6.6 Theorem  Any 4-plat has a diagram as above with all ci's zero.

Proof   The idea is to turn c's into a's by flyping. A flype moves a crossing across a part of a diagram at the expense of turning the part over:

Notice that if we introduce two crossings using R2 then move one over with a flype we get:
Suppose c > 0 and consider the sequence of flypes:

Repeat c times to get:

Similarly if  c < 0 If  b is odd K will have been rotated through pi about an horizontal axis. Use this procedure to eliminate cn , cn-1 , ... ,c2 in turn. Then finally:

Notice that a double flype leaves R undisturbed:

We can be more efficient in unwrapping a daisy chain. For example:

After n - 2 double flypes we have:

We now use flips to eliminate the unshaded crossings. Here is a flip:

Think of this as turning R over to change the sign of a crossing between middle and outer string at the expense of adding a crossing between middle and other outer string.

Combine a flip with R2 to get 3 new crossings all of the same sign as follows:

Apply  R2 flips:

If n odd get:

which is isotopic to

and if n even get:

which is isotopic to:

Remark  It follows that the classes of knots defined by Daisy chains, 4-plats and 2-bridge knots are identical.

See the first fourteen prime knots added together. Notice that they are all 2-bridge knots.

Previous lecture   Next lecture     An application of braids to Public Key Cryptosystems      An application of braids to Quantum computers.   Another reference.     See also the April 2006 edition of Scientific American, Computing with Quantum Knots            An old version of lecture 14 with proof of 6.6 in a special case