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

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

In dimension four, the two positive roots are $\alpha ={\epsilon }_1-{\epsilon }_2$ and $\beta ={\epsilon }_1+{\epsilon }_2$. Using the matrix root basis of the even orthogonal Lie algebra, defineThe matrix unit identity $[E_{ij},E_{kl}]={\delta }_{jk}{E}_{il}-{\delta }_{li}{E}_{kj}$ gives $[e_{\gamma },f_{\gamma }]=h_{\gamma }$, $[h_{\gamma },e_{\gamma }]=2e_{\gamma }$, and $[h_{\gamma },f_{\gamma }]=-2f_{\gamma }$ for $\gamma =\alpha ,\beta$. Thus each triple spans an sl2 Lie algebra.

The two triples commute. Their Cartan actions on one another vanish because $\alpha (h_{\beta })=\beta (h_{\alpha })=0$; their mixed root-vector brackets vanish because none of $\pm \alpha \pm \beta$, namely $\pm 2{\epsilon }_1$ and $\pm 2{\epsilon }_2$, is a root. This can equally be checked from the displayed matrix units. The six vectors are linearly independent, occupying the four distinct root spaces and two independent Cartan directions, so they exhaust the six-dimensional so4 Lie algebra. The two commuting spans are consequently ideals, andThis gives an explicit Chiral decomposition of the complexified so4 Lie algebra, including generators for both factors.

In the defining representation $V={\mathbb{C}}^{2n}$, the basis vectors $v_i$ and $v_{\bar{i}}$ have weights ${\epsilon }_i$ and $-{\epsilon }_i$, respectively, each with multiplicity one. Thus its formal character isThe vector $v_1$ is killed by every positive root space, since all positive-root matrices are upper triangular, and is a highest-weight vector. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is irreducible for $n\ge 2$. In terms of the fundamental weights of Dn, its highest weight is ${\omega }_1$ for $n\ge 3$ and ${\omega }_1+{\omega }_2$ for $n=2$.

The flip of the two tensor factors is equivariant, henceUse the invariant symmetric bilinear form $Q(u,v)=u^{T}Jv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the trivial Lie algebra representation. Its inverse-form tensoris invariant and satisfies $C(\tau )=2n$. ConsequentlyThe traceless summand contains the highest-weight vector $v_1\otimes v_1$ of weight $2{\epsilon }_1$. The Weyl complete reducibility theorem states that every finite-dimensional complex semisimple Lie algebra representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2{\epsilon }_1)$. The Weyl dimension formula verifies that it is the entire traceless space: only positive roots involving ${\epsilon }_1$ change their factors, andThis equals $\dim {\mathrm{Sym}}^{2}V-1=n(2n+1)-1$. Thus the traceless symmetric square of the defining even orthogonal representation is irreducible, including $n=2$.

For the other summand, the exterior square of the defining orthogonal representation has the equivariant identificationThe image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the Adjoint representation of a Lie algebra. For $n\ge 3$, the connected Dn Dynkin diagram gives a simple Lie algebra (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is ${\epsilon }_1+{\epsilon }_2$, and $v_1\wedge v_2$ is a highest vector. The tensor-square decomposition of the defining even orthogonal representation is consequentlyThe three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2{\omega }_1$, ${\omega }_2$ and zero for $n\ge 4$; for $n=3$ the middle weight is ${\omega }_2+{\omega }_3$.

For $n=2$, the Chiral decomposition of the complexified so4 Lie algebra splits the adjoint module into its two three-dimensional ideals. ThusThe weights are $2{\omega }_1+2{\omega }_2$, $2{\omega }_1$, $2{\omega }_2$ and zero. Equivalently, labelling highest weights by the two sl2 Lie algebra factors, this is $(2,2)\oplus (2,0)\oplus (0,2)\oplus (0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L({\epsilon }_1)\oplus L(-{\epsilon }_1)$ and $V\otimes V=L(2{\epsilon }_1)\oplus 2L(0)\oplus L(-2{\epsilon }_1)$; there is then no single highest weight for an irreducible natural module.

Use the standard sl2 Lie algebra relations $[h,e]=2e$, $[h,f]=-2f$ and $[e,f]=h$. The classification of finite-dimensional sl2 representations says that every finite-dimensional complex module is a direct sum of $L(m)$, $m\in {\mathbb{Z}}_{\ge 0}$, where $L(m)$ has weights $m,m-2,\dots ,-m$. The raising operator $e$ and lowering operator $f$ move weights by two and minus two. On $L(m)$ each therefore has $(m+1)$st power zero. There is consequently one integer $N$ such that $e^N=f^N=0$ on all of $V$.

The matrix exponentials are thus finite polynomials in nilpotent operators:Each is a well-defined endomorphism, and its inverse is obtained by negating the argument. Their product is therefore a linear automorphism,This Weyl reflection lift in an sl2 representation is algebraically defined by finite sums; no choice of topology or convergence of an infinite series is needed.

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.

Realize the defining sl2 Lie algebra Lie algebra representation as $\mathbb{C}x\oplus \mathbb{C}y$, with $H=x{\partial }_x-y{\partial }_y$, $E=x{\partial }_y$, $F=y{\partial }_x$. A basis of the symmetric power is $e_i=x^{4-i}{y}^i$ for $0\le i\le 4$, with $H{e}_i=(4-2i)e_i$. Hence the exterior square basis $w_{ij}=e_i\wedge e_j$, $i<j$, has weight $8-2(i+j)$.

Collecting equal weights gives the complete weight-space decomposition:Every one of the ten exterior basis vectors appears exactly once. Thus $U={⨁}_{m=6,4,2,0,-2,-4,-6}{U}_m$, with the displayed bases and dimensions.

Here $A_1(K)$ denotes the adjoint Chevalley group of type A1; the isogeny form matters, since the simply connected form is $S{L}_2(K)$ itself. Over $\mathbb{C}$, take the sl2 Lie algebra with Chevalley basisTheir Lie brackets satisfy $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Conjugation defines a group homomorphism $\rho :S{L}_2(\mathbb{C})\to \mathrm{Aut}({\mathfrak{sl}}_2(\mathbb{C}))$. Since $e^2=f^2=0$, $u(t)=\exp (te)$ and $l(s)=\exp (sf)$. Differentiating conjugation, or expanding its finite polynomial on this basis, givesThe adjoint Chevalley group is generated by these two families. Part (i) therefore proves that it is exactly the image of $\rho$. A matrix commuting with $e$ has the form $\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)$; also commuting with $f$ forces $b=0$. Thus the kernel of a group homomorphism $\ker \rho$ consists of scalar matrices of determinant one, namely $\{I,-I\}$. The first isomorphism theorem now gives $A_1(\mathbb{C})\cong S{L}_2(\mathbb{C})/\{\pm I\}=PS{L}_2(\mathbb{C})$.

For a general field, first compute the Chevalley basis root operators as integral polynomials, and only then reduce their coefficients to $K$. Explicitly,They are still precisely conjugation by $u(t),l(s)$ on the three-dimensional trace-zero matrix space. These formulas remain valid in characteristic two, where $h=I$ but $e,h,f$ remain linearly independent. Part (i) still identifies the generated image with the conjugation image of $S{L}_2(K)$. The same computation with $e,f$ gives kernel of a group homomorphism $\{aI:a^2=1\}$, exactly its scalar center. ConsequentlyIn characteristic two the scalar center is trivial. The integral formulas avoid the invalid operation of dividing by factorials inside a field of small positive characteristic.

Assume $0\le k\le n$ with $n,k$ integers, as required for the quantum binomial expression. Let $L_{n-1}$ be the $n$-dimensional irreducible sl2 Lie algebra representation, with one-dimensional weight spaces of weights $n-1-2j$, $0\le j<n$. The exterior-power Lie algebra representation on ${\Lambda }^k{L}_{n-1}$ has characterThis is already a Laurent polynomial with nonnegative integer coefficients.

To identify it with the requested quotient, introduce $t=q^{-2}$ and the Gaussian binomial coefficientSplitting the subsets according to whether $n-1$ is included provesThe product ${\prod }_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence: after factoring common numerator and denominator terms, the needed identity is $(1-t^{n-k})+t^{n-k}(1-t^k)=1-t^n$. Induction on $n$ therefore proves the product identity, including its polynomiality.

Since the quantum integer is $[a{]}_q=q^{a-1}(1-q^{-2a})/(1-q^{-2})$, taking the quotient of the products givesThus this is the symmetric quantum binomial coefficient. All its weights have parity $k(n-k)$, because the exponent in the exterior-power sum differs from $k(n-k)$ by an even integer. Applying part i to this finite-dimensional representation proves symmetric unimodality in steps of two. Equivalently, the ordinary polynomial $G_{n,k}(t)$ has a unimodal coefficient sequence at consecutive powers of $t$.

Again, the full integer-exponent coefficient sequence in $q$, including the absent parity, need not be unimodal: for $n=2,k=1$ the answer is $q+q^{-1}$. This counterexample makes precise the convention under which the claimed quantum unimodality holds. The cases $k=0$ or $k=n$ give the constant one.

Let $h$ be the standard Cartan element of the sl2 Lie algebra, with $[h,e]=2e$ and $[h,f]=-2f$. For a finite-dimensional representation, its formal character records its weight multiplicities:The classification of finite-dimensional sl2 representations and the Weyl complete reducibility theorem give $V={⨁}_{m\ge 0}{a}_m{L}_m$, with finitely many nonzero $a_m$. Each $L_m$ has weights $m,m-2,\dots ,-m$, each of multiplicity one. ThereforeIn particular $c_j=c_{-j}$, and for $j\ge 0$ one has $c_j-c_{j+2}=a_j\ge 0$. This proves parity unimodality of an sl2 character: within each parity, the coefficients increase towards zero and decrease away from zero. For a representation whose weights have a single parity, deleting the intervening zero coefficients gives a symmetric unimodal sequence.

The parity convention is necessary for the printed assertion. If unimodality means the full coefficient sequence at every consecutive integer exponent, it is false: the standard two-dimensional representation has character $q+q^{-1}$, whose coefficients at exponents $-1,0,1$ are $1,0,1$. The statement established above is unimodality on cosets of the root lattice for ${\mathfrak{sl}}_2$, and is the version needed in part ii. Infinite-dimensional representations need not even have Laurent-polynomial characters.

The induced raising operator and lowering operator act bySince $E{w}_{01}=0$ and $w_{01}$ has weight $6$, it generates the seven-dimensional irreducible sl2 Lie algebra module ${\mathrm{Sym}}^{6}V$. By complete reducibility of semisimple Lie algebra representations, subtracting its weights $6,4,2,0,-2,-4,-6$, each of multiplicity one, from the weight diagram leaves weights $2,0,-2$ once each. HenceThese are the only two irreducible isomorphism types occurring as submodules.

For the three-dimensional summand, calculate $E{w}_{03}=3w_{02}$ and $E{w}_{12}=w_{02}$. Thereforeis a nonzero highest-weight vector of weight $2$. The classification of finite-dimensional sl2 representations shows that the submodule $W$ it generates is the irreducible module ${\mathrm{Sym}}^{2}V$, of dimension three.

Take a basis $x,y$ of the defining representation of the sl2 Lie algebra, with $Hx=x$, $Hy=-y$, $Ey=x$, $Ex=0$, $Fx=y$ and $Fy=0$. A monomial basis of the symmetric power ${\mathrm{Sym}}^{4}V$ isFor the exterior power $U={\bigwedge }^2{\mathrm{Sym}}^{4}V$, put $w_{ij}=v_i\wedge v_j$ for $i<j$. These ten elements form a basis, and their weights of a representation add:Thus the complete weight-space decomposition, with explicit bases, isThe weight diagram therefore has multiplicities $1,1,2,2,2,1,1$ at weights $6,4,2,0,-2,-4,-6$, respectively.

Start with a nonzero highest-weight vector $v_0=|n⟩$ and set $v_r=E_{-}^r{v}_0$. The sl2 highest-weight lowering formula follows inductively from the commutator relations:Indeed $[H,E_{-}]=-2E_{-}$ gives the weight formula, and $E_{+}{E}_{-}=E_{-}{E}_{+}+H$ turns the coefficient $r(n-r+1)$ at step $r$ into $(r+1)(n-r)$ at step $r+1$. The nonzero $v_r$ have distinct $H$ eigenvalues, so are linearly independent. Finite dimensionality therefore forces a last nonzero vector $v_N$. Applying $E_{+}$ to $v_{N+1}=0$ gives $(N+1)(n-N)v_N=0$, hence $N=n$. This also proves that no earlier vector can vanish. Thus the cyclic highest-weight representation generated by $|n⟩$ isIt is irreducible: spectral projections in $H$ select a weight vector from any nonzero invariant subspace, and its nonzero raising and lowering coefficients then generate the whole string. This gives the classification of finite-dimensional sl2 representations used for each root direction below.

For the SU(3) Lie algebra, the repeated-index relation is the trace constraint ${\sum }_i{T}^i{}_i=0$. The displayed generators are a basis of its complexification, not nine independent generators or three individually vanishing diagonal elements. Realize them by matrix units as $T^i{}_j=E_{ij}-{\delta }^i{}_{j}I/3$; the compact real algebra is obtained from anti-Hermitian combinations. Thenand $E_{1+}=E_{12}$, $E_{1-}=E_{21}$, $E_{2+}=E_{23}$, $E_{2-}=E_{32}$. Substituting into the given commutators givesThus both root pairs obey the required sl2 Lie algebra relations. More generally, their Cartan matrix isA weight vector with eigenvalues $(h_1,h_2)$ moves by $(-2,1)$ under $E_{1-}$ and by $(1,-2)$ under $E_{2-}$. The third negative-root operator is $[E_{2-},E_{1-}]=E_{31}$ and moves it by $(-1,-1)$. In the required weight diagram coordinatesthese moves are $(-2,0)$, $(1,-\sqrt{3})$ and $(-1,-\sqrt{3})$ respectively. Successive lowerings from a highest-weight vector, with linear dependencies removed, construct the following bases.

For Dynkin labels $(1,0)$, use the defining action of SU(3) on ${\mathbb{C}}^3$. Its highest vector is $e_1$; lowering gives $e_2=E_{21}{e}_1$ and $e_3=E_{32}{e}_2$. There are no further independent vectors. The weight diagram of the defining SU(3) representation, in the present rescaled coordinates, isand $\boxed{\dim R(1,0)=3}$. Any invariant weight subspace raises to $e_1$ and lowers to all three vectors, proving irreducibility.

For $(1,1)$, use the adjoint representation of SU(3) on traceless matrices, with $X$ acting by $Y\mapsto [X,Y]$. The highest vector is $E_{13}$: both $[E_{12},E_{13}]$ and $[E_{23},E_{13}]$ vanish, while $[H_i,E_{13}]=E_{13}$. Lowering gives $[E_{21},E_{13}]=E_{23}$ and $[E_{32},E_{13}]=-E_{12}$. Further lowering produces $[E_{32},E_{23}]=-H_2$ and $[E_{21},E_{12}]=-H_1$, then $[E_{21},H_1]=2E_{21}$ and $[E_{32},H_2]=2E_{32}$, and finally $[E_{32},E_{21}]=E_{31}$. Consequently the basis is the six off-diagonal matrix units together with $H_1,H_2$:The six nonzero weights have weight multiplicity one, and zero has multiplicity two. Thus $\boxed{\dim R(1,1)=6+2=8}$. Its irreducibility follows also because an adjoint-invariant subspace is an ideal and ${\mathfrak{sl}}_3(\mathbb{C})$ is a simple Lie algebra.

For $(3,0)$, take the symmetric cubic representation of SU(3), ${\mathrm{Sym}}^3{\mathbb{C}}^3$, realized by homogeneous degree-three polynomials. Hereso $z_1^3$ is a highest-weight vector with eigenvalues $(3,0)$. For $a+b+c=3$,which is nonzero and constructs every monomial. These monomials form a basis and have distinct weights $(a-b,b-c)$. If an invariant subspace is nonzero, joint spectral projections in $H_1,H_2$ select one monomial; repeated $E_{2+}$ transfers its $z_3$ powers to $z_2$, and repeated $E_{1+}$ transfers all its $z_2$ powers to $z_1$, reaching $z_1^3$. Lowering then generates the whole basis, proving irreducibility without merely assuming the dimension formula. The plotted weights arewith multiplicity one at every point. Hence $\boxed{\dim R(3,0)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10}$.

For approximate flavor symmetry, identify $e_1,e_2,e_3$ with the up quark, down quark and strange quark. The defining representation is their flavour triplet. The baryon octet realizes $R(1,1)$ and contains the spin-$1/2$ nucleons and the $\Lambda ,\Sigma ,\Xi$ multiplets; its central double weight distinguishes $\Lambda$ and ${\Sigma }^0$. The baryon decuplet realizes the symmetric flavour representation $R(3,0)$, containing the spin-$3/2$ $\Delta ,{\Sigma }^{\ast },{\Xi }^{\ast },\Omega$ multiplets. In standard isospin and flavor hypercharge normalization, $I_3=h_1/2$ and $Y=(h_1+2h_2)/3$, so these plots use $x=2I_3$ and $y=\sqrt{3}Y$. Three-quark spin and flavour combine subject to the Pauli constraint on three-quark flavour multiplets and the antisymmetric three-quark colour singlet. The adjoint flavour representation also describes a meson octet. A distinct, exact colour SU(3) gauge symmetry places quarks in its defining representation and gluons in its adjoint octet; colour and flavour here are different physical group actions.

Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.

Choose a Cartan subalgebra, simple roots ${\alpha }_1,\dots ,{\alpha }_r$, and root vectors $e_i,f_i$ normalized by $[e_i,f_i]=h_i={\alpha }_i^{\vee }$. LetHere the coroots lie in the Cartan subalgebra, and ${\alpha }_i({\rho }^{\vee })=1$. This last identity is the Weyl vector identity applied to the dual root system. For distinct simple roots, ${\alpha }_i-{\alpha }_j$ is not a root, hence $[e_i,f_j]=0$. ConsequentlyThese are exactly the sl2 Lie algebra relations. For nonzero $\mathfrak{g}$ the resulting homomorphism from ${\mathfrak{sl}}_2$ is nonzero and therefore injective, since ${\mathfrak{sl}}_2$ is simple. Its image is a Principal sl2 subalgebra. Equivalently its raising element is a principal nilpotent element, a nilpotent element with Lie algebra centralizer of dimension equal to the rank. The construction works for a semisimple Lie algebra with several simple factors by taking the triple in every factor, giving one diagonally embedded ${\mathfrak{sl}}_2$. This is the construction of a principal sl2 triple.

For a representation $V$, restrict to this subalgebra and take its formal character. If $V_{\mu }$ is the weight space for the original Cartan subalgebra, the resulting q-character of a highest-weight representation, or the analogous q-character for any finite-dimensional representation, isIt is independent of conjugating the chosen principal triple. In particular a root $\beta ={\sum }_i{k}_i{\alpha }_i$ has grading $\beta (h)=2{\sum }_i{k}_i$. Some conventions use $q^{h/2}$ instead; that convention divides all exponents here by two. The normalization above agrees with the ordinary sl2 Lie algebra character in the preceding part.

Use the generators $e,f,h$ of the sl2 Lie algebra with $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. For a finite-dimensional Lie algebra representation, $h$ is diagonalizable with integral weights. Its formal character of a weight module isHere $q$ is a formal variable; equivalently the expression is ${\mathrm{tr}}_V(q^h)$, or the formal sum ${\sum }_m(\dim V_m)e^m$ with $e^m=q^m$. This records every weight multiplicity, rather than only the dimension.

The classification of finite-dimensional sl2 representations says that $L_d$, the irreducible of highest weight $d\ge 0$, has the weight string $d,d-2,\dots ,-d$, each with multiplicity one. HenceThe quotient denotes the displayed Laurent polynomial and has the removable value $d+1$ at $q=1$. Formal characters are additive on direct sums, so these formulas determine the character of every finite-dimensional module.

Choose a positive system of a root system and the corresponding triangular decomposition of a Lie algebra $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ with $hv=\omega (h)v$ for every $h\in H$ and ${\mathfrak{n}}^{+}v=0$. Thus it is a highest-weight vector; the choice of positive roots is part of this definition. A highest-weight representation is generated by such a vector.

Let $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ be the corresponding Borel subalgebra. Define its one-dimensional Lie algebra representation ${\mathbb{C}}_{\omega }$ by the scalar $\omega (h)$ on $H$ and zero on ${\mathfrak{n}}^{+}$. This respects the Lie bracket, since $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. The induced Verma moduleis nonzero: the Poincaré-Birkhoff-Witt theorem identifies its underlying vector space with $U({\mathfrak{n}}^{-})$, and $v=1\otimes 1$ is a primitive element of weight $\omega$. Its weight spaces are finite dimensional, its top weight space is the line $\mathbb{C}v$, and all other weights are $\omega -\sum n_i{\alpha }_i$ with $n_i\ge 0$ and at least one positive coefficient.

A proper submodule cannot contain $v$, because $v$ generates $M(\omega )$. More strongly it has no component of weight $\omega$: any finite sum of distinct weight vectors can be projected onto its individual components by a polynomial in a generic element of $H$. Therefore the sum $N$ of all proper submodules still misses the top weight line and is proper. It contains every proper submodule, so the irreducible quotient of a Verma moduleis irreducible and retains the nonzero primitive element $v+N$. This constructs the requested representation for every $\omega \in H^{\ast }$. It is not asserted to be finite dimensional for arbitrary $\omega$.

For the sl2 Lie algebra, use $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The classification of finite-dimensional sl2 representations gives one irreducible $V_n$ for every integer $n\ge 0$. In a basis $v_0,\dots ,v_n$ its action iswhere $v_{-1}=v_{n+1}=0$. These formulas satisfy the three Lie brackets. Any nonzero invariant subspace contains a weight vector by polynomial projection using $h$; repeated application of $e$ gives $v_0$, and applications of $f$ then give the entire basis. Thus $V_n$ is irreducible.