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In the notation of part (i), the tensor raising operator and lowering operator act bywith terms beyond the stated index range omitted. Both $v_4=A_0$ and $v_2=B_0-A_1$ are killed by $e$, and have respective weights four and two. They are the desired highest-weight vectors.

For $v_4$, successive lowering givesThus a basis of the five-dimensional summand, in descending weight order, isFor $v_2$, the successive vectors are $f{v}_2=2(B_1-A_2)$, $f^2{v}_2=2(B_2-A_3)$ and $f^3{v}_2=0$. HenceThese spans are invariant under $h,e,f$. The sl2 highest-weight lowering formula gives $e{f}^r{v}_n=r(n-r+1)f^{r-1}{v}_n$, so all interior raising coefficients in either chain are nonzero. Any nonzero submodule therefore contains the highest vector and then the whole chain; each span is irreducible.

At the shared weights $2,0,-2$, the two displayed summand vectors are independent: their coefficient determinants in the original weight bases are $4,2,4$. At weights $4,-4$ only the five-dimensional summand occurs. Consequently the eight displayed vectors form a basis of $U$, proving that these are complementary submodules, not only modules with the right dimensions.

The displayed identity is valid for a highest-weight vector $v$, satisfying $Xv=0$ and $Hv=kv$, not for arbitrary $v\in V$. For example, in the defining module with $k=1$, take $v=y$: then $XYv=0$ but the proposed right side for $n=1$ is $v\ne 0$.

For a highest-weight vector, the sl2 triple relations give $H{Y}^{j}v=(k-2j)Y^{j}v$. The operator identitycomes from expanding a commutator with a product. Apply it to $v$, using $Xv=0$. Each term becomes $(k-2(n-1-j))Y^{n-1}v$, so the sum isThis proves the intended sl2 highest-weight lowering formula. The identity valid for an arbitrary vector is instead

On the basis $v,Yv,\dots ,Y^{k}v$ of the irreducible module, $XY$ has eigenvalue $(j+1)(k-j)$ on $Y^{j}v$, for $0\le j\le k$. Summing gives the trace of a raising-lowering product in an irreducible sl2 module:The last equality follows by the formulas for the sums of the first $k$ integers and their squares.

Here representations mean finite-dimensional rational representations of $S{L}_2(\mathbb{C})$. Put $W={\mathbb{C}}^2$. The irreducibles areThe invariant alternating form on $W$ identifies $W$ with $W^{\ast }$, so these can equally be realized as binary forms of degree $d$. With basis $e_j=x^{d-j}{y}^j$, the standard Lie algebra generators act asThe distinct torus weights and these nonzero lowering and raising coefficients show irreducibility: any nonzero invariant subspace contains a weight vector, and repeated applications of $E,F$ obtain the entire basis.

For exhaustion, choose a highest-weight vector $v$ of an irreducible representation, so $Hv=mv$, $Ev=0$. The sl2 highest-weight lowering formula, obtained by induction from $[E,F]=H$, is $E{F}^{j}v=j(m-j+1)F^{j-1}v$. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, applying $E$ to the latter equality gives $(r+1)(m-r)F^{r}v=0$, hence $m=r$. The vectors $v,Fv,\dots ,F^{r}v$ have distinct weights and span an invariant submodule, so irreducibility makes it the whole representation. After rescaling basis vectors it is $V_r$. Complete reducibility from Question 2 now describes every representation as a finite direct sum of these symmetric powers. Their central element $-I$ acts by $(-1{)}^d$.

The symbolic method for binary-form invariants works as follows. Write a degree-$d$ form as $f(x,y)={\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{d}{j}\right)f_j{x}^{d-j}{y}^j$ and introduce a formal letter $a=(a_0,a_1)$. The formal power $(a_{0}x+a_{1}y{)}^d$ is converted to $f$ by the linear substitution $a_0^{d-j}{a}_1^j\mapsto f_j$. This is an umbral linear rule, not a claim that every form is an actual power of a linear form.

For a coefficient polynomial of degree $r$, polarize it into $r$ copies of the form, represented by letters $a,b,\dots$. Each letter must occur with degree $d$. In characteristic zero, polarization and evaluation on equal forms recover the original coefficient polynomial. The basic invariant expressions in the letters are bracketsThey are invariant because $\det g=1$. Invariant tensors on copies of a two-dimensional space are spanned by pairwise alternating contractions; equivalently, the invariant polynomials in the letters are generated by these brackets. One can see the tensor statement from the two-row Young-symmetrizer description of the determinant powers in tensor space. Their basic relations include $(ab)=-(ba)$ and $(ab)(cd)-(ac)(bd)+(ad)(bc)=0$. Multiply brackets, ensure the required degree in each letter, symmetrize letter copies as needed, and apply the umbral substitutions. This gives all coefficient invariants, and also works for joint invariants of several forms.

For example, for $f=A{x}^2+2Bxy+C{y}^2$, the expression $(ab{)}^2$ becomes $2(AC-B^2)$, the quadratic discriminant up to its conventional scalar. Contractions leaving variable indices unpaired give covariants. A useful systematic construction is the transvectant of binary forms:For forms of degrees $m,n$, this is an equivariant form of degree $m+n-2r$. The corresponding tensor decomposition isIts character identity follows by multiplying the weight sums; complete reducibility then gives the decomposition, and alternating contractions supply the displayed maps onto its summands.

Apply the lowering operator $F$ to the highest-weight vector from part (ii). Since $F{w}_{03}=4w_{13}+w_{04}$ and $F{w}_{12}=2w_{13}$, we obtainThis is nonzero because $w_{04},w_{13}$ are distinct elements of the exterior power basis. Both have weight zero. As a consistency check, the raising operator sends this vector to $2v_{+}$, as required by the sl2 highest-weight lowering formula.

Let $v$ be a highest-weight vector of weight $\lambda$. For a simple root ${\alpha }_i$, choose its sl2 subalgebra associated with a root, with operators $E_i,F_i,H_i$ satisfying $[E_i,F_i]=H_i$, $[H_i,F_i]=-2F_i$, and $H_i={\alpha }_i^{\vee }$. We have $E_{i}v=0$ and $H_{i}v=\lambda (H_i)v$.

Since $F_i$ decreases the weight by ${\alpha }_i$ and there are only finitely many weights of a representation, let $n\ge 0$ be maximal with $F_i^{n}v\ne 0$. Induction using the two bracket relations gives the sl2 highest-weight lowering formulaApplying this at $k=n+1$, the left side vanishes, and the nonzero vector $F_i^{n}v$ on the right forces $\lambda (H_i)=n$. This holds for every simple root, soThus the highest weight is a dominant integral weight, in particular an element of the Closed dominant Weyl chamber.

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.

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.

Conversely, in any finite-dimensional irreducible module, start with an eigenvector of $h$ and apply $e$ until reaching a nonzero vector $v$ with $ev=0$. This process terminates because $e$ increases the eigenvalue by two and only finitely many eigenvalues occur. Write $hv=av$. The sl2 highest-weight lowering formula givesLet $f^{n}v\ne 0$ and $f^{n+1}v=0$, which again follows from the finite set of weights. Applying $e$ to the latter identity yields $(n+1)(a-n)f^{n}v=0$, so $a=n$. The resulting $n+1$ distinct weight vectors span an invariant subspace, hence the entire irreducible module. Therefore $\dim V_n=n+1$ and its primitive weight is ${\omega }_n(h)=n$, with $n=0,1,2,\dots$; its weights are $n,n-2,\dots ,-n$.

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.