2.2. Normal Forms, Cyclings and Decyclings
It is well-known that Garside groups have solvable word problem, as one can compute a normal form for each element.
Let us first define the right complement
of a simple element
. That is,
is the only element
. Let us see that
is also a simple element. Recall that the simple elements are the positive prefixes of
(by definition of Garside group), we have that
is positive. Now
, which implies that t
is a positive prefix of
, that is,
. It follows that we have a map
. Notice that, by definition,
Given two simple elements , we say that the decomposition is left weighted if s is the biggest possible simple element (with respect to ≼) in any decomposition of the element as a product of two simple elements. This condition can be restated as , i.e., and t have no non-trivial prefixes in common.
]). The left normal form of an element is the unique decomposition so that , , for , and is a left weighted decomposition, for .
Given such a decomposition, we define the infimum, supremum
and canonical length
, respectively. Equivalently, the infimum and supremum of x
can be defined as the maximum and minimum integers p
It is important to notice that conjugation by preserves the Garside structure of . Hence, if the left normal form of a braid x is , then the left normal form of is . We will make use of this property later.
Garside groups also have solvable conjugacy problem. One of the main tools to solve problems related to conjugacy in braid groups are the summit sets, which are subsets of the conjugacy class of a braid. Throughout this article we are going to use two of them: the super summit set
] and the ultra summit set
]. Let us first introduce some concepts:
Definition 2. Letbe in left normal form, with
. Notice that we can write:
We define the initial factor of x as
, and the final factor of x as
. We can then write:
, we setand
Notice that, as is the identity, we actually have either if p is even, or if p is odd. This happens in braid groups, but not in other Garside groups in which the order of is bigger.
]). Let be in left normal form, with
. The cycling and decycling of x are the conjugates of x defined, respectively, as
Thus is the conjugate of x by , and that is the conjugate of x by .
Cyclings and decyclings were defined in [12
] in order to try to simplify the braid x
by conjugations. Usually, if
, the decomposition
the left normal form of
could a priori have a shorter normal form (with less factors). A similar situation happens for
If is actually the left normal form of (when ), we say that the braid x is rigid. This happens if and only if (that is, ) is a left weighted decomposition. We can extend this definition to every case, when :
We say that is rigid if is a left weighted decomposition.
is rigid, neither cycling nor decycling can simplify its normal form
. Actually, the normal forms of the iterated cyclings of x
are, if p
in this case. In the case when p
is odd we have:
in this case.
In the same way, if x
is rigid we have, for p
in this case. If p
is odd we get:
in this case. We then see that, if x
is rigid, iterated cyclings and decyclings correspond to cyclic permutations of the factors in the normal form of x
(possibly conjugated by
, if p
is odd); moreover, when applied to rigid braids,
are inverses of each other.
2.3. Summit Sets
be an arbitrary braid (not necessarily rigid). Consider the conjugacy class of x
, and write
) for the maximal infimum (resp. the minimal supremum) of an element in
. These numbers are known to exist [12
], and are called the summit infimum
and the summit supremum
, respectively. Set
, the summit length
. It is shown in [12
] that the elements in
having the shortest possible normal form are those whose canonical length is precisely
, and they coincide with the elements whose infimum and supremum are equal to
, respectively. The set formed by these elements is called the supper summit set
of the braid x
Starting from x
, it is possible to obtain an element in
by applying cyclings and decyclings iteratively. It is known [12
] that if
then the infimum of x
can be increased by iterated cycling. Actually, in this case
]). Hence, every
iterations either the infimum has increased, or one is sure to have an element whose infimum is the summit infimum.
In the same way, if
, then the supremum of x
can be decreased by iterated decycling [12
], and in that case
]. Hence, every
iterations either the supremum has decreased, or we are sure to have an element whose supremum is the summit supremum. Since decycling can never decrease the infimum of an element, it follows that starting with any
and applying iterated cycling (until summit infimum is obtained) followed by iterated decycling (until summit supremum is obtained) yields an element
The super summit set
is a finite set, but it is usually huge, so smaller subsets of the conjugacy class of x
were defined in order to solve the conjugacy problem of x
more efficiently. Namely, the ultra summit set
, denoted by
, is a subset of
defined as follows [4
Since is finite, the subset is also finite. It is then clear that one obtains an element is by iterated application of cycling, starting from an element in , when a repeated element is obtained. Actually, the whole orbit under cycling of an element in belongs to . So is a finite set of orbits under cycling.
Notice that every rigid braid belongs to its ultra summit set, as cylings and decyclings are basically cyclic permutations of its factors. It is shown in [15
] that, if x
is conjugate to a rigid braid and
coincides with the set of rigid conjugates of x
There is actually a simpler way, in the general case, to obtain an element in starting from x. Instead of using cyclings and decyclings, one can use the following single type of conjugation:
]). Given , the cyclic sliding of x is defined as , where .
]). Given , there are integers such that . For every such pair of integers, one has .
By the above result, one can obtain an element in
by iterated cyclic sliding starting form x
. Furthermore, if x
is conjugate to a rigid element (this will be the generic situation, as we will see in Section 2.4
), iterated cyclic sliding yields the shortest
positive conjugating element from x
to a rigid element.
]). Let and suppose that x is conjugate to a rigid braid. Then there is an integer such that is rigid. Moreover, the conjugating element α from x to
, that is
is the smallest positive element (with respect to ≼) conjugating x to a rigid element, meaning that for every positive element β such that is rigid, one has
After obtaining one element in , it is possible to compute all elements in together with conjugating elements connecting them. In this way, one solves the conjugacy problem in , as two elements x and y are conjugate if and only if or, equivalently, if . Then, in order to check whether x and y are conjugate, one can compute the whole set , and one element . Then, x and y are conjugate if and only if . By construction, one can even compute a conjugating element from x to y.
In order to understand the forthcoming proofs in this paper, we will need to describe some conjugating elements connecting the elements of .
]). Let and . A simple non-trivial element is said to be a
minimal simple elementfor y if and , for every .
], Gebhardt showed that for any two elements
there exists a sequence
is a minimal simple element for
. Moreover, he introduced an algorithm to compute all minimal simple elements for a given
. This allows to construct a directed graph
, whose vertices correspond to elements of
, and whose arrows correspond to minimal simple elements, in such a way that for every minimal simple element s
, there is an edge with label s
. By the above discussion, it follows that
is a connected graph, and this is why
can be computed starting with a single vertex, iteratively computing the minimal simple elements corresponding to each known vertex, until all vertices are obtained.
We will later see that, generically, ultra summit sets are really small. Actually, they usually have a very simple structure, that we explain now.
]). Let with and let s be a minimal simple element for y. Then, s is a prefix of either or , or both.
The above lemma allows us to classify the arrows in into two groups: a directed edge labelled by s starting at is black (resp. grey), if s is a prefix of (resp. of ). In principle, an edge could be of both colors at the same time (a bi-colored arrow, whose label is a prefix of both and ), but not in the case of rigid braids, as if x is rigid. Actually, this is a necessary and sufficient condition:
]). A braid with is rigid if and only if none of the edges starting at y is bi-colored.
Given a braid , its associated isminimalif and, for every vertex y in the graph , there are exactly two directed edges starting at y, a black one labeled and a grey one labeled .
Notice that, as a consequence of Lemma 2, if is minimal then all elements in are rigid. Moreover, the arrow labeled corresponds to a cycling of y, and the arrow labeled corresponds to a twisted decycling of y, meaning a decycling followed by the automorphism . This implies that, if is minimal, the elements of are obtained from y by applying and in every possible way. Since y is rigid, cyclings and decyclings basically correspond to cyclic permutations of the factors. Therefore, if is minimal, it consists of either two orbits under cycling (conjugate to each other by ), or one orbit under cycling (conjugate to itself by ). If the infimum of y is even, the orbit of y has at most elements, so the size of is at most . If the infimum of y is odd, the orbit of y has at most elements, and it is conjugate to itself by , so it is the only orbit. Therefore, in any case, if is minimal it has at most elements.
Remark 1. In order to see whether is minimal, one should a priori check the condition in Definition 7 for every element in . But it is actually shown in (, Theorem 4.6) that, given , the set is minimal if and only if and the minimal simple elements for y are precisely and . Hence, one just needs to compute the minimal elements for a single arbitrary element .
Let us see that this case, in which is so small and has such a simple structure, is generic.
2.4. Generic Braids
Since is an infinite set, it is necessary to explain what we mean by `picking a random braid’ or by saying that a braid is `generic’. Even if we fix the subset of braids of a given length, we must specify if we choose braids from the subset with a uniform distribution, or if we pick braids by choosing a random walk in the Cayley graph, which are the two usual situations.
We will consider the Cayley graph of the braid group
, taking as generators the simple braids, and assume that each edge of the Cayley graph has length 1, so it becomes a metric space. Let us point out that left normal forms of braids are closely related to geodesics in this Cayley graph [18
Now let denote the ball of radius r centered at the trivial braid 1. As the number of simple braids is finite, the set is a finite subset of . We will consider the uniform distribution within this set. It turns out that `most’ elements in have a very simple ultra summit set:
]). The proportion of braids in whose ultra summit set is minimal tends to 1 exponentially fast, as r tends to infinity.
This is why we can say that the ultra summit set of a `generic braid’ is minimal. Moreover, the above result was obtained by refining the following theorem, which gives some important information concerning the elements in . We have simplified the statement to adapt it to our situation:
]). The proportion of braids x in which are conjugate to a rigid braid , in such a way that α is a positive braid with , tends to 1 exponentially fast, as r tends to infinity.
Therefore, not only generic braids have minimal ultra summit sets (made of rigid braids), but one can also obtain a rigid conjugate of a generic braid x very fast, applying iterated cyclic sliding to x. By Theorem 2, the obtained conjugating element will be the smallest possible positive conjugator, so its canonical length will be smaller than . Once that a rigid conjugate y (which belongs to ) is obtained, one can compute the whole very fast, as it consists of at most elements, connected by cyclings and twisted decyclings. This is why solving the conjugacy problem in braid groups is generically very fast.
We will also be interested in the centralizer of a braid x. Notice that if , then . Therefore, knowing is equivalent to knowing , via . We will then be interested in for .
Definition 8. Letand
, and let t be the smallest positive integer such that
. Denotethe positive element conjugating to
. Then the preferred cycling conjugator of y is defined as
In other words
, corresponds to the conjugating element along the whole cycling orbit of y. By construction
, commutes with y
In the generic case (when is minimal), it turns out that is isomorphic to , and one can describe the generators of for any (and thus of ) in a very explicit way:
]). Let and
. Let as above. If is minimal, then all elements in are rigid,
, and one of the following conditions holds:
has two orbits under cycling, conjugate to each other by Δ, and .
has one orbit under cycling, conjugate to itself by Δ, and:
If, then t is even and.