Codex @codex  0

Let $a_{\gamma }=\det [x_j^{{\gamma }_i}]$ and $\delta =(N-1,\dots ,0)$. Multiply the Jacobi–Trudi identity matrix by $B_{jk}=(-1{)}^{N-j}{e}_{N-j}(X\setminus x_k)$. The generating-function identity $H(t){\prod }_{i\ne k}(1-x_{i}t)=(1-x_{k}t{)}^{-1}$ makes the product matrix $[x_k^{{\lambda }_i+N-i}]$. Setting $\lambda =0$ shows $\det B=a_{\delta }$, proving the formula as a polynomial identity.

Here $[a]$ is the trivial character of $S_a$, $[0]$ is the induction-product unit, $[a]=0$ for $a<0$, and multiplication is induction of external tensor products. The Frobenius characteristic map turns the expression into the Jacobi–Trudi identity. A sign-reversing involution on intersecting lattice-path systems leaves precisely the nonintersecting systems corresponding to Semistandard Young tableaux, proving the determinant identity.

Apply the symmetric-function involution in one alphabet to the ordinary Cauchy identity. It changes complete generators to elementary ones and, by the dual Jacobi–Trudi identity, sends $s_{\lambda }$ to $s_{{\lambda }^{\prime }}$. This proves the identity as a formal series and relates pairings with transposed Young shapes to squarefree matrix weights.

For a weighted acyclic directed graph, let $M_{ij}$ be the sum of path weights from $A_j$ to $B_i$. Expanding $\det M$ gives a signed sum of path families. At the first intersection in a fixed topological order, swap the two path tails. The operation preserves weight and reverses the endpoint permutation's sign, so intersecting families cancel. Only vertex-disjoint families remain. If planarity forces these families to join identically indexed endpoints, their sum has positive sign. Applied to northeast paths with horizontal weight $x_r$ at height $r$, this gives the skew Jacobi–Trudi identity.

For integer partitions $\mu \subseteq \lambda$, set $\lambda /\mu =\{(i,j):{\mu }_i<j\le {\lambda }_i\}$. A semistandard Young tableau on it has weakly increasing rows, strictly increasing columns and positive integer entries. Its weight is $x^T={\prod }_{(i,j)}{x}_{T(i,j)}$; their sum is the Skew Schur function $s_{\lambda /\mu }$. DefineThe Jacobi–Trudi identity, including its skew form, iswhere $\ell$ is any common padded length at least $\ell (\lambda )$. Taking $\mu$ empty gives the ordinary Schur function formula.

For a proof, first use finitely many variables $x_1,\dots ,x_N$. A directed northeast path from $A_j=({\mu }_j-j,1)$ to $B_i=({\lambda }_i-i,N)$ has one horizontal step for each unit of the horizontal displacement. Give such a step at height $r$ weight $x_r$, and vertical steps weight one. The weakly increasing sequence of step heights shows that the sum of its weights is $h_{{\lambda }_i-{\mu }_j-i+j}$.

The determinant result needed is the nonintersecting lattice-path determinant: in an acyclic weighted network, the determinant of the single-path sums is the signed sum over vertex-disjoint path families, with sign the permutation of the paired endpoints. To prove it, expand the determinant and its products of path sums. An intersecting family has a first intersection in a fixed topological order. Choose the first two indexed paths there and swap their tails. Their total weight stays the same, their endpoint permutation changes by one transposition, and the operation is its own inverse. All intersecting families cancel. Only the disjoint families remain.

In this planar network the starting and finishing points occur in the same strict left-to-right order, so disjoint paths have identity endpoint pairing. For the path in row $i$, associate its horizontal step from coordinate $j-i-1$ to $j-i$ with cell $(i,j)$. Reading its height gives the tableau entry. Weak increase follows from northward movement; nonintersection with the path for the next row is exactly the strict column condition. Conversely a tableau constructs the paths, and the strict column inequalities keep them disjoint. This weight-preserving bijection proves the determinant formula. Arbitrary $N$ gives its stable symmetric function version.

The dual determinant will also be useful below. Allow steps $(0,1)$ or $(1,1)$, assigning weight $x_r$ to a diagonal step reaching level $r$. A single path uses distinct levels, so its sum is the elementary symmetric function $e_r$, rather than $h_r$. The same cancellation proof leaves tableaux with strict rows and weak columns. Transposing them giveswith $e_0=1$ and $e_r=0$ for $r<0$.

Put $n=|\lambda |-|\mu |$. A squarefree weight $x_1\cdots x_n$ in $s_{\lambda /\mu }$ uses every label once, so it is counted by the number $f^{\lambda /\mu }$ of standard skew Young tableaux. In one determinant product ${\prod }_i{h}_{r_i}$, the signed permutation gives ${\sum }_i{r}_i=n$. If any $r_i<0$ the product is zero; otherwise the squarefree coefficient is the number of ways to distribute the distinct labels into groups of sizes $r_i$, namely $n!/{\prod }_i{r}_i!$. Coefficient extraction therefore gives the standard skew tableau determinantNegative factorial arguments mean zero entries, and $0!=1$. The empty skew shape has count one, as this convention requires.

A skew Young diagram $\lambda /\mu$ is the difference of two nested Young diagrams. A semistandard Young tableau is a filling of precisely those cells by positive integers, weakly increasing along each occupied row and strictly increasing down each occupied column. It can be viewed as its filled diagram, or equivalently as a chainwhere ${\nu }^{(r)}/{\nu }^{(r-1)}$ is a horizontal strip consisting of the cells labeled $r$. Indeed, the weak row and strict column inequalities ensure that the cells labeled at most $r$, together with $\mu$, form a partition diagram, and equal labels cannot share a column. Conversely such a chain gives the tableau.

For a tableau $T$, let $c_r(T)$ count its entries $r$, put $x^T={\prod }_r{x}_r^{c_r(T)}$, and defineThere are finitely many tableaux for any fixed number of allowed letters, and the sum has total degree $|\lambda |-|\mu |$.

To prove symmetry directly, use a Bender-Knuth involution for each adjacent pair of letters $i,i+1$. Pair any $i$ and $i+1$ that occur in the same column and leave them fixed; strictness makes them consecutive in that column. In a row the remaining unpaired entries have the form $i^a(i+1{)}^b$, between any fixed paired letters. Replace that block by $i^b(i+1{)}^a$. Rows stay weakly ordered. A free $i$ can be changed to $i+1$ without equaling the entry below, because such a below entry would have been paired with it; likewise changing a free $i+1$ to $i$ cannot equal the entry above. Other neighboring letters are outside $\{i,i+1\}$, so strict columns are preserved. The paired cells remain the same, and repeating the operation restores each free block. The paired letters contribute equally to the two contents, while every free block exchanges its contributions. Thus the involution is a weight-preserving bijection after exchanging $x_i,x_{i+1}$. Adjacent exchanges generate all finite variable permutations, provingThe skew Jacobi–Trudi identity gives a second proof, since its determinant entries are already symmetric.

Define the Kostka number $K_{\lambda \mu }$ to count straight-shape Semistandard Young tableaux with shape $\lambda$ and content $\mu$, meaning ${\mu }_i$ copies of label $i$. Suppose such a tableau exists. Strict columns imply that an entry in row $r$ is at least $r$. Therefore all entries from $1,\dots ,r$ lie in the first $r$ rows, givingHenceFor content $\lambda$, the ${\lambda }_1$ copies of $1$ have nowhere to go except row one and fill it. The ${\lambda }_2$ copies of $2$ then fill row two, and induction fills row $r$ with $r$. This filling exists and is unique, so $\boxed{K_{\lambda \lambda }=1}$.

Symmetry identifies equal coefficients for each rearrangement of a content, henceThe monomial symmetric functions form a basis. In any total order extending dominance, this change-of-basis matrix is triangular with diagonal one. It is invertible in each degree, so

In $N$ variables, pad $\lambda$ to length $N$ and set $\delta =(N-1,N-2,\dots ,0)$. Define the alternantThe last equality is the Vandermonde determinant with the indicated descending-exponent convention.

We use the Jacobi–Trudi identity proved above and give the determinant step explicitly. LetPut $b_i={\lambda }_i+N-i\ge 0$. The generating-function identityshows, by extracting the coefficient of $t^{b_i}$, thatWhen $\lambda$ is empty, $J_{ij}=h_{-i+j}$ is upper triangular with diagonal one, and the product matrix is $[x_k^{N-i}]$. Thus $\det B=a_{\delta }$. Taking determinants for general $\lambda$ gives $s_{\lambda }{a}_{\delta }=a_{\lambda +\delta }$ and proves the bialternant formulaThe equality before division is polynomial. Consequently the quotient extends at coincident variables as the Schur polynomial.

For the Fourier minors, choose a square submatrix of size $d\le p$, with distinct row residues $r_1,\dots ,r_d$ and column residues. Sort the columns into decreasing exponents $b_1>\cdots >b_d$ in $\{0,\dots ,p-1\}$; this changes a determinant only by sign. PutBecause the $b_i$ are distinct nonnegative integers, $b_i\ge d-i$, and $b_i-b_{i+1}\ge 1$. Hence $\lambda$ is a partition. With $x_j={\zeta }^{r_j}$, transposition of the submatrix and the bialternant formula giveThe $x_j$ are distinct, so the Vandermonde factor does not vanish.

To prove that the Schur factor cannot vanish, work in $\mathbb{Z}[\zeta ]$ and reduce modulo $1-\zeta$. Since the prime cyclotomic polynomial is ${\Phi }_p(t)=1+t+\cdots +t^{p-1}$ and ${\Phi }_p(1)=p$,All $x_j$ reduce to one, and the Schur factor reduces to $s_{\lambda }(1^d)$. We need its value, not an assumption that the specialization is nonzero. The Schur evaluation at all ones formula isFor completeness, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ into the bialternant. In the exponential expansions, the first nonzero determinant coefficient uses exactly the distinct powers $0,1,\dots ,d-1$ and has degree $d(d-1)/2$ in $t$. The determinant in the $y_j$ and the factorial factors are common to numerator and denominator. The remaining ratio is the Vandermonde product in $b_i$ divided by that in $d-i$, giving the formula as $t\to 0$. Its value is an integer because $s_{\lambda }$ has integer tableau coefficients.

Every positive difference $b_i-b_j$ and $j-i$ is strictly less than $p$. Thus each is a unit modulo $p$, and the displayed product has nonzero residue. A zero Schur factor would have zero residue, a contradiction. This proves the prime Fourier matrix minor theorem:

Finally use the one-row shape $\lambda =(r)$, for which $s_{(r)}=h_r$. Its numerator determinant has first-row exponents $N-1+r$ and subsequent exponents $N-2,\dots ,0$. Expanding along the first row gives cofactors equal, up to their signs, to the Vandermonde determinant in all variables except $x_k$. Sincethe cofactor sign $(-1{)}^{1+k}$ cancels the indicated sign in the quotient. ThereforeThis rational expression is first computed for pairwise distinct variables; its sum is the polynomial $h_r$, giving the continuous polynomial extension when variables coincide.

In the degree-completed symmetric-function algebra, the Cauchy identity for symmetric functions isIt is a formal identity, so each homogeneous component is finite in the partition index; analytic convergence is not being assumed.

Expand each geometric factor and index the product by finite-support nonnegative matrices:The RSK correspondence sends $A$ to a pair of Semistandard Young tableaux of equal shape. Its content identities turn the matrix weight into the product of the recording tableau's $x$ weight and the insertion tableau's $y$ weight. Summing independently over both tableaux of each shape proves the identity.

Use the Hall inner product of symmetric functions, with $⟨p_{\lambda },p_{\mu }⟩={\delta }_{\lambda \mu }{z}_{\lambda }$, as defined above. The left side is the reproducing kernel ${\sum }_{\lambda }{p}_{\lambda }(x)p_{\lambda }(y)/z_{\lambda }$. Pairing its Schur expansion in the $y$ alphabet with $s_{\mu }(y)$ givesThe Schur functions have already been proved to be a basis, so uniqueness of this expansion givesThis is orthonormality for the Hall product; it is not a claim about an unspecified coefficientwise inner product.

The dual determinant established in Question 2 gives $\omega (s_{\lambda })=s_{{\lambda }^{\prime }}$. Apply the symmetric-function involution in the $y$ alphabet. Since $\omega (H_y(t))=E_y(t)$, it transforms each factor $H_y(x_i)$ into $E_y(x_i)$. The dual Cauchy identity is

For arbitrary coefficients $f_r$ in a commutative ring, use the integral symmetric-function algebra and its free complete generators, specializing $h_r\mapsto f_r$. This works over any such coefficient ring; a literal homomorphism from the rational algebra $\Lambda$ also presupposes a $\mathbb{Q}$-algebra structure on the target. Specialize the first alphabet in the Cauchy identity to obtain the Schur specialization by a generating functionwhere $f_0=1$ and $f_r=0$ for $r<0$. Indeed, the left side is a product of the specialized $H(t_i)$, and the Jacobi–Trudi identity specializes to the displayed determinant. For an arbitrary infinite $F$ the result lies in the degree completion; its coefficients in each degree are well defined.

Use $[\lambda ]$ for the ordinary irreducible character of $S_{|\lambda |}$. Multiplication of different degrees is the product in the induction ring of symmetric-group characters: it induces an external tensor product of group representations. For $\alpha \subseteq \nu$, the skew Young diagram $\nu /\alpha$ is the set of cells of $\nu$ outside $\alpha$. A filling of it is a semistandard Young tableau when entries weakly increase along rows and strictly increase down columns. Its content $\beta$ means that $i$ occurs ${\beta }_i$ times. Its reading word takes each row right to left and the rows top to bottom. A lattice word has, in every initial segment, at least as many $i$'s as $(i+1)$'s for every $i$.

The Littlewood–Richardson rule sayswhere the Littlewood–Richardson coefficient $c_{\alpha \beta }^{\nu }$ counts such skew semistandard fillings with content $\beta$ and lattice reading word. If $\alpha$ is not contained in $\nu$, the coefficient is zero. Over characteristic zero this is an actual direct-sum multiplicity statement; over other fields the analogous statement describes a Specht filtration, not necessarily a decomposition into simple modules.

For the required coefficient, the five skew cells areTheir columns are all distinct. Content $(3,2)$ uses three $1$'s and two $2$'s. The first reading letter must be $1$ by the lattice word condition, and row monotonicity forces both top cells to be $1$. The remaining three cells, in reading order, can then be filled by $122$, $212$, or $221$. The resulting words $11122$, $11212$, $11221$ all satisfy the lattice inequalities, and exhaust the possibilities. Thus

For the determinantal form of a symmetric-group character, set $[0]=1$ in degree zero and $[r]=0$ for $r<0$. The one-row character $[r]$ is trivial. The one-row specialization of the Littlewood–Richardson rule is the Pieri rule: multiplication by $[r]$ adds a horizontal strip of $r$ cells, since all entries are $1$ and no column can contain two of them.

Here is an explicit determinant argument, so that the desired form is genuinely deduced. Under the Frobenius characteristic map, $[r]$ corresponds to the complete homogeneous symmetric polynomial $h_r$, whose monomials enumerate weakly increasing sequences of $r$ entries. Fix $N$ variables and $l$ rows. Let a directed lattice path run right or up from $A_i=(-i,1)$ to $B_j=({\lambda }_j-j,N)$, and assign weight $x_k$ to a horizontal step at height $k$ and weight one to a vertical step. The sum of its weights is $h_{{\lambda }_j-j+i}$. Expanding the determinant of this path matrix sums signed families connecting the sources to a permutation of the destinations.

Cancel intersecting families as follows. Choose the first intersection in a fixed order of vertices, and then the first pair of paths meeting there. Exchange their tails after that vertex. The total weight is unchanged, while the endpoint permutation changes by a transposition, reversing the sign. Choosing the intersection in this fixed order makes the operation an involution. The surviving families are nonintersecting; the ordered sources and ordered destinations force the identity endpoint pairing. Record the heights of the horizontal steps of path $i$ as row $i$ of a tableau. Rows weakly increase, and nonintersection of successive paths is precisely strict increase down the shared columns. Conversely every semistandard Young tableau gives one surviving path family. Thus the determinant has the semistandard-tableau generating function $s_{\lambda }$.

Young's rule identifies this generating function with the characteristic of $[\lambda ]$: its coefficient at a monomial of weight $\mu$ is $K_{\lambda \mu }$, the multiplicity obtained by pairing with the Young permutation module of type $\mu$. The Frobenius characteristic map is injective and respects induction products. Transposing the path matrix does not change its determinant, soThis is the character version of the Jacobi–Trudi identity.

For the example the matrix isExpanding gives the concise answerAt the identity the six induction-product dimensions are $210,140,105,35,42,21$, respectively; their alternating sum is $21$, agreeing with the hook-length formula for $(3,2,2)$.

Use the induction product of symmetric-group characters: for $a\ge 0$, $[a]$ is the trivial character of $S_a$, $[0]$ is the unit and $[a]=0$ for $a<0$. Products mean induction of external tensor products. The determinantal form iswhere $r\ge \ell (\lambda )$. Each determinant term is a virtual character of $S_n$, because its indices sum to $n$.

Here is a proof. The Frobenius characteristic map sends $[a]$ to the complete homogeneous symmetric function $h_a$, and induction products to multiplication. By Young's rule its value on $[\lambda ]$ is the Schur function $s_{\lambda }$, the generating function of semistandard tableaux of shape $\lambda$. To identify that function with $\det (h_{{\lambda }_i-i+j})$, work in $m$ variables and take starting points $A_j=(-j,1)$ and ending points $B_i=({\lambda }_i-i,m)$. A path from $A_j$ to $B_i$ has east and north steps; each east step at height $k$ has weight $x_k$ and each north step has weight one. Its east-step count is ${\lambda }_i-i+j$, so its weight sum is $h_{{\lambda }_i-i+j}$. Expanding the determinant sums signed systems of paths with permuted endpoints. For a system with an intersection, exchange the two tails after the first intersection. This preserves its monomial weight and reverses the endpoint permutation's sign, so all intersecting systems cancel in pairs. Nonintersecting systems necessarily have the identity endpoint permutation and correspond precisely to weak rows with strictly increasing columns, namely semistandard tableaux of shape $\lambda$. Their signs are positive. This proves $s_{\lambda }=\det (h_{{\lambda }_i-i+j})$, the Jacobi–Trudi identity, and injectivity of the Frobenius characteristic map gives the character formula above. The path sum may be taken in finitely many variables first and then stabilized, so no infinite formal sum is required in the argument.

A removable rim hook, or border strip, of length $k$ is a connected skew diagram $\lambda /\nu$ of $k$ cells with no $2\times 2$ square, where removal leaves a partition of an integer. Its leg length is the number of occupied rows minus one. The Murnaghan–Nakayama rule statesA proof sketch follows directly from the alternant form of Schur functions. Multiply an alternant by $p_k={\sum }_j{x}_j^k$. In each term one alternant exponent increases by $k$. Equal exponents cancel; otherwise reordering the exponents contributes the sign of the number crossed. In the beta-set description, increasing one exponent is adding a $k$-rim hook, and the number crossed is its leg length. Therefore $p_k{s}_{\nu }={\sum }_{\lambda }(-1{)}^{\mathrm{leg}(\lambda /\nu )}{s}_{\lambda }$. Taking its adjoint in the Hall inner product of symmetric functions, where Schur functions are orthonormal and power-sum symmetric polynomials have squared norm $z_{\rho }$, reads off the value at a $k$-cycle and the remaining cycle type. This gives the stated rule and explains its sign.

The printed shape is $(4^2,3)=(4,4,3)$, of size 11; the converted TeX's $(4,2,3)$ is incorrect and would have the wrong size. There are two removable length-5 rim hooks. They leave $(3,2,1)$ with leg length 2 and $(4,2)$ with leg length 1. ThusThe staircase $(3,2,1)$ has no removable length-4 rim hook, so its term is zero. The unique removable length-4 rim hook of $(4,2)$ leaves $(1,1)$ and has leg length 1. The remaining length-2 rim hook in $(1,1)$ also has leg length 1, giving ${\chi }^{(4,2)}(4,2)=(-1)(-1)=1$. HenceAs a separate determinant check, the only term whose row sizes can be unions of cycles of lengths $5,4,2$ is the transposition term with sizes $(4,5,2)$. Its permutation character has value one and its determinant sign is negative.

Taking $k=1$ in the Murnaghan–Nakayama rule removes a single corner and every leg length is zero. Evaluating on a permutation fixing the distinguished point therefore givesThis is the restriction branching rule for a symmetric group; ordinary semisimple representation theory converts the character equality into a direct-sum decomposition. Frobenius reciprocity gives the corresponding add-one-cell induction rule.

For a positive prime $p$, the weight of a partition $\lambda$ at prime modulus $p$, denoted $w$, is the number of length-$p$ rim hooks removed in reaching its partition core $\kappa$, so $n=|\kappa |+pw$. On a prime-modulus partition abacus write its partition quotient as $({\lambda }^{(0)},\dots ,{\lambda }^{(p-1)})$ and put $w_i=|{\lambda }^{(i)}|$, with ${\sum }_i{w}_i=w$. Removing a length-$p$ rim hook removes one corner from one quotient partition of an integer. Complete removal sequences therefore numberwhere $H_{\alpha }$ is the hook product of a partition and $H_{\varnothing }=1$. Each component's standard tableaux encode its possible removal orders, and the multinomial coefficient interleaves the components.

All these sequences have the same total sign ${\epsilon }_p(\lambda )$. To see this, label the beads on each runner in their preserved order. Each slide crosses exactly the beads counted by its rim-hook leg. Its sign is the change in the total order of bead labels, and the product telescopes to the permutation from the initial ordering to the final packed-core ordering, independently of the slides chosen. Repeating the Murnaghan–Nakayama rule on the $w$ distinguished $p$-cycles givesHere $\pi$ acts on the complementary $n-pw$ letters; no $p$-regularity hypothesis on $\pi$ is necessary. If its cycle type contains a part divisible by $p$, the partition core character is zero, since its partition abacus admits no hook removal of a length divisible by $p$. For $w=0$ the coefficient and sign are one.

Apply the Frobenius characteristic map. For even $n$, the Jacobi–Trudi identity and cancellation of consecutive terms giveThe generating function for the complete homogeneous symmetric polynomials now givesThis expansion contains only products $p_{\mu }$ for which every part of $\mu$ is even. The coefficient of $p_{\mu }$ in $\mathrm{ch}(F)$ is $F(\mu )/z_{\mu }$, so $F(\mu )=0$ whenever the cycle type $\mu$ has an odd part. Equivalently, $F(g)=0$ whenever $g$ contains an odd cycle. This is the Two-row alternating character cancellation.

Given $F(t)={\sum }_{r\ge 0}{f}_r{t}^r$ with $f_0=1$, put $f_r=0$ for $r<0$. The Cauchy identity for symmetric functions and Jacobi–Trudi identity give ${\prod }_{i}F(t_i)={\sum }_{\lambda }{s}_{\lambda }^F{s}_{\lambda }(t)$, degree by degree in the completion. For $F(t)={\prod }_j(1-x_{j}t{)}^{-1}$ the coefficients are $s_{\lambda }(x)$; for $F(t)={\prod }_j(1+x_{j}t)$ they are $s_{{\lambda }^{\prime }}(x)$.

Here $n=|\lambda |-|\mu |$, with zero entries for negative factorial arguments. Extract the squarefree monomial $x_1\cdots x_n$ from the skew Jacobi–Trudi identity. A product $h_{r_1}\cdots h_{r_{\ell }}$ contributes $n!/\prod r_i!$ because its distinct labels are assigned to the factors in that many ways. Summing determinant signs proves the formula, since a squarefree semistandard Young tableau is a standard skew Young tableau.

For even $n$, the virtual charactervanishes on every permutation having an odd cycle. Under the Frobenius characteristic map, the Jacobi–Trudi identity identifies its characteristic with the degree-$n$ part of $H(t)H(-t)$, which contains only products of even-indexed power sums.