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The Adjoint representation of a Lie algebra acts on the Lie algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity givesThus it is a Lie algebra representation. In the basis $T_a$ with $[T_a,T_b]=c^c{}_{ab}{T}_c$, its matrices areThe column index labels the basis vector being acted on; this convention fixes the matrix signs.

Work in finite dimension over characteristic zero, as in the usual real or complex gauge algebras. A simple Lie algebra is nonabelian and has no nonzero proper Lie algebra ideal. A semisimple Lie algebra has no nonzero solvable ideal; equivalently it is a direct sum of simple ideals. The nondegenerate Killing form is $\kappa (X,Y)=\mathrm{tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. Its invariance follows from the representation identity and trace cyclicity:

For the reduction, take a nonzero proper ideal $\mathfrak{a}$, if one exists, and define its Killing-orthogonal complement ${\mathfrak{a}}^{\perp }$. Invariance makes it an ideal: for $y\in {\mathfrak{a}}^{\perp }$, $x\in L$, $z\in \mathfrak{a}$, $\kappa ([x,y],z)=-\kappa (y,[x,z])=0$. Moreover $[\mathfrak{a},{\mathfrak{a}}^{\perp }]=0$, because for $u\in \mathfrak{a}$, $v\in {\mathfrak{a}}^{\perp }$, and any $x\in L$,and the global Killing form is nondegenerate. Their intersection is consequently an abelian ideal: its elements lie in both commuting ideals. Semisimplicity forces that intersection to be zero. The dimension formula for a nondegenerate bilinear form then givesThe summands commute and are semisimple, since a solvable ideal in either summand would also be an ideal of $L$. Repeating on smaller summands terminates in simple ideals. This proves the orthogonal ideal splitting for a nondegenerate Killing form and describes the required reduction; abelian one-dimensional factors do not occur in a semisimple algebra.

If $\kappa (T_a,T_b)=\ell {\delta }_{ab}$ with $\ell \ne 0$, invariance givesso $c_{abc}=-c_{acb}$ after lowering the output index with $\delta$. The bracket already gives $c_{abc}=-c_{bac}$. These two adjacent transpositions generate all permutations, proving that the structure constants of a Lie algebra are completely antisymmetric. This is the antisymmetry of Killing-lowered structure constants; the scalar $\ell$ cancels in the chosen normalization.

For the gauge calculation write $\Lambda ={\lambda }_a{t}_a$, $A_{\mu }=A_{\mu a}{t}_a$, and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The representation generators here obey $[t_a,t_b]=c_{abc}{t}_c$; factors of $i$ and the gauge coupling are absorbed in this connection convention. Expanding the varied derivative givesFor this to equal $\Lambda D_{\mu }\varphi$, choose the gauge transformation in the derivative-plus-connection conventionThen all extra terms cancel, proving covariance of the gauge covariant derivative for arbitrary space-dependent parameters.

Apply both derivatives to a test field. The second derivatives commute and the terms involving one derivative of the field cancel, leaving the Yang-Mills field strengthThus its components areThe already obtained variation can be written $\delta D_{\mu }=[\Lambda ,D_{\mu }]$. The operator Jacobi identity then yieldsorFinally, with fixed space-time metric,The contracted product is symmetric in $a,c$, while $c_{abc}$ is antisymmetric in those indices. Hence the Yang-Mills action density is gauge invariant, with the sign convention derived rather than assumed.

For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

Let $V$ be a nonzero finite-dimensional Irreducible Lie algebra representation. Since $c$ is central, its action is a scalar $\lambda I$ by the Schur lemma. Taking the trace of $[\rho (p),\rho (q)]=\rho (c)$ gives $0=\lambda \dim V$, so $\lambda =0$.

The actions of $p,q$ now commute. Choose an eigenvalue $a$ of $\rho (p)$. Its nonzero eigenspace is preserved by $\rho (q)$, which has an eigenvector there, with some eigenvalue $b$. The span of this common eigenvector is invariant under all of $\mathfrak{g}$. Irreducibility forces it to be the whole representation. ConsequentlyConversely each such assignment is a one-dimensional Lie algebra representation and hence irreducible. Distinct pairs give nonisomorphic representations. This classifies all finite-dimensional irreducible representations; none is faithful because all annihilate $c$.

On ${\mathbb{C}}^3$ take the matrix unitsTheir products give $[E_{12},E_{23}]=E_{13}$, while $E_{13}$ commutes with each of the other two. Thus this is a Lie algebra representation. The three images are linearly independent, so its kernel is zero: it is faithful and finite-dimensional. It is reducible, as expected from the classification in part ii; for example, the first coordinate line is invariant.

For a finite-dimensional Lie algebra, the Killing form is defined using the Adjoint representation on the algebra itself, rather than the given Lie algebra representation $d$:The Jacobi identity gives ${\mathrm{ad}}_{[X,Y]}=[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]$. Using the cyclic property of the trace,The same trace identity makes the Killing form symmetric. For a basis $(t_a)$, the lowered structure constants of a Lie algebra are $c_{abc}=\kappa ([t_a,t_b],t_c)$. Antisymmetry of the Lie bracket gives $c_{abc}=-c_{bac}$, while invariance gives $c_{abc}=\kappa (t_a,[t_b,t_c])=-c_{acb}$. The two adjacent transpositions generate every permutation, proving the antisymmetry of Killing-lowered structure constants: $c_{abc}$ is totally antisymmetric, even if the Killing form is degenerate.

A semisimple Lie algebra has no nonzero solvable ideal; equivalently, its solvable radical is zero, or, in characteristic zero, its Killing form is nondegenerate. A real compact Lie algebra is the algebra of some compact Lie group, equivalently an algebra admitting a positive-definite invariant bilinear form on a Lie algebra. For a real semisimple Lie algebra the compactness criterion from the Killing form says precisely that $\kappa$ is negative-definite. A compact algebra with a nonzero abelian center has a degenerate Killing form, so the semisimplicity qualification matters.

For a semisimple Lie algebra, let ${\kappa }^{ab}$ denote the inverse matrix of ${\kappa }_{ab}=\kappa (t_a,t_b)$. Its Casimir operator in $d$ isThis contraction is independent of the chosen basis. To prove centrality directly, put $[X,t_b]=A^a{}_b{t}_a$. Invariance gives $A^{\mathsf{T}}\kappa +\kappa A=0$, hence $A{\kappa }^{-1}+{\kappa }^{-1}{A}^{\mathsf{T}}=0$. The Lie algebra representation property and the product rule for a commutator now giveIndeed, the coefficient of $d(t_c)d(t_b)$ in this expression is $(A{\kappa }^{-1}+{\kappa }^{-1}{A}^{\mathsf{T}}{)}^{cb}$. Thus $[d(X),C]=0$ for every $X$. No irreducibility of $d$ is required.

The matrix units obey $E_{ij}{E}_{pq}={\delta }_{jp}{E}_{iq}$. Consequently the general linear Lie algebra $\mathcal{M}={\mathfrak{gl}}_n(\mathbb{R})$ hasFor its Killing form, write $L_X(Z)=XZ$ and $R_X(Z)=ZX$, so that ${\mathrm{ad}}_X=L_X-R_X$. Left multiplication acts identically on each of the $n$ columns, and right multiplication identically on each of the $n$ rows. Thus $\mathrm{Tr}(L_X{L}_Y)=\mathrm{Tr}(R_X{R}_Y)=n\mathrm{tr}(XY)$. The diagonal coefficient of $L_X{R}_Y$ on $E_{ij}$ is $X_{ii}{Y}_{jj}$, giving $\mathrm{Tr}(L_X{R}_Y)=\mathrm{tr}X\mathrm{tr}Y$; the other mixed trace is identical. This proves the Killing form of the general linear Lie algebra:The scalar matrices form a nonzero central abelian ideal, so $\mathcal{M}$ is not semisimple for any $n\ge 1$. More explicitly, the radical of the Killing form is $\mathbb{R}I$, since vanishing against every $Y$ forces $nX-(\mathrm{tr}X)I=0$ by nondegeneracy of the trace pairing. For $n\ge 2$, take $H=E_{11}-E_{22}$. Then $\kappa (H,H)=4n>0$, whereas a positive-definite invariant inner product would make every adjoint map skew and hence give $\kappa (X,X)\le 0$. Therefore $\mathcal{M}$ is not compact when $n\ge 2$. For the allowed exceptional value $n=1$, the abstract abelian algebra $\mathbb{R}$ is a compact Lie algebra, being isomorphic to the algebra of $U(1)$; the particular matrix group $G{L}_1(\mathbb{R})$ is nevertheless noncompact. This distinguishes compactness of an algebra from that of a chosen group realizing it.

For the final algebra, the printed basis specifies strictly upper triangular matrices. If $i<j$ and $p<q$, the first term in $[E_{ij},E_{pq}]$ can be nonzero only when $i<j=p<q$, and then $E_{iq}$ is strictly upper triangular; the second can be nonzero only when $p<q=i<j$, and then $E_{pj}$ is strictly upper triangular. Bilinearity proves closure under the Lie bracket. To compute its own Killing form, grade $E_{ij}$ by the positive integer $j-i$. Bracketing with any basis element increases that degree whenever the result is nonzero. In a basis ordered by degree, all adjoint maps are strictly triangular, as are their products. Every such product has zero trace, so the Killing form of a nilpotent Lie algebra vanishes here. Therefore the subspace denoted by a prime in the PDF isThis is the defined radical of the Killing form, not the derived algebra; for $n=1$ both $\mathcal{L}$ and this radical are zero.

A highest-weight vector is a nonzero vector $v_0$ satisfying $J_3{v}_0=j{v}_0$ and $J_{+}{v}_0=0$; its eigenvalue $j$ is the highest weight. Existence does not require the Lie algebra representation to be antihermitian. Since $V$ is a nonzero finite-dimensional complex vector space, $J_3$ has an eigenvector $v$ with some eigenvalue $\lambda$. The commutator with $J_{+}$ shows that each nonzero $J_{+}^{k}v$ has eigenvalue $\lambda +k$. Infinitely many such vectors would have distinct eigenvalues and be linearly independent, which is impossible. The last nonzero vector in this raising chain is therefore a highest-weight vector $v_0$.

Put $v_{\ell }=J_{-}^{\ell }{v}_0$. The ladder operator relations give $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. The raising coefficient follows by expanding the commutator of $J_{+}$ with a power of $J_{-}$:The factor $1/2$ is essential: these ladder operators have $[J_{+},J_{-}]=J_3$, rather than the convention with $2J_3$. The same finite-dimensionality argument gives a last nonzero lowered vector $v_r$, with $J_{-}{v}_r=v_{r+1}=0$. Applying $J_{+}$ to $v_{r+1}=0$ yieldsso $2j=r\in {\mathbb{Z}}_{\ge 0}$. All $v_0,\dots ,v_r$ are nonzero and have distinct eigenvalues. Their span is invariant under $J_3,J_{+},J_{-}$ and therefore under the original three generators. Irreducibility of the Lie algebra representation makes this span equal to $V$. HenceIn particular, the initial construction produces the largest eigenvalue of $J_3$, since its full spectrum is now $j,j-1,\dots ,-j$. This establishes the highest weight without assuming beforehand that all eigenvalues are real or that $J_3$ is diagonalizable.

For the additional antihermitian Lie algebra representation, the given definitions imply $J_3^{†}=J_3$ and $J_{+}^{†}=J_{-}$. Thus distinct $J_3$ eigenvalues give orthogonal weight vectors. With $\Vert v_0\Vert =1$, write $N_{\ell }=\Vert v_{\ell }{\Vert }^2$. The adjoint relation and the raising coefficient giveMultiplying this recursion proves the normalized highest-weight lowering formula in the present normalization:Every factorial argument is an integer, even for half-integral $j$. The displayed states have unit norm, mutually distinct eigenvalues, and span $V$, so they form an orthonormal basis. Their phase convention also fixes the positive ladder amplitudeswith the endpoint actions interpreted as zero. These amplitudes will determine the Clebsch-Gordan coefficients.

The SU(3) Lie algebra is the compact real algebra of traceless skew-Hermitian matrices; its complexification is ${\mathfrak{sl}}_3(\mathbb{C})$, a special linear Lie algebra. In an antihermitian Lie algebra representation, the two commuting Cartan operators $H_1,H_2$ are Hermitian matrices, and the root operators satisfy $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA weight diagram plots the pairs $\mu$, recording the weight multiplicity $\dim V_{\mu }$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the roots of a root systemThe raising operators $E_{+}^a$ move a nonzero weight vector by $+{\alpha }_a$ and the lowering operators move it by $-{\alpha }_a$. These six root directions form the A2 root system. Each root and its negative generate an SU(2) representation ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt{3}{H}_2/2$ constrain the two simple-root strings.

Choose a linear functional positive on the three positive roots. A maximal weight for this ordering is killed by every raising operator. In an Irreducible Lie algebra representation it generates the entire representation by lowering, and its weight space has dimension one. The two SU(2) representation strings show that its Dynkin labels are nonnegative integers:Conversely, each such pair specifies exactly one finite-dimensional irreducible highest-weight representation. Write it as $R(p,q)$. The fundamental weights in these Cartesian coordinates are ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe dimension follows from the three positive-root factors in the Weyl dimension formula: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every weight negated. The SU(3) highest-weight coordinates here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.

The weight diagram is invariant under the Weyl group, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the highest weight. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda +$ root lattice. A point in a different weight lattice coset is not a weight merely because it lies inside the polygon.

The shell multiplicities in an SU(3) weight diagram can be stated geometrically. The outer boundary has weight multiplicity one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual weight multiplicity by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min (p,q)+1$; when $p=q$ it is just the central point. A triangular weight diagram has multiplicity one everywhere. Thus $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, place a copy of the second weight diagram at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining dominant weight, identify its irreducible highest-weight representation, subtract that representation's whole diagram with its weight multiplicities, and repeat. This highest-weight character subtraction must exhaust every weight, not merely the corners. Unlike the Clebsch-Gordan decomposition for SU2, an SU(3) product may contain repeated copies of an irreducible representation.

The two three-dimensional Fundamental representations of sl3 are $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, the largest weight is $2{\omega }_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $3\otimes \overline{3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the octet and singlet in a fundamental SU3 tensor product and the two-quark product areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation:The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.

For the quark model, the relevant approximate flavour symmetry rotates the three light quarks $u,d,s$ in the fundamental $3$. The up quark, down quark, and strange quark basis states have the weight diagram of the defining SU(3) representation. In physical coordinates define isospin $I_3=H_1$ and flavor hypercharge $Y=2H_2/\sqrt{3}$. ThenThe last row follows from the Gell-Mann--Nishijima formula $Q=I_3+Y/2$, in units of the positive elementary electric charge. The antiquarks transform in $\overline{3}$ and have the negatives of these additive quantum numbers. Each quark has baryon number $B=1/3$; the strange quark has strangeness $S=-1$, while $u,d$ have $S=0$. Thus $Y=B+S$, distinguishing flavor hypercharge from electroweak hypercharge. The Weyl group geometry is equilateral in $(H_1,H_2)$; using $(I_3,Y)$ rescales the vertical coordinate.

A conventional meson has valence content $q\bar{q}$, so its flavour symmetry gives a meson octet and a singlet. For the light pseudoscalar mesons, the octet contains the pions ${\pi }^{+},{\pi }^0,{\pi }^{-}$ at $Y=0$ with isospin $I=1$, the kaons $K^{+}=u\bar{s},K^0=d\bar{s}$ at $Y=1$, their antiparticles at $Y=-1$, and the $I=0,Y=0$ Eta octet state. The central weight multiplicity two is realized byThe orthogonal eta singlet state is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The eta and eta prime mesons are mixtures of ${\eta }_8,{\eta }_1$, rather than exact octet and singlet mass eigenstates; this gives the pseudoscalar meson nonet. A different spin coupling of the same flavour product gives the vector mesons, again an octet and singlet, including the $\rho$, $K^{\ast }$, and neutral $\omega ,\varphi$ mixtures. The lowest orbital states have $J^P=0^{-}$ and $1^{-}$ respectively.

A conventional baryon has three valence quarks, so the tensor-cube decomposition supplies the flavour possibilities. The observed lowest baryon octet has spin $1/2$ and positive parity. Its isospin rows are $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$: the nucleons $p,n$, the three Sigma baryons, the Lambda baryon, and the two Xi baryons. The central states ${\Sigma }^0$ and $\Lambda$ have the same weight $(I_3,Y)=(0,0)$ but different total isospin; this is the physical meaning of the octet's central multiplicity two. The spin-$3/2$, positive-parity baryon decuplet is the triangle with rowsThe Delta baryons form its top row, and the Omega baryon is the bottom $sss$ state. All ten weights have multiplicity one. Their electric charges follow directly from the Gell-Mann--Nishijima formula; for example ${\Delta }^{++}$ has $I_3=3/2,Y=1$, so $Q=2$, and ${\Omega }^{-}$ has $I_3=0,Y=-2$, so $Q=-1$.

The two algebraic octets do not mean that two independent ground-state octets must occur. A three-quark colour singlet has a completely antisymmetric colour wavefunction. For a symmetric spatial ground state, the Pauli exclusion principle therefore requires a symmetric combined spin-flavour wavefunction. The symmetric flavour decuplet pairs with symmetric spin $3/2$; mixed flavour octets pair with mixed spin $1/2$ in one symmetric combination. Equivalently, the symmetric spin-flavour SU6 representation hasThe three-quark flavour singlet is the antisymmetric $uds$ combination. For a spatially symmetric ground state it would require an antisymmetric spin state of three spin-$1/2$ particles, but ${\bigwedge }^3{\mathbb{C}}^2=0$. The Pauli constraint on three-quark flavour multiplets thus excludes it from this ground-state sector; orbital excitation permits other multiplets.

Finally, this flavour symmetry is approximate: the unequal light-quark masses, especially the larger strange-quark mass, and electromagnetic interactions split masses within the multiplets and permit octet-singlet mixing. The weight diagrams classify flavour states, rather than asserting exact equality of their masses. Colour SU(3) is a separate gauge action on each flavour and supplies the colour-singlet constraint; it is not the approximate flavour action used in the eightfold way. The light conventional mesons organize into flavour octets and singlets, while the spatial ground-state three-quark baryons form one spin-$1/2$ octet and one spin-$3/2$ decuplet.

A highest-weight vector is a nonzero weight vector $v$ such that ${\mathfrak{g}}_{\alpha }v=0$ for every positive root $\alpha$. Its weight of a representation is a highest weight. For an irreducible finite-dimensional Lie algebra representation, this is the unique highest weight of a representation, and $v$ generates the module; a reducible module can have several highest weights of different summands.

Here is an existence argument for every nonzero finite-dimensional module $V$. For each simple root, its root operators and coroot form an sl2 subalgebra associated with a root. The classification of finite-dimensional sl2 representations shows that this coroot acts diagonalizably with integral eigenvalues. These commuting coroots span the Cartan subalgebra, so simultaneous diagonalization gives a finite weight-space decompositionChoose a real linear functional $\ell$ on the weight lattice span that is positive on every simple root, hence on every positive root, and choose a weight $\lambda$ maximizing $\ell$ among the finitely many weights of $V$. If $v\in V_{\lambda }$ and $X_{\alpha }\in {\mathfrak{g}}_{\alpha }$, thenso $X_{\alpha }v$ belongs to $V_{\lambda +\alpha }$ if nonzero. For positive $\alpha$, that would contradict maximality because $\ell (\lambda +\alpha )>\ell (\lambda )$. Thus every nonzero $v\in V_{\lambda }$ is a highest-weight vector. Every nonzero finite-dimensional module therefore has a highest weight. The zero module has no nonzero vector or weight, so it is the vacuous exception to the literal unrestricted wording.

A complex semisimple Lie algebra has zero centre, so its Adjoint representation is faithful. Its absolute Jordan decomposition is the unique expression $x=x_s+x_n$ with $[x_s,x_n]=0$, $\mathrm{ad}{x}_s$ diagonalizable and $\mathrm{ad}{x}_n$ nilpotent. Existence follows by applying part (ii) to the adjoint embedding, and uniqueness follows from uniqueness of the matrix Jordan–Chevalley decomposition. The representation-compatibility theorem states that for every finite-dimensional Lie algebra representation $\rho$, $\rho (x_s)$ and $\rho (x_n)$ are respectively the semisimple and nilpotent parts of $\rho (x)$. Thus these parts are intrinsic to the Lie algebra, independent of a chosen matrix embedding, and their semisimplicity and nilpotence are preserved in every finite-dimensional representation.

We prove closure using complete reducibility of semisimple Lie algebra representations, rather than assuming that a Lie subalgebra is closed under polynomials in its elements. Write the matrix Jordan–Chevalley decomposition as $X=S+N$. On $\mathrm{End}(V)$, the commutator operator $\mathrm{ad}S$ is diagonalizable: if $S$ has eigenspaces $V_{\lambda }$, then its eigenvalue on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ is $\mu -\lambda$. The operator $\mathrm{ad}N$ is nilpotent: if $N^r=0$, thenThe two operators commute because $[S,N]=0$. Uniqueness of the Jordan–Chevalley decomposition therefore says that $\mathrm{ad}S$ is the semisimple part of $\mathrm{ad}X$. Since a matrix's semisimple part is a polynomial in that matrix, and $\mathfrak{g}$ is invariant under $\mathrm{ad}X$, it follows thatNow regard $\mathrm{End}(V)$ as a Lie algebra representation of $\mathfrak{g}$ under commutators. By complete reducibility of semisimple Lie algebra representations, there is an invariant complement $M$ with $\mathrm{End}(V)=\mathfrak{g}\oplus M$. Write $S=Y+C$, with $Y\in \mathfrak{g}$ and $C\in M$. For $A\in \mathfrak{g}$, the normalization property gives $[C,A]=[S,A]-[Y,A]\in \mathfrak{g}$, while invariance of $M$ gives $[C,A]\in M$. Thus $[C,A]=0$: $C$ commutes with the whole Lie algebra.

Decompose $V={⨁}_j{V}_j$ into irreducible $\mathfrak{g}$-modules. Each $V_j$ is preserved by $X$, hence by its polynomial parts $S,N$, and also by $Y$; therefore it is preserved by $C$. By the Schur lemma, $C{|}_{V_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every element is a sum of Lie brackets and has trace zero on every finite-dimensional representation. Consequently $\mathrm{tr}(X{|}_{V_j})=\mathrm{tr}(Y{|}_{V_j})=0$. Also $N{|}_{V_j}$ is nilpotent, so it has trace zero; henceThus every $c_j$ is zero, $C=0$, and $S=Y\in \mathfrak{g}$. Finally $N=X-S\in \mathfrak{g}$. We have proved

Let $R_a=\mathrm{ad}(T_a)$, where $\mathrm{ad}(X)Y=[X,Y]$. In the given basis, the Adjoint representation of a Lie algebra isThe Jacobi identity gives $[\mathrm{ad}X,\mathrm{ad}Y]=\mathrm{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}{R}_c$, as required for a Lie algebra representation. The Killing form is the symmetric Trace form of a Lie algebra representationCyclicity of the trace proves its invariance directly:In components this is precisely $\boxed{{\kappa }_{dc}{f}^d{}_{ab}+{\kappa }_{bd}{f}^d{}_{ac}=0}$, so the Killing form is an invariant bilinear form on a Lie algebra.

Choose the infinitesimal gauge transformation convention $\delta \varphi =\omega \varphi$, with $\omega (x)={\omega }^a(x)t_a$, and write $A_{\mu }=A_{\mu }^a{t}_a$. A gauge covariant derivative must satisfy $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$ for $U=1+\omega +O({\omega }^2)$; henceEquivalently $\delta A_{\mu }^a=-{\partial }_{\mu }{\omega }^a+f^a{}_{bc}{\omega }^b{A}_{\mu }^c$. Expanding $\delta (D_{\mu }\varphi )$ makes the cancellation transparent:Thus $\boxed{\delta (D_{\mu }\varphi )={\omega }^a{t}_a{D}_{\mu }\varphi }$: it transforms in the same representation as $\varphi$. In adjoint covariant derivative notation, $\delta A_{\mu }=-D_{\mu }^{\mathrm{ad}}\omega$.