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For $v\in V_{\lambda }$ and $w\in W_{\mu }$, the tensor product of Lie algebra representations acts byThus its weight space of weight $\nu$ isSubstituting this into the formal character and using $e(\lambda +\mu )=e(\lambda )e(\mu )$ provesEvery sum is finite for the Lie algebra representations under consideration.

The ambient group ring of a weight lattice isAddition is coefficientwise, and multiplication uses addition in the weight lattice. For a finite-dimensional Lie algebra representation $V={⨁}_{\lambda }{V}_{\lambda }$, its formal character isThe formal character records the dimensions of all weight spaces. The ambient lattice ring should be distinguished from the representation ring of a semisimple Lie algebra, which identifies with its Weyl-invariant subring.

For the ${\mathfrak{sl}}_3$ module ${\Gamma }_{1,2}$, $L_1+L_2+L_3=0$ and the highest weight is $L_1-2L_3={\omega }_1+2{\omega }_2$. The decomposition $V\otimes {\mathrm{Sym}}^2{V}^{\ast }={\Gamma }_{1,2}\oplus V^{\ast }$ givesCollecting equal weights yieldsThe six first weights and three second weights have multiplicity one; each of the three weights $-L_i$ has multiplicity two. Their total is $6+3+6=15$, as required.

Use the standard sl2 triple $H=x{\partial }_x-y{\partial }_y$, $X=x{\partial }_y$, $Y=y{\partial }_x$. In the tensor product of Lie algebra representations, each operator acts on both factors. PutThese twelve vectors form a weight basis, and $H{u}_{ij}=(5-2(i+j))u_{ij}$. Thus the bases of the weight spaces areThere are no other weights. The weight diagram shows their multiplicities, with the decomposition also displayed for comparison.

The raising operator acts bywith out-of-range terms omitted. At weights five, three and one its kernels are respectively spanned byAt weight one the kernel equations are $2a+b=0$ and $b+2c=0$ for $a{u}_{20}+b{u}_{11}+c{u}_{02}$. At weight minus one they are $3a+b=0$, $2b+2c=0$, $c=0$, which force zero; the maps on the two remaining negative-weight spaces are injective as well. Thus these three lines give all nonzero highest-weight vectors up to scalar, at their respective weights. A linear combination of different lines is killed by $X$ but is not a weight vector, so is not a highest-weight vector.

The lowering operator isIn particular,is an explicit basis for the submodule ${\Gamma }_1$. Indeed $Y^2{h}_1=0$, $X{h}_1=0$, $X(Y{h}_1)=h_1$, and the two vectors have weights $1,-1$. The analogous strings from $h_5,h_3$ have dimensions six and four. Their distinct irreducible highest weights and total dimension twelve give the sl2 tensor product of cubic and quadratic symmetric powers

Let $f_1,f_2,f_3$ be the dual basis of $W^{\ast }$, and use $L_1+L_2+L_3=0$. The elementary tensors $e_i\otimes e_j\otimes f_k$ have weight $L_i+L_j-L_k$. Grouping them gives every weight space explicitly:The counts are $6+3\cdot 2+3\cdot 5=27$, so this table exhausts the tensor product. The diagrams below use equilateral coordinates $L_1=(1,0)$, $L_2=(-1/2,\sqrt{3}/2)$, $L_3=(-1/2,-\sqrt{3}/2)$; numbers at points are weight multiplicities.

To decompose the module, first split the first two factors into their symmetric and alternating parts. Put $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, including $t_{ii}=e_i\otimes e_i$. The equivariant contractionis onto, since $C(t_{ii}\otimes f_i)=2e_i$. Its kernel $K$ has dimension fifteen and contains the highest-weight vector $t_{11}\otimes f_3$ of weight $2L_1-L_3$.

Here is an explicit weight basis of the trace-free symmetric-square dual tensor module, and thus the requested fifteen-vector basis:Every displayed vector contracts to zero. The first two families occupy distinct one-dimensional weights; at each weight $L_i$ the two vectors in the last family are independent because their respective $t_{ij}\otimes f_j$ terms are distinct. Thus all fifteen are independent and span $K$.

For completeness, $K$ is irreducible, not just an invariant kernel of the right dimension. On elementary tensors the raising action isAt the six weights $2L_i-L_k$, both $E_{12}$ and $E_{23}$ kill the vector only for $(i,k)=(1,3)$. At each weight $-2L_k$, at least one of those operators is nonzero. At weight $L_1$, write a vector as $a(t_{12}\otimes f_2-t_{11}\otimes f_1/2)+b(t_{13}\otimes f_3-t_{11}\otimes f_1/2)$. The two raising equations are $3a+b=0$ and $-a+b=0$, forcing zero. On the two-dimensional weight-$L_2$ space $E_{12}$ is injective, and on the weight-$L_3$ space $E_{23}$ is injective, as substitution in the same action formula shows. Therefore the common raising kernel in $K$ is the single line $\mathbb{C}(t_{11}\otimes f_3)$. By Weyl complete reducibility theorem, every irreducible summand supplies a highest-weight line, so $K$ has only one summand and is irreducible.

The symmetric-factor contraction splits off a defining copy $W$; explicitly $e_i\mapsto {\sum }_j{t}_{ij}\otimes f_j$ is equivariant and its contraction is $4e_i$. For the alternating factor, the invariant volume form identifies ${\Lambda }^{2}W\cong W^{\ast }$, sending $e_1\wedge e_2$ to $f_3$, $e_2\wedge e_3$ to $f_1$ and $e_3\wedge e_1$ to $f_2$. ThereforeThe symmetric dual module is irreducible of dimension six and highest weight $-2L_3$; its symmetric-monomial basis has weights $-2L_i$ and $-L_i-L_j=L_k$, all of multiplicity one. In the original tensor product its highest-weight vector isThe raising action above annihilates it, and its weight is $L_1+L_2-L_3=-2L_3$.

Altogether the Triple tensor decomposition for the defining sl3 representation isThe fifteen-dimensional module has multiplicity one at each $2L_i-L_k$ and $-2L_i$, and multiplicity two at each $L_i$. The six-dimensional module has multiplicity one at $-2L_i$ and $L_i$. Each defining module has the three weights $L_i$, once each. These add to the original table of multiplicities.

The two defining summands are not intrinsically distinguished: other diagonal copies inside their isotypic component have the same three-point weight diagram. Thus the displayed diagrams account for every irreducible submodule type, as well as showing both copies in this decomposition.

Choose simple roots ${\alpha }_1,\dots ,{\alpha }_r$. For the orbit statement one must use the Closed dominant Weyl chamberIts interior, defined using strict inequalities, is the open Weyl chamber. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.

Choose $\rho$ strictly inside the chamber, for example the sum of the fundamental weights. The finite Weyl group orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda ,\rho )$. If $(\lambda ,{\alpha }_i)<0$ for some simple root, thencontradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda \in \mathcal{W}$. Consequently some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber. The argument in fact works for every vector in the real weight space.

Put $V={\mathbb{C}}^n$. We use complete reducibility of compact-group representations, and the highest-weight classification: the multiplicity of $V_{\lambda }$ in a completely reducible representation is the dimension of its weight-$\lambda$ subspace killed by all positive root vectors. Each irreducible summand contributes just its one-dimensional highest-weight line to that subspace.

Symmetrizing and antisymmetrizing the three tensor factors give invariant subspaces ${\mathrm{Sym}}^{3}V$ and ${\Lambda }^{3}V$ of $V^{\otimes 3}$. The vectors $e_1^{\otimes 3}$ and $e_1\wedge e_2\wedge e_3$ are highest-weight vectors of weights $(3,0,\dots ,0)$ and $(1,1,1,0,\dots ,0)$ respectively. The unitary version of the Weyl dimension formula isIt givesThese are exactly the dimensions of the third symmetric power and exterior power, so each is irreducible and occurs once in those subspaces.

The weight $(2,1,0,\dots ,0)$ in $V^{\otimes 3}$ has basisThe only positive root vector that acts nontrivially on this weight space is $E_{12}$. Acting on the tensor product as the sum of its actions on the three factors, it sends each of $a,b,c$ to $e_1^{\otimes 3}$. Consequently the highest-weight vectors in this space are preciselyThis is a two-dimensional space, so $V_{(2,1)}$ occurs exactly twice. Its dimension isThe dimensions of the summands already found add toThus no other irreducible summand can remain. We have proved the full decompositionThere are four summands counted with multiplicity and three distinct isomorphism types.

For $n=2$, the exterior cube vanishes. The third symmetric power still has highest weight $(3,0)$ and dimension $4$, while the same two-dimensional highest-vector calculation gives two copies of the representation of weight $(2,1)$ and dimension $2$. It is $\det \otimes V$, since a determinant twist shifts the standard highest weight $(1,0)$ to $(2,1)$. Hence

Use the comultiplication $\Delta E=E\otimes K+1\otimes E$, $\Delta F=F\otimes 1+K^{-1}\otimes F$, $\Delta K=K\otimes K$ of the quantum enveloping algebra of sl2. Choose a basis $u_0,u_1$ for $V_{1,1}$ with $F{u}_0=u_1$, $E{u}_1=u_0$, and $K{u}_i=q^{1-2i}{u}_i$. Choose a basis $w_0,w_1,w_2$ for $V_{1,2}$ withand all terminal raising or lowering actions zero. This is one of the corrected bases from part (a).

In the tensor product of modules, putThe vector $z_0$ is killed by $E$ and has $K$ eigenvalue $q^3$. Direct application of the comultiplication givesand the string formula gives $E{z}_1=[3]z_0$, $E{z}_2=[2]z_1$, $E{z}_3=z_2$. Also $K{z}_i=q^{3-2i}{z}_i$. Thus $W_3=⟨z_0,z_1,z_2,z_3⟩$ is a submodule isomorphic to $V_{1,3}$.

The other highest vector and its lowered partner areIndeed $E{h}_0=([2]q^2-q^2[2])u_0\otimes w_0=0$, $K{h}_0=q{h}_0$, $K{h}_1=q^{-1}{h}_1$, $E{h}_1=h_0$ and $F{h}_1=0$. Therefore $W_1=⟨h_0,h_1⟩$ is a submodule isomorphic to $V_{1,1}$.

To verify that this is a direct sum, in the $q$ weight space the determinant of the coefficients of $z_1,h_0$ is $-q^2-q^{-1}[2]=-[3]\ne 0$. In the $q^{-1}$ weight space, the coefficient determinant of $z_2,h_1$ is $-q[2]-q^{-2}=-[3]\ne 0$. The extreme weight spaces are spanned by $z_0,z_3$. Hence these six vectors form a basis of the entire tensor product of modules. Each summand is a simple module: projecting a nonzero vector onto one of its distinct $K$ weight spaces and applying the nonzero raising and lowering coefficients recovers its entire basis. Consequently

At a primitive cube root of unity, $[1]=1$, $[2]=-1$, $[3]=0$. In the degree-three quantum plane module, put $z_0=x^3$, $z_1=x^{2}y$, $z_2=x{y}^2$, $z_3=y^3$. The complete actions areHence $W=⟨z_0,z_3⟩$ is a two-dimensional trivial submodule; every line in $W$ is a simple module. The quotient by $W$ is a two-dimensional simple module, with distinct $K$ eigenvalues $q,q^{-1}$ and nonzero maps connecting its two weight spaces.

Let $S$ be a simple module that is a submodule of this cubic component. If $S\cap W\ne 0$, simplicity makes $S\subseteq W$, so $S$ is one of its lines. If $S\cap W=0$, the quotient map injects $S$ into the simple quotient. Its nonzero image is the entire quotient, so $\dim S=2$. Since $K$ is diagonalizable, such an $S$ must contain the unique $q$ weight space $\mathbb{C}{z}_1$. But $E{z}_1=z_0\ne 0$, which then belongs to $S\cap W$, a contradiction. Therefore the cubic-root quantum-plane socle gives the complete answer:All simple submodules lie in the two-dimensional $W$, so their sum cannot equal the four-dimensional cubic component. It is not a direct sum of simple modules. The obstruction is the nonzero extension to the simple two-dimensional quotient, not a failure of that quotient to be simple.

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.

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 vector in a rational representation is an unstable vector in a rational representation when $0\in \overline{Gv}$, with Zariski closure. For $G=S{L}_n(\mathbb{C})$, the Hilbert-Mumford criterion for the affine null cone isEvery algebraic one-parameter subgroup is conjugate to $\mathrm{diag}(t^{a_1},\dots ,t^{a_n})$ with integer $a_j$ summing to zero. This follows by decomposing the standard representation into characters of the multiplicative group, then imposing determinant one. In its weight decomposition, a limit of zero means that only strictly positive weights occur.

The reverse implication in the criterion is immediate: the subgroup lies in $G$ and supplies points of the orbit tending to zero. For the harder implication, the complex orbit is locally closed and Zariski open in its closure. A nonempty Zariski-open subset of a complex irreducible variety is dense also in the usual topology. Thus choose $g_{j}v\to 0$ in the ordinary vector-space topology. Singular value decomposition, with determinant phases adjusted, givesUse an $S{U}_n$-invariant Hermitian inner product on $V$. By compactness, after a subsequence ${\ell }_j\to \ell$. Put $w=\ell v$, and decompose it into the weights of the diagonal torus, $w={\sum }_{\chi }{w}_{\chi }$. These weight spaces are orthogonal because the unitary diagonal torus preserves the inner product. HenceFor every $\chi$ with $w_{\chi }\ne 0$, its final norm is bounded away from zero for large $j$, forcing $\chi (b_j)\to -\infty$. There are only finitely many such weights, so choose a single real vector $b$ with sum zero and $\chi (b)<0$ for all of them. The inequalities are strict with integer coefficients. A nearby rational vector still satisfies them; multiply its negative by a common denominator to obtain integers $a_r$ of sum zero with $\chi (a)>0$ throughout this support. Then $\mu (t)=\mathrm{diag}(t^{a_r})$ takes $w$ to zero. Conjugating gives $\lambda (t)={\ell }^{-1}\mu (t)\ell$ taking $v$ to zero, completing the proof. The zero vector itself satisfies the criterion trivially.

For degree-six binary forms, a nontrivial algebraic one-parameter subgroup of $S{L}_2$ is, after conjugation and choosing its direction, $\mathrm{diag}(t^a,t^{-a})$ with $a>0$. On $x^{6-j}{y}^j$ the weight is $a(6-2j)$. Strictly positive weights permit only $j=0,1,2$, so the form is divisible by $x^4$. Conversely every such form tends to zero under that subgroup. Changing basis replaces $x$ by any linear factor. The root-multiplicity criterion for unstable binary forms therefore givesA triple root alone is insufficient: $x^3{y}^3$ has weight zero and is not unstable. Nor does merely having a repeated root suffice.

An affine algebraic group is a linearly reductive algebraic group if every finite-dimensional rational representation is completely reducible; equivalently the functor of invariants is exact. It is a geometrically reductive algebraic group if, whenever $0\ne v\in V^G$, there is a homogeneous invariant polynomial $F\in k[V{]}^G$ of positive degree with $F(v)\ne 0$. Linear reductivity permits degree one; geometric reductivity allows higher degree.

Over the algebraically closed field, an algebraic torus is $T\cong ({\mathbb{G}}_m{)}^r$. Its coordinate ring is the Laurent polynomial ring. A rational action writes the coaction of a vector as a finite sum $\delta (v)={\sum }_{\chi }{v}_{\chi }\otimes t^{\chi }$. The coaction identity and linear independence of Laurent monomials give $\delta (v_{\chi })=v_{\chi }\otimes t^{\chi }$ and $v={\sum }_{\chi }{v}_{\chi }$. ThusEach weight space has scalar action, so any vector-space complement within each weight space is invariant. This proves complete reducibility, and hence tori are linearly reductive in every characteristic. The argument uses characters themselves, not their Lie algebra differentials, which could coincide modulo $p$.

We now prove geometric reductivity of $G=S{L}_2(k)$ by determinant separation for positive-characteristic SL2. Fix a rational $G$-module $V$ and a nonzero fixed vector $v$. For the diagonal torus, choose a weight-zero functional $l\in V^{\ast }$ with $l(v)=1$, using the torus decomposition just proved. Define matrix coefficientsThese are regular functions on $G$, and $\Phi (hw)(g)=\Phi (w)(h^{-1}g)$ makes $\Phi$ equivariant for left translations. Moreover $\Phi (w)(gt)=\Phi (w)(g)$ because $l$ is torus-invariant. Thus $\Phi$ maps into $k[G{]}^T$ for the right torus action, and $\Phi (v)=1$.