The cofibration
$$
S(E)\longrightarrow D(E)\longrightarrow\operatorname{Th}(E)
$$
gives the long exact sequence of a pair in <Topological K-theory>. Identify $D(E)$ with $X$ by deformation retraction and use multiplication by the <K-theory Thom class>
$$
\lambda_E\in\widetilde K^0(\operatorname{Th}(E))
$$
to identify the relative term with $K^*(X)$. Pullback along the zero section sends $\lambda_E$ to the <K-theory Euler class>
$$
e^K(E)=\Lambda_{-1}(\overline E).
$$
The map from the relative term to $K^*(D(E))$ is therefore multiplication by $e^K(E)$, giving the <K-theory Gysin sequence of a sphere bundle>
$$
\cdots\to
K^i(X)\xrightarrow{\cdot e^K(E)}K^i(X)
\xrightarrow{p^*}K^i(S(E))
\xrightarrow{p_!}K^{i+1}(X)
\to\cdots.
$$

For
$$
Y=S(\gamma_{\mathbb C}^{1,n+1}\oplus\gamma_{\mathbb C}^{1,n+1})
$$
over $\mathbb{CP}^n$, put $t=1-[\overline\gamma]$. The <Complex K-theory of complex projective space> is
$$
K^0(\mathbb{CP}^n)=\mathbb Z[t]/(t^{n+1}),
\qquad
K^{-1}(\mathbb{CP}^n)=0,
$$
and
$$
e^K(\gamma\oplus\gamma)
=(1-[\overline\gamma])^2=t^2.
$$
The Gysin sequence consequently identifies
$$
K^{-1}(Y)
\cong\ker\left(t^2:\mathbb Z[t]/(t^{n+1})\to
\mathbb Z[t]/(t^{n+1})\right)
=\mathbb Z\{t^{n-1},t^n\}
\cong\mathbb Z^2
$$
for $n\geq1$. This is the <Odd K-theory of the sphere bundle of two tautological lines>.

If $n=0$, then the base is a point and $Y=S^3$, so $K^{-1}(Y)\cong\mathbb Z$ by <Bott periodicity>.

The <cannibalistic class> is defined by the identity
$$
\psi^k(\lambda_E)=\rho^k(E)\lambda_E
$$
for the <Adams operation> $\psi^k$. The Thom class of a direct sum is the product of the pulled-back Thom classes. Applying the ring homomorphism $\psi^k$ gives
$$
\rho^k(E\oplus E')=\rho^k(E)\rho^k(E').
$$
If $L$ is a line bundle, restriction along the zero section gives
$$
(1-\overline L^k)
=\rho^k(L)(1-\overline L),
$$
so
$$
\rho^k(L)
=1+\overline L+\cdots+\overline L^{k-1}.
$$

Let $\delta:K^{-1}(S(E))\to\widetilde K^0(\operatorname{Th}(E))$ be the boundary map. By definition of $p_!$,
$$
\delta x=\lambda_Ep_!(x).
$$
The natural operation $\psi^k$ commutes with $\delta$, and therefore
$$
\lambda_Ep_!(\psi^kx)
=\psi^k(\lambda_Ep_!(x))
=\rho^k(E)\lambda_E\psi^k(p_!(x)).
$$
Cancelling the Thom class proves the <Adams operation and the boundary pushforward of a sphere bundle> formula
$$
p_!(\psi^kx)=\rho^k(E)\psi^k(p_!(x)).
$$

Choose the basis $a,b$ of $K^{-1}(Y)$ characterized by
$$
p_!(a)=t^{n-1},
\qquad
p_!(b)=t^n.
$$
For $E=\gamma\oplus\gamma$,
$$
\rho^2(E)=(1+\overline\gamma)^2=(2-t)^2,
\qquad
\psi^2(t)=1-\overline\gamma^2=2t-t^2=t(2-t).
$$
Modulo $t^{n+1}$, this gives
$$
\begin{aligned}
p_!(\psi^2a)
&=(2-t)^2\bigl(t(2-t)\bigr)^{n-1}\\
&=2^{n+1}t^{n-1}-(n+1)2^nt^n,\\
p_!(\psi^2b)
&=(2-t)^2\bigl(t(2-t)\bigr)^n
=2^{n+2}t^n.
\end{aligned}
$$
Since $p_!$ identifies $K^{-1}(Y)$ with this kernel, the <Second Adams operation on the odd K-theory of the sphere bundle of two tautological lines> is
$$
\psi^2(a)=2^{n+1}a-(n+1)2^nb,
\qquad
\psi^2(b)=2^{n+2}b.
$$
For $n=0$, the single generator of $K^{-1}(S^3)$ is multiplied by $4$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 142 / 4 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-142/4/solution