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Advanced School and Conference on Representation Theory and Related Topics (9 - 27 January 2006) Integral Quadratic Forms and the Representation Type of an Algebra (Lecture 4) José Antonio de la Peña Universidad Nacional Autonoma de Mexico - UNAM Instituto de Matemáticas Mexico City D.F., Mexico

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Page 1: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

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Advanced School and Conference on Representation

Theory and Related Topics

(9 - 27 January 2006)

Integral Quadratic Forms and the Representation Type of an

Algebra

(Lecture 4)

José Antonio de la PeñaUniversidad Nacional Autonoma de Mexico - UNAM

Instituto de Matemáticas

Mexico City D.F., Mexico

Page 2: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

INTEGRAL QUADRATIC FORMSAND THE REPRESENTATION TYPE

OF AN ALGEBRA

LECTURE 4

Dr. Jose' Antonio de la PeñaInstituto de Matemáticas, UNAM

Page 3: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Tame algebras and modules on tubes.

Standard tubes in Auslander-Reiten quivers.Let A be a finite dimensional k-algebra. We say that an A-module X is a brick if

EndA(X) = k. In particular, a brick is always an indecomposable module.

We recall that two modules X1, X2 are said to be orthogonal if HomA(Xl, X2) 0HomA(X2, X1).

Let El,..., E8 be a family of pairwise orthogonal bricks. Define e(E1, ..., E5) as thefull subcategory of mod A whose objects X admit a filtration X = X0 3 X1 3 ... 3Xm = 0 for some in e N, with X/X+1 isomorphic to some E1, for any 1 i n.

Lemma. The category c = e(E1, ..., E8) is an abeliart category, with E1, ..., E5 beingthe simple objects of E. LIII

Page 4: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

An abelian category is said to be serial provided any object in E has finite lenghtand any indecomposable object in has a unique composition series.

Proposition. Let E1, ..., E be pairwise orthogonal bricks in some module cate-gory mod A. Assume that (a) E_1 for 1 i s with E0 = E3 and (b)EXt2A(Ei, E) = 0 for all 1 <i,j <n. Then E = ..., E3) is serial, it is a standardcomponent of FA of the form ZA/(n).

With the notation of the Proposition above: we denote by E [t] the unique modulein the serial category E which has socle E and lenght t.

A family T = (TA),\EL of the Auslander-Reiten quiver of an algebra A is a standardstable tubular family if each TA is a standard component of the form ZA/(nA) forsome nA and for \ a the components TA and T are orthogonal.

Corollary. Let T = (TA)AEL be a standard stable tubular family in the Auslander-Reiten quiver of A. Then the additive closure add T of T in mod A is an abeliancategory which is serial and is closed under extensions in mod A. El

Page 5: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

A standard stable tubular family Y = (TA)AEL is said to be separating if there arefull subcategories P and I of mod A satisfying the following conditions:

(i) each indecomposable A-module belongs to one of P, T or I;

(ii) for modules X E P, Y E T and Z e I we have HomA(Z, Y) = 0 =HomA(Z, X) and HomA(Y, X) = 0.

(iii) each non zero morphism f E HomA(X, Z) for indecomposable modules X EP, Z E I, factorizes through each component TA.

Page 6: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Example: Let A be the algebra given by the quiver with relations below

. S

S S " S

where 'PA,, is a preprojective component, 'A0 a preinjective component and TA" is aseparating tubular family of tubular type (2, 3, 3). In T = (T,),\ almost all tubesare of rank one with a module on the mouth with

w/A= /" A0:

0

dim E.:1

11 1

1 1

Then A is the one-point extension A0[E0] as follows[A0 E01A0 [E0] [o k]

with the usual matrix operations and where E0 is considered as an A0 - k-bimodule.Moreover rad P0 = E0.The algebra A0 is tame hereditary with an Auslander-Reiten quiver of the shape

Page 7: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

1

dim: 2

2 3

2

1 1

The tubes of rank 2 and rank 3 have modules on the mouths with the uniqueindecomposable A0-modules with the indicated dimension vectors:

0 1

1E1:

1

1 1 1 1 2

1

1 1 0 0

00 1

X0:1X1:0

X2:1

1 1 01 1 10 1

1

0 01 00 1

0 0 1

Z0:1

Z1:0

Z2:1

0 1 11 1 11 1 0

0 0 0 1 1 0

and where the Auslander-Reiten translation is given by 7-AO E1 = E0, TAOX = X_1and TAOZi Z_1 cyclically.

Page 8: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

The structure of FA is given as follows:

2 . . S S S S S S S S "

S

Too7-07-8

where To = V TA VT2 [E0] is the family of tubes T with the exception of the tube ofA2

rank 2 which appears now 'inserted' with the new projective at the extension vertex0.

For each positive rational number 5 = , (a, b), 7 is a separating family of tubes oftubular type (3,3,3) with all homogeneous tubes but 2 of rank 3. The homogeneoustubes have modules on the mouths of vector dimension

az0 ± bzoowhen Zoo is given by

.1 1

A " > z 1 2 100 dim Eoo: 0 1 000" 4SNI 'SI/I"( 0 >1 1 10 0 0

Observe that Aoo is tame concealed and A [Eoo]Aoo is a one-point coextensionwhere the module Eoo lies on a regular tube of FA. The algebra A is typical tubularalgebra as defined by Ringel.

Page 9: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Proposition. Let T = (T,\),\ be a standard separating tubular family for the modulecategory mod A. Then

a) For almost every A, the tube TA is homogeneous.b) Let TA be a homogeneous tube of the family T. Let X be a module in the mouth

of TA and v = dim X. Then qA(v) = 0.

Proof of (b): Let X be a module in the mouth of a homogeneous tube TA in Y.Let B be the convex closure in A of LJsupp X with X E TA. Since B is convex in Aand gldim B 2, then

qA(dim X) = qB(dim X) = dimk EndA(X) - dimk Ext(X, X).Since TA is standard and X X, then Ext(X, X) DH0mA(X, TX) and we get

qA(dim X)=0.

Page 10: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Notation: Let T = (TA)A be a standard separating stable tubular family in modA. Let r(A) be the period (or rank) of the tube TA. Consider those r(A'),. . . , r(.A)which are strictly bigger than 1 (finite number by (1.4)). We define the star diagramTT of the family T as the diagram with a unique ramification point and s branchesof lengths r (\i), . . . , r(,\). For example, the tame concealed algebra of tubular type(2, 3,3) has the star diagram depicted below.

./

Theorem. [Ringel, Lenzing-de la Peña]Let A = kQ/I be a k-algebra. Let ri be the number of vertices ofQ. Let T =

be a standard separating stable sincere tubular family in mod A. Let r(A) be the rankof the tube TA. Then

E(r(A)- 1) ~ n - 2.

AcLMoreover, A is a tame algebra if and only if the star diagram Tr is a Dynkin or

extended Dynkin diagram.

Page 11: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Tubes and isotropic roots of the Tits form.We say that a property P is satisfied by almost every indecomposable if for each

d e N, the set of indecomposable A-modules of dimension d which do not satisfy Pform a finite set of isomorphism classes. The following is a central fact about thestructure of the representation-quiver FA of a tame algebra A.

Theorem [Crawley-Boevey]Let A be a tame algebra. Then almost every indecomposable lies in a homogeneous

tube. In particular, almost every indecomposable X satisfies X X.

Open problem: Is it true that an algebra is of tame type if and only if almost everyindecomposable module belongs to a homogeneous tube?

Page 12: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Proposition. Let A be an algebra such that almost every irtdecomposable lies in astandard tube. Then A is tame.

Proof. Our hypothesis implies that almost every indecomposable X satisfies dimkEndA(X) < dirnX. We show that this condition implies the tameness of A.

Indeed, assume that A is wild and let Al be a A - k(u, v)-birnodule which is finitelygenerated free as right k('u, v)-module and the functor M®k()-insets indecompos-ables. Consider the algebra B given by the quiver ti

t2 and with radical J

satisfying J2 = 0. Then there is a A-B-birnodule N such that NB is free and N®B-mod B mod A is fully faithful. Therefore the composition F = Al ®A (N ®B -)is faithful and insets indecomposables. Moreover, dirnkFX < 'in dimkX for any X Emod B if we set m = dimk(M ®A N).

Consider also the functor H : mod A > mod B sending X to the space X' XXwith endornorphisms

XI(ti) =[

Xew)],X/(t2) =

[0X(v)] and X'(t3) =

[ ']This functor insets indecomposables. For the simple A-modules X of dimension

n, we get indecomposable A-modules FH(X) withdimkFH(X) <m dimkH(X) = 2mn

anddimk EndA(FH(X)) ˆ dimk EndB(H(X)) ,2 ± dimk EndA(X) ± 1.

Page 13: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Let A = kQ/I be a triangular algebra. In case A is tame, we would like to find thedimensions z E N° where indecomposable modules X with dim X = z and X in ahomogeneous tube exist.

Proposition. Assume that A is tame and qA(z) = 0. Then there is a decompositionz = w1 +... + w3 with w E NQO and an open subset U of rnodA(z) satisfying:

(a) dim U = dim modA(z).

(b) Every X e U has an indecornposable decomposition X = X, T ... X8 suchthat dim X = 'w and the module X1 lies in the mouth of a homogeneous tube.Moreover, dirnkHomA(X,X) = 5jj = Ext(X,X) for 1 <i,j s.

Page 14: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Hypercritical algebras.

Let q =4 +axx be a unit form. Let M be the symmetric matrix associ-i=1 ij

ated with q.

Proposition:The following are equivalent:(a) q is weakly non negative(b) Every critical restriction q'

v°M > 0.of q with v the positive generator of rad q', satisfies

Proof: a) b): Assume that q' is critical and v°M has its j-th componentnegative. Then 0 2v° ± e E Z" and q(2v° ± e) = 2v°Me ± 1 <0

b) = a): Assume q satisfies (b) but not (a). By induction, we may suppose thatq(i) satisfies (a), 1 i n. Let 0 < z be such that q(z) < 0. Let q' be a criticalrestriction. Let v be the positive generator of rad q'. We can find a number a 0such that 0 z ± av° and (z ± av°)(j) = 0 for some 1 <j <ui Then

0 <q(z ± av°) <av0IVIzt <0a contradiction.

Corollary:The unit form q is weakly non negative if and only if 0 q(z) for every z E [0, 12].

Page 15: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

Conjeture:Let A be a good algebra. Then A is tame if and only if its Tits form q is weakly

non negative.

Which are the good algebras?

Theorem [Brüstle]Let A be tree algebra. Then A is tame if and only if q is weakly non-negative.

Proposition:Let A be a tree algebra. Then q is weakly non negative if and only if A has no

convex hypercritical subalge bras.

Let A be a tree algebra and assume that q is not weakly non negative. Therefore,there is a hypercritical restriction q. Then I J U {x} such that q is critical andqA(v°, e) < 0, where v is the positive generator of radq.

Hence, q = q, where B is a convex critical subalgebra of A. Since qA(v°, e) <0,there is an arrow connecting x and A. Let C be the full (convex !) subcategory of Awith vertices x and those of B. Therefore q = q is hypercritical.

Page 16: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

We recall that a triangular algebra A = kQ/I is said to satisfy the separationcondition if every vertex x E Qo has separated radical. The algebra A is stronglysimply connected if every convex subcategory B of A satisfies the separation condition.Theorem [Brüstle-de la Peña-Skowroñski].

Let A be a strongly simply connected algebra, then the following are equivalent.'

(a) A is tame

(b) q is weakly non-negative

(c) A does not contain a full convex subcategory which is hypercritical.

Page 17: INTEGRAL QUADRATIC FORMS AND THE REPRESENTATION …webusers.imj-prg.fr/.../lecturenotes/delapena4.pdf · Theorem. [Ringel, Lenzing-de la Peña] LetA=kQ/I be ak-algebra. Letri be the

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