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We saw in the first chapter a way to describe the braid group:
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we specified some elementary braids, the generators
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and then we found the relations,
moves that can transform a word without changing the braid it describes
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But we still don't know all about this group!
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For example, the relations don't say how
to check if two words are equivalent
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Here we have two words and we construct
a sequence of relations that connect them
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At each stage we just replace
the part of the word inside the yellow box with an equivalent one
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We use the two kinds of relations
and the insertion or deletion of a generator and its inverse
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In each level we change the word,
but the braid it represents remains the same
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It is not so easy to note immediately, just looking at the two words,
that they are equivalent
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This becomes evident when we construct the sequence
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What about these two words?
We try to construct a sequence as before to connect them
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It can be a hard job to construct such a sequence of words using the relations
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We could spend our whole life looking for one, without ever finding it
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we will never know whether no sequence exists
or we haven't been lucky enough to find it
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That's why we need another way to handle the question
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More precisely, we wish to construct a machinery
that takes two arbitrary words as input,
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makes some operations and computations
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and gives as output the answer "yes"
if the two words are equivalent
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Following the same procedure, the answer will be "no"
if the two words represent different braids
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Such a procedure is called an algorithm
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We are lucky: for braids we have such an algorithm
Let's look inside it
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First of all we simplify the problem:
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two words represent the same braid if and only if one,
composed with the inverse of the other, is equivalent to the trivial braid
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Therefore, we can reformulate our problem:
given an arbitrary word, decide if it is equivalent to the identity
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Or in other words: is there a way to transform the given word into "1"?
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Look at this braid: the color of the bubbles is enough to know where each strand ends
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If the bubbles at the same level, on the left and on the right,
have the same colors, then the braid is called pure
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Of course two equivalent braids will have
the bubbles with the same color arrangement
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So a braid that is equivalent to the trivial one has to be pure
That's why we consider only pure braids in this chapter
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Each pure braid can be decomposed into blocks in this way:
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in the first block, all the strands but the last are straight.
The green strand can link to the others
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In the second block, all the strands but the second-to-last are straight.
The yellow strand can link only to the strands below
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And so on
until the last block, in which the second strand can link to the first one
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We know that for each braid there are lots of representatives
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We would like to choose a specific one of them to describe the braid
It will be called the normal form
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Now, we need an algorithm that puts any arbitrary word in its normal form
Artin's braid combing is such an algorithm
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Here we show it on a non trivial braid:
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first copy the braid
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and delete the last strand, replacing it with a trivial one
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Compose this braid with its inverse, getting the identity braid
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So, composing the original braid with it, we don't change the braid
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On the left side we have a braid that can be put into the desired form,
with all the strands straight except the top one
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This is the first block
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We deform the strands so that the green one encircles a strand
passing behind the others,
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goes back to the initial position, encircles another strand, and so on
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This piece is combed
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Do the same procedure on the right side:
copy the braid,
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replace the yellow strand with a trivial one,
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compose this braid with its inverse
getting a trivial braid
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compose this trivial braid with the original one
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and tidy up the second block
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We put it in a form such that
the yellow strand links only with the blue one and the red one
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We have combed the second block
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Now the last block is easy to rearrange
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We have put the braid in its normal form
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Artin proved that a braid is trivial if and only if,
once it is combed, every block is trivial,
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meaning that it can be represented with no crossings at all
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So our braid is not trivial
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And we have an algorithm to check it:
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We start with a pure braid
Comb it into blocks
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If one of the blocks is not trivial, terminate, answering "no"
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Otherwise continue with the next block
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If all the blocks are trivial, we get the trivial braid and the answer is
"yes, the input word is equivalent to the identity"
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This algorithm works but Artin himself was not satisfied with it
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In his paper "Theory of braids" he writes:
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Why did he write so?
Combing is an extremely slow method
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The complexity of the algorithm is very high:
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if we draw a graph showing the mean time needed to comb a braid
as a function of the number of crossings, the curve is very steep
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Increasing the input size by a small amount
the time grows very fast and may be too long even for a computer
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This is why mathematicians look for other, quick, algorithms
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The fastest found so far is probably the one proposed
about 15 years ago by Dehornoy, a French mathematician
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It is based on manipulating braids and is called handle reduction
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This is a handle: he red strand passes in front of the other strand just above
and then in front of the strand just below
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A handle can also pass twice behind the other strands:
this is another handle
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This is not a handle because it faces downwards
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Neither this is a handle:
the blue strand passes once in front of the strand below and once behind
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Reducing handles means to move these pieces of string
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In this way we change the word
but the braid it represents remains the same
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A theorem of Dehornoy ensures that a word with no handles is
either empty or it represents a nontrivial braid
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Let's sketch the handle reduction algorithm:
take a pure braid
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If there are no handles and the braid has at least one crossing,
terminate, answering "no"
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Otherwise, if there are handles,
find the one that ends first and reduce it
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Continue like this until there are no more handles:
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if the braid is not trivial, the answer is "no"
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If after reducing all the handles, the braid is trivial,
then the answer will be "yes"
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But there is a problem: what if the algorithm does not end?
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It could maybe go into a loop:
if it comes back to a word that it has already met,
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the steps after that one will always be the same
and the algorithm will never stop
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In fact this is never the case:
it has been shown that the algorithm always terminates
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We don't yet know the complexity of this algorithm
but we have some experimental estimates
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and it seems much faster than the other known algorithms
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Combing, handles:
the formalization of braids in terms of words is really powerful!