the reduced expressions of a permutation. with an ... · Let (S;R) be a semigroup presentation. Then two words w;w0on S represent the same element of the monoid hSjRi+ if and only

The Subword Reversing Method

Patrick Dehornoy

Laboratoire de MathematiquesNicolas Oresme, Universite de Caen

• A strategy for constructing van Kampen diagrams for semigroups,with an application to the combinatorial distance between

the reduced expressions of a permutation.


Patrick Dehornoy





Patrick Dehornoy





Patrick Dehornoy





Patrick Dehornoy


• A strategy for constructing van Kampen diagrams for semigroups,

with an application to the combinatorial distance betweenthe reduced expressions of a permutation.


Patrick Dehornoy




Plan :

• The general case:

- Subword reversing as a strategyfor constructing van Kampen diagrams

- Subword reversing as a syntactic transformation

- A cancellativity criterion

• The case of permutations:

- bounds for the combinatorial distancebetween reduced expressions of a permutation

- recognizing the optimality of a van Kampen diagram

Plan :








Plan :








Plan :








Plan :








Plan :








Van Kampen diagrams

all relations of the form u = v with u, v nonempty words on S↓

• Let (S,R) be a semigroup presentation. Then two words w, w′ on Srepresent the same element of the monoid 〈S |R〉+

if and only if there exists an R-derivation from w to w′.

• Proposition (van Kampen, ?): If (S,R) is a semigroup presentation,two words w, w′ on S represent the same element of the monoid 〈S |R〉+

if and only if there exists a van Kampen diagram for (w, w′).

↑a tesselated disk with (oriented) edges labeled by elements of S and

faces labelled by relations of R, with boundary paths labelled w and w′.

Van Kampen diagrams








Van Kampen diagrams








Van Kampen diagrams

• Example: Let B+n =

Ds1, ..., sn−1

˛ sisjsi = sjsisj for |i− j| = 1sisj = sjsi for |i− j| > 2

E+

(the n-strand Artin braid monoid).

Then

s1

s3

s2

s2

s2 s2

s1 s3

s3 s1

s3 s1

s1 s3

s2

s2s2

s2

s1

s3

s3

s1

is a van Kampen diagram for (s1s2s1s3s2s1, s3s2s3s1s2s3).

Van Kampen diagrams


Ds1, ..., sn−1


E+(the n-strand Artin braid monoid).

Then

s1

s3

s2

s2

s2 s2

s1 s3

s3 s1

s3 s1

s1 s3

s2

s2s2

s2

s1

s3

s3

s1


Van Kampen diagrams


Ds1, ..., sn−1


E+(the n-strand Artin braid monoid).

Then

s1

s3

s2

s2

s2 s2

s1 s3

s3 s1

s3 s1

s1 s3

s2

s2s2

s2

s1

s3

s3

s1


A building strategy

• How to build a van Kampen diagram (when it exists)?

↑∼= solve the word problem: decide w ≡+R w′

• Subword reversing = the left strategy: starting with two words w, w′,

- look at the leftmost pending pattern

s

t

- choose a relation sv = tu of R to close it intos

t

v

u

, and repeat.

• Facts: - May not be possible (no relation s... = t...);- May not be unique (several relations s... = t...);- May never terminate (when u, v have length more than 1);- May terminate but boundary words are longer than w, w′

(certainly happens if w, w′ are not R-equivalent).

A building strategy

• How to build a van Kampen diagram (when it exists)?↑∼= solve the word problem: decide w ≡+

R w′



s

t


t

v

u

, and repeat.



A building strategy


R w′

• Subword reversing = the left strategy:

starting with two words w, w′,


s

t


t

v

u

, and repeat.



A building strategy


R w′



s

t


t

v

u

, and repeat.



A building strategy


R w′



s

t


t

v

u

,

and repeat.



A building strategy


R w′



s

t


t

v

u

, and repeat.



A building strategy


R w′



s

t


t

v

u

, and repeat.

• Facts: - May not be possible (no relation s... = t...);

- May not be unique (several relations s... = t...);- May never terminate (when u, v have length more than 1);- May terminate but boundary words are longer than w, w′


A building strategy


R w′



s

t


t

v

u

, and repeat.

• Facts: - May not be possible (no relation s... = t...);- May not be unique (several relations s... = t...);

- May never terminate (when u, v have length more than 1);- May terminate but boundary words are longer than w, w′


A building strategy


R w′



s

t


t

v

u

, and repeat.

• Facts: - May not be possible (no relation s... = t...);- May not be unique (several relations s... = t...);- May never terminate (when u, v have length more than 1);

- May terminate but boundary words are longer than w, w′


A building strategy


R w′



s

t


t

v

u

, and repeat.



The subword reversing strategy

• At least: deterministic whenever R is a complemented presentation:

for each pair of letters s, t in S, there is exactly one relation s... = t... in R.


Ds1, ..., sn−1

˛ E+.

sisjsi = sjsisj for |i− j| = 1sisj = sjsi for |i− j| > 2

Applying the reversing strategy to s1s2s1s3s2s1 and s3s2s3s1s2s3:

s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2

So, on this particular example, the reversing strategy works.


• At least: deterministic whenever R is a complemented presentation:for each pair of letters s, t in S, there is exactly one relation s... = t... in R.


Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2





Ds1, ..., sn−1

˛ E+.



s1

s2

s1 s3

s2

s1

s3

s2

s3 s1

s2

s3

s3

s1

s2s3

s3 s2

s1

s1

s2

s3

s1

s1

s3

s2


Reversing diagrams

• Another way of drawing the same diagram:

s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1

only vertical and horizontal edges,plus dotted arcs connecting vertices that are to be identified

in order to get an actual van Kampen diagram.

Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3

s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams


s3

s2

s1

s3

s2

s3

s1 s2 s1 s3 s2 s1

s3

s1

s1 s2

s2

s1

s2 s3

s3

s2

s3s1

s1s3

s1

s3s2 s3

s3s2

s1 s2

s2

s1



Reversing diagrams

• In this way, a uniform pattern:

s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Reversing diagrams


s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Reversing diagrams


s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Reversing diagrams


s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Reversing diagrams


s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Reversing diagrams


s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Reversing diagrams


s

t

becomes s

t

u

v

for sv = tu in R

• More exactly:

s

t

becomess

t

u

v

for sv = tu in R

including

s

s

becomess

s

.

Syntactic description

• Introduce two types of letters:

- S for horizontal edges, S−1 for vertical edges;- read words the Mull of Kintyre to the Pentland Fifth (SW to NE).

• Basic step:

s

t

7→s

t

u

v

reads: s−1t 7→ vu−1,

including

s

s

7→s

s

reads: s−1s 7→ ε.↑

the empty word

• Syntactically, “subword reversing”: replacing −+ with +−.


• Introduce two types of letters:- S for horizontal edges,

S−1 for vertical edges;- read words the Mull of Kintyre to the Pentland Fifth (SW to NE).

• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word



• Introduce two types of letters:- S for horizontal edges, S−1 for vertical edges;

- read words the Mull of Kintyre to the Pentland Fifth (SW to NE).

• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word



• Introduce two types of letters:- S for horizontal edges, S−1 for vertical edges;- read words the Mull of Kintyre to the Pentland Fifth (SW to NE).

• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word




• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word




• Basic step:

s

t

7→s

t

u

v

reads:

s−1t 7→ vu−1,

including

s

s

7→s

s


the empty word




• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word




• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s

reads:

s−1s 7→ ε.↑

the empty word




• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word




• Basic step:

s

t

7→s

t

u

v


including

s

s

7→s

s


the empty word


Reversing sequences

• Definition: For w, w′ words on S ∪ S−1, declare w y(1)R

w′ if

∃s, t, u, v (sv = tu lies in R and w = ...s−1t... and w′ = ...vu−1...).

Declare w yR w′ if there exist w0, ..., wp s.t.w0 = w, wp = w′, and wi y(1)

Rwi+1 for each i.

• Terminal words: w′w−1 with w, w′ words on S (no letter s−1).

• Lemma: If w, w′, v, v′ are words on S and w−1w′ yR v′v−1,

i.e., w

w′

v

v′

yR , then wv′ ≡+R

w′v.

• In particular, if w−1w′ yR ε, i.e., if w

w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if

∃s, t, u, v (sv = tu lies in R

and w = ...s−1t... and w′ = ...vu−1...).


Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if


Declare w yR w′ if there exist w0, ..., wp s.t.

w0 = w, wp = w′, and wi y(1)R

wi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′

yR ,

then wv′ ≡+R

w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR ,

then w ≡+R

w′.↑

the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.

↑the empty word

Reversing sequences


w′ if



Rwi+1 for each i.



i.e., w

w′

v

v′


w′v.


w′

yR , then w ≡+R

w′.↑

the empty word

Completeness

• Conversely, does w ≡+R

w′ implies w−1w′ yR ε?

• Definition: A presentation (S,R) is called complete (w.r.t. subword re-versing)

if w ≡+R

w′ implies w−1w′ yR ε.

↑hence ... is equivalent to ...

• Remark: Completeness implies the solvability of the word problemonly if one knows that reversing always terminates.

• Two questions:

- How to recognize completeness?

- What to do with a complete presentation?

Completeness



• Definition: A presentation (S,R) is called complete (w.r.t. subword re-versing) if w ≡+

Rw′ implies w−1w′ yR ε.



• Two questions:



Completeness







• Two questions:



Completeness






• Remark: Completeness implies the solvability of the word problem

only if one knows that reversing always terminates.

• Two questions:



Completeness







• Two questions:



Completeness







• Two questions:



Completeness







• Two questions:



Completeness







• Two questions:



The cube condition

• Theorem: (D., ’97) Assume that (S,R) is a homogeneous complementedpresentation.

Then (S,R) is complete if, and only if,for each triple r, s, t in S, the cube condition for r, s, t is satisfied.

• homogeneous: ∃ R-invariant λ : S∗ → N (λ(sw) > λ(w)).

• cube condition for a tripleof positive words u, v, w: u

v

w

...hence checkable (for one triple)

The cube condition

• Theorem: (D., ’97) Assume that (S,R) is a homogeneous complementedpresentation. Then (S,R) is complete if, and only if,

for each triple r, s, t in S, the cube condition for r, s, t is satisfied.



v

w


The cube condition





v

w


The cube condition



• homogeneous:

∃ R-invariant λ : S∗ → N (λ(sw) > λ(w)).


v

w


The cube condition




• cube condition for a tripleof positive words u, v, w:

uv

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


The cube condition





v

w


A cancellativity criterion

• Proposition: Assume that (S,R) is a complete complemented presenta-tion. Then the monoid 〈S |R〉+ is left-cancellative.

↑sa = sa′ implies a = a′

• Proof: Assume sw ≡+Rsw′. Want to prove w ≡+

Rw′.

Completeness implies: (sw)−1(sw′) yR ε, i.e., w−1s−1sw′ yR ε.

s

w

s w′

The first step must be w−1s−1sw′ yR w−1w′,so the sequel must be w−1w′ yR ε, hence w ≡+

Rw′. �





Rw′.


s

w

s w′


Rw′. �




• Proof: Assume sw ≡+Rsw′.

Want to prove w ≡+R

w′.


s

w

s w′


Rw′. �





Rw′.


s

w

s w′


Rw′. �





Rw′.

Completeness implies: (sw)−1(sw′) yR ε,

i.e., w−1s−1sw′ yR ε.

s

w

s w′


Rw′. �





Rw′.


s

w

s w′


Rw′. �





Rw′.


s

w

s w′


Rw′. �





Rw′.


s

w

s w′


Rw′. �





Rw′.


s

w

s w′

The first step must be w−1s−1sw′ yR w−1w′,

so the sequel must be w−1w′ yR ε, hence w ≡+R

w′. �





Rw′.


s

w

s w′


Rw′. �





Rw′.


s

w

s w′


Rw′. �

Application to the word problem(s)

• Proposition: Assume that (S,R) is a complete complemented presenta-

tion and there exists a finite set bS including S and closed under reversing.Then the word problem of 〈S |R〉+ is solvable in quadratic time, and so isthat of 〈S |R〉 if 〈S |R〉+ is right-cancellative.

↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )

• Proof: Reversing terminatesin quadratic time: constructan bS-labeled grid:

∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS

∈bS ∈bS ∈bS ∈bS ∈bS• For w, w′ words on S:w ≡+

Rw′ iff w−1w′ yR ε.

• For w a word on S ∪ S−1:assume w yR v′v−1;then w ≡R ε iff v ≡R v′ iff v ≡+

Rv′

iff v−1v′ yR ε (double reversing).

w

yR

v

v′

v′

yR

?



tion

and there exists a finite set bS including S and closed under reversing.Then the word problem of 〈S |R〉+ is solvable in quadratic time, and so isthat of 〈S |R〉 if 〈S |R〉+ is right-cancellative.

↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?



tion and there exists a finite set bS including S and closed under reversing.

Then the word problem of 〈S |R〉+ is solvable in quadratic time, and so isthat of 〈S |R〉 if 〈S |R〉+ is right-cancellative.

↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?



tion and there exists a finite set bS including S and closed under reversing.Then the word problem of 〈S |R〉+ is solvable in quadratic time,

and so isthat of 〈S |R〉 if 〈S |R〉+ is right-cancellative.

↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )

• Proof: Reversing terminatesin quadratic time:

constructan bS-labeled grid:

∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS

∈bS ∈bS ∈bS ∈bS ∈bS

• For w, w′ words on S:w ≡+



Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS

∈bS ∈bS ∈bS ∈bS ∈bS• For w, w′ words on S:

w ≡+R

w′ iff w−1w′ yR ε.


Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS



• For w a word on S ∪ S−1:

assume w yR v′v−1;then w ≡R ε iff v ≡R v′ iff v ≡+

Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS



• For w a word on S ∪ S−1:assume w yR v′v−1;

then w ≡R ε iff v ≡R v′ iff v ≡+R

v′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS



• For w a word on S ∪ S−1:assume w yR v′v−1;then w ≡R ε iff v ≡R v′

iff v ≡+R

v′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?




↑∀w,w′∈bS ∃v,v′∈bS ( w−1w′ yR v′v−1 )


∈S ∈S ∈S ∈S ∈S

∈S

∈S

∈bS∈bS

∈bS∈bS

∈bS∈bS




Rv′


w

yR

v

v′

v′

yR

?

Subword reversing as a tool

Range

• For semigroups: in principle, all are eligible: completion procedure(when the cube condition fails).

• For groups: unknown; at least: classical and dual presentations of(generalized) braid groups (and all Garside groups) —but certainly more.

Uses

• Cancellativity criterion;

• Existence of least common multiples, identification of Garside structures;

• Computation of the greedy normal form;

• (with Y. Lafont) Construction of explicit resolutions (whence homology);

• (with B. Wiest) Solution to the word problem (complexity issues);

• (with M. Autord) Combinatorial distance betweenthe reduced expressions of a permutation.


Range

• For semigroups: in principle, all are eligible:

completion procedure(when the cube condition fails).


Uses








Range



Uses








Range


• For groups: unknown;

at least: classical and dual presentations of(generalized) braid groups (and all Garside groups) —but certainly more.

Uses








Range


• For groups: unknown; at least: classical and dual presentations of(generalized) braid groups

(and all Garside groups) —but certainly more.

Uses








Range


• For groups: unknown; at least: classical and dual presentations of(generalized) braid groups (and all Garside groups)

—but certainly more.

Uses








Range



Uses








Range



Uses








Range



Uses








Range



Uses








Range



Uses








Range



Uses








Range



Uses








Range



Uses







Reduced expressions of a permutation

• Every permutation of {1, ...,n} is a product of transpositions:

Sn =Ds1, ..., sn−1

˛sisjsi = sjsisj for |i− j| = 1

sisj = sjsi for |i− j| > 2, s21 = ...=s2n−1 =1

E.

of minimal length↓

• Proposition (“Exchange Lemma”): Any two reduced expressions of apermutation are connected by braid relations (no need of using s2i = 1).

• Combinatorial distance: d(u, v) = minimal number of braid relationsneeded to transform u into v.

• Question: Bounds on d(u, v)? (The standard proof of the ExchangeLemma gives an exponential upper bound.)

• Proposition (folklore ?): There exist positive constants C,C′ s.t.- d(u, v) 6 Cn4 holds for every permutation f of {1, ...,n}

and all reduced expressions u, v of f ,- d(u, v) > C′n4 holds for some permutation f of {1, ...,n}

and some reduced expressions u, v of f .



Sn =Ds1, ..., sn−1



E.










Sn =Ds1, ..., sn−1



E.










Sn =Ds1, ..., sn−1



E.


• Proposition (“Exchange Lemma”):

Any two reduced expressions of apermutation are connected by braid relations (no need of using s2i = 1).








Sn =Ds1, ..., sn−1



E.


• Proposition (“Exchange Lemma”): Any two reduced expressions of apermutation are connected by braid relations

(no need of using s2i = 1).








Sn =Ds1, ..., sn−1



E.










Sn =Ds1, ..., sn−1



E.



• Combinatorial distance:

d(u, v) = minimal number of braid relationsneeded to transform u into v.







Sn =Ds1, ..., sn−1



E.










Sn =Ds1, ..., sn−1



E.




• Question: Bounds on d(u, v)?

(The standard proof of the ExchangeLemma gives an exponential upper bound.)






Sn =Ds1, ..., sn−1



E.










Sn =Ds1, ..., sn−1



E.





• Proposition (folklore ?): There exist positive constants C,C′ s.t.

- d(u, v) 6 Cn4 holds for every permutation f of {1, ...,n}and all reduced expressions u, v of f ,

- d(u, v) > C′n4 holds for some permutation f of {1, ...,n}and some reduced expressions u, v of f .



Sn =Ds1, ..., sn−1



E.






and all reduced expressions u, v of f ,

- d(u, v) > C′n4 holds for some permutation f of {1, ...,n}and some reduced expressions u, v of f .



Sn =Ds1, ..., sn−1



E.








Naming crossings

• Here: lower bounds; more specifically:

• Aim: Recognize whether a given Van Kampen diagramor reversing diagram is possibly optimal.

↑# faces = combinatorial distance between the bounding words

• Associate a braid diagram with every (reduced) s-word and use thenames (or the colors) of the strands that cross

(i.e., use a “position vs. name” duality):

s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)

a sequence N(w) of pairs of integers in {1, ...,n}.

Naming crossings


• Aim:

Recognize whether a given Van Kampen diagramor reversing diagram is possibly optimal.




s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings




• Associate a braid diagram with every (reduced) s-word

and use thenames (or the colors) of the strands that cross


s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1

7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s1

1

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3

{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Naming crossings






s1s2s1 7→

s1 s2 s11

2

3{1,2}{1,3}{2,3}

← w

← N(w)


Lower bounds

• For S,S′ sequences of pairs of integers in {1, ...,n}:- I3(S,S′) = # triples {p, q, r} s.t.

{p, q}, {p, r} and {q, r} appear in different orders in S,S′.

- I2,2(S,S′) = # pairs of pairs {{p, q}, {p′, q′}} s.t.{p, q} and {p′, q′} appear in different orders in S,S′.

• Lemma: If w, w′ are two reduced expressions of some permutation, then

d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).

• Proof: Each type I braid relation (“hexagon”) contributes at most 1 to I3,each type II braid relation (“square”) contributes at most 1 to I2,2. �

• Example: w = s1s2s1s3s2s1, w′ = s3s2s3s1s2s3.

Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.

• Question (Conjecture?): Is the above inequality an equality?

Lower bounds

• For S,S′ sequences of pairs of integers in {1, ...,n}:- I3(S,S′)

= # triples {p, q, r} s.t.{p, q}, {p, r} and {q, r} appear in different orders in S,S′.



d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds



- I2,2(S,S′)

= # pairs of pairs {{p, q}, {p′, q′}} s.t.{p, q} and {p′, q′} appear in different orders in S,S′.


d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).

• Proof: Each type I braid relation (“hexagon”) contributes at most 1 to I3,

each type II braid relation (“square”) contributes at most 1 to I2,2. �


Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),

N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Lower bounds





d(w, w′) > I3(N(w),N(w′)) + I2,2(N(w),N(w′)).



Then N(w) = ({1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3, 4}),N(w′) = ({3, 4}, {2, 4}, {2, 3}, {1, 4}, {1, 3}, {1, 2}).

Hence d(w, w′) > 4 + 2 = 6.


Naming faces

• Back to van Kampen diagrams with the aim of recognizing optimality.

↑# faces = combinatorial distance between bounding words

• Having given names to the generators si (= the edges of the diagram),give names to the faces:

type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

{p,q,r} type II:{p,q}

{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}

• Criterion 1: A van Kampen diagram in which different faceshave different names is optimal.

Naming faces

• Back to van Kampen diagrams with the aim of recognizing optimality.↑

# faces = combinatorial distance between bounding words


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces



• Having given names to the generators si (= the edges of the diagram),

give names to the faces:

type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:

{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

{p,q,r}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

{p,q,r} type II:

{p,q}

{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}

{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces




type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}


{p′,q′} {p,q}

{p′,q′}{{p,q},{p′,q′}}


Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1

{2,3,4}

{1,2,3}

{{1,3},{2,4}} {{1,2},{3,4}}

{1,3,4}

{1,2,4}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1

{2,3,4}

{1,2,3}

{{1,3},{2,4}} {{1,2},{3,4}}

{1,3,4}

{1,2,4}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1

{2,3,4}

{1,2,3}

{{1,3},{2,4}} {{1,2},{3,4}}

{1,3,4}

{1,2,4}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1

{2,3,4}

{1,2,3}

{{1,3},{2,4}}

{{1,2},{3,4}}

{1,3,4}

{1,2,4}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1

{2,3,4}

{1,2,3}

{{1,3},{2,4}} {{1,2},{3,4}}

{1,3,4}

{1,2,4}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1

{{1,2},{3,4}} {{1,3},{2,4}}

{2,3,4}

{1,2,4}

{1,3,4}

{1,2,3}

{{1,4},{2,3}}

{{1,4},{2,3}}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1

{{1,2},{3,4}} {{1,3},{2,4}}

{2,3,4}

{1,2,4}

{1,3,4}

{1,2,3}

{{1,4},{2,3}}

{{1,4},{2,3}}

Naming faces (2)

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1

{{1,2},{3,4}} {{1,3},{2,4}}

{2,3,4}

{1,2,4}

{1,3,4}

{1,2,3}

{{1,4},{2,3}}

{{1,4},{2,3}}

Separatrices

• (Again in a van Kampen diagram) connect the edges with the same name:

type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}

for each pair {p, q}, an (oriented) curve that connect all {p, q}-edges:the {p, q}-separatrix Σp,q.

type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:

{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}

for each pair {p, q}, an (oriented) curve that connect all {p, q}-edges:

the {p, q}-separatrix Σp,q.

type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices


type I:{p,q}

{q,r}

{q,r}

{p,q}

{q,r}

{p,q}

type II:{p,q}

{p′,q′} {p,q}

{p′,q′}


type I:

Σp,q Σq,r

Σq,r

type II:

Σp,q Σp′,q′

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1

Σ1,2

Σ1,3

Σ2,3 Σ1,4

Σ2,4

Σ3,4

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1Σ1,2

Σ1,3

Σ2,3 Σ1,4

Σ2,4

Σ3,4

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s2s3

s1

s2

s2

s1

s3

s2

s1

s3

s1

s3

s2

s2

s3

s1Σ1,2

Σ1,3

Σ2,3 Σ1,4

Σ2,4

Σ3,4

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1

Σ1,2

Σ1,3 Σ2,4

Σ3,4

Σ2,3 Σ1,4

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1Σ1,2

Σ1,3 Σ2,4

Σ3,4

Σ2,3 Σ1,4

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1Σ1,2

Σ1,3 Σ2,4

Σ3,4

Σ2,3 Σ1,4

Separatrices

• Example:

s3

s1

s2

s2

s3

s1

s3

s1

s1

s3

s2

s3

s2s1

s1

s3

s2 s2

s1

s3

s3

s1

s2

s2

s3

s1Σ1,2

Σ1,3 Σ2,4

Σ3,4

Σ2,3 Σ1,4

An optimality criterion

• Criterion 2: A van Kampen diagram in whichany two separatrices cross at most once is optimal.

• Question: Is the condition necessary, i.e., do any two separatricescross at most once in an optimal van Kampen diagram?

• Remark: Compare with “a s-word is reduced iffany two strands in the associated braid diagram cross at most one”.









Separatrices and reversing

• Applies in particular to reversing diagrams(viewed as particular van Kampen diagrams):

s1

s2

s3

s1 s1s1

s2

s2

s1 s1

s2

s3

s3

s3 s2 s1 s3 s2

s2s3 s1

s3 s2 s1

s2 s3 s1 s2 s1

s1 s1

s2

s3

s2

s2

s1

Σ2,3

Σ1,4


• How are separatrices in a reversing diagram?

Three types of faces:

type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ

• Criterion 3: A reversing diagram containing no type III face is optimal.

• Proof: In order that two separatrices cross twice,one has to go from horizontal to vertical. �


• How are separatrices in a reversing diagram? Three types of faces:

type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ





type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ





type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ





type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ





type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ





type I:

Σ

Σ′Σ′′

type II:

Σ

Σ′

type III:

Σ



A lower bound result

• An improvement: Same argument when reversing steps are grouped:

replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,

corresponding to and .

• An application:

• Proposition: For each `, there exist length ` reduced s-words w, w′

satisfying w−1w′ yR v′v−1 and d(wv′, w′v) > `4/8.

By contrast: for fixed n, Garside’s theory gives an upper bound in O(`2).



replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,

corresponding to

and .

• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:






replace sisisj

sisj

sj si

with si

sj si

sj si

sj for |i− j| = 1,


• An application:




• Two conclusions:

• Even in the simple(?) case of braids and permutations, many open questions.

• Importance of having van Kampen diagrams included in a grid.










• P. Dehornoy, Deux proprietes des groupes de tressesC. R. Acad. Sci. Paris 315 (1992) 633–638.

• F.A. Garside, The braid group and other groupsQuart. J. Math. Oxford 20-78 (1969) 235–254.

• K. Tatsuoka, An isoperimetric inequality for Artin groups of finite typeTrans. Amer. Math. Soc. 339–2 (1993) 537–551.

• R. Corran, A normal form for a class of monoids including the singular braidmonoids J. Algebra 223 (2000) 256–282.

• P. Dehornoy, Complete positive group presentations;J. Algebra 268 (2003) 156–197.

• P. Dehornoy & Y. Lafont, Homology of Gaussian groupsAnn. Inst. Fourier 53-2 (2003) 1001–1052.

• P. Dehornoy & B. Wiest, On word reversing in braid groupsInt. J. for Algebra and Comput. 16-5 (2006) 941–957.

• P. Dehornoy & M. Autord, On the combinatorial distance between expressions ofa permutation in preparation.

www.math.unicaen.fr/∼dehornoy


















































































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the reduced expressions of a permutation. with an ... · Let (S;R) be a semigroup presentation. Then two words w;w0on S represent the same element of the monoid hSjRi+ if and only