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The drawn component $L_1$ is an unknot. Its exterior is a solid torus $V$, in which $L_2$ is a pattern of a satellite knot. A meridian of a solid torus of $V$ is ${\ell }_1$, whereas its longitudinal direction is $m_1$. Thus the prescribed gluing sends these directions to $m$ and $\ell$, respectively. This identifies $V$ with a tubular neighborhood of $K$ with its zero Seifert framing.

Removing the pattern before making this identification givesEquivalently, meridionally filling the remaining boundary first removes $L_2$ from the construction and leaves $E(K)\cup V=S^3$; its filling core is the required satellite knot $C(K)$.

In the original link diagram, the two parallel passages through a meridional disk of $V$ run in the same direction; closing them uses a single interchange. The winding number of a satellite pattern is therefore two. Untwisting the surrounding disk shows that $P(U)$ is the unknot (the pattern is the $(2,1)$ cable knot, up to the harmless sign of its single interchange). In particular ${\Delta }_{P(U)}=1$. The Satellite formula for the Alexander polynomial now gives the concise answerHere $\doteq$ allows multiplication by $\pm t^j$. Reversing the winding orientation replaces $t^2$ by $t^{-2}$, giving the same Alexander polynomial of a knot up to such a unit.

Use the meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

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.

Apply the given skein relation to a positive and a negative kink in a diagram of the unknot. Both crossed link diagrams represent the unknot, whereas the oriented smoothing represents the two-component unlink. ThusFactor the left-hand side and cancel the nonzero factor in the Laurent polynomial ring:The positive sign follows from the skein relation as printed in this paper. It differs from the frequently used Jones polynomial convention whose two-component unlink has value $-(q^{1/2}+q^{-1/2})$. With the printed convention, an $r$-component link has invariant $(-1{)}^{r-1}$ times that convention's Jones polynomial.

Insert a kink on any component of an oriented link diagram for $L$. Switching the kink's crossing gives another link diagram of $L$; its oriented smoothing separates off a small unknot in a ball disjoint from the rest of $L$. The skein relation therefore givesCancel the nonzero factor $q^{1/2}-q^{-1/2}$ in the Laurent polynomial ring to obtain

Let $H_{+}$ be the positively oriented Hopf link obtained by closing a two-strand braid with two positive crossings. Switching one crossing gives the two-component unlink; the oriented smoothing gives the unknot. The printed skein relation givesRepresent the right-handed trefoil knot by the closure of the two-strand braid with three positive crossings. Switching one crossing gives the unknot, while its oriented smoothing gives $H_{+}$. Consequentlyso the two Jones polynomials areThe second expression follows from the mirror of a link identity. They are unequal Laurent polynomials, so invariance under link isotopy proves that the right-handed and left-handed trefoils are not isotopic.

Take the untwisted Whitehead double of a trefoil knot. The Whitehead pattern has winding number of a satellite pattern zero and becomes the unknot when its companion is the unknot. The Satellite formula for the Alexander polynomial therefore givesThe Whitehead pattern is geometrically essential in its solid torus, so the satellite knot with nontrivial companion $T_{2,3}$ is nontrivial. Hence $K_4$ is not isotopic to the unknot despite having trivial Alexander polynomial of a knot.

Induct on the Seifert genus $g_s(K)$. A genus-zero knot is the unknot. If $K$ is prime there is nothing to prove. Otherwise write $K=K_1\#K_2$ with both summands nontrivial. Part c givesso each summand has strictly smaller genus than $K$. Apply the induction hypothesis to both. Since the genus drops at every nontrivial split, the process terminates after finitely many steps and expresses $K$ as a finite connected sum of prime knots.

If $K=K_1\#K_2$ with both summands nontrivial, the sphere separating the two punctured three-balls in the connected-sum construction is a splitting sphere of a knot. If it were trivial, the corresponding one-string tangle would be boundary-parallel and one of $K_1,K_2$ would be the unknot. The sphere is therefore nontrivial.

Conversely, a splitting sphere $S$ divides $S^3$ into three-balls $B_1,B_2$, and $K\cap B_i$ is a properly embedded arc. Join its endpoints by a fixed arc on $S$ and push that joining arc slightly into $B_i$; this closes the two tangles to knots $K_1,K_2$. Reversing the construction showsIf either $K_i$ were unknotted, an innermost-disk argument for a spanning disk of $K_i$ would make the corresponding tangle boundary-parallel, which is exactly the stated triviality condition for $S$. A nontrivial splitting sphere therefore makes both summands nontrivial, so $K$ is composite.

The exterior of an unknot is a solid torus. Inclusion of its boundary kills exactly the primitive Seifert longitude, so every essential simple closed curve bounding a disk in that exterior is isotopic, as an unoriented curve, to that longitude. The bounding disk need not avoid other link components.

In a complete decomposing sphere system for a knot, a disk inside a prime-factor ball that meets the knotted arc once cuts its factor into a connected sum of knots. One side must be an unknot. Replacing the ball by the side carrying the prime factor therefore preserves the unordered factor list. A disk missing the arc cuts off an empty ball and gives the same conclusion. After a small transverse perturbation, such a replacement removes an innermost intersection circle with another sphere system. If a chosen innermost disk contains both marked points, another innermost disk, or its complementary disk when it is the only intersection, contains neither. Iterating produces a prime-factor ball disjoint from both systems. The replacement changes only the location of a joining arc in the intervening punctured ball; capping the other boundaries still recovers the same oriented factors.

The quotient of the three-sphere by its antipodal map is a closed oriented three-manifold. Equivalently it is the lens space obtained by integer surgery of coefficient two on the unknot. Its fundamental group is cyclic of order two, from the sphere double covering. This makes it a basic example for Jones polynomial surgery invariants.

// Target: knot-theory.bigb

The trefoil is the torus knot $T_{2,3}$. It is the simplest nontrivial knot by crossing number of a knot: $c(T_{2,3})=3$. A standard three-crossing alternating knot diagram is reduced, so the Tait crossing-number theorem establishes minimality. The four-crossing knot diagram shown below also represents the trefoil knot, illustrating that a particular projection need not achieve that minimum. Its Seifert genus is one.

Here is a small algebraic certificate that a trefoil knot cannot be deformed into an unknot. One Seifert matrix for a trefoil knot isTaking the determinant gives the Alexander polynomial of a knotwhere $\doteq$ allows multiplication by a Laurent unit $\pm t^n$ in the Laurent polynomial ring. No such multiplication turns this three-term Alexander polynomial of a knot into the unknot's value $1$. Since the Alexander polynomial of a knot is a knot invariant, this computation rules out every possible ambient isotopy between the two knots; it does not merely rule out the deformations we happened to try.

The knot group supplies another certificate. The presentation $⟨a,b\mid a^2=b^3⟩$ maps onto the symmetric group $S_3$ by $a\mapsto (12)$ and $b\mapsto (123)$, since both sides of the relation map to the identity. These images generate a non-abelian group, so the knot group cannot be the unknot's infinite cyclic group. However, a trefoil knot and its mirror of a link have isomorphic knot groups: the abstract knot group alone does not distinguish all knots.

The unknot $U$ is the knot type represented by the planar circleEquivalently, an unknot is a knot that bounds a smoothly embedded disk in the three-sphere. The disk must itself be embedded; merely filling a loop with a map that is allowed to intersect itself does not establish that the knot is an unknot. Flattening a neighborhood of an embedded spanning disk explains why this criterion recovers the planar circle.

A complicated drawing can still represent an unknot. Starting with a crossing-free knot diagram of $U$, insert a small kink by a type I Reidemeister move. The resulting knot diagram has one crossing, but removing the kink exhibits the same unknot. Repeating in separate small regions gives knot diagrams of the unknot with arbitrarily many crossings, although the crossing number of a knot is $c(U)=0$. A drawing's apparent complexity is therefore not a certificate of knotting.

The normalized Alexander polynomial of a knot and Jones polynomial of the unknot are both $1$, and its knot group is the infinite cyclic group $\mathbb{Z}$, because the knot exterior is a solid torus. These are useful comparison values. In particular, a nonconstant Alexander polynomial of a knot certifies that a knot is not an unknot, but the converse fails: nontrivial untwisted Whitehead doubles also have Alexander polynomial of a knot $1$.