By Konrad Schöbel

Konrad Schöbel goals to put the rules for a consequent algebraic geometric remedy of variable Separation, that's one of many oldest and strongest how to build designated options for the basic equations in classical and quantum physics. the current paintings unearths a shocking algebraic geometric constitution at the back of the well-known record of separation coordinates, bringing jointly an outstanding variety of arithmetic and mathematical physics, from the past due nineteenth century idea of separation of variables to trendy moduli area conception, Stasheff polytopes and operads.

"I am relatively inspired by way of his mastery of quite a few recommendations and his skill to teach sincerely how they have interaction to supply his results.” (Jim Stasheff)

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**Example text**

Q − 1)! e a b ··· c d .. 9) e and the same formula holds for its adjoint. The isomorphism class of the irreducible representation deﬁned by a Young tableau is labelled by the corresponding Young frame, which is the Young tableau with the labels of its boxes erased. On the level of isomorphism classes, the decomposition of tensor products of irreducible representations is given by the Littlewood-Richardson rule. For example, according to this rule, the tensor product of a symmetric and an antisymmetric representation decomposes into two irreducible components, each of hook symmetry: p p q ..

23c). The proof of the remaining part of (ii) is completely analogous to the proof of (i), so we leave it to the reader. 23c). This ﬁnishes the proof. 3 The 2nd integrability condition The proceeding for the second integrability condition is similar. 3b): j j i ¯ δ Kδα = g¯ij S i N βγ a 2 b 1 b 2 S c 2 d 1 d 2 + S c2 b 1 b 2 S d 1 a 2 d 2 x b 1 x b 2 x d 1 ∇ δ x a 2 ∇ β x c2 ∇ γ x d 2 Se 1 e 2 f 1 f2 x e 1 x e 2 ∇ δ x f1 ∇ α x f2 . 4 and omit the terms that vanish due to the Bianchi identity: j j i ¯ δ Kδα = g¯ij g¯a2 f1 S i N βγ a2 b 1 b 2 S c 2 d 1 d 2 + S c2 b 1 b 2 S d 1 a 2 d 2 S e 1 e 2 f 1 f 2 x b 1 x b 2 x d 1 x e 1 x e 2 ∇ β x c2 ∇ γ x d 2 ∇ α x f 2 .

25d) = b 2 b1 d 1 c2 d2 a2 g¯ij 3S ia2 b1 b2 S jc 2 d1 d2 . 21b). To continue, antisymmetrise 0= c2 d2 a2 =2 c2 d2 a2 b2 b1 d 1 g¯ij S ia2 b1 b2 S jc g¯ij S ia2 b1 b2 S jc 2 d1 d2 2 d1 d2 + S ia2 b2 d1 S jc 2 b1 d2 + S ia2 d1 b1 S jc 2 b2 d2 in a2 , b2 , c2 , d2 . Then the last term vanishes by the symmetry of S j c2 b2 d2 in b2 , d2 and yields 0= a2 b2 c2 d2 g¯ij S ia2 b1 b2 S jc 2 d1 d2 + S ia2 d1 b2 S jc 2 b1 d2 . Both sum terms are equal under antisymmetrisation in a2 , b2 , c2 , d2 and contraction with g¯ij .