"Braid group"의 두 판 사이의 차이
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imported>Pythagoras0 |
Pythagoras0 (토론 | 기여) |
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| (사용자 2명의 중간 판 3개는 보이지 않습니다) | |||
| 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> | ** <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)]] | * [[Yang-Baxter equation (YBE)]] | ||
| − | * For a solution of the YBE | + | * 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}) | \rho : B_n \to \rm{Aut}(V^{\otimes n}) | ||
| − | + | </math> where <math>\rho(\sigma_i)=\bar{R}_i</math> | |
| − | There is also a natural surjective morphism from | + | 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]] | [[파일:Braid.png]] | ||
* read the braid word from left to right accordingly. | * read the braid word from left to right accordingly. | ||
| − | * For instance, the braid word corresponding to the braid above is | + | * 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== | ==Markov moves== | ||
| 68번째 줄: | 68번째 줄: | ||
[[분류:개인노트]] | [[분류:개인노트]] | ||
[[분류:math and physics]] | [[분류:math and physics]] | ||
| + | [[분류:migrate]] | ||
| + | |||
| + | ==메타데이터== | ||
| + | ===위키데이터=== | ||
| + | * ID : [https://www.wikidata.org/wiki/Q220409 Q220409] | ||
| + | ===Spacy 패턴 목록=== | ||
| + | * [{'LOWER': 'braid'}, {'LEMMA': 'group'}] | ||
2021년 2월 17일 (수) 02:36 기준 최신판
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
- 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
encyclopedia
expositions
- Abad, Camilo Arias. 2014. “Introduction to Representations of Braid Groups.” arXiv:1404.0724 [math], April. http://arxiv.org/abs/1404.0724.
메타데이터
위키데이터
- ID : Q220409
Spacy 패턴 목록
- [{'LOWER': 'braid'}, {'LEMMA': 'group'}]