Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 142 (Characteristic Classes and K-theory)

For a compact space $X$, $K^0(X)$ is the Grothendieck group of isomorphism classes of finite-rank complex vector bundles under direct sum. Tensor product descends to this group and makes it a ring, with $[{\mathbb{C}}_X]$ as its unit.

The hypothesis $E_0\oplus {\mathbb{C}}_X\cong E_1\oplus {\mathbb{C}}_X$ says that $E_0$ and $E_1$ define the same stable class. Rank-$d$ bundles are classified by maps to $BU(d)$, and the stabilization $BU(d)\to BU$ is $2d$-connected. Since the finite CW complex $X$ has dimension $k\le 2d$, stabilization is injective on $[X,BU(d)]$. Thus stable cancellation for complex vector bundles gives

Now let $E_0,E_1$ have rank $d$ over Complex projective space ${\mathbb{CP}}^d$ and have the same Chern classes. Their Chern characters agree because each component of $\mathrm{ch}(E)$ is a universal rational polynomial in the Chern classes. The ringis torsion-free, while the Chern character becomes an isomorphism after tensoring with $\mathbb{Q}$; it is therefore injective. Hence $[E_0]=[E_1]$ in K-theory, so after adding trivial bundles they are isomorphic. Since ${\mathbb{CP}}^d$ has real dimension $2d$, cancellation applies once more:

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 $E\to X$, the K-theory Euler class is the zero-section pullback of its K-theory Thom class:The cofibration of the disk and sphere bundles gives the K-theory Gysin sequence of a sphere bundle

Let $\gamma ={\gamma }_{\mathbb{C}}^{1,n+1}$ be the tautological bundle. The Euler sequence on complex projective space gives the bundle isomorphismPut $t=1-[\gamma ]$, so $K^0({\mathbb{CP}}^n)=\mathbb{Z}[t]/(t^{n+1})$. Fromand evaluation at $s=-1$, or polynomial division followed by differentiation at the removable root, we obtain

Because $K^{-1}({\mathbb{CP}}^n)=0$, the Gysin sequence identifies even K-theory with the cokernel and odd K-theory with the kernel of multiplication by $(n+1)t^n$. Thereforeand, since multiplication only detects the constant coefficient,

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