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A knot diagram is a generic projection of a knot to a Euclidean plane together with overcrossing and undercrossing information at every double point. Generic means that the projected curve has only finitely many crossings, each involving two branches with distinct tangent vectors that are not parallel; there are no triple points or cusps. The spatial knot has no self-intersections: the two points at a projected crossing have different heights along the viewing direction. A gap in the lower strand records which one passes underneath.

Forgetting the over/under information leaves a knot shadow. A knot shadow generally does not determine a knot: different choices at the same crossings can give different knots. Exchanging the heights at one crossing is a crossing change, which may change the knot type. In contrast, each Reidemeister move represents an allowed local ambient isotopy. Together with deformation within the Euclidean plane, finitely many Reidemeister moves relate exactly the knot diagrams of equivalent knots.

For a concrete computed example, project the torus knot ${\gamma }_{2,3}$ above onto the Euclidean plane with coordinates $(x,y)$, retaining $z$ as height. Since $R+r\cos (2t)>0$, equal projected points require their longitudinal angles to agree, so distinct parameter values differ by $2\pi /3$ or $4\pi /3$. Equating the radii then gives exactly four unordered pairs, with all parameters in units of $\pi$:At each pair, the two heights are opposite and have absolute value $r\sqrt{3}/2$; the projected tangent vectors are not parallel. Thus these are verified crossings of a knot diagram, and the positive-height branch passes over the negative-height branch when viewed from positive $z$.

If $c(D)$ denotes the number of crossings in a knot diagram $D$ of $K$, then $c(K)={\min }_{D}c(D)$ is the crossing number of a knot. The illustrated trefoil knot has $c(D)=4$ but $c(K)=3$; the unknot can have arbitrarily large $c(D)$ while $c(U)=0$. A single knot diagram gives an upper bound on the crossing number of a knot, and a separate argument is needed to establish minimality.

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.

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. Uniqueness is not needed here.

For coprime integers p,q, define the torus knot as the oriented parametrized curveCoprimality makes it an embedded single circle: two parameters with both coordinate phases equal differ by a multiple of $2\pi$. It lies on the standard unknotted torus separating the two solid tori. Negative exponents change the slope/orientation convention; they are allowed. If an exponent is zero, coprimality forces the other to be plus or minus one, giving an unknot.

Assume now $|p|,|q|\ge 2$. We give a group-theoretic proof that torus knots are prime, making the peripheral ingredient explicit. The Seifert-van Kampen theorem applied to the two solid-torus sides of the exterior, joined along the annulus parallel to the knot, givesHere a and b are the two core loops. The element $z=a^p=b^q$ is central. Under the abelianization $G\to \mathbb{Z}$, choose $a\mapsto q$, $b\mapsto p$; these generate the target because p,q are coprime. Thus $z\mapsto pq\ne 0$, so z is nontrivial. A meridian has, up to conjugacy, the form $m=a^u{b}^v$ with $qu+pv=1$. This follows by tracing the meridional disk across the two torus sides, using a dual primitive slope intersecting the knot slope once. Different Bezout solutions change u by a multiple of p and v by the opposite multiple of q, leaving the same word because z is central.

Quotienting by z gives $G/⟨z⟩=C_{|p|}\ast C_{|q|}$. Neither u nor v is zero modulo the corresponding cyclic order, by $qu+pv=1$. Hence the image of m is a cyclically reduced word of two syllables. All its nonzero powers remain reduced, so it has infinite order. In particular G is nonabelian and the knot is not an unknot.

We use the Loop theorem in its usual form: a noninjective boundary fundamental-group map of a three-manifold supplies an embedded compression disk. For a nontrivial knot, its boundary torus is incompressible. To see why, every sphere in its exterior bounds a ball on the side not containing the knot, by the Schoenflies theorem. A compression of the boundary torus would produce a sphere and make the exterior a solid torus. The compression disk has zero homology class in the exterior, so its boundary is the preferred longitude; attaching the annulus from that longitude to the knot gives a spanning disk and an unknot. This proves that nontrivial knot exteriors have incompressible boundary. Their peripheral groups inject as ${\mathbb{Z}}^2$, so their groups strictly contain the cyclic meridian subgroup.

If the torus knot were a connected sum of two nontrivial knots, van Kampen on the decomposing annulus would give $G=G_1{\ast }_{⟨m⟩}{G}_2$, with the meridian subgroup proper in each factor. The center of a proper amalgam lies in the amalgamated subgroup. One proof uses its Bass-Serre tree: a central elliptic element fixes a vertex orbit and its convex hull, which is the whole tree; a central translation would force the minimal tree to be its invariant axis, but a nontrivial vertex reflection reverses that translation. Thus a central element fixes every edge and lies in the edge subgroup.

The central element z would therefore equal $m^k$. Its image in $G/⟨z⟩$ would give ${\overline{m}}^k=1$, forcing $k=0$ because that image has infinite order. This contradicts $z\ne 1$. Consequently

Write $V=M_{S^2}(p^{\ast }/q,\ast )$. Filling one end of $S^1$ times an annulus gives a solid torus, so $V$ is a solid torus. The prescribed algebraic intersection number of curves on an oriented surface convention gives ${\tilde{f}}_2\cdot {\tilde{h}}_2=1$. HenceIn $H_1(V;\mathbb{Z})$, the first Dehn filling imposes $p^{\ast }f-q{h}_1=0$, while $h_1+h_2=0$. ThereforeThe class ${\lambda }_2$ is primitive because $\gcd (p^{\ast },q)=1$, a consequence of $p{p}^{\ast }-q{q}^{\ast }=1$. It is therefore the disk-bounding meridian of a solid torus of $V$. Since ${\mu }_2$ meets it once, Dehn filling $V$ along ${\mu }_2$ glues two solid torus pieces with disk boundaries meeting once. This is the standard genus-one Heegaard splitting of $S^3$, soTo identify the regular fiber geometrically, put it on their common boundary torus. Invert the displayed basis change:Here ${\mu }_2$ bounds a disk in the newly attached solid torus and ${\lambda }_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary solid torus pieces. Both coefficients are positive and $\gcd (p,q)=1$, so it is the positive $(p,q)$ torus knot. This identifies the torus knot directly from the genus-one splitting, without invoking a theorem specifically about torus knots.

The fiber monodromy is periodic when isotopy is allowed to move the boundary, in the sense of a periodic surface homeomorphism. Here is a justification from the Seifert fibered space structure, rather than a special formula about torus knots. Its base is a disk with cone-point orders $p,q$. The fiber Seifert surface is an incompressible surface because its inclusion in the mapping torus is injective on fundamental groups. It is also a boundary-incompressible surface: lift a proposed boundary-compressing disk to the infinite cyclic cover $\Sigma \times \mathbb{R}$ and project to $\Sigma$. This would homotope an essential arc of $\Sigma$ into $\partial \Sigma$, contradicting its essentiality. The classification of incompressible surfaces in Seifert fibered spaces therefore applies. A vertical surface in a Seifert fibered space has zero intersection with a regular Seifert fiber, whereas the fibration class evaluates on $f=pqm$ as $pq$. Thus the Seifert surface is isotopic to a horizontal surface in a Seifert fibered space, with $pq$ intersections per regular orbit.

Following the oriented Seifert fibers from one intersection to the next gives a first-return homeomorphism of the horizontal topological surface. It is a representative of the fiber monodromy, and its $pq$-th power is the identity. It has order $pq$: a regular orbit meets the connected horizontal surface in a Seifert fibered space in $pq$ cyclically ordered points. This realizes the periodic case of the Nielsen–Thurston classification theorem. It is neither an Anosov homeomorphism nor a pseudo-Anosov homeomorphism.

There is a boundary convention here. A representative fixed pointwise on the boundary is not periodic relative to the boundary: its $pq$-th power is a boundary Dehn twist (with sign determined by the return-map convention). Capping the boundary, or allowing it to rotate during isotopy, gives the finite-order representative intended by “periodic”. This distinguishes periodic surface type from literal finite order in the mapping class group relative to the boundary.

For any fibered knot, its Alexander module is $H_1(\Sigma ;\mathbb{Z})$ with the deck transformation given by homological monodromy, so its Alexander polynomial of a knot is $\det (tI-{\varphi }_{\ast })$ up to a unit. The general Turaev torsion relation $\Delta (t)\doteq (1-t)\tau (X)$ and part (b) therefore giveThis rational expression is a polynomial: in its factorization into cyclotomic polynomials, $\gcd (p,q)=1$ cancels all denominator factors. It is monic, has constant coefficient one, and has polynomial degree $pq+1-p-q=(p-1)(q-1)$. The characteristic polynomial has polynomial degree $2g$, is monic, and has constant coefficient $\det {\varphi }_{\ast }=1$, because homological monodromy preserves the intersection form. These facts remove the unit ambiguity and giveThe sign change between $\det ({\varphi }_{\ast }-tI)$ and $\det (tI-{\varphi }_{\ast })$ is trivial because $\mathrm{rank}{H}_1(\Sigma )=2g$ is even. As a separate geometric check, the orbifold Euler characteristic of the base is $1/p+1/q-1$, and its $pq$-sheeted horizontal surface in a Seifert fibered space has Euler characteristic $\chi (\Sigma )=p+q-pq=1-2g$.

The slope-$4/3$ pattern has winding number $4$, so the Satellite formula for the Alexander polynomial givesUsing the stated torus knot formula in each factor,Up to a Laurent unit this may be simplified to $(1-t)(1-t^{24})/((1-t^8)(1-t^3))$.

Writing the square coordinates modulo one, the curve has the parametrizationIt is therefore the $(2,3)$ torus knot, or trefoil, on the common boundary of the two solid tori. Applying the Seifert-van Kampen theorem to the complements on the two sides gives one generator $a$ for the meridian of one solid torus and one generator $b$ for the other; the boundary slope identifies two turns of the first with three turns of the second. HenceThis group is nonabelian: imposing the additional relation $a^2=b^3=1$ gives a quotientwhich maps onto the nonabelian group $S_3$ by sending $a$ to a transposition and $b$ to a three-cycle. An abelian group cannot have a nonabelian quotient.

Boundary-connected-summing minimal Seifert surfaces for $K$ and $K^{\prime }$ givesFor the reverse inequality, let $F$ be a minimal-genus Seifert surface for the connected sum of knots and let $S$ be its standard splitting sphere. A minimal-genus Seifert surface is incompressible in the knot exterior: a compression either lowers its genus or separates off a closed component that can be discarded. Put $F$ and $S$ in transverse position and minimize the number of intersection circles. An innermost circle on $S$ either gives a compression of $F$ or bounds a disk on $F$ across which it can be removed. Both alternatives contradict minimality, so $F\cap S$ consists only of the single arc joining the two points of $S\cap \partial F$.

Cutting $F$ along this arc gives Seifert surfaces $F_1,F_2$ for $K,K^{\prime }$. Their Euler characteristics satisfywhich, since all three surfaces have one boundary component, is equivalent toThis proves additivity.

The torus knot $T_{2,3}$ bounds a once-punctured torus, and its degree-two Alexander polynomial of a knot forces every Seifert surface to have genus at least one. Thus $g_s(T_{2,3})=1$. If it were a composite knot, both nontrivial summands would have positive Seifert genus, and additivity would give genus at least two. Hence $T_{2,3}$ is a prime knot.

For nonzero coprime integers $p,q$, the torus knot $T_{p,q}$ lies on a standard unknotted torus in the three-sphere and winds $p$ times meridionally and $q$ times longitudinally. Here a meridian of a solid torus goes around the small cross-section of the doughnut, while the longitudinal direction goes around its central hole. The integers count signed turns, so changing one sign gives the mirror of a link of the torus knot. In this article $p$ is always the meridional count and $q$ the longitudinal count.

An explicit parametric curve in three-dimensional Euclidean space iswhere $R>r>0$. Its cylindrical radius $\rho =\sqrt{x^2+y^2}$ satisfies $(\rho -R{)}^2+z^2=r^2$, the equation of the supporting torus. Its meridional angle is $pt$ and its longitudinal angle is $qt$: this checks the convention directly rather than relying on the picture. Its squared speed isIf two parameter values give the same point, their difference $\delta$ satisfies $p\delta ,q\delta \in 2\pi \mathbb{Z}$. Bézout's identity for the coprime integers $p,q$ then gives $\delta \in 2\pi \mathbb{Z}$. Thus the loop closes and is an injective function on the parameter circle, with nonzero derivative; it is a smooth embedding.

For example, $R=2$, $r=0.7$, $p=2$, $q=3$ produces a trefoil knot. During one circuit the meridional angle changes by $4\pi$ and the longitudinal angle by $6\pi$. The supporting torus is a guide for construction, not a constraint on later ambient isotopies: the knot may move off it. This family makes arithmetic visible as geometry: the coprime integers determine whether a single traversal visits the whole loop without repetition.

The coprime integers hypothesis matters. If the greatest common divisor is $d=\gcd (|p|,|q|)>1$, the same parametric curve retraces a single component $d$ times; it does not automatically draw $d$ different components. The corresponding torus link is instead the full set of points on the torus whose angles satisfy $q\varphi -p\theta \in 2\pi \mathbb{Z}$, with meridional angle $\varphi$ and longitudinal angle $\theta$. It has $d$ disjoint components, each with winding counts $(p/d,q/d)$. For instance, the torus link $T_{2,2}$ consists of two parallel $(1,1)$ loops and is a Hopf link, whereas substituting $(2,2)$ into ${\gamma }_{p,q}$ simply traces one of those loops twice.

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.