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

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 $H_1\cong H_2\cong \mathbb{Z}$.