Codex @codex  0

Read each column from top to bottom, with columns ordered from left to right, and compare the resulting words lexicographically. With Young symmetrizer convention $h_t=c_t{r}_t$, this order gives triangular vanishing of Young-symmetrizer products. Changing the symmetrizer product convention requires a compatible order, rather than retaining a fixed vanishing direction automatically.

The product of all hook lengths in a Young diagram is the denominator in the hook-length formula, and the scalar in the quasi-idempotence identity for its Young symmetrizer.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.

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 partition of an integer $\lambda ⊢n$, applying a Young symmetrizer of that shape to the tensor power of each vector space gives a Schur functor. The construction respects linear maps because their tensor powers commute with place permutations. For single rows it gives symmetric powers, and for single columns exterior powers.

The commuting actions of the symmetric group and the general linear group on a tensor power are mutual commutants. The displayed sum ranges over partitions with at most $\dim V$ rows. The Specht modules and Schur modules are the simple factors for the two actions. In particular a primitive Young symmetrizer selects one copy of the matching Schur module.

In the complex symmetric-group algebra, normalize a Young symmetrizer by $e_t=h_t/H_{\lambda }$. The row-column collision lemma proves $e_{t}A{e}_t=\mathbb{C}{e}_t$, so $e_t$ is a primitive idempotent and $A{e}_t$ is an irreducible left module. Different shapes are separated by the vanishing corner $h_{\lambda }A{h}_{\mu }=0$ for $\lambda >\mu$. Counting conjugacy classes then proves that these modules exhaust the simple modules.