| > restart:with(plots): Since our pentafoil knot is toroidal it can be easily drawn (on a torus and without crossings). We first plot the torus. > torus:=plot3d([cos(s)*(5+2*cos(t)), sin(s)*(5+2*cos(t)),2*sin(t)],s=0..2*Pi, t=0..2*Pi,scaling=constrained,axes=boxed): > torus; | a pentafoil diagram [an analysis of its colorability can be found at the bottom of this page] |
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| > | lat:=conformal(z, z=0+0*I..10+4*I, grid=[6,3], numxy=[50,20]): a:=seq([-1+2*n mod 10,-1+2*n mod 4],n=1..10): mid:=pointplot([a],color=blue,symbol=(DIAMOND,14)): for j from 1 to 10 do p[j]:=textplot([-1+2*j mod 10,-1+2*j mod 4,j], align={RIGHT,BELOW},font=[TIMES,BOLD,20]) od: ps:=convert(p,set): |
- I have used Rob Scharein's powerful Knot drawing program KnotPlot , mentioned above (and free for personal use) to produce, on the fly, some smooth pictures of the [torus 2 5] knot. A single specimen can be found below to the right
| > | display({lat,mid} union ps); |
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- If we now successively draw the diagonals (parallel to the vector [1,1]) of the ten small "squares" of size 2 x 2 contained in the rectangle to the left, and do this in the order from 1 to 10, as indicated in blue font, we not only have constructed a pentafoil knot (recall that parallel side of the rectangle are identified), but we have also regained the geometrical argument for the additive group isomorhism:
- A chess player would say: "A single bishop can reach all ten squares of a five by two toroidal (unicolored) chessboard."
- Finally Maple will plot us the two pentafoil versions:
- To get a better display, the second knot picture is placed "by hand" next to the first. Maple can do this also, automatically, but then the images will be packed in a single box reducing their size considerably. If you zoom in, to restore to original size, your fonts will grow accordingly, like in the following two pdf files phi (62 kB) and theta (65 kB). They contain nine and four frames of a 3D animation of the (2,5) torus knot rotating around the y axis and the z axis, respectively.
| > | pentafoil_(5,2):=tubeplot([cos(5*t)*(5+2*cos(2*t)), sin(5*t)*(5+2*cos(2*t)),2*sin(2*t)], t=0..2*Pi,scaling=constrained, radius=0.5, numpoints=100, tubepoints=10, orientation=[45,45]): | > | pentafoil_(2,5):=tubeplot([cos(2*s)*(5+2*cos(5*s)), sin(2*s)*(5+2*cos(5*s)),2*sin(5*s)], s=0..2*Pi,scaling=constrained, radius=1, numpoints=80,tubepoints=20,orientation=[0,0]): | |
| > | pentafoil_(5,2); | > | pentafoil_(2,5); |
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the pentafoil page in Dror Bar-Nathan's Knot Atlas
- A diagram of a (mathematical) knot K is a projection of K onto a plane such that no multiple crossings occur and no arcs are tangent or cusped. The picture to the right shows a diagram of the pentafoil. It has five arcs labelled a, b, c, d, e and five crossings or, rather, fly-overs, numbered 1 to 5. A knot K is called n-colorable (n > 2) if for some diagram D of K the arcs of D can be colored by the elements of Zn in such a way that:
(1) At each crossing the sum of the colors of the underpass arcs equals the double of the color of the overpass arc.
(2) Not all colors are the same [a constant coloring trivially satisfies (1)].
- Since n-colorability of diagrams remains unaltered under Reidermeister moves, n-colorability is a knot property and provides a simple invariant to distinguish between (some) inequivalenct knots. The pentafoil is, not unexpectedly, 5-colorable. In fact, if A is the coefficient matrix of the set of five linear equations in a, b, ...,e corresponding to the five crossing conditions from (1), then A is a 5 x 5 circulant matrix with first row equal to [2,-1,0,0,-1] and so A[1,2,3,4,0]T=[0,0,0,5,-5]T, which is congruent to the all-null vector modulo 5. So a=1, b=2, c=3, d=4, e=0 is a proper 5-coloring.
- On the otherhand we know from graph theory that A=LLT, where L is the incidence matrix of the circulant (or cyclic) graph C5 on five points, and that hence the adjoint of A is equal to a constant matrix with all entries equal to the complexity of C5, the number of its spanning trees (which is 5 in this case, of course). It follows that the rank of A modulo n is less than 4 iff n = 0 mod 5. Finally, A has two eigenvalues unequal to zero, namely (5+sqrt(5))/2 and (5-sqrt(5))/2, both with multiplicity two. Their product is 5, what else.