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A knot is the image $K=f(S^1)$ of a smooth embedding $f:S^1\hookrightarrow {\mathbb{R}}^3$, considered up to ambient isotopy. The circle $S^1$ supplies a closed loop with no free ends; the smooth embedding condition rules out self-intersections and requires a nonzero tangent vector in a regular parametrization. The equivalent convention $S^1\hookrightarrow S^3$ uses the three-sphere: removing a point outside the knot identifies the remaining space with three-dimensional Euclidean space by stereographic projection.

More precisely, knots $K_0,K_1$ are equivalent if there is a smooth family of diffeomorphisms $F_s$ of the ambient space, for $0\le s\le 1$, such that $F_0$ is the identity and $F_1(K_0)=K_1$. Every intermediate $F_s(K_0)$ remains a smooth embedding. This ambient isotopy allows bending, stretching and moving a loop, but neither cutting it nor letting strands cross through each other. We compare the embedded images rather than their parametrizations; an oriented knot additionally remembers a direction of travel.

For a worked example, start with $U(t)=(\cos t,\sin t,0)$, where $t$ is taken modulo $2\pi$, and useBoth scale factors stay positive, so each $F_s$ is a diffeomorphism. At $s=1$ the circle becomes the ellipse $(2\cos t,\frac{1}{2}\sin t,0)$, yet its knot type is still the unknot. Exact lengths and angles are therefore irrelevant to knot type; self-intersection is the obstruction that the deformation must avoid.

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.

Knot theory studies closed curves in three-dimensional Euclidean space that cannot intersect themselves, and asks which can be deformed into one another without cutting or passing one strand through another. Its basic object is a knot: a smooth embedding of a circle, considered up to ambient isotopy. Several disjoint knots form a link. More generally, knot theory studies such embeddings in other manifolds and in higher dimensions.

The central surprise is that a flexible circle can carry permanent information about how it sits in space. A trefoil knot and an unknot each consist of just one circle, but no ambient isotopy turns one into the other. A knot diagram translates this spatial question into a finite drawing; Reidemeister moves describe the changes of drawing that preserve the knot. Knot invariants translate it into computations: unequal Alexander polynomials of a knot, Jones polynomials, or knot groups certify different knots. Equal values of a particular knot invariant generally leave the question open; even an isomorphism of knot groups alone is not a complete classification.

An early foundation for working with knot diagrams is Kurt Reidemeister's 1927 paper Elementare Begründung der Knotentheorie, published in volume 5 of Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg, pages 24–32.^[ref] Marc Lackenby's authored notes Elementary knot theory, Section 2.4, explain the Reidemeister move theorem and its role in searching for a certificate that two knot diagrams represent the same knot.^[ref]

A link isotopy is an isotopy of the ambient three-sphere carrying one link to another. It preserves the number and individual knot types of the components.

Here ambient isotopy is applied to the embedded link inside the three-sphere.

The two-dimensional Schoenflies theorem says that a simple closed piecewise linear curve in $S^2$ separates it into two components whose closures are piecewise linear balls of dimension two. In particular, a homeomorphism of the curve to the equator extends to a piecewise linear homeomorphism of the whole sphere.

A piecewise linear ball is a space piecewise linearly homeomorphic to a standard simplex. Write $D=B_1\cap B_2$. The complementary disk to $D$ in each $\partial B_j$ is a disk by the two-dimensional Schoenflies theorem. Thus each pair $(\partial B_j,D)$ is piecewise linearly homeomorphic to a sphere with a chosen hemisphere. Choose these boundary maps to agree on $D$. Extend them to the interiors by coning from an interior point, after compatible subdivisions. The resulting maps identify $B_1$ and $B_2$ with the two halves of a standard ball, glued along its equatorial disk. ConsequentlyThe same boundary parametrization proves a useful companion fact: if a ball is indented by removing a smaller ball attached along a boundary disk, the closure of what remains is again a ball. In the standard boundary-disk model, move the indented disk through a collar neighbourhood to a flat disk.

A handle decomposition is a filtration obtained by attaching handles $D^r\times D^{n-r}$ along $\partial D^r\times D^{n-r}$ in the boundary of the preceding stage. The number $r$ is the handle index; index-zero handles are disjoint balls, and index-$n$ handles close the remaining spherical boundary components. Starting with a triangulation of $M$, thicken its vertices, then its edges, and successively its higher-dimensional simplexes. In a barycentric subdivision these thickenings have precisely the indicated handle and attaching-region product structures. Adding collars between successive attachments gives a collared handle decomposition.

For a critical level handle embedding, assign successive distinct heights $t_j$ to the handles. At height $t_j$ the horizontal slice contains the $j$th handle and its attaching region. Between handle heights, the embedded boundary of the previously attached handles travels along its product collar; there is no topological change there. The lower boundary collar meets the attaching region of the next handle, and the upper collar leaves its new boundary. The stated ambient isotopy therefore separates all topology changes into horizontal handles and all intervening motion into collars. For an embedded surface, the local changes are the appearance of a disk, attachment of a band, and disappearance of a disk.

Here is the three-dimensional conclusion, with the distinction between the bounded and unbounded regions made explicit: a piecewise linear embedded $S^2$ in ${\mathbb{R}}^3$ bounds a piecewise linear three-ball; in the one-point compactification $S^3$, both complementary closures are three-balls. The unbounded closure in ${\mathbb{R}}^3$ itself is not a compact ball.

Put the embedded sphere in critical level handle embedding position. At a regular horizontal level its intersection with the plane is a finite disjoint union of simple closed curves. An innermost such curve bounds a planar disk whose interior misses the sphere. Cut the sphere along this curve and insert two slightly displaced copies of the disk. This replaces one sphere by two spheres. Repeat at regular levels between the critical heights, using innermost curves first. The number of curves at these chosen levels decreases at each operation, so the process stops. Each resulting sphere then lies in a slab containing at most one horizontal handle.

These terminal spheres bound balls directly. With no handle the surface is a disk transported through a collar and capped at its ends, the boundary of $D^2\times I$. A minimum or maximum merely supplies one of those caps. At a band level there are two planar boundary circles on one side and one on the other. Cap the two circles, straighten each planar disk using the two-dimensional Schoenflies theorem, and straighten the joining band in its planar collar. If the circles are not nested, this gives two balls joined along a boundary disk; if they are nested, it gives a ball with one boundary-attached ball removed. The gluing three-balls along a boundary disk argument and its indentation version show in both cases that the terminal sphere bounds a ball. This also explains why the planar position of the band matters: no assertion about an arbitrary knotted band is being made.

Reverse the disk surgeries. Suppose the two resulting spheres bound balls $A$ and $B$. Their disjoint boundaries imply that the balls are disjoint or one contains the other. Shrink the small displacement of the two surgery caps back to the original planar disk. In the disjoint case, the pre-surgery sphere bounds the union of two balls along that disk. In the nested case, it bounds the larger ball with the smaller boundary-attached ball removed. Both are balls by the preceding elementary facts. Induction through the finitely many surgeries proves the Three-dimensional Schoenflies theorem. To treat the other region in $S^3$, remove a point inside the ball just constructed and apply the same argument in the resulting copy of ${\mathbb{R}}^3$.

A 3-manifold is an irreducible three-manifold when every piecewise linear embedded two-sphere bounds a ball. Suppose $p:{\mathbb{R}}^3\to M$ is a universal cover, and lift an embedded sphere $S$ to $\tilde{S}$. Since $S$ is simply connected, its lift maps homeomorphically to $S$. By the Three-dimensional Schoenflies theorem, $\tilde{S}$ bounds a compact ball. The inverse image of $S$ is locally finite, so only finitely many other lifted spheres lie inside this ball. Choose an innermost one, bounding a ball $B$ whose interior meets no lift of $S$.

The interior of $B$ maps to one component $U$ of $M\setminus S$. This restriction is a covering map onto $U$: a path in $U$ starting in the image lifts into $\mathrm{int}B$, since it cannot cross a lifted boundary sphere. The covering is proper, because a compact subset of $U$ has preimage closed in the compact ball and cannot meet $\partial B$. Near $S$, the boundary map $\partial B\to S$ is one-to-one, so this connected covering has one sheet. The map on $B$ is therefore a homeomorphism onto $U\cup S$. Thus every embedded sphere in $M$ bounds a ball, and $M$ is an irreducible three-manifold.

For a real or complex vector bundle $E$, choose a fiber metric. Its Thom space is the pointed quotientcollapsing the sphere bundle to one basepoint. When the base is compact, this is the one-point compactification of the total space: fiberwise radial compression identifies the total space with the interior of $D(E)$, and all directions approaching its boundary become the single point at infinity.

Decompose ${\mathbb{C}}^{n+m+2}=V\oplus W$ with $V={\mathbb{C}}^{n+1}$, $W={\mathbb{C}}^{m+1}$, so the subspace being collapsed is $\mathbb{P}(V)={\mathbb{CP}}^n$. Every line outside it has a representative $(v,w)$ with $w\ne 0$. Its projection to $W$ gives the line $\ell =\mathbb{C}w\in \mathbb{P}(W)={\mathbb{CP}}^m$, and there is a unique complex-linear map $A:\ell \to V$ such that $A(w)=v$. The line in $V\oplus W$ is precisely the graph of $A$. This construction is unaffected by replacing $(v,w)$ with $(\lambda v,\lambda w)$.

Let $\gamma ={\eta }_1$ be the tautological line over $\mathbb{P}(W)$. The graph construction and its inverse are continuous, indeed smooth in the usual affine charts, and identifyA Hermitian metric gives a complex-linear bundle isomorphism $\bar{\gamma }\cong {\gamma }^{\ast }$. With the Hermitian form linear in its first argument, it sends $\bar{v}$ to the functional $w\mapsto ⟨w,v⟩$. Linearity in the conjugate vector is exactly what makes this a complex, rather than merely real, isomorphism. Thus the bundle is $({\bar{\eta }}_1{)}^{\oplus (n+1)}$, with the conjugation bars present in the original PDF.

Collapsing the closed subspace ${\mathbb{CP}}^n$ in compact Hausdorff space ${\mathbb{CP}}^{n+m+1}$ gives the one-point compactification of its open complement: neighborhoods of the collapsed point have compact complements, and conversely every compact subset of the complement is closed in the ambient space. The graph homeomorphism therefore extends over the added point. This proves the projective-space quotient as a Thom space of conjugate tautological lines:

The graph coordinates also determine the normal complex structures requested in the second part. The copy $\mathbb{P}(W)$ corresponds to the zero section of the graph bundle, whose normal bundle is that bundle itself. For a coordinate embedding ${\mathbb{CP}}^k\subset {\mathbb{CP}}^n$, use $W={\mathbb{C}}^{k+1}$ and its complementary coordinate space $V={\mathbb{C}}^{n-k}$. Then the normal bundle of a linear complex-projective embedding isThe complex structure is the linear structure in these graph coordinates. Equivalently, applying the preceding homeomorphism with its parameters $n-k-1$ and $k$ identifies the Thom space of this normal bundle with ${\mathbb{CP}}^n/{\mathbb{CP}}^{n-k-1}$, after reordering coordinates. Define ${\nu }_m$ in the identical way. The Hermitian-metric identification with the conjugate line does not change the complex normal orientation.

The original coordinate embeddings may be nested, so first move one into transverse position. Keep ${\mathbb{CP}}^k=\mathbb{P}(A)$ for the first $k+1$ coordinates, and move ${\mathbb{CP}}^m$ to $\mathbb{P}(B)$ for the last $m+1$ coordinates. A unitary coordinate permutation accomplishes this. Every unitary matrix can be joined to the identity by diagonalizing it and varying its eigenvalue phases continuously, so this move is an ambient isotopy. Its trace gives a cobordism carrying the transported normal complex structure, and hence leaves the input cobordism class unchanged.

If $k+m\ge n$, thenTheir projectivizations are transverse and meet in ${\mathbb{CP}}^r$, where $r=k+m-n$. The direct-sum normal structure from question 1 restricts toThe quotient map defining the transverse normal-bundle isomorphism is complex-linear here, so it agrees with the standard normal complex structure and contributes no orientation sign. If $k+m<n$, the first and last coordinate subspaces have no nonzero vector in common, so their projectivizations are disjoint.

Consequently the product of linear projective subspaces in complex cobordism isThe product degree is $2(2n-k-m)$, equal to the real codimension of the stated intersection when it exists. In the equality case $k+m=n$, the representative is one point with its complex-oriented normal space.

For tame or smooth embeddings $f_0,f_1:S^1\hookrightarrow S^3$, ambient isotopy means there is a continuous family $F_t:S^3\to S^3$, $0\le t\le 1$, of homeomorphisms, with $F_0=\mathrm{id}$ and $F_1\circ f_0=f_1$. Equivalently the map $(x,t)\mapsto (F_t(x),t)$ is a homeomorphism of $S^3\times [0,1]$ preserving the time coordinate. In the smooth category one uses diffeomorphisms and a smooth isotopy. If only unparametrized images are compared, the last condition is $F_1(f_0(S^1))=f_1(S^1)$; for oriented knots it must also preserve the strand orientation.

A link is an embedding of a finite disjoint union of circles. Choose a point of $S^3$ outside it and identify the complement of that point with ${\mathbb{R}}^3$. A generic planar projection gives an immersion with only finitely many transverse double points, no triple points and no singularities of the projected strands. A link diagram is this planar immersed union together with the over/under information at each double point, recording which strand has the larger height. An oriented link diagram also records each component's orientation. The crossing data are indispensable: the planar immersion alone does not specify the link.

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.