We provide a complete classification of Fano threefolds X having canonical Gorenstein singularities and anticanonical degree (−K X ) 3 = 64.Lemma 5.9. Let Z ⊂ Y ′ be an irreducible rational curve such that dim |Z| > 0 and E Z ≃ O P 1 (d 1 ) ⊕ O P 1 (d 2 ) for some d 1 , d 2 ∈ Z. Then |d 1 − d 2 | 2 + (Z 2 ).Proof. This follows from exactly the same arguments as in the proof of [30, Lemma 10.6] (recall that the linear system | − nK Y | is basepoint-free for n large).Hence we may assume that both c i are integers. We have two cases:(1) c 1 is even. Then we may assume that c 1 = 0 up to twisting E by a line bundle (cf. [8]). In this case, from (5.7) we obtain the equalitywhich implies that c 2 = −5/4 ∈ Z, a contradiction.
9(2) c 1 is odd. Then, as above, we may assume that c 1 = −K Y ′ . In this case, from (5.5) we get the equalityIdentify D with a generic element in |D|. This is a weak del Pezzo surface with at worst Du Val singularities (see [35]). Then from the adjunction formula we obtainand since (−K Y ) 3 = 64, we get K 2 D = 8. Now, since Y ′ ≃ P 2 , one easily gets that either D ≃ P 2 or F 1 (see e.g. [2, Theorem 8.1.5]). But the former case is impossible forBut then either ext * (D) = 0 or Z is a fiber of ext, respectively, thus all being absurd. Hence we get Z = ∅. On the other hand, dim E f = 2 (see Lemma 5.2) and D = F 1 contains a fiber of ext by construction, which implies that D ∩ E f = ∅, a contradiction.Lemma 5.11. Let Y ′ = P 1 × P 1 . Then X is isomorphic to the cone from Example 1.3.Proof. Recall that Y ′ is smooth and the locus E f ′ is of pure codimension 1 for X ′ = P(6, 4, 1, 1) (see Example 3.13 and Remark 3.14). Then we have the following two results: Lemma 6.10. f ′ (ext(E)) is a point.Proof. Suppose that f ′ (ext(E)) is a curve. Then C := ext(E) is also a curve and we have E f ′ ∩C = ∅ by Lemma 6.3. Further, the map p : X ′ X is the linear projection from a subspace Λ ⊂ P 38 such that X ′ ∩ Λ = f ′ (C) (see Lemma 6.4). More precisely, since g = 33 and g ′ = 37, we find that dim Λ = 3, which gives −K X ′ · f ′ (C) 4 (for X ′ is the intersection of quadrics). But on X ′ = P(6, 4, 1, 1) we have O X ′ (−K X ′ ) ≃ O X ′ (12) (see Example 3.13), and hencewhich implies that the curve f ′ (C) passes through a singular point on X ′ . This contradicts E f ′ ∩ C = ∅. 13 Lemma 6.11. ext(E) is a point.Proof. Suppose that C := ext(E) is a curve. Then Lemmas 6.10 and 6.4 show that p is the linear projection from the point f ′ (C). In particular, we get g ′ − g = 1, which contradicts g = 33, g ′ = 37.Further, Lemmas 6.11 and 6.5 imply that p is the linear projection from the tangent space at the smooth point f ′ (ext(E)) on X ′ .Conversely, linear projection from the tangent space at a smooth point on X ′ = P (6, 4, 1, 1) is birational (see Lemma 3.16) and leads to a Fano threefold X := p(X ′ ) with (−K X ) 3 = 64 (cf. Remark 3.10 and the construction in [28, Section 7]). Proposition 6.9 is completely proved. Proposition 6.12. Let X ′ = X 70 . Then p is the linear projection from a plane Π. More precisel...