Codex @codex  0

A Heegaard diagram marks the boundaries of compressing disks from the two sides of a Heegaard surface. Attaching two-handles along the two collections, and capping resulting spherical boundary components, reconstructs the three-manifold.

The least genus of a Heegaard surface among all Heegaard splittings of a closed orientable three-manifold. Having a splitting of genus $g$ only proves $g_H(M)\le g$; stabilization supplies larger-genus splittings of the same manifold.

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.

Label the four crossings $A,B,C,D$ in top, left-middle, right-middle, bottom order. Label the bounded regions $d$ (left), $e$ (right), $a$ (upper middle), $b$ (central bigon), $c$ (lower middle), and the unbounded region $o$. The drawing preserves the crossings and the orientation of the supplied figure-eight knot diagram. Its overpassing strands are the outer-left/upper-right strand at $A$, the upper-left/lower-central strand at $B$, the upper-central/lower-right strand at $C$, and the lower-left/outer-right strand at $D$.

Here is the Dehn presentation Heegaard diagram specified by this drawing. Replace the underlying four-valent planar graph $G$ by a small three-dimensional regular neighborhood $N(G)$ and put $\Sigma =\partial N(G)$. Since $G$ has four vertices and eight edges, $\Sigma$ has genus $8-4+1=5$. On $\Sigma$, take the five $\alpha$ curves bordering the bounded faces, pushed slightly onto the boundary of $N(G)$. They bound a complete disk system in the complementary handlebody $S^3\setminus \mathrm{int}N(G)$; their labels are ${\alpha }_d,{\alpha }_e,{\alpha }_a,{\alpha }_b,{\alpha }_c$.

The four $\beta$ curves bound the disks dual to the four short crossing tunnels in $N(G)\setminus \mathrm{int}\nu K_8$. Lift the crossing circles shown in red onto the upper and lower sheets of $\Sigma$, routing them along the corresponding face boundaries; the crossing's overpassing strand determines the lift. Orient the $\alpha$ disk normals and the $\beta$ curves so that their signed successive intersections with the $\alpha$ curves areAt $A,D$ the missing fourth corner is the unbounded region $o$, whose generator is omitted. This sheet-lifting prescription, together with the projected graph, specifies the curves on the Heegaard surface; the red circles alone are only their local projections.

To recover the knot exterior from $\Sigma \times I$, attach five two-handles along the $\alpha$ curves on one side and cap the resulting sphere with a three-handle. On the other side attach four two-handles along the $\beta$ curves. Compressing that side leaves the boundary torus of the knot exterior, which is retained. In particular, the counts are five $\alpha$ compressions and four $\beta$ compressions, rather than two complete genus-five disk systems.

For intersection generators $\mathbf{x},\mathbf{y}$, a Whitney disk is a map of a disk into the symmetric product, with its two boundary semicircles mapped to the two product tori and its endpoints mapped to $\mathbf{x},\mathbf{y}$. Its homotopy class belongs to ${\pi }_2(\mathbf{x},\mathbf{y})$. Projecting it to the Heegaard surface gives an integral domain. With the surface oriented, the convention is ${\partial }_{\alpha }D=\mathbf{y}-\mathbf{x}$. The obstruction to a disk is the class $ϵ(\mathbf{x},\mathbf{y})\in H_1(Y)$ obtained by joining the points along alpha arcs and returning along beta arcs. In the Heegaard Floer setup, its vanishing is equivalent to equality of the generators' Spin-c structures; for the rational homology sphere diagrams this is equivalent to ${\pi }_2(\mathbf{x},\mathbf{y})\ne \varnothing$.