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The commutant of $A\subseteq \mathrm{End}(T)$ is $A\prime =\{b:ba=ab\text{ for all }a\in A\}$. This centralizer is an operator algebra. The double-centralizer theorem for semisimple operator algebras and Schur–Weyl duality use this meaning, which is distinct from the centralizer of a single element of a group.

For a finite-dimensional semisimple algebra acting faithfully on $T$, write $T={⨁}_i{S}_i\otimes M_i$. Then the algebra acts by the full matrix algebra on each simple factor, while its commutant acts by the full matrix algebra on each multiplicity factor. Taking the commutant twice recovers the original image. This is the algebraic basis of Schur–Weyl duality.

Invariant theory studies functions and tensors unchanged by a group action. For a linear representation, the polynomial invariant ring records invariant polynomial functions on the representation space. Tensor invariants and commuting algebra actions connect it with Schur–Weyl duality.

On $T=V^{\otimes n}$, define the permutation and diagonal actions byThe inverse in the first formula gives a left action. Applying $g$ to every tensor factor and then reordering produces the same tensor as reordering and then applying $g$. Thus the two actions commute. The Schur algebra isIt is the commutant of an operator algebra of the permutation action.

To identify this commutant, use the multilinear isomorphismConjugation by a permutation reorders the factors on the right. Therefore the invariant subspace is the space of symmetric tensors of degree $n$ in $\mathrm{End}(V)$. It is spanned by $a^{\otimes n}$. Indeed, the polarization identityexpresses every symmetrized elementary tensor as a linear combination of pure powers. Those symmetrized elementary tensors span the invariant subspace in characteristic zero.

One may restrict $a$ to invertible endomorphisms without changing the span. For fixed $a$, the vector-valued polynomial $(a+tI{)}^{\otimes n}$ has degree at most $n$. Choose $n+1$ distinct values of $t$ away from the finitely many roots of $\det (a+tI)$. Polynomial interpolation expresses its constant term $a^{\otimes n}$ as a linear combination of those invertible tensor powers. ConsequentlyThe span is an algebra, since products are $(gh{)}^{\otimes n}$.

The image $A_T$ of $\mathbb{C}{S}_n$ is semisimple, as a quotient of a semisimple algebra. The double-centralizer theorem for semisimple operator algebras gives $A_T^{\prime \prime }=A_T$. Since the preceding computation says that $A_T^{\prime }$ is the span of the general linear action, the two actions are mutual commutants. More explicitly, semisimple module decomposition givesHere $D_{\lambda }(V)={\mathrm{Hom}}_{S_n}(S^{\lambda },T)$ is the multiplicity space, with its natural general linear action. The commutant algebra is the product of the full endomorphism algebras of these multiplicity spaces. Hence each nonzero $D_{\lambda }$ is irreducible, and distinct spaces have nonisomorphic general linear representations, because the operators $g^{\otimes n}$ span that commutant. A primitive $e_t$ picks a one-dimensional factor from $S^{\lambda }$, so $e_{t}T$ is an equivalent realization of $D_{\lambda }$ as a Schur module.

The range of shapes is exact. If $\lambda$ has more than $m$ rows, a column antisymmetrizes more than $m$ vectors and its action vanishes by ${\Lambda }^{m+1}V=0$. Conversely, for at most $m$ rows place the $i$th basis vector in every tensor position belonging to row $i$ of $t$. Row symmetrization multiplies this tensor by ${\prod }_i{\lambda }_i!$. Column antisymmetrization is nonzero, because all vectors within each column are distinct, and its different permutations give distinct basis tensors. Thus $h_{t}T\ne 0$. This proves the length bound for a Schur module and the stated Schur–Weyl duality. For $n=0$, the empty partition and $V^{\otimes 0}=\mathbb{C}$ give the trivial version.

For a partition $\nu$ with at most $m$ parts, let $D_{\nu }(V)$ be the Schur module constructed in Question 3, equivalently the image of a Young symmetrizer on $V^{\otimes |\nu |}$. For a weakly decreasing integer tuple $\lambda$, put $k=\max (0,-{\lambda }_m)$ and $\nu =\lambda +k(1,\dots ,1)$. Define the rational Schur module byHere ${\det }^{-k}$ is the one-dimensional representation $g\mapsto \det (g{)}^{-k}$. This is rational, and it is irreducible under the permitted irreducibility assumption; for nonnegative $\lambda$ it is the original polynomial Schur module. Larger shifts give the same module, as will also follow from the character formula below.

Write $s_r(x)={\sum }_{i=1}^m{x}_i^r$ and $p_{\alpha }(x)={\prod }_{r=1}^n{s}_r(x{)}^{{\alpha }_r}$. For a permutation $\sigma$ of the given cycle type, the trace of a permuted tensor power isIn a tensor basis, the trace contracts the matrix entries of $\xi$ around each cycle of $\sigma$; a cycle of length $r$ contributes $\mathrm{tr}({\xi }^r)$. This proof applies to nondiagonalizable endomorphisms as well. The eigenvalues give $\mathrm{tr}({\xi }^r)=s_r(x)$.

The Schur–Weyl duality decomposition, on which $\sigma$ and ${\xi }^{\otimes n}$ act on the two respective factors, gives the same trace asFor a partition, the character extends polynomially to all endomorphisms because the tensor-power action does. This extension is not asserted for determinant-twisted modules at singular matrices.

We now derive the alternant character formula for the general linear group. Put $\delta =(m-1,m-2,\dots ,0)$ andUse the permitted symmetric-group character result, the Frobenius alternant character formula:It concerns characters of $S_n$, rather than assuming the character formula we seek for the general linear group. Substitute the proved trace identity and setThen ${\chi }^{\lambda }(\alpha )={\sum }_{\mu }{c}_{\lambda \mu }{\chi }^{\mu }(\alpha )$ for every conjugacy class. Independence of the irreducible symmetric-group characters forces $c_{\lambda \mu }={\delta }_{\lambda \mu }$.

Each character ${\varphi }_{\mu }$ on diagonal matrices is a symmetric homogeneous polynomial of degree $n$, since conjugation by permutation matrices permutes its arguments. Thus $a_{\delta }{\varphi }_{\mu }$ is alternating of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Every alternating polynomial of this degree has a unique expansion in alternants $a_{\lambda +\delta }$: a monomial with repeated exponents has zero coefficient, while each strictly decreasing nonnegative exponent vector is uniquely $\lambda +\delta$ for a partition $\lambda$ of $n$ with at most $m$ parts. Its coefficient at $x^{\lambda +\delta }$ is the coefficient of that alternant. The identities for $c_{\lambda \mu }$ therefore give $a_{\delta }{\varphi }_{\mu }=a_{\mu +\delta }$. We have derived the Weyl character formulaFor partitions the quotient is the Schur polynomial, with removable apparent singularities when eigenvalues coincide. For arbitrary dominant integer tuples, multiply the formula for $\nu$ by $(x_1\cdots x_m{)}^{-k}$; this shifts every numerator exponent by $-k$ and yields the same boxed formula. The variables must then be nonzero. The expression also shows independence of the shift used to define the determinant twist. Equality of characters identifies these irreducible modules: the group-algebra image on a direct sum of two irreducibles is finite-dimensional and semisimple, and its span of group operators detects the traces on every simple block.

Finally, the character of a dual representation evaluates the original character at $x_i^{-1}$. Reverse the numerator's columns after making this substitution. The exponents become $-{\lambda }_{m+1-j}+1-j$. Factoring $(x_1\cdots x_m{)}^{-(m-1)}$ converts these into ${\mu }_j+m-j$ for ${\mu }_j=-{\lambda }_{m+1-j}$. The denominator undergoes the identical column reversal and factor, so both signs and factors cancel. Hence ${\varphi }_{\lambda }(x^{-1})={\varphi }_{\mu }(x)$, andThis tuple is again weakly decreasing, so it is precisely the required dominant label.

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.

The Schur algebra is the commutant of place permutations on a tensor power. It is the linear span of the diagonal general linear group action, by Schur–Weyl duality. Modules over $S(m,n)$ correspond to homogeneous degree-$n$ polynomial representations. Over the complex numbers it is a semisimple algebra: decompose the tensor power as a module for the semisimple group algebra $\mathbb{C}{S}_n$, and take its endomorphism algebra.

If $\sigma$ has ${\alpha }_r$ cycles of length $r$, contraction of matrix entries around its cycles gives $\mathrm{tr}(\sigma {\xi }^{\otimes n})={\prod }_r\mathrm{tr}({\xi }^r{)}^{{\alpha }_r}$. The formula holds for every endomorphism, not only diagonalizable ones. The Schur–Weyl duality decomposition equates it with a sum of products of Specht and Schur characters.