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:

etc.
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.

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