Codex @codex  0

For the B2 root system with ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ long and ${\alpha }_2={\epsilon }_2$ short, the spin fundamental weight is ${\Lambda }_2=({\epsilon }_1+{\epsilon }_2)/2$. Its four-dimensional highest-weight representation has the minuscule weight orbit $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$, each of multiplicity one. The crystal basis is the chain $++\stackrel{2}{\to }+-\stackrel{1}{\to }-+\stackrel{2}{\to }--$, where signs denote the coordinates. Each Kashiwara operator subtracts its indicated simple root; the sl2 Lie algebra strings have length one. This fixes every edge and gives a connected highest-weight graph with the correct character.

Give the real root span an inner product with ${\epsilon }_1,{\epsilon }_2$ orthonormal. The B2 root system consists of four short roots on the coordinate axes and four long roots on the diagonals. Choose the long-first simple rootsThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2={\epsilon }_1$, and ${\alpha }_1+2{\alpha }_2={\epsilon }_1+{\epsilon }_2$. The corresponding simple reflections areThey generate every signed permutation of the two coordinates. Thus the Weyl group has order eight and is the symmetry group of a square; every element sends $(x,y)$ to $(\pm x,\pm y)$ or $(\pm y,\pm x)$, with independent signs.

Since ${\alpha }_1,{\alpha }_2$ form an integral basis of ${\mathbb{Z}}^2$, the root lattice is $Q=\mathbb{Z}{\epsilon }_1\oplus \mathbb{Z}{\epsilon }_2$. The simple coroots are ${\alpha }_1^{\vee }={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2^{\vee }=2{\epsilon }_2$. Therefore the weight lattice isSolving $⟨{\Lambda }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsThe cone of dominant weights isIndeed $a=x-y$ and $b=2y$ are exactly the two simple-coroot pairings. This describes representations of the Lie algebra, or of its simply connected Spin group; half-integral weights are allowed. For representations descending to $\mathrm{SO}(5)$ itself, the weight must lie in ${\mathbb{Z}}^2$, equivalently $b$ must be even.

For a real inner product space $V={\mathbb{R}}^5$, defineThis map takes values in the Special orthogonal Lie algebra: its matrices are skew-symmetric. It sends $e_i\wedge e_j$ to $E_{ij}-E_{ji}$, so it is an isomorphism between the two ten-dimensional spaces. For $g\in SO(5)$, invariance of the inner product givesThis proves the exterior square realization of the orthogonal adjoint representation, including equivariance, rather than merely matching dimensions.

After complexification, part (a) is therefore the weight decomposition of the Adjoint representation. Its zero weight space is the complex centralizer of $\mathfrak{t}$. The two zero-weight wedges are nonzero multiples of $e_1\wedge e_2$ and $e_3\wedge e_4$, which map under $\Phi$ to the two infinitesimal rotation generators of $T$. HenceIf a larger torus contained $T$, its Lie algebra would commute with $\mathfrak{t}$ and therefore lie in $\mathfrak{t}$. Equality of Lie algebras and connectedness of both tori force equality of the groups. Thus $T$ is a maximal torus.

Let ${\epsilon }_j$ denote the real coordinate functional $x_j$. In the real weight convention, with the common factor $i$ removed from the differentiated torus weights, the root system isThese are the roots of a root system of type B2 root system. In the imaginary-valued compact Lie-algebra convention each displayed functional is multiplied by $i$. Every root space is one-dimensional, as part (a) explicitly shows.

The root hyperplanes are $x_1=0$, $x_2=0$, $x_1=x_2$ and $x_1=-x_2$. Their reflections are coordinate sign changes, coordinate interchange and a signed interchange. They generate all signed permutations of the two coordinates:Geometrically this is the dihedral symmetry group of a square, with order eight. It preserves both the four short roots and the four long roots of the B2 root system.

Representatives in $SO(5)$ can be given explicitly. The matrixhas determinant one and conjugates the first rotation block to its inverse, giving $(x_1,x_2)\mapsto (-x_1,x_2)$. The matrix $D_2=\mathrm{diag}(1,1,1,-1,-1)$ similarly reverses $x_2$. Flipping the fifth coordinate compensates for the determinant of the planar reflection.

The permutation matrixswaps the two coordinate planes. Its permutation is $(13)(24)$, so its determinant is also one. Conjugation gives $(x_1,x_2)\mapsto (x_2,x_1)$, the reflection in $x_1=x_2$. These elements normalize $T$, and $P,D_1$ generate all eight Weyl actions. For example, $P{D}_1$ induces $(x_1,x_2)\mapsto (x_2,-x_1)$, a quarter-turn, and $D_1{D}_2$ induces the half-turn. The representatives supply the specified reflections and their typical products inside the actual special orthogonal group.

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 double-bond arrow points to the middle node in the PDF. Since $\Vert \alpha \Vert =1$ and $\Vert \beta \Vert =\sqrt{2}$, the short root $\alpha$ must occupy that middle node and the long root $\beta$ the left node. The right node $\gamma$, joined to $\alpha$ by a single bond, is another short root. Thus the diagram is C3 root system type, labelled as follows.

The bonds require $\gamma \cdot \beta =0$, $\gamma \cdot \alpha =-1/2$ and $\Vert \gamma {\Vert }^2=1$. Writing $\gamma =(a,b,c)$ gives $a=-1/2$, $b=a$, and $c^2=1/2$. One choice is thereforeThe opposite choice of the last coordinate reflects the entire realization across the first two coordinate axes' plane and is equally valid.

To enumerate exhaustively, set $f_1=(0,0,1)$, $f_2=(1,1,0)/\sqrt{2}$ and $f_3=(-1,1,0)/\sqrt{2}$. These are orthonormal, andThe complete C3 root system is $\{\pm \sqrt{2}{f}_i,(\pm f_i\pm f_j)/\sqrt{2}:i<j\}$, with eighteen roots. Its nine roots with nonnegative coefficients in this simple basis are exactlyThere are six short and three long positive roots, and the other nine roots are their negatives.

The single-bond rank-two diagram is A2 root system type. Three suitable positive simple pairs for its root subsystems areEach pair $p,q$ has squared lengths one and inner product $-1/2$, so its angle is $120$ degrees. Its plane meets the full root set in $\pm p,\pm q,\pm (p+q)$, with $p+q$ respectively $\alpha +\gamma$, $2\alpha +\beta +\gamma$ and $\alpha +\beta +\gamma$. These exhaust the roots of the plane, and their coordinates in the pair have one sign. Thus each pair is genuinely a simple basis of an $A_2$ subsystem.

The double-bond diagram is rank-two B2 root system type, equivalently $C_2$. Three suitable pairs, written short root first, areFor each pair $s,l$, $\Vert s{\Vert }^2=1$, $\Vert l{\Vert }^2=2$ and $s\cdot l=-1$. The remaining positive roots in its plane areTogether with their negatives these are precisely the eight plane roots, and all four positive roots have nonnegative integral pair-coordinates. Hence the displayed pairs are simple bases of the required subsystems. The rank-two subsystems of a C3 root system description explains these as the three coordinate-plane subsystems in the orthonormal $f_i$ realization; the chosen $A_2$ pairs lie in three different diagonal planes.

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.

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.

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.

The diagonal Cartan subalgebra isPut ${\epsilon }_1=L_1$ and ${\epsilon }_2=L_2$; then $L_3=-{\epsilon }_1$, $L_4=-{\epsilon }_2$ and $L_5=0$ on $\mathfrak{h}$. For the matrix unit $E_{ij}$, $[H,E_{ij}]=(L_i-L_j)(H)E_{ij}$. Applying this to the general matrix in part (i) gives the following one-dimensional root spaces and chosen root vectors:Each row satisfies $[H,X_{\alpha }]=\alpha (H)X_{\alpha }$ and belongs to the Special orthogonal Lie algebra. The two diagonal parameters and eight root vectors span the ten-parameter matrix from part (i), giving the explicit root-space decompositionThis is the B2 root system, with short roots on the coordinate axes and long roots on the diagonals.

The special orthogonal group in five dimensions isIts Lie algebra consists of real skew-symmetric matrices. There are $5\cdot 4/2=10$ independent entries above the diagonal, so $\dim SO(5)=10$. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the determinant condition chooses a component without changing the dimension.

Fix the fifth coordinate. The matrices $\mathrm{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as ${\mathbb{R}}^4\oplus \mathbb{R}{\mathbf{e}}_5$, so its branching rule is $5\downarrow SO(4)=4\oplus 1$. An element of the so5 Lie algebra can be written uniquely asConjugation by $\mathrm{diag}(S,1)$ sends $B$ to $SB{S}^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional Adjoint representation of a Lie algebra of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the SO5 to SO4 branching gives

For the left SU(2) subgroup of SO(4), identify ${\mathbb{R}}^4$ with the quaternions. Left multiplication by a unit quaternion is a real orthogonal transformation and gives an embedded SU(2) group. More generally, $q\mapsto aq{b}^{-1}$ gives the double cover $SU(2{)}_L\times SU(2{)}_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(2,2)$ and the Adjoint representation is $(3,1)\oplus (1,3)$. Restricting to the left factor turns the right factor into a multiplicity space. ThereforeThese are decompositions into complex irreducible representations; the real $4$ is the underlying real representation of a quaternionic doublet. Combining the branching rules gives

Write a weight as $\lambda =(x,y)$. The integrality conditions for the B2 root system give $2x\in \mathbb{Z}$ and $y-x\in \mathbb{Z}$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in \mathbb{Z}$. ThusThis is the B2 weight lattice, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight roots of a root system areThe short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha +\beta$, $2\alpha +\beta$.

The integrality conditions determine the weight lattice of the Lie algebra, equivalently of the simply connected Spin group $\mathrm{Spin}(5)$. For the global special orthogonal group $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine group representation requires integer $x,y$. Thus the half-integer coset contains spin representations that do not descend to $SO(5)$. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.

The root system of the displayed $SO(4)$ subgroup is $\{\pm (1,1),\pm (1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta =(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its coroot pairs with a weight asThis demonstrates the diagonal-root SU(2) embedding in SO(5) directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.

For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb{C}$, the two planes give opposite pairs of weights, while the fifth coordinate is fixed. Hence its weight diagram iswith every weight multiplicity equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two SU(2) representations of dimension two and one singlet.

For the Adjoint representation, the root-space decomposition has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight Cartan subalgebra. ThusThe coroot values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm 2$ states to make a triplet; the $\pm 1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier branching rule and accounts for all ten dimensions.

Use the wiki's Cartan matrix conventionLabel the long simple root first when the lengths differ. The matrix of an irreducible rank-two root system has the form $A=\left(\begin{array}{cc}2& -a\\ -b& 2\end{array}\right)$, where $a,b$ are positive integers. Integrality and the signs follow from the Cartan integers for distinct simple roots, and irreducibility rules out a zero off-diagonal pair. Its positive-definite symmetrization gives $ab<4$, equivalently $ab=4{\cos }^2\theta <4$. Moreover $a/b=\Vert {\alpha }_1{\Vert }^2/\Vert {\alpha }_2{\Vert }^2\ge 1$. Thus $(a,b)=(1,1),(2,1),(3,1)$, giving the entire classification of rank-two root systems relevant to a simple algebra:These correspond to ${\mathfrak{sl}}_3(\mathbb{C})$, ${\mathfrak{so}}_5(\mathbb{C})\cong {\mathfrak{sp}}_4(\mathbb{C})$, and the exceptional algebra ${\mathfrak{g}}_2$. Here $B_2=C_2$ denotes the Isomorphism between so5 and sp4, not an additional case. The disconnected $A_1\times A_1$ system corresponds to a semisimple but nonsimple algebra and is excluded.

The Dynkin diagrams, with nodes in the same order as the matrices, areThe second and third diagrams have two and three bonds respectively, with the arrowhead pointing to the short root ${\alpha }_2$. In the last diagram $\equiv$ draws the three bonds and $>$ their arrowhead.

To enumerate roots, use the standard root string theorem: for $\beta \ne \pm \alpha$, the roots $\beta +n\alpha$ are consecutive from $\beta -p\alpha$ through $\beta +q\alpha$, with $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. Also use the standard results that the root system is reduced, every root is conjugate under the Weyl group to a simple root, and each root space has dimension one.

Write $m=a\in \{1,2,3\}$. Since a difference of simple roots is not a root, the ${\alpha }_2$-string starting at ${\alpha }_1$ has $p=0$ and $q=m$. It produces ${\alpha }_1,{\alpha }_1+{\alpha }_2,\dots ,{\alpha }_1+m{\alpha }_2$. The ${\alpha }_1$-string starting at ${\alpha }_2$ has $q=1$. In type $G_2$, the further string through $\beta ={\alpha }_1+3{\alpha }_2$ hasand $\beta -{\alpha }_1=3{\alpha }_2$ is not a root, so it also produces $2{\alpha }_1+3{\alpha }_2$. The resulting positive roots areAll roots are these and their negatives. To check completeness, the simple root reflections act on coordinates byEach listed set together with its negatives is stable under both reflections. It contains the simple roots and consists of roots already forced by strings. Since every root lies in a Weyl orbit of a simple root, no other roots can occur. This proves the lists for the A2 root system, B2 root system, and G2 root system. The root-space decomposition then gives