Two links with the same link diagram are equivalent
I really recommend that you just accept this as obvious and
continue, since the proof is long, nowhere near complete, and is not
referred to anywhere else. I never found a reference to cite for
this, although I can't imagine how one could prove without this that
the three Reidemeister moves are sufficient for describing legal
transformations to link diagrams.
First, one needs definitions.
- A knot is a polygon embedded in threespace, that is, a set
of line segments forming a cycle with line segments overlapping only
at endpoints, and exactly two line segments meeting at each endpoint.
- A directed knot is a knot where the cycle is directed, that
is, each endpoint has one input and one output.
- A link is a set of disjoint knots.
- A directed link is a link in which all the knots are
- link1 and link2 are equivalent if there is a continuous
mapping f:R^3x[0,1] -\gt R^3 such that f(link1,0) = link1 and f(link1,1) =
link2 and f(*,x):R^3 -\gt R^3 is bijective for all x in [0,1].
- A projection is, um, the standard mathematical projection of
3-space onto 2-space, except we require that no more than two line
segments be projected to any single point in 2-space. This is possible
because any three line segments form a one-dimensional set of projections
that maps some point of all three to a single point; there are a
finite number of such one-dimensional sets but a 2-space of
projections to choose from.
- A crossing in a projection is a point in 2-space that two
points in the link map to. Note that the two points will be on the
interiors of two distinct line segments.
- An overpass, first, choose "up" and "down" in the direction
that was eliminated in the projection. Given a crossing in a
projection, the overpass is the string segment in the link with
the top point, the underpass is the string segment in the link
with the bottom point.
Claim: Any two links with the same projection and the same
overpass information are equivalent.
- Given the common projection, at each crossing, choose a point in
3-space at height 1 for the overpass and height 0 for the underpass.
- Given the string segment from the projection connecting crossings
x and y with projected length s, for a point at distance t from x
along the projected string segment, choose the point at height
height(x)+((height(y)-height(x))*(t/s)), where height(x) is 1 if the
string segment is an overpass at x, 0 otherwise, similarly for y.
- You now have a third link with the same projection (and link
diagram) as the first two.
- Define f(link1,0)=link1, f(link1,.5) to be this third link, and
f(link1,1)=link2. Bleah, I need to specify the smooth interpolation
of the link edges and also of the surrounding 3-space. There is some
minimal distance that all line segments are apart. This can all be
done, it's just ugly.
Claim: If two links have projections with the same crossings,
the same overpass information, and crossings are connected the same
way, then the two links are equivalent.
This is the hardest part. The connecting arcs could be all over
the plane, mixed up like kneaded dough. I believe a reasonable
strategy is to unwind both knots until the connecting arcs are going
more or less directly to where they ought to go, then to transform one
knot into the other. I've never spelled out this part of the proof.
Claim: If two links have the same link diagram, then they can be
converted to having projections with the same crossings, the same
overpass information, and the crossings connected in the same way, so
they are equivalent.
This requires arguing about embedding planar graphs in the plane,
since the link diagram is a planar graph (plus crossing information),
but it is not an embedded planar graph. It also requires a definition
of a link diagram.
Claim: The three Reidemeister moves are sufficient.
This is similar to the argument that you can choose projections
where no more than two line segments are mapped to any point. The
Reidemeister moves are a list of ways that the link diagram can change
by moving through a projection that has one point with three line
segments mapping to it. This can happen in only a few ways:
- One line segment being on end (type I)
- The endpoint of two line segments colliding with the middle of
another line segment (type I or II)
- Two endpoints colliding (type I or II or nothing)
- The interiors of three line segments colliding (type III)