MEMOIRS

8.4. A RELATED TUBULAR CURVE OF INDEX TWO 109We determine the image of Φ: Consider the tubular mutations σ L and σ S whichare associated to the tube containing L (of slope 0) and to the tube containing anexceptional simple object S (of slope ∞), respectively. By [57, 6.] thesemapsonslopes induce the actions q ↦→ q/(1−2q) andq ↦→ q + 2, respectively. It follows thatthe image of Φ is given by 〈t, s 2 ,l 2 〉 (with t =(sl) 3 ), which is a subgroup in B 3of index 6. It follows from [100] that〈s 2 ,l 2 〉 is isomorphic to F 2 and t is central,hence we get Im Φ ≃ F 2 × Z. It is easy to see, that the induced exact sequence ofgroups1 −→ Pic 0 (X) ⋊ Aut(X) −→ Aut(D b Φ(X)) −→ Im Φ −→ 1splits. Then the result follows.□8.3.2. We keep the notation of the proposition. There are two companioncurves which are derived equivalent to X, namely the curves X〈0〉 and X〈1〉 (in theslopes q =0andq = 1, respectively). Moreover, X, X〈0〉 and X〈1〉 are pairwisenon-isomorphic.This corresponds to the fact, that there are two (further) tilting bundles in H,such that the endomorphism rings are the canonical algebras Λ〈0〉 and Λ〈1〉 givenby the speciesQF ∗ KM ∗ KF(where F ∗ and M ∗ denote the dual bimodules) andFFKQK(plus relations), respectively (for some bimodule N), see [59]. The algebras Λ,Λ〈0〉, andΛ〈1〉 are tilting equivalent.It follows from 1.7.12 that for X〈1〉 one gets as projective coordinate algebra agraded algebra, arising by inserting the weight p = 2 into some prime element ofQ〈X, Y, Z〉/(XY − YX,XZ− ZX, Y Z + ZY, Z 2 − 3Y 2 − 2X 2 ).8.3.3. In [73] the following is shown over an algebraically closed field: Two finitedimensional algebras Λ and Λ ′ which are derived equivalent to the same tubularexceptional curve and having the same Cartan matrix are isomorphic.This is not true in general over arbitrary fields, since the tilting equivalenttubular canonical algebras Λ and Λ〈1〉 as above have the same Cartan matrix, butare obviously not isomorphic.8.4. A related tubular curve of index twoThe next example shows that the index is not a K-theoretic invariant.Proposition 8.4.1. There is a tubular exceptional curve X ′ over the field k =Q(i) such that the following holds:(1) With the tubular curve X from 8.3.1, the Grothendieck groups K 0 (X ′ ) andK 0 (X), equipped with the Euler form, are isomorphic.(2) The index of X ′ is two.(3) There is a graded factorial coordinate algebra of X ′ which arises by insertionof the weight p =2into the central prime element X 4 −Y 4 in the twisted polynomialalgebra K[X; Y,α], whereK = k( 4√ 2) and α is the k-automorphism 4√ 2 ↦→ i 4√ 2.N