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braid closures

asked 2019-01-11 06:45:51 -0500

danieleC gravatar image

Hi! I've noticed that when taking the closure of a braid on $n$ strands, the resulting link is obtained by ignoring any "unused" strand:

B = BraidGroup(3)
print B([1]).components_in_closure(), Link(B([1])).number_of_components()

outputs 2 1, instead of 2 2. In the documentation for the braid group I've read "The behavior of removing unused strands from an element of a braid group may change without notice in the future. Do not rely on this feature."

Is there a workaround that allows to have the same number of components in a braid and in its closure? Note that braids given as an array of generators (as in the example above) are automatically simplified, so closing the braid [2,-2] (which should give the $2$-component unlink) produces the unknot. (In particular, the empty braid on $n$ strands should close to the $n$-component unlink.)

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answered 2019-01-12 06:20:31 -0500

mmarco gravatar image

You can conjugate the braid so there are no unused strands anynore. In order to prevent the simplification, you can use two expressions of the same braid.

In your example:

cb1 = B([1,2,1]) cb2 = B([-2,-1,-2])

mybraid = B([1])

myconjbraid = cb1mybraidcb2

myconjbraid produces the same link as mybraid, but since it has no unused strands, it will contain also the trivial components.

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Thanks, that works perfectly!

danieleC gravatar imagedanieleC ( 2019-01-12 08:47:54 -0500 )edit

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Asked: 2019-01-11 06:45:51 -0500

Seen: 111 times

Last updated: Jan 12