루트 시스템 (root system)과 딘킨 다이어그램 (Dynkin diagram)

수학노트
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개요



정의

  • E를 내적이 주어진 유클리드 벡터공간이라 하자.
  • 다음 조건을 만족시키는 E의 유한인 부분집합 \(\Phi\)를 루트 시스템이라 한다.
    • \(\Phi\)는 E를 스팬(span)하며 \(0 \not \in \Phi\)
    • (reduced) \(\alpha \in \Phi\), \(\lambda \alpha \in \Phi \iff \lambda=\pm 1\)
    • \(\alpha,\beta \in \Phi\)이면 \(\sigma_\alpha(\beta) =\beta-2\frac{(\beta,\alpha)}{(\alpha,\alpha)}\alpha \in \Phi\)
    • \(\langle \beta, \alpha \rangle = 2 \frac{(\beta,\alpha)}{(\alpha,\alpha)} \in \mathbb{Z}\)
  • 마지막 조건을 crystallographic 또는 integrality 조건이라 한다
  • a subgroup of \(GL(V)\) is crystallographic if it stabilizes a lattice L in V
  • e.g. the Weyl group of a Lie algebra stabilizes the root lattice or the weight lattice

딘킨 다이어그램 (Dynkin diagram)

  • first draw the simple roots as nodes
  • draw \(4(e_i, e_j)^2\)lines for two roots \(e_i, e_j\)
  • \(\frac{\pi}{2}\) , \(\frac{\pi}{3}\), \(\frac{\pi}{4}\), \(\frac{\pi}{6}\) 0,1,2,3 lines




2차원 루트 시스템의 분류

  • \(A_1\times A_1\), \(A_2\), \(B_2\), \(G_2\)

A1 x A1

http://www.wolframalpha.com/input/?i=r%3D1%2Bcos+(4theta)

A2

http://www.wolframalpha.com/input/?i=r%3D1%2B+cos+(6theta)

B2

http://www.wolframalpha.com/input/?i=r%3D1-+(sqrt2+%2B1)^2+cos+(4theta)

G2

http://www.wolframalpha.com/input/?i=r%3D1-(sqrt+3+%2B1)^2cos+(6theta)/2

[1]

2696052-MSP45719773453e5409bcd000043c1iebh17cda58g.gif

2696052-MSP402197733f5dbe80g5d000056hb767e4digb412.gif

2696052-MSP132719772cfcfe659i75000064ieda8fh9d30h5e.gif

2696052-MSP98119772g2ig5gid8he000031i1h30a8gacdi00.gif




ADE 의 분류

(0) G cannot contain affine A_n, D_n, E_n

(1) G is a tree (contains no cycles = affine A_n)

(2) G has \leq 1 branch point (does not contain affine D_5, D_6,D_7, )

(3) branch point has order \leq 3 (affine D_4) What are length of legs of G?

Leg of length 0 -> G=A_n

so assume legs have length \geq 1

(4) Not all legs have length \geq 2 : cannot contain affine E_6

so one leg has length 1

2 legs of length 1 : G is D_n

so can assume 2 other legs have length \geq 2

(5) cannot have 2 legs length \geq 3 because of affine E_7

So G has 1 leg length 1, 1 of length 2, one of length \geq 2

length is \leq 4, as G does not contain affine E_8

So G is E6,E7, E8




일반적인 경우

  • how to classify all connected admissible diagrams
    • subdiagram is also admissible
    • there are at most (n-1) pairs of nodes
    • no node has more than 3 lines
    • study double lines and triple nodes




reflection groups

  • B_n, C_n, BC_n -> same reflection group (Z/nZ).S_n




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