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> $W(\widehat L_1)$ are
$$
Q_1=J_n=
\begin{pmatrix}
1&\cdots&1\\
\vdots&\ddots&\vdots\\
1&\cdots&1
\end{pmatrix}.
$$
Its <Smith normal form> is $\operatorname{diag}(1,0,\ldots,0)$. The surgery exact sequence, equivalently the kernel and cokernel of $Q_1$, gives
$$
H_i(S^3_{\widehat L_1};\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,3,\\
\mathbb Z^{n-1},&i=1,2,\\
0,&\text{otherwise}.
\end{cases}
$$

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. Hence
$$
W(\widehat L_1)\cong(\mathbb{CP}^2\setminus\operatorname{int}B^4)\,\natural\,\mathop{\natural}_{n-1}(S^2\times D^2),
\qquad
\boxed{S^3_{\widehat L_1}\cong\mathop{\#}_{n-1}(S^1\times S^2),}
$$
where $\natural$ denotes <boundary connected sum>.

Solution

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

{title2=$M\mathbin{\natural}N$}
{wiki=Boundary_connected_sum}

The boundary connected sum of two manifolds with boundary removes a boundary half-ball from each and glues the resulting hemispherical faces. Its boundary is the ordinary connected sum of the original boundaries.


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