Corrigenda to Chapter

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Corrigenda to Chapter 11 K. Iohara and Y. Koga August 31, 2011 1 Step VI For Lemma 11.10 of [IK], the proof given there in Case B with ?0 = 1 does not work and requires the following modifications. 1. Lemma 11.10.2 should be modified as follows: Case B: Suppose that (c, h) = (c[m], h?0,?0 [m]). Then (?0, ?0) ? B(m) ?? { m > ?0 + ?0 ? 1 if ?0 > 1, m ≥ 2 if ?0 = 1. For its proof, 2. Modify the sentence just after (11.14) in pp. 389 as follows: Furthermore, for ?0 > 1, by Figure 11.4 and (11.14), we have 3. Before the sentence ‘Since m ≥ 2, we obtain the conclusion.', insert the next sentence: For ?0 = 1, there is nothing to prove since this is our assumption (cf. Step III). 4. After the end of the proof of Lemma 11.10, insert For Case B and ?0 = 1, we have Lemma 11.10 12 If m < ?0 + 1, then (·|·)c,h is not positive semi-definite. Proof. Since the lemma is clear for ?0 = 1, we may assume that ?0 > 1.

• virasoro algebra


• dimensional lie


• sentence


• implies propo- sition


• representation theory


• world sci


• ?0 ?

Sujets

• Virasoro algebra
• Sentence
• Representation theory

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 18
• Langue: English

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Corrigenda to Chapter 11

K. Iohara and Y. Koga

August 31, 2011

1 StepVI For Lemma 11.10 of [IK], the proof given there in CaseBwithβ 0 = 1 does not work and requires the following modifications.

1.Lemma 11.10.2 should be modified as follows: Case B: Suppose that (c, h) = (c[m], h α 0 ,β 0 [m]). Then ( m > α 0 +β 0 −1 ifβ 0 >1, (α 0 , β 0 )∈` B (m)⇐⇒ m≥2 ifβ 0 = 1. For its proof, 2.Modify the sentence just after (11.14) in pp.389 as follows: Furthermore,forβ 0 >1,by Figure 11.4 and (11.14), we have 3.Before the sentence ‘Sincem≥2, we obtain the conclusion.’, insert the next sentence: Forβ 0 Step III).= 1, there is nothing to prove since this is our assumption (cf. 4.After the end of the proof of Lemma 11.10, insert For Case B andβ 0 = 1, we have 1 Lemma 11.10Ifm < α 0 + 1,then (∙|∙) c,h is not positive semi-definite. 2 Proof.Since the lemma is clear forα 0 = 1, we may assume thatα 0 >1. As we have the equivalence m < α 0 + 1⇐⇒m(α 0 + 1)−(m+ 1)∙1< M, ◦ this condition onmis equivalent to (α 0 + 1,1)∈E(c, h). Atthe same time, we also have Φ 1,1 (c[m], h α 0 ,1 [m] +α 0 )>0 which together with Proposition 11.5 implies detH(c[m], h α 0 ,1 [m]) α 0 +1 <0.✷ 5.The line 3∼4↓of pp.389 should be modified as follows: ( m > α 0 +β 0 −1 CaseA∨(Case B∧β 0 >1), m≥α 0 Case B+ 1∧β 0 = 1.

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