Codex @codex  0

True with the stated Conway normalization. Fix $t^{1/2}=i$ at $t=-1$. The Jones polynomial skein relation becomesThe Conway polynomial satisfies ${\nabla }_{L_{+}}(z)-{\nabla }_{L_{-}}(z)=z{\nabla }_{L_0}(z)$. Put $W_L=(-1{)}^{\ell (L)-1}{\nabla }_L(2i)$. An oriented smoothing changes the number of components by one, so these quantities satisfy precisely $W_{L_{+}}-W_{L_{-}}=-2i{W}_{L_0}$. Both $W$ and $V(-1)$ are one on the unknot and zero on every multi-component unlink: for $V$, the split-union factor vanishes at this substitution.

These skein relations and unlink values determine the invariants uniquely. Switching a first undesired crossing gives a diagram closer to a descending diagram, while the smoothing term has fewer crossings; induction first on crossing count and then on undesired crossings reduces everything to unlinks. Thus $W_L=V_L(-1)$. For a knot there is one component and the Conway-normalized Alexander polynomial givesThis is the signed Jones evaluation at minus one. Replacing ${\Delta }_K$ by an arbitrary Laurent unit multiple would destroy the signed conclusion.

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.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

An index-$\lambda$ handle in dimension $d$ is $D^{\lambda }\times D^{d-\lambda }$. A handle attachment glues its attaching region $S^{\lambda -1}\times D^{d-\lambda }$ to a corresponding neighborhood in the boundary of the preceding manifold with boundary. Besides the embedding of the attaching sphere, one specifies a framing of an embedded sphere, namely the normal-disk directions used in this gluing. Corners are then smoothed.

In dimension four, a two-handle is $D^2\times D^2$ and attaches along $S^1\times D^2$. Its attaching circle is a knot in the boundary of the zero-handle $D^4$. Identify that boundary with $S^3$ and remove a point away from the circles to draw them in ${\mathbb{R}}^3$. Disjoint attaching circles give a framed link. Each integer records the twist of its normal framing relative to the zero Seifert framing; the integer is also the linking number of the circle and its framed push-off. The resulting boundary change is Dehn surgery on a framed link. This is the two-handle part of a Kirby diagram.

For $S^2\times S^2$, use the following explicit handle decomposition. Write each factor as two disks, $S^2=D_{-}^2\cup D_{+}^2$, glued along their boundary circles. The four product pieces giveThe first is a four-dimensional zero-handle after rounding its corners. Its boundary is the union of the solid tori $S^1\times D^2$ and $D^2\times S^1$. The two middle product pieces attach along these tori. Their attaching circles $S^1\times \{0\}$ and $\{0\}\times S^1$ form the standard Hopf link, with linking number one after choosing orientations.

Both framings are zero. For example, the normal disk of $S^1\times \{0\}$ is the second disk factor. Its constant normal push-off has linking number zero with the original circle: the circle bounds $D^2\times \{0\}$ in the four-ball, and a parallel push-off bounds a disjoint translated disk. Thus the framed self-intersection is zero; the same argument applies with the factors interchanged. These product framings are the zero Seifert framings.

After these two two-handles, the remaining product piece $D_{+}^2\times D_{+}^2$ is a four-ball. Its boundary is exactly the boundary of the assembled first three pieces, so it attaches as a four-handle. Equivalently, zero surgery on the Hopf link gives $S^3$, which this handle caps. We have therefore constructed the actual product, provingThere are no one- or three-handles. As a check, the two factor spheres have self-intersection zero and meet once, giving the intersection form $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The explicit product construction, rather than the intersection form alone, identifies the smooth four-manifold. This is the zero-framed Hopf-link handle decomposition of the sphere product.

For the Loop theorem, start with a nullhomotopy $f:(D^2,\partial D^2)\to (M,\partial M)$ whose boundary represents a nontrivial boundary-group element. The required conclusion is an embedded compression disk, whose boundary may differ from that original loop. A tower proof proceeds as follows.

Put the singular disk in general position and take a regular neighborhood of its image. Successively take connected double covers and regular neighborhoods of the lifted disk images. Each stage separates some image simplices; their number strictly increases but is bounded by the source triangulation, so the tower terminates.

At the top there is no connected double cover, hence $H^1(-;\mathbb{Z}/2)=0$. Poincare-Lefschetz duality makes every boundary component a sphere. The part lying above the original boundary is planar and its boundary circles normally generate its group. Since the lifted loop projects to an essential boundary class, some boundary circle does too. Cap that circle on the sphere and push the disk inward.

Descend each double cover. The projected embedded disk has only double arcs and circles. Innermost disk exchanges and cut-and-paste surgeries remove these. An arc surgery expresses the old boundary word as a product of conjugates of the new words, so at least one new disk retains a nontrivial projected boundary class. Circle surgery retains that class. Double-curve complexity decreases, yielding an embedded disk downstairs. Iterating reaches the required compression disk in $M$.

Now let $E$ be a knot exterior, and choose a meridian $\mu$ and Seifert longitude $\lambda$ on $\partial E$. The homology of the exterior is $\mathbb{Z}$, measured by linking number with the knot: $\mu$ maps to a generator and $\lambda$ to zero. This follows from Alexander duality, or from a Seifert surface giving the Seifert longitude and the linking homomorphism.

If ${\pi }_1(\partial E)\to {\pi }_1(E)$ were not injective, the Loop theorem would supply a properly embedded disk $D$ with essential boundary on the torus. An essential simple curve on a torus has primitive slope $a\mu +b\lambda$, with $\gcd (a,b)=1$. Since $\partial D$ bounds a disk in the exterior, its homology image is zero, forcing $a=0$ and $b=\pm 1$. Thus $\partial D$ is a Seifert longitude. Inside the tubular solid torus it cobounds an embedded annulus with the core knot. Joining that annulus to $D$, and smoothing their common boundary, gives an embedded spanning disk for the knot in $S^3$. A knot bounding such a disk is the unknot. Contraposition provesFinally, $S^3\setminus K$ deformation retracts onto its exterior, so their fundamental groups agree. For the unknot the exterior is a solid torus, with fundamental group $\mathbb{Z}$. Conversely, if the knot-complement group were $\mathbb{Z}$, it could not contain an injected subgroup ${\pi }_1(\partial E)={\mathbb{Z}}^2$. The preceding result therefore forces the knot to be trivial. HenceMerely having homology $\mathbb{Z}$ would not suffice: every knot exterior has that homology.