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Taking the matrix trace of the adjugate expansion and using part (b) givesHence $\mathrm{Tr}(B_i)=(n-i)c_i$. Taking traces in $B_i=A{B}_{i-1}+c_{i}I$ gives, for $1\le i<n$,For $i=n$, the same formula follows by taking the trace of $-A{B}_{n-1}=c_{n}I$. Iterating the recursion from part (a) also givesConsequently the Faddeev–LeVerrier algorithm isStarting with $c_0=1$, this recursively expresses every $c_i$ using only $\mathrm{Tr}(A),\dots ,\mathrm{Tr}(A^i)$; these are the Newton identities for the characteristic polynomial.

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

For a complex vector bundle $E$, define the kth Adams operation by the kth Newton polynomial in its exterior powers:After applying the splitting principle for complex vector bundles and writing $E=L_1\oplus \cdots \oplus L_r$, the Newton identities giveThis formula proves additivity. Applied simultaneously to splittings of $E$ and $F$, it also givesThus ${\psi }^k$ extends to a natural ring endomorphism of $K^0(X)$ and, for a line bundle,

Write $X={\mathbb{CP}}_k^{k+2}$ and let $x=[\overline{\gamma }]-1$. Its reduced K-theory is freely generated by $x^k,x^{k+1},x^{k+2}$, with $x^{k+3}=0$, and restriction to the bottom cell sends $x^k$ to a Bott element $u\in {\tilde{K}}^0(S^{2k})$ and the two higher powers to zero. If $r:X\to S^{2k}$ retracts the bottom-cell inclusion, thenfor some integers $a,b$. Naturality and the Adams operation on a Bott class giveSince ${\psi }^2(x)=2x+x^2$, comparison modulo $x^{k+3}$ first in degree $k+1$ and then in degree $k+2$ givesThe first equality makes $k$ even. Then $3k+5$ is odd, so integrality of $b$ forces $8\mid k$; it is also congruent to $2$ modulo $3$, so $3\mid k$. Consequently

Let $V_n$ be the space of homogeneous degree-$n$ polynomials in $x,y$. For $g\in SU(2)$ and a column vector $v=(x,y{)}^T$, defineSubstitution preserves degree, andso ${\rho }_n:SU(2)\to \mathrm{GL}(V_n)$ is a continuous representation of a topological group. This is the homogeneous polynomial representation of SU2.

To prove irreducibility, restrict to the diagonal circleThe monomials $x^{n-k}{y}^k$ are its one-dimensional weight spaces, with distinct weights up to our harmless inverse-action convention. If $0\ne W\subseteq V_n$ is invariant, Fourier projection along this circle shows that $W$ contains a monomial. Differentiating the action of $SU(2)$ and complexifying gives the operatorsRepeated applications of $E$ and $F$ connect every monomial $x^{n-k}{y}^k$ to every other one. Hence $W$ contains the entire monomial basis, so $W=V_n$.

Every $g\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. Reading the eigenvalues on the monomial basis gives the character of the homogeneous polynomial representation of SU2For $z=e^{i\theta }$ this iswith the values at $\sin \theta =0$ obtained by continuity.

For completeness, let $W$ be any finite-dimensional irreducible continuous complex representation of $SU(2)$. Its restriction to the diagonal circle splits into integral weight spaces. Choose a vector of largest weight $n$. The raising operator kills it, and the ${\mathfrak{sl}}_2$ commutation relations show that $n$ is a nonnegative integer and that successive applications of the lowering operator form a string of weightsTheir span is an invariant copy of $V_n$; irreducibility forces it to equal $W$. Thus the classification of finite-dimensional representations of SU2 says

If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, then the eigenvalues on ${\bigwedge }^{2}V$ are ${\lambda }_i{\lambda }_j$ for $i<j$. Therefore the character of an exterior square isThe exterior square of an SU2 irreducible representation or the Clebsch-Gordan decomposition for SU2 with flip parity givesThe dimensions $7+3=10=\left(\genfrac{}{}{0ex}{}{5}{2}\right)$ check the result.

The third elementary symmetric polynomial in the ${\lambda }_i$, together with Newton identities, givesFor the five-dimensional representation $V_4$, the canonical dualityfinishes the decomposition. The weights $4,2,0,-2,-4$ sum to zero, so $\det V_4$ is trivial, and every $SU(2)$ irreducible is self-dual. Consequently

The splitting principle for complex vector bundles says that for every complex vector bundle $E\to X$ there is a map $p:F(E)\to X$ such that $p^{\ast }$ is injective on cohomology andsplits into complex line bundles. Write $x_i=c_1(L_i)$ for the formal Chern roots.

Define the Chern character after this injective pullback byEach homogeneous component is a symmetric polynomial in the $x_i$ with rational coefficients, hence a polynomial in the elementary symmetric functions $c_j(E)$. It therefore descends uniquely to $H^{\mathrm{ev}}(X;\mathbb{Q})$ and depends only on $E$. Seton the Grothendieck group $K^0(X)$; additivity under direct sums makes this well defined.

If $E$ has roots $x_i$ and $F$ has roots $y_j$, then $E\otimes F$ has roots $x_i+y_j$. ConsequentlyIt also sends the trivial line to $1$, so it is a unital ring homomorphism.

For $S^{2n}$, a generator of ${\tilde{K}}^0(S^{2n})$ is the $n$-fold exterior product of the degree-two Bott element. The Chern character respects exterior products, and the degree-two Bott element has Chern character equal, up to sign, to the integral generator of ${\tilde{H}}^2(S^2;\mathbb{Z})$. Its $n$-fold product maps to the integral top-dimensional generator. Hence the Chern character on an even-dimensional sphere is integral.

Let the formal Chern roots of $E\to S^{2n}$ be $x_1,\dots ,x_r$ and write $p_n={\sum }_i{x}_i^n$. Sinceall lower Chern classes $c_1(E),\dots ,c_{n-1}(E)$ vanish. The Newton identities then reduce toThe degree-$2n$ term of the Chern character is thereforeIts evaluation on the fundamental class is an integer by integrality of the reduced Chern character. Thusis divisible by $(n-1)!$, proving the Divisibility of the top Chern number on an even-dimensional sphere.