Blocks with a generalized quaternion defect group and three simple modules over a 2-adic ring
aa r X i v : . [ m a t h . R T ] J un Blocks with a generalized quaternion defect group andthree simple modules over a -adic ring Florian EiseleApril 5, 2018
Abstract
We show that two blocks of generalized quaternion defect with three simple modules overa sufficiently large -adic ring O are Morita-equivalent if and only if the corresponding blocksover the residue field of O are Morita-equivalent. As a corollary we show that any two blocksdefined over O with three simple modules and the same generalized quaternion defect groupare derived equivalent. Let ( K, O , k ) be a -modular system. We assume that K is complete and that k is algebraicallyclosed. The aim of this article is to prove the following result: Theorem 1.1.
Assume n ∈ N and K ⊇ Q ( ζ n − + ζ − n − ) , where ζ n − denotes a n − -th rootof unity. Let Λ be a block of O G and Γ be a block of O H for finite groups G and H . If the defectgroups of Λ and Γ are both isomorphic to the generalized quaternion group Q n and Λ and Γ bothhave three isomorphism classes of simple modules, then the following hold:1. Λ and Γ are Morita-equivalent if and only if the k -algebras k ⊗ Λ and k ⊗ Γ are Morita-equivalent.2. Λ and Γ are derived equivalent. The problem of classifying blocks of quaternion defect over O arises naturally from a well-knownclassification result of Erdmann in [4], where such blocks are classified over k . If the structureof these blocks is as narrowly restricted over k as it is revealed to be in [4], does it also followthat their structure is equally restricted over O ? A priori it is not even clear that the number ofMorita-equivalence classes of blocks over O reducing to a single Morita-equivalence class of blocksover k is finite. However, Theorem 1.1 tells us that in our case this number is not merely finite,but in fact equal to one. In a way this was to be expected, considering the result of [7], whichproves the first part of Theorem 1.1 for one of the three possible Morita-equivalence classes ofblocks of defect Q . The results of [10] concerning character values of blocks of quaternion defectand the later refinement of those results to perfect isometries between such blocks in [2] also hinttowards Theorem 1.1 being true, as they already show that all of the blocks over O which areclaimed to be isomorphic respectively derived equivalent in Theorem 1.1 do at the very least haveisomorphic centers. Moreover, a derived equivalence between the principal block of O SL ( q ) andits Brauer correspondent has been shown to exist in [5].The proof of Theorem 1.1 builds mainly upon the results in [4] and [2]. For the case of defect Q we also make use of the result in [7], as it can be seen already from the decomposition matrix(see (19)) that this case is somewhat different as there are more symmetries. Here is a roughoutline of the proof: 1. First we prove that one of the algebras in Erdmann’s classification [4], namely the basicalgebra of the principal block of k SL ( q ) for q ≡ , has a unique (or, rather, “atmost one”) symmetric O -order with split semisimple K -span reducing to it, provided thecenter of the order is prescribed. Roughly speaking, we do this by showing that the endo-morphisms of the projective indecomposable modules can be recovered as the projection ofthe (prescribed) center to certain Wedderburn components. Moreover, the homomorphismsbetween different projective indecomposables viewed as bimodules over the endomorphismrings of those projective modules can also be obtained in this way. Which projections wehave to take can be read off from the decomposition matrix, which we know. After that itcomes down to conjugation and exploiting the symmetry of the order.2. The result of [2] gives us a perfect isometry between any one of the blocks we are interestedin and the principal block of O SL ( q ) for appropriately chosen q ≡ . In particular,this determines the center of the block in question, which is needed to apply the result of thefirst step. However, this is where a technical difficulty arises: above we assume that boththe center of the block and the decomposition matrix are known. The K -span of the centeris a semisimple commutative K -algebra whose Wedderburn components can be associatedwith the rows of the decomposition matrix. In order to perform the above step properlywe need to know which Wedderburn component of the center is associated with which rowof the decomposition matrix. The problem we are facing here boils down to the following:given a basic O -order Λ which reduces to the k -algebra treated in the first step as well as thebasic order Γ of the principal block of O SL ( q ) , we have two separate isometries betweenthe Grothendieck groups of their K -spans. On the one hand an isometry coming from theperfect isometry between blocks, which induces an isomorphism between the centers. On theother hand an isometry which preserves the decomposition matrix, and therefore induces analgebra isomorphism between the K -spans of Λ and Γ . We need to show that we can choosethese two isometries equal to each other. The first isometry is determined up to perfect self-isometries of Γ , while the second isometry is determined up to self-isometries preserving thedecomposition matrix, which come from algebra automorphisms of the K -span of Γ . Whatboth isometries have in common is that they map the sublattice of the Grothendieck groupof K ⊗ Λ generated by the K -spans of projective Λ -modules onto the analogously definedsublattice of the Grothendieck group of K ⊗ Γ . Therefore they differ by a self-isometry ofthe Grothendieck group of K ⊗ Γ which preserves the sublattice generated by the projective Γ -modules. Thus, in order to show that we can choose the above two isometries equal toone another, we merely have to show that the group of perfect self-isometries of Γ and thegroup of self-isometries of Γ stabilizing the decomposition matrix taken together generatethe group of self-isometries of Γ which preserve the lattice generated by projective modules.This is a fairly concrete problem, and proving this comes down to explicitly determiningthese groups of self-isometries.3. In the last step we generalize the result of the first two steps to all blocks of quaternion defectwith three simple modules using the fact that each such block (over k ) is derived equivalentto the principal block of k SL ( q ) for some q ≡ . The latter block is of course thesame block we studied in the first two steps. Now we exploit the fact that a one-sided tiltingcomplex T over a k -algebra lifts to a tilting complex b T over an O -order (which we may choosefreely) reducing to said k -algebra. The endomorphism ring of the lifted tilting complex b T is also reasonably well behaved, and in particular reduces to the endomorphism ring of T .Hence we can start with an arbitrary block (over O ) of quaternion defect with three simplemodules, to obtain, as the endomorphism ring of some tilting complex, an O -order reducingto the basic algebra of the principal block of k SL ( q ) . This O -order is then determined upto isomorphism by the first two steps. We can use this to determine the original block up toMorita-equivalence. Obviously there are a lot of technical pitfalls here that we just skimmedover in this short explanation.Of course there still remain some important open questions surrounding tame blocks, and blocks2f quaternion defect in particular. For blocks with three simple modules one might ask whetherthe algebra Q (3 B ) c from [4] actually occurs as a block of some group ring for each c . The answerto this, unfortunately, does not follow from our result. Moreover, the classification of blocks ofquaternion defect with two simple modules is still not entirely satisfactory, even over k . Q (3 K ) c In this section we are going to look at a specific basic algebra of a block of defect Q n overan algebraically closed field of characteristic two, namely the algebra Q (3 K ) c from [4], where c = 2 n − . This algebra is the basic algebra of the principal block of k SL ( q ) for q ≡ ,where q depends on n (more about that later). The point of this section is to extract from thepresentation of that algebra given in [4] the properties that we are going to need to prove Theorem4.2. Therefore a lot of what we are going to prove here will be elementary.Let ¯ Λ denote the algebra Q (3 K ) c from Erdmann’s classification (see the annex of [4]) withparameters a = b = 2 and c = 2 n − for some n ∈ N . That is, ¯ Λ = kQ/I , where Q is thefollowing quiver Q = • • • γ s s β δ (cid:10) (cid:10) η J J κ (cid:30) (cid:30) λ ^ ^ (1)and I is the ideal generated by the following relations: βδ = κλκηγ = λκλδλ = γβγκη = βγβλβ = ( ηδ ) n − − ηγκ = ( δη ) n − − δγβδ = 0 δηγ = 0 λκη = 0 (2) Proposition 2.1.
The following elements are contained in I : λβγ, βγκ, γβδ, ηγβ, βδη, δηγ, ηδλ, κηδ, δλκ, λκη, κλβ, γκλ (3) Proof.
This is just a calculation (in each case we highlight the part of the path that we willsubstitute in the next step using one of the relations given in (2)): λβ γ = ( ηδ ) c − ηγ = ( ηδ ) c − η δηγ = 0 (4) β γκ = β ( δη ) c − δ = βδ ( ηδ ) c − = κλκ ( ηδ ) c − = κ λκη δ ( ηδ ) c − = 0 (5) γβδ = 0 (6) ηγ β = λ κλβ = 0 (uses (14)) (7) βδ η = κλ κη = κ λβγ β = 0 (uses (4)) (8) δηγ = 0 (9) η δλ = ηγ βγ = λκ λβγ = 0 (uses (4)) (10) κη δ = β γβδ = 0 (11) δλ κ = γ βγκ = 0 (uses (5)) (12) λκη = 0 (13) κ λβ = κ ( ηδ ) c − η = κηδ ( ηδ ) c − η = 0 (uses (11)) (14) γκ λ = ( δη ) c − δλ = ( δη ) c − γβγ = ( δη ) c − δηγ βγ = 0 (15)3 emark 2.2. Think of the quiver Q as a triangle and of paths in Q as walks along the edges ofthat triangle. Then Proposition 2.1 tells us that going two steps forward in any direction and thengoing one step back in the opposite direction gives us a path which is zero in ¯ Λ . And, in the samevein, going one step in any direction followed by two steps in the opposite direction also yieldszero. This implies that there are two kinds of non-zero paths in ¯ Λ :1. Paths going back and forth between two vertices.2. Paths walking around the triangle without changing direction. Proposition 2.3.
The elements βγ + γβ , δη + ηδ and λκ + κλ lie in the center of ¯ Λ .Proof. Proposition 2.1 implies that the product of βγ + γβ with any arrow other than β and γ must be zero, regardless of whether we multiply from the left or from the right. Moreover, thefollowing holds already in the quiver algebra kQ , independently of the relations in I : β · ( βγ + γβ ) = βγβ = ( βγ + γβ ) · β and γ · ( βγ + γβ ) = γβγ = ( βγ + γβ ) · γ (16)The element βγ + γβ also commutes with the idempotents e , e and e , also already in kQ . Soclearly βγ + γβ is central, and the exact same reasoning implies that δη + ηδ and λκ + κλ arecentral as well. Proposition 2.4.
Assume x is a path in Q . Then one of the following three possibilities holds: x + I = γ d · ( βγ ) m · β d + I for some m ∈ Z > and d , d ∈ { , } or x + I = η d · ( δη ) m · δ d + I for some m ∈ Z > and d , d ∈ { , } or x + I = κ d · ( λκ ) m · λ d + I for some m ∈ Z > and d , d ∈ { , } (17) In other words, any path in Q is equivalent mod I to a path going back and forth between twovertices.Proof. We will prove this by induction. For paths of length strictly less than two there is nothingto show. For paths of length two the claim follows from the first six relations given in (2). Usingthe induction hypothesis a path of length at least three may be written as a path alternatingbetween two vertices composed with a single arrow. Either the resulting path is also alternatingbetween two vertices, in which case there is nothing to show, or the last three arrows occurringin the path must lie in I by the “one step in one direction, two steps in the opposite direction”criterion explained in Remark 2.2. Proposition 2.5.
We have e i ¯ Λe i = e i Z ( ¯ Λ ) e i for all i ∈ { , , } (18) Proof.
Proposition 2.3 guarantees that e Z ( ¯ Λ ) e contains βγ and κλ , e Z ( ¯ Λ ) e contains γβ and δη and that e Z ( ¯ Λ ) e contains ηδ and λκ . Hence in order to prove (18) it suffices to show thatfor each i the given two elements generate e i ¯ Λe i as a k -algebra. But this follows directly fromProposition 2.4. Proposition 2.6.
Let i, j ∈ { , , } with i = j . Then e i ¯ Λe j is generated by a single element asa left e i ¯ Λe i -module, namely by the arrow in Q going from e i to e j .Proof. By Proposition 2.4, e i ¯ Λe j is spanned as a k -vector space by paths going back and forthbetween the vertices e i and e j , starting in e i and ending in e j . But necessarily the last arrowinvolved in such a path is the arrow going from e i to e j . Therefore the path is a product of a pathstarting and ending in e i , that is, an element of e i ¯ Λe i , and the arrow going from e i to e j . Thisproves that the latter arrow generates the module.4 Generalities on the principal block of O SL ( q ) In this section we gather some basic facts on the principal -block of SL ( q ) , in particular the case q ≡ . A Sylow -subgroup of SL ( q ) is isomorphic to Q n , where n is the -valuation of q − . We should first remark that for every n > there exists a q such that a Sylow -subgroupof SL ( q ) is isomorphic to Q n . Remark 3.1.
Assume n > . Then the condition q ≡ n − − n ensures that q + 1 is divisible by two exactly n − times, and q − is divisible by two exactly once, and therefore q − q − q + 1) has -valuation n . Moreover, such a q satisfies q ≡ . ByDirichlet’s theorem on arithmetic progressions there exists an infinite number of primes q satisfyingthe condition q ≡ n − + 1 mod 2 n .Similarly there is a prime q with q ≡ n − + 1 mod 2 n . This q will of course satisfy q ≡ and a Sylow -subgroup of SL ( q ) is isomorphic to Q n . According to [1, Table 9.1 on page 107] the decomposition matrix of B (SL ( q )) for q ≡ is as follows e e e χ χ χ χ χ χ χ r exactly once for each r = 1 , . . . , l ] (19)where l = 2 n − − . By comparing this matrix with the possible decomposition matrices given inthe appendix of [4] (or by comparing Cartan matrices), one deduces that B ( k SL ( q )) is Moritaequivalent to Q (3 K ) c for c = 2 n − (i. e. the algebra we looked at in the previous sections) if q ≡ . Proposition 3.2 (Decomposition matrix) . Assume n > . If D ∈ { , } (6+ l ) × is a matrixsatisfying the equation D ⊤ · D = n − n − n − n − =: C (20) and if every row of D is non-zero, then D is equal to the decomposition matrix given in (19) , upto permutation of rows and columns.Proof. For i ∈ { , , } denote by c i the set of all indices j ∈ { , . . . , l } such that D j,i = 1 . Sinceevery row of D is non-zero we have | c ∪ c ∪ c | = 6 + l . Moreover | c i | = C i,i and | c i ∩ c j | = C i,j for all i, j ∈ { , , } . The inclusion-exclusion principle implies that | c ∩ c ∩ c | = | c ∪ c ∪ c | − | c | − | c | − | c | + | c ∩ c | + | c ∩ c | + | c ∩ c | = (5 + 2 n − ) − − (2 + 2 n − ) − (2 + 2 n − ) + 2 + 2 + 2 n − = 1 (21)The number of rows of D equal to a given row can now easily be computed. For instance thenumber of rows equal to (1 , , is equal to | c ∩ c ∩ c | = 1 , and the number of rows equal to (1 , , is equal to | c |− | c ∩ c |− | c ∩ c | + | c ∩ c ∩ c | = 1 . It is clear that the inclusion-exclusionprinciple determines for each vector in { , } × how often it occurs as a row in D . Proposition 3.3.
Assume that ¯ Λ is a k -algebra, and that e , . . . , e n ∈ ¯ Λ is a full set of orthogonalprimitive idempotents. If e i Z ( ¯ Λ ) = e i ¯ Λe i for each i , then every O -order Λ in a split-semisimple K -algebra A with k ⊗ Λ ∼ = ¯ Λ and rank O Z ( Λ ) = dim k Z ( ¯ Λ ) has decomposition numbers less thanor equal to one. roof. Since Z ( Λ ) is a pure sublattice of Λ (meaning r · x ∈ Z ( Λ ) implies x ∈ Z ( Λ ) for any r ∈ O )we have an embedding k ⊗ Z ( Λ ) ֒ → k ⊗ Λ . The image of this embedding lies in the center of k ⊗ Λ , and equality of dimensions implies that Z ( k ⊗ Λ ) = k ⊗ Z ( Λ ) . Now we have a commutativediagram Z ( Λ ) / / (cid:15) (cid:15) (cid:15) (cid:15) b e i Λ b e i (cid:15) (cid:15) (cid:15) (cid:15) Z ( ¯ Λ ) / / / / e i ¯ Λe i (22)where the surjectivity of all except the upper horizontal arrow has either been shown above (for thevertical arrow on the left), holds by assumption (for the lower horizontal arrow) or is a general fact(for the rightmost vertical arrow). If we consider this as a diagram of O -modules, then it followsfrom the Nakayama lemma that the top arrow has to be surjective too. But that implies that b e i Λ b e i is commutative (and hence so is b e i A b e i ), which implies that the column in the decompositionmatrix belonging to b e i has entries . Since this holds for all i , the statement is proven.Note that the previous two propositions, in conjunction with Proposition 2.5, show that any O -order reducing to Q (3 K ) c which has semisimple K -span and the right dimension of the centerhas the matrix given in (19) as its decomposition matrix. Proposition 3.4 (Splitting field in characteristic zero) . Let Λ be a block of O G for some finitegroup G , and assume that the defect group of Λ is isomorphic to Q n for some n > . Assumemoreover that K ⊇ Q ( ζ n − + ζ − n − ) . Then K ⊗ Λ is split.Proof. By [11, Corollary 31.10] every finite-dimensional division algebra D over Q has a splittingfield E which is finite-dimensional and unramified over Z ( D ) . We can write E as F · Z ( D ) , where F is an unramified extension of Q . Since we assume the residue field of O to be algebraicallyclosed each unramified extension of Q is contained in K , and therefore K · Z ( D ) ⊇ E . Hence K ⊗ Λ is split if and only if its center is split (which means that K contains each field occurring inthe Wedderburn decomposition of Z ( K ⊗ Λ ) ), which happens if and only if K contains all valuesof all characters of the block Λ . By [10, Proposition 4.1] the character values of Λ are containedin the extension of some unramified extension of Q by ζ n − + ζ − n − , which is contained in K byassumption. ¯ Λ Theorem 4.2 below shows that the algebra ¯ Λ = Q (3 K ) c lifts uniquely to a symmetric O -order ifwe prescribe the center. That is a key ingredient in proving the main theorem of this article. Proposition 4.1.
Let A be a finite-dimensional semisimple K -algebra, let Λ ⊂ A be a symmetric O -order and let e, f ∈ Λ be two idempotents with ef = f e = 0 . Then eΛf = { x ∈ eAf | x · f Λe ⊆ eΛe } Proof.
Set M := { x ∈ eAf | x · f Λe ⊆ eΛe } . Clearly eΛf ⊆ M , and it suffices to prove M ⊆ Λ .Let T : A × A −→ K be an associative K -bilinear form on A such that Λ is self-dual with respectto T (associativity means T ( ab, c ) = T ( a, bc ) , and symmetry means T ( a, b ) = T ( b, a ) ). Such aform exists for any symmetric order. Then T ( M, Λ ) = T ( eM f, Λ ) = T ( M, f Λe ) = T ( M · f Λe, ⊆ T ( eΛe, ⊆ O (where associativity and symmetry of T has been used to pull e and f across). So M is containedin the dual of Λ with respect to T , which is of course again Λ . That is, M ⊆ Λ .6 heorem 4.2. Let ¯ Λ denote the algebra Q (3 K ) c from section 2 with c = 2 n − ( n ∈ N arbitrary) and let A = K ⊕ K ⊕ K ⊕ K × ⊕ K × ⊕ K × ⊕ l M i =7 K × (23) where l = 2 n − − . Denote by ε , . . . , ε l the primitive idempotents in Z ( A ) with the naturalchoice of indices.Let Λ ⊂ A and Γ ⊂ A be two O -orders such that all of the following holds:1. k ⊗ Λ ∼ = k ⊗ Γ ∼ = ¯ Λ Λ and Γ are symmetric3. Λ and Γ both have the decomposition matrix given in (19) (note: the order of the rows isfixed by the inclusion of Λ resp. Γ into A )4. Z ( Λ ) = Z ( Γ ) Then Λ and Γ are conjugate in A .Proof. First let us fix an isomorphism ψ : k ⊗ Λ −→ ¯ Λ . That fixes labels for the simple k ⊗ Λ -modules, since the simple ¯ Λ -modules correspond to the vertices of the quiver Q from section 2,which where labeled , and . Without loss of generality we may choose ψ in such a way thatthe decomposition matrix of Λ with respect to this labeling is equal to (19) with the order of bothrows and columns being fixed. This can be done since ¯ Λ has sufficiently automorphisms (that iseasy to check). Choose orthogonal primitive idempotents b e , b e , b e in Λ such that ψ (1 k ⊗ b e i ) = e i for i ∈ { , , } . Do the analogous things for Γ to obtain ψ ′ , b f , b f and b f . The equality of thedecomposition matrices implies that the systems of orthogonal idempotents b e , b e , b e and b f , b f , b f are conjugate within A (as the entries of the decomposition matrix determine the ranks of thoseidempotents in each matrix ring summand of A ). We can therefore replace Λ by u Λu − for anappropriately chosen unit u ∈ A , and assume that b e i = b f i for i ∈ { , , } .The fact that e i ¯ Λe i = e i Z ( ¯ Λ ) e i implies that b e i Λ b e i = b e i Z ( Λ ) b e i (and the same for Γ ). Tosee this first note that for any order there is an embedding k ⊗ Z ( Λ ) ֒ → Z ( k ⊗ Λ ) . We have dim k k ⊗ Z ( Λ ) = rank O Z ( Λ ) = dim K Z ( A ) = 6+ l , which is equal to dim k ( Z ( k ⊗ Λ )) = dim k Z ( ¯ Λ ) ,and therefore the aforementioned embedding is in fact an isomorphism. Since we assume that Z ( Λ ) = Z ( Γ ) (as subsets of Z ( A ) ), we can conclude that b e i Λ b e i = b e i Z ( Λ ) b e i = b f i Z ( Γ ) b f i = b f i Γ b f i (24)for all i ∈ { , , } .Let i = j ∈ { , , } . We know that e i ¯ Λe j is generated by a single element as a left e i ¯ Λe i -module. Hence the same is true for b e i Λ b e j and b e i Γ b e j as left b e i Λ b e i -modules respectively b e i Γ b e i -modules. Let v ij respectively w ij denote generators for b e i Λ b e j respectively b e i Γ b e j .Since all decomposition numbers are zero or one, it follows that the K -vector space ε m b e i A b e j isat most one-dimensional for each central primitive idempotent ε m ∈ Z ( A ) and all i, j ∈ { , , } .Since ε b e i A b e j ∼ = K for all i, j ∈ { , , } , and because we can replace v ij by o · v ij for a unit o ∈ O × , we can ask that ε v ij be equal to π d ij · ε w ij for certain numbers d ij ∈ Z , where π denotesa (fixed) generator of the maximal ideal of O .Consider an element of the form u := l X i =1 3 X j =1 c ij · ε i b e j where the c ij are parameters in K × yet to be determined. Note that some of those parameters aresuperfluous since some of the products ε i b e j are zero (we include those superfluous parameters only7ecause it simplifies notation). We will try to choose the c ij in such a way that (1 − ε ) · u v ij u − =(1 − ε ) · w ij for all pairs ( i, j ) ∈ { (1 , , (2 , , (3 , } .We will take care of the fourth Wedderburn component later on, and therefore we set c = c = c = 1 . Note that for each Wedderburn-component except for the fourth there is at mostone pair ( i, j ) ∈ { (1 , , (2 , , (3 , } such that v ij and w ij projected to that component is non-zero, as all rows of the decomposition matrix (19) except for the fourth have at least one zeroin them. Seeing how ε m · u · v ij · u − = c mi c mj · ε m · v ij for each m ∈ { , . . . , l } , it followsthat we can choose the c mi such that ε m · u · v ij · u − = ε m · w ij for all = m ∈ { , . . . , l } and for all ( i, j ) ∈ { (1 , , (2 , , (3 , } (where the pair ( i, j ) is uniquely determined by m , as wejust discussed). Without loss of generality we will replace Λ by u Λu − . Hence, we now have (1 − ε ) · v ij = (1 − ε ) · w ij for all ( i, j ) ∈ { (1 , , (2 , , (3 , } .Now consider the product βδλ ∈ e ¯ Λe . This is non-zero, which implies that b e Λ b e Λ b e Λ b e π · Λ (25)On the other hand v · v · v generates this (left or right) b e Λ b e -module, since b e Λ b e Λ b e Λ b e = b e Λ b e v · b e Λ b e v · b e Λ b e v = b e Z ( Λ ) b e v · b e Z ( Λ ) b e v · b e Z ( Λ ) b e v = Z ( Λ ) · v v v The analogous statement holds for w · w · w . By construction we have v · v · v = π d + d + d · w · w · w (by looking at the decomposition matrix we see that this element hasa non-zero entry only in the fourth Wedderburn component, so this follows immediately from thedefinition of the d ij ). It follows that d + d + d = 0 , since if it were greater than zero, then v · v · v would lie in π · Λ , as w · w · w already lies in b e Γ b e = b e Λ b e ⊂ Λ . And byswapping the roles of Λ and Γ in this argument, we can also conclude that d + d + d cannotbe smaller than zero.Now we can use conjugation by the element u = (1 − ε ) + ε · ( b e + π d b e + π d + d b e ) By construction (1 − ε ) · u v ij u − = (1 − ε ) · v ij for all i, j , and we have already shown that (1 − ε ) · v ij = (1 − ε ) · w ij for all ( i, j ) ∈ { (1 , , (2 , , (3 , } . Moreover, ε · u v u − = ε w , ε · u v u − = ε w and ε · u v u − = π d + d + d · ε w = ε w . Hence we can concludethat u · v ij · u − = w ij for all ( i, j ) ∈ { (1 , , (2 , , (3 , } . Replace Λ by u Λu − .Now we have b e i Λ b e i = b e i Γ b e i for all i ∈ { , , } and b e i Λ b e j = b e i Γ b e j for all ( i, j ) ∈{ (1 , , (2 , , (3 , } . By the assumption that Λ and Γ are symmetric, and using Proposition4.1 it follows that b e i Λ b e j = b e i Γ b e j for all ( i, j ) ∈ { (2 , , (3 , , (1 , } as well. Therefore we get Λ = Γ , which finishes the proof.We should remark at this point that the condition “ Z ( Λ ) = Z ( Γ ) ” in the previous theoremis quite strong, and certainly stronger than merely asking that the centers should be isomorphic.Hence, in order to apply Theorem 4.2 to blocks of quaternion defect, we must first study to whichextent the perfect isometries between them provide us with information on the embedding of thecenter into the Wedderburn-decomposition of the block. That is what the next section is about. B ( O SL ( q )) for q ≡ In this section we will study self-isometries preserving projectivity and perfect self-isometries of B ( O SL ( q )) for q ≡ . For the most part we restrict our attention to the case n > ( n being the -valuation of q − ), but in the case n = 3 the main result of this section, Corollary5.6, is just the same as [7, Proposition 1.1]. Since later on we will have to deal with orders whichare not a priori known to be blocks of groups rings, we are going to formulate Proposition 5.1below in a slightly more general setting than needed for the purposes of this section.8 roposition 5.1 (Isometries preserving projectivity) . Let Λ be an O -order with split-semisimple K -span and decomposition matrix as given in (19) . We index the elements of Irr K ( K ⊗ Λ ) by thenumbers { , . . . , l } using the same ordering as for the rows of the decomposition matrix (19) .We write self-isometries of the Grothendieck group K ( K ⊗ Λ ) as signed permutations, using thefollowing convention for a signed cycle: ( i , . . . , i k ) denotes the permutation that sends | i j | to i j +1 for each j ∈ Z /k Z .Denote the projective indecomposable Λ -modules belonging to the first, second and third columnof (19) by P , P and P , respectively. Let ϕ : K ( K ⊗ Λ ) −→ K ( K ⊗ Λ ) be an isometry thatmaps the Z -lattice h [ K ⊗ P ] | P projective Λ -module i Z onto itself. Then ϕ ∈ h± id , (2 , , , ( − , − , − − , (2 , − , − − i · Sym( { , . . . , l } ) (26) Proof.
Clearly ϕ induces a permutation of ± Irr( K ⊗ Λ ) . If [ V ] ∈ Irr( K ⊗ Λ ) , then ( ϕ ([ P i ]) , [ V ]) =([ P i ] , ϕ − ([ V ])) ∈ {− , , } , where we made use of the fact that all decomposition numbers of Λ are . For each [ P i ] there is a [ V i ] ∈ Irr( K ⊗ Λ ) which occurs with multiplicity in [ P i ] andwith multiplicity in each [ P j ] with i = j . If we write ϕ ([ P i ]) = a i, [ P ] + a i, [ P ] + a i, [ P ] , then a i,j = ([ V j ] , ϕ ([ P i )]) = ( ϕ − ([ V j ]) , [ P i ]) ∈ {− , , } . In the same vein, for each j = j ′ ∈ { , , } wehave a [ V ] ∈ Irr( K ⊗ Λ ) such that ([ P l ] , [ V ]) = 1 if and only if l ∈ { j, j ′ } and ([ P l ] , [ V ]) = 0 for theunique l ∈ { , , }−{ j, j ′ } . Then a i,j + a i,j ′ = ([ V ] , ϕ ([ P i ])) = ( ϕ − ([ V ]) , [ P i ]) ∈ {− , , } . Hencethe only possibilities for ϕ ([ P i ]) are linear combinations of [ P ] , [ P ] and [ P ] with coefficients , and − , where either only one coefficient is non-zero, or one coefficient is equal to zero, one isequal to +1 and one is equal to − . Let [ P ] = a [ P ] + b [ P ] + c [ P ] . Then ([ P ] , [ P ]) = (cid:0) a b c (cid:1) · n − n − n − n − · abc = 4 a + (2 + 2 n − ) · ( b + c ) + 4 ab + 4 ac + 2 n − bc (27)By symmetry we only have to check the cases ( a, b, c ) = (1 , − , and ( a, b, c ) = (0 , , − . Weget ([ P ] , [ P ]) = 2 + 2 n − respectively ([ P ] , [ P ]) = 4 . As ( ϕ ([ P i ]) , ϕ ([ P i ])) = ([ P i ] , [ P i ]) is equal to if i = 1 and equal to n − if i ∈ { , } it follows that ± ([ P ] − [ P ]) and ± ([ P ] − [ P ]) aresuitable images for both [ P ] and [ P ] , and ± ([ P ] − [ P ]) is a suitable image for [ P ] .Now we list all possible images of the triple [ P ] , [ P ] and [ P ] which stabilize the Cartanmatrix, which means that they may be induced by an isometry ϕ , but we will still have to finda signed permutation on ± Irr K ( K ⊗ Λ ) that induces them. We give such a permutation in eachcase. ϕ ([ P ]) ϕ ([ P ]) ϕ ([ P ]) Signed permutation on { , . . . , l }± [ P ] ± [ P ] ± [ P ] ± id ± [ P ] ± [ P ] ± (2 , , ± ([ P ] − [ P ]) ± ([ P ] − [ P ]) ± (1 , − − , − id { ,..., l } ) ± ([ P ] − [ P ]) ± ([ P ] − [ P ]) ± (1 , − , − − id { ,..., l } ) ± ([ P ] − [ P ]) ± [ P ] ∓ ([ P ] − [ P ]) ± (2 , − , − − ∓ [ P ] ± ([ P ] − [ P ]) ± (1 , , − , − , − − id { ,..., l } ) ∓ ([ P ] − [ P ]) ± [ P ] ± ( − , − , , , − ± ([ P ] − [ P ]) ∓ [ P ] ± (1 , − , − − − id { ,..., l } ) (28)Note that the given signed permutations are unique up to composition with permutations whichfix [ P ] , [ P ] and [ P ] . That are precisely all the permutations that fix { , . . . , } point-wise. Thisproves the lemma (it is easy to verify that the elements given in (26) are generators for the groupwe just determined).Now we are interested in the question which of these self-isometries are actually perfect if wepick Λ = B ( O SL ( q )) . For the rest of the section we will fix some prime q with q ≡ andassume that K ⊗ B ( O SL ( q )) is split, which is equivalent to asking that K ⊇ Q ( ζ n − + ζ − n − ) ( n is the -valuation of q − ). 9 emark 5.2 (Character table) . We are going to need the character table of SL ( q ) , which canbe found in [1, Table 5.4], to check perfectness of characters. The table below is the part weare interested in, namely the characters which lie in the principal -block of SL ( q ) . We alreadyreplaced those character values for which there is a more explicit description due to our constraintson q (concretely: q = − q , θ ( ε ) = 1 , α ( ε ) = ε ). εI d( a ) d ′ ( ξ ) εu τ ε ∈ {± } a ∈ µ q − ξ ∈ µ q +1 ε, τ ∈ {± }| C SL ( q ) ( g ) || q ( q − q − q + 1 2 qo ( g ) o ( ε ) o ( a ) o ( ξ ) q · o ( ε )1 χ R ′ + ( θ ) χ ( q −
1) 0 − θ ( ξ ) ( − τ √− q ) R ′− ( θ ) χ ( q −
1) 0 − θ ( ξ ) ( − − τ √− q )St χ q − R + ( α ) χ ( q + 1) ε α ( a ) 0 ε (1 + τ √− q ) R − ( α ) χ ( q + 1) ε α ( a ) 0 ε (1 − τ √− q ) R ′ ( θ ) χ i ( q − · θ ( ε ) 0 − θ ( ξ ) − θ ( ξ ) − − θ ( ε ) (29) The top row and the leftmost column contains the names for the conjugacy classes respectivelycharacters used in [1]. The symbol µ j denotes the group of j -th roots of unity in the algebraicclosure of F q (i. e., a cyclic group of order j if gcd( q, j ) = 1 ). The symbol θ denotes the uniqueordinary character of order two of the group µ q +1 , and the symbol α denotes the unique ordinarycharacter of order two of the group µ q − . The parameter θ in the last row ranges over all charactersof order i of the group µ q +1 for i > , although different θ may still yield the same character R ′ ( θ ) . Namely, we have R ′ ( θ ) = R ′ ( θ ) for θ = θ if and only if θ is the complex conjugate of θ . We should recall that, due to our choice of q , the -valuation of q − is one, and the -valuationof q +1 is n − . In particular θ ranges over (2 n − − / n − − different values. The followingwill be useful later on:1. Since q − has -valuation one, the -valuation of the order of an element a ∈ µ q − is either or . We have α ( a ) = (cid:26) − if a has even order if a has odd order (30)
2. Since q + 1 has -valuation n − , the -valuation of the order of an element ξ ∈ µ q +1 is atmost n − . We have θ ( ξ ) = (cid:26) − if the -valuation of o ( ξ ) is n − otherwise (31) Proposition 5.3.
1. The involution g g − on O SL ( q ) induces a self-isometry (2 , , of the principal block. This self-isometry is also induced by an automorphism (see Proposition6.3 below).2. Alvis-Curtis duality (see [1, Chapter 8.4]) swaps the trivial character and the Steinbergcharacter, which correspond to the first and fourth row of the decomposition matrix (19) .By inspecting the rows of (28) we see that the signed permutation it induces on irre-ducible characters is either in (1 , − − , − id { ,..., l } ) · Sym( { , . . . , l } ) or in (1 , − , − − id { ,..., l } ) · Sym( { , . . . , l } ) . Z ( B ( O SL ( q ))) hasautomorphisms which induce the corresponding (unsigned) permutations on the Wedderburn com-ponents (or, equivalently, the primitive idempotents of Z ( K ⊗ B ( O SL ( q ))) ).It is now quite natural to ask whether the self-isometry inducing the permutation (2 , − , − − is also perfect, since then all three generating (signed) permutations in equa-tion (26) would come from perfect isometries. It turns out that this is false. But, as we will seein a moment, we can find an element σ ∈ Sym( { , . . . , l } ) such that (2 , − , − − ◦ σ is perfect. Nevertheless, we will start by looking at the permutation (2 , − , − − , and thecorresponding character of G × G : ι ( g, h ) := − χ ( g ) χ ( h ) − χ ( g ) χ ( h ) + χ ( g ) χ ( h ) + χ ( g ) χ ( h ) + χ ( g ) χ ( h ) − χ ( g ) χ ( h ) + P li =7 χ i ( g ) χ i ( h ) (32)Since perfect characters form an additive group, and we know that the character of G × G repre-senting the identity permutation is perfect, we may just as well study the character µ ( g, h ) = (cid:16)P li =1 χ i ( g ) χ i ( h ) (cid:17) − ι ( g, h )= χ ( g ) χ ( h ) + χ ( g ) χ ( h ) + χ ( g ) χ ( h ) + χ ( g ) χ ( h ) + χ ( g ) χ ( h )+ χ ( g ) χ ( h ) − χ ( g ) χ ( h ) − χ ( g ) χ ( h ) + 2 χ ( g ) χ ( h ) (33)Now we will look at how exactly µ ( g, h ) fails to be perfect, and how we can rectify that. Unfortu-nately, checking that a character is perfect is usually a somewhat tedious computation, and whatfollows is no exception. Lemma 5.4.
The character µ ( g, h ) defined in (33) has the following properties:1. If g has even order, and h has odd order, then µ ( g, h ) = µ ( h, g ) = 0 .2. If g and h both have odd order, then µ ( g, h ) | C SL ( q ) ( g ) | ∈ O and µ ( g, h ) | C SL ( q ) ( h ) | ∈ O (34)
3. If g and h both have even order, then µ ( g, h ) | C SL ( q ) ( g ) | ∈ O and µ ( g, h ) | C SL ( q ) ( h ) | ∈ O (35) except possibly if g and h belong to the conjugacy classes labeled by d ′ ( ξ ) and d ′ ( ξ ) in thecharacter table (29) , and ξ and ξ both have order divisible by n − .Proof. Note that only the characters χ through χ are involved in µ , and the entries in the rowsof the character table (29) depend only slightly on the concrete parameters ε , a , ξ , ε and τ thatspecify the actual conjugacy class (note: there are in fact two different parameters called “ ε ”).In fact, we only need to know ε for the first column, α ( a ) for the second, θ ( ξ ) for the thirdand ε as well as τ for the fourth. If we are looking at the character values of an element of oddorder, then ε = 1 , α ( a ) = 1 , θ ( ξ ) = 1 and ε = 1 respectively (only τ still depends on theactual conjugacy class). If we are looking at the character values of an element of even order,then ε = − , α ( a ) = − , θ ( ξ ) still depends on the order of ξ and ε = − (again, τ depends onthe actual conjugacy class). Here we used the last part of Remark 5.2. Hence we can write thevalues of µ ( g, h ) for g ∈ { ε I , d( a ) , d ′ ( ξ ) , ε u τ } of odd order and h ∈ { ε I , d( a ) , d ′ ( ξ ) , ε u τ } of even order into a × -matrix, whose entries will only depend on τ , θ ( ξ ) , and τ . One cancompute this × -matrix by hand or using a computer (which seems more sensible seeing howthis is a rather lengthy computation), and one obtains the zero matrix, which proves the first partof our assertion. 11n the same vein we can put the values of µ ( g,h ) | C SL2( q ) ( g ) | for g ∈ { ε I , d( a ) , d ′ ( ξ ) , ε u τ } and h ∈ { ε I , d( a ) , d ′ ( ξ ) , ε u τ } both of odd order into a × -matrix. Note that | C SL ( q ) ( h ) | and | C SL ( q ) ( g ) | and do not depend on ε , a , etc. at all. We obtain the following matrix: ε I d( a ) d ′ ( ξ ) ε u τ ε I q +1( q − q q − q − − τ √− q ( q − q d( a ) 2 q +1 q − q − − − τ √− qq − d ′ ( ξ ) 0 0 0 0 ε u τ − ( τ √− q − q )2 q − − τ √− qq ( τ √− q − τ √− q − q (36)To see that all of the entries lie in O it suffices to know the following: divides q − and q + 1 , q − divides both and q + 1 , and both and q − divide ± √− q . The latter is owed to thefact that ±√− q is integral due to our choice of q . Since µ ( g, h ) = µ ( h, g ) for all g and h the proofof the second part of our assertion is complete.To finish the proof let us look at the values of µ ( g,h ) | C SL2( q ) ( g ) | for g ∈ { ε I , d( a ) , d ′ ( ξ ) , ε u τ } and h ∈ { ε I , d( a ) , d ′ ( ξ ) , ε u τ } both of even order. We get almost the same matrix as above: ε I d( a ) d ′ ( ξ ) ε u τ ε I q +1( q − q q − q − − τ √− q ( q − q d( a ) 2 q +1 q − q − − − τ √− qq − d ′ ( ξ ) 0 0 θ ( ξ ) θ ( ξ ) − θ ( ξ ) − θ ( ξ )+1) q +1 ε u τ − ( τ √− q − q )2 q − − τ √− qq ( τ √− q − τ √− q − q (37)The only entry that is different is the (3 , -entry, which will in general not lie in O . However,this entry is zero unless both θ ( ξ ) and θ ( ξ ) are equal to − , which, according to (31), happensonly if ξ and ξ both have order divisible by n − . That finishes the proof. Lemma 5.5.
Let ι ( g, h ) := X θ ( R ′ ( θ )( g ) R ′ ( θ )( h ) − R ′ ( θ )( g ) R ′ ( θ · θ )( h )) (38) using the notation for the irreducible characters from (29) . The summation index θ ranges overthe same n − − characters of µ q +1 as in the character table.Then ι − ι is a perfect isometry, and clearly the induced signed permutation on irreduciblecharacters is (2 , − , − − ◦ σ for an element σ ∈ Sym( { , . . . , l } ) of order two.Proof. It is clear by definition that ι − ι is an isometry, we only need to check that it is perfect.To do that we may just as well prove that µ + ι is perfect.Let us first note that ι ( g, h ) = 0 if and only if both g and h have order divisible by n − . Itis clear that if h has order not divisible by n − , then ι ( g, h ) = 0 because R ( θ )( h ) = R ( θ · θ )( h ) which means that each individual summand in the definition of ι is zero. The fact that ι ( g, h ) is zero whenever g has order not divisible by n − can be seen by rearranging the sum. To beprecise, we apply an index shift to obtain the following equality: ι ( g, h ) = P θ R ′ ( θ )( g ) R ′ ( θ )( h ) − P θ R ′ ( θ )( g ) R ′ ( θ · θ )( h )= P θ ( R ′ ( θ )( g ) R ′ ( θ )( h ) − P θ R ′ ( θ · θ )( g ) R ′ ( θ )( h ))= ι ( h, g ) (39)12e conclude that µ + ι satisfies all of the conditions 1 through 3 from the statement of Lemma5.4 just as µ does (because the relevant values of µ + ι are actually equal to those of µ ).It remains to prove that if g and h both have order divisible by n − (which we will assumefrom here on out), then µ ( g, h ) + ι ( g, h ) | C SL ( q ) ( g ) | = µ ( g, h ) + ι ( g, h ) q + 1 ! ∈ O (40)To show this we can simply evaluate µ and ι : µ ( g, h ) = 8 (from (37)), and since θ ( h ) = − theformula for ι ( g, h ) simplifies to the following: ι ( g, h ) = 2 · X θ ( θ ( g ) + θ ( g ) − )( θ ( h ) + θ ( h ) − ) (41)The latter is technically abuse of notation, since θ is not really defined on g (or h , for that matter),but rather on ξ ∈ µ q +1 , where d ′ ( ξ ) is the “standard” representative (from (29)) of the conjugacyclass that g (or h ) is an element of.Recall that if we take all characters of µ q +1 of two-power order except those of order ,and partition these characters into sets of cardinality two containing a character and its complexconjugate, then the summation index θ in (41) ranges over representatives of those sets. Since thevalues of R ′ ( θ ) and R ′ (¯ θ ) are actually equal, we might as well let θ range over all characters oftwo-power order > , and then divide the resulting sum by two. Using the fact that the charactersof µ q +1 of two-power order are obtained by mapping g respectively h to all possible n − -st rootsof unity, we get the following: ι ( g, h ) = P n − i =2 P σ ∈ Gal( Q ( ζ i ) / Q ) ( ζ σ i + ( ζ σ i ) − )(( ζ z i ) σ + (( ζ z i ) σ ) − )= P n − i =2 P σ ∈ Gal( Q ( ζ i ) / Q ) ( ζ z i + ζ − z i + ζ − z i + ζ − − z i ) σ = . . . (continued below) (42)Here ζ i denotes a primitive i -th root of unity, and z ∈ Z is chosen such that θ ( h ) = θ ( g z ) forall θ of two-power order (i.e. the two-part of h is conjugate to the -part of g z ; in particular, z isodd). Note that for any j ∈ Z we have n − X i =0 X σ ∈ Gal( Q ( ζ i ) / Q ) ( ζ j i ) σ = (cid:26) if j n − n − if j ≡ n − (43)This follows for instance from column orthogonality in the character table of the cyclic group oforder n − . This allows us to simplify (42) as follows (note: we need to pay attention to thedifferent ranges for the summation index i in (42) and (43)): . . . = N ( z ) − P i =0 P σ ∈ Gal( Q ( ζ i ) / Q ) ( ζ z i + ζ − z i + ζ − z i + ζ − − z i ) σz odd = N ( z ) − (44)Here N ( z ) denotes n − · ( δ z, + δ − z, + δ − z, + δ − − z, ) (where δ a,b = 1 if a ≡ b mod 2 n − and δ a,b = 0 otherwise). We can now conclude that µ ( g, h ) + ι ( g, h ) = N ( z ) (45)which is divisible by n − , and therefore also by q + 1 (in O ). Corollary 5.6.
Let ϕ : K ( K ⊗ B ( O SL ( q ))) −→ K ( K ⊗ B ( O SL ( q ))) (46)13 e an isometry that maps the Z -lattice h [ K ⊗ P ] | P projective B ( O SL ( q )) -module i Z onto itself.Then ϕ can be written as ϕ ◦ ϕ , where ϕ ∈ Sym( { , . . . , l } ) (we identify isometries andsigned permutations as in Proposition 5.1) and ϕ is a perfect isometry.Proof. For n = 3 the assertion has been proved in [7, Proposition 1.1]. For n > the claim followsfrom Proposition 5.1, which yields a factorization of ϕ as ϕ ◦ ϕ with ϕ ∈ Sym( { , . . . , l } ) .Proposition 5.3 and Lemma 5.5 then show that ϕ composed with an appropriate element of Sym( { , . . . , l } ) is a perfect isometry. Throughout this section we will assume the following:1. Λ is a symmetric O -order with split semisimple K -span.2. k ⊗ Λ is basic.3. The decomposition matrix of Λ is the one given in (19), up to permutation of rows andcolumns.4. We fix a prime q ≡ and an isomorphism (which we assume to exist) Φ : Z ( Γ ) ∼ −→ Z ( Λ ) (47)where Γ is the basic order of B ( O SL ( q )) . We assume moreover that there is an isometry b Φ : K ( K ⊗ Γ ) −→ K ( K ⊗ Λ ) (48)with the following properties:(a) if ε ∈ Z ( K ⊗ Γ ) is a central primitive idempotent, V is the simple K ⊗ Γ -moduleassociated with ε , and W is the simple K ⊗ Λ -module associated with the centralprimitive idempotent (id K ⊗ Φ )( ε ) ∈ Z ( K ⊗ Λ ) , then b Φ ([ V ]) = ± [ W ] .(b) b Φ ( h [ K ⊗ P ] | P is a projective Γ -module i Z ) = h [ K ⊗ Q ] | Q is a projective Λ -module i Z Note that since Γ is the basic order of B ( O SL ( q )) we may identify K ( K ⊗ Γ ) with K ( K ⊗ B ( O SL ( q ))) . The following diagram visualizes the situation we are looking at: Z ( Γ ) (cid:15) (cid:15) ∼ Φ (cid:18) r wed Γ | Z ( Γ ) $ $ ❏❏❏❏❏❏❏❏❏❏❏ Z ( Λ ) (cid:31) (cid:127) wed Λ | Z ( Λ ) / / (cid:127) _ (cid:15) (cid:15) Z ( A ) = K ⊕ K ⊕ K ⊕ K ⊕ K ⊕ K ⊕ l copies z }| { K ⊕ . . . ⊕ K (cid:127) _ (cid:15) (cid:15) Λ (cid:31) (cid:127) wed Λ / / A = K ⊕ K ⊕ K ⊕ K × ⊕ K × ⊕ K × ⊕ K × ⊕ . . . ⊕ K × | {z } l copies ✚❩ (cid:9)(cid:9) (49)where the maps wed Λ : Λ −→ A and wed Γ : Γ −→ A denote Wedderburn embeddings whoseimages have the decomposition matrix (19) with the order of the rows being fixed. The potentialnon-commutativity of the top part of this diagram is the main obstacle in showing there is onlyone Morita equivalence class of quaternion blocks over O reducing to the Morita equivalence classof B ( k SL ( q )) . We will now investigate this non-commutativity in greater detail.14 emark 6.1.
1. Let us denote the primitive idempotents of Z ( A ) by ε , . . . , ε l . A K -algebraautomorphism of Z ( A ) is given by a permutation of these idempotents, i. e. each automor-phism is of the form α σ for some σ ∈ Sym( { , . . . , l } ) , where α σ ( ε i ) = ε σ ( i ) (50) Since A is split the permutation σ determines the automorphism uniquely.2. A self-isometry of K ( A ) determines an automorphism of Z ( A ) . To be more specific, we canconstruct the corresponding permutation σ by forgetting the signs in the signed permutationacting on ± Irr( A ) . Similarly, an isometry between K ( K ⊗ Γ ) and K ( A ) determines anembedding Z ( Γ ) −→ Z ( A ) .3. Two Wedderburn embeddings of a commutative O -order into Z ( A ) differ only by an auto-morphism of Z ( A ) . Proposition 6.2.
There is a permutation σ ∈ Sym( { , . . . , l } ) such that wed Λ ( Z ( Λ )) = α σ (wed Γ ( Z ( Γ ))) (51)Remark : Note that α σ extends to an automorphism of A , and α − σ ◦ wed Λ can be regarded asanother Wedderburn embedding of Λ , whose image has the same decomposition matrix as theimage of wed Λ (where the order of the rows is fixed).Proof. We may (canonically) identify K ( A ) with Z l (equipped with the usual euclidean scalarproduct), since the Wedderburn components of A are ordered. With this identification the iso-morphisms id K ⊗ wed Λ : K ⊗ Λ −→ A and id K ⊗ wed Γ : K ⊗ Γ −→ A come from isometries d wed Λ : K ( K ⊗ Λ ) −→ Z l respectively d wed Γ : K ( K ⊗ Γ ) −→ Z l (52)that send the equivalence classes of simple K ⊗ Λ - respectively K ⊗ Γ -modules to the stan-dard basis of Z l . Due to our assumption on the decomposition matrices of the images of wed Λ and wed Γ , these isometries send the Z -lattice h [ P ] | P projective Λ -module i Z respectively h [ P ] | P projective Γ -module i Z onto the Z -sublattice of Z l generated by the columns of thedecomposition matrix (19).Now let us compare the maps wed Γ | Z ( Γ ) and wed Λ | Z ( Λ ) ◦ Φ . They are induced by the isometries d wed Γ and d wed Λ ◦ b Φ , both of which send h [ K ⊗ P ] | P a projective Γ -module i Z to the Z -sublattice of Z l which is generated by the columns of the decomposition matrix (here we use the assumptionswe made on b Φ ). By Corollary 5.6 we can conclude that d wed Γ and d wed Λ ◦ b Φ differ by a self-isometry of K ( K ⊗ Γ ) which is the composition of a perfect self-isometry of Γ and an element of Sym( { , . . . , l } ) .Assume γ is an automorphism of Z ( Γ ) which comes from a self-isometry b γ of K ( K ⊗ Γ ) .Then wed Γ ◦ γ comes from the isometry d wed Γ ◦ b γ ., In light of the previous paragraph, we canchoose a b γ = b γ ◦ b γ , with b γ ∈ Sym( { , . . . , l } ) and b γ being a perfect isometry, such that d wed Λ ◦ b Φ = d wed Γ ◦ b γ . Now we can pull b γ through d wed Γ , to obtain b γ ′ ◦ d wed Γ ◦ b γ for a certain self-isometry b γ ′ of K ( A ) . Note that there is an automorphism of Z ( A ) induced by the self-isometry b γ ′ , and this automorphism is equal to α σ with σ ∈ · Sym( { , . . . , } ) . By Remark 6.1 (2) wecan conclude that wed Λ | Z ( Λ ) ◦ Φ = α σ ◦ wed Γ | Z ( Γ ) ◦ γ (53)where γ is the automorphism of Z ( Γ ) induced by the perfect isometry b γ . Now we can simplytake the images of the maps on both sides of this equation, and our claim immediately follows. Proposition 6.3.
Assume q ≡ ± . There is a non-trivial outer automorphism of SL ( q ) induced by conjugation with an appropriately chosen element of GL ( q ) . This outer automorphisminduces a non-trivial permutation of the simple B ( k SL ( q )) -modules. roof. By looking at the parametrization of the conjugacy classes of SL ( q ) given in [1], we seethat there are two conjugacy classes of unipotent matrices in SL ( q ) , while there is only one suchclass in GL ( q ) . Hence GL ( q ) acts non-trivially on the conjugacy classes of SL ( q ) , and thereforealso on its (absolutely) irreducible characters. It can be seen by inspection of the charactertable [1, Table 5.4] (see also [1, Excercise 4.3]) that the unique non-trivial outer automorphism of SL ( q ) induced by an element of GL ( q ) swaps the characters R + ( α ) and R − ( α ) as well as thecharacters R ′ + ( θ ) and R ′− ( θ ) . The reduction of the characters R ′ + ( θ ) and R ′− ( θ ) to -regularconjugacy classes gives the Brauer characters belonging to the simple B ( k SL ( q )) -modules St k + and St k − (this is proven in [1, Section 9.4.4]). This shows that twisting by α swaps the two simplemodules St k + and St k − .Note that an automorphism of B ( O SL ( q )) gives rise to an automorphism of the basic order of B ( O SL ( q )) inducing the same permutation on isomorphism classes of simple modules. Thereforewe get the following: Corollary 6.4.
Assume q ≡ ± . By n we denote the -valuation of the order of SL ( q ) ,and we assume n > . Again, let Γ be the basic algebra of B ( O SL ( q )) and let S , S and S denote its simple modules. Then for every automorphism α of k ⊗ Γ there exists an automorphism b α of Γ such that (id k ⊗ b α ) ◦ α − fixes all simple modules, that is, S (id k ⊗ b α ) ◦ α − i ∼ = S i for each i ∈ { , , } Note: S (id k ⊗ b α ) ◦ α − i denotes the module obtained from S i by letting k ⊗ Γ act on it through theautomorphism (id k ⊗ b α ) ◦ α − . Later on we will also use the analogous notation for bimodules.Proof. We know that the Cartan matrix of B (SL ( q )) looks as follows: n − n − n − n − | {z } if q ≡ or n n − n − n − n − n − n − n − n − | {z } if q ≡ (54)If α is an automorphism of k ⊗ Γ , then the dimension of the endomorphism ring of the projectivecover of S αi is the same as the dimension of the endomorphism ring of the projective cover of S i (foreach i ). If either n > or q ≡ , then the diagonal entries of the above Cartan matricesare not all equal, which implies that α needs to fix one isomorphism class of simple modules, andit might swap the other two. Hence, if α is non-trivial, then it necessarily needs to induce thesame permutation on simple modules as the automorphism of Γ coming from Proposition 6.3.The case q ≡ and n = 3 is special, since then the Cartan matrix imposes no restric-tion on the permutation of the simple modules induced by α . However, in that case, [7, TheoremA] implies that Γ ∼ = O ˜ A = O Q ⋊ C , and [7, Lemma 1.2] implies that this O -order has an auto-morphism which induces a permutation of order three on isomorphism classes of simple modules.It follows that this automorphism together with the automorphism from Proposition 6.3 generatesa full symmetric group on three points, which implies our assertion. In this section we are going to use a technique reminiscent of the one used in [3] to get a theoremsimilar to Theorem 4.2 for arbitrary blocks of quaternion defect. Note that, technically, Theorem4.2 cannot be applied to any block of quaternion type yet, but only to their basic algebras if theyhappen to be isomorphic to Q (3 K ) c . The fact that Theorem 4.2 remains valid if one replaces thealgebra ¯ Λ = Q (3 K ) c by an algebra Morita-equivalent to it would be a side-note at best. The mainidea in [3] was that, up to technicalities, one can in fact replace ¯ Λ = Q (3 K ) c by an algebra derivedequivalent to it, instead of just Morita-equivalent. In our case the technical side of this argumentis in fact much simpler than in [3], and hence we will only have to use well-known facts aboutderived equivalences. 16 efinition 7.1 (Admissible lifts of quaternion blocks) . We call an O -order Λ admissible if thefollowing three conditions hold:1. K ⊗ Λ is split semisimple2. Λ is symmetric3. For some prime q there is an isometry b Φ : K ( K ⊗ Λ ) −→ K ( K ⊗ B ( O SL ( q ))) andan isomorphism Φ : Z ( Λ ) −→ Z ( B ( O SL ( q ))) which satisfy the assumptions made at thebeginning of section 6.If ¯ Λ is a finite-dimensional k -algebra, then an admissible lift is an admissible O -order Λ with k ⊗ Λ ∼ = ¯ Λ Lemma 7.2 (Admissible lifts of Q (3 K ) c ) . Let Λ be an admissible lift of Q (3 K ) c ( c = 2 n − , n arbitrary). Assume that K ⊗ B ( O SL ( q )) is split, where q ≡ is a prime such that q + 1 has -valuation n − . Then Λ is isomorphic to the basic order of B ( O SL ( q )) .Proof. Let Γ be the basic order of B ( O SL ( q )) . We need to show that Λ and Γ are isomorphic.Proposition 2.5 and Proposition 3.3 imply that both Λ and Γ have decomposition numbers .Now Proposition 3.2 implies that the decomposition matrices of Λ and Γ are both equal to theone given in (19), up to permutation of rows and columns. Let A := K ⊕ K ⊕ K ⊕ K × ⊕ K × ⊕ K ⊗ ⊕ l M i =7 K × (55)From the decomposition matrices of Λ and Γ we know that this algebra A is isomorphic to both K ⊗ Λ and K ⊗ Γ . We can choose Wedderburn embeddings wed Λ : Λ −→ A and wed Γ : Γ −→ A such that the decomposition matrices of their images are both equal to (19) (the order of therows now being fixed). Moreover, Proposition 6.2 implies that there is an automorphism α of A permuting the Wedderburn components , . . . , l such that α (wed Λ ( Z ( Λ ))) = wed Γ ( Z ( Γ )) .Note that wed Λ ( Z ( Λ )) = Z (wed Λ ( Λ )) (and the same for Γ ). Now we may apply Theorem 4.2 tothe O -orders α (wed Λ ( Λ )) and wed Γ ( Γ ) . It follows that α (wed Λ ( Λ )) and wed Γ ( Γ ) are conjugatein A , which implies that Λ and Γ are isomorphic.At this point we have to look at the other two Morita equivalence classes of -blocks withdefect group Q n . These are the algebras Q (3 A ) c and Q (3 B ) c from the appendix of [4] (note:we parametrize these algebras as in [6], using only a single parameter “ c ”, as that is the onlyundetermined parameter in the context of blocks). The article [6] gives us an explicit (one-sided)two-term tilting complex in K b ( proj − Q (3 A ) c ) with endomorphism ring Q (3 B ) c and an explicittwo-term tilting complex in K b ( proj − Q (3 A ) c ) with endomorphism ring Q (3 K ) c (the algebra wehave been looking at exclusively so far).We will need a few well-known results on derived equivalences in order to get a version ofLemma 7.2 for the algebras Q (3 A ) c and Q (3 B ) c . The first one is a theorem of Rickard which tellsus that a derived equivalences between two k -algebras give rise to derived equivalences betweentwo O -orders reducing to these respective k -algebras. The caveat of this is that only one of thetwo O -orders can be chosen freely, while the other one is then determined up to isomorphism bythis choice. Note that while our notation is mostly standard, there is one peculiarity that may beworth pointing out: we consider one-sided tilting complexes as complexes of right modules, and theendomorphism ring has the usual composition as its multiplication (that is, the endomorphism ringof a module acts on the module from the left). With this convention, a ring is derived equivalentto the endomorphism ring of a tilting complex, rather than the opposite ring thereof. Theorem 7.3 (see [13, Theorem 3.3]) . If Λ is an O -order and T ∈ K b ( proj − k ⊗ Λ ) is a tiltingcomplex, then there is a tilting complex b T ∈ K b ( proj − Λ ) (unique up to isomorphism) with k ⊗ b T ∼ = T . Moreover, End D b ( Λ ) ( b T ) is an O -order, and k ⊗ End D b ( Λ ) ( b T ) ∼ = End D b ( k ⊗ Λ ) ( T ) .
17 second fact we will need is that two-term tilting complexes are determined by their terms.
Theorem 7.4 (see [8, Corollary 8]) . Let A be a k -algebra and let P and P be projective A -modules. Then there is at most one tilting complex of the form −→ P −→ P −→ , up toisomorphism in K b ( proj − A ) . On top of that we are going to use some facts from [12], [14] and [9], which are formulated for an R -algebra A which is projective as an R -module, where R is an arbitrary commutative ring. Thesefacts can therefore be applied to finite dimensional k -algebras and O -orders equally. Namely, ifwe have a one-sided tilting complex T ∈ K b ( proj − A ) whose endomorphism ring is isomorphic to B , then there is a two-sided tilting complex X ∈ D b ( mod − B op ⊗ R A ) whose restriction to A isisomorphic to T in the derived category (see [12, Corollary 3.5]). Moreover, by [14, Lemma 2.2]such an X can be chosen in such a way that the restriction of each term of X to both A and B is projective (without necessarily being projective as a B - A -bimodule). An equivalence between D b ( B ) and D b ( A ) is then afforded by the functor − ⊗ L B X . If we choose X is such a way that allits terms are projective as B -modules, then we may replace the left derived tensor product by theordinary tensor product of complexes. We denote the inverse of X by X ∨ . We have X ⊗ L A X ∨ ∼ = B in D b ( B op ⊗ R B ) and X ∨ ⊗ L B X ∼ = A in D b ( A op ⊗ R A ) . For symmetric algebras A the complex X ∨ can be computed as Hom R ( X, R ) (see [9, Section 9.2.2]).In the case of self-injective algebras it is fairly easy to check whether a derived equivalenceis actually a Morita-equivalence. A one-sided tilting complex T over a self-injective algebra A isalways isomorphic (in D b ( A ) ) to a tilting complex whose highest and lowest degree non-zero termsare in the same degree as its highest and lowest degree non-zero homologies (as both epimorphismsonto projectives and embeddings of projectives split in this case). In particular, a one-sided tiltingcomplex which has non-zero homology only in a single degree is isomorphic to the stalk complexof a projective module (which has to be a progenerator), and its endomorphism ring in D b ( A ) is isomorphic to the endomorphism ring of that projective module. It follows that A is Moritaequivalent to End D b ( A ) ( T ) . If there is a two-sided tilting complex for two algebras which hashomology concentrated in a single degree, then these algebras are Morita-equivalent because therestriction of said tilting complex to either side is isomorphic to a one-sided tilting complex, andisomorphisms in D b ( A ) preserve homology (pretty much by definition).We should also note that if A is a symmetric k -algebra, and B is the endomorphism ring ofa two-term tilting complex over A , then A is also the endomorphism ring of a two-term tiltingcomplex over B . This follows simply from the fact that X ∨ can be computed as Hom k ( X, k ) , andthe fact that we can choose a one-sided tilting complex in such a way that its non-zero terms areconcentrated between the highest and lowest degree non-zero homology.If we have a one-sided tilting complex T over a symmetric O -order Λ , then its endomorphismring Γ is an O -order by [15], and by [12, Theorem 2.1 and Corollary 2.2] the fact that Γ is an O -order implies that k ⊗ T is a tilting complex over k ⊗ Λ with endomorphism ring k ⊗ Γ . If k ⊗ T has homology concentrated in a single degree then it is isomorphic to the stalk complexof a progenerator P in mod − k ⊗ Λ . We know that there is a progenerator b P of mod − Λ with k ⊗ b P ∼ = P . Hence the stalk complex associated with b P is a tilting complex over Λ reducing tothe stalk complex associated with P . By Theorem 7.3 such a complex is unique up to (quasi-)isomorphism, and therefore T must be isomorphic to the stalk complex associated with b P , whichmeans that Γ is Morita-equivalent to Λ .Assuming A is a symmetric R -algebra we can also give an explicit description of two-sidedtilting complexes with homology concentrated in a single degree. Namely, such a complex X isquasi-isomorphic to the stalk-complex of its non-zero homology, which we will denote by M . Sincewe also know that its restriction to either side is quasi-isomorphic to a one-sided tilting complex,which under the assumptions made is quasi-isomorphic to the stalk complex of a progenerator,it follows that M is projective as a left and as a right module. Of course we can do the samefor X ∨ , which must be quasi-isomorphic to some A - B -bimodule M ∨ , also projective from the leftand from the right. Due to projectivity it follows that we do not have to bother with the derivedtensor product (as discussed above), and we can conclude that M ⊗ A M ∨ ∼ = B as a B - B -bimoduleand M ∨ ⊗ B M ∼ = A as an A - A -bimodule. That is, M is an invertible bimodule. In the case where18 is equal to B and A is a basic k -algebra or a basic O -order we can go even further: in that case M ∼ = A α for some automorphism α of A . Proposition 7.5.
Let Λ be an admissible O -order, and let T ∈ K b ( proj − Λ ) be a tilting complex.Then Γ := End D b ( Λ ) ( T ) is an admissible O -order.Proof. Since Λ is symmetric, Γ is both an O -order and symmetric (see [15]). Moreover, K ⊗ T is a tilting complex over K ⊗ Λ with endomorphism ring K ⊗ Γ (this is elementary), and since K ⊗ Λ is assumed to be split semisimple, so is K ⊗ Γ (since two semisimple algebras are derivedequivalent if and only if they are Morita equivalent).It remains to find an isometry b Ψ : K ( K ⊗ Γ ) −→ K ( K ⊗ Λ ) mapping the sublattice of K ( K ⊗ Γ ) generated by K -spans of projective Γ -modules onto the analogously defined sublatticeof K ( K ⊗ Λ ) . Moreover there should be an isomorphism Ψ : Z ( Γ ) −→ Z ( Λ ) whose K -linearextension maps a primitive idempotent ε V in Z ( K ⊗ Γ ) to ε W , the primitive idempotent in Z ( K ⊗ Λ ) that belongs to [ W ] = ± b Ψ ([ V ]) . Once we have found these maps, it is immediatethat the composition of b Ψ respectively Ψ with the maps b Φ respectively Φ from the definition ofadmissibility (of Λ ) yields the maps needed for Γ to be admissible. First let us choose a two-sidedtilting complex X with terms that are projective as left Γ -modules and as right Λ -modules suchthat − ⊗ Γ X affords an equivalence between D b ( mod − Γ ) and D b ( mod − Λ ) . We choose b Ψ to bethe induced map from K ( K ⊗ Γ ) to K ( K ⊗ Λ ) , that is, [ V ] [ V ⊗ K ⊗ Γ ( K ⊗ X )] . This map b Ψ isan isometry since K ⊗ X is a two-sided tilting complex, and it maps the sublattice of K ( K ⊗ Λ ) spanned by the projective modules into the sublattice of K ( K ⊗ Γ ) spanned by projectives. Thereason for the latter is simply that if P is projective, then all terms of P ⊗ Γ X are projective, andthe homomorphism between the Grothendieck groups is defined by applying − ⊗ Γ X and thentaking the alternating sum of the terms. Since the same argument applies to − ⊗ K ⊗ Λ ( K ⊗ X ∨ ) ,it also follows that b Ψ maps the sublattice generated by projective Γ -modules surjectively onto thesublattice generated by projective Λ -modules.To get the required isomorphism between the centers we first note that it is well known that thecenters of derived equivalent algebras are isomorphic. Concretely, an isomorphism Ψ : Z ( Γ ) −→ Z ( Λ ) can be obtained by identifying both Z ( Γ ) and Z ( Λ ) with the ring of endomorphisms of X (in D b ( Λ ⊗ Γ op ) ). That means in particular that x · Ψ ( z ) = z · x for all x ∈ H i ( X ) ( i arbitrary)and all z ∈ Z ( Γ ) . This implies that if an idempotent ε ∈ Z ( K ⊗ Γ ) acts non-trivially on a simple K ⊗ Γ -module V , then the image of ε under the K -linear extension of Ψ acts non-trivially on V ⊗ K ⊗ Γ ( K ⊗ X ) , which by definition becomes ± b Ψ ([ V ]) in the Grothendieck group. This showsthat b Ψ and Ψ have the required properties. Proposition 7.6.
Assume that ¯ Λ is a k -algebra and Λ is an admissible lift of ¯ Λ such that thefollowing hold:1. If Λ is an arbitrary admissible lift of ¯ Λ , then Λ ∼ = Λ .2. Every automorphism of K ( k ⊗ Λ ) which is induced by some element of Aut k ( k ⊗ Λ ) isalso induced by an element of Aut O ( Λ ) .Then the following holds for every basic k -algebra ¯ Λ ′ which is derived equivalent to ¯ Λ by meansof a two-term tilting complex: There is an admissible lift Λ ′ of ¯ Λ ′ such that if Λ ′ is an arbitraryadmissible lift of ¯ Λ ′ , then Λ ′ ∼ = Λ ′ .Proof. Let T be a two-term tilting complex over ¯ Λ ′ with End D b ( ¯ Λ ′ ) ( T ) ∼ = ¯ Λ . Then, for anyadmissible lift Λ ′ of ¯ Λ ′ there exists a tilting complex b T ∈ K b ( proj − Λ ) such that k ⊗ b T ∼ = T ϕ ,where ϕ : k ⊗ Λ ′ ∼ −→ ¯ Λ ′ is an isomorphism. Define Λ to be End D b ( Λ ′ ) ( b T ) , and let X denote atwo-sided tilting complex in D b ( Λ op ⊗ Λ ′ ) whose restriction to the right is isomorphic to b T . Weassume without loss that all terms of X are projective as left Λ -modules and as right Λ ′ -modules.We let Λ ′ and Λ ′ be arbitrary admissible lifts of ¯ Λ ′ . Let ϕ , ϕ , X , X , Λ and Λ beconstructed as above. Then Λ and Λ are both admissible lifts of ¯ Λ by Proposition 7.5, and19herefore they are both isomorphic to Λ by assumption. For i ∈ { , } let α i : Λ ∼ −→ Λ i bean isomorphism. Then α i X i for i ∈ { , } are two-sided tilting complexes in D b ( Λ op0 ⊗ Λ ′ i ) . Therestriction to the right of k ⊗ α i X i is of course still isomorphic to T ϕ i , and therefore we have thefollowing isomorphisms in D b ( k ⊗ Λ )( k ⊗ α X ) ⊗ k ⊗ Λ ′ ϕ ¯ Λ ′ ϕ | {z } ∼ = T ϕ ∼ = k ⊗ α X ⊗ k ⊗ Λ ′ ( k ⊗ α X ) ∨ ∼ = k ⊗ Λ (56)Of course this is only a (quasi-)isomorphism of complexes of left modules, but we can still deducethat the left hand side has homology concentrated in a single degree. This implies that ( k ⊗ α X ) ⊗ k ⊗ Λ ′ ϕ ¯ Λ ′ ϕ ⊗ k ⊗ Λ ′ ( k ⊗ α X ) ∨ is isomorphic to ( k ⊗ Λ ) β for some automorphism β of k ⊗ Λ . Hence we get ϕ ¯ Λ ′ ϕ ∼ = ( k ⊗ α X ) ∨ ⊗ k ⊗ Λ ( k ⊗ Λ ) β ⊗ k ⊗ Λ ( k ⊗ α X ) (57)which is now a quasi-isomorphism of two-sided complexes. By assumption there exists an au-tomorphism γ ∈ Aut( Λ ) which induces the same action on K ( k ⊗ Λ ) as β . In particular ( k ⊗ α X ) ∨ ⊗ k ⊗ Λ ( k ⊗ Λ ) β ∼ = ( k ⊗ α X ) ∨ ⊗ k ⊗ Λ ( k ⊗ Λ γ ) in D b ( Λ ) (i. e. again forgettingabout the left action), since ( k ⊗ α X ) ∨ restricted to the right is a two-term tilting complex, andthese are determined by their terms (and β and id k ⊗ γ act on these terms in the same way bydefinition). This implies that if we replace ( k ⊗ Λ ) β by k ⊗ Λ γ in the right hand side of (57), westill get a complex with homology concentrated in a single degree. Moreover, since all involvedcomplexes have terms which are projective as both left and right modules, we have ( k ⊗ α X ) ∨ ⊗ k ⊗ Λ ( k ⊗ Λ γ ) ⊗ k ⊗ Λ ( k ⊗ α X ) ∼ = k ⊗ (( α X ) ∨ ⊗ Λ Λ γ ⊗ Λ α X ) (58)in D b ( Λ ′ op1 ⊗ Λ ′ ) . We conclude that Y := ( α X ) ∨ ⊗ Λ Λ γ ⊗ Λ α X is a two-sided tilting complexall of whose terms are projective from the left and from the right such that k ⊗ Y is isomophic toa stalk complex. By the properties of tilting complexes we revisited above we can conclude that Y actually affords a Morita equivalence. As Λ ′ and Λ ′ are basic this implies Λ ′ ∼ = Λ ′ . Proof of Theorem 1.1.
First we need to show that (the basic order of) a block of quaternion defectof O G with three simple modules is admissible in the sense of Definition 7.1. It is clearly symmetricand it has split semisimple K -span by Proposition 3.4. That takes care of the first two propertiesrequired for admissibility.In [10, Definition following Corollary 2.7], the blocks of quaternion defect are divided intothree different cases, labeled “(aa)”, “(ab)” and “(bb)”. The blocks with three simple modulescorrespond to the case “(aa)” (see [10, table on page 231]). By [2, Theorem 1] there is a perfectisometry between any two blocks of quaternion defect with the same label (i. e. either “(aa)”,“(ab)”, or “(bb)”). In particular, there is a perfect isometry between any block of quaterniondefect with three simple modules and the principal block of O SL ( q ) for an appropriately chosen q . A perfect isometry gives rise to an isomorphism between centers and an isometry betweenGrothendieck groups satisfying the required properties of Φ and b Φ . This takes care of the thirdproperty required in the definition of admissibility. It follows that Λ and Γ are both admissible,and so are their basic orders.By [4, Chapter IX] there are only three possible basic algebras for k ⊗ Λ respectively k ⊗ Γ for any fixed generalized quaternion defect group. These are the algebras Q (3 A ) c , Q (3 B ) c and Q (3 K ) c . Hence the basic algebras of k ⊗ Λ and k ⊗ Γ each are isomorphic to one of those (forappropriate c ). Any two admissible lifts of Q (3 K ) c are isomorphic by Lemma 7.2. The algebra Q (3 A ) c is derived equivalent to Q (3 K ) c by means of a two-term tilting complex, and Corollary 6.4implies that the second condition of Proposition 7.6 is satisfied for the (unique) admissible lift of Q (3 K ) c , and hence Proposition 7.6 yields that any two admissible lifts of Q (3 A ) c are isomorphic.By comparing Cartan matrices one sees that B ( O SL ( q )) for an appropriately chosen q ≡ is a lift of Q (3 A ) c , and hence Corollary 6.4 and Proposition 7.6 can be applied again. Itfollows that any two admissible lifts of Q (3 B ) c are isomorphic. This completes the proof of thefirst assertion. 20he second assertion follows from the fact that the algebras k ⊗ Λ and k ⊗ Γ are derivedequivalent, together with Theorem 7.3, Proposition 7.5 and the uniqueness of admissible lifts wejust showed. Acknowledgments
This research was supported by the Research Foundation Flanders (FWO- Vlaanderen) project G.0157.12N.
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