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For finite-dimensional rational representations of complex $G{L}_m$ or $S{L}_m$, average a Hermitian inner product over the compact group $U(m)$ or $SU(m)$ using normalized Haar measure. Orthogonal complements become invariant under the compact group and its complexified Lie algebra. The latter generates the complex group, giving invariant complements and hence complete reducibility. This conclusion does not hold for arbitrary affine algebraic groups, such as the additive group.

Tensoring a rational representation of the general linear group with a determinant power shifts every highest-weight coordinate by the same integer $r$. Positive twists can clear determinant denominators; negative twists supply rational representations that do not extend to singular matrices. Twisting preserves irreducibility and dimension.

For every finite-dimensional rational representation $V$ and every nonzero $v\in V^G$, a geometrically reductive algebraic group admits a homogeneous invariant polynomial $F\in k[V{]}^G$ of positive degree with $F(v)\ne 0$. A linearly reductive group can take degree one. In positive characteristic, higher degree is essential: a power $p^r$ can separate a fixed vector even when an invariant linear functional cannot.

An affine algebraic group is linearly reductive when every finite-dimensional rational representation is completely reducible. Equivalently its invariants functor is exact. Over a field of characteristic zero, reductive algebraic groups are linearly reductive; in positive characteristic this stronger property fails for many reductive groups. An algebraic torus is linearly reductive in every characteristic because its rational representations decompose into character spaces.

For a real Lie group $G$, its Lie algebra is the real tangent space $T_{e}G$ at the identity. Each tangent vector extends uniquely to a left-invariant vector field, and the commutator of those fields defines the Lie bracket. For a matrix group this bracket is $[X,Y]=XY-YX$.

For a complex affine algebraic group, the Lie algebra of an affine algebraic group is the complex Zariski tangent space $T_{e}G={\mathrm{Hom}}_{\mathbb{C}}({\mathfrak{m}}_e/{\mathfrak{m}}_e^2,\mathbb{C})$. Its bracket is the commutator of left-invariant derivations. In a faithful matrix realization, one can compute the tangent space using dual numbers: $X$ belongs to it exactly when $I+\epsilon X$ satisfies the group equations modulo ${\epsilon }^2$. This is a complex Lie algebra, whereas compact groups such as $U_n$ and $S{U}_n$ have real Lie algebras.

Differentiate $g^{\ast }g=I$ at the identity to get $X^{\ast }+X=0$. Conversely $e^{tX}$ is unitary for every skew-Hermitian $X$. Differentiating the determinant gives $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$, while $\det e^{tX}=e^{t\mathrm{tr}X}$ proves sufficiency of the trace condition. Finally, $G{L}_n(\mathbb{C})$ is an open subset of all complex matrices, so its tangent space has no linear constraint. ThereforeTheir dimensions are respectively $n^2,n^2,n^2-1,n^2-1$, with real dimensions for the first and third and complex dimensions for the second and fourth. In every case the Lie bracket is the matrix commutator.

An irreducible complex affine algebraic group is connected. It is a reductive algebraic group when its unipotent radical $R_u(G)$, the largest connected normal unipotent subgroup, is trivial. This is an algebraic condition; in characteristic zero it also implies complete reducibility of finite-dimensional rational representations, as proved in Question 2.

Examples are algebraic tori, $G{L}_n(\mathbb{C})$ and $S{L}_n(\mathbb{C})$, and products of these groups. A diagonal matrix that is unipotent must be the identity, so a torus has no nontrivial unipotent subgroup. For the two matrix groups, let $N$ be a normal unipotent subgroup. The Kolchin theorem supplies a nonzero fixed vector in their standard representation. The subspace fixed by $N$ is invariant under the whole group: for $g\in G$, $n(gv)=g(g^{-1}ng)v=gv$. The standard representation is irreducible, so every vector is fixed by $N$; its faithfulness makes $N$ trivial. A normal unipotent subgroup of a product projects to such a subgroup of each factor, proving the product assertion. By contrast the additive group and a nontrivial upper-unitriangular group are not reductive: their unipotent radicals are the whole group.

For a compact Lie group, take any positive-definite Hermitian inner product $h_0$ on $V$ and average using probability Haar measure:It remains positive definite, because the integrand is positive for a nonzero vector, and translation invariance makes it $G$-invariant. If $W\subset V$ is an invariant subspace, then $W^{\perp }$ is invariant too: $h(gv,w)=h(v,g^{-1}w)=0$ when $v\in W^{\perp }$. Thus $V=W\oplus W^{\perp }$. Induction on dimension proves complete reducibility. This uses the usual continuous finite-dimensional representations of the compact group.

For a connected complex reductive algebraic group, representations are understood to be rational representations. Use a compact real form $K\subset G$ with $\mathrm{Lie}(G)=\mathrm{Lie}(K){\otimes }_{\mathbb{R}}\mathbb{C}$. The structural construction takes a compact real form of the semisimple derived factor and the unit circles in the central torus, then passes through their finite central quotient. This is the compact-form structure of a complex reductive group, rather than an assumption that its given representation is already unitary.

The group $K$ is Zariski dense: the Lie algebra of its Zariski closure contains the complex span of $\mathrm{Lie}(K)$, hence all of $\mathrm{Lie}(G)$; connectedness then makes that closure $G$. Average over $K$ as above. The orthogonal complement of a $G$-submodule is $K$-invariant. Its stabilizer is a Zariski-closed subgroup, since preserving a fixed subspace is a system of algebraic matrix conditions. The stabilizer contains dense $K$, so it is all of $G$. This again gives an invariant complement to every submodule and provesIf reductivity is defined to allow finitely many components, first obtain a $G^0$-equivariant projection $P:V\to W$. Average its conjugates over the finite component group. Since $P$ commutes with $G^0$, the average is independent of coset representatives; it is $G$-equivariant and remains the identity on $W$. Its kernel supplies the required complement. Thus the result holds in that convention too.

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.

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 $\Phi (v)=1$.

Write a matrix's two columns as $a=(a_0,a_1)$ and $b=(b_0,b_1)$, with $\Delta =a_0{b}_1-a_1{b}_0=1$. Right torus multiplication scales these columns by $t$ and $t^{-1}$. Every right-torus-invariant regular function has a weight-zero polynomial lift, hence a sum of polynomials of bidegrees $(j,j)$ in $(a,b)$. To see that a weight-zero lift exists, take any lift and retain its zero-weight component; the ideal $(\Delta -1)$ is torus-stable and the quotient map preserves weights. Multiply each piece by ${\Delta }^{n-j}$ to put the entire sum in bidegree $(n,n)$ without changing its restriction to $G$.

Let $P_n$ be the space of degree-$n$ polynomials in one column. Restriction gives an injective equivariant map $P_n\otimes P_n\to k[G{]}^T$. Indeed if a bidegree-$(n,n)$ polynomial vanishes on determinant-one matrices, scale the first column of any invertible matrix to make its determinant one. Homogeneity then makes the polynomial vanish on every invertible matrix, a dense open subset, so it is zero. Denote the image by $R_n$. These spaces increase because multiplication by $\Delta$ realizes $R_{n-1}\subset R_n$, and their union is $k[G{]}^T$. The finite-dimensional image $\Phi (V)$ therefore lies in one $R_n$, and in any sufficiently larger filtration piece.

Choose $q=p^s$ with $q-1$ at least that bound, and set $n=q-1$. The identity $(1+z{)}^q=1+z^q$ in characteristic $p$ givesThe invariant tensortherefore defines an invertible equivariant linear map $D:P_n^{\ast }\to P_n$: its matrix is anti-diagonal with all entries nonzero. Interpret any tensor $B\in P_n\otimes P_n$ as a map $P_n^{\ast }\to P_n$ and send it to $B{D}^{-1}\in \mathrm{End}(P_n)$. Equivariance of $D$ makes this identification intertwine the diagonal tensor action with conjugation on endomorphisms. Crucially, $D$ restricts to the constant function $1$ on $G$, and is sent to the identity endomorphism.

Under the resulting equivariant linear map $V\to \mathrm{End}(P_n)$, the vector $v$ is sent to $I$. Since $\dim P_n=n+1=q$, the endomorphism determinant is a homogeneous polynomial of positive degree $q$, invariant under conjugation. Pull it back to $V$. The resulting polynomial satisfiesThis constructs the required invariant for every fixed nonzero vector and proves $S{L}_2(k)$ is geometrically reductive, including characteristic two. No averaging by a group order or division by $q$ is used; those operations would fail in the field's characteristic.

For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

For a decreasing integer tuple $\lambda =({\lambda }_1,\dots ,{\lambda }_m)$, let ${\mu }_i={\lambda }_i-{\lambda }_m$. Then $\mu$ is a partition and the highest-weight classification of rational GL representations definesThis is a determinant twist of a Schur module. Every irreducible rational $G{L}_m$ representation becomes polynomial after multiplication by a sufficiently large positive determinant power, which clears all matrix-entry denominators. The polynomial degree decomposition and Schur–Weyl duality then identify it with a Schur module. Undoing the twist gives exactly one decreasing integer tuple $\lambda$. Distinct tuples have distinct highest torus weights, so these are the complete pairwise nonisomorphic irreducible rational representations.

The Weyl character formula specializes toIt is a symmetric Laurent polynomial in the eigenvalues. Equality extends from the dense set of diagonalizable matrices to all invertible matrices: both the character and the expression in the characteristic-polynomial coefficients are regular functions on $G{L}_m$. For a polynomial representation this also extends to every endomorphism of $V$. For a general rational representation, the printed claim at singular endomorphisms needs this qualification: for example ${\det }^{-1}$ is undefined at a singular matrix. The displayed formula is valid on $G{L}_m$, and on all of $\mathrm{End}(V)$ when ${\lambda }_m\ge 0$.

To compute the degree, set $x_j=e^{t{c}_j}$ with distinct $c_j$ and take $t\to 0$. For ${\ell }_i={\lambda }_i+m-i$ and $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, the leading coefficient of an exponential alternant isIndeed, expand every exponential in powers of $t$. The first nonzero determinant uses the distinct powers $0,1,\dots ,m-1$; its coefficient is the product of the two Vandermonde determinants divided by $\prod k!$. Taking the same expansion in the denominator cancels the powers and the $c$ factors, giving the Weyl dimension formulaThis proof works for negative ${\lambda }_m$ as well, since determinant twists have dimension one.

Every finite-dimensional rational module is completely reducible. The characters of its irreducible constituents are linearly independent: each Schur Laurent character has its highest dominant monomial $x^{\lambda }$ with coefficient $1$, and only lower weights besides it. In a finite relation, choose a lexicographically highest remaining weight; its coefficient must vanish, and iterate. Therefore equal characters give equal multiplicities of every irreducible constituent, proving rational modules with the same character are isomorphic.

Finally the symmetric algebra of $V\oplus {\Lambda }^{2}V$ has the formal torus characterEach factor sums the symmetric powers of a one-dimensional weight space; the exterior square has weights $x_i{x}_j$ for $i<j$. The permitted Schur identity makes this ${\sum }_{\lambda }{s}_{\lambda }(x)$. In each fixed scalar degree there are only finitely many terms, so complete reducibility and character independence apply degree by degree without a convergence assumption. Thus the multiplicity-free symmetric-algebra model for polynomial GL representations contains each irreducible polynomial representation exactly once. The word irreducible is necessary: arbitrary reducible polynomial modules, such as two copies of the trivial module, do not each occur once in a multiplicity-free sum.

A rational representation is a group homomorphism whose matrix coefficients are regular functions on the affine algebraic group. For the general linear group, those functions are polynomials in matrix entries and the inverse determinant. The word rational does not permit poles at points of the group.