Divisibility of the top Chern number on an even-dimensional sphere
= Divisibility of the top Chern number on an even-dimensional sphere
{c}
For a complex vector bundle $E\to S^{2n}$,
$$
\left\langle c_n(E),[S^{2n}]\right\rangle
$$
is divisible by $(n-1)!$. Since the lower Chern classes vanish, the Newton identity gives
$$
\operatorname{ch}_n(E)
=(-1)^{n+1}\frac{c_n(E)}{(n-1)!},
$$
and the <Chern character on an even-dimensional sphere is integral>.