MEMOIRS

5.2. THE STRUCTURE OF Aut(H) 77right Λ-module N to N φ −1 (observe the −1 here, ensuring that the assignmentφ ↦→ φ ∗ is covariant). This induces an autoequivalence of the triangulated categoryD b (Λ), and since D b (Λ) = D b (X) this finally induces an autoequivalence (whichwe also denote by φ ∗ )ofH. Now φ ∗ fixes L since it is easy to see that the correspondingindecomposable summand of Λ is preserved under the automorphismφ :Λ−→ Λ. Moreover, φ is inner if and only if φ ∗ on mod(Λ) is isomorphic to theidentity (compare [8, II.5]). These constructions are mutually inverse. □Remark 5.1.5. Let M = F M G be a tame bimodule over k. Denote by K itscentre (compare 0.5.5). Clearly, Aut K (M) ⊂ Aut k (M) andInn K (M) =Inn k (M).Moreover, there is an exact sequenceρ1 −→ Aut K (M) −→ Aut k (M) −→ Gal(K/k),where ρ is defined by restricting ϕ F and ϕ G to K. It follows that the factor groupAut k (X)/ Aut K (X) can be embedded into Gal(K/k).5.2. The structure of Aut(H)5.2.1 (Orbit cases IIIa and IIIb). Let X be a homogeneous exceptional curvewith hereditary category H. We would like to analyse the structure of Aut(H). Inorder to do this we have to refine the definition of orbit case III. By definition ofthis case there is no σ ∈ Aut 0 (H) withσ(L) =L (see 1.1.5). But is it possiblethat there is a σ ∈ Aut(H) with this property? If there is such an automorphism,we call this case orbit case IIIb, if not, we call it case IIIa. In other words, incase IIIa each efficient automorphism is transitive. Of course, if in case III theendomorphism skew fields of line bundles in two different Auslander-Reiten orbitsare non-isomorphic, then we are in case IIIa.Problem 5.2.2. Is orbit case IIIb non-empty, that is, does there exist a tamebimodule over some field belonging to this case?Recall that O is the Aut(H)-orbit and O 0 is the Aut 0 (H)-orbit of L. Notethatonly in orbit case IIIb there is a difference between O and O 0 . Obviously, in anyorbit case there exists a transitive automorphism: in case IIIb by definition, in allother cases there is even a point fixing one.Fix a transitive σ ∈ Aut(H) (which is unique up to an automorphism of X).Let φ ∈ Aut(H). Then there is a (unique) n ∈ Z such that σ n (φ(L)) ≃ L, thatis,σ n ◦ φ ∈ Aut(X). In other words, each φ ∈ Aut(H) is a composition of a power ofσ and some element of Aut(X). Therefore, Aut(H) consists of the following typesof automorphisms:• the geometric automorphisms of the curve• the tubular shifts• the ghosts• one transitive automorphism (if not already in the Picard group).As usual, denote by G the ghost group of X. Recall that Pic 0 (X) =Pic(X) ∩Aut(X), which is Pic(X) ∩G.Proposition 5.2.3. Let X be a homogeneous exceptional curve. Assume, thatthere is an exhaustive automorphism σ lying in Pic(X), and assume that we are notin orbit case IIIb. Then there are split exact sequences of groups1 −→ Pic(X)/ Pic 0 (X) −→ Aut(H)/G −→Aut(X)/G −→1,