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The core disk of a $k$-handle attachment $D^k\times D^{n-k}$ is $D^k\times \{0\}$. Together with its cap in the preceding stage, it often represents a homology class; for a surgery trace, capped two-handle cores give the basis of its surgery linking matrix.

For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix $H$ and vector $d$ are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix H and vector d are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

For real projective 3-space the surgery diagram is a single unknot of framing two. Thus $m=1$, $\sigma =1$, and its colored evaluation is ${\sum }_j{d}_j^2{\theta }_j^2$. Since $\mathcal{D}\kappa =p_{+}$, the explicit answer isThe denominator is nonzero by the proved Gauss-sum identity. For instance at $r=3$ the twists are $1,-i$, the squared dimensions are both one, and the numerator vanishes, so this normalized invariant is zero for ${\mathbb{RP}}^3$ at that level.

There are no one-handles, so attaching the three two-handles to a four-ball cannot create a fundamental group generator: ${\pi }_1(W)=1$. The handle chain complex has $C_2={\mathbb{Z}}^3$, $C_1=0$ and $C_0=\mathbb{Z}$, givingTo calculate the boundary, orient the two long components down their central strands, and orient the small component clockwise. The six mutual crossings of the long components have negative sign, giving linking number $-3$. Each long component has two positive crossings with the small component, giving linking number $1$. The surgery linking matrix, in the order of framings $4,3,2$, is consequentlyChanging a component orientation changes the corresponding row and column signs, and does not change the resulting homology. The long exact sequence of a pair and Poincare-Lefschetz duality identify $H_2(\partial W)=\ker Q$ and $H_1(\partial W)=\mathrm{coker}Q$. The determinant is nonzero, so the kernel vanishes. The gcd of matrix entries is one; the two-by-two minors include $-6$ and $7$, so their gcd is also one. The Smith normal form therefore has diagonal $1,1,7$, andIn particular the sign of the braid crossings matters: replacing $-3$ by $+3$ while leaving both meridian linkings positive would represent a different link and would give the wrong determinant.

The three handle attachments give $H_2(W;\mathbb{Z})\cong {\mathbb{Z}}^3$: there are no one-handles or three-handles. Use the capped core disks of handles as the homology basis, ordered by the framing labels $-2,-1,+1$. The surgery linking matrix gives the intersection form. Orient the three components so that the outer components have linking number $3$ and the middle component has linking number $1$ with each outer component. There are six crossings between the outer components, and two crossings between the middle component and each outer component; counting their signs gives these linking numbers. Reversing a component changes the corresponding row and column signs and does not change the invariants.

Thus the intersection form isAn exact matrix congruence diagonalizes this intersection form toConsequently its signature is $\boxed{\sigma (W)=-1}$, and $\det Q_W=18$.

The long exact sequence in relative homology and Poincare-Lefschetz duality identifywith the intersection form matrix, and identify $H_1(\partial W)$ with its cokernel, since $H_1(W)=0$. The Smith normal form has first invariant factor $1$, because the entries have greatest common divisor $1$, and second invariant factor $1$, because the upper-left two-by-two matrix minor has determinant $1$. The last invariant factor is $18$. Hence

Give the three Hopf fibration components coherent orientations, so their pairwise linking numbers are one. In the link exterior, their longitudes satisfy $[{\lambda }_i]={\sum }_{j\ne i}{\mu }_j$. Filling along $p_i{\mu }_i+q_i{\lambda }_i$ imposes that relation on first homology. For the three rational coefficients the presentation matrix isIts determinant is $30$, and the greatest common divisor of its two-by-two minors is one (for example, $-5$ and $-3$ occur). Its Smith normal form is $\mathrm{diag}(1,1,30)$, soA nonzero rational coefficient with numerator one does not by itself make a multi-component surgery an integral homology sphere: the nonzero linking numbers must be included.

For the integer filling, use the Seifert fibered space structure $P\times S^1$. Its central regular fiber is $h=abc$, and each preferred longitude is $h{\mu }_i^{-1}$. The three filling relations are consequentlySubstituting $b=h$ into $abc=h$ gives $c=a^{-1}$. The remaining equations say $a^2=a^4$, hence $a^2=1$ and $h=1$. Thus its fundamental group is cyclic of order two.

Geometrically, a filling coefficient $r$ attaches a Seifert fibered space solid torus with multiplicity $|r-1|$, the distance of $r\mu +\lambda$ from the regular fiber $\mu +\lambda$. The multiplicities are $2,1,4$; the middle filling creates no exceptional fiber. The result is a Seifert fibered space over the sphere with at most two exceptional fibers, hence a union of two solid tori, or a lens space. A lens space with fundamental group of order two is $L(2,1)$. ThereforeAs a separate arithmetic check, the integer surgery linking matrix has determinant $-2$, in agreement with its first homology of order two.

Take a Hopf link with its two unknot components labeled by integer surgery coefficients $2,3$.

The slam-dunk move eliminates a meridional component with coefficient $b$ and replaces the other coefficient $a$ by $a-1/b$. Eliminating the component with coefficient three givesThus this Dehn surgery on a framed link becomes $5/3$ Dehn filling on the unknot, with the required orientation. As a check, its surgery linking matrix iswhose determinant is five.

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