Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 77 (Symmetric Functions)

Use nonnegative integer matrix entries, including zero, and English Young diagram conventions: rows run left to right and columns run down. Replace $A$ by a two-line array containing $a_{ij}$ copies of $\left(\genfrac{}{}{0ex}{}{i}{j}\right)$, sorted lexicographically by upper letter $i$ and then lower letter $j$. Finite support makes the array finite.

In the RSK algorithm, insert each lower letter $j$ into $P$ by row insertion. In the first row replace the leftmost entry strictly greater than $j$, carrying the replaced entry into the next row; if there is no such entry, append $j$ and stop. Repeat in subsequent rows until a new cell is made. Put the upper letter $i$ in that new cell of the recording tableau $Q$.

The insertion-path facts used here are the following: row insertion preserves weak rows and strict columns; inserting $u$ followed by $v\ge u$ puts the latter new cell strictly to the right of the former; reverse insertion undoes insertion; and the reverse paths from a right-to-left horizontal strip recover its input letters in weakly decreasing order. These standard path facts ensure that $P$ is a semistandard Young tableau. The upper letters are nondecreasing, so the entries of $Q$ are weakly increasing in rows. Equal upper letters have sorted lower letters, and hence create a horizontal strip; they cannot occupy the same column. Thus $Q$ is also a semistandard Young tableau, of exactly the same shape as $P$.

Here is the inverse on any such pair. Select a largest entry $i$ in $Q$, and among the cells carrying it select the rightmost one. It is an outer corner: there can be no larger entry below it, and any entry to its right would be another maximal entry farther right. Delete that cell from $Q$ and reverse-insert the value at its cell in $P$. In each preceding row replace the rightmost entry strictly smaller than the carried value and carry that old entry upward. The value emerging from the first row is $j$. Record $\left(\genfrac{}{}{0ex}{}{i}{j}\right)$ and repeat. Equal maximal entries form a horizontal strip, so the reverse-path fact gives their $j$'s in weakly decreasing order. Reversing the recovered list therefore gives exactly a lexicographically sorted two-line array. The procedures are mutually inverse, proving the RSK correspondence bijection, including the zero matrix and the pair of empty tableaux.

Insertion only moves previous entries and adds one copy of its new letter; recording adds one copy of its upper letter. Consequently

For the trace assertion, use the insertion-path form of the RSK growth-diagram local rule. Let $\rho ,\mu ,\nu ,\lambda$ be the shapes from the northwest, northeast, southwest and southeast matrix prefixes around entry $a_{ij}$. Padding row lengths with zeros, the rule isThis local rule counts the two merging insertion paths row by row: the entry supplies the initial carry, the longer extension supplies the current row, and the overlap of the extensions beyond $\rho$ is bumped to the next row. It is a standard insertion-path fact being used in addition to the preceding monotonicity statements.

Define $o(\gamma )={\gamma }_1-{\gamma }_2+{\gamma }_3-\cdots$. Each column of the Young diagram contributes its alternating vertical sum, which is one for odd length and zero for even length. Thus $o(\gamma )$ counts odd-length columns. For symmetric $A$, transposition symmetry of RSK gives equal shapes for the two prefixes off the diagonal, so $\mu =\nu$ at entry $a_{ii}$. The rule becomesTaking alternating sums makes the two sums involving $\mu$ cancel, leavingStarting from the empty northwest corner and proceeding along the diagonal proves the trace and odd columns in symmetric RSK identityThis also accommodates arbitrary diagonal entries, rather than only permutation matrices.

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.

For a finite group action $G\curvearrowright X$, Burnside's lemma statesCount pairs $(g,x)$ with $gx=x$ in two ways. Counting first by $g$ gives the displayed sum. Counting first by $x$ gives ${\sum }_x|G_x|$. For an orbit $\mathcal{O}$, the orbit-stabilizer theorem gives $|\mathcal{O}||G_x|=|G|$, so each orbit contributes exactly $|G|$. Division proves the formula.

Use colors $1,\dots ,m$, with variables $x_1,\dots ,x_m$, on the underlying finite set $S$. The weight of a coloring $f:S\to \{1,\dots ,m\}$ is $w(f)={\prod }_{s\in S}{x}_{f(s)}$. It is constant on each coloring orbit. The pattern inventory is $F_G(x)={\sum }_{\mathcal{O}}w(\mathcal{O})$, one weight for each orbit, including its multiplicity when different orbits have the same content.

If $c_r(g)$ is the number of length-$r$ cycles of $g$ on $S$, define the cycle indexIn the symmetric-function cycle-indicator convention, write $Z_G(x)$ for its substitution $t_r=p_r(x)={\sum }_{i=1}^m{x}_i^r$. A fixed coloring must be constant on every cycle of $g$. Coloring one length-$r$ cycle therefore has weight sum $p_r(x)$, independently of other cycles. The weighted Burnside lemma, proved by exactly the same orbit-stabilizer double count with weights, yieldsThis defines both sides in the same alphabet; the raw cycle index in the independent variables $t_r$ is specialized before comparison. The identity holds for all finite $m$ and hence stably for arbitrarily many colors.

For the last assertion, the permutation character on $T$ satisfies $\chi (g)=|{\mathrm{Fix}}_T(g)|$. A pair $(s,t)$ is fixed in the diagonal action on $T\times T$ exactly when both components are fixed. ThereforeThe character is real, so its character inner product with itself and another application of Burnside's lemma giveThis is the rank of a permutation action formula, and in fact does not require transitivity. If the action is transitive and $|T|\ge 2$, the diagonal is one orbit; rank two means exactly one further orbit on distinct ordered pairs, which is double transitivity.

Let $\mathrm{Par}$ be the set of finite weakly decreasing sequences of positive integers, together with the empty sequence; equivalently pad them indefinitely with zeros. The size is $|\lambda |={\sum }_i{\lambda }_i$, the length counts positive parts, and $\mathrm{Par}(n)$ consists of size-$n$ integer partitions. Their Young diagrams have cells $(i,j)$ with $1\le j\le {\lambda }_i$.

Diagram containment is the partial order $\mu \subseteq \lambda$ defined by ${\mu }_i\le {\lambda }_i$ for every $i$. Componentwise minimum and maximum give its meet and join, so this is Young's lattice. Its covers add one cell. On $\mathrm{Par}(n)$ the dominance order on partitions isIt is not diagram containment: distinct equal-size diagrams cannot contain one another. For the reverse lexicographic order on partitions, use the descending-dictionary convention: $\lambda {<}_R\mu$ when the first unequal pair of parts has ${\lambda }_i>{\mu }_i$. Thus for size four its increasing list is $(4),(3,1),(2,2),(2,1,1),(1,1,1,1)$. This is a total order opposite to increasing dictionary order on integer partitions, not the last-differing-exponent order on polynomial monomials.

Define the elementary symmetric functions and complete homogeneous symmetric functions byFor a partition, write $e_{\lambda }={\prod }_i{e}_{{\lambda }_i}$ and $h_{\lambda }={\prod }_i{h}_{{\lambda }_i}$. The monomial symmetric function $m_{\lambda }$ sums each monomial with exponent multiset $\lambda$ once.

Here is a proof of the Fundamental theorem of symmetric polynomials sufficient for the stable algebra. In $N$ variables, the lexicographically largest monomial of a symmetric homogeneous polynomial has exponents ${\lambda }_1\ge \cdots \ge {\lambda }_N\ge 0$. Otherwise exchanging an inverted pair of variables produces a larger monomial with the same coefficient. The polynomialhas that monomial as its largest one, with coefficient one. Subtract a suitable multiple and continue; there are only finitely many monomials of the fixed degree, so the procedure terminates and proves generation. Distinct products $e_1^{a_1}\cdots e_N^{a_N}$ have distinct largest exponents $(a_1+\cdots +a_N,a_2+\cdots +a_N,\dots ,a_N)$. The largest such monomial in any nonzero polynomial relation cannot cancel, proving algebraic independence. Stabilizing in each degree, where only $e_r$ with $r$ no greater than that degree can occur, gives

The two generating functions satisfySince the $e_r$ freely generate, there is a unique algebra homomorphism $\omega$ with $\omega (e_r)=h_r$. Apply it to the last identity:It follows that $\omega (h_r)=e_r$ and ${\omega }^2(e_r)=e_r$. Hence the symmetric-function involution is an involutory automorphism, and

For the forgotten coefficients, it is useful to derive the required duality explicitly. Put $p_r={\sum }_i{x}_i^r$. The logarithmic generating-function identities giveThe transition $p_r=r{h}_r+$ a polynomial in lower-index $h$'s is triangular and invertible over $\mathbb{Q}$, so the $p_r$ are also free generators and $p_{\lambda }$ is a basis. Define the Hall inner product of symmetric functions byExpand the same formal kernel in two ways:The last expansion is a reproducing kernel for this inner product. Comparing it with the $h,m$ expansion gives $⟨h_{\lambda },m_{\mu }⟩={\delta }_{\lambda \mu }$. One can see this without assuming any Schur orthogonality: pair either kernel with $h_{\nu }(y)$; the reproducing property gives $h_{\nu }(x)$, and uniqueness of its $h$ expansion proves the duality. The signs in $\omega (p_{\lambda })$ show that $\omega$ is an isometry; together with ${\omega }^2=1$ this makes it self-adjoint.

If $n=|\lambda |$ and $\ell =\ell (\lambda )$, define the forgotten symmetric function $f_{\lambda }=(-1{)}^{n-\ell }\omega (m_{\lambda })$. Its monomial coefficient is consequentlyIt remains to compute this coefficient rather than merely invoke duality. Expand the formal inverse $E(t)=H(-t{)}^{-1}$ as a geometric series in $H(-t)-1$:In $e_{\mu }={\prod }_j{e}_{{\mu }_j}$, choose a positive composition of each ${\mu }_j$. A term is $h_{\lambda }$ precisely when the concatenated list is a distinct rearrangement $\alpha$ of the parts of $\lambda$. Its boundaries at the end of each chosen composition are exactly the prescribed prefix sums of $\mu$. Conversely such a rearrangement uniquely determines the compositions by cutting at those prefix sums. Every one of its signs is $(-1{)}^{\sum {\mu }_j-\ell }=(-1{)}^{n-\ell }$, canceled by the prefactor in $a_{\lambda \mu }$. ThusIdentical parts are not artificially distinguished. For example $f_{(2,1)}=m_{(2,1)}+2m_{(3)}$: only the order $(2,1)$ refines $(2,1)$, while both orders refine the single block $(3)$.

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.

For $F(t)={\prod }_i(1-x_{i}t{)}^{-1}$, its coefficients are precisely the complete homogeneous symmetric functions $f_r=h_r(x)$. Thus the homomorphism sends every abstract generator $h_r$ to its ordinary value in the alphabet $x$. Equivalently the general product expansion is the ordinary Cauchy identity for symmetric functions. ThereforeFor every symmetric function $u$, the same specialization gives $u^F=u(x)$.

For $F(t)={\prod }_i(1+x_{i}t)$ the coefficients are the elementary symmetric functions, $f_r=e_r(x)$. The specialization is evaluation in $x$ after the symmetric-function involution, because $\omega (h_r)=e_r$. HenceOne can also read this coefficient directly from the dual Cauchy identity, after exchanging its alphabets. For a general symmetric function, $u^F=(\omega u)(x)$; it is not simply $u(x)$.

A skew Young diagram is connected when any two of its cells can be joined by a sequence of cells sharing edges; touching only at a corner does not connect them. A nonempty border strip, or rim hook, is a connected skew diagram containing no $2\times 2$ square. Its height is its number of occupied rows minus one.

For a weak composition $\alpha =({\alpha }_1,\dots ,{\alpha }_{\ell })$, a border-strip tableau of shape $\lambda /\mu$ and type $\alpha$ is a chainwhose $i$th difference is a border strip of size ${\alpha }_i$. Filling that strip with $i$ gives the equivalent weakly increasing row-and-column filling. A zero part corresponds to an empty difference and has height zero. Put $\mathrm{ht}(T)={\sum }_i\mathrm{ht}({\lambda }^{(i)}/{\lambda }^{(i-1)})$. For positive $r$, $p_r={\sum }_j{x}_j^r$; write $p_{\alpha }={\prod }_{{\alpha }_i>0}{p}_{{\alpha }_i}$. Thus zero type parts are ignored: one does not multiply by the finite-variable value $p_0=N$.

We first prove the one-strip multiplication formula using alternants. Choose $N$ sufficiently large and let ${\beta }_i={\mu }_i+N-i$ be the decreasing beta numbers of a partition. Expanding the determinant by permutations showsIndeed, in each determinant monomial every variable occurs in exactly one row factor; multiplying that variable by its $r$th power is equivalent to raising that row exponent.

A term vanishes if the new exponent equals another exponent. Otherwise suppose the changed exponent moves from position $i$ to position $j\le i$ when restored to decreasing order. Its determinant sign is $(-1{)}^{i-j}$. The resulting partition $\lambda$ hasand all other rows unchanged. The new diagram contains $\mu$, and its size increases by $r$. Between successive added rows, the overlap is exactly the single column ${\mu }_{t-1}+1$. Thus the added cells form a connected strip without a $2\times 2$ square, of height $i-j$. Conversely every added border strip has consecutive occupied rows and exactly this single-column overlap, forcing these formulas and the corresponding exponent shift. The correspondence is therefore bijective. Divide by the unchanged Vandermonde determinant and use the bialternant formula to get the power-sum border-strip multiplication identity

Apply this formula successively to each positive part of $\alpha$. Each sequence of choices is exactly a border-strip tableau, and its product of signs is $(-1{)}^{\mathrm{ht}(T)}$. ThereforeCommutativity of the power sums also shows that the resulting coefficient does not depend on the order of the positive parts of the type.

Take $\mu$ empty and restrict to $N\ge \ell (\lambda )$ variables. Multiply by $a_{\delta }$, with $\delta =(N-1,\dots ,0)$:In an alternant $a_{\nu +\delta }$, the exponents are distinct and their decreasing arrangement is $\nu +\delta$. Consequently the monomial $x^{\lambda +\delta }$ occurs only for $\nu =\lambda$, and in that determinant its coefficient is one. This proves the Frobenius alternant character formula

To identify these coefficients with symmetric-group characters, recall the Frobenius characteristic mapThe representation-theoretic identification $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ can be justified by Young's rule: the permutation character on ordered blocks of sizes $\nu$ is ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$. Its characteristic is $h_{\nu }$, because its fixed block decompositions allocate whole cycles to blocks, and the exponential generating series for those allocations is ${\prod }_{j}H(t_j)$. On the symmetric-function side, Hall duality and Schur orthonormality give $h_{\nu }={\sum }_{\lambda }{K}_{\lambda \nu }{s}_{\lambda }$. Inverting the unitriangular Kostka matrix establishes the identification. Thus the coefficient of $s_{\lambda }$ in $p_{\rho }$ is the irreducible character value on cycle type $\rho$. We have supplied the border-strip computation rather than assuming the character rule.

Grouping border-strip chains by their first removal of a cycle length $r$ now yields the Murnaghan–Nakayama ruleOne may choose the $r$-cycle to be last in a strip-building chain because the power sums commute. The empty-shape value on the empty type is one, providing the recursion's initial condition.

For the staircase $\delta =(m-1,m-2,\dots ,1)$, use $m$ beta numbers including its padded zero row. Its beta set isRemoving a rim hook of length $r$ lowers one beta number by $r$ into an unoccupied nonnegative position, reversing the exponent-shift description above. For a positive even $r$, this move either hits another occupied even position or goes below zero. Therefore there are no removable even-size border strips.

Under the Hall inner product of symmetric functions, the adjoint of multiplication by $p_r$ is $r\partial /\partial p_r$: this follows on the power-sum basis from $z_{\nu }/z_{\nu \setminus r}=r{m}_r(\nu )$. Taking adjoints of the one-strip multiplication rule expresses this derivative on $s_{\delta }$ as the signed sum over removable size-$r$ strips. That sum is empty for even $r$, soThe rational symmetric-function algebra is freely generated by the power sums, and its characteristic is zero. Vanishing of each of these derivatives therefore proves the staircase Schur functions use only odd power sums statement:

Finally, the cycle type of an $n$-cycle has one strip of size $n$. A straight diagram is a border strip exactly when its second row has length at most one: otherwise it contains the upper-left $2\times 2$ square; conversely the first row with a single-column tail is connected and contains none. These diagrams are exactly the hook partitions $(n-s,1^s)$, of height $s$. There is one strip filling in this case and none otherwise, giving