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The Casimir operator formed from the Killing form acts on an Irreducible Lie algebra representation with highest weight $\lambda$ by the scalar $(\lambda ,\lambda +2\rho )$, where $\rho$ is the half-sum of positive roots. It is zero on the trivial Lie algebra representation and nonzero on every nontrivial finite-dimensional irreducible complex representation of a semisimple Lie algebra.

For a finite-dimensional Irreducible Lie algebra representation of a complex semisimple Lie algebra with highest weight $\lambda$, this recursion computes its weight multiplicities from $m_{\lambda }(\lambda )=1$. Here $\rho$ is the half-sum of positive roots and the inner product is induced by the Killing form. Write the Casimir operator on the weight-$\mu$ space as $(\mu ,\mu +2\rho )I+2{\sum }_{\alpha >0}{F}_{\alpha }{E}_{\alpha }$. The cyclic trace identity between adjacent weight spaces gives $\mathrm{tr}(F_{\alpha }{E}_{\alpha }{|}_{V_{\mu }})={\sum }_{j\ge 1}(\mu +j\alpha ,\alpha )m_{\lambda }(\mu +j\alpha )$. Taking the trace and using the Casimir eigenvalue proves the formula. Only finitely many terms are nonzero.

For a complex semisimple Lie algebra and a choice of positive roots, every finite-dimensional Irreducible Lie algebra representation has a unique highest weight, which is a dominant integral weight. Conversely each dominant integral weight labels exactly one such irreducible module up to isomorphism. Its highest-weight vector spans the highest-weight space and generates the module. This classification and Weyl complete reducibility theorem reduce decomposing any finite-dimensional module to its irreducible highest-weight summands.

The solvable Lie algebra $\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, has an infinite-dimensional Irreducible Lie algebra representation on $\mathbb{C}[t]$: $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. Indeed $[x,y]f=yf$. An invariant subspace is an ideal $p(t)\mathbb{C}[t]$ because it is invariant under multiplication by $t$. Shift invariance gives $p(t)\mid p(t-1)$, hence $p(t-1)=p(t)$ and $p$ is constant. This shows why the Lie theorem needs finite-dimensional representations, even when the algebra itself is finite-dimensional.

The weight diagram has weights $6,4,2,0,-2,-4,-6$ with multiplicities $1,1,2,2,2,1,1$. An Irreducible Lie algebra representation of the sl2 Lie algebra with highest weight $n$ has weights $n,n-2,\dots ,-n$, all of multiplicity one. The top weight therefore forces one $L(6)$ summand. Subtracting its formal character leaves weights $2,0,-2$, each once, which is $L(2)$. By Weyl complete reducibility theorem,Their dimensions $7+3=10$ account for the entire module.

The Weyl group acts on the weight lattice by its root reflections and extends to ring automorphisms of the group ring of a weight lattice:Weights and their multiplicities in a finite-dimensional Lie algebra representation of a semisimple Lie algebra are permuted by this action, so its formal character is Weyl-invariant.

Choose a set ${\Phi }^{+}$ of positive roots and putHere ${\omega }_i$ are the fundamental weights, and $\det (w)=(-1{)}^{\ell (w)}$, where the Coxeter length $\ell (w)$ counts positive roots carried to negative ones. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote the finite-dimensional Irreducible Lie algebra representation of highest weight $\lambda$. The Weyl character formula isThe second denominator expression is the Weyl denominator formula. Both alternants change sign under each root reflection; their quotient is invariant. Although written as a fraction, the theorem asserts that it is a finite element of $\mathbb{Z}[{\Lambda }_W]$ with the nonnegative coefficients given by weight multiplicities. The highest-weight classification specifies which $\lambda$ label the irreducibles, while the root system and choice of positive roots determine every other term.

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 $c$.

Label the weight vertices by $1,\dots ,n,0,\bar{n},\dots ,\bar{1}$, with weights $e_1,\dots ,e_n,0,-e_n,\dots ,-e_1$. The crystal of the defining odd-orthogonal representation is the chainEach arrow of color $i$ subtracts the simple root ${\alpha }_i$. The two arrows through zero form the length-two short-root string.

Here is a complete grid description of the tensor product of crystals, valid for arbitrary $n$. Place the vertex $a\otimes b$ in row $a$, column $b$, in the above chain order; there are $(2n+1{)}^2$ vertices. Let ${\epsilon }_i(a)$ and ${\phi }_i(a)$ count incoming and outgoing steps along the color-$i$ string. For $i<n$, the two strings are $i\to i+1$ and $\overline{i+1}\to \bar{i}$, each of length one. For $i=n$, the string is $n\to 0\to \bar{n}$, with pairs $({\epsilon }_n,{\phi }_n)$ equal to $(0,2),(1,1),(2,0)$. Other vertices have both counts zero.

Using the crystal tensor-product rule, draw every color-$i$ edge byThe first case is a downward arrow to the next row and the second a rightward arrow to the next column; omit an arrow when its indicated factor has no outgoing edge. Thus the rule specifies every vertex and every arrow of the general tensor-square diagram, including the boundary nodes. The figure displays the resulting entire grid for $n=3$, alongside the vector crystal; the generator can draw any positive rank.

The raising rule chooses the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$ and the second otherwise. For $n\ge 2$, its only highest vertices areIndeed a highest tensor must have highest first factor, hence $a=1$. For colors $i>1$, the second factor cannot have an incoming edge. For color one, its incoming count must be at most ${\phi }_1(1)=1$. Inspecting the strings leaves precisely $b=1,2,\bar{1}$.

These three connected components encode the Irreducible Lie algebra representationsHere $S_0^{2}V$ is the traceless symmetric square, ${\Lambda }^{2}V$ is the Adjoint representation, and the invariant symmetric form supplies the trivial summand. Their dimensions are $n(2n+3)$, $n(2n+1)$, and one, summing to $(2n+1{)}^2$. In fundamental-weight notation the highest weights are $2{\omega }_1,{\omega }_2,0$ for $n\ge 3$, but $2{\omega }_1,2{\omega }_2,0$ for $n=2$.

For $n=1$ the vector crystal is $1\stackrel{1}{\to }0\stackrel{1}{\to }\bar{1}$. The same tensor rule gives highest vertices $1\otimes 1,1\otimes 0,1\otimes \bar{1}$, of weights $2e_1,e_1,0$, or $4{\omega }_1,2{\omega }_1,0$, respectively. Its summand dimensions are $5,3,1$.

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.

The derived series of a Lie algebra isThe algebra is a solvable Lie algebra if ${\mathfrak{b}}^{(m)}=0$ for some $m$. Here a bracket of subspaces means the linear span of their Lie brackets.

The Lie theorem states that a nonzero finite-dimensional representation of a complex solvable Lie algebra has a common eigenvector; equivalently, its operators can be simultaneously upper triangularized. Applied to an Irreducible Lie algebra representation, the common eigenvector gives a nonzero invariant line, which must be the entire module. A scalar action $\rho (x)=\lambda (x)$ respects the Lie bracket exactly when $\lambda ([x,y])=0$. Thus, in the finite-dimensional setting,Every functional displayed gives a one-dimensional irreducible, and two such modules are isomorphic exactly when their functionals agree.

The finite-dimensional qualification is essential here, although the printed wording does not repeat it. For example, let $[x,y]=y$ and act on $\mathbb{C}[t]$ by $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. These operators satisfy $[x,y]f=yf$. A nonzero submodule stable under $x$ is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ forces $p(t)\mid p(t-1)$, so the two polynomials, having the same degree and leading coefficient, are equal. In characteristic zero a polynomial invariant under a nonzero shift is constant. The module is therefore an infinite-dimensional simple module for the two-dimensional affine Lie algebra. It rules out extending the one-dimensional classification to unrestricted representations.

Use the derived series of a Lie algebraA solvable Lie algebra has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional Lie algebra is abelian, so $[[L,L],[L,L]]=0$. Every two-dimensional complex Lie algebra is solvable, with derived length at most two. This dimension argument actually works over any field with the alternating definition of a Lie bracket.

There is a genuine qualification in the last request: the assertion about all Irreducible Lie algebra representations is true for finite-dimensional representations. Without that restriction, the infinite-dimensional simple module for the two-dimensional affine Lie algebra is a counterexample. Take $L=\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, and let it act on $V=\mathbb{C}[t]$ byThen $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a Lie algebra representation. Any nonzero submodule is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two polynomials have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. The literal unrestricted assertion is false.

Here is a proof of the intended finite-dimensional assertion, including the essential Lie theorem argument. We prove that a nonzero finite-dimensional representation $\rho :L\to \mathrm{End}(V)$ of a complex solvable algebra has a common eigenvector, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus \mathbb{C}x$. Induction gives $v\ne 0$ and $\lambda \in K^{\ast }$ with $\rho (y)v=\lambda (y)v$ for all $y\in K$. Write $X=\rho (x)$ and suppress $\rho$ on elements of $K$.

Let $v_j=X^{j}v$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\dots ,v_{m-1}$. The space $W=\mathrm{span}(v_0,\dots ,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively thatThus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda (y)$ in this basis. The trace of a commutator is zero, soSince we work over $\mathbb{C}$, $\lambda ([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous eigenspaceis $X$-invariant: $yXw=\lambda (y)Xw+\lambda ([y,x])w=\lambda (y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an eigenvector. An eigenvector of $X{|}_{V_{\lambda }}$ is consequently a common eigenvector for $L$.

Its span is a one-dimensional submodule. If $V$ is irreducible, that span must be all of $V$. Hence every finite-dimensional irreducible representation of a finite-dimensional complex solvable Lie algebra has dimension one. Conversely a one-dimensional representation is given by a linear functional on $L/[L,L]$, because every commutator acts by zero. Passing repeatedly to quotients also gives the simultaneous triangularization of a Lie algebra representation. Neither the eigenvector step nor the trace argument extends to the infinite-dimensional counterexample above.

The Weyl complete reducibility theorem states that every finite-dimensional Lie algebra representation of a finite-dimensional complex semisimple Lie algebra is a direct sum of Irreducible Lie algebra representations. Equivalently, every invariant vector subspace has an invariant complement.

We use the permitted Casimir operator properties in the following precise form. There is a central quadratic operator $C$ commuting with the action on every module; it is zero on the trivial Lie algebra representation, and on every nontrivial finite-dimensional Irreducible Lie algebra representation it is a nonzero scalar $c$. This follows from the Schur lemma and the Casimir eigenvalue $c_{\lambda }=(\lambda ,\lambda +2\rho )$, using the Killing form normalization. We also use the permitted one-dimensional-representation fact: a complex semisimple Lie algebra has only trivial one-dimensional representations. Equivalently, it is a perfect Lie algebra, $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so a character $\mathfrak{g}\to \mathbb{C}$ annihilating brackets must vanish. Neither fact assumes complete reducibility of the module being proved reducible.

First prove that every finite-dimensional short exact sequencewith trivial quotient splits. If $W$ is nontrivial irreducible, the Casimir operator $C_E$ has image in $W$ and restricts to $c{I}_W$ there. Consequently $\ker C_E$ is a one-dimensional invariant complement to $W$. If $W$ is trivial irreducible, $E$ has a basis in which every action is $\left(\begin{array}{cc}0& a(x)\\ 0& 0\end{array}\right)$. The Lie algebra representation identity makes $a([x,y])=0$, so the one-dimensional-representation fact gives $a=0$ and again the sequence splits.

For general $W$, induct on $\dim W$. Choose an irreducible submodule $W_1\subseteq W$. The induced sequence with kernel $W/W_1$ and middle term $E/W_1$ splits by induction. The inverse image $E_1\subseteq E$ of its invariant complement is a submodule fitting into $0\to W_1\to E_1\to \mathbb{C}\to 0$. The irreducible-kernel case gives an invariant line in $E_1$ mapping isomorphically to the quotient. It is also an invariant complement to $W$ in $E$. The case $W=0$ starts this induction. This proves splitting of a trivial quotient for a semisimple Lie algebra, including kernels that are not assumed completely reducible.

Now let $U\subseteq V$ be any invariant subspace. On the Hom representation $\mathrm{Hom}(V,U)$ the action isLet $E$ consist of the maps whose restriction to $U$ is a scalar multiple of $I_U$. This is a submodule, and restriction givesThe right-hand map is surjective because an ordinary linear projection $V\to U$ exists; its quotient action is trivial because a commutator with $I_U$ is zero. The splitting just proved supplies an invariant $T\in E$ with $T{|}_U=I_U$. Thus $T$ intertwines the actions, $T^2=T$, andThe cases $U=0$ and $U=V$ are immediate. Choosing an irreducible submodule and repeating this complement construction proves the Weyl complete reducibility theorem. This last step is invariant complement from an equivariant projection.

Dropping finite dimensionality gives an example with the complex simple Lie algebra ${\mathfrak{sl}}_2$. Its Verma module of highest weight zero has a basis $v_0,v_1,\dots$ withwhere $e{v}_0=0$. These actions obey $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The span of $v_1,v_2,\dots$ is a proper submodule and the quotient is trivial. It has no invariant complement: such a complement would be a trivial line, while $f$ is injective on the entire module. Thus a simple Lie algebra can have an infinite-dimensional representation that is not completely reducible, even over $\mathbb{C}$.

Use the sl2 Lie algebra relations $[H,X]=2X$, $[H,Y]=-2Y$ and $[X,Y]=H$. A highest-weight vector satisfies $Xv=0$ and $Hv=mv$, and $H{Y}^{j}v=(m-2j)Y^{j}v$. For $k=0$, $XYv=YXv+Hv=mv$. Assuming the formula at $k-1$, one getsThus the sl2 highest-weight lowering formula isIn particular a finite-dimensional Irreducible Lie algebra representation has highest weight $m\in {\mathbb{Z}}_{\ge 0}$ and weight vectors $v,Yv,\dots ,Y^{m}v$, as in the classification of finite-dimensional sl2 representations.

Let $V_{\lambda }$ be a finite-dimensional Irreducible Lie algebra representation of a complex semisimple Lie algebra, with highest weight $\lambda$. Write $m_{\lambda }(\mu )=\dim (V_{\lambda }{)}_{\mu }$, taking it to be zero when $\mu$ is not a weight, and let $\rho$ be the half-sum of positive roots. The Killing form induces an inner product on the real span of weights. Freudenthal multiplicity formula statesThe sums are finite. For a weight $\mu \ne \lambda$, the coefficient on the left is positive: move $\mu$ into the dominant Weyl chamber, use that a weight is below $\lambda$ in dominance order, and note that $(\mu ,\rho )$ does not increase on moving back out of that chamber. The resulting recursion starts from $m_{\lambda }(\lambda )=1$.

Here is a proof using the allowed Casimir operator. Choose root vectors $E_{\alpha },F_{\alpha }$ for positive and negative roots, normalized by $B(E_{\alpha },F_{\alpha })=1$, and let $t_{\alpha }\in \mathfrak{h}$ satisfy $B(t_{\alpha },h)=\alpha (h)$. Invariance of a bilinear form on a Lie algebra gives $[E_{\alpha },F_{\alpha }]=t_{\alpha }$. If $h_i,h^i$ are dual bases of the Cartan subalgebra under $B$, the Casimir operator isOn the weight space of $\mu$, the first sum acts by $(\mu ,\mu )$. Replacing $E_{\alpha }{F}_{\alpha }$ with $F_{\alpha }{E}_{\alpha }+t_{\alpha }$ givesDefine $T_{\alpha }(\mu )=\mathrm{tr}(F_{\alpha }{E}_{\alpha }{|}_{(V_{\lambda }{)}_{\mu }})$. The cyclic trace identity between adjacent weight spaces and the commutator relation yieldIterate upwards until the weight spaces vanish to obtain $T_{\alpha }(\mu )={\sum }_{j\ge 1}(\mu +j\alpha ,\alpha )m_{\lambda }(\mu +j\alpha )$. Finally take the trace of the displayed restriction of $C$. Its Casimir eigenvalue is $(\lambda ,\lambda +2\rho )$, so subtracting $(\mu ,\mu +2\rho )m_{\lambda }(\mu )$ proves the formula. This trace argument handles weight multiplicities greater than one without choosing a separate sl2 Lie algebra string through each vector.

Work over $\mathbb{C}$. Decompose $V$ as the direct sum of the generalized eigenspaces $V_{\lambda }$ of $X$. Define $X_s$ to be $\lambda I$ on $V_{\lambda }$ and set $X_n=X-X_s$. Then $X_s$ is a diagonalisable endomorphism, $X_n$ is a nilpotent endomorphism, and they commute. For uniqueness, any commuting decomposition $X=S+N$ has $S$ and $N$ commuting with $X$, hence preserving each $V_{\lambda }$. Decompose $V_{\lambda }$ further into eigenspaces of $S$. On a nonzero such space with eigenvalue $\sigma$, the operator $X=\sigma I+N$ has only the eigenvalue $\sigma$, so $\sigma =\lambda$. Since $S$ is diagonalizable, $S=\lambda I$ throughout $V_{\lambda }$. Thus $S=X_s$ and $N=X_n$. This is the Additive Jordan decomposition.

The Chinese remainder theorem for the pairwise coprime polynomials $(t-\lambda {)}^{r_{\lambda }}$ also gives a polynomial $p$ with $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^{r_{\lambda }})$, where $r_{\lambda }$ is the largest Jordan block size. Consequently $X_s=p(X)$ and $X_n=X-p(X)$. This polynomial description shows that both parts preserve every $X$-invariant subspace.

On $\mathrm{Hom}(V_{\lambda },V_{\mu })$ the operator $\mathrm{ad}{X}_s$ is the scalar $\mu -\lambda$, so it is diagonalizable. The nilpotence of commutation by a nilpotent endomorphism makes $\mathrm{ad}{X}_n$ nilpotent. They commute, since $[\mathrm{ad}{X}_s,\mathrm{ad}{X}_n]=\mathrm{ad}[X_s,X_n]=0$. Uniqueness of the Additive Jordan decomposition therefore gives the adjoint compatibility of additive Jordan decomposition:Now let $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ be a complex semisimple Lie algebra and $X\in \mathfrak{g}$. Since the semisimple part of $\mathrm{ad}X$ is a polynomial in $\mathrm{ad}X$, it preserves $\mathfrak{g}$. Thus $[X_s,\mathfrak{g}]\subseteq \mathfrak{g}$. By the Weyl complete reducibility theorem, the Adjoint representation of $\mathfrak{g}$ on $\mathrm{End}(V)$ has a decomposition $\mathrm{End}(V)=\mathfrak{g}\oplus M$ into invariant subspaces. Write $X_s=y+z$ with $y\in \mathfrak{g}$ and $z\in M$. For $a\in \mathfrak{g}$, the vector $[z,a]=[X_s,a]-[y,a]$ lies in $\mathfrak{g}$, and invariance of $M$ puts it in $M$ as well. Hence $[z,a]=0$.

Decompose $V={⨁}_j{U}_j$ into Irreducible Lie algebra representations. Each $U_j$ is preserved by $X_s=p(X)$ and by $y$, hence by $z$. The Schur lemma makes $z{|}_{U_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every representing element of $\mathfrak{g}$ has zero trace on every $U_j$. Also $\mathrm{tr}(X_s{|}_{U_j})=\mathrm{tr}(X{|}_{U_j})$, because $X_n$ is nilpotent there. It follows that $c_j\dim U_j=\mathrm{tr}(z{|}_{U_j})=0$. In characteristic zero, $c_j=0$, so $z=0$. Therefore $X_s\in \mathfrak{g}$ and $X_n=X-X_s\in \mathfrak{g}$. This proves that semisimple matrix Lie algebras are closed under additive Jordan decomposition, including representations with repeated isomorphic irreducible summands.

For a complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra consisting of elements whose Adjoint representation matrices are diagonalizable. Equivalently it is a nilpotent self-normalizing Lie subalgebra. The root-space decomposition iswhere the roots are the nonzero weights of that Adjoint representation.

We use these properties of the Killing form $B$: it is nondegenerate on $\mathfrak{g}$ and on $\mathfrak{h}$, it is invariant, distinct root spaces are orthogonal unless their roots sum to zero, and $B$ pairs ${\mathfrak{g}}_{\alpha }$ nondegenerately with ${\mathfrak{g}}_{-\alpha }$. Define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$, and choose $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)=1$. Since weights add, $[e,f]\in \mathfrak{h}$, and invariance givesThe nonisotropic root lemma shows $\alpha (t_{\alpha })\ne 0$. Here is its short proof: if this number were zero, the span of $e,f,t_{\alpha }$ would be a solvable Lie algebra with $t_{\alpha }$ central. Apply the Lie theorem to its action on $\mathfrak{g}$. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}e,\mathrm{ad}f]$ would be strictly upper triangular and hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$ makes it diagonalizable. It would therefore be zero, putting $t_{\alpha }$ in the zero center of a Lie algebra of $\mathfrak{g}$, contrary to its definition. SetThenso their span is the sl2 subalgebra associated with a root.

The abstract reduced crystallographic root system axioms are as follows. In a finite-dimensional real inner product space $E$, the set $R$ is finite, consists of nonzero vectors and spans $E$; for every $\alpha \in R$, one has $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; the root reflection $s_{\alpha }(\beta )=\beta -2(\beta ,\alpha )\alpha /(\alpha ,\alpha )$ preserves $R$; and $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for every pair of roots. We verify the positive-definite real form as well as these axioms, rather than assuming the complex Killing form is already positive.

First $R$ is finite and has no zero element by its definition. The roots span ${\mathfrak{h}}^{\ast }$ over $\mathbb{C}$: any $h\in \mathfrak{h}$ annihilated by all roots commutes with the whole root-space decomposition and is central, hence zero. Therefore the $t_{\alpha }$, and also the $H_{\alpha }$, span $\mathfrak{h}$ over $\mathbb{C}$. For every root $\beta$, the classification of finite-dimensional sl2 representations applied to the Adjoint representation of the root subalgebra gives $\beta (H_{\alpha })\in \mathbb{Z}$. Put ${\mathfrak{h}}_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\{H_{\alpha }:\alpha \in R\}$. Every root takes real values on this space, andThe strict inequality follows because the roots span ${\mathfrak{h}}^{\ast }$. This also shows that the complexification of ${\mathfrak{h}}_{\mathbb{R}}$ injects into $\mathfrak{h}$: an equality $h_1+i{h}_2=0$ with real $h_j$ would contradict positivity of $B(h_1,h_1)$ and $B(h_2,h_2)$. Since its complex span is all of $\mathfrak{h}$, it is a real form. Moreover $B(H_{\alpha },H_{\alpha })=4/\alpha (t_{\alpha })>0$, so $t_{\alpha }$ is a real scalar multiple of $H_{\alpha }$. The dual inner product thus makes $E={\mathrm{span}}_{\mathbb{R}}R$ a Euclidean space, as in the Euclidean subspace of a Cartan subalgebra.

To prove reducedness without assuming it, consider the root-subalgebra modulewith absent root spaces understood to be zero. Its $H_{\alpha }$ weights are even, so every nontrivial Irreducible Lie algebra representation in it has even positive highest weight and one-dimensional weight-zero space. The action of $X_{\alpha }$ on its weight-zero space $\mathfrak{h}$ has image exactly $\mathbb{C}{X}_{\alpha }$, of dimension one. Therefore there is precisely one nontrivial irreducible summand, the already embedded adjoint ${\mathfrak{sl}}_2$ module of highest weight $2$. Thus $\dim {\mathfrak{g}}_{\alpha }=1$ and $2\alpha \notin R$. If $\beta =c\alpha$ is any root on the same real line, the integral numbers $\beta (H_{\alpha })=2c$ and $\alpha (H_{\beta })=2/c$ have product $4$. Hence $c$ is one of $\pm \frac{1}{2},\pm 1,\pm 2$; the half and double cases are excluded by applying the preceding argument to the appropriate root. This proves the root-space reducedness lemma and $R\cap \mathbb{R}\alpha =\{\pm \alpha \}$.

For a root $\beta$ not parallel to $\alpha$, the sum of root spaces ${⨁}_{j\in \mathbb{Z}}{\mathfrak{g}}_{\beta +j\alpha }$ is stable under the root ${\mathfrak{sl}}_2$. Its integer $H_{\alpha }$ weights are symmetric under sign in each irreducible summand. Therefore the weight $-\beta (H_{\alpha })$ also occurs, at the root $\beta -\beta (H_{\alpha })\alpha$. The Killing form normalization givesso this root is exactly $s_{\alpha }(\beta )$. For $\beta =\pm \alpha$ the reflection just swaps the two roots. This proves reflection invariance and the Cartan integer condition. All axioms of the reduced crystallographic root system have now been verified.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.