No Quadrilateral/Pentagon Knots: Simple Closed Polygons are Trivial

[Bacle]

Hi, everyone:

I would appreciate any help with the following:

I am trying to refresh my knot theory--it's been a while. I am trying to answer
the following:

1) Every simple polygonal knot P in R^2 is trivial.:

I have tried to actually construct a homeomorphism h:R^3 -->R^3 , with

h|_P = S^1 , i.e., the restriction of the automorphism h to the poly. knot P gives

us S^1

I actually tried going in the opposite direction by creating a homeo. between a

polygonal knot P and S^1 (and seeing if I can extend it to h:R^3 -->R^3 ):

find a circle C going through the vertices of P (So that C an P are coplanar)

and then smoothly deform P to S^1. But I cannot see how to extend this to

an homeo. h:R^3 -->R^3 that restricts to this map.


2) Show there are no knotted quadrilateral nor pentagonal knots.

I have no clue here. Any Ideas.?.

Thanks.

 

[wofsy]

I don't know anything about knots but if you explain what knot equivalence is and what a polygonal knot and a trivial knot are - maybe we could figure it out.

 

[Bacle]

Sorry, Wofsy, I completely missed your reply:

i) A knot is just an embedding of S^1 in R^3 (there are other types of knots, but

these are the ones I am referring to. )*. For this reason, any two knots are

homeomorphic, and ( I think ) isotopic (see below). So we distinguish knots

by using:

ii) Knots K,K' are equivalent if there exists an ambient isotopy between them.

(An isotopy is a homotopy thru isomorphisms, i.e., each of the maps f(s,t)

in a homotopy is not just continuous, but a homeomorphism. )

Assumming the knots are embedded in R^3 , an ambient isotopy is an

isotopy from R^3-->R^3 that restricts to an isotopy from K to K'


iii) A trivial knot is any knot in the equivalence class of the trivial embedding

i:S^1-->R^3 (inclusion.).

iv) A polygonal knot is one that is the union of a finite number of closed

straight-line segments; the endpoints of these segments are the vertices

of the knot.


Like it often happens, I did prove this; I presented my proof to a prof., then

I forgot it.




* More generally, given a pair (A,S) , then A is knotted in S if there are different

classes of isotopic embeddings of A in S.

 

[Tinyboss]

Any projection of a quadrilateral "knot" can have at most one crossing, so that rules them out. And I think that a pentagonal knot could be projected onto a plane perpendicular to one of the segments, and in this projection it would appear to be a quadrilateral knot, and again have only one crossing.

 

[Bacle]

Tiny Boss:

Sorry, I wrong said a quadrilateral. I meant a pentagon --maybe kremlin --
or a hexagon. I think I can show a pentagonal knot can have at most two
crossings and a hexagonal can have at most three (alternating ones.) . I
don't know if my argument is correct, but I am basing it on this:

we need to use up two vertices before doing a crossing. After that,
we can do crossings with the third and forth vertices. After that, we
complete the loop (joining the 5th vertex with the 1st vertex), so
we can have at most two crossings. Similarly for the hexagonal knot:
we can have at most three alternating crossings. I know that we cannot
have an unknot with two crossing alone, and I think it should not be
too hard to show that we can unknot a graph with three alternate crossings.


But of course, that is not a proof.