"Braid group"의 두 판 사이의 차이

2020년 11월 16일 (월) 04:30 판

review of symmetric groups

원소의 개수가 n인 집합의 전단사함수들의 모임
\(n!\) 개의 원소가 존재함
대칭군의 부분군은 치환군(permutation group)이라 불림

presentation of symmetric groups

\(S_n\)
generators \(\sigma_1, \ldots, \sigma_{n-1}\)
relations
- \({\sigma_i}^2 = 1\)
- \(\sigma_i\sigma_j = \sigma_j\sigma_i \mbox{ if } j \neq i\pm 1\)
- \(\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i\sigma_{i+1}\)

presentation of braid groups

\(B_n\)
generators \(\sigma_1,...,\sigma_{n-1}\)
relations (known as the braid or Artin relations):
- \(\sigma_i\sigma_j =\sigma_j \sigma_i\) whenever \(|i-j| \geq 2 \)
- \(\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i \sigma_{i+1}\) for \(i = 1,..., n-2\)
Yang-Baxter equation (YBE)
For a solution of the YBE \(\bar{R}\), we can construct a representation \(\rho\) of the braid group by

\[ \rho : B_n \to \rm{Aut}(V^{\otimes n}) \] where \(\rho(\sigma_i)=\bar{R}_i\)

There is also a natural surjective morphism from \(B_n\) to the symmetric group \(\mathfrak{S}_n\), given on the generators by \(B_n\ni\sigma_i\mapsto s_i\in \mathfrak{S}_n\), \(i=1,\dots,n-1\). For a braid \(\beta\in B_n\), we denote \(p_{\beta}\) its image in \(\mathfrak{S}_n\), and refer to \(p_{\beta}\) as to the underlying permutation of \(\beta\).

examples

in a braid diagram, read from bottom to top and we number all strands of the braid with the indices it starts at the bottom

파일:Braid.png

read the braid word from left to right accordingly.
For instance, the braid word corresponding to the braid above is \(\sigma_1^{-1}\sigma_2\sigma_1^{-1}\sigma_2\sigma_1^{-1}\)

Markov moves

braid group version of Reidemeister moves

computational resource

https://docs.google.com/file/d/0B8XXo8Tve1cxZ3NjMGpGUWI0QkE/edit

related items

encyclopedia

http://en.wikipedia.org/wiki/Braid_group

expositions

Abad, Camilo Arias. 2014. “Introduction to Representations of Braid Groups.” arXiv:1404.0724 [math], April. http://arxiv.org/abs/1404.0724.

@@ 26번째 줄: / 26번째 줄: @@
 ** <math>\sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i \sigma_{i+1}</math> for <math>i = 1,..., n-2</math>
 * [[Yang-Baxter equation (YBE)]]
-* For a solution of the YBE $\bar{R}$, we can construct a representation $\rho$ of the braid group by
+* For a solution of the YBE <math>\bar{R}</math>, we can construct a representation <math>\rho</math> of the braid group by
-$$
+:<math>
 \rho : B_n \to \rm{Aut}(V^{\otimes n})
-$$ where $\rho(\sigma_i)=\bar{R}_i$
+</math> where <math>\rho(\sigma_i)=\bar{R}_i</math>
-There is also a natural surjective morphism from $B_n$ to the symmetric group $\mathfrak{S}_n$, given  on the generators by $B_n\ni\sigma_i\mapsto s_i\in \mathfrak{S}_n$, $i=1,\dots,n-1$. For a braid $\beta\in B_n$, we denote $p_{\beta}$ its image in $\mathfrak{S}_n$, and refer to $p_{\beta}$ as to the underlying permutation of $\beta$.
+There is also a natural surjective morphism from <math>B_n</math> to the symmetric group <math>\mathfrak{S}_n</math>, given  on the generators by <math>B_n\ni\sigma_i\mapsto s_i\in \mathfrak{S}_n</math>, <math>i=1,\dots,n-1</math>. For a braid <math>\beta\in B_n</math>, we denote <math>p_{\beta}</math> its image in <math>\mathfrak{S}_n</math>, and refer to <math>p_{\beta}</math> as to the underlying permutation of <math>\beta</math>.
@@ 39번째 줄: / 39번째 줄: @@
 [[파일:Braid.png]]
 * read the braid word from left to right accordingly.
-* For instance, the braid word corresponding to the braid above is $\sigma_1^{-1}\sigma_2\sigma_1^{-1}\sigma_2\sigma_1^{-1}$
+* For instance, the braid word corresponding to the braid above is <math>\sigma_1^{-1}\sigma_2\sigma_1^{-1}\sigma_2\sigma_1^{-1}</math>
 ==Markov moves==