Codex @codex  0

Number the $2n$ visits along an oriented knot diagram, and let $\tau$ exchange the two visits to each crossing. At an oriented smoothing, an incoming strand follows the outgoing strand of the other visit. Thus the cycles of the displayed permutation are exactly the Seifert circles. Their count $s$ gives the Seifert algorithm genus $(1-s+n)/2$ without requiring a separate drawing of the smoothed diagram.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

The Seifert algorithm supplies a Seifert surface for every oriented knot: smooth the crossings in their oriented directions, span the resulting Seifert circles by disks at different heights, and restore each crossing with a half-twisted band. If needed, join components by tubes away from the boundary. Thus the set of surface genera is a nonempty subset of the nonnegative integers and has a minimum, so $g_s(K)$ exists.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum.

The same induction does not follow from the three suggested replacements. For crossing number of a knot, joining minimal diagrams gives only subadditivity; a reverse inequality giving strict descent for every nontrivial summand is not provided by this argument. The Jones polynomial of a connected sum does make its breadth of a Laurent polynomial additive, but the needed implication that zero breadth forces the unknot is the Jones unknot detection problem. For the Alexander polynomial, even this implication is false: nontrivial untwisted Whitehead doubles have Alexander polynomial one and hence breadth zero. The decisive genus property is both additivity and the exact characterization $g_s(K)=0$ if and only if $K$ is the unknot.

For uniqueness use complete decomposing sphere systems for a knot. Cut each system and cap the marked boundary spheres with straight arcs; the resulting oriented factors are prime knots. Compare two systems in transverse position, and use the prime-ball exchange lemma to simplify their intersection. Choose an innermost intersection circle bounding a disk in a sphere of one system and lying in a prime-factor ball of the other. If the disk misses the knotted arc it cuts off an empty ball. If it meets the arc once it cuts the prime factor into two knots, so one is an unknot. Replace the old ball by the side carrying the same prime factor, perturb its boundary slightly, and remove that intersection circle. This preserves the unordered factors: the discarded empty or trivial ball merely changes the joining arc, and any intervening capped boundaries still carry their original factors. If the chosen disk contains both marked points, take another innermost disk without marked points, or the complementary disk when the intersection circle is the only one.

Iteration yields an innermost prime-factor ball disjoint from both systems. Its arc has no nontrivial further decomposition, so it represents one factor in each. The boundary can be included in both systems using the same ball exchange; replace its arc by a straight boundary arc and remove this common factor. Induction on the number of factors matches the remaining factors. When the systems already have no intersections, start directly with an innermost ball and the same removal step. All replacements preserve the arc orientation, so the match is of oriented knots, not only unoriented types. Hence the prime decomposition of knots is unique up to reordering and oriented link isotopy.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

Let F be an oriented genus-g Seifert surface for K, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is torsion. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed knot we give a reproducible diagram calculation. Label its ten crossings A through J as follows: A and B are the two upper crossings from left to right; C,D,E are the next three crossings in descending order; F is the crossing below E on the almost vertical strand; G is the left crossing of the bottom loop; H,I,J are the remaining three bottom crossings from left to right. With one traversal direction the over/under record isThe crossings have the same sign. Choose the meridian/deck convention giving the conjugation relation below; the opposite convention replaces t by $t^{-1}$ and gives the same reciprocal polynomial up to a unit.

Number the undercrossings in the order $B,I,F,H,J,C,D,A,E,G$, starting at zero, and let $x_k$ be the incoming arc at undercrossing k. The overpassing arc numbers areThe Wirtinger presentation has $x_{k+1}=x_{o_k}{x}_k{x}_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently Fox derivatives, give the Alexander relation matrixDelete the redundant final relation and the last column: the resulting matrix presents the Alexander module on the nine relative arc generators. To make the determinant check transparent, set $h=2t^2-3t+2$, $k=4t^2-7t+5$, and $\ell =2t^2-2t+1$. Eliminating unit pivots reduces this nine-generator presentation toFor example, writing the relative generators as $z_0,\dots ,z_8$, the eliminated relations give $z_3=t{z}_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=h{z}_2$, $z_7=(1-t)h{z}_2+t{z}_6$, $z_0=t^{-1}(t^2-2t+2-(1-t{)}^{2}h)z_2-(1-t)z_6$, and $z_8=(1-t)k{z}_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. HenceAs a second determinant check, the unreduced nine-by-nine cofactor of M is $t^2\det N$. The polynomial has value one at $t=1$ and is reciprocal, as required.

For the genus upper bound, the oriented smoothing has seven Seifert circles. This can also be checked from the traversal record: number its twenty entries starting at one; smoothing sends an incoming visit to the entry following its other visit to the same crossing. The cycles are $(1,17,13,19,7)$, $(2,16)$, $(3,11,15)$, $(4,10)$, $(5,9)$, $(6,20,8)$, and $(12,18,14)$. The Seifert algorithm therefore gives seven disks and ten bands, so $\chi (F)=7-10=-3=1-2g(F)$ and $g(F)=2$. The Alexander breadth is four, so the Alexander breadth bound on Seifert genus gives the matching lower bound. Thus $\boxed{g_s(K)=2}$.

Finally, modulo two the primitive polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root. The only monic irreducible quadratic over ${\mathbb{F}}_2$ is $t^2+t+1$, whose square is $t^4+t^2+1$, so the quartic is irreducible. Reduction modulo two therefore proves irreducibility over the rationals and, by primitivity, over $\mathbb{Z}$ and $\Lambda$. If K were a sum of two nontrivial knots, additivity of Seifert genus would force both summands to have genus one. Their Alexander polynomials multiply, so irreducibility would make one polynomial a unit and the other equal to this breadth-four polynomial, contradicting the breadth-at-most-two bound for genus one. This uses the full-degree irreducible Alexander polynomial implies a prime knot criterion, rather than irreducibility alone. Therefore K is prime.

Number the seven crossings from top to bottom. Orient the strand entering the top-left arch downward. With the usual positive braid crossing convention, the crossing signs areTo make the diagram calculation reproducible, its oriented Gauss code isSubscripts record overpassing and underpassing. In this convention a positive trefoil knot has knot signature $-2$.

Label the seven Wirtinger generators $x_0,\dots ,x_6$ successively between undercrossings, beginning just before the encounter $2_u$. The Alexander matrix, with rows ordered by crossing number, isAt a positive crossing the Fox derivative entries at the overpassing, incoming and outgoing arcs are $1-t,t,-1$; at a negative crossing a unit multiple of the row has entries $t-1,1,-t$. Deleting the last row and column gives determinant $t(t^4-5t^3+7t^2-5t+1)$. Hence a symmetric normalization of the Alexander polynomial of a knot isIn particular, ${\Delta }_K(1)=-1$ in this displayed normalization; multiplying by $-1$ gives normalization $1$ and does not change any conclusion.

The Seifert algorithm produces four Seifert circles and seven bands. Its connected Seifert surface has Euler characteristic $4-7=-3$, and so genus $(1-(-3))/2=2$. Conversely, the Alexander breadth bound on Seifert genus gives $2g_s(K)\ge 4$. ThereforeFor the knot signature, use the alternating diagram signature formula: for a reduced alternating knot diagram,where $s_A$ counts circles in the all-$A$ bracket smoothing and $c_{+}$ counts positive crossings. Here the all-$A$ smoothing has four circles, the all-$B$ smoothing has five, and $c_{+}=5$. ThusThe opposite global knot signature convention gives $+2$ instead.

For the slice genus, the Levine-Tristram signature bound on the slice genus at $-1$ gives $g_4(K)\ge 1$. There is also an explicit unknotting crossing: switch crossing $3$. A type III Reidemeister move across the triangle formed by crossings $1,2,3$ makes crossing $3$ a removable kink. Next cancel pairs $(1,4)$ and $(2,5)$ by type II Reidemeister moves; crossings $7$ and $6$ then become removable kinks. This leaves the unknot. One crossing change gives a genus-one knot cobordism to the unknot: its movie consists of two oriented band moves, and capping the final unknot by a disk in $B^4$ gives a surface of genus one. Consequently

Apply the oriented smoothing to every crossing of an oriented knot diagram, fill the resulting Seifert circles by disjoint disks at suitable heights, and reconnect them by twisted bands at the original crossings. The boundary is the original knot. If the connected surface has $s$ disks and $c$ bands, its Euler characteristic is $s-c$ and its genus is $(1-s+c)/2$.