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For simple roots $(1,0)$ and $(-1,1)$, coroot integrality gives the weight lattice $P={\mathbb{Z}}^2\cup (\mathbb{Z}+1/2{)}^2$. This is the lattice for the Spin group $\mathrm{Spin}(5)$ and its Lie algebra. Representations descending to $SO(5)$ use the integer coset, because a $2\pi$ rotation in either coordinate plane must act trivially.

For the C3 root system, take ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=2e_3$. With columns indexing coroots, the Cartan matrix isThe convention with rows indexing coroots gives $A^T$. Both describe a double bond between the short root ${\alpha }_2$ and long root ${\alpha }_3$, with arrow toward ${\alpha }_2$.

Choose simple root vectors with $[e_i,f_i]=h_i={\alpha }_i^{\vee }$, and let ${\rho }^{\vee }$ be the half-sum of positive coroots. It satisfies ${\alpha }_i({\rho }^{\vee })=1$. Expand $h=2{\rho }^{\vee }={\sum }_i{c}_i{h}_i$. For $i\ne j$, $[e_i,f_j]=0$, since the difference of distinct simple roots is not a root of a root system. Thus $[e,f]=h$, while $[h,e]=2e$ and $[h,f]=-2f$. These identities construct an embedded sl2 Lie algebra in any nonzero complex semisimple Lie algebra. Its raising vector is the principal nilpotent element with a nonzero component in each simple-root space.

The long roots of a root system $\pm (1,1)$ and $\pm (1,-1)$ give the commuting $SU(2)$ factors of the standard $SO(4)$ subgroup. Choosing $(1,1)$ for the left factor pairs a weight $(x,y)$ with its coroot as $x+y$. The five-dimensional vector then gives two doublets and a singlet, whereas a short-axis root would give a triplet and two singlets.

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.

With simple roots ${\alpha }_i={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n={\epsilon }_{n-1}+{\epsilon }_n$, the fundamental weights are ${\omega }_i={\epsilon }_1+\cdots +{\epsilon }_i$ for $i\le n-2$, followed by ${\omega }_{n-1}=({\epsilon }_1+\cdots +{\epsilon }_{n-1}-{\epsilon }_n)/2$ and ${\omega }_n=({\epsilon }_1+\cdots +{\epsilon }_n)/2$. All roots have squared length two, so their pairings with coroots verify the dual-basis identities. The last two weights are the half-spin highest weights; for $n=2$, the defining highest weight is ${\omega }_1+{\omega }_2$.

For the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, the fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$ and ${\omega }_2=3{\alpha }_1+2{\alpha }_2$. The positive coroots, relative to the simple coroots, have coordinates $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. The Weyl dimension formula therefore gives, for nonnegative integers $a,b$,The denominator is the product of the pairings of the Weyl vector with those coroots. The dimensions at ${\omega }_1,{\omega }_2,2{\omega }_1$ are seven, fourteen and twenty-seven.

The coordinates of a highest weight record its eigenvalues on a chosen basis of a Cartan subalgebra. They depend on that basis. Dynkin labels instead use the simple coroots; orthonormal Cartesian coordinates and Dynkin labels must be converted before applying a dimension or Casimir eigenvalue formula.

Restrict a finite-dimensional representation of a complex semisimple Lie algebra to the sl2 subalgebra associated with a root $\alpha$. Its Cartan element is the coroot $h_{\alpha }$. The classification of finite-dimensional sl2 representations gives only integer eigenvalues for this element, so every weight of a representation has integral value on every coroot. This proves containment of all such representation weights in the weight lattice; it does not hold for unrestricted infinite-dimensional modules.

Let $\mathcal{B}(H)=B(H,\cdot )$ be the Killing form identification of a Cartan subalgebra with its dual. The root-space decomposition gives $B(H,K)={\sum }_{\beta \in \Phi }\beta (H)\beta (K)$, and therefore the displayed root expansion. On the real coroot span its coefficients are real, proving $\mathcal{B}({\mathfrak{h}}_{\mathbb{R}})\subseteq \mathbb{R}P$. For ${\mathfrak{sl}}_3$ and $H_{12}=E_{11}-E_{22}$, the formula gives $\mathcal{B}(H_{12})=6(L_1-L_2)$. The normalization is $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$, not a rescaled trace form.

// Target: semisimple-lie-algebra.bigb

For a root system with $N$ positive roots, put $A_{\nu }(H)={\sum }_{w\in W}\det (w)e^{(w\nu ,H)}$. Its degree-$m$ Taylor coefficient is alternating separately in $\nu$ and $H$, and hence divisible in each set of variables by the product of positive-root linear forms. Thus it vanishes for $m<N$ and its degree-$N$ term is $C{\prod }_{\alpha >0}(\nu ,\alpha ){\prod }_{\alpha >0}(H,\alpha )$. The Weyl denominator formula at $\nu =\rho$ determines the nonzero constant $C$. Taking the ratio $A_{\lambda +\rho }(tH)/A_{\rho }(tH)$ for regular $H$ and letting $t\to 0$ gives the Weyl dimension formula $\dim L(\lambda )={\prod }_{\alpha >0}(\lambda +\rho ,\alpha )/(\rho ,\alpha )$. Factors converting roots to coroots cancel.

Let $N=|R^{+}|$ and identify the real weight and Cartan spaces using a Weyl group-invariant positive inner product. For a regular $H$, meaning $(\alpha ,H)\ne 0$ for every root, specialize $e^{\mu }$ to $e^{t(\mu ,H)}$. Then the value of the formal character at $t=0$ is $\dim L(\lambda )$, but both Weyl alternants in the Weyl character formula vanish there. We compute their leading terms instead of substituting into that zero-over-zero expression.

ExpandFor every root reflection $s$, replacing $\nu$ by $s\nu$ changes the sign of $P_m$, by replacing $w$ with $ws$ in the sum. Replacing $H$ by $sH$ also changes its sign, by replacing $w$ with $sw$. Thus the polynomial vanishes whenever $\nu$ or $H$ lies on any reflecting hyperplane. It is divisible by ${\prod }_{\alpha >0}(\nu ,\alpha )$ in its first variables and by ${\prod }_{\alpha >0}(H,\alpha )$ in its second variables. Distinct positive roots give distinct linear factors; each product has degree $N$.

The polynomial $P_m$ has degree $m$ in each group of variables. Therefore $P_m=0$ for $m<N$, and for $m=N$ it has the formwith a constant $C$ independent of both variables. At $\nu =\rho$, the Weyl denominator formula givesConsequently $C{\prod }_{\alpha >0}(\rho ,\alpha )=1$. For strictly dominant $\nu =\lambda +\rho$, its leading coefficient is nonzero, and the ratio of the two Weyl alternants tends toThe coroot is ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, so the root-length factors cancel in each ratio. This proves the Weyl dimension formula by the lowest-degree alternant proof of the Weyl dimension formula, including the limiting step.

Let $E$ be the real span of the roots in ${\mathfrak{h}}^{\ast }$. The weight lattice iswhere $h_{\alpha }$ is the coroot in the root's sl2 subalgebra associated with a root, normalized by $\alpha (h_{\alpha })=2$. It suffices to test the simple coroots. Equivalently, if ${\omega }_i(h_{{\alpha }_j})={\delta }_{ij}$ defines the fundamental weights, then $P={⨁}_i\mathbb{Z}{\omega }_i$.

To prove integrality of representation weights on coroots, restrict a finite-dimensional module $V$ to the root's sl2 Lie algebra. By the classification of finite-dimensional sl2 representations, the eigenvalues of its Cartan element $h_{\alpha }$ are integers. Thus every $\mu \in {\Lambda }_V$ has $\mu (h_{\alpha })\in \mathbb{Z}$. The simple coroots form a basis of the real Cartan space, so these integral values also put $\mu$ in $E$. Therefore $\boxed{{\Lambda }_V\subseteq {\Lambda }_W}$ for every finite-dimensional representation in this setting. The weight lattice need not coincide with the smaller root lattice.

Work first in a finite-dimensional SU(2) representation, with its invariant positive inner product. In the complexified Lie algebra, choose $H^{†}=H$ and $(E^{-}{)}^{†}=E^{+}$. Let $v_0$ be a highest-weight vector with $H{v}_0=m{v}_0$ and $E^{+}{v}_0=0$, and put $v_r=(E^{-}{)}^r{v}_0$. The commutator with $H$ givesInduction also givesFor $r=1$ this is $E^{+}{E}^{-}{v}_0=H{v}_0=m{v}_0$. For the induction step, use $E^{+}{E}^{-}=E^{-}{E}^{+}+H$ on $v_r$: the coefficient becomes $r(m-r+1)+m-2r=(r+1)(m-r)$. In particularwhich proves the assumed formula rather than merely invoking it.

The norm identity $\Vert E^{-}{v}_r{\Vert }^2=(r+1)(m-r)\Vert v_r{\Vert }^2$ shows that, for nonnegative integer $m$, the ladder is nonzero for $0\le r\le m$ and stops with $E^{-}{v}_m=0$. Distinct eigenvalues make these vectors independent. Their span is stable under $H,E^{\pm }$, and in the irreducible module it is the full space. ThusFor normalized states, $E^{-}|n⟩=\frac{1}{2}\sqrt{(m+n)(m-n+2)}|n-2⟩$, with the last state killed. General finite-dimensional SU(2) representations are direct sums of these ladders.

The representation category matters here. The Lie-algebra relations alone allow an infinite Verma module even at integral highest weight: its nonzero vector $v_{m+1}$ is singular. The finite representation can instead be constructed algebraically by taking the basis $v_0,\dots ,v_m$ with the displayed actions and declaring $E^{-}{v}_m=0$; all three Lie brackets then hold, including at the endpoints. Thus the finite-dimensional assertion uses the integrable SU(2) module, or this explicit finite quotient, rather than assuming that a singular vector must vanish in every highest-weight module.

For rank two, take a Cartan subalgebra $\mathfrak{h}$ with commuting basis elements. A root $\alpha$ is a nonzero linear functional on $\mathfrak{h}$ for which a root space contains an element $E_{\alpha }$ satisfyingChoose simple roots ${\alpha }_1,{\alpha }_2$ and their coroots $H_i={\alpha }_i^{\vee }$. In the PDF's index convention, the Cartan matrix isNormalize $E_i^{\pm }=E_{\pm {\alpha }_i}$ so $[E_i^{+},E_i^{-}]=H_i$. The difference of distinct simple roots is not a root, so the mixed positive-negative bracket is zero. This givesThe transpose in the last formula is essential and is printed correctly in the original PDF.

A highest-weight vector of labels $({\overline{n}}_1,{\overline{n}}_2)$ is killed by both $E_i^{+}$ and is an eigenvector of the two $H_i$. Repeated words in $E_1^{-},E_2^{-}$ generate its highest-weight representation; each lowers the weight by the corresponding simple root. Restricting to the sl2 subalgebra associated with a root for either simple root gives the preceding integer ladder, so finite-dimensional highest labels must be nonnegative integers. The finite irreducible module is the highest-weight quotient imposing $(E_i^{-}{)}^{{\overline{n}}_i+1}v=0$ and the submodules generated by these relations, together with the Serre relations. Choosing independent nonzero lowering words gives a simultaneous weight basis; an extra multiplicity index is needed where a weight space has dimension greater than one.

For the specified B2 root system, the simple-root squared lengths are $1,2$ and their inner product is $-1$, givingThe positive root ${\alpha }_1+{\alpha }_2$ is generated by $[E_1^{+},E_2^{+}]\ne 0$, and $2{\alpha }_1+{\alpha }_2$ by $[E_1^{+},[E_1^{+},E_2^{+}]]\ne 0$. No further positive sums occur, so $(\mathrm{ad}{E}_1^{+}{)}^3{E}_2^{+}=0$ and $(\mathrm{ad}{E}_2^{+}{)}^2{E}_1^{+}=0$. These are the relevant Serre relations. Applying the Cartan commutators to a weight vector proveswhenever the image is nonzero.

To enumerate the requested representation without guessing when arrows stop, realize it concretely as the Adjoint representation. For a Euclidean weight $(x,y)$, its labels are $(n_1,n_2)=(2x,-x+y)$. The highest root $2{\alpha }_1+{\alpha }_2=(1,1)$ has labels $(2,0)$, so it generates the irreducible adjoint module of this simple rank-two Lie algebra. Its nonzero weights are exactly its eight roots, and its zero weight space is the two-dimensional Cartan subalgebra. This gives the B2 adjoint representation weight diagram and the nine distinct states labels:Their actual production by lowering can be seen fromAn arrow labelled $i$ denotes a nonzero action of $E_i^{-}$ between the indicated weight spaces. The two paths into zero produce independent Cartan vectors. Hence there are nine distinct pairs, as requested when degeneracy is disregarded, but the full representation has dimension ten, not nine.

Put $\bar{i}=2n+1-i$ and let $E_{ij}$ denote a matrix unit. The antidiagonal matrix satisfies $J^{-1}=J$, so the orthogonal condition is $A=-J{A}^{T}J$, or entrywise $A_{ij}=-A_{\bar{j}\bar{i}}$. Consequently the diagonal Cartan subalgebra isWe first take $n\ge 2$. Since $[H,E_{ab}]=(H_{aa}-H_{bb})E_{ab}$, the matrix root basis of the even orthogonal Lie algebra givesThe zero weight space is $\mathfrak{t}$; its basis is $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. No vector of weight $2{\epsilon }_i$ occurs, because the corresponding matrix unit is forced to be its own negative. These spaces account for $n+2n(n-1)=n(2n-1)$ dimensions, exhausting the Special orthogonal Lie algebra. Thus its decomposition as a $\mathfrak{t}$-module and its Dn root system areEach nonzero root space has dimension one; the zero weight has multiplicity $n$.

The upper triangular root spaces are precisely those of weights ${\epsilon }_i-{\epsilon }_j$ and ${\epsilon }_i+{\epsilon }_j$ with $i<j$. ThereforeThese $n$ roots form the simple roots. To check that they generate the chosen positive system with nonnegative coefficients, write ${\epsilon }_i-{\epsilon }_j={\sum }_{k=i}^{j-1}{\alpha }_k$. For $j<n$,and for $j=n$ use ${\epsilon }_i+{\epsilon }_n={\sum }_{k=i}^{n-2}{\alpha }_k+{\alpha }_n$. Empty sums are zero. Their linear independence and these expansions identify the base of the root system.

Equip the real span of the roots in ${\mathfrak{t}}^{\ast }$ with the inner product making the ${\epsilon }_i$ orthonormal. All roots then have squared length two. The coroots of ${\epsilon }_i\pm {\epsilon }_j$ are $H_i\pm H_j$, so the fundamental weights of Dn, characterized by ${\omega }_i({\alpha }_j^{\vee })={\delta }_{ij}$, areTaking their pairings with $H_j-H_{j+1}$ and $H_{n-1}+H_n$ verifies the defining identities directly. For each pair $i<j$, the sum of its two positive roots is $2{\epsilon }_i$. Hence the half-sum of positive roots is

The Dynkin diagram has a chain ending at ${\alpha }_{n-2}$, which is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. Every edge is a single bond because the root lengths agree; the nonzero off-diagonal simple-root pairings are minus one. The labelled Dn Dynkin diagram and its low-rank versions are shown below.

For $n=2$, ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_1+{\epsilon }_2$ are orthogonal, giving two disconnected A1 root systems. For $n=3$, the diagram is the chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$, of type A3 root system; the displayed weight formulas still apply. If $n=1$ is allowed, ${\mathfrak{so}}_2=\mathfrak{t}$ is one-dimensional and abelian: there are no roots or fundamental weights, and $\rho =0$.

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.

Use the standard coroot normalization $\alpha (H_{\alpha })=2$. The symbols $H_{\alpha }$ are vectors in the Cartan subalgebra, not the root-space vectors $X_{\alpha }$. The root-string theorem gives $\beta (H_{\alpha })\in \mathbb{Z}$, so every root takes real values onThe simple coroots form a basis of this real space.

In the root-space decomposition, $\mathrm{ad}H$ is zero on $\mathfrak{h}$ and scalar multiplication by $\beta (H)$ on the one-dimensional space ${\mathfrak{g}}_{\beta }$. Therefore, for $H,K\in \mathfrak{h}$,For $H,K\in {\mathfrak{h}}_{\mathbb{R}}$ every term is real, so the restricted form is real. Moreover,If it is zero, all $\beta (H)$ vanish. Then $H$ commutes with every root space and with $\mathfrak{h}$, hence is central in $\mathfrak{g}$. The center of a Lie algebra vanishes for a semisimple Lie algebra, so $H=0$. Thus the Euclidean subspace of a Cartan subalgebra hasEquivalently, positivity follows because the roots span ${\mathfrak{h}}^{\ast }$.

The weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$. The coroots use the normalization fixed above.

The Killing form defines the complex-linear mapBy nondegeneracy of the Killing form on a Cartan subalgebra, this is an isomorphism. For $H\in {\mathfrak{h}}_{\mathbb{R}}$, expand its image in the dual fundamental-weight basis:The coefficients are real by the preceding positivity calculation, and the ${\omega }_i$ are in ${\Lambda }_W$. ConsequentlyIn fact both spaces have real dimension $r$, so this inclusion is equality.

Work with a finite-dimensional complex semisimple Lie algebra. The Dynkin diagram is obtained from its intrinsic root system, with several structure theorems entering the construction.

First choose a Cartan subalgebra $\mathfrak{h}$. The semisimple structure theory gives the root-space decompositionThe nonzero roots span ${\mathfrak{h}}^{\ast }$, root spaces have dimension one, and every root has its negative. The rank is $r=\dim \mathfrak{h}$.

Next use the Killing form. Its nondegeneracy on $\mathfrak{h}$ and its positive-definite restriction to the real coroot span identify the real root span $E=\mathbb{R}\Phi$ with a Euclidean space. Explicitly, if $t_{\lambda }={\mathcal{B}}^{-1}(\lambda )$, put $(\lambda ,\mu )=B(t_{\lambda },t_{\mu })$. The sl2 subalgebra associated with a root supplies its normalized coroot $H_{\alpha }$ and the integral pairingThe root-string theorem makes the reflection $s_{\alpha }(\beta )=\beta -\beta (H_{\alpha })\alpha$ permute $\Phi$. Together with the absence of multiples other than $\pm \alpha$, this shows that $\Phi$ is a finite reduced root system and a crystallographic root system. These properties come from semisimple Lie algebra theory, not from the diagram's definition.

Choose a vector in $E$ that is not perpendicular to any root. Define the positive roots by positive inner product with this vector. The simple roots are those positive roots that cannot be expressed as sums of two positive roots. Root-system theory shows that they form a basis $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, and each root is an integer combination of them with coefficients all of one sign. Distinct simple roots have nonpositive inner product.

Use the standard Cartan matrix conventionThere is one diagram vertex for each simple root. For $i\ne j$, the integral off-diagonal entries are nonpositive andConnect the two vertices with respectively zero, one, two or three edges. The corresponding angles are $90^{\circ },120^{\circ },135^{\circ },150^{\circ }$. A multiple edge carries an arrow toward the shorter root. For joined vertices, the length ratio is determined by $A_{ij}/A_{ji}=({\alpha }_i,{\alpha }_i)/({\alpha }_j,{\alpha }_j)$; hence the diagram records both angle and relative length.

For example, ${\mathfrak{sl}}_3$ has simple roots $L_1-L_2$ and $L_2-L_3$, giving two equal-length vertices joined by one edge: type $A_2$. For ${\mathfrak{sp}}_4$, take ${\alpha }_1=L_1-L_2$ and ${\alpha }_2=2L_2$. Thenso the two vertices have a double edge directed toward ${\alpha }_1$, the shorter root: type $C_2$.

Finally, orthogonal irreducible components of the root system correspond to the simple ideals of a Lie algebra in the semisimple Lie algebra, so its diagram is the disjoint union of their connected diagrams. Conjugacy of Cartan subalgebras and the Weyl group action on choices of positive roots make the resulting diagram independent of these choices up to isomorphism. Thus the vertices, edge multiplicities, arrows and connected components encode the simple-root geometry and the simple-ideal decomposition.