Codex @codex  0

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 $\alpha ,\beta \in \mathrm{Spec}(n)$, define $\alpha \sim \beta$ when their Gelfand–Tsetlin basis lines occur in the same irreducible representation. This is well defined: $G{Z}_n$ contains the centre of $\mathbb{C}{S}_n$, whose primitive central idempotents separate the irreducible modules, and part 1(ii) says these central elements are polynomials in the $X_j$. Consequently the same weight cannot label two different irreducible modules.

For content vectors of standard Young tableaux, an admissible adjacent swap of a standard Young tableau exchanges coordinates $i,i+1$ whose difference has absolute value greater than $1$. Define $\alpha \approx \beta$ as the equivalence relation generated by such swaps within $\mathrm{Cont}(n)$.

Equivalently, after reconstructing the tableaux, $\alpha \approx \beta$ means that the two tableaux have the same final Young diagram. Indeed, an admissible swap exchanges consecutive entries in incomparable cells, preserves the diagram and leaves a standard Young tableau. Conversely, tableaux of a fixed shape are linear extensions of a partially ordered set, with the cells ordered by their row and column requirements; any two linear extensions are connected by adjacent swaps of incomparable elements. We explain this connectivity and its spectral consequence in part (v).

The joint spectrum $\mathrm{Spec}(n)$ consists of the tuples of simultaneous eigenvalues of $X_1,\dots ,X_n$ occurring in irreducible complex $S_n$-modules. A Gelfand–Tsetlin basis is a Gelfand–Tsetlin basis: its vectors lie in the one-dimensional joint eigenspaces selected by successive restrictions. The weight of $v$ is $\alpha =(a_1,\dots ,a_n)$ when $X_{j}v=a_{j}v$ for every $j$.

Define $\mathrm{Cont}(n)$ to be the integer tuples satisfying the content vector of a standard Young tableau conditions: $a_1=0$; for $j>1$, at least one of $a_j-1,a_j+1$ occurs earlier; and between any two occurrences of the same integer $a$, both $a-1$ and $a+1$ occur. We will justify their tableau interpretation in part (iv).

Here is the local spectral rules for Young–Jucys–Murphy elements calculation used in this question. Average a positive definite Hermitian inner product over $S_n$, making the representation unitary. The $X_j$ are commuting self-adjoint operators because transpositions are self-adjoint involutions. Put $s_i=(ii+1)$ and $d=a_{i+1}-a_i$. The group-algebra relations areFor a unit weight vector $v$, the first relation gives $d⟨v,s_{i}v⟩=1$. Thus $d\ne 0$, $|d|\ge 1$, andhas the weight obtained by swapping $a_i,a_{i+1}$, with $\Vert w{\Vert }^2=1-d^{-2}$. If $d=\pm 1$, then $w=0$ and $s_{i}v=\pm v$. If $|d|>1$, then $w\ne 0$, providing an admissible spectral swap in the same irreducible module.

For the proposed three consecutive weights $(a,a+1,a)$, these rules give $s_{i}v=v$ and $s_{i+1}v=-v$. But the braid relation in a Coxeter group would implya contradiction. Hence the proposed tuple is not in $\mathrm{Spec}(n)$. The same argument, with the signs reversed, excludes $(a,a-1,a)$ as well.

Use the tableau representation supplied by Young seminormal form, stated explicitly in Question 5, or its normalized Young orthogonal form. It constructs the complex Specht module $S^{\lambda }$ with one basis vector for each standard Young tableau of shape $\lambda$. Its adjacent-transposition matrices imply the content eigenvalues directly.

Indeed, start with $X_1=0$ and use $X_{i+1}=s_i{X}_i{s}_i+s_i$. On an admissible pair $T,R=s_{i}T$, set $a=c_T(i)$, $b=c_T(i+1)$, $d=b-a$, and $q=\sqrt{1-d^{-2}}$. In the orthogonal basis, $X_i$ is $\mathrm{diag}(a,b)$ andMultiplication givesFor a nonadmissible swap, $s_i=\pm 1$ and the next content differs by the same $\pm 1$, giving the same conclusion. Induction therefore proves $X_i{w}_T=c_T(i)w_T$. Every content vector of a standard Young tableau, reconstructed as a standard Young tableau in part (iv), is consequently a spectral weight. Together with the reverse inclusion,

An admissible adjacent swap has nonzero off-diagonal coefficient in Young orthogonal form, and the local spectral calculation in part (i) keeps it inside one irreducible representation. Thus $\alpha \approx \beta$ implies $\alpha \sim \beta$.

Two standard Young tableaux of one shape are connected by admissible swaps. Regard each as a linear extension of a partially ordered set of cells. Move the first cell of the desired extension to the front of the other extension: every cell it crosses is incomparable with it, since otherwise their order would be forced in both extensions. Repeat after fixing that first cell. Each step swaps consecutive labels in incomparable cells. Such cells lie strictly northeast and southwest of each other, so their contents differ by at least $2$ in absolute value. These are exactly admissible swaps.

This also establishes irreducibility of the constructed tableau representations. A submodule is invariant under the commuting $X_j$ and hence under their joint spectral projections; since their weights are distinct, it is a sum of tableau lines. A nonzero off-diagonal coefficient propagates any included line along every admissible edge, and connectivity then includes all lines. Different shapes give nonisomorphic modules, since their content-vector sets are disjoint and content reconstruction determines the shape. Finally, the Robinson–Schensted correspondence gives ${\sum }_{\lambda ⊢n}{f}_{\lambda }^2=n!$, and the sum of squares of irreducible degrees leaves room for no other irreducible representations.

Thus within each irreducible module the weights are exactly the tableaux of one shape, so $\alpha \sim \beta$ implies $\alpha \approx \beta$. Hence the two equivalence relations coincide.

The Clebsch-Gordan decomposition for SU2 statesRealize the tensor product as polynomials of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The determinant $D=x_1{y}_2-y_1{x}_2$ is invariant under the simultaneous SU2 action. For each $r$, the vectorhas torus weight $q=m+n-2r$ and is killed by $E=x_1{\partial }_{y_1}+x_2{\partial }_{y_2}$. Applying $F=y_1{\partial }_{x_1}+y_2{\partial }_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}{v}_r=0$. The relations $[E,F]=H$ and $H{v}_r=q{v}_r$ give $E{F}^j{v}_r=j(q-j+1)F^{j-1}{v}_r$, so their span is an irreducible copy of $V_q$.

These irreducible submodules have distinct highest weights and form a direct sum: a sum of earlier ones cannot contain an irreducible module inequivalent to all its summands. Equivalently use complete reducibility, obtained by averaging an inner product over the compact group. Assuming $m\le n$, their total dimension iswhich equals the dimension of the tensor product. Hence they exhaust it, proving the theorem.

No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

A module is semisimple, or completely reducible, when it is a direct sum of simple modules. Equivalently, every submodule has a direct-sum complement.