(1.) For any subset P S we denote by L(P ) the abelian group of integral relations between elements of P, i.e. L(P ) := ker Z P! span Z P S S : For ea

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1 Torsion of dierentials on toric varieties Klaus Altmann Institut fur reine Mathematik, Humboldt-Universitat zu Berlin Ziegelstr. 13a, D Berlin, Germany. Abstract We introduce an invariant for semigroups with cancellation property. When the semigroup equals the set of lattice points in a rational, polyhedral cone, then this invariant describes the torsion of the dierential sheaf on the associated toric variety. Finally, as an example, we present the case of two-dimensional cones (corresponding to two-dimensional cyclic quotient singularities). 1 An invariant for semigroups (1.1) Let S be a commutative semigroup with 0 and cancellation property (i.e. a + s = b + s implies a = b for a; b; s S). In particular, S can be embedded into a group, and the notion a for a S makes sense. Assume that (inside this group) S \ ( S) = f0g; then via S turns also into a partially ordered set. a b :() a b S ; For each  S we will dene a certain abelian group T. Their direct sum T := S T or the T's itself should be considered an invariant of the original semigroup. If S is nitely generated, then so will the T's. In case S equals the set of lattice points of some rational, polyhedral cone (i.e. generated by a nite set of rays through rational points) in IR n, we will show that T IC coincides with the torsion of the dierential sheaf 1 Y of the ane toric variety Y = Spec IC[S] ( IC[S] denotes the semigroup algebra of S over IC). This paper was written at M.I.T. and supported by a DAAD-fellowship. 1

2 (1.) For any subset P S we denote by L(P ) the abelian group of integral relations between elements of P, i.e. L(P ) := ker Z P! span Z P S S : For each q L(P ) we dene its support as supp q := fs P j q s 6= 0g ; it is a nite subset of P. Then, every q L(P ) is uniquely representable as a dierence q = q + q with q + ; q IN P and supp q + \ supp q = ;. In particular, ss q + s s = ss q s s span IN P S ; and we denote this element by q (the \bar-value" of q). Denition: For every  S let S := fs S j s g. Then, we dene T := L(S). hq L(S) j q i = hq L(S) j supp q i. hq L(S) j q i : (hq L(S) j q i means the subgroup generated by those q's, and \supp q " is just an abbreviation for \s  for all s supp q".) (1.3) Proposition: Let E S be a subset that generates the semigroup S. Denoting E := E \ S = fs E j s g, we obtain for  S T = L(E). hq L(E) j q i = hq L(E) j supp q i. hq L(E) j q i : Proof: We regard the canonical map ' : L(E). hq L(E) j q i! L(S). hq L(S) j q i : ' is surjective: Suppose a relation q L(S) involves s 1 ;... ; s N S (i.e. supp q = fs 1 ;... ; s N g). Each s v can be represented as a sum of elements from E providing a special relation q(s v ) with q(s v ) = s v . Then, if q equals the relation [ P v q v s v = 0], we obtain q P v q v q(s v ) L(E). (By the way, ~q := q P v q v q(s v ) can be regarded as similar to q, but each occurrence of s v is replaced by an appropriate sum of elements from E. In particular, the bar-value of ~q is less or equal than that of q.) P ' is injective: Let q L(E) such that '(q) = 0, i.e. it splits into a sum q = i qi with q i L(S); q i . Treating each of the summands as we did q previously, we obtain q i = ~q i + q i v q(s v ) with ~q i L(E); ~q i ; and s v S: v

3 Hence, up to relations from hq L(E) j q i, we may assume that q = s v SnE g v q(s v ) L(S) Z S for some integer coecients g v. However, since q is actually contained in L(E) Z E, evaluating this equation at each of the sv S n E yields always g v = 0. Corollary: If S is a nitely generated semigroup, then the T are nitely generated abelian groups. (1.4) We conclude this general section with a trivial, but important lemma. Lemma: Let  S, and dene S 0 S as the semigroup generated by S. The relation \ S 0" associated to S 0 is not the restriction of that of S; in general, \ S 0" is stricter than \ S ". However, (i) (S 0 ) = S, and (ii) the abelian groups T coincide for S and S 0. Proof: For (i) assume that s S fullls s S . By denition, there is an t S such that s + t = . In particular, s; t S S 0, and we obtain s S 0 . The same trick works for (ii): If q L(S) is a relation with q S , then there exists a t S such that q + t = . In particular, t S S 0, hence q S 0 . Hence, to compute T, we may always change the semigroup and assume that E = E. Rational, polyhedral cones (.1) For this section we assume that S Z n equals the set of lattice points of some rational, polyhedral cone in IR n. In particular, it denes an ane toric variety Y := Spec IC[S]. (For general facts about toric varieties see [Da] or [Od]). Denition: Let Y = Spec B be an arbitrary ane scheme over the complex numbers, i.e. let B be an arbitrary (commutative) IC-algebra. Then, the B-module of Kahler dierentials 1 Y together with the canonical dierential d : B! 1 Y are dened by the following universal property: (i) d is additive, kills the constants, and it fullls the rule d(f g) = f d(g) + g d(f) for f; g B: 3

4 (ii) For each B-module Q all maps d 0 : B! Q meeting the properties similar to (i) factorize through d by a uniquely determined B-linear map 1 Y! Q. Indeed, the Kahler dierentials 1 Y do exist, and they are uniquely determined by the previous denition. For a proof of these facts and basic properties (for instance the two fundamental exact sequences) see Chapter 10 of [Ma]. Theorem: In the toric situation the IC[S]-module of Kahler dierentials 1 Y torsion submodule tors ( 1 Y ) are Z n -graded. For  S we have and its tors ( 1 Y )() = T Z IC ; and tors ( 1 Y ) vanishes in the remaining degrees. Moreover, if s S, then the canonical map T! T+s describes the multiplication with x s IC[S]. (.) Proof: Step 1: Let E = fs 0 ;... ; s w g be a generating set of the P semigroup S; in particular, we have a surjective map : IN w+1! S ; a 7! i a i s i. Let m := f(a; b) j a; b IN w+1 ; (a) = (b) in Sg. Then, Y can be regarded as the closed subset of IC w+1 dened by the ideal I = (z a z b j (a; b) m) IC[z 0 ;... ; z w ]. We obtain the standard exact sequence (cf. Theorem 58 in [Ma]) inducing I=I d! 1 IC w+1 IC[z 0 ;...;z w] IC[S]! 1 Y! 0 ; IC[S] m! IC[S] w+1! 1 Y! 0 via composing d with the canonical surjection IC[S] m! I=I. The maps in the latter exact sequence could be described as follows: Denote by fe ab g and fe i g the standard bases of IC[S] m and IC[S] w+1, respectively; and for an s S denote by x s IC[S] the corresponding element in the semigroup algebra, e.g. x si z i (mod I). Then, P the image of e ab in I=I is the equation z a z b, and this maps onto d(z a z b w ) = (a i=0 i b i ) x (a) si d z i 1 IC IC[S]. In particular, w+1 (e ab ) = w i=0 (a i b i ) x (a) si e i : (At the rst glance, might not always map into IC[S] w+1. However, if (a) s i = S, then this would imply a i = b i = 0.) Finally, e i IC[S] w+1 maps onto d x si 1 Y by the second map. Step : The IC-linear map IC[S]! Z n Z IC[ Z n ] ; x s 7! s x s is a IC-derivation: It kills the constants, and for s; t S we have x s (t x t ) + x t (s x s ) = (s + t) x s+t. Hence, by denition of the Kahler dierentials, we obtain a IC[S]-linear map 1 Y! Z n Z IC[ Z n ] ; d x s 7! s x s : 4

5 On the other hand, Z n Z IC[ Z n ] can be identied with the module of Kahler dierentials on the torus ( IC ) n = Spec IC[ Z n ], and the previous map corresponds to the restriction of dierentials from Y onto the open subset ( IC ) n Y. Since 1 ( IC is just a localisation of 1 ) n Y, we obtain that this map has exactly tors ( 1 Y ) as its kernel. Putting this fact together with the information from the rst step, we obtain the sequence of IC[S]-modules () IC[S] m! IC[S] w+1! Z n Z IC[ Z n ] and, moreover, tors ( 1 Y ) = ker =im. e i 7! s i x si Step 3: Dening deg(e ab ) := (a) = (b) S for IC[S] m, deg(e i ) := s i S for IC[S] w+1, and deg(s x t ) := t Z n for Z n Z IC[ Z n ], the sequence () turns out to be Z n -graded. In particular, to calculate ker =im, we can deal with each degree  S separately (degrees from Z n n S do not appear). Claim: T Z IC! (ker =im ) ; q 7! P s i E q i x si e i is an isomorphism of IC-vector spaces. Proof: For q L(E), the sum P s i E q i x si e i denes an element of ( IC[S] w+1 ). Applying, we obtain ( P s i E q i x si e i ) = P s i E q i x si (s i x si ) = ( P s i E q i s i ) x = 0 : P Obviously, each element of (ker ) can be obtained that way: If s i E q i(x) e i IC[S] w+1 is of degree , we obtain q i (x) = q x si i (q i IC with q i = 0 P unless  s i ). Moreover, we have just seen that kills this element if and only if s i E q i s i = 0, i.e. it comes from a relation q L(E). On the other hand, q L(E) with q  can be written as q = a b (a; b IN w+1 ; (a) = (b) ). Then, s i E (a i b i ) x si e i = x (a) (e ab ) ; and those elements generate (im ). (.3) Remark: We never used the fact that Y = Spec IC[S] is a normal variety. Hence, Theorem (.1) remains true, if S Z n is just supposed to be a nitely generated semigroup with S \ ( S) = f0g. In particular, over IR, S generates a rational, polyhedral cone IR n, but there may exist nitely many points of \ Z n not belonging to S. 5

6 Example: If S IN, then Y is a curve singularity. For instance, S := IN n f1g is generated by the integers and 3, and Y IC is the cusp given by the equation y x 3 = 0. Let us compute the torsion of 1 Y in this case: With E = fs 0 := ; s 1 := 3g we have the only relation q := [3s 0 s 1 = 0]. Since q S 5, on the one hand, but q = 6, on the other hand, we obtain T = T 5 = Z q, i.e. tors ( 1 Y ) is a one-dimensional vector space concentrated in the singular point. 3 Two-dimensional, cyclic quotient singularities (3.1) Two-dimensional, cyclic quotient singularities coincide with the twodimensional, ane toric varieties; they were rst investigated by Riemenschneider, [Ri]. We want to compute our invariant for this special case. Let S be the set of lattice points in the two-dimensional cone spanned by some primitive vectors s 0 ; s w Z. Then, following x1.6 in [Od], E = fs 0 ;... ; s w g is built from all lattice points of the compact part of the boundary of conv (S n f0g); in (S n f0g)) consists of w primitive edges (containing no interior lattice point). Every pair of adjacent elements s i 1 ; s i (i = 1;... ; w) provides a Z-basis for the lattice Z, and there are relations s i 1 + s i+1 = a i s i (a i IN; a i ; i = 1;... ; w 1) : rs w r E EEE r s 4 This is a rough draft of S; in fact it cannot exactly be shaped as shown in the gure. E r s 3 r s s r EJJaaaaahhhhhhhhhr 1 Remark: (cf. Lemma 1.0 in [Od]) The coecients a i can be obtained from the continued fraction n n q = a 1 1 a 1 a s 0

7 with n := det(s 0 ; s w ) and q f1;... ; n 1g such that nj(qs 0 + s w ). (3.) Since a i (i = 1;... ; w 1), the sets E E are \connected" (i.e. they are shaped as E = fs p ; s p+1 ;... ; s p+k g for some p; k) for every  S. Hence, to compute T, we may assume that E = fs 0 ;... ; s w g (cf. Lemma (1.4)). Lemma: Let  S such that E = fs 0 ;... ; s w g. Then, (i)  s 0 + s w, and (ii) T = 0 unless  = s 0 + s w. Proof: (i) The vectors s 0 and s w form a IR-basis of IR, and, for an s Z, the condition \s S" is equivalent to having non-negative coordinates only. Hence, the inequalities  s 0 ; s w imply that both coordinates of  are not smaller than 1, i.e.  s 0 s w has non-negative coordinates. (ii) Let  > s 0 + s w be given; we will show that T = 0. We distinguish two cases: Case 1:  (s 0 + s w ) s 0 (or similarly  (s 0 + s w ) s w ). Then, since s i s 0 + s w, we know that s 0 + s i s 0 + s w  for i = ;... ; w. Summing up the equations s j 1 + s j+1 = a j s j (1 j i 1) provides relations q i (q i ) s 0 + s i = (s 1 + s i 1 ) + i 1 j=1 (a j ) s j (i = ;... ; w) having exactly s 0 + s i  as their bar-values. In particular, they yield zero in T. On the other hand, if we are given an arbitrary relation q L(E), then we can use q w ;... ; q to eliminate s w ;... ; s step by step from the support of q (without changing its value in T). Since s 0 and s 1 are linearly independent, q has to be trivial then. Case : Not Case 1; in particular,  (s 0 + s w ) s i for some i = 1;... ; w 1. P Then, we have  (s 0 + s w w 1 ) = g j=1 j s j with non-negative integers g j and, moreover, g i 1. (There are no summands involving s 0 or s w, because this would t in the rst case.) Again, we start P with an arbitrary q L(E). First, we use the relation s 0 + s w = (s 1 + s w 1 w 1 ) + (a j=1 j ) s j (bar-value s 0 + s w ) to eliminate s 0 (if i = 1) or s w (if i ) from supp q. If w =, then we are already done. Otherwise, we use the similar relations expressing s i + s w (if i = 1) or s 0 + s i (if i ) by the generators in between to eliminate s w or s 0, respectively. The result is a relation q with supp q fs 1 ;... ; s w 1 g. Finally, we know that w 1 w 1  = (s 0 + s w ) + j=1 g j s j = (s 1 + s w 1 ) + 7 j=1 (g j + a j ) s j :

8 Hence, the fact T = 0 for the cone spanned by s 1 and s w 1 (induction by w) tells us that L(s 1 ;... ; s w 1 ) is spanned by relations with bar-value not greater than . In particular, our q can be reduced to zero. (3.3) Lemma: Assume  = s 0 + s w (including E = fs 0 ;... ; s w g). Then, T = 8 < : 0 for w Z [(w 1)=] for a 1 =... = a w 1 = Z otherwise : (The second case (a 1 =... = a w 1 = ) means that the points s 0 ;... ; s w are sitting on an ane line; the corresponding cyclic quotient singularity equals the cone over the rational normal curve of degree w.) Proof: The case w is obvious. Assuming a 1 =... = a w 1 =, the entire collection of relations with bar-value not greater than  is given by the [w=] equations s 0 + s w = s 1 + s w 1 =... = s [w=] + s w [w=] ; and we obtain (w 1) [w=] = [(w 1)=] for the rank of T. Finally, if w 3 and not a 1 =... = a w 1 =, then  = s 0 + s w = s 1 + s w 1 + w 1 i=1 (a i ) s i > hs 1 ;s w 1 i s 1 + s w 1 shows (cf. Lemma (3.)) that L(s 1 ;... ; s w 1 ) hq L(E) j q i. On the other hand, the only relations involving s 0 or s w and having a bar-value  are those representing s 0 + s w as linear combination of s 1 ;... ; s w 1. (Indeed, if for instance P P P s 0 w + g i=0 i s i w = h i=1 i s i P is such a relation (g i ; h i 0; g i h i = 0), then s 0 w + g i=0 i s i  = s 0 + s w implies i g i s i s w. Hence g w = 0 or 1, and g i = 0 for i 6= w. Moreover, since s 0 is not representable by other generators, this implies that s 0 + s w forms the left hand side of the relation - and the right hand side has to be built from s 1 ;... ; s w 1 then.) (3.4) We are gathering our results and obtain the following description of the invariants T for a two-dimensional cyclic quotient singularity: Theorem: Let S and a 1 ;... ; a w 1 as in (3.1); dene a 0 := a w := 3. (1) Let s p ; s p+k E (0 p < p + k w) be elements such that (i) k 3, and (ii) at least one of the numbers a p ;... ; a p+k is greater than two. 8

9 Then,  := s p + s p+k uniquely determines p and k, and we have T = Z [(k 1)=] for a p+1 =... = a p+k 1 = Z otherwise : The abelian group T vanishes in the remaining degrees. () For T = T we obtain T = Z (w 1)(w )=. In particular, w 1 dim (tors ( 1 Y )) = : (w + 1 equals the embedding dimension of the cyclic quotient singularity Y ). Proof: For (1) assume that we are given some  S. Then, E is shaped as E = fs p ;... ; s p+k g, and by Lemma (3.) we know that T = 0 unless  = s p + s p+k. On the other hand, if  = s p + s p+k, then Lemma (3.3) tells us about T. The only thing being left is asking the other way around: What is the condition for an  := s p + s p+k to yield exactly E = fs p ;... ; s p+k g? Obviously, E does always contain fs p ;... ; s p+k g, and we show that it is exactly the condition a p =... = a p+k = saying that both sets are not equal: Assume E = fs p i ;... ; s p ;... ; s p+k ;... ; s p+k+j g (w.l.o.g. i j 0, i 6= 0), then we obtain a chain of inequalities s p + s p+k =  s p i + s p+k+j s p i+1 + s p+k+j 1... s p i+j + s p+k : If i > j, this would imply that two dierent elements of E (s p and s p i+j ) would be comparable in S. Hence, i = j, and all signs in the previous chain turn into equalities implying a p i+1 =... = a p =... = a p+k =... = a p+k+i 1 =. The reversed direction is easy; the equalities a p =... = a p+k = imply  = s p 1 + s p+k+1, hence s p 1 ; s p+k+1 E. To prove the second part of the theorem, we have to count dimensions. Assume that the compact part of the (S n f0g)) consists of P m edges, each m containing w i 1 (i = 1;... ; m) interior lattice points. In particular, w i=1 i = w. Then, we have (a) w+1 possibilities of choosing two dierent points sp and s p+k from E; (b) w 1 of those pairs with 1 k ; w (c) i +1 possibilities of choosing two dierent points sp and s p+k from the i-th edge; (d) w i 1 of those pairs with 1 k. 9

10 Hence, we obtain w + 1 (w 1) m i=1 wi m i=1 w 1 (w i 1) = m i=1 wi 1 possibilities of choosing pairs s p ; s p+k E with k 3 and such that at least one of the numbers a p+1 ;... ; a p+k 1 is greater than two. Those pairs yield T = Z. On the other hand, let fs q ;... ; s q+w ig form the i-th edge. Then, its only pairs (s p ; s p+k ) meeting the assumption that at least one of the numbers a p ;... ; a p+k is greater than two are (s q ; s q+k ) and (s q+wi k ; s q+w i ) with 1 k w i. Each of them providing a contribution of Z [(k 1)=] (which is automatically zero if k = 1; ), we obtain for the entire i-th edge w i 1 k 1 wi 1 wi = k=3 dimensions for T. Acknowledgement: I would like to thank the referee for many useful hints and suggestions. References [Da] Danilov, V.I.: The Geometry of Toric Varieties. Russian Math. Surveys 33/ (1978), [Ma] Matsumura, H.: Commutative Algebra. W.A.Benjamin, New York [Ri] Riemenschneider, O.: Deformationen von Quotientensingularitaten (nach zyklischen Gruppen). Math. Ann. 09 (1974), [Od] Oda, T.: Convex bodies and algebraic geometry. Ergebnisse der Mathematik und ihrer Grenzgebiete (3/15), Springer-Verlag,

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