"Braid group"의 두 판 사이의 차이
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imported>Pythagoras0 |
imported>Pythagoras0 |
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| 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 $\bar{R}$, we can construct a representation $\rho$ of the braid group by | ||
| + | $$ | ||
| + | \rho : B_n \to \rm{Aut}(V^{\otimes n}) | ||
| + | $$ where $\rho(\sigma_i)=\bar{R}_i$ | ||
| − | |||
==computational resource== | ==computational resource== | ||
* https://docs.google.com/file/d/0B8XXo8Tve1cxZ3NjMGpGUWI0QkE/edit | * https://docs.google.com/file/d/0B8XXo8Tve1cxZ3NjMGpGUWI0QkE/edit | ||
2013년 5월 5일 (일) 03:04 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 $\rho(\sigma_i)=\bar{R}_i$
computational resource