Seifert surfaces and knot genus

Parametrize the circle by a smooth regular embedding $\gamma :S^1\to {\mathbb{R}}^2$, and let $n$ be its smooth unit normal obtained by rotating and normalizing its tangent. The map $T(s,r)=\gamma (s)+rn(s)$ has invertible derivative at $r=0$. By [F1] and compactness it is a local diffeomorphism on a common strip $\mathrm{\mid }r\mathrm{\mid }<a$. It is injective after shrinking $a$: otherwise collisions for widths tending to zero would have a subsequence of their circle parameters converging; equality of the limiting images and injectivity of $\gamma$ give the same limiting parameter, contrary to a local inverse there. Thus $T$ is a smooth two-sided annular coordinate.

We prove that a rounded polygonal region is a smooth disk. Take the finite triangulation in [F3]. The dual adjacency graph of its triangles, with an exterior root adjacent across every boundary edge, is connected. Indeed the triangles at any one vertex form a consecutive cyclic fan or a consecutive interval fan, so all their sectors are joined through shared edges, and connectedness of the triangulated region joins all fans. Choose a finite spanning tree of this dual graph rooted at the exterior. Process triangles in increasing distance from that root. Remove each triangle and its edge toward its parent: this edge is free, since its only other incident triangle has already been removed or is exterior. Removing a triangle with its free side leaves its two other sides and vertices, preserves connectedness, and preserves $V-E+F$. Plane Euler in [F3], with the exterior counted as one extra face, gives initial value $V-E+F=1$. At the end the remaining connected graph $K$ therefore has $V-E=1$, so it is a tree: a spanning tree already has $V-1$ edges, leaving no edge outside that spanning tree.

Inscribe a polygon with a sufficiently fine cyclic mesh in $\gamma$. Uniform continuity of ${\gamma }^{\mathrm{\prime }}$ gives positive scalar product of each chord direction with the tangent at every nearby normal projection, and the chord error tends uniformly to zero. Hence the normal projection of each chord is strictly increasing in its cyclic coordinate; the whole polygon has degree one under normal projection and is a normal graph. In particular it is embedded. Smooth its finitely many corners in disjoint small graph charts: convolution of the local graph function with a smooth compactly supported kernel on a smaller interval, joined by a smooth cutoff on a larger interval, leaves it unchanged outside that chart and can be made arbitrarily small in height and tangent direction. The smoothed curve $P^{\mathrm{\prime }}$ is still a normal graph $r=f(s)$, with $f$ smooth and arbitrarily small. Choose a smooth cutoff $\chi$ supported in $(-a,a)$ and equal to one near zero. For $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\chi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$, the map $(s,r)\mapsto (s,r+\chi (r)f(s))$ is a diffeomorphism, because its normal derivative is positive, and equals the identity near the ends of the strip. It extends by the identity to a smooth plane diffeomorphism carrying $\gamma$ onto the rounded polygon $P^{\mathrm{\prime }}$. The same graph construction compares any two sufficiently small choices of corner rounding. By [F2], the polygon has bounded and unbounded regions; rounding preserves them by local graph homeomorphisms: in a graph chart, a continuous vertical displacement, cut off in the transverse direction with positive transverse derivative, carries the original piecewise affine graph to the rounded graph, and transporting them by the displayed plane diffeomorphism proves the corresponding separation and smooth boundary charts for $\gamma$.

Realize these finite collapses on smooth rounded neighborhoods. In affine coordinates on a triangle whose free side is to be removed, retain thin strips along its other two sides and a rounded disk at their common vertex. The missing sector can be chosen as a convex triangle with its corners rounded. Its attaching part runs along the inner sides of the two retained strips, joined around their common vertex, and is one boundary interval. Near the other two vertices use disjoint vertex disks and the corresponding short collar intervals; outside this triangle the neighborhood is unchanged. No other simplex meets its open interior. These choices give compatible rounded neighborhoods before and after the collapse whose difference is one smooth disk attached along one embedded boundary interval. To check that this added piece is a standard disk, a convex rounded triangle has a positive smooth radial boundary function $a(\theta )$ about any interior center. On the standard disk the radial map $r\mapsto r\exp (\eta (\log r)\log a(\theta ))$ is smooth and linear near zero when $\eta =0$ there, agrees with $r\mapsto ra(\theta )$ near $r=1$ when $\eta =1$ there, and has positive radial derivative if $\mathrm{\mid }{\eta }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{\parallel }\log a{\mathrm{\parallel }}_{\mathrm{\infty }}<1$; choose its transition over a sufficiently long interval of $\log r$. Thus it is a disk parametrization. The corner models match the retained strip collars before [F4] is applied. Consequently each collapse preserves the smooth diffeomorphism type of the rounded neighborhood. A rounded neighborhood of the final tree is a disk: remove a leaf disk and its band successively, each a disk attached along one interval, until one vertex disk remains. Applying [F4] finitely often proves the assertion for the initial rounded polygon. Step 2.1 then proves it for the original smooth circle.

Choose orientation-preserving disk parametrizations $p_i:\bar{B}(0,1)\to D_i$. The boundary map $p_1^{-1}b{p}_0$ has a smooth increasing angular lift $g:\mathbb{R}\to \mathbb{R}$, with $g(\theta +2\pi )=g(\theta )+2\pi$. Let $\rho (r)$ be smooth, zero near zero and one near one. The map $(r,\theta )\mapsto (r,\theta +\rho (r)(g(\theta )-\theta ))$ is the identity near the center and has angular derivative $1+\rho (r)(g^{\mathrm{\prime }}(\theta )-1)>0$. It is consequently a smooth disk diffeomorphism extending the boundary map, and is a product in the radial collars. Conjugate by $p_0,p_1$ to obtain a preliminary extension.

We can replace either radial collar by its prescribed collar without changing the boundary. More generally, express two collars with the same boundary parametrization in the normal coordinates of step 1.1. Their relative coordinate map is $(s,r)\mapsto (S(s,r),R(s,r))$, where $S(s,0)=s$, $R(s,0)=0$, and ${\mathrm{\partial }}_{r}R(s,0)>0$. Use the periodic lift of $S-s$ and interpolate to the identity by $(s+u(S-s),(1-u)r+uR)$ for $0\le u\le 1$. At $r=0$ the Jacobian is triangular with positive diagonal entries $1,(1-u)+u{\mathrm{\partial }}_{r}R(s,0)$. Smooth collar maps extend across $r=0$ in the normal chart, by their local smooth-extension convention and finitely many chart cutoffs. Compactness and the local-inverse and separation argument of step 1.1 give a common narrow two-sided closed annulus on which every interpolated map is an embedding. Reparametrize $u$ smoothly to be constant near its endpoints. Apply [F5] to this compact two-dimensional annulus and the ambient plane, with support in a thin boundary neighborhood. The prescribed isotopy fixes its middle circle, so its ambient extension fixes every boundary point and preserves the bounded side. Its final diffeomorphism identifies the two collar germs exactly. Perform this correction on both sides of the preliminary extension of step 4.1. The resulting extension agrees with $c_0(p,r)\mapsto c_1(b(p),r)$ on a common smaller collar. Its smooth inverse and orientation follow from the same properties of all the factors.

For the interior prescription, first move $B(p_0)$ to $p_1$ by a diffeomorphism supported in $\mathrm{int}{D}_1$. A smooth path exists there by the disk parametrization of step 3.1. Cover it by finitely many interior coordinate balls and subdivide it so each successive pair lies in one smaller convex coordinate ball. For each pair, translation of a sufficiently small closed coordinate disk along their segment is an isotopy of full-dimensional disk embeddings whose track lies in that ball. Apply [F5], with a time parameter constant near its ends, and compose the resulting supported ambient diffeomorphisms. This moves the point as stated and leaves a neighborhood of the boundary fixed. Rename the resulting extension $B$. The germ $k=q\circ B^{-1}$ at $p_1$ fixes $p_1$ and preserves orientation. In centered coordinates its rescalings $k_t(x)=k(tx)\mathrm{/}t$ for $t>0$ extend smoothly at $t=0$ to $Ax$, where $A=Dk(0)$, since $k(tx)\mathrm{/}t={\int }_0^{1}Dk(stx)xds$. Each positive linear matrix $A$ is smoothly joined to the identity: write $A=RP$ with $P$ positive definite and $R$ a rotation, interpolate $P$ linearly to the identity and the rotation angle linearly to zero. Concatenate the resulting path from the identity to $A$ with $k_t$ from $A$ to $k$, and make the time parameter constant near the endpoints and all joins. On one common sufficiently small closed disk all slices are embeddings with track in an interior coordinate ball, by compactness and the inverse-function and separation argument of step 1.1. [F5] extends this disk isotopy to an ambient isotopy supported in that ball. Its terminal map $K$ agrees with $k$ near $p_1$, so $K\circ B=q$ near $p_0$. The boundary collars stay fixed. Only [F4] and [F5] use Countable Choice; all selections and compositions are finite.

At each point of $K$, the inward unit vector in $TF$ perpendicular to $TK$ is a smooth nonzero normal vector. The orientation of $S^3$, the orientation of $K$, and the metric orient the normal plane; its positive quarter-turn supplies a smooth second unit vector. These vectors give an oriented orthonormal frame of the normal plane along all of $K$.

Normalize any nonzero normal field to unit length and express it in this reference frame as $(\cos 2\pi \theta (p),\sin 2\pi \theta (p))$. This defines a map $r:K\to S^1$. The frame identifies homotopies through nonzero normal fields exactly with free homotopies of these circle maps, since normalization is continuous and the frame is defined at every point.

Fix a point of $K$. Any free loop $r$ can be made based by a constant rotation of the target sending its value at that point to $1$; the rotation is homotopic to the identity because $S^1$ is path-connected. By [F2], based homotopy classes are classified by degree and every integer occurs. Undoing the constant rotations shows that free homotopy classes are also classified by degree. Therefore the winding parameter is a bijection to $\mathbb{Z}$, and multiplication by the degree-$n$ loop adds $n$, giving a free transitive twist action.

By [F1], the smoothing gives finitely many disjoint embedded circles in the projection sphere. Choose a point $q$ outside them. By [F3], each circle is nullhomotopic; [F2] and [A1] supply both complementary disk regions, so for each circle choose the region not containing $q$ and orient it to induce the Seifert-circle orientation. Disjoint Jordan circles have disk regions that are nested or disjoint, so their finite strict-inclusion order admits a linear extension. Place nested disks at distinct nearby heights in that order; their interiors are then pairwise disjoint as subsets of $S^3$.

Choose pairwise disjoint crossing neighborhoods small enough to avoid all other Seifert-circle boundaries. Put the two projected tangent lines on the coordinate axes. Up to reversing both orientations and exchanging axes, the strand directions are east-and-north or east-and-south. In the first case the orientation-compatible noncrossing pairings are west-to-north and south-to-east; in the second they are north-to-east and west-to-south. Thus in each case the two smoothing arcs are locally parallel and equally directed. Any disk region for a third circle whose interior meets this crossing neighborhood contains the neighborhood and both affected circles, so it is an ancestor of each; its assigned height is lower than their disk heights. Raise narrow collars of both attaching intervals into a local slab above every disk sheet meeting the neighborhood. Attach a rectangle along those intervals with one half-twist. Of the two twist directions, choose the one whose long sides project to the original overpassing and underpassing branches. The equal arrow directions on the smoothing arcs permit opposite boundary parameterizations at the two attaching edges of the rectangle, so its orientation extends the orientations of both disks across the gluing. The crossing neighborhoods are disjoint; finiteness lets all collars and bands be thin enough to avoid every other disk sheet and one another.

Gluing the finitely many rectangles to disjoint boundary intervals of the disks and rounding the collar corners gives a smooth embedded compact two-manifold. The orientations agree along each attaching interval, so the surface is oriented. Outside a crossing neighborhood its boundary follows the original diagram arcs; inside each neighborhood the two long sides of the half-twisted band restore the specified crossing and its orientation. Hence the boundary is exactly the original oriented knot.

Every disk has an unused boundary arc because only finitely many short attaching intervals were chosen; each band attachment replaces two boundary intervals by its two long sides, so no component of the resulting disk-and-band surface becomes closed. Each connected surface component therefore contributes at least one boundary component. Step 3.1 gives a total boundary consisting of the one knot component, so the surface has exactly one connected component and satisfies [F5].

By [F2], $D$ has a finite crossing set of cardinality $c$. The first loop in [F1] visits each crossing once and performs one smoothing, so it ends after $c$ iterations. No smoothing introduces a new crossing to be processed.

By [F3], the completed smoothing has a finite set of Seifert circles of cardinality $s$. The second loop in [F1] records one disk per circle and one band per original crossing, so it also has finitely many iterations and returns exactly $(s,c)$ with the prescribed records. No topological correctness claim follows from this finite termination argument.

By [F1], move $K$ by an orientation-preserving ambient isotopy into ${\mathbb{R}}^3$. Apply [F2] and [A1] to choose a regular projection direction. The resulting decorated projection is a finite regular oriented diagram $D$ of a single knot.

Apply [F3] to $D$ to obtain a connected compact smooth oriented embedded surface with oriented boundary $D$. Pull the surface back by the inverse ambient isotopy from step 1.1; the inverse diffeomorphism preserves embeddedness, compactness, connectedness, orientation, and the boundary orientation, giving the required $F$ for the original $K$.

Subdivide the disk boundaries so all band-attachment intervals are subcomplexes. Start with $s$ disjoint disks, each with Euler characteristic $1$. Relative to the two attaching intervals, a rectangular band has two free long sides and one square $2$-cell; collapsing the square across one free long side leaves one relative $1$-cell joining the intervals. Thus the band changes the alternating cell count by $-1$. By [F1], after $c$ bands $\chi (F)=s-c$.

The single boundary component gives $b=1$. By [F2] and [A1], $s-c=\chi (F)=2-2g(F)-1=1-2g(F)$, so $g(F)=(1-s+c)\mathrm{/}2$.

For the algorithm output on $D$, [F4] identifies the counts with its $s$ circles and $c$ crossings, and [F5] makes the output an admissible Seifert surface. By [F3], the minimum defining knot genus is at most this surface's genus, yielding $g(K)\le (1-s+c)\mathrm{/}2$.

By [A1], the ${\mathrm{A}\mathrm{C}}_{\omega }$ hypotheses in [F1], [F3], and [F6] hold. The given $F$ is a Seifert surface by [F1]. Use its inward unit conormal as the reference normal field. By [F2], this Seifert framing has relative winding integer zero.

Let $s$ be the inward unit conormal, viewed in the round-metric normal bundle. By [F8], choose a smooth collar $c:K\times [0,{\epsilon }_0)\to F$ and orient its levels by the original $K$. For sufficiently small $\delta >0$, set $J_0=c(K,\delta )$. It is embedded, lies in $\mathrm{int}F$, and is disjoint from $K$. No prescribed collar velocity is assumed: since $c$ is an embedding onto a boundary neighbourhood, ${\mathrm{\partial }}_{t}c(p,0)$ is inward transverse to $K$ in $F$. Its normal quotient is therefore $a(p)s(p)$ for a smooth function $a(p)>0$; a tangential component is allowed.

Use a normalized tubular chart $\mathrm{\Phi }$ of $K$ as in [F4] and [F9], and write ${\mathrm{\Phi }}^{-1}(c(p,t))=(b_t(p),v_t(p))$, with $v_t(p)$ in the normal fibre over $b_t(p)$. Compactness permits one small interval of $t$ on which these expressions are defined for every $p$. At $t=0$ they are $(p,0)$, and normalization identifies the vertical derivative with $a(p)s(p)$. The maps $b_t$ are $C^1$-close to the identity. In a smooth oriented circle coordinate their lifts have the form $u+r_t(u)$ with $r_t$ periodic and $\mathrm{\parallel }{r}_t^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}2$ for small $t$; thus the lifts are increasing degree-one diffeomorphisms. Hence each $b_t$ is an orientation-preserving diffeomorphism. Reparametrize the level over $x=b_t(p)$, obtaining a smooth normal section $w_t(x)=v_t(b_t^{-1}(x))$. Taylor expansion in finitely many bundle charts gives, uniformly on $K$, $w_t(x)=ta(x)s(x)+O(t^2)$. Because $a$ has positive minimum and $s$ is unit, after decreasing $\delta$ this implies $⟨w_{\delta }(x),s(x)⟩>0$ for every $x$.

Since $F$ is a compact embedded surface, it is closed and has empty interior in $S^3$; hence its complement contains a point $q$. Choose an orientation-preserving stereographic chart from $q$. It places $K\cup J_0$ and $F$ in ${\mathbb{R}}^3$. Let $F_0$ be the closure in $F$ of the complement of the open boundary collar $c(K\times [0,\delta ))$. Then $F_0$ is compact, is disjoint from $K$, and has boundary set $J_0$. Give $F_0$ its orientation inherited from $F$; at its new boundary the outward collar-normal points back toward $K$, so the induced boundary orientation is the transported orientation of $J_0$, parallel to $K$. Apply [F6] and [F7] to choose a regular oriented diagram $D$ of $K\cup J_0$, with the projection-plane orientation induced from $S^3$.

For $0\le u\le 1$, interpolate in each normal fibre by $z_u(x)=(1-\chi (u))w_{\delta }(x)+\chi (u)\delta s(x)$, where $\chi (u)=\sigma (3u-1)$ is the plateau from [F10]. Every $z_u$ has positive pairing with $s$, so it never meets the zero section. Uniformly these vectors have norm $O(\delta )$; a common disk bundle about the compact zero section lies in the tubular domain, so all interpolating graphs lie in that domain when $\delta$ is small. Each graph projects identically to $K$, hence is an embedded circle, and $\mathrm{\Phi }$ carries it to an embedded circle disjoint from $K$. With $K$ as the fixed first component this is a smooth two-component link isotopy, stationary near its parameter endpoints, from $K\cup J_0$ to $K\cup \mathrm{\Phi }(\delta s)$. The orientation on $J_0$ agrees with the graph orientation because $b_{\delta }$ preserves orientation. By [F9] the family extends to an ambient isotopy and its linking number is unchanged. The endpoint $\mathrm{\Phi }(\delta s)$ is exactly a sufficiently small normalized-chart push-off in [F3]–[F4]. Consequently the self-linking in the statement equals $\mathrm{lk}(K,J_0)$, without identifying the collar level literally with a chart graph of $\delta s$.

Apply [F5] with its first component $J$ equal to the original knot $K$, its second (boundary) component equal to $J_0$, and the oriented spanning surface $\mathrm{\Sigma }=F_0$. The curve $K$ is disjoint from $F_0$, so it is transverse to $F_0$ with empty intersection and has signed intersection number zero. Thus [F5] gives ${\mathrm{lk}}_D(K,J_0)=\mathrm{lk}(K,J_0)=0$. This computes the linking number in the same order as the self-linking convention. The Axiom of Choice is the hypothesis used by [F5]; [A1] supplies the countable-choice hypotheses of the other cited suppliers.

By [F3] and step 3.1, $\mathrm{lk}(K,J_0)$ is the self-linking of the Seifert framing; step 3.2 makes it zero. Step 1.1 gives relative framing integer zero.