For a complex vector bundle $E\to X$, the <K-theory Euler class> is the zero-section pullback of its <K-theory Thom class>:
$$
e^K(E)=\lambda_{-1}(\overline E)
=\sum_{j=0}^{\operatorname{rank}E}(-1)^j[\Lambda^j\overline E].
$$
The cofibration of the disk and sphere bundles gives the <K-theory Gysin sequence of a sphere bundle>
$$
\cdots\to K^i(X)\xrightarrow{\cdot e^K(E)}K^i(X)
\longrightarrow K^i(S(E))\longrightarrow K^{i+1}(X)\to\cdots.
$$

Let $\gamma=\gamma_{\mathbb C}^{1,n+1}$ be the <tautological bundle>. The <Euler sequence on complex projective space> gives the bundle isomorphism
$$
T\mathbb{CP}^n\oplus\mathbb C\cong(n+1)\overline\gamma.
$$
Put $t=1-[\gamma]$, so $K^0(\mathbb{CP}^n)=\mathbb Z[t]/(t^{n+1})$. From
$$
\lambda_s(\overline{T\mathbb{CP}^n})
=\frac{(1+s\gamma)^{n+1}}{1+s}
$$
and evaluation at $s=-1$, or polynomial division followed by differentiation at the removable root, we obtain
$$
e^K(T\mathbb{CP}^n)
=(n+1)[\gamma](1-[\gamma])^n
=\boxed{(n+1)t^n}.
$$

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 $(n+1)t^n$. Therefore
$$
\boxed{K^0(S(T\mathbb{CP}^n))
\cong\mathbb Z[t]/(t^{n+1},(n+1)t^n)
\cong\mathbb Z^n\oplus\mathbb Z/(n+1),}
$$
and, since multiplication only detects the constant coefficient,
$$
\boxed{K^{-1}(S(T\mathbb{CP}^n))\cong(t)\cong\mathbb Z^n.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 142 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-142/3/solution