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For a permutation group, $c_r(g)$ is the number of length-$r$ cycles of $g$. The cycle index records the average cycle structure. On coloring a finite underlying set, a coloring fixed by $g$ is constant on each cycle. Replacing $t_r$ by the power-sum symmetric polynomial $p_r(x)$ therefore gives the pattern inventory by the weighted Burnside lemma.

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.

For an integer tuple $\ell =({\ell }_1,\dots ,{\ell }_m)$, the monomial alternant isNegative exponents require nonzero coordinates; all exponents in the character expansion below are nonnegative. The power-sum symmetric polynomial is $s_k(x)={\sum }_{i=1}^m{x}_i^k$. Put $\delta =(m-1,m-2,\dots ,0)$, so $A_{\delta }(x)={\prod }_{i<j}(x_i-x_j)$ is the Vandermonde determinant.

Let a conjugacy class of $S_n$ have ${\alpha }_k$ cycles of length $k$, with $\sum k{\alpha }_k=n$, and put $p_{\alpha }(x)={\prod }_k{s}_k(x{)}^{{\alpha }_k}$. The product $p_{\alpha }{A}_{\delta }$ is an alternating polynomial homogeneous of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. In an alternating polynomial, a monomial with two equal exponents has zero coefficient, since interchanging those variables fixes the monomial and reverses its sign. Grouping the remaining monomials by their permutation orbits gives a unique expansion in alternants $A_{\ell }$ with ${\ell }_1>\cdots >{\ell }_m\ge 0$.

Such tuples of the indicated total degree are exactly ${\ell }_i={\lambda }_i+m-i$ for partitions of an integer $\lambda$ of $n$ with at most $m$ parts. Define the class functionThe displayed monomial occurs with coefficient $1$ in $A_{\lambda +\delta }$ and in no other ordered alternant, soThis proves the expansion and explicitly defines its coefficients. They depend only on the cycle counts and hence are class functions. Identifying these coefficients with Specht module characters is the Frobenius alternant character formula, which the remaining parts allow us to assume.

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

For variables $x_1,\dots ,x_m$, the elementary symmetric polynomial and power-sum symmetric polynomial arePut $E(t)={\prod }_i(1+x_{i}t)={\sum }_{r\ge 0}{e}_r{t}^r$. Its logarithmic derivative isComparing the coefficient of $t^{n-1}$ in $E^{\prime }(t)=E(t)E^{\prime }(t)/E(t)$ gives the Newton identities

The given Bockstein homomorphism is the first Steenrod square. A real line bundle is pulled back from the universal line bundle over ${\mathbb{RP}}^{\infty }$, where the degree-one generator $x$ satisfies $\beta (x)=x^2$. By naturality,

Apply the splitting principle for real vector bundles, writing the pulled-back bundle as $L_1\oplus \cdots \oplus L_r$ and $x_i=w_1(L_i)$. The pullback in cohomology is injective, $w_j(E)=e_j(x_1,\dots ,x_r)$, and the Bockstein derivation rule givesIn $e_1{e}_j$, the terms for which the index from $e_1$ already lies in the $j$-element subset give this sum, while each square-free monomial of degree $j+1$ occurs $j+1$ times. Over ${\mathbb{F}}_2$ this yields

The boundary can recover more than a count. Under the hypotheses of the argument principle, let $h$ be a holomorphic function on a neighborhood of the contour and its interior. If the enclosed zeros of a function are $a_j$ with zero multiplicities $m_j$ and the enclosed poles are $b_k$ of orders $n_k$, thenMultiplying the local logarithmic derivative by $h$ changes its residue from $m_j$ to $m_{j}h(a_j)$, or from $-n_k$ to $-n_{k}h(b_k)$. The residue theorem then gives the identity. Choosing $h=1$ recovers the argument principle; choosing $h(z)=z$ gives the sum of the enclosed zero locations minus the sum of the enclosed pole locations, both counted with zero multiplicities or pole orders.^[ref]

For the rational function in argument principle for a rational function, the $h(z)=z$ integral is $2(1/4)-(-1/2)=1$. Adding the known pole location $-1/2$ back gives a zero-location sum of $1/2$. Dividing by the zero count two gives the arithmetic mean of the enclosed zeros, with multiplicity, as $1/4$. Choosing $h(z)=z^k$ similarly gives the difference between the power-sum symmetric polynomials evaluated on the enclosed zero and pole locations, repeating each location according to its multiplicity or order. These identities let boundary information constrain both the number and aggregate locations of enclosed zeros of a function.