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The finite negative continued fraction isFor $a_i\ge 2$, it describes a negative-framed chain of unknots: the diagonal framings are $-a_i$ and adjacent linking numbers are one. Repeated slam-dunk moves give rational surgery coefficient $-[a_1,\dots ,a_r{]}^{-}$. The corresponding chain matrix is negative definite. To check this, the successive determinants of its negative have recurrence $d_j=a_j{d}_{j-1}-d_{j-2}$, starting with $d_0=1$, $d_1=a_1$; they are positive and increasing, so Sylvester's criterion applies.

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

A Heegaard splitting of a closed oriented three-manifold is a decomposition $Y=H_0{\cup }_{\Sigma }{H}_1$ into two handlebodies, with their common boundary the Heegaard surface $\Sigma$. With boundary present, the corresponding pieces are compression bodies, whose negative boundaries account for $\partial Y$.

For the given triangulation, take the barycentric subdivision. A regular neighbourhood $H_0$ of the original one-skeleton is a handlebody: thicken vertices to balls and edges to one-handles, then contract a spanning tree. The complementary region is a regular neighbourhood $H_1$ of the dual one-skeleton, whose vertices are tetrahedron centres and whose edges cross triangular faces. It too is a handlebody. Both graphs are connected, and their common boundary supplies the Heegaard splitting. If the original triangulation has $v$ vertices and $e$ edges, its Heegaard surface has genus $e-v+1$; the dual count gives the same number by $v-e+f-t=0$.

Choose an oriented meridian of a knot $\mu$ and the Seifert longitude $\lambda$ supplied by a Seifert surface for the null-homologous knot. For relatively prime integers $p,q$, rational Dehn surgery removes the interior of a tubular neighborhood and attaches a solid torus with its meridian of a solid torus on the unoriented slope $p\mu +q\lambda$. The choices $(p,q)$ and $(-p,-q)$ describe the same slope; $q=0$ is the original meridional filling. The boundary gluing reverses boundary orientation so that the oriented three-manifold extends across the filling.

For integral coefficients $n_i$, attach two-handles to $B^4$ along the components of the framed link, with their indicated Seifert framing shifts. The boundary operation removes $S^1\times D^2$ and inserts $D^2\times S^1$, with meridian $n_i{\mu }_i+{\lambda }_i$. Thus the compact oriented surgery trace satisfiesFor a finite rational coefficient, use a negative continued fractionReplace that component by an integrally framed chain of successive meridians with coefficients $a_0,\dots ,a_k$. Repeated slam-dunk moves give back $p/q$. Perform this replacement for every rationally framed component, leaving meridional fillings out. The resulting integral framed link has the same filled boundary, so its surgery trace proves that every such rational filling bounds a compact oriented four-manifold.

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.

For a null-homologous knot, choose a meridian of a knot $\mu$ and Seifert longitude $\lambda$. Rational surgery with relatively prime numerator $p$ and denominator $q$ fills the knot exterior along $p\mu +q\lambda$. A negative continued fraction for $p/q$ replaces this operation by integer surgery on a chain of meridians, using slam-dunk moves.