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The real span of the simple coroots inside a complex Cartan subalgebra carries a positive-definite restriction of the Killing form. Equivalently it consists of the $h$ for which all roots have real values. The formula $\kappa (h,h)={\sum }_{\alpha \in \Phi }\alpha (h{)}^2$ is positive for $h\ne 0$, since the roots span the dual. The induced dual inner product makes the root system Euclidean and makes root reflections orthogonal.

For a Cartan-Weyl basis, the Jacobi identity givesThus the bracket has joint Cartan weight $\alpha +\beta$. If $\beta =-\alpha$, it belongs to the zero-weight space, which is the Cartan subalgebra, and so is a linear combination ${\tilde{\alpha }}_i{H}_i$. If $\alpha +\beta$ is a nonzero root, its root space is one-dimensional and the bracket is proportional to $E_{\alpha +\beta }$. If that nonzero sum is not a root, the bracket vanishes. This proves the asserted forms without assuming every pair of roots has a root sum.

With the stated normalization, put $H_{\alpha }=2{\alpha }_i{H}_i/{\alpha }^2$. The commutators becomeThe adjoint relations make $J_3=H_{\alpha }/2$, $J_{+}=E_{\alpha }$, $J_{-}=E_{-\alpha }$ the usual SU(2) Lie algebra generators. Hence every finite-dimensional SU(2) irreducible component has $J_3$ eigenvalues $m=-j,-j+1,\dots ,j$, with $2j$ an integer. Therefore $H_{\alpha }$ has integer eigenvalues $2m$, including in the finite-dimensional Adjoint representation. Algebraically this is the sl2 subalgebra associated with a root.

If $[H_{\alpha },X_{\lambda }]=\lambda X_{\lambda }$, another use of the Jacobi identity givesThe bracket is therefore zero or lies in the eigenvalue-$\lambda +2$ space. Within one irreducible SU(2) ladder it is proportional to the next chosen vector $X_{\lambda +2}$. If the full eigenspace has multiplicity, the weight statement is the general one; proportionality to one named vector presupposes the ladder choice. Similarly $E_{-\alpha }$ lowers this eigenvalue by two, and the ladder weights run from $-n$ to $n$ in steps of two.

For a root vector $E_{\beta }$, its $H_{\alpha }$ weight isThe adjoint SU(2) ladder therefore provesFor nonparallel $\alpha ,\beta$, raising and lowering $E_{\beta }$ move its joint Cartan weight through the root string $\beta +k\alpha$. The finite irreducible ladder has weights symmetric about zero. Starting at $m_{\alpha \beta }$, reaching its partner $-m_{\alpha \beta }$ requires $k=-m_{\alpha \beta }$. The corresponding nonzero vector lies in the root space of $\beta -m_{\alpha \beta }\alpha$, proving the Weyl reflection property:There is no zero joint root in this string when $\alpha ,\beta$ are nonparallel. Its one-dimensional root spaces prevent two irreducible ladders with the same weight parity from overlapping, so the ladder argument applies to the entire string. For $\beta =\pm \alpha$, reflection simply interchanges the two opposite roots, which already exist by the adjoint relation.

For angle and length restrictions, set $p=2\alpha \cdot \beta /{\alpha }^2$ and $q=2\alpha \cdot \beta /{\beta }^2$. These Cartan integers have the same sign when nonzero and obeyFor nonparallel, nonorthogonal roots, $pq$ is an integer strictly between zero and four. The possibilities are consequently:

Orthogonal roots have $p=q=0$ and $\theta =\pi /2$. Parallel roots have $pq=4$; the integer pairs would allow only length multiples $1,2,1/2$. The stated reducedness condition excludes the latter two, so parallel roots are equal or opposite, with angles zero or $\pi$ and equal lengths. ThusNot every root system realizes every listed angle.

The orthogonal pair alone gives no ratio through $q/p$. To finish the length restriction for a simple Lie algebra, use irreducibility of its root system. Let $U$ span the Weyl group orbit of $\alpha$. Both $U$ and $U^{\perp }$ are invariant under the orthogonal root reflections. If a root had nonzero projections in both, reflection of its projection in $U$ would have a nonzero $U^{\perp }$ component, violating invariance. Hence every root is entirely in one of these subspaces. A proper nonzero $U$ would split the root system into two orthogonal parts, contradicting irreducibility. This proves that the Weyl orbit spans an irreducible root space.

In particular, some orbit root $\gamma$ is nonorthogonal to any chosen $\beta$, and ${\gamma }^2={\alpha }^2$. Applying the previous integer-pair restriction to $\gamma ,\beta$ therefore also handles orthogonal $\alpha ,\beta$:There can be at most two different root lengths: three lengths would have ratios two and three relative to the shortest, but then the longest-to-middle squared-length ratio would be $3/2$, which is forbidden. A reducible collection of orthogonal rank-one factors could have independently chosen scales; the simple-algebra assumption excludes that exception.

Choose a generic linear functional to define positive roots. The simple roots are those positive roots which cannot be written as sums of two positive roots; they form a basis in which all root coefficients are integers of one sign. Distinct simple roots have nonpositive inner product: if their inner product were positive, their root string would contain their difference, and whichever difference is positive would decompose one of them. In rank two, irreducibility excludes the orthogonal case. The allowed simple-root angles are therefore $120^{\circ },135^{\circ },150^{\circ }$, giving the three irreducible diagrams in the classification of rank-two root systems: A2 root system, B2 root system and G2 root system. Reflections in the two simple roots generate their respective six, eight and twelve roots.

Here are explicit simple-root choices and positive roots; adding their negatives gives each complete diagram.

The nonnegative coefficients in each positive-root list verify the chosen simple roots explicitly. The orthogonal four-root system $A_1\times A_1$ is rank two but reducible, so it is not a diagram for the simple algebra assumed here.

Define the Cartan matrix in the column-coroot convention:Using the ordered roots above givesThe alternative row-coroot convention transposes these matrices. Stating the convention and root ordering avoids a spurious disagreement in the unequal-length cases.

The drawn simple roots have $\Vert \alpha {\Vert }^2=1$, $\Vert \beta {\Vert }^2=2$ and $\alpha \cdot \beta =-1$. Their Cartan integers are therefore $⟨\beta ,{\alpha }^{\vee }⟩=-2$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$, identifying a B2 root system. The corresponding root reflections act on coordinates byIn particular $s_{\beta }(\alpha )=\alpha +\beta =(0,1)$ and $s_{\alpha }(\beta )=2\alpha +\beta =(1,1)$, so both new positive roots are forced. Reflection in any root also forces its negative. ThusThese eight roots are drawn below, with the given simple pair highlighted.

For completeness, no further roots can occur. In a finite reduced root system, every root is a Weyl group translate of a simple root: reflect a nonsimple positive root in a simple root with positive inner product to decrease its height, and iterate. This standard simple-root descent may also be quoted here. The eight-vector set just listed is stable under both generating reflections, and the orbits of the two simple roots already fill it. Hence no simple-root orbit, and therefore no root, lies outside that set. This establishes both the necessity of every drawn root and exhaustion of the root system.

In the semisimple Lie algebra convention, an abstract root system is a finite subset $\Phi$ of a finite-dimensional real Euclidean space $E$ with the following properties: it spans $E$, does not contain zero, is reduced, is invariant under each of its root reflections, and satisfies crystallographic integrality. More explicitly, for every $\alpha ,\beta \in \Phi$,Here $(,)$ is a positive-definite inner product. The first condition is reducedness, the second gives closure under root reflections, and the last number is a Cartan integer. The rank is $\dim E$. A simple root basis is a basis of $E$ consisting of roots such that each root has integral coordinates that are either all nonnegative or all nonpositive. The two signs distinguish positive and negative roots. The diagrams in this problem concern these reduced crystallographic root systems; broader definitions sometimes omit one of these conditions.

Identify $\mathfrak{t}$ with the coordinate vector $t=(t_1,\dots ,t_n)$ through $H(t)$. A root reflection is $s_{\alpha }(H)=H-\alpha (H){\alpha }^{\vee }$. Since $({\epsilon }_i\pm {\epsilon }_j{)}^{\vee }=H_i\pm H_j$, for $i<j$ the two possibilities areEvery other coordinate is unchanged, and $s_{-\alpha }=s_{\alpha }$, so these formulas cover every root. The complete diagonal matrix is reconstructed with its last $n$ entries the reversed negatives of the first $n$ entries.

The Weyl group of Dn consists of all permutations of the $n$ coordinates together with an even number of coordinate sign changes. Equivalently,Its order is $2^{n-1}n!$. The difference-root reflections give ordinary transpositions; composing a sum-root reflection with the corresponding difference-root reflection negates precisely the two selected coordinates. These describe the generators of all even signed permutations. For $n=2$ the group has four elements; for the toral case $n=1$ it is trivial.

The Weyl group acts on the weight lattice by its root reflections and extends to ring automorphisms of the group ring of a weight lattice:Weights and their multiplicities in a finite-dimensional Lie algebra representation of a semisimple Lie algebra are permuted by this action, so its formal character is Weyl-invariant.

Choose a set ${\Phi }^{+}$ of positive roots and putHere ${\omega }_i$ are the fundamental weights, and $\det (w)=(-1{)}^{\ell (w)}$, where the Coxeter length $\ell (w)$ counts positive roots carried to negative ones. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote the finite-dimensional Irreducible Lie algebra representation of highest weight $\lambda$. The Weyl character formula isThe second denominator expression is the Weyl denominator formula. Both alternants change sign under each root reflection; their quotient is invariant. Although written as a fraction, the theorem asserts that it is a finite element of $\mathbb{Z}[{\Lambda }_W]$ with the nonnegative coefficients given by weight multiplicities. The highest-weight classification specifies which $\lambda$ label the irreducibles, while the root system and choice of positive roots determine every other term.

Write $u={\sum }_{\lambda }{a}_{\lambda }e(\lambda )$ with finite support. The root reflection is an involution, and $s_{\alpha }u=-u$ implies $a_{s_{\alpha }\lambda }=-a_{\lambda }$. Any fixed weight has coefficient zero. For each nonfixed orbit choose its representative $\lambda$ withIntegrality of $m$ follows because $\lambda$ belongs to the weight lattice; positivity is ensured by our choice of orbit representative. Pairing the orbit coefficients givesFor each pair, elementary finite telescoping showsThere are finitely many chosen orbits, and each inner sum is finite. This proves the root-binomial divisibility of reflection anti-invariants:

For precision, the geometric-series expression is interpreted coefficientwise. Its coefficient at a weight $\nu$ is ${\sum }_{n\ge 0}{a}_{\nu -n\alpha }$, which is a finite sum because $u$ has finite support. On each coset of $\mathbb{Z}\alpha$, reflection pairs the coefficients with opposite signs, so their total is zero. These cumulative sums vanish both before the smallest occupied weight and after the largest occupied weight. Only finitely many cosets are occupied, hence the product of the formal series with $u$ has finite support and equals the boxed quotient.

The individual terms $e(n\alpha )u$ need not vanish for large $n$. For example, $u=e(\alpha )-e(-\alpha )$ is anti-invariant, and its quotient is $-1-e(-\alpha )$. Thus “finite sum” means a finite result after coefficientwise cancellation, not an actually truncated geometric series. No analytic convergence assumption is involved.

Choose a fundamental system of a root system $\{\alpha ,\beta \}$. Distinct simple roots have nonpositive inner product. Their Cartan matrix has diagonal entries two and off-diagonal entries $-p,-q$, where either $p=q=0$, or $p,q$ are positive integers. The root-system finiteness lemma gives $pq<4$. After exchanging the simple roots, the possibilities areTheir angles and relative lengths determine the two simple roots up to an orthogonal transformation and common scaling, except that the two orthogonal components can be scaled separately.

To see that no extra roots remain unspecified, write a positive root $\gamma =m\alpha +n\beta$ with nonnegative integer coefficients. If it is not simple, some simple root $\delta$ has $(\gamma ,\delta )>0$: otherwise $\Vert \gamma {\Vert }^2=m(\gamma ,\alpha )+n(\gamma ,\beta )\le 0$. The root reflection in $\delta$ subtracts a positive integer multiple of $\delta$. The other coordinate stays nonnegative; since $\gamma$ is not proportional to $\delta$, it is positive, so the reflected root must still be positive by the defining sign property of a fundamental system of a root system. Its height decreases. Repeating reduces $\gamma$ to a simple root. Thus every root is in a Weyl group orbit of a simple root.

For the four matrices, applying their simple reflections gives, respectively, the following positive roots:where in the last two rows $\alpha$ is long, $\beta$ short. For example, $s_{\beta }(\alpha )=\alpha +q\beta$ and $s_{\alpha }(\beta )=\beta +\alpha$ for the displayed matrices when the convention is $a_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. The sets comprising these positive roots and their negatives are invariant under both reflections, as direct substitution verifies. The height argument therefore shows they are the whole root system. This proves the classification of rank-two root systems:

Identify the real span of the roots with its dual using the invariant inner product, and put ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$. The weight lattice isThe Weyl group is generated by the root reflections $s_{\alpha }$. Each reflection permutes the finite set of roots and preserves their lengths, hence also permutes their coroots. Thus, for $\lambda \in {\Lambda }_W$,Every generating reflection therefore maps the weight lattice to itself, and so does every product. Since the inverse product has the same property, every Weyl-group element acts bijectively on ${\Lambda }_W$. The roots span the ambient space, so the action on the finite root set is faithful; in particular the Weyl group is finite.

A reduced crystallographic abstract root system in a finite-dimensional real inner product space $E$ is a finite set $R$ such that: it spans $E$ and does not contain zero; the only scalar multiples of $\alpha \in R$ belonging to $R$ are $\pm \alpha$; every root reflectionpermutes $R$; and each Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is integral. These are the crystallographic and reduced conventions appropriate to roots of a complex semisimple Lie algebra. Without crystallographic integrality, the numerical restrictions in the following subparts would not hold.

Identify the Cartan subalgebra with its coordinate vector $t=(t_1,\dots ,t_n)$ using the invariant inner product. A root reflection isFor every root of the Bn root system, the explicit formulas areAll unlisted coordinates stay fixed, and opposite roots define the same reflection. These reflections give all coordinate permutations and independent sign changes. Therefore the Weyl group isIts action on $\mathfrak{t}$ is the group of signed permutation matrices.

Identify the Cartan subalgebra with the coordinate space $(t_1,\dots ,t_n)$ used above. The root reflection in any root of a root system fixes the perpendicular hyperplane. For every root of the Cn root system, its complete coordinate action isEvery unlisted coordinate is fixed, and changing the sign of the root leaves its reflection unchanged. These formulas follow directly from $s_{\alpha }(t)=t-\alpha (t){\alpha }^{\vee }$, with the coroot regarded as a vector of the Cartan subalgebra.

The Cn Weyl group is the signed symmetric group: its elements permute the $n$ coordinates and independently change their signs. The Weyl reflections associated with the simple roots interchange consecutive coordinates for $i<n$ and negate the last coordinate for $i=n$. Thus

The appropriate abstract object is a finite reduced crystallographic root system $R$ in a real inner product space $E$. Its axioms are: $R$ is finite, spans $E$, and does not contain zero; for $\alpha \in R$, $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; each root reflectionpermutes $R$; and every Cartan integer $n_{\alpha \beta }=2(\alpha ,\beta )/(\beta ,\beta )$ is an integer. The restriction to a reduced root system and crystallographic integrality distinguishes roots of complex semisimple Lie algebras from more general reflection configurations.

For any two roots of a root system, the Cauchy-Schwarz inequality givesIf the inner product is zero, both Cartan integers vanish. Otherwise their signs agree, and the absolute value of each is a positive integer. Dividing their product by an integer of absolute value at least one provesFor nonproportional roots the product is strictly less than four. In a reduced root system, proportional roots are just $\pm \alpha$ and have Cartan integers $\pm 2$, so the printed bound is intentionally looser than the resulting bound of three.

A fundamental system of a root system is a basis $\Delta$ of $E$ made of roots, such that each root is an integer combination of $\Delta$ with either all coefficients nonnegative or all nonpositive. Its members are the simple roots. Suppose distinct $\alpha ,\beta \in \Delta$ had $n_{\alpha \beta }>0$. Thenhas a positive coefficient of $\alpha$ and a negative coefficient of $\beta$, contradicting the defining sign condition. Thus $n_{\alpha \beta }\le 0$. Distinct simple roots are linearly independent, so their Cartan-integer product is strictly less than four. Combining integrality and the sign condition givesHere nonpositive is the intended sense of the printed convention that includes zero among “negative” numbers; orthogonal simple roots really do give zero.

To form a Dynkin diagram, place a vertex at each simple root. Join two vertices by $n_{\alpha \beta }{n}_{\beta \alpha }$ bonds, hence zero, one, two or three. A multiple bond has an arrow toward the short root. Indeed $n_{\alpha \beta }/n_{\beta \alpha }=(\alpha ,\alpha )/(\beta ,\beta )$ determines the squared length ratio, and the diagram with the Cartan matrix reconstructs the angles and relative lengths. A single bond joins equal-length roots.

The connected finite Dynkin diagrams are the following. The descriptions include bond multiplicities and arrow directions, so distinguish dual diagrams:

There are no other connected finite Dynkin diagrams. Low-rank conventions also identify $B_1=C_1=A_1$ and $D_3=A_3$; $D_2$ is disconnected, so introduces no further connected type. Affine diagrams are outside this finite classification.

Finally suppose the underlying graph contained a cycle on $m\ge 3$ distinct simple roots ${\alpha }_1,\dots ,{\alpha }_m$. Put $u_i={\alpha }_i/\Vert {\alpha }_i\Vert$. Any bonded pair hasand all other distinct pairs have nonpositive inner products. The cycle contributes at least $m$ bonded pairs, soBut the simple roots, and hence these normalized vectors, are linearly independent, making the displayed sum nonzero. Positive definiteness gives a contradiction. Thus the underlying graph of a finite Dynkin diagram has no cycle. This acyclicity of a finite Dynkin diagram argument also excludes cycles with extra chords or multiple bonds; multiple bonds themselves are not treated as two-edge cycles.

For a complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra consisting of elements whose Adjoint representation matrices are diagonalizable. Equivalently it is a nilpotent self-normalizing Lie subalgebra. The root-space decomposition iswhere the roots are the nonzero weights of that Adjoint representation.

We use these properties of the Killing form $B$: it is nondegenerate on $\mathfrak{g}$ and on $\mathfrak{h}$, it is invariant, distinct root spaces are orthogonal unless their roots sum to zero, and $B$ pairs ${\mathfrak{g}}_{\alpha }$ nondegenerately with ${\mathfrak{g}}_{-\alpha }$. Define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$, and choose $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)=1$. Since weights add, $[e,f]\in \mathfrak{h}$, and invariance givesThe nonisotropic root lemma shows $\alpha (t_{\alpha })\ne 0$. Here is its short proof: if this number were zero, the span of $e,f,t_{\alpha }$ would be a solvable Lie algebra with $t_{\alpha }$ central. Apply the Lie theorem to its action on $\mathfrak{g}$. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}e,\mathrm{ad}f]$ would be strictly upper triangular and hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$ makes it diagonalizable. It would therefore be zero, putting $t_{\alpha }$ in the zero center of a Lie algebra of $\mathfrak{g}$, contrary to its definition. SetThenso their span is the sl2 subalgebra associated with a root.

The abstract reduced crystallographic root system axioms are as follows. In a finite-dimensional real inner product space $E$, the set $R$ is finite, consists of nonzero vectors and spans $E$; for every $\alpha \in R$, one has $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; the root reflection $s_{\alpha }(\beta )=\beta -2(\beta ,\alpha )\alpha /(\alpha ,\alpha )$ preserves $R$; and $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for every pair of roots. We verify the positive-definite real form as well as these axioms, rather than assuming the complex Killing form is already positive.

First $R$ is finite and has no zero element by its definition. The roots span ${\mathfrak{h}}^{\ast }$ over $\mathbb{C}$: any $h\in \mathfrak{h}$ annihilated by all roots commutes with the whole root-space decomposition and is central, hence zero. Therefore the $t_{\alpha }$, and also the $H_{\alpha }$, span $\mathfrak{h}$ over $\mathbb{C}$. For every root $\beta$, the classification of finite-dimensional sl2 representations applied to the Adjoint representation of the root subalgebra gives $\beta (H_{\alpha })\in \mathbb{Z}$. Put ${\mathfrak{h}}_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\{H_{\alpha }:\alpha \in R\}$. Every root takes real values on this space, andThe strict inequality follows because the roots span ${\mathfrak{h}}^{\ast }$. This also shows that the complexification of ${\mathfrak{h}}_{\mathbb{R}}$ injects into $\mathfrak{h}$: an equality $h_1+i{h}_2=0$ with real $h_j$ would contradict positivity of $B(h_1,h_1)$ and $B(h_2,h_2)$. Since its complex span is all of $\mathfrak{h}$, it is a real form. Moreover $B(H_{\alpha },H_{\alpha })=4/\alpha (t_{\alpha })>0$, so $t_{\alpha }$ is a real scalar multiple of $H_{\alpha }$. The dual inner product thus makes $E={\mathrm{span}}_{\mathbb{R}}R$ a Euclidean space, as in the Euclidean subspace of a Cartan subalgebra.

To prove reducedness without assuming it, consider the root-subalgebra modulewith absent root spaces understood to be zero. Its $H_{\alpha }$ weights are even, so every nontrivial Irreducible Lie algebra representation in it has even positive highest weight and one-dimensional weight-zero space. The action of $X_{\alpha }$ on its weight-zero space $\mathfrak{h}$ has image exactly $\mathbb{C}{X}_{\alpha }$, of dimension one. Therefore there is precisely one nontrivial irreducible summand, the already embedded adjoint ${\mathfrak{sl}}_2$ module of highest weight $2$. Thus $\dim {\mathfrak{g}}_{\alpha }=1$ and $2\alpha \notin R$. If $\beta =c\alpha$ is any root on the same real line, the integral numbers $\beta (H_{\alpha })=2c$ and $\alpha (H_{\beta })=2/c$ have product $4$. Hence $c$ is one of $\pm \frac{1}{2},\pm 1,\pm 2$; the half and double cases are excluded by applying the preceding argument to the appropriate root. This proves the root-space reducedness lemma and $R\cap \mathbb{R}\alpha =\{\pm \alpha \}$.

For a root $\beta$ not parallel to $\alpha$, the sum of root spaces ${⨁}_{j\in \mathbb{Z}}{\mathfrak{g}}_{\beta +j\alpha }$ is stable under the root ${\mathfrak{sl}}_2$. Its integer $H_{\alpha }$ weights are symmetric under sign in each irreducible summand. Therefore the weight $-\beta (H_{\alpha })$ also occurs, at the root $\beta -\beta (H_{\alpha })\alpha$. The Killing form normalization givesso this root is exactly $s_{\alpha }(\beta )$. For $\beta =\pm \alpha$ the reflection just swaps the two roots. This proves reflection invariance and the Cartan integer condition. All axioms of the reduced crystallographic root system have now been verified.

An abstract root system, in the reduced crystallographic convention, is a finite subset $\Phi$ of a finite-dimensional real Euclidean space $E$ satisfying the following conditions:

The root system is irreducible if it is not a union of two nonempty mutually orthogonal root systems. The integrality condition is essential for the finite angle list below; arbitrary finite reflection systems need not satisfy it.

The possible angles are $0$, $\pi /6$, $\pi /4$, $\pi /3$, $\pi /2$, $2\pi /3$, $3\pi /4$, $5\pi /6$, and $\pi$. For roots $\alpha ,\beta$, putBoth are integers, they have the same sign when nonzero, andIf the roots are linearly independent, strict Cauchy-Schwarz inequality gives $0\le ab<4$. Hence $ab\in \{0,1,2,3\}$. These values give, respectively,For the last column, the possible absolute pairs $(|a|,|b|)$ are $(1,1)$, $(1,2)$, and $(1,3)$, up to interchange, and $a/b=(\beta ,\beta )/(\alpha ,\alpha )$. Parallel roots are equal or opposite by reducedness, giving $0$ or $\pi$. Thus the list is exhaustive, and the rank-two root systems realize each nonparallel possibility.

Now suppose two simple roots $\alpha ,\beta$ have angle $5\pi /6$. Distinct simple roots have nonpositive inner product. Here is a proof from their defining sign property: if $(\delta ,ϵ)>0$, then $s_{\delta }(ϵ)=ϵ-n\delta$ with a positive Cartan integer $n$. This would be a root whose coefficients in the root basis include both a positive and a negative coefficient, which is impossible. Therefore, after normalizing the simple roots to unit length, any nonzero off-diagonal inner product is at most $-1/2$, by the angle list.

Let $u=\alpha /\Vert \alpha \Vert$, $v=\beta /\Vert \beta \Vert$, so $(u,v)=-\sqrt{3}/2$. If there were a third simple root with unit vector $w$, write $r=(u,w)\le 0$ and $s=(v,w)\le 0$. Independence of the simple roots makes their Gram matrix positive definite, soIf either $r$ or $s$ were nonzero, its absolute value would be at least $1/2$. Since $rs\ge 0$, the displayed determinant would then be nonpositive. Thus $r=s=0$. Every other simple root is orthogonal to both $\alpha$ and $\beta$: a triple bond isolates a G2 component.

Indeed, the root reflections in $\alpha ,\beta$ preserve their plane and fix its orthogonal complement; the root reflections in the other simple roots preserve that complement and fix the plane. The given facts that these reflections generate the Weyl group and that every root is Weyl group-conjugate to a simple root imply that every root lies entirely in one of those two orthogonal subspaces. If there were any other simple root, this would make $\Phi$ reducible. Therefore irreducibility forces rank two.

Up to interchanging the simple roots, take $\alpha$ short and $\beta$ long. The squared-length ratio is $3$. Choose a scale and orthonormal coordinates so thatThe Cartan integers are $⟨\beta ,{\alpha }^{\vee }⟩=-3$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$. In coefficient coordinates $m\alpha +n\beta$, the two root reflections areTheir orbits of $\alpha$ and $\beta$ are, respectively,These sets are closed under both displayed reflections; for example $s_{\beta }(\alpha )=\alpha +\beta$, $s_{\alpha }(\alpha +\beta )=2\alpha +\beta$, $s_{\alpha }(\beta )=3\alpha +\beta$, and $s_{\beta }(3\alpha +\beta )=3\alpha +2\beta$. The assumption that every root is conjugate to a simple root now determines all twelve roots, with no further ones possible.

For completeness, this twelve-element set really is a root system. It is finite, spans the plane, and is reduced. The six short roots have length $1$ and directions spaced by $\pi /3$; the six long roots have length $\sqrt{3}$ and interleaved directions, also spaced by $\pi /3$. Their mutual Cartan integers are integral: within either length class they are $0$, $\pm 1$, or $\pm 2$ as applicable, and between the two length classes they are $0$, $\pm 1$, or $\pm 3$. Closure under all root reflections follows by conjugating the two simple reflections, since every listed root lies in a simple-root orbit. The nonorthogonal simple roots make it irreducible. We have therefore proved existence as well as uniqueness, up to isometry and overall scale: