-bonded nodes existed. Then $0<\big\|\sum_{i\in C}\epsilon_i\big\|^2=|C|+2\sum_{i<j\in C}\langle\epsilon_i,\epsilon_j\rangle\le|C|+2\cdot|C|\cdot(-\tfrac12)=0$ , a contradiction. So Dynkin diagrams contain no loops. Why: a cycle supplies $|C|$ linked pairs, each pulling the norm down by $1$ — just enough to cancel the diagonal and violate positivity.

Step 4. A similar estimate at a single node bounds its incident bond strength, $\sum_{j\text{ linked to }i}\langle\epsilon_i,\epsilon_j\rangle^2<1$ , permitting at most three single bonds (or one triple, giving $G_2$ ) and one branch point per connected diagram. Why: this outlaws a fourth arm at a vertex, restricting branch nodes to the $D$ and $E$ shapes.

Step 5. Solving the remaining finite set of positive-definite diagram constraints yields exactly $A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4, G_2$ . Why: with loops forbidden and branching limited, only these shapes stay positive-definite — the classification theorem.

\boxed{\ \Big\|\sum_i \epsilon_i\Big\|^2>0\ \Rightarrow\ \text{no loops, one branch}\ \Rightarrow\ A_n,B_n,C_n,D_n,E_6,E_7,E_8,F_4,G_2\ }

Limiting cases

The rank-1 and -2 systems make the machinery concrete; the classical families show how it scales.

Case	Result	Meaning
Rank 1, $A_1=\mathfrak{su}(2)$	$\Delta=\{\pm\alpha\},\ A=(2)$	one root pair — the ladder $J_\pm$ and one Cartan $J_z$ ; the atom of the whole subject
Rank 2, $A_2=\mathfrak{su}(3)$	angle $120^\circ$ , $\|\alpha_1\|=\|\alpha_2\|$ , $\dim 8$	six roots in a regular hexagon — colour and flavour $SU(3)$
Rank 2, $B_2=C_2$	angle $135^\circ$ , ratio $\sqrt2$ , $\dim 10$	eight roots; $\mathfrak{so}(5)\cong\mathfrak{sp}(4)$ , the first double bond
Rank 2, $G_2$	angle $150^\circ$ , ratio $\sqrt3$ , $\dim 14$	twelve roots, triple bond — the smallest and most lopsided exceptional
Rank $n$ , $A_n=\mathfrak{su}(n{+}1)$	$A_{ij}=2\delta_{ij}-\delta_{i,j+1}-\delta_{i,j-1}$	a straight chain of $n$ equal-length roots each $120^\circ$ to its neighbour
Rank 8, $E_8$	$\|\Delta\|=8\times30=240$ , $\dim 248$	the maximal exceptional; its root lattice is the densest packing in eight dimensions

Common trapthe misconception

The most common error is drawing the Cartan matrix as symmetric. It is symmetric only when all simple roots share a length ( $A,D,E$ , the simply-laced cases); with a double or triple bond ( $B,C,F,G$ ) it is merely symmetrizable, and $A_{ij}\ne A_{ji}$ records which root is longer — confusing the two swaps long and short and turns $B_n$ into $C_n$ . A second slip places roots in $\mathfrak{h}$ rather than the dual $\mathfrak{h}^*$ : roots are linear functionals whose inner product is imported from the Killing form via $\mathfrak{h}\cong\mathfrak{h}^*$ . Finally, roots are the weights of just one representation — the adjoint — so not every weight is a root; forgetting this conflates the root lattice with the finer weight lattice of general multiplets.

Worked examples

1 · Building su(3) = A₂ from scratch (confidence builder)

Start from two equal-length simple roots at $120^\circ$ . Construct the Cartan matrix, count the roots, get the dimension, and use the Cartan data to size the smallest irreducible representations.

Step 1. With $|\alpha_1|=|\alpha_2|$ and $\theta=120^\circ$ , the off-diagonal entries are $A_{12}=A_{21}=2\cos120^\circ=2(-\tfrac12)=-1$ , so $A=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ . The product $A_{12}A_{21}=1$ sits in the allowed set $\{0,1,2,3\}$ , confirming a legal single-bond diagram.

Step 2. The positive roots are $\alpha_1$ , $\alpha_2$ , and $\alpha_1+\alpha_2$ (allowed because $\langle\alpha_1,\alpha_2^\vee\rangle=-1$ gives a string of length one above $\alpha_1$ ); with negatives, $|\Delta|=6$ . Hence $\dim\mathfrak{su}(3)=n+|\Delta|=2+6=\mathbf{8}$ — the eight Gell-Mann generators.

Step 3. Irreducible representations are labelled by Dynkin labels $(n_1,n_2)$ , and for $\mathfrak{su}(3)$ the Weyl dimension formula collapses to $d(n_1,n_2)=\tfrac12(n_1+1)(n_2+1)(n_1+n_2+2)$ . The fundamental $(1,0)$ gives $d=\tfrac12(2)(1)(3)=\mathbf{3}$ — the quark triplet.

Step 4. The adjoint $(1,1)$ gives $d=\tfrac12(2)(2)(4)=\mathbf{8}$ , matching $\dim\mathfrak{g}$ as it must, since the adjoint acts on the algebra itself. The symmetric $(3,0)$ gives $d=\tfrac12(4)(1)(5)=\mathbf{10}$ — the baryon decuplet.

Result. From two arrows at $120^\circ$ : Cartan matrix $\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ , 6 roots, dim 8, and the multiplets 3, 8, 10 that organize the eightfold way.

2 · Reading an algebra off its Cartan matrix

A rank-2 algebra has Cartan matrix $A=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$ . Identify it, and find the angle, length ratio, root count, and dimension.

Step 1. Form the diagnostic product $A_{12}A_{21}=(-1)(-3)=3$ . Then $4\cos^2\theta=3$ , so $\cos^2\theta=\tfrac34$ and $\theta=\arccos(-\tfrac{\sqrt3}{2})=\mathbf{150^\circ}$ — a triple bond.

Step 2. The asymmetry $A_{12}=-1$ versus $A_{21}=-3$ fixes the length ratio: $A_{12}/A_{21}=\langle\alpha_2,\alpha_2\rangle/\langle\alpha_1,\alpha_1\rangle$ , giving $|\alpha_1|^2/|\alpha_2|^2=3$ , so one root is $\sqrt3$ times longer than the other.

Step 3. The unique rank-2 system with a triple bond is $G_2$ , the smallest exceptional, with $|\Delta|=\mathbf{12}$ roots (six long, six short).

Step 4. The dimension follows from the master count: $\dim G_2=n+|\Delta|=2+12=\mathbf{14}$ , consistent with $|\Delta|=n\,h=2\times6=12$ using the Coxeter number $h=6$ .

Result. The matrix is $G_2$ : angle $150^\circ$ , length ratio $\sqrt3$ , 12 roots, dimension 14 — every feature extracted from four integers.

Physics of the portrait

The classification carries three names welded together. Wilhelm Killing found the roots and the list in a burst of papers he himself distrusted; Élie Cartan, in a doctoral thesis of legendary precision, corrected the errors, proved the exceptional cases genuinely exist, and made the whole edifice rigorous. It is Cartan's name the classification bears.

Origins · 1 minÉlie Cartan and how this was discovered

ÉC

Élie Cartan

1869–1951 · structure theory of Lie groups and modern differential geometry

Élie Cartan was the son of a village blacksmith in the French Alps, lifted out of poverty by a state scholarship spotted by a travelling inspector. His 1894 doctoral thesis, Sur la structure des groupes de transformations finis et continus, took Wilhelm Killing's brilliant but error-strewn 1888–1890 classification and rebuilt it on solid ground — supplying the correct definition of the Cartan subalgebra, proving that the four families and five exceptional algebras really are the complete list, and showing that the mysterious $G_2, F_4, E_6, E_7, E_8$ genuinely exist rather than being mere combinatorial possibilities. Cartan went on to reshape much of modern mathematics: the exterior calculus of differential forms, the method of moving frames underlying modern differential geometry, the theory of symmetric spaces, and — in 1913, purely from the representation theory of the orthogonal algebras — the spinors that Dirac would later need for the electron. His work on connections produced Einstein–Cartan gravity with torsion. Recognition came late, cemented when his son Henri became a leading light of the Bourbaki generation. The classification that opens every course in Lie theory is his monument: a proof that continuous symmetry, for all its variety, is finite and knowable.

History

1888

— Killing classifies the simple Lie algebras and introduces roots and the characteristic equation

1894

— Cartan's thesis corrects and completes the classification and proves the exceptional algebras exist

1925

— Weyl proves complete reducibility and derives the character and dimension formulas

1934

— Coxeter classifies the finite reflection groups underlying the Weyl groups

1947

— Dynkin introduces the labelled diagrams that now bear his name

Going further

Texts. Humphreys, Introduction to Lie Algebras and Representation Theory (the standard proof of the classification, §10–11) · Fulton & Harris, Representation Theory: A First Course (root systems worked example by example) · Georgi, Lie Algebras in Particle Physics, 2nd ed. (the physicist's operational treatment) · Stone & Goldbart, Mathematics for Physics, ch. on Lie groups and algebras · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed. · Hassani, Mathematical Physics, chapters on group theory. Key results to keep in mind: integrality $\langle\beta,\alpha^\vee\rangle=p-q$ from the embedded- $\mathfrak{sl}(2)$ root string; $A_{ij}A_{ji}=4\cos^2\theta\in\{0,1,2,3\}$ and the four rank-2 Cartan matrices; the dimensions $\mathfrak{su}(3)=8$ , $\mathfrak{so}(5)=10$ , $G_2=14$ , $F_4=52$ , $E_8=248$ via $\dim=n+|\Delta|$ with the cross-check $|\Delta|=nh$ ; and the $\mathfrak{su}(3)$ multiplets $3,8,10$ from $d(n_1,n_2)=\tfrac12(n_1+1)(n_2+1)(n_1+n_2+2)$ . The golden-ratio $E_8$ mass ratio $m_2/m_1=2\cos(\pi/5)=1.618\ldots$ is the result of Zamolodchikov (1989) and Coldea et al., Science 327, 177 (2010).

Prerequisites. Lie Algebras and the Exponential Map · Lie Groups · Group Representations · Reducible and Irreducible Representations · The Rotation Group SO(3). Leads to. Tensor Products and Clebsch–Gordan Decomposition · Group Representations.

Ask the tutor: Ask me to walk you through drawing the $G_2$ root system by hand, or to derive the $\mathfrak{su}(5)$ Cartan matrix and show exactly which node you delete to recover the Standard Model.

What exactly is a root, in one sentence?

A root is a nonzero weight of the adjoint representation: a vector $\alpha\in\mathfrak{h}^*$ with a generator $E_\alpha$ satisfying $[H,E_\alpha]=\alpha(H)E_\alpha$ for every Cartan element $H$ . Concretely it is the amount by which the ladder operator $E_\alpha$ shifts the simultaneous Cartan eigenvalues. In $\mathfrak{su}(3)$ those eigenvalues are isospin $I_3$ and hypercharge $Y$ , so each root is the $(\Delta I_3,\Delta Y)$ step of one of the six raising and lowering operators. Roots live in the same space as representation weights, which is why root and weight diagrams share one plane.

Why does the whole classification reduce to su(2)?

Because every pair $\pm\alpha$ generates a copy of $\mathfrak{sl}(2)=\mathfrak{su}(2)_{\mathbb C}$ , and letting it act on the rest of the algebra imports the one rigid fact of $\mathfrak{su}(2)$ theory: its weights are integers, spaced by two and symmetric about zero. Applied to the string through any other root, that symmetry forces $\langle\beta,\alpha^\vee\rangle$ to be an integer (Derivation A). Integers admit only finitely many configurations of angles and lengths, and enumerating them is the classification. In a real sense there is only one simple Lie algebra — $\mathfrak{su}(2)$ — and the rest are consistent ways of gluing copies of it together.

What is the difference between simple roots and positive roots?

A hyperplane through the origin that misses every root splits $\Delta$ into positive roots (one side) and negative roots (their mirror images). A positive root is simple if it is not a sum of two other positive roots. The simple roots are the minimal generating set: exactly $n=\mathrm{rank}$ of them, forming a basis of $\mathfrak{h}^*$ , with every positive root a non-negative integer combination of them. So simple roots are few — one per Dynkin node — while positive roots are many; $\mathfrak{su}(3)$ has two simple roots ( $\alpha_1,\alpha_2$ ) but three positive roots ( $\alpha_1,\alpha_2,\alpha_1+\alpha_2$ ).

How can there be only five exceptional algebras and not, say, six?

The exceptionals are the connected Dynkin diagrams that survive the positive-definiteness constraints of Derivation C but escape the four regular families. The branch-and-loop bounds allow only a handful of shapes with a branch node or multiple bond: the triple bond gives $G_2$ ; a central double bond on four nodes gives $F_4$ ; and the three " $E$ -type" trees with one short arm give $E_6,E_7,E_8$ , after which lengthening the arm destroys positive-definiteness. Try to build $E_9$ and the Gram determinant hits zero — the diagram turns affine, an infinite-dimensional Kac–Moody algebra, not a finite simple one. Five is where finiteness runs out.

Is the Cartan matrix always symmetric?

No, and assuming so is a classic trap. It is symmetric only for the simply-laced algebras $A_n,D_n,E_6,E_7,E_8$ , where all simple roots share a length. Whenever a double or triple bond appears — $B_n,C_n,F_4,G_2$ — the roots come in two lengths, and $A_{ij}=2\langle\alpha_i,\alpha_j\rangle/\langle\alpha_j,\alpha_j\rangle$ divides by $|\alpha_j|^2$ , so swapping $i$ and $j$ changes the entry. It is still symmetrizable: a positive diagonal $D$ (the squared root lengths) makes $DA$ symmetric. The asymmetry is information, not a defect — it is how the diagram remembers which root is long.

How do root systems turn into particle multiplets?

Through the highest-weight construction. Attach non-negative integers $n_i$ — the Dynkin labels — to the diagram's nodes; this names an irreducible representation with highest weight $\lambda=\sum_i n_i\omega_i$ . From $\lambda$ you generate the full weight set by repeatedly subtracting simple roots, the allowed steps dictated by the Cartan matrix, until the diagram closes under the Weyl group. Those weights are the states' quantum numbers, and their count is the dimension. In $\mathfrak{su}(3)$ this turns $(1,1)$ into the eight-state adjoint (meson octet) and $(3,0)$ into the ten-state decuplet — the eightfold way, generated mechanically from root data.