blown away: what knot to do when sailing sir randolph bacon iii

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Blown Away: What Knot to Do When Sailing Sir Randolph Bacon III

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Page 1: Blown Away: What Knot to Do When Sailing Sir Randolph Bacon III

Blown Away: What Knot to Do When Sailing

Blown Away: What Knot to Do When Sailing

Sir Randolph Bacon IIISir Randolph Bacon III

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How can we tell if a How can we tell if a knot is really knotted?knot is really knotted?

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Theorem:Theorem: Two projections represent the same knot if Two projections represent the same knot if and only if they are related through a sequence ofand only if they are related through a sequence of Reidemeister moves.Reidemeister moves.

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?

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Theorem.Theorem. A projection of the trivial knot with n A projection of the trivial knot with n crossings can be turned into the trivial projection incrossings can be turned into the trivial projection in no more than 2no more than 2100,000,000,000n100,000,000,000n Reidemeister moves. Reidemeister moves.

((““The number of Reidemeister moves needed for unknottingThe number of Reidemeister moves needed for unknotting””,,Joel Hass and Jeff Lagarias, Jour. A.M.S. 14 (2001), 399-428.)Joel Hass and Jeff Lagarias, Jour. A.M.S. 14 (2001), 399-428.)

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Theorem.Theorem. A projection of the trivial knot with n A projection of the trivial knot with n crossings can be turned into the trivial projection incrossings can be turned into the trivial projection in no more than 2no more than 2100,000,000,000n100,000,000,000n Reidemeister moves. Reidemeister moves.

((““The number of Reidemeister moves needed for unknottingThe number of Reidemeister moves needed for unknotting””,,Joel Hass and Jeff Lagarias, Jour. A.M.S. 14 (2001), 399-428.)Joel Hass and Jeff Lagarias, Jour. A.M.S. 14 (2001), 399-428.)

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< 2 700,000,000,000

Theorem.Theorem. A projection of the trivial knot with n A projection of the trivial knot with n crossings can be turned into the trivial projection incrossings can be turned into the trivial projection in no more than 2no more than 2100,000,000,000n100,000,000,000n Reidemeister moves. Reidemeister moves.

((““The number of Reidemeister moves needed for unknottingThe number of Reidemeister moves needed for unknotting””,,Joel Hass and Jeff Lagarias, Jour. A.M.S. 14 (2001), 399-428.)Joel Hass and Jeff Lagarias, Jour. A.M.S. 14 (2001), 399-428.)

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Tricoloration:Tricoloration:

A projection is tricolorable if its strands can be colored with three colors such that:

A projection is tricolorable if its strands can be colored with three colors such that:

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Tricoloration:Tricoloration:

A projection is tricolorable if its strands can be colored with three colors such that:

1. At each crossing, one or three colors meet.

A projection is tricolorable if its strands can be colored with three colors such that:

1. At each crossing, one or three colors meet.

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Tricoloration:Tricoloration:

A projection is tricolorable if its strands can be colored with three colors such that:

1. At each crossing, one or three colors meet.

2. At least two colors are utilized.

A projection is tricolorable if its strands can be colored with three colors such that:

1. At each crossing, one or three colors meet.

2. At least two colors are utilized.

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Examples:Examples:

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Examples:Examples:

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Examples:Examples:

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Non-examples:Non-examples:

EExamples:xamples:

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Non-examples:Non-examples:

EExamples:xamples:

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Thm. A projection of a knot is tricolorable if and only if every projection of that knot is tricolorable.

Thm. A projection of a knot is tricolorable if and only if every projection of that knot is tricolorable.

Pf. It is enough to show that the Reidemeister moves preserve tricolorability.

Pf. It is enough to show that the Reidemeister moves preserve tricolorability.

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No tricoloration, so perhaps trivial, perhaps not!No tricoloration, so perhaps trivial, perhaps not!

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14 =2(mod12)

3 =27(mod12)

4 = -8(mod 12)

14 =2(mod12)

3 =27(mod12)

4 = -8(mod 12)

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a=b(mod p) if p divides a-b

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e.g. 7 = 2 (mod5) since 5 divides 7-2.e.g. 7 = 2 (mod5) since 5 divides 7-2.

a=b(mod p) if p divides a-b

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e.g. 7 = 2 (mod5) since 5 divides 7-2.

9= 0 (mod 3) since 3 divides 9-0.

e.g. 7 = 2 (mod5) since 5 divides 7-2.

9= 0 (mod 3) since 3 divides 9-0.

a=b(mod p) if p divides a-b

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e.g. 7 = 2 (mod5) since 5 divides 7-2.

9= 0 (mod 3) since 3 divides 9-0.

1=7(mod 6) since 6 divides 1-7.

e.g. 7 = 2 (mod5) since 5 divides 7-2.

9= 0 (mod 3) since 3 divides 9-0.

1=7(mod 6) since 6 divides 1-7.

a=b(mod p) if p divides a-b

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Theorem. p-colorations are preserved by Reidemeister moves.

Theorem. p-colorations are preserved by Reidemeister moves.

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Theorem. p-colorations are preserved by Reidemeister moves.

Corollary. If one projection of a knot is p-colorable for some p, then every projection of that knot is p-colorable.

Theorem. p-colorations are preserved by Reidemeister moves.

Corollary. If one projection of a knot is p-colorable for some p, then every projection of that knot is p-colorable.

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Theorem. p-colorations are preserved by Reidemeister moves.

Corollary. If one projection of a knot is p-colorable for some p, then every projection of that knot is p-colorable.

Corollary. If K is p-colorable for some p ≥ 3, then K is nontrivial.

Theorem. p-colorations are preserved by Reidemeister moves.

Corollary. If one projection of a knot is p-colorable for some p, then every projection of that knot is p-colorable.

Corollary. If K is p-colorable for some p ≥ 3, then K is nontrivial.

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So it is nontrivial!So it is nontrivial!

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A 7-coloringA 7-coloring

So this is nontrivial!So this is nontrivial!

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And thanks to Danny DeVito and especiallyAngelina Jolie.

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And thanks to Danny DeVito and especiallyAngelina Jolie.

Who could ever forget the champagne, the walk on the beach and the moonlit swim.

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And thanks to Danny DeVito and especiallyAngelina Jolie.

Who could ever forget the champagne, the walk on the beach and the moonlit swim.

Just the three of us.

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And thanks to Danny DeVito and especiallyAngelina Jolie.

Who could ever forget the champagne, the walk on the beach and the moonlit swim.

Just the three of us.