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First show that the Cartan subalgebra is orthogonal, for the Killing form, to every nonzero root space. For $H\in \mathfrak{h}$ and $X\in {\mathfrak{g}}_{\alpha }$, choose $H_0\in \mathfrak{h}$ with $\alpha (H_0)\ne 0$. Invariance and the abelian nature of $\mathfrak{h}$ giveHence $B(H,X)=0$.

Now let $H$ be in the radical of the restriction to $\mathfrak{h}$, so $B(H,H^{\prime })=0$ for every $H^{\prime }\in \mathfrak{h}$. The preceding argument also gives $B(H,{\mathfrak{g}}_{\alpha })=0$ for every root. By the root-space decomposition, $H$ is orthogonal to all of $\mathfrak{g}$. The assumed nondegeneracy on $\mathfrak{g}$ forces $H=0$. Therefore the Killing form restricted to the Cartan subalgebra is nondegenerate.

Here the Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra is a maximal toral subalgebra: it is abelian, its adjoint operators are simultaneously diagonalizable, and its centralizer is itself. These standard properties give the root-space decompositionThe roots of a root system $\alpha$ are the nonzero linear functionals in ${\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }\ne 0$; $\Phi$ is their finite set. Each root space is one-dimensional. The zero weight space is the centralizer of $\mathfrak{h}$, hence exactly $\mathfrak{h}$. Thus the displayed sum is a direct sum of simultaneous eigenspaces of the Adjoint representation. It is the Cartan root-space decomposition meant here.

The bracket respects weight addition: $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, where ${\mathfrak{g}}_0=\mathfrak{h}$ and a missing weight means the zero space. This follows by applying the Jacobi identity with $H\in \mathfrak{h}$ to the bracket of two root vectors. It explains how the vector-space decomposition encodes the Lie algebra structure.

For the usual diagonal Cartan subalgebra, write ${ϵ}_i(H)=H_{ii}$. The roots are ${ϵ}_i-{ϵ}_j$ for $i\ne j$. The three diagonal entries of the specified $H_0$ have orderingTherefore the positive roots selected by $\ell (\mu )=\mu (H_0)$ areTheir positive evaluations are respectively $2-\sqrt{2}$, $2\sqrt{2}-1$ and $1+\sqrt{2}$. For this nonstandard ordering the simple roots are ${\alpha }_1={ϵ}_2-{ϵ}_3$ and ${\alpha }_2={ϵ}_3-{ϵ}_1$; the third positive root is their sum. These simple roots differ from the usual choice used in Question 1.

Identify the real weight plane with triples $(u_1,u_2,u_3)$ of sum zero. Pairing with these simple coroots gives the differences $u_2-u_3$ and $u_3-u_1$. Consequently the fundamental chamber of a root system isIts closure replaces the inequalities by weak inequalities. The fundamental weights for this ordering are ${\omega }_1={ϵ}_2$ and ${\omega }_2=-{ϵ}_1$, represented by $(-1/3,2/3,-1/3)$ and $(-2/3,1/3,1/3)$. Their simple-coroot pairings are $(1,0)$ and $(0,1)$, so the two boundary rays areThe sketch uses ${ϵ}_1=(1,0)$, ${ϵ}_2=(-1/2,\sqrt{3}/2)$ and ${ϵ}_3=(-1/2,-\sqrt{3}/2)$ in a Euclidean plane. The shaded sector is the sixty-degree closed chamber; its interior is the open chamber. The ordering vector $H_0$, identified with its dual direction, lies inside it.

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.

Write the affine modes as $J_n^a=T_n^a$, and put $Q=Q^{\mathfrak{g}}$. The Sugawara construction requires a restricted module, meaning that sufficiently large positive current modes annihilate each fixed vector, and a noncritical level $2k+Q\ne 0$. A positive-energy highest-weight representation has the required restriction. DefineNormal ordering puts positive modes to the right, making each sum finite on a fixed finite-energy vector. The invariant sum over the current index $a$ is the quadratic Casimir operator construction for the affine current algebra.

Here is the normal ordering calculation responsible for its denominator. In commuting $S_m$ with $J_n^b$, the two affine central terms supply $-2kn{J}_{m+n}^b$. The remaining terms contain $i{f}^{abc}$ times the two ways to replace a current by $J_{p+n}^c$. Relabeling the summation index and using antisymmetry of $f^{abc}$ cancels the un-reordered terms. Reordering the modes that cross the creation/annihilation boundary leaves $n$ signed terms, whose two structure constants contract by $f^{abc}{f}^{abd}=Q{\delta }^{cd}$. Their sum is $-Qn{J}_{m+n}^b$. ThusThe adjoint-Casimir contribution is genuinely quantum. For an explicit check of its coefficient, let $|0⟩$ be the affine vacuum, with all $J_n^a|0⟩=0$ for $n\ge 0$. ThenThe first term acts by the adjoint Casimir operator $Q$, and only $p=1$ contributes $2k$ to the second. This also verifies that a current has conformal weight one under the displayed normalization.

Now commute the quadratic generators with one another. Acting on either current factor produces $(m-n)L_{m+n}^{\mathfrak{g}}$; the residual full contractions are scalar and vanish unless $m+n=0$. The Jacobi identity fixes their cubic dependence on $m$, and the plane vacuum convention $L_{-1}|0⟩=L_0|0⟩=L_1|0⟩=0$ fixes the combination $m^3-m$. To determine its coefficient explicitly, noteCommuting a single $J_1^a$ through the numerator gives $(2k+Q)J_{-1}^a|0⟩$: the two central contractions give $2k{J}_{-1}^a$, and the zero-mode commutator in the structure-constant term gives $Q{J}_{-1}^a$. The remaining $J_1^a$ contraction gives $k$ for each adjoint index. ThereforeFor an algebraic module without a positive inner product, the same calculation is its vacuum matrix element and fixes the universal central term. ConsequentlyThis is a representation of the Virasoro algebra, not merely its centerless part.

For a complex semisimple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra of semisimple elements; equivalently it is the complexification of a maximal torus algebra in the compact real form. Its dimension is the rank of a semisimple Lie algebra, $r$. Choose an orthonormal Cartan basis with currents $H_n^i$. They have no structure-constant term:For $k\ne 0$, their abelian Sugawara construction isEach Cartan current has weight one under both $L^{\mathfrak{g}}$ and $L^{\mathfrak{h}}$, so $K_m=L_m^{\mathfrak{g}}-L_m^{\mathfrak{h}}$ commutes with all $H_n^i$, hence also with $L_n^{\mathfrak{h}}$. Subtracting the two commuting stress tensors yields the Coset constructionIn the intended nontrivial compact unitary theory at positive admissible level, this coset acts on a positive inner-product space, with $K_m^{†}=K_{-m}$. Its vacuum satisfies $K_n|0⟩=0$ for $n\ge -1$, soThus $c^{\mathfrak{g}}\ge r$. Also $Q>0$ for a compact simple Lie algebra and $k>0$, so $2k/(2k+Q)<1$, which proves the upper bound directly. Hence, under these physical representation hypotheses,The upper inequality is strict at finite positive level; $c^{\mathfrak{g}}$ tends to $\dim \mathfrak{g}$ as the level grows.

The positivity hypothesis is necessary and is not implied by the bare affine commutation relation. For example, at $k=0$ its trivial representation has all currents zero and gives $c^{\mathfrak{g}}=0<\mathrm{rank}\mathfrak{g}$. Even positive arbitrary levels need not satisfy the bound: with $\mathfrak{g}=\mathfrak{su}(2)$, $f^{abc}={ϵ}^{abc}$, $Q=2$ and $k=1/4$, the formula gives $c^{\mathfrak{g}}=3/5<1$. Such a level does not give the intended compact unitary positive-energy theory. Thus the last bound is a statement about nontrivial unitary admissible-level representations; it cannot hold for every representation in the unrestricted algebraic formulation. At $2k+Q=0$ the displayed Sugawara construction itself is undefined.

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 $\rho =0$.

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, choose a Cartan subalgebra $\mathfrak{t}$ and a set $R^{+}$ of positive roots, and write $W$ for the Weyl group. Let $\lambda$ be a dominant integral weight and $L(\lambda )$ the finite-dimensional irreducible highest-weight representation with that highest weight. If $L(\lambda {)}_{\mu }$ denotes its weight space of weight $\mu$, its formal character is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The symbols $e^{\mu }$ belong to the group algebra of the weight lattice and multiply by $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$; they are formal symbols, not matrix exponentials.

Put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the half-sum of positive roots, and let $\ell (w)$ be the Coxeter length of $w$, the least number of simple reflections expressing it. Then $(-1{)}^{\ell (w)}=\det (w)$ in the real reflection representation. The Weyl character formula statesBy the Weyl denominator formula, the denominator is $e^{\rho }{\prod }_{\alpha \in R^{+}}(1-e^{-\alpha })$, giving the equivalent expressionThe quotient can be interpreted in the fraction field of the formal group algebra; the theorem says it simplifies to the finite formal character above. In particular the apparent denominator does not mean that the representation has infinitely many weights.

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.

Let $e_1,e_2,e_3$ be the basis of the defining representation and $y_1,y_2,y_3$ its dual. On a diagonal element of the Cartan subalgebra their weights are $L_i$ and $-L_i$, with $L_1+L_2+L_3=0$. Use the tensor product and symmetric square basisIts weight is $L_i-L_j-L_k$. For the three repeated weights putIf $\{i,l\}=\{1,2,3\}\setminus \{j\}$, the complete weight-space decomposition isThere are six spaces of the first type, three of the second, and three of the third. Their dimensions sum to $6+3+9=18$. The weight diagram below labels every space with all of its basis vectors.

For the Adjoint representation, the root-space decomposition gives the eight nonzero weightseach of multiplicity one, and weight zero of multiplicity two from the Cartan subalgebra. Thus its weight diagram is the root diagram with a double zero weight, and its total dimension is $8+2=10$.

In the defining representation, the diagonal element of the Cartan subalgebra acts on the four standard basis vectors with eigenvalues $r,s,-r,-s$. ThereforeTheir weight diagram is the four-point cross below.

For $H=\mathrm{diag}(r,s,-r,-s)$, put $L_1(H)=r$, $L_2(H)=s$. Then $L_3=-L_1$, $L_4=-L_2$. The matrix unit identityreads off the root spaces directly. The complete list isThus the C2 root system isAll eight root spaces have dimension one. The zero-weight space is the two-dimensional Cartan subalgebra $\mathbb{C}(E_{11}-E_{33})\oplus \mathbb{C}(E_{22}-E_{44})$; zero itself is not a root.

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.