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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.

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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 CaseBwithβ0= 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], hα0,β0[m]). Then ( m > α0+β0−1 ifβ0>1, (α0, β0)∈`B(m)⇐⇒ m≥2 ifβ0= 1. For its proof, 2.Modify the sentence just after (11.14) in pp.389 as follows: Furthermore,forβ0>1,by Figure 11.4 and (11.14), we have 3.Before the sentence ‘Sincem≥2, we obtain the conclusion.’, insert the next sentence: Forβ0Step 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 andβ0= 1, we have 1 Lemma 11.10Ifm < α0+ 1,then (∙|∙)c,his not positive semi-deﬁnite. 2 Proof.Since the lemma is clear forα0= 1, we may assume thatα0>1. As we have the equivalence m < α0+ 1⇐⇒m(α0+ 1)−(m+ 1)∙1< M, ◦ this condition onmis equivalent to (α0+ 1,1)∈E(c, h). Atthe same time, we also have Φ1,1(c[m], hα0,1[m] +α0)>0 which together with Proposition 11.5 implies detH(c[m], hα0,1[m])α0+1<0.✷ 5.The line 3∼4↓of pp.389 should be modiﬁed as follows: ( m > α0+β0−1 CaseA∨(Case B∧β0>1), m≥α0Case B+ 1∧β0= 1.

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