Alternating normal forms of braids
Alternating normal forms of braids
Patrick Dehornoy
Laboratoire de Mathématiques Nicolas Oresme
Université de Caen
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
but also (presumably): random walks
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
but also (presumably): random walks
• Plan:
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
but also (presumably): random walks
• Plan:
- 1. A normal form for positive braids
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
but also (presumably): random walks
• Plan:
- 1. A normal form for positive braids
- 2. Connection with the braid order
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
but also (presumably): random walks
• Plan:
- 1. A normal form for positive braids
- 2. Connection with the braid order
- 3. Application: an unprovability statement involving braids
• The Garside normal form of braids is excellent,
but there exist other normal forms with equally good properties
that can be more suitable for specific questions.
↑
here: order properties
but also (presumably): random walks
• Plan:
- 1. A normal form for positive braids
- 2. Connection with the braid order
- 3. Application: an unprovability statement involving braids
(Surprising that so simple things were not considered earlier)
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
+
• The monoid Bn is a Garside monoid with Garside element ∆n .
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
+
• The monoid Bn is a Garside monoid with Garside element ∆n .
+
Every x in Bn
has a unique expression sp ...s1
with si maximal right divisor of sp ...si lying in Div(∆n ).
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
+
• The monoid Bn is a Garside monoid with Garside element ∆n .
+
Every x in Bn
has a unique expression sp ...s1
with si maximal right divisor of sp ...si lying in Div(∆n ).
−1
Every x in Bn has a unique expression sp ...s1 t−1
1 ...tq
with s1 , ..., sp and t1 , ..., tq as above and gcd(s1 , t1 ) = 1.
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
+
• The monoid Bn is a Garside monoid with Garside element ∆n .
+
Every x in Bn
has a unique expression sp ...s1
with si maximal right divisor of sp ...si lying in Div(∆n ).
−1
Every x in Bn has a unique expression sp ...s1 t−1
1 ...tq
with s1 , ..., sp and t1 , ..., tq as above and gcd(s1 , t1 ) = 1.
• Theorem: This normal form gives a bi-automatic structure on Bn .
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
+
• The monoid Bn is a Garside monoid with Garside element ∆n .
+
Every x in Bn
has a unique expression sp ...s1
with si maximal right divisor of sp ...si lying in Div(∆n ).
−1
Every x in Bn has a unique expression sp ...s1 t−1
1 ...tq
with s1 , ..., sp and t1 , ..., tq as above and gcd(s1 , t1 ) = 1.
• Theorem: This normal form gives a bi-automatic structure on Bn .
Proof: Defined by a local condition + computable by a grid.
The greedy normal form
+
• The group Bn is the group of fractions of the monoid Bn
.
+
• The monoid Bn is a Garside monoid with Garside element ∆n .
+
Every x in Bn
has a unique expression sp ...s1
with si maximal right divisor of sp ...si lying in Div(∆n ).
−1
Every x in Bn has a unique expression sp ...s1 t−1
1 ...tq
with s1 , ..., sp and t1 , ..., tq as above and gcd(s1 , t1 ) = 1.
• Theorem: This normal form gives a bi-automatic structure on Bn .
Proof: Defined by a local condition + computable by a grid.
the Garside theory of braids
Other normal forms
+
• There are other normal forms on Bn or Bn
that are not—or not
directly—connected with the Garside structure.
Other normal forms
+
• There are other normal forms on Bn or Bn
that are not—or not
directly—connected with the Garside structure.
• Type 1: Normal forms coming from combing
(Artin, Markov–Ivanovsky).
Other normal forms
+
• There are other normal forms on Bn or Bn
that are not—or not
directly—connected with the Garside structure.
• Type 1: Normal forms coming from combing
(Artin, Markov–Ivanovsky).
• Type 2: Normal forms coming from relaxation strategies
(Dynnikov–Wiest, Bressaud).
Other normal forms
+
• There are other normal forms on Bn or Bn
that are not—or not
directly—connected with the Garside structure.
• Type 1: Normal forms coming from combing
(Artin, Markov–Ivanovsky).
• Type 2: Normal forms coming from relaxation strategies
(Dynnikov–Wiest, Bressaud).
• Type 3 Normal forms coming from an order on braid words:
NF(x) defined to be the least word representing x;
Other normal forms
+
• There are other normal forms on Bn or Bn
that are not—or not
directly—connected with the Garside structure.
• Type 1: Normal forms coming from combing
(Artin, Markov–Ivanovsky).
• Type 2: Normal forms coming from relaxation strategies
(Dynnikov–Wiest, Bressaud).
• Type 3 Normal forms coming from an order on braid words:
NF(x) defined to be the least word representing x;
Example 1 (Bronfman): lexicographical order of braid words;
Other normal forms
+
• There are other normal forms on Bn or Bn
that are not—or not
directly—connected with the Garside structure.
• Type 1: Normal forms coming from combing
(Artin, Markov–Ivanovsky).
• Type 2: Normal forms coming from relaxation strategies
(Dynnikov–Wiest, Bressaud).
• Type 3 Normal forms coming from an order on braid words:
NF(x) defined to be the least word representing x;
Example 1 (Bronfman): lexicographical order of braid words;
Example 2 (Burckel): associate with every braid word w
a certain finite tree T (w), and use a well-ordering on trees.
(exists because of well-order but not easily computed for n > 4)
The S-tail of a braid
+
• Lemma: Let S be a subset of Bn
, closed under left lcm and right
+
divisor. Then each x in Bn admits a unique decomposition
x = x0 x1
s.t. x1 is the maximal right divisor of x lying in S.
The S-tail of a braid
+
• Lemma: Let S be a subset of Bn
, closed under left lcm and right
+
divisor. Then each x in Bn admits a unique decomposition
x = x0 x1
s.t. x1 is the maximal right divisor of x lying in S.
+
If S generates Bn
, iterating gives a unique normal form, e.g.,
S = Div(∆n ) gives the standard (right) greedy NF.
The S-tail of a braid
+
• Lemma: Let S be a subset of Bn
, closed under left lcm and right
+
divisor. Then each x in Bn admits a unique decomposition
x = x0 x1
s.t. x1 is the maximal right divisor of x lying in S.
+
If S generates Bn
, iterating gives a unique normal form, e.g.,
S = Div(∆n ) gives the standard (right) greedy NF.
+
• Lemma (variant): Let S be a submonoid of Bn
, closed under
+
left lcm and right divisor. Then each x in Bn admits a unique
decomposition
x = x0 x1
s.t. the only right divisor of x0 lying in S is 1: “ x0 ⊥ S ”.
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
1
x
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
+
Use two submonoids S2 , S1 that, together, generate Bn
,
+
+
+
typically: S1 = Bn−1 = hσ1 , ..., σn−2 i , S2 = hσ2 , ..., σn−1 i .
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
+
Use two submonoids S2 , S1 that, together, generate Bn
,
+
+
+
typically: S1 = Bn−1 = hσ1 , ..., σn−2 i , S2 = hσ2 , ..., σn−1 i .
n
..
.
2
1
x
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
+
Use two submonoids S2 , S1 that, together, generate Bn
,
+
+
+
typically: S1 = Bn−1 = hσ1 , ..., σn−2 i , S2 = hσ2 , ..., σn−1 i .
n
..
.
2
1
x
x1
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
+
Use two submonoids S2 , S1 that, together, generate Bn
,
+
+
+
typically: S1 = Bn−1 = hσ1 , ..., σn−2 i , S2 = hσ2 , ..., σn−1 i .
n
..
.
2
1
x
x2
x1
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
2
x0
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
+
Use two submonoids S2 , S1 that, together, generate Bn
,
+
+
+
typically: S1 = Bn−1 = hσ1 , ..., σn−2 i , S2 = hσ2 , ..., σn−1 i .
n
..
.
2
1
x3
x
x2
x1
Using several submonoids
+
• Example: S = Bn−1
:
n
..
.
x0
2
x
x1
1
| {z }
+
maximal right divisor of x in Bn−1
... stuck after one step.
+
Use two submonoids S2 , S1 that, together, generate Bn
,
+
+
+
typically: S1 = Bn−1 = hσ1 , ..., σn−2 i , S2 = hσ2 , ..., σn−1 i .
n
..
.
2
1
x4
x3
x
x2
x1
The flip splitting of a braid
unique normal form in B3+ : x = ...σ2e4 σ1e3 σ2e2 σ1e1
with maximal e1 , then maximal e2 , etc.
The flip splitting of a braid
unique normal form in B3+ : x = ...σ2e4 σ1e3 σ2e2 σ1e1
with maximal e1 , then maximal e2 , etc.
+
+
+
+
• More generally, using Bn
= hφn Bn−1
, Bn−1
i ,
The flip splitting of a braid
unique normal form in B3+ : x = ...σ2e4 σ1e3 σ2e2 σ1e1
with maximal e1 , then maximal e2 , etc.
+
+
+
+
• More generally, using Bn
= hφn Bn−1
, Bn−1
i ,
the flip automorphism σi 7→ σn−i
The flip splitting of a braid
unique normal form in B3+ : x = ...σ2e4 σ1e3 σ2e2 σ1e1
with maximal e1 , then maximal e2 , etc.
+
+
+
+
• More generally, using Bn
= hφn Bn−1
, Bn−1
i ,
the flip automorphism σi 7→ σn−i
+
• Proposition: For each x in Bn
, there is a unique sequence
+
(xp , ..., x1 ) in Bn−1 s.t.
p−1
x = φn
xp · .... · φn x4 · x3 · φn x2 · x1
p−i
and the only σj right dividing φn
xp · ... · φn xi+1 · xi is σ1 .
The flip splitting of a braid
unique normal form in B3+ : x = ...σ2e4 σ1e3 σ2e2 σ1e1
with maximal e1 , then maximal e2 , etc.
+
+
+
+
• More generally, using Bn
= hφn Bn−1
, Bn−1
i ,
the flip automorphism σi 7→ σn−i
+
• Proposition: For each x in Bn
, there is a unique sequence
+
(xp , ..., x1 ) in Bn−1 s.t.
p−1
x = φn
xp · .... · φn x4 · x3 · φn x2 · x1
p−i
and the only σj right dividing φn
xp · ... · φn xi+1 · xi is σ1 .
• Examples:
∆23 = φ3 (σ1 ) · σ12 · φ3 (σ1 ) · σ12 ,
The flip splitting of a braid
unique normal form in B3+ : x = ...σ2e4 σ1e3 σ2e2 σ1e1
with maximal e1 , then maximal e2 , etc.
+
+
+
+
• More generally, using Bn
= hφn Bn−1
, Bn−1
i ,
the flip automorphism σi 7→ σn−i
+
• Proposition: For each x in Bn
, there is a unique sequence
+
(xp , ..., x1 ) in Bn−1 s.t.
p−1
x = φn
xp · .... · φn x4 · x3 · φn x2 · x1
p−i
and the only σj right dividing φn
xp · ... · φn xi+1 · xi is σ1 .
• Examples:
∆23 = φ3 (σ1 ) · σ12 · φ3 (σ1 ) · σ12 ,
∆24 = φ4 (σ1 ) · σ2 σ12 · φ4 (σ2 σ1 ) · ∆23 .
The flip normal form
• Iterate to obtain a unique normal form
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
σ1
σ2 σ12
σ2 σ1
∆23
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
σ1
σ2 σ12
φ4
σ2 σ1
∆23
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
σ1
σ2 σ12
φ4
∆23
σ2 σ1
φ3
σ1
φ3
σ12
σ1
σ12
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
σ1
φ4
σ2 σ12
∆23
σ2 σ1
φ3
σ1
φ3
σ1
σ1
φ3
σ12
σ1
σ12
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
φ4
σ2 σ12
σ1
φ3
σ1
φ3
σ12
∆23
σ2 σ1
σ1
φ3
σ1
σ1
φ3
σ12
σ1
σ12
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
φ4
σ2 σ12
σ1
φ3
σ1
σ1
φ3
σ12
∆23
σ2 σ1
σ1
φ3
σ1
σ1
φ3
σ12
σ1
σ12
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
φ4
σ2 σ12
σ1
φ3
∆23
σ2 σ1
φ3
φ3
φ3
σ1
σ1
σ12
σ1
σ1
σ1
σ12
σ1
σ12
after flipping: σ3
σ2
σ12
σ2
σ3
σ2
σ12
σ2
σ12
The flip normal form
• Iterate to obtain a unique normal form
= construct a tree for each braid
∆24
φ4
φ4
σ2 σ12
σ1
φ3
∆23
σ2 σ1
φ3
φ3
φ3
σ1
σ1
σ12
σ1
σ1
σ1
σ12
σ1
σ12
after flipping: σ3
σ2
σ12
σ2
σ3
σ2
σ12
σ2
σ12
The flip normal form of ∆24 is σ3 σ2 σ12 σ2 σ3 σ2 σ12 σ2 σ12 .
The flip normal form
• Proposition:
The flip normal form
+
(hence in Bn ) admits a unique
• Proposition: - Every braid in Bn
normal form, which can be computed in quadratic time;
The flip normal form
+
(hence in Bn ) admits a unique
• Proposition: - Every braid in Bn
normal form, which can be computed in quadratic time;
- Normal words are recognized by a finite state automaton.
The flip normal form
+
(hence in Bn ) admits a unique
• Proposition: - Every braid in Bn
normal form, which can be computed in quadratic time;
- Normal words are recognized by a finite state automaton.
• Step by step computation: choose the right divisor σi that is
minimal w.r.t. some order that is updated at each step, according
to a position in a reference binary tree, e.g., for B4 :
σ2
σ3
σ2
σ1
The flip normal form
+
(hence in Bn ) admits a unique
• Proposition: - Every braid in Bn
normal form, which can be computed in quadratic time;
- Normal words are recognized by a finite state automaton.
• Step by step computation: choose the right divisor σi that is
minimal w.r.t. some order that is updated at each step, according
to a position in a reference binary tree, e.g., for B4 :
σ2
σ3
σ2
σ1
• The flip normal form is not connected with an automatic structure.
The flip normal form
+
(hence in Bn ) admits a unique
• Proposition: - Every braid in Bn
normal form, which can be computed in quadratic time;
- Normal words are recognized by a finite state automaton.
• Step by step computation: choose the right divisor σi that is
minimal w.r.t. some order that is updated at each step, according
to a position in a reference binary tree, e.g., for B4 :
σ2
σ3
σ2
σ1
• The flip normal form is not connected with an automatic structure.
• Fact: Works in every monoid that is locally Garside on the right,
The flip normal form
+
(hence in Bn ) admits a unique
• Proposition: - Every braid in Bn
normal form, which can be computed in quadratic time;
- Normal words are recognized by a finite state automaton.
• Step by step computation: choose the right divisor σi that is
minimal w.r.t. some order that is updated at each step, according
to a position in a reference binary tree, e.g., for B4 :
σ2
σ3
σ2
σ1
• The flip normal form is not connected with an automatic structure.
• Fact: Works in every monoid that is locally Garside on the right,
hence in particular in every Artin–Tits monoid.
The standard braid order
• Definition: For x, y in B∞ , say that x < y holds if, among all
words representing x−1 y, at least one is such that the generator σi
with highest index appears positively only (σi occurs, σi−1 does not).
The standard braid order
• Definition: For x, y in B∞ , say that x < y holds if, among all
words representing x−1 y, at least one is such that the generator σi
with highest index appears positively only (σi occurs, σi−1 does not).
e.g., σ2 < σ2 σ1 holds, because σ2−1 σ1 σ2
The standard braid order
• Definition: For x, y in B∞ , say that x < y holds if, among all
words representing x−1 y, at least one is such that the generator σi
with highest index appears positively only (σi occurs, σi−1 does not).
e.g., σ2 < σ2 σ1 holds, because σ2−1 σ1 σ2 = σ1 σ2 σ1−1 ,
and, in the latter word, σ2 appears positively only.
The standard braid order
• Definition: For x, y in B∞ , say that x < y holds if, among all
words representing x−1 y, at least one is such that the generator σi
with highest index appears positively only (σi occurs, σi−1 does not).
e.g., σ2 < σ2 σ1 holds, because σ2−1 σ1 σ2 = σ1 σ2 σ1−1 ,
and, in the latter word, σ2 appears positively only.
• Theorem: (i) The relation < is a left-invariant total order on B∞ ,
and Bn is the interval (σn−1 , σn ) of (B∞ , <).
The standard braid order
• Definition: For x, y in B∞ , say that x < y holds if, among all
words representing x−1 y, at least one is such that the generator σi
with highest index appears positively only (σi occurs, σi−1 does not).
e.g., σ2 < σ2 σ1 holds, because σ2−1 σ1 σ2 = σ1 σ2 σ1−1 ,
and, in the latter word, σ2 appears positively only.
• Theorem: (i) The relation < is a left-invariant total order on B∞ ,
and Bn is the interval (σn−1 , σn ) of (B∞ , <).
+
(ii) (Laver) The restriction of < to B∞
is a well-order;
The standard braid order
• Definition: For x, y in B∞ , say that x < y holds if, among all
words representing x−1 y, at least one is such that the generator σi
with highest index appears positively only (σi occurs, σi−1 does not).
e.g., σ2 < σ2 σ1 holds, because σ2−1 σ1 σ2 = σ1 σ2 σ1−1 ,
and, in the latter word, σ2 appears positively only.
• Theorem: (i) The relation < is a left-invariant total order on B∞ ,
and Bn is the interval (σn−1 , σn ) of (B∞ , <).
+
(ii) (Laver) The restriction of < to B∞
is a well-order;
+
(iii) (Burckel) The restriction of < to Bn
—which is the initial inn−2
+
ω
terval [1, σn ) of (B∞ , <)—has length ω
.
Connection order vs. greedy normal form
• The well-order property gives a distinguished element
+
in every nonempty subset of Bn
(e.g., in each conjugacy class)
Connection order vs. greedy normal form
• The well-order property gives a distinguished element
+
in every nonempty subset of Bn
(e.g., in each conjugacy class)
but cannot be computed in practice (?).
Connection order vs. greedy normal form
• The well-order property gives a distinguished element
+
in every nonempty subset of Bn
(e.g., in each conjugacy class)
but cannot be computed in practice (?).
• Typically: < is not well connected with the greedy normal form:
If x, y are divisors of ∆n , then x < y iff perm(x) <Lex perm(y),
(good!)
Connection order vs. greedy normal form
• The well-order property gives a distinguished element
+
in every nonempty subset of Bn
(e.g., in each conjugacy class)
but cannot be computed in practice (?).
• Typically: < is not well connected with the greedy normal form:
If x, y are divisors of ∆n , then x < y iff perm(x) <Lex perm(y),
(good!)
... but does not extend to arbitrary positive braids,
viewed as sequences of divisors of ∆n .
(bad!)
Connection order vs. flip normal form
+
• Theorem: The order < on Bn
is a twisted ShortLex-extension
+
of < on Bn−1 .
Connection order vs. flip normal form
+
• Theorem: The order < on Bn
is a twisted ShortLex-extension
+
+
of < on Bn−1 . Let x, y in Bn admit the flip splittings
q−1
x = φpn−1 xp · ... · x3 · φn x2 · x1 and y = φn
yq · ... · y3 · φn y2 · y1 .
Then x < y holds iff either p < q,
Connection order vs. flip normal form
+
• Theorem: The order < on Bn
is a twisted ShortLex-extension
+
+
of < on Bn−1 . Let x, y in Bn admit the flip splittings
q−1
x = φpn−1 xp · ... · x3 · φn x2 · x1 and y = φn
yq · ... · y3 · φn y2 · y1 .
Then x < y holds iff either p < q, or p = q and
there exists r s.t. xi = yi for i > r and xr < yr .
Connection order vs. flip normal form
+
• Theorem: The order < on Bn
is a twisted ShortLex-extension
+
+
of < on Bn−1 . Let x, y in Bn admit the flip splittings
q−1
x = φpn−1 xp · ... · x3 · φn x2 · x1 and y = φn
yq · ... · y3 · φn y2 · y1 .
Then x < y holds iff either p < q, or p = q and
there exists r s.t. xi = yi for i > r and xr < yr .
Proof: The flip normal form coincides
with the Burckel normal form.
Case of the Birman-Ko-Lee monoid
Same situation with BKL+
n?
Case of the Birman-Ko-Lee monoid
Same situation with BKL+
n?
+
.
• Let φn = conjugation by δn in BKLn
• Proposition (Fromentin): For each x in BKL+
n , there is a unique
sequence (xp , ..., x1 ) in BKL+
s.t.
n−1
p−1
2
x = φn
xp · .... · φn
x3 · φn x2 · x1
and φp−k
xp · ... · xk is right divisible by no ai,j with i, j 6= n − 1.
n
Case of the Birman-Ko-Lee monoid
Same situation with BKL+
n?
+
.
• Let φn = conjugation by δn in BKLn
• Proposition (Fromentin): For each x in BKL+
n , there is a unique
sequence (xp , ..., x1 ) in BKL+
s.t.
n−1
p−1
2
x = φn
xp · .... · φn
x3 · φn x2 · x1
and φp−k
xp · ... · xk is right divisible by no ai,j with i, j 6= n − 1.
n
• Conjecture (Fromentin): Let x, y in BKL+
n admit decompositions
q−1
x = φpn−1 xp · .... · φn x2 · x1 and y = φn
yq · .... · φn y2 · y1 .
Then x < y holds iff
Case of the Birman-Ko-Lee monoid
Same situation with BKL+
n?
+
.
• Let φn = conjugation by δn in BKLn
• Proposition (Fromentin): For each x in BKL+
n , there is a unique
sequence (xp , ..., x1 ) in BKL+
s.t.
n−1
p−1
2
x = φn
xp · .... · φn
x3 · φn x2 · x1
and φp−k
xp · ... · xk is right divisible by no ai,j with i, j 6= n − 1.
n
• Conjecture (Fromentin): Let x, y in BKL+
n admit decompositions
q−1
x = φpn−1 xp · .... · φn x2 · x1 and y = φn
yq · .... · φn y2 · y1 .
Then x < y holds iff either p < q, or p = q and
there exists r s.t. xi = yi for i > r and xr < yr .
Long sequences of braids
(joint with A.Bovykin, L.Carlucci, and A.Weiermann)
Long sequences of braids
(joint with A.Bovykin, L.Carlucci, and A.Weiermann)
• Construct (very) long descending sequences of braids
using a simple inductive rule.
Long sequences of braids
(joint with A.Bovykin, L.Carlucci, and A.Weiermann)
• Construct (very) long descending sequences of braids
using a simple inductive rule.
Reminiscent of Goodstein’s sequences and Hydra battles:
“battle against a malevolent braid”: try to get rid of all crossings;
at step t, one chops off one crossing,
but t new crossings reappear in general.
Long sequences of braids
(joint with A.Bovykin, L.Carlucci, and A.Weiermann)
• Construct (very) long descending sequences of braids
using a simple inductive rule.
Reminiscent of Goodstein’s sequences and Hydra battles:
“battle against a malevolent braid”: try to get rid of all crossings;
at step t, one chops off one crossing,
but t new crossings reappear in general.
• Here in the 3 strand version—but exists for each n.
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
σ2
σ1
|
{z
4
}|
{z
3
} |{z} ↑
2
1
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
σ2
σ1
|
{z
4
}|
{z
3
} |{z} ↑
2
1
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
σ2
σ1
|{z} |
5
{z
4
}|
{z
3
} |{z} ↑
2
1
Critical position
e
Lemma: The word σ[p]p ...σ2e2 σ1e1 is normal iff
1 or 2 according to p (mod 2)
we have ep > 1, ep−1 > 2, ..., e3 > 2, e2 > 1, and e1 > 0.
• The critical position: smallest (= rightmost) k s.t. ek does not
have the minimal legal value, if it exists, p otherwise.
σ2
σ1
|{z} |
5
{z
4
}|
{z
3
} |{z} ↑
2
1
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
• Example: σ22 σ12 ,
...
...
|
{z
}
next block (if it exists):
add t crossings
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
• Example: σ22 σ12 , σ22 σ1 ,
...
...
|
{z
}
next block (if it exists):
add t crossings
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
• Example: σ22 σ12 , σ22 σ1 , σ22 ,
...
...
|
{z
}
next block (if it exists):
add t crossings
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
...
| {z }
critical block:
remove 1 crossing
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 ,
...
|
{z
}
next block (if it exists):
add t crossings
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
...
| {z }
critical block:
remove 1 crossing
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 ,
...
|
{z
}
next block (if it exists):
add t crossings
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 , σ15 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 , σ15 ,
σ14 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 , σ15 ,
σ14 , σ13 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 , σ15 ,
σ14 , σ13 , σ12 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 , σ15 ,
σ14 , σ13 , σ12 , σ1 ,
G3 -sequences
• The G3 -sequence from a positive 3-braid x:
- Start with the alternating normal form of x;
- At step t: remove 1 crossing in the critical block;
add t new crossings in the next block, if it exists;
- The sequence stops when (if) one reaches the braid 1.
...
...
| {z }
critical block:
remove 1 crossing
...
...
|
{z
}
next block (if it exists):
add t crossings
• Example: σ22 σ12 , σ22 σ1 , σ22 , σ2 σ13 , σ2 σ12 , σ2 σ1 , σ2 , σ17 , σ16 , σ15 ,
σ14 , σ13 , σ12 , σ1 , 1.
An unprovability statement
• More examples:
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 9
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,1
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,15
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,9
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,95
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,4
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,47
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,6
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,63
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
Nevertheless:
• Proposition 1: Every G3 -sequence (resp. G∞ -sequence) is finite.
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
Nevertheless:
• Proposition 1: Every G3 -sequence (resp. G∞ -sequence) is finite.
Proof: It is descending in the braid well-order.
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
Nevertheless:
• Proposition 1: Every G3 -sequence (resp. G∞ -sequence) is finite.
Proof: It is descending in the braid well-order.
But:
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
Nevertheless:
• Proposition 1: Every G3 -sequence (resp. G∞ -sequence) is finite.
Proof: It is descending in the braid well-order.
But:
• Theorem: Proposition 1 cannot be proved in IΣ1 (resp. IΣ2 ).
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
Nevertheless:
• Proposition 1: Every G3 -sequence (resp. G∞ -sequence) is finite.
Proof: It is descending in the braid well-order.
But:
• Theorem: Proposition 1 cannot be proved in IΣ1 (resp. IΣ2 ).
↑
↑
the subsystem of Peano arithmetic in which induction
is restricted to formulas with one ∃ quantifier (resp. ∃∀)
An unprovability statement
• More examples:
- starting with σ1 σ2 σ1 requires 30 steps;
- starting with σ12 σ22 σ12 requires 90,159,953,477,630 steps...
Nevertheless:
• Proposition 1: Every G3 -sequence (resp. G∞ -sequence) is finite.
Proof: It is descending in the braid well-order.
But:
• Theorem: Proposition 1 cannot be proved in IΣ1 (resp. IΣ2 ).
↑
↑
the subsystem of Peano arithmetic in which induction
is restricted to formulas with one ∃ quantifier (resp. ∃∀)
in contrast with the folklore result:
• All usual (algebraic) properties of braids can be proved in IΣ1 .
P. Dehornoy,
Alternating normal forms for braids and locally Garside monoids
math.GR/0702592.
P. Dehornoy,
Alternating normal forms for braids and locally Garside monoids
math.GR/0702592.
P. Dehornoy, I. Dynnikov, D. Rolfsen, B. Wiest,
Why are braids orderable?
Panoramas & Synthèses vol. 14, Soc. Math. France (2002).
P. Dehornoy,
Alternating normal forms for braids and locally Garside monoids
math.GR/0702592.
P. Dehornoy, I. Dynnikov, D. Rolfsen, B. Wiest,
Why are braids orderable?
Panoramas & Synthèses vol. 14, Soc. Math. France (2002).
http://www.math.unicaen.fr/∼dehornoy