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  7. Coxeter Groups are Biautomatic: Lecture 2 by Piotr Przytycki & Damian Osajda

Coxeter Groups are Biautomatic: Lecture 2 by Piotr Przytycki & Damian Osajda

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Learn about Coxeter Groups being biautomatic in this insightful lecture by Piotr Przytycki and Damian Osajda. Understand the concept of groups being biautomatic and their regular language representation for every group element.
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Video Transcript

0:11
first talk of today is uh P's second
0:14
talk on Cox to groups being biomatic
0:17
thank
0:18
you so first a small recap from last
0:22
time a group is by automatic if there is
0:24
a generating set s and the regular
0:27
language over s such that the map from
0:31
the language to the group is subjective
0:33
every word has a represent every group
0:35
element has a representative in this
0:38
language and fellow traveler property
0:40
holds so for every pair of words in the
0:44
language that are equal in the group or
0:47
the fair by a generator on the right
0:51
these words feral travel and second part
0:54
if they defer on the left in the group
0:56
then the words W and s w Prime below
0:59
travel
1:01
and we Pro the corollary last time which
1:03
is true even for automatic groups so
1:05
just for groups satisfying part one that
1:10
one implies
1:13
that any word any group
1:16
element in
1:18
the uh automatic group represented by a
1:21
word that might not necessarily be in
1:23
the language can be still represented
1:25
efficiently can be represented by a word
1:27
in the language of length at most some
1:30
constant times the length of the
1:31
original word plus and
1:33
zero and today we are supposed to prove
1:36
the theorem that every automatic group
1:38
so satisfying just part
1:41
one is finally presented we assume it's
1:44
finally generated but for the moment we
1:45
don't know it's Finly presented and
1:49
moreover it satisfies quadratic
1:51
isoparametric inequality which means
1:53
that there is a constant C such that for
1:55
any word representing the trial element
1:58
in the group this word can be
2:00
represented efficiently as a product of
2:03
conjugates of
2:04
relators in the free group namely the
2:07
number of these conjugates it say most
2:10
quadratic in the length of the
2:13
word any questions before we start the
2:15
proof okay so let's do the
2:23
proof so we have a

Video Summary & Chapters

No chapters for this video generated yet.

Video Transcript

0:11
first talk of today is uh P's second
0:14
talk on Cox to groups being biomatic
0:17
thank
0:18
you so first a small recap from last
0:22
time a group is by automatic if there is
0:24
a generating set s and the regular
0:27
language over s such that the map from
0:31
the language to the group is subjective
0:33
every word has a represent every group
0:35
element has a representative in this
0:38
language and fellow traveler property
0:40
holds so for every pair of words in the
0:44
language that are equal in the group or
0:47
the fair by a generator on the right
0:51
these words feral travel and second part
0:54
if they defer on the left in the group
0:56
then the words W and s w Prime below
0:59
travel
1:01
and we Pro the corollary last time which
1:03
is true even for automatic groups so
1:05
just for groups satisfying part one that
1:10
one implies
1:13
that any word any group
1:16
element in
1:18
the uh automatic group represented by a
1:21
word that might not necessarily be in
1:23
the language can be still represented
1:25
efficiently can be represented by a word
1:27
in the language of length at most some
1:30
constant times the length of the
1:31
original word plus and
1:33
zero and today we are supposed to prove
1:36
the theorem that every automatic group
1:38
so satisfying just part
1:41
one is finally presented we assume it's
1:44
finally generated but for the moment we
1:45
don't know it's Finly presented and
1:49
moreover it satisfies quadratic
1:51
isoparametric inequality which means
1:53
that there is a constant C such that for
1:55
any word representing the trial element
1:58
in the group this word can be
2:00
represented efficiently as a product of
2:03
conjugates of
2:04
relators in the free group namely the
2:07
number of these conjugates it say most
2:10
quadratic in the length of the
2:13
word any questions before we start the
2:15
proof okay so let's do the
2:23
proof so we have a
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