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The connected Dynkin diagrams of finite reduced crystallographic root systems are $A_r$, $B_r$, $C_r$, $D_r$, $E_6$, $E_7$, $E_8$, $F_4$ and $G_2$. Bonds encode products of off-diagonal Cartan integers; multiple-bond arrows point toward shorter roots. The simply laced branching diagrams have arm lengths $(1,1,r-3)$ for $D_r$ and $(1,2,2)$, $(1,2,3)$, $(1,2,4)$ for the three exceptional $E$ types. The $B$ and $C$ chains have an end double bond, $F_4$ a middle double bond, and $G_2$ a triple bond. This edge weighting is distinct from the vertex labels used to classify nilpotent orbits.

A finite root system in a real inner product space is crystallographic when every Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer. The roots of a complex semisimple Lie algebra satisfy this because each number is a weight of its sl2 subalgebra associated with a root. Reducedness is a separate axiom; the root-space reducedness lemma establishes it for these Lie-algebra roots.

Over a field of characteristic zero, in a finite-dimensional Lie algebra representation of an sl2 triple, start with a nonzero vector $X_0$ satisfying $f{X}_0=0$ and $h{X}_0=-\lambda X_0$, and set $X_n=e^n{X}_0$. The commutator identities give the displayed weight and lowering formulas by induction. If $X_N\ne 0$ and $X_{N+1}=0$, then $f{X}_{N+1}=(N+1)(\lambda -N)X_N$ forces $\lambda =N$, a nonnegative integer. The nonzero terminal vector is essential: choosing a later index after the ladder has already vanished gives no constraint. Applied to the Adjoint representation of a semisimple Lie algebra, this argument proves integrality of Cartan integers.

For nonproportional roots in a reduced root system which is a crystallographic root system, the negative Cartan integers $m=2(\alpha ,\beta )/(\alpha ,\alpha )$ and $n=2(\alpha ,\beta )/(\beta ,\beta )$ satisfy $mn=4{\cos }^2\theta \in \{1,2,3\}$. At least one of $m,n$ equals $-1$, so reflecting one root in the other produces their sum. The exclusion of opposite roots is essential: their inner product is negative but their sum is zero, which is not a root.

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.

In the finite-dimensional complex semisimple Lie algebra setting, the rank of a semisimple Lie algebra is the dimension of a Cartan subalgebra $\mathfrak{h}$, equivalently the number of independent commuting Cartan generators. For a compact real algebra, use its complexification and the corresponding maximal torus. Simultaneously diagonalizing the commuting adjoint actions of $\mathfrak{h}$ gives the root-space decompositionThe nonzero linear functionals $\alpha$ for which this space is nonzero are the roots of a root system. An invariant inner product identifies them with vectors in a real Euclidean root space. These are the usual semisimple assumptions behind the rank and root language here.

The displayed generators $H,E^{+},E^{-}$ form an sl2 triple. Start with a nonzero lowest-weight element $X_0$ and define $X_{n+1}=[E^{+},X_n]$. The Jacobi identity givesInduction starting with $[H,X_0]=-\lambda X_0$ consequently gives the finite sl2 lowest-weight ladder weightsNow commute the lowering equation with $E^{+}$ and again use the Jacobi identity:The left side is $q_n{X}_n$. Whenever $X_n\ne 0$, coefficient comparison givesThe initial condition is $q_0=0$, meaning $[E^{-},X_0]=0$; directly $[E^{-},X_1]=[-H,X_0]=\lambda X_0$, so $q_1=\lambda$. Summing the recurrence from $0$ to $n-1$ yieldsIn fact the vector identity $[E^{-},X_n]=n(\lambda -n+1)X_{n-1}$ follows inductively from the same Jacobi identity without needing a separately assumed lowering coefficient.

If $X_{n_0}$ is the last nonzero element and $[E^{+},X_{n_0}]=0$, apply $E^{-}$ to this zero next element. The lowering identity givesso $\lambda =n_0$, a nonnegative integer. The nonzero endpoint is necessary. Literally allowing $X_{n_0}=0$ would make the printed conclusion false: take $X_0=E^{-}$ in the adjoint representation of ${\mathfrak{sl}}_2$. Then $\lambda =2$, $X_1=H$, $X_2=-2E^{+}$, $X_3=0$. Choosing $n_0=3$ gives a vanishing next bracket but not $\lambda =3$. The intended endpoint is the first terminating bracket after the nonzero ladder.

For the plus-sign nested commutators, use $H=H_1$, $E^{\pm }=E_1^{\pm }$ and $X_0=E_2^{+}$. The cross relation gives $[E_1^{-},E_2^{+}]=0$, while $[H_1,E_2^{+}]=K_{21}{E}_2^{+}$. Thus the lowest-weight label is $\lambda =-K_{21}$. If the $n$-fold raised element is nonzero and the $(n+1)$-fold element vanishes, the preceding endpoint argument gives $\lambda =n$ and henceFor the minus-sign ladder, use the sl2 triple $H^{\prime }=-H_1$, $E^{\prime +}=E_1^{-}$, $E^{\prime -}=E_1^{+}$ and start at $X_0=E_2^{-}$. Its lowest-weight equation is again $[H^{\prime },X_0]=K_{21}{X}_0$, and its lowering bracket vanishes by the cross relation. The same argument proves exactly the same result, including the endpoint assumption that is explicit in the nested-commutator question.

For arbitrary roots of a root system $\alpha ,\beta$, choose the normalized sl2 subalgebra associated with a root $\beta$, whose generators obey $[H_{\beta },E_{\beta }^{\pm }]=\pm 2E_{\beta }^{\pm }$ and $[E_{\beta }^{+},E_{\beta }^{-}]=H_{\beta }$. The coroot normalization givesThe normalization has a concrete construction. Use the nondegenerate Killing form $B$, and define $t_{\beta }\in \mathfrak{h}$ by $B(t_{\beta },H)=\beta (H)$. Invariance makes root spaces orthogonal unless their roots sum to zero, so there exist $e\in {\mathcal{L}}_{\beta }$ and $f\in {\mathcal{L}}_{-\beta }$ with $B(e,f)=1$. Invariance then gives $B([e,f],H)=B(e,[f,H])=\beta (H)$, hence $[e,f]=t_{\beta }$. With $(\beta ,\beta )=\beta (t_{\beta })$, take $E_{\beta }^{+}=e$, $E_{\beta }^{-}=2f/(\beta ,\beta )$ and $H_{\beta }=2t_{\beta }/(\beta ,\beta )$. Their brackets have the required normalization and $\alpha (H_{\beta })=2(\alpha ,\beta )/(\beta ,\beta )$. Let $v$ be a nonzero $\alpha$ root vector with $H_{\beta }$ weight $w=\alpha (H_{\beta })$. Repeatedly apply $\mathrm{ad}{E}_{\beta }^{-}$ until reaching a last nonzero vector $v_0=(\mathrm{ad}{E}_{\beta }^{-}{)}^{p}v$. Its weight is $w-2p$. This process terminates because different weights are linearly independent and the Lie algebra is finite-dimensional. Starting at $v_0$, repeatedly raise; this also terminates for the same reason. The proven lowest-weight endpoint formula says $w-2p=-N$ for a nonnegative integer $N$. ThereforeThis proves integrality of Cartan integers directly from the finite ladder, rather than assuming the crystallographic axiom.

Finally use the indices actually printed in the PDF: $E_1^{+}=R^1{}_2$, $E_2^{+}=R^2{}_3$, $E_1^{-}=R^2{}_1$, $E_2^{-}=R^3{}_2$. The converted TeX has corrupted these indices. The prescribed bracket is that of trace-free matrix units. The sum of the diagonal generators is zero, leaving two independent diagonal elements, and their pair brackets giveFor example $[R^1{}_2,R^2{}_1]=R^1{}_1-R^2{}_2$, and the analogous second pair gives $H_2$. The cross raising-lowering brackets vanish. A diagonal $H$ with entries $h_r$ has $[H,R^r{}_s]=(h_r-h_s)R^r{}_s$. Thus $H_1$ acts on $E_1^{+},E_2^{+}$ with weights $2,-1$, and $H_2$ acts on them with weights $-1,2$. The negative generators have the opposite weights. In the question's index ordering,This is the Cartan matrix of the A2 root system, and the Lie algebra generated by these matrix units has rank two. As an additional bracket check, $[E_1^{+},E_2^{+}]=R^1{}_3\ne 0$ whereas $[E_1^{+},R^1{}_3]=0$, so the nested ladder has $n=1$ and indeed $K_{21}=-1$. The negative ladder likewise terminates after one nonzero bracket.

If the roots are parallel, their Euclidean norms are equal by reducedness. For nonparallel, nonorthogonal roots, the Cartan integers $a,b$ above are nonzero integers of the same sign and have product $1,2$ or $3$. AlsoThe absolute pairs $(|a|,|b|)$ are $(1,1)$, $(1,2),(2,1)$, or $(1,3),(3,1)$. HenceMore precisely, angles $\pi /3,2\pi /3$ require equal lengths; angles $\pi /4,3\pi /4$ require a long-to-short ratio $\sqrt{2}$; and angles $\pi /6,5\pi /6$ require a ratio $\sqrt{3}$. The exclusion of orthogonality matters: orthogonal irreducible components can be scaled independently.

Use the course convention that a root system is finite and is a reduced root system satisfying the crystallographic root system condition. These hypotheses are essential for the angle restriction. For roots $r,s$, the two Cartan integerssatisfy $ab=4{\cos }^2\theta$. If $r,s$ are not parallel, the strict Cauchy-Schwarz inequality and integrality give $ab\in \{0,1,2,3\}$. These values give angles $\pi /2$, $\pi /3$ or $2\pi /3$, $\pi /4$ or $3\pi /4$, and $\pi /6$ or $5\pi /6$, respectively. If the roots are parallel, reducedness implies $s=\pm r$, giving $0$ or $\pi$. ThereforeHere are examples accounting for every angle. In the A1 root system, the pairs $(r,r)$ and $(r,-r)$ give $0,\pi$. Two roots from different summands of $A_1\oplus A_1$ give $\pi /2$. The six unit vectors at angles $k\pi /3$ form an A2 root system: taking one root at angle zero and another at $\pi /3$ or $2\pi /3$ gives those two angles. In the B2 root system $\{\pm e_1,\pm e_2,\pm e_1\pm e_2\}$, pair $e_1$ with $e_1+e_2$ or $-e_1+e_2$ to obtain $\pi /4,3\pi /4$. In a G2 root system, take short unit roots at angles $k\pi /3$ and long roots of length $\sqrt{3}$ at angles $\pi /6+k\pi /3$. A short root at zero paired with long roots at $\pi /6,5\pi /6$ gives the remaining angles.

A Dynkin diagram has one vertex for each simple root. Vertices $i,j$ are joined by $a_{ij}{a}_{ji}$ bonds, where $a_{ij}$ are the off-diagonal Cartan integers. No bond means orthogonality; one, two and three bonds mean angles $2\pi /3,3\pi /4,5\pi /6$. A multiple-bond arrow points to the shorter root. Thus the Dynkin diagram determines the Cartan matrix, including the relative lengths inside each connected component.

An irreducible root system corresponds to a connected Dynkin diagram. The classification of finite crystallographic Dynkin diagrams givesHere $C_2$ is already represented by $B_2$, and $B_1,C_1$ by $A_1$. General systems are orthogonal direct sums of these, with separate component scales. The following diagrams specify all the types; dots denote a continued chain, not additional vertices.

For orientation, the An Dynkin diagram is a single chain, the Bn Dynkin diagram has a terminal double bond pointing to its terminal short root, and the Cn Dynkin diagram reverses that arrow. The Dn Dynkin diagram has arm lengths $1,1,l-3$. The En Dynkin diagram has arm lengths $1,2,2$; $1,2,3$; or $1,2,4$. The F4 Dynkin diagram has a central double bond and the G2 Dynkin diagram a triple bond.

The positivity behind this finite list is the positive-definite Gram matrix of the simple roots. In particular the underlying graph is a tree: if it contained a cycle, the sum of its normalized roots would have squared norm at most zero, since each bonded inner product is at most $-1/2$. This contradicts linear independence. Restricting the possible bonds and branches by positive definiteness gives precisely the listed chains and trees. Conversely their Cartan matrices are positive definite after symmetrization and their reflection-generated root sets give the indicated finite root systems.

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.

If $\beta =-\alpha$, then both Cartan integers are $-2$. This opposite-root case satisfies the printed inner-product hypothesis and must be included.

Otherwise reducedness makes the roots nonproportional. Writing the angle as $\theta$, strict Cauchy-Schwarz inequality givesBoth factors are negative integers, so their product is $1,2$ or $3$. The full list is thereforeThe nonproportional pairs correspond respectively to angles $120^{\circ }$, $135^{\circ }$ and $150^{\circ }$; reversing the order of the roots reverses the unequal pairs.

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: