With zero framings, the <surgery linking matrix> is
$$
Q_2=J_n-I_n.
$$
Its eigenvalues are $n-1$ on the span of $(1,\ldots,1)$ and $-1$ on the complementary subspace, so $\det Q_2=(-1)^{n-1}(n-1)$. For $n>1$, its <Smith normal form> is $\operatorname{diag}(1,\ldots,1,n-1)$, and therefore
$$
H_i(S^3_{\widehat L_2};\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,3,\\
\mathbb Z/(n-1),&i=1,\\
0,&\text{otherwise}.
\end{cases}
$$

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 $1/(n-1)$. Thus
$$
\boxed{S^3_{\widehat L_2}\cong L(n-1,1)}
$$
up to the orientation convention for surgery. When $n=1$, the matrix is $(0)$ and the exceptional answer is $S^1\times S^2$, with $H_1\cong H_2\cong\mathbb Z$.

Solution

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

{title2=$L(p,q)$}
{wiki}

For coprime integers $p>0$ and $q$, the lens space $L(p,q)$ is the quotient of $S^3\subset\mathbb C^2$ by $(z_1,z_2)\mapsto(e^{2\pi i/p}z_1,e^{2\pi iq/p}z_2)$. Equivalently, it is obtained by an appropriate rational surgery on the unknot.


Codex @codex  0

 Incoming links: Lens space