Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 112 / 5

A framing of an embedded sphere $C:S^{k-1}\hookrightarrow N^{n-1}$ is a trivialization of its rank-$(n-k)$ normal bundle. The standard complex line ${\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ has normal bundle of Euler number $+1$, so this embedded $S^2$ has no framing.

If one framing $f_0$ exists, every other orientation-compatible framing is obtained from it by a map $S^{k-1}\to SO(n-k)$. Consequently the set of homotopy classes is a torsor forFor $k=2$ and $n=4$, this is ${\pi }_1(SO(2))\cong \mathbb{Z}$; the integer is the winding number of one framing relative to $f_0$.

Distinct components of the positively oriented torus link $T_{n,n}$ have linking number one. Since every component also has framing $+1$ relative to the Seifert framing, the surgery linking matrix and hence the intersection form of the surgery trace $W({\hat{L}}_1)$ areIts Smith normal form is $\mathrm{diag}(1,0,\dots ,0)$. The surgery exact sequence, equivalently the kernel and cokernel of $Q_1$, gives

In Kirby calculus, sliding the components over one chosen component diagonalizes the framed link to a split $+1$-framed unknot and $n-1$ zero-framed unknots. Hencewhere $♮$ denotes boundary connected sum.

With zero framings, the surgery linking matrix isIts eigenvalues are $n-1$ on the span of $(1,\dots ,1)$ and $-1$ on the complementary subspace, so $\det Q_2=(-1{)}^{n-1}(n-1)$. For $n>1$, its Smith normal form is $\mathrm{diag}(1,\dots ,1,n-1)$, and therefore

Handle slides reduce this surgery diagram to the standard surgery diagram of the lens space $L(n-1,1)$; equivalently, the generator of the cokernel has linking pairing $1/(n-1)$. Thusup to the orientation convention for surgery. When $n=1$, the matrix is $(0)$ and the exceptional answer is $S^1\times S^2$, with $H_1\cong H_2\cong \mathbb{Z}$.