Almost Split Morphisms, Preprojective Algebras and Multiplication Maps of Maximal Rank

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1 Syracuse University SURFACE Mathematics Faculty Scholarship Mathematics Almost Split Morphisms, Preprojective Algebras an Multiplication Maps of Maximal Rank Steven P. Diaz Syracuse University Mark Kleiner Syracuse University Follow this an aitional works at: Part of the Mathematics Commons Recommene Citation Diaz, Steven P. an Kleiner, Mark, "Almost Split Morphisms, Preprojective Algebras an Multiplication Maps of Maximal Rank" (2005). Mathematics Faculty Scholarship This Article is brought to you for free an open access by the Mathematics at SURFACE. It has been accepte for inclusion in Mathematics Faculty Scholarship by an authorize aministrator of SURFACE. For more information, please contact

2 arxiv:math/ v1 [math.rt] 30 Dec 2005 ALMOST SPLIT MORPHISMS, PREPROJECTIVE ALGEBRAS AND MULTIPLICATION MAPS OF MAXIMAL RANK STEVEN P. DIAZ AND MARK KLEINER Abstract. With a graing previously introuce by the secon-name author, the multiplication maps in the preprojective algebra satisfy a maximal rank property that is similar to the maximal rank property proven by Hochster an Laksov for the multiplication maps in the commutative polynomial ring. The result follows from a more general theorem about the maximal rank property of a minimal almost split morphism, which also yiels a quaratic inequality for the imensions of inecomposable moules involve. 1. Introuction Let k be a fiel an let R be the polynomial ring in n commuting variables over k. Let R i be its i th grae piece consisting of homogeneous polynomials of egree i. A result of Hochster an Laksov [4] says that if i 2 an V R i is a general subspace then the natural multiplication map from V R 1 to R i+1 has maximal rank, that is is either injective or surjective, an it is not known what happens if one replaces R 1 by R for > 1. One may woner which other grae rings have a similar property. In [5] a new graing on the preprojective algebra was introuce. In this paper we show that with this graing, the preprojective algebra of a finite quiver without oriente cycles satisfies a property analogous to the Hochster-Laksov property for polynomial rings, an much of our proof is quite similar to their proof. At one point the proof for preprojective algebras becomes easier than the proof for polynomial rings: some of the more complicate imension counts neee for polynomial rings are not neee for preprojective algebras. This allows us to obtain a result for preprojective algebras that is stronger than the analogous result for polynomial rings. The key to making things work is the fact that the multiplication-by-arrow maps into a fixe homogeneous component of the infinite imensional (in general) preprojective algebra give rise to a minimal right almost split morphism of moules over the finite imensional path algebra of the quiver [5], which implies the maximal rank property. In fact we show that a minimal right almost split morphism g : B C of finite imensional moules over a k-algebra satisfies a maximal rank property analogous to the Hochster-Laksov property for polynomial rings, an if C is not projective an B 1,..., B l are the nonisomorphic inecomposable summans of B then im k C < (im k B 1 ) 2 + +(im k B l ) 2. We o not know what happens if multiplication by arrows is replace by multiplication by paths of fixe length greater than one. There is a natural ual to the Hochster-Laksov maximal rank property, an the two properties always occur simultaneously. We give two explanations of this fact, one general homological an the other base on the vector space uality D = Hom k (, k). As a consequence, the multiplicationby-arrow maps out of a fixe homogeneous component of the preprojective algebra satisfy the ual Hochster-Laksov maximal rank property Mathematics Subject Classification. Primary 13D02, 16G30, 16G70. Key wors an phrases. Almost split morphism, preprojective algebra, linear map of maximal rank. The secon-name author is partially supporte by a grant from NSA. 1

3 2 STEVEN P. DIAZ AND MARK KLEINER The organization of the paper is as follows. In Section 2 we prove a theorem that gives a general situation in which one can obtain a maximal rank property analogous to the Hochster-Laksov property for polynomial rings. This general situation oes not inclue the polynomial ring as a special case. In Section 3 we review some facts about almost split morphisms an preprojective algebras an then show that almost split morhisms in general an the preprojective algebra in particular fit into the general set up of Section 2. In Section 4 we use the material in Sections 2 an 3 to obtain results for the preprojective algebra that look very analogous to the Hochster- Laksov result for polynomial rings. Then we conclue with some examples to illustrate the results. In this paper for simplicity we work over a fixe algebraically close fiel k an im always means im k. For unexplaine terminology we refer the reaer to [1]. 2. The General Theorem Let V 1, V 2,..., V l, W 1, W 2,..., W l, U be finite imensional vector spaces. Let T be a linear transformation from the irect sum of the tensor proucts V 1 W 1, V 2 W 2,...,V l W l to U. l (2.1) T : (V i W i ) U Definition 2.1. We say that T satisfies the right omnipresent maximal rank property if an only if for every choice of subspaces W i W i for i = 1,..., l the restriction of T to the irect sum of the tensor proucts V 1 W 1, V 2 W 2,..., V l W l has maximal rank, that is, is either injective or surjective. Notice that if T satisfies the right omnipresent maximal rank property then so oes its restriction to l V i W i, an T must itself have maximal rank. Every injective T satisfies the right omnipresent maximal rank property. The interesting case is when T is surjective but not injective. Denote by En(V i ) the k-algebra of linear operators on V i. The tensor prouct V i W i is a left En (V i )-moule by means of ϕ i (v i w i ) = ϕ i (v i ) w i, ϕ i En (V i ), v i V i, w i W i. Applying this to each term of the irect sum one obtains a bilinear evaluation map l l l e : En (V i ) (V i W i ) (V i W i ). Denote l En (V i) by B. The map e efines a structure of a left B-moule on l (V i W i ). Notice that B has imension Σ(imV i ) 2. Let P i be the projective space of one imensional subspaces of V i W i an let P be the projective space of one imensional subspaces of l (V i W i ). Notice that P has imension Σ(imV i imw i ) 1. We shall stuy the prouct B P together with its two projection maps π 1 onto B an π 2 onto P. Since the evaluation map e is bilinear, we may conclue that the inverse image uner e of KerT is a Zariski close subset of the omain of e. Furthermore using bilinearity again we see that e 1 (KerT) is the affine cone over a Zariski close subset of B P. We enote this subset by Y. For each i from 1 to l let X i be an irreucible quasiprojective subset of P i an let C(X i ) be its corresponing affine cone in V i W i. Let X be the irreucible quasiprojective subset of P corresponing to C(X 1 ) C(X 2 )... C(X l ). Notice that imx = l im C(X i) 1. Theorem 2.1. Assume that T satisfies the right omnipresent maximal rank property an that l im C(X i) imu. Then π 1 (π2 1 (X) Y ) is containe in a proper Zariski close subset of B. If T is injective then Y is empty an the result trivially follows. Thus we may assume that T is surjective. To procee with the proof we shall ivie Y into two pieces base on the following

4 PREPROJECTIVE ALGEBRAS AND MULTIPLICATION MAPS OF MAXIMAL RANK 3 easy statement, an then eal with each piece separately. Denote by D the contravariant functor Hom k (, k). Lemma 2.2. Let V an W be k-vector spaces an let α : V W Hom k (D V, W) be the k-linear map given by α(v w)(f) = f(v)w, v V, w W, f DV (α is an isomorphism if imv < ). For x V W enote by En(V )x the cyclic En(V )-submoule of V W generate by x. Then En(V )x = V Im α(x). Proof. We have x = s v i w i. If s is the smallest possible, the sets of vectors {v 1,...,v s } an {w 1,...,w s } are linearly inepenent [2, Theorem (1.2a), p. 142] so Imα(x) is the span of {w 1,..., w s } an the rest is clear. Definition 2.2. Let α i : V i W i Hom k (D V i, W i ) be the k-linear map escribe in Lemma 2.2, i = 1,...,l. If x i V i W i an 0 c k, then Im α i (x i ) = Im α i (cx i ), so if p P is represente by [x 1,..., x l ], x i V i W i, then for each i the subspace Im α i (x i ) of W i is inepenent of the choice of representative for p. We set Y 1 = {(b, p) Y : l (im V i)(rankα i (x i )) < imu} an Y 2 = Y Y 1. Lemma 2.3. Y = Y 1 Y 2 where Y 1 is a close subset of Y, an Y 2 is an open subset of Y. Proof. That Y = Y 1 Y 2 is obvious. For the other two statements consier the projection onto the secon factor π 2 : B P P an note that Y 1 (Y 2 ) is the intersection of Y with the inverse image of the close (open) subset of P consisting of points corresponing to tuples of tensors satisfying l (imv i)(rankα i (x i )) < ( )imu. Lemma 2.4. Assume that T satisfies the right omnipresent maximal rank property. Suppose that (b, p) Y 1 an b = [ϕ 1, ϕ 2,..., ϕ l ]. Then for some i, ϕ i is not an isomorphism. Proof. If x = [x 1,...,x l ] represents p, then T(bx) = T ([ϕ 1 x 1,..., ϕ l x l ]) = 0 because (b, p) Y 1. By Lemma 2.2, Bx = l En(V i)x i = l V i Im α i (x i ), so the restriction of T to Bx is injective because (b, p) Y 1 an T satisfies the right omnipresent maximal rank property. Since T(bx) = 0 then bx = [ϕ 1 x 1,..., ϕ l x l ] = 0 whence ϕ i x i = 0 for all i. Because p is a point in a projective space, at least one x i is not equal to 0. For this i, ϕ i is not an isomorphism. Lemma 2.5. Assume that T satisfies the right omnipresent maximal rank property an that l im C(X i) imu. Suppose π2 1 (X) Y 2 is nonempty. Then π2 1 (X) Y 2 has Krull imension at most Σ(im V i ) 2 1, one less than the imension of B. Proof. As with Lemma 2.3 we consier the projection map onto the secon factor π 2 : B P P. Let X 2 X be the set of points p such that l (im V i)(rankα i (x i )) imu for any x = [x 1,..., x l ] representing p. Since X 2 is open in X an by assumption nonempty, imx 2 = imx. Pick any point p in X 2 an, ientifying B with B {p}, consier the composite T : B U of T an the k-linear map B l (V i W i ) sening b to bx. Clearly KerT = π2 1 (p) Y 2 an Im T = T(Bx) = T( l (V i Im α i (x i ))) (use Lemma 2.2). From the assumptions on the ranks of the α i (x i ) an that T satisfies the right omnipresent maximal rank property, we conclue that T is surjective. We then conclue that im(π2 1 (p) Y 2) = imb imu. Having compute the imensions of the fibers of π2 1 (X) Y 2 over X 2 we then see that the imension of π2 1 (X) Y 2 equals imx +imb imu = l imc(x i) 1+imB imu imb 1 = Σ(im V i ) 2 1. The proof of the theorem is now easy.

5 4 STEVEN P. DIAZ AND MARK KLEINER Proof of theorem. By Lemma 2.4 the image of π2 1 (X) Y 1 in B will be containe in the proper close subset of B consisting of points where at least one ϕ i is not an isomorphism. By Lemma 2.5 π2 1 (X) Y 2 has imension less than that of B an so its closure also oes. Since π 1 is a projective morphism, the image in B of the closure of π2 1 (X) Y 2 will be close an have imension less than that of B. Since by Lemma 2.3 Y = Y 1 Y 2 we are one. Corollary 2.6. Assume that T satisfies the right omnipresent maximal rank property. Make a choice of subspaces Z i V i W i, i = 1,..., l. Then there exists a ense Zariski open subset A B such that if [ϕ 1,..., ϕ l ] A then the restriction of T to the irect sum of the ϕ i (Z i ) has maximal rank. Proof. We first o the case where l im (Z i) im(u). In Theorem 2.1 set Z i = C(X i ). Choose A to be the complement of any proper Zariski close subset of B containing π 1 (π2 1 (X) Y ). For [ϕ 1,..., ϕ l ] A, l ϕ i(z i ) intersects the kernel of T only in 0. Thus the restriction of T to l ϕ i(z i ) is injective. When l im (Z i) = im(u) it is also surjective. For the case where l im(z i) > im(u) choose subspaces Z i Z i such that l im (Z i ) = im(u). By the previous case we fin A such that if [ϕ 1,..., ϕ l ] A then the restriction of T to l ϕ i(z i ) is surjective, so the restriction of T to l ϕ i(z i ) is also surjective. Corollary 2.7. Assume that T is surjective an satisfies the right omnipresent maximal rank property. Fix integers a i, 0 a i (imv i )(im W i ), i = 1,..., l, such that a i = imu. For each i choose a i linearly inepenent elements m(i, j), 1 j a i, of V i W i. Then there exists a ense Zariski open subset A B such that if [ϕ 1,..., ϕ l ] A then the elements T(ϕ i (m(i, j))) form a basis for U. Proof. In Corollary 2.6 set Z i equal to the span of the m(i, j) s. Definition 2.3. We say that T satisfies the left general maximal rank property if an only if for a general choice of subspaces V i V i for i = 1,..., l the restriction of T to l (V i W i) has maximal rank, that is, is either injective or surjective. By a general choice of subspaces we mean the following. Once the imensions of the V i s to be chosen are fixe, the set of all possible choices of V i s can be ientifie with a prouct of Grassmanians. We mean that there exists a Zariski open ense subset of that prouct such that if the choice of V i s comes from that set, then the restriction of T has maximal rank. Similar to Definitions 2.1 an 2.3 one can efine what it means for the map T of (2.1) to satisfy the left omnipresent or right general maximal rank property. With these efinitions, we leave it to the reaer to interchange appropriately the wors left an right in the above assertions an obtain true statements. Of course, this comment also applies to the remainer of the section. Corollary 2.8. If T satisfies the right omnipresent maximal rank property then T satisfies the left general maximal rank property. Proof. Make a choice of subspaces V i V i for i = 1,...l. In Corollary 2.6 set Z i = V i W i. Notice that ϕ i (V i W i) = ϕ i (V i ) W i. A general tuple of enomorphisms [ϕ 1,..., ϕ l ] applie to a specific tuple of subspaces [V 1,..., V l ] gives a general tuple of subspaces. In Section 4 we will give examples to show that the right omnipresent maximal rank property oes not imply the left omnipresent maximal rank property an the right general maximal rank property oes not imply the left general maximal rank property.

6 PREPROJECTIVE ALGEBRAS AND MULTIPLICATION MAPS OF MAXIMAL RANK 5 We now inicate how to ualize the above results of this section. Let V 1, V 2,..., V l, W 1, W 2,..., W l, Q be finite imensional vector spaces an let (2.2) S : Q be a linear transformation. l (V i W i ) Definition 2.4. We say that S satisfies the right omnipresent maximal rank property if an only if for every choice of subspaces W i W i for i = 1,..., l the composition of S with the linear transformation l (1 Vi τ i ) : l (V i W i ) l (V i (W i /W i)) has maximal rank, that is, is either injective or surjective, where τ i : W i W i /W i is the natural projection. An we say that S satisfies the left general maximal rank property if an only if for a general choice of subspaces V i V i for i = 1,..., l the composition of S with the linear transformation l (σ i 1 Wi ) : l (V i W i ) l ((V i/v i ) W i) has maximal rank, where σ i : V i V i /V i is the natural projection. The following lemma shows that the question of whether a map of the type (2.1) satisfies the omnipresent or general maximal rank property is equivalent to the same question for a map of the type (2.2). Lemma 2.9. Let 0 0 A f B i B q g C 0 B 0 be an exact iagram in an abelian category. Then gi is monic (epi) if an only if qf is monic (epi). Proof. By the 3 3 lemma the following commutative iagram is exact.

7 6 STEVEN P. DIAZ AND MARK KLEINER Kergi B gi Im gi 0 h i j 0 A p f r B q g C 0 0 Cokerh B Cokerj Hence gi is monic if an only if Kergi = 0, if an only if p is iso, if an only if qf is monic. The rest of the proof is similar. Corollary (a) If a linear transformation T : l (V i W i ) U is surjective, it satisfies the left omnipresent (general) maximal rank property if an only if so oes the inclusion KerT l (V i W i ). (b) If a linear transformation S : Q l (V i W i ) is injective, then it satisfies the left omnipresent (general) maximal rank property if an only if so oes the projection l (V i W i ) Coker S. Proof. The statement follows immeiately from Lemma 2.9. We note that if the map T of Corollary 2.10(a) is injective, it satisfies both the right omnipresent an left general maximal rank property, an if T is neither surjective nor injective then it satisfies neither of the properties. A similar remark applies to the map S of Corollary 2.10(b). A ifferent way to relate the maps of the types (2.1) an (2.2) is through the vector space uality D. Proposition A linear transformation T : l (V i W i ) U satisfies the left omnipresent (general) maximal rank property if an only if so oes its ual DT : D U l (DV i DW i ). Proof. For i = 1,...,l let X i be a subspace of V i an f i : X i V i the inclusion map, then Df i : DV i D X i is an epimorphism with KerDf i = Xi = {φ DV i : φ(x i ) = 0} so that DX i = DV i /Xi. Therefore T ( l (f i 1 Wi )) is monic (epi) if an only if ( l (D f i 1 D Wi )) DT is epi (monic), if an only if ( l (ψ i 1 D Wi )) D T is epi (monic), where ψ i : DV i DV i /Xi is the natural projection. Note that X i runs through the set of all subspaces of V i if an only if Xi runs through the set of all subspaces of D V i. Hence T satisfies the left omnipresent maximal rank property if an only if so oes DT. For a fixe sequence of nonnegative integers i n i = imv i, the l-tuple (X 1,...,X i,..., X l ) runs through a ense open set of the prouct of Grassmanians l G( i, V i ) if an only if (X1,...,X i,..., X l ) runs through the corresponing ense open set of the prouct of Grassmanians l G(n i i, DV i ) uner the isomorphism that is the prouct of the natural isomorphisms D : G( i, V i ) G(n i i, DV i ), see [3, p. 200]. Therefore T satisfies the left general maximal rank property if an only if so oes DT.

8 PREPROJECTIVE ALGEBRAS AND MULTIPLICATION MAPS OF MAXIMAL RANK 7 We en this section with a lemma showing that the right omnipresent maximal rank property puts a restriction on the relative sizes of the vector spaces involve. Lemma (a) If T : l V i W i U satisfies the right omnipresent maximal rank property an T is surjective but not injective, then imu < l (imv i) 2. (b) If S : Q l V i W i satisfies the right omnipresent maximal rank property an S is injective but not surjective, then imq < l (imv i) 2. Proof. (a) Suppose to the contrary that imu l (im V i) 2. Let {v i1,..., v iki } be a basis for V i. Express some nonzero ( element of KerT in the form k1 j=1 v 1j w 1j,..., k i j=1 v ij w ij..., ) k l j=1 v lj w lj an let W i equal the span of {w i1,..., w iki }. Then im l (V i W i ) imu but the restriction of T to l (V i W i ) is neither surjective nor injective because its kernel is not zero. (b) Follows from (a) an Proposition Almost Split Morphisms an Preprojective Algebras We apply the results of Section 2 to representations of algebras which provie a large supply of linear transformations of the form l (V i W i ) U or Q l (V i W i ). Let Λ be an associative k-algebra, let mo Λ be the category of finite imensional left Λ-moules, an let g : B C an f : A B be morphisms in mo Λ. Replacing B with an isomorphic moule if necessary, we may assume that B = V n1 1 V n l l where V 1,...,V l are nonisomorphic inecomposable Λ-moules, l, n 1,...,n l are nonegative integers, an V m stans for the irect sum of m copies of V. For i = 1,...,l enote by W i the k-space with a basis e i1,..., e ini, an for each j = 1,...,n i enote by h ij : V i V i ke ij the isomorphism of Λ-moules sening each v V i to v e ij. Let (3.1) h : B l (V i W i ) be the isomorphism in mo Λ inuce by the h ij s. Denote by g ij : V i C an f ij : A V i the morphisms in mo Λ inuce by g an f, respectively, an consier the morphisms T i : V i W i C an S i : A V i W i efine by T i (v e ij ) = g ij (v), v V i, an S i (a) = (f ij (a) e ij ), a A, respectively. Let (3.2) T : l (V i W i ) C an S : A l (V i W i ) be the morphisms in mo Λ inuce by the T i s an S i s, respectively. It is straight forwar to check that (3.3) g = Th an S = hf. Proposition 3.1. Let g : B C an f : A B be morphisms in moλ with B = l V ni i where V 1,..., V l are nonisomorphic inecomposable Λ-moules. Let h be the isomorphism in (3.1), let T an S be the morphisms in (3.2) constructe from g an f, respectively. If g is a minimal right almost split morphism in mo Λ then: (a) T satisfies the right omnipresent maximal rank property. (b) For a general choice of k-subspaces U i V i, the restriction of g to l Uni i rank. (c) If g is surjective then im C < l (imv i) 2. If f is a minimal left almost split morphism in mo Λ then: () S satisfies the right omnipresent maximal rank property. has maximal

9 8 STEVEN P. DIAZ AND MARK KLEINER (e) For a general choice of k-subspaces U i V i, enote by σ i : V i V i /U i the natural projection. Then the linear transformation ( l σni i ) f has maximal rank. (f) If f is injective then ima < l (im V i) 2. If 0 A f B g C 0 is an almost split sequence in mo Λ then: (g) imb < 2 l (imv i) 2 1. Proof. (a) Since g is minimal right almost split, so is T by (3.3). If W i is a subspace of W i, the Λ-moule V i W i is a irect summan of V i W i. Hence the restriction of T to l (V i W i ) is an irreucible morphism an thus is either a monomorphism or an epimorphism [1, Ch. V, Theorem 5.3(a) an Lemma 5.1(a)], so (a) hols. Accoring to Corollary 2.8, T satisfies the left general maximal rank property. In view of the structure of the isomorphisms h ij constructe above, we conclue that (b) hols. Part (c) is a irect consequence of (a), formula (3.3), an Lemma 2.12(a). () The proof is similar to that of (a) using the analogous properties of minimal left almost split morphisms. (e) If f is surjective, the statement is clear. If f is not surjective, it is injective, an so is S in view of formulas (3.3). By () an Corollary 2.10(b), the projection l (V i W i ) CokerS satisfies the right omnipresent maximal rank property. By Corollary 2.8, it satisfies the left general maximal rank property, an so oes S by Corollary 2.10(b). Then f satisfies the esire property in view of formulas (3.3). Another way to prove () an (e) is to note that both Df an D S are minimal right almost split morphisms in mo Λ op, an then use (a), Corollary 2.8, an Proposition 2.11 together with formulas (3.3). (f) The proof is similar to that of (c), using Lemma 2.12(b). (g) The formula follows from (c) an (f). Remark 3.1. (a) Lemma 2.12 hols when k is an arbitrary fiel. Hence so o parts (a), (c), (), (f), an (g) of Proposition 3.1; moreover, they hol if mo Λ is replace by any full subcategory of an abelian category close uner extensions an irect summans where the objects an morphism sets are finite imensional k-vector spaces an composition of morphisms is k-bilinear. (b) Parts (a), (b), an (c) of Proposition 3.1 hol if g : B C is an irreucible morphism with C inecomposable, an parts (), (e), an (f) hol if f : A B is an irreucible morphism with A inecomposable. This follows from the observation after Definition 2.1 that the right omnipresent maximal rank property of a linear transformation is inherite by its appropriate restrictions, an from the ual statement. (c) Parts (c), (f), an (g) of Proposition 3.1 imply that for a fixe number of nonisomorphic inecomposable summans of the mile term of an almost split sequence, the summans cannot be much smaller than the en terms of the sequence, i.e., the multiplicities of the summans cannot be too large, an that there is a balance between the sizes of the en terms. Part (c) is false if the morphism g is not surjective, an part (f) is false if the morphism f is not injective. We will apply this in particular to the preprojective algebra where the graing introuce in [5] allows us to interpret the multiplication-by-arrow maps into (from) a fixe homogeneous component as a minimal right (left) almost split morphism of moules over the path algebra of the quiver. We recall some facts from the latter paper. For the remainer of this paper we fix a finite quiver Γ = (Γ 0, Γ 1 ) without oriente cycles with the set of vertices Γ 0 an the set of arrows Γ 1. Let Γ = ( Γ 0, Γ 1 ) be a new quiver with Γ 0 = Γ 0 an Γ 1 = Γ 1 Γ 1, where Γ 1 Γ 1 = an the elements of Γ 1 are in the following one-to-one corresponence with the elements of Γ 1 : for each γ : t v in Γ 1, there is a unique element γ : v t in Γ 1. To turn the path algebra k Γ of Γ over a fiel k into a grae k-algebra, we assign

10 PREPROJECTIVE ALGEBRAS AND MULTIPLICATION MAPS OF MAXIMAL RANK 9 egree 0 to each trivial path e t, t Γ 0, an each arrow γ Γ 1 ; egree 1 to each arrow γ Γ 1 ; an compute the egree of a nontrivial path q = δ 1... δ r as eg q = r egδ i. Clearly, kγ is the k-subalgebra of k Γ comprising the elements of egree 0. Let N be the set of nonnegative integers. For all t Γ 0, N, let W t be the span of all those paths in Γ of egree that start at t. Note that W t mo kγ so (3.4) k Γ = W t N t Γ 0 is a ecomposition of k Γ as a irect sum of its left kγ-submoules Let now a an b be any two functions Γ 1 k satisfying a(γ) 0 an b(γ) 0 for all γ Γ 1. If s(γ) is the starting point an e(γ) is the en point of γ Γ 1, for each t Γ 0 set m t = a(γ)γ γ b(γ)γγ γ Γ 1 s(γ)=t γ Γ 1 e(γ)=t an enote by J the two-sie ieal of k Γ generate by the element m t = [γ, γ] a,b t Γ 0 γ Γ 1 where [γ, γ] a,b = a(γ)γ γ b(γ)γγ is the (a, b)-commutator of γ an γ. The factor algebra P k (Γ) a,b = k Γ/J is the (a, b)-preprojective algebra of Γ. Since the elements m t are homogeneous of egree 1, J is a homogeneous ieal containing no nonzero elements of egree 0. Hence P k (Γ) a,b is a grae k-algebra, an the restriction to kγ of the natural projection π : k Γ P k (Γ) a,b is an isomorphism of kγ with the subalgebra of P k (Γ) a,b comprising the elements of egree 0; we view the isomorphism as ientification. From (3.4) we get P k (Γ) a,b = V t N t Γ 0 where V t = π(w t) mo kγ. If γ Γ 1 we write β = π(γ) an β = π(γ ). If q is a path in Γ starting at t an ening at v, we call π(q) a path in P k (Γ) a,b starting at t an ening at v. Then V t is the span of all paths of egree in P k(γ) a,b starting at t. Since we ientify kγ with π(kγ), we in particular ientify e t with π(e t ), t Γ 0 ; γ with β = π(γ), γ Γ 1 ; W0 t with V 0 t ; an we set W 1 t = V 1 t = 0. We nee the following statement. When appropriate, the map (c) : X Y enotes the right multiplication by c. Theorem 3.2. Suppose V t 0 where t Γ 0, N. (a) V t t is inecomposable in mo kγ, an V = V s t = s an = c. (b) The map g t : ( V e(γ) s(γ)=t ) ( V s(γ) 1 e(γ)=t c in mo kγ, s Γ 0, c N, if an only if ) V t inuce by the right multiplications (a(γ)β) : V e(γ) V t, s(γ) = t, an ( b(γ)β ) : V s(γ) 1 V t, e(γ) = t, where γ Γ 1, is a minimal right almost split ( morphism) in( mo kγ. ) (c) The map f t : V t V e(γ) +1 V s(γ) inuce by the right multiplications s(γ)=t e(γ)=t (β ) : V t V e(γ) t +1, s(γ) = t, an (β) : V V s(γ), e(γ) = t, where γ Γ 1, is a minimal left almost split morphism in mo ( kγ. ) ( ) () If V+1 t 0 then 0 V t f t V e(γ) +1 V s(γ) g t +1 V+1 t 0 is an almost s(γ)=t e(γ)=t split sequence in mo kγ.

11 10 STEVEN P. DIAZ AND MARK KLEINER Proof. These are parts of [5, Theorem 1.1 an Corollary 1.3] combine with well-known properties of preprojective moules, see [1, VIII.1]. Applying parts (b) an () of Proposition 3.1 to Theorem 3.2, we obtain the following statement. Corollary 3.3. (a) In the setting of Theorem 3.2(b), for a general choice of k-subspaces U e(γ) U e(γ) ) ( U s(γ) ) has maximal rank. V e(γ) an U s(γ) 1 V s(γ) 1, the restriction of gt to ( 1 s(γ)=t e(γ)=t (b) In the setting of Theorem 3.2(c), for a general choice of k-subspaces U e(γ) +1 V e(γ) U s(γ) V s(γ), enote by σ e(γ) +1 : V e(γ) +1 V e(γ) +1 /Ue(γ) +1 an σs(γ) : V s(γ) σ e(γ) +1 ) ( projections. Then the linear transformation (( s(γ)=t σ s(γ) e(γ)=t V s(γ) )) f t /U s(γ) +1 an the natural has maximal rank. Remark 3.2. As follows from Remark 3.1(b), if one leaves out any number of summans in the irect sum of part (b) of Theorem 3.2 an replaces the map g t by its restriction to the sum of the remaining summans, Corollary 3.3(a) will still hol. Likewise, if one leaves out any number of summans in the irect sum of part (c) of Theorem 3.2 an replaces the map f t by its composition with the projection onto the sum of the remaining summans, Corollary 3.3(b) will still hol. The results of this section have ealt with left moules over a k-algebra Λ an with the right multiplication-by-arrow maps in the preprojective algebra. One may ask if analogous results are true for right Λ-moules an for the left mulitplication-by-arrow maps. We leave it to the reaer to state the analog of Proposition 3.1, an note that [5, Theorem 1.1 an Corollary 1.3] aress left multiplication by arrows in P k (Γ) a,b by replacing W t an V t with W t,, the span of all those paths in Γ of egree that en at t, an V t, = π(w t, ), respectively. Since V t, is a finite imensional right kγ-moule for all t an, with the appropriate replacements the analogs of Theorem 3.2 an Corollary 3.3 hol. These remarks also apply to the consierations of Section Corollaries an Examples In this section we strengthen Corollary 3.3 in a form that is analogous to the result of Hochster an Laksov [4]. To help the reaer see the analogy we shall first state their result. Set R = k[x 1, x 2,..., x r ], the commutative polynomial ring grae by egree, an enote by R its homogeneous piece of egree. Let N(r, ) be the imension of R as a vector space over k. The following is then the result of Hochster an Laksov [4]. Theorem 4.1. Given an integer 2, we etermine an integer n by the inequalities (n 1)r < N(r, + 1) nr an let s = N(r, + 1) (n 1)r. Then if F 1, F 2,..., F n are n general forms in R we have that the (n 1)r forms x j F i for j = 1,..., r an i = 1, 2,..., n 1 together with the s forms x j F n for j = 1, 2,...s (in total N(r, + 1) forms) are a k-vector space basis for R +1. By general forms they mean that there exists a ense Zariski open subset of the affine space (R ) n such that if the n-tuple (F 1, F 2,..., F n ) is chosen from that open set, then the conclusion follows. We wish to apply Corollary 2.7 to the maps g t of Theorem 3.2, which is possible accoring to Proposition 3.1(a). To make the result clearly analogous to the result of Hochster an Laksov we must set up our notation properly. Fix a vertex t Γ 0 an a nonnegative integer. Let s 1, s 2,..., s m be the istinct vertices that have in Γ 1 arrows from them to t, an let u 1, u 2,...u n be the istinct vertices with arrows in Γ 1 going from t to them. To match things up with the set up in Section 2, for i = 1, 2,..., m

12 PREPROJECTIVE ALGEBRAS AND MULTIPLICATION MAPS OF MAXIMAL RANK 11 let V i = V si 1, let W i be the k-linear span of the arrows βi,j in Γ 1 going from t to s i, an set w i,j = b(β i,j )βi,j. For i = m + 1, m + 2,..., m + n = l let V i = V ui m +1, let W i be the k-linear span of the arrows β i,j in Γ 1 going from t to u i m, an set w i,j = a(β i,j )β i,j. For all i, we choose {w i,j } as a basis for W i an put the w i,j s in a column vector x i. Set U = V t. Let M i be the vector space of imv i imw i matrices with elements in k. Let B be the affine space l im Vi Vi. An element b of B is an l-tuple [b 1, b 2,..., b l ] where each b i is a im V i-tuple of elements of V i, written as a row vector. For b B an m(i) M i, using orinary matrix multiplication an the multiplication an aition in the preprojective algebra, we see that b i m(i)x i is an element of U = V t. Corollary 4.2. Let > 0 an V t 0. Fix integers a i satisfying 0 a i (im V i )(im W i ), i = 1,..., l, an a i = imu. For each i choose a i linearly inepenent elements of M i an call them m(i, j), 1 j a i. There exists a Zariski open ense subset E of B such that if b = [b 1, b 2,..., b l ] E, then the elements b i m(i, j)x i, 1 i l, 1 j a i, form a basis for U = V t. Proof. We alreay have a chosen basis for each W i. Suppose we also choose a basis for each V i. The pairwise tensor proucts of these basis elements give a basis for V i W i, so we may ientify im Vi V i W i with M i. We may also ientify En(V i ) with Vi by matching ϕ i En(V i ) with the image uner ϕ i of the chosen basis. Uner these ientifications the elements T(ϕ i (m(i, j))) appearing in Corollary 2.7 become ientifie with the elements b i m(i, j)x i appearing in Corollary 4.2. Thus Corollary 4.2 is a particular case of Corollary 2.7. With the proper choice of the m(i, j) we can get a corollary that souns even more like the result of Hochster an Laksov. Corollary 4.3. Let > 0 an V t 0. For each i satisfying 1 i m, let β i,j, 1 j imw i, be the new arrows going from t to s i. For each i satisfying m + 1 i l, let β i,j, 1 j imw i, be the ol arrows going from t to u i m. Choose positive integers n i, 1 i l, satisfying 1 n i imv i an l (n i 1)imW i < imv t l n i imw i, an set c = imv t l (n i 1)imW i. Write c as a sum of nonnegative integers c = c 1 +c c l, 0 c i imw i. For a general choice of l where F i,k V si 1 for 1 i m an F i,k V ui m elements form a basis for V t : F i,k β i,j for 1 i m, 1 k n i 1, 1 j imw i ; F i,ni β i,j for 1 i m, 1 j c i; F i,k β i,j for m + 1 i l, 1 k n i 1, 1 j imw i ; F i,ni β i,j for m + 1 i l, 1 j c i. n i elements F i,k, 1 i l, 1 k n i, for m + 1 i l, the following imv t Here in an inequality giving the range of possible j or k, if the number on the right is less than 1, we simply mean there are no such j or k. Proof. Choose the m(i, j) s as follows. Note that a i = (n i 1)imW i + c i. For a fixe i, the a i elements m(i, j) will be the (n i 1)imW i istinct matrices having a 1 in one place among the (n i 1)imW i positions available in the first (n i 1) rows of the im V i imw i matrices involve an zeros elsewhere. The remaining c i elements m(i, j) have a 1 in one of the first c i places in the n i -th rows, an zeros elsewhere. Corollary 4.4. (a) If > 0 an V t t 0 then imv < n j=1 ( u imv j ) 2 m ( + im V s i 2. 1)

13 12 STEVEN P. DIAZ AND MARK KLEINER If 0 an V+1 t 0 then: (b) 0 < im V t < n ( u j=1 imv j ) m si (imv ( ) ( ) (c) 0 < + < 2 where γ Γ 1. imv e(γ) +1 s(γ)=t imv s(γ) e(γ)=t )2. ( n j=1 ( u imv j ) 2 m +1 + si (im V )2) 1 Proof. This is a irect consequence of Theorem 3.2 an parts (c), (f), an (g) of Proposition 3.1. Example 4.1. This example shows that the F i,j of Corollary 4.3 must be chosen generically. In other wors the right omnipresent maximal rank property oes not imply the left omnipresent maximal rank property. Let the quiver Γ have two vertices labele 1 an 2 an one arrow β going from 1 to 2. Γ then has in aition one new arrow β going from 2 to 1. For any choice of nonzero functions a an b the relations become ββ = β β = 0. In Theorem 3.2 set = 1 an t = 2. The map becomes V0 1 V1 2 where V0 1 has basis {e 1, β}, an V1 2 has basis {β }. The map is multiplication by β so e 1 goes to β an β goes to 0. Consier one imensional subspaces of V 1 The one spanne by β maps to 0 an so oes not surject onto V1 2, all others o surject onto V 2 Example 4.2. Here we show that if in Theorem 2.1 the hypothesis that T satisfies the right omnipresent maximal rank property is weakene to the right general maximal rank property, then the conclusion might not follow. In other wors the right general maximal rank property oes not imply the left general maximal rank property. Let V be a vector space of imension 3 with basis {v 1, v 2, v 3 }. Let W be a vector space of imension 2 with basis {w 1, w 2 }. Let U be the quotient of V W by the subspace spanne by {v 1 w 1, v 2 w 1 }. Finally let T : V W U be the quotient map. The only one-imensional subspace W of W such that V W has nonzero intersection with the kernel of T is the span of w 1. Thus T satisfies the right general maximal rank property. Any subspace V of V of imension 2 must have nonzero intersection with the span of {v 1, v 2 }. Thus V W must have nonzero intersection with the span of {v 1 w 1, v 2 w 1 }. This means that the restriction of T to V W cannot have maximal rank. References [1] M. Auslaner, I. Reiten an S. O. Smalø, Representation Theory of Artin Algebras, Cambrige Stuies in Avance Mathematics, vol. 36, Cambrige University Press, New York, [2] J. A. Green, Locally Finite Representations, J. Alg. 41 (1976), [3] P. Griffiths an J. Harris, Principles of algebraic geometry, Pure an Applie Mathematics, Wiley-Interscience [John Wiley & Sons], New York, [4] M. Hochster, D. Laksov, The Linear Syzygies of Generic Forms, Comm. Alg. 15 (1987), no. 1-2, [5] M. Kleiner, The Grae Preprojective Algebra of a Quiver, Bull. Lonon Math. Soc. 36 (2004), Department of Mathematics, Syracuse University, Syracuse, New York , Unite States of America aress: spiaz@syr.eu aress: mkleiner@syr.eu 0. 1.