 Solution

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.