Lie groups and Lie algebras [Lecture notes] by Bas Edixhoven

By Bas Edixhoven

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A2 a2 x .. a1 a2 a3 1 1 1 + + ··· + x d1 − x dn − x . ⎞ . . an . . an ⎟ ⎟ . . an ⎟ ⎟ .. ⎟ .. . ⎠ ... x = (x + a1 + a2 + · · · + an )(x − a1 )(x − a2 ) · · · (x − an ). 3. Let b(t) = x + t + t −1 . Prove that Dn (b) = x n − 4. Let ωn = e2π i/n . ⎛ 1 ⎜ 1 ⎜ ⎜ det ⎜ 1 ⎜ .. ⎝ . Prove that 1 5. Prove that ⎛ 1 ⎜ 1 ⎜ 2 det ⎜ ⎜ .. ⎝ . 1 n n − 1 n−2 n − 2 n−4 x + x − +··· . 1 2 1 ωn ωn2 .. 1 ωn2 ωn4 .. ... 1 . . ωnn−1 . . ωn2(n−1) .. ωnn−1 ωn2(n−1) ... 1 2 1 3 1 3 1 4 .. .. 1 n+1 1 n+2 ... ⎟ ⎟ ⎟ ⎟ = nn/2 i −(n−1)(n+2)/2 .

Zi − zk )⎟ ⎠. 18) i,k∈M k=i The coefficients Bi of the partial fraction decomposition are Bi = − (zi − z ). =i Hence ⎛ ⎞ ⎜ Bi ⎠ ⎜ ⎝ ⎟ (zi − zk )⎟ ⎠ ⎞ ⎛ ⎝ i∈M i,k∈M k=i (zi − zj )−1 = (−1)r (zi − z )−1 i, ∈M =i i∈M j ∈M (zi − zk ) i,k∈M k=i (zi − zj )−1 . 19), we arrive at the equality ⎛ ⎝ Dn (b) = bsn (−1)rn a0r+n M i∈M ⎞ ⎛ ⎞ ⎜ zi−n−r ⎠ (−1)r−rs ⎜ ⎝ ⎟ (zi − zj )−1 ⎟ ⎠. 5. Trench’s Formula buch7 2005/10/5 page 41 ✐ 41 Since a0 = (−1)r+s z1 · · · zr+s , it follows that ⎛ ⎞ ⎝ Dn (b) = bsn (−1)rn+(r+s)(r+n)+r+rs j ∈M M ⎞ ⎛ ⎟ (zi − zj )−1 ⎟ ⎠.

Zjs )r+n = zjr+n 1 .. zjr+n s ... (zkα − zkβ ) α>β M ... zjr+n+s−1 1 .. zjr+n+s−1 s (zjγ − zjδ ). 22) γ >δ Permuting the rows 1, 2, . . , r + s of G0 to k1 , . . , kr , j1 , . . , js shows that G0 equals (−1)(k1 −1)+(k2 −2)+···+(kr −r) (zkα − zkβ ) α>β (zjγ − zjδ ) γ >δ (zjγ − zkα ). 6. 23) we get Gn = G0 (zjγ − zkα )−1 (zj1 . . zjs )r+n M ⎛ ⎝ = zj ⎠ j ∈M M γ ,α ⎞n (zj − zk )−1 zjr j ∈M j ∈M k∈M ((−1)s bs−1 wM )n CM = (−1)sn bs−n = M n CM wM , M which completes the proof in the case of simple zeros.

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