(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, HumboldtUniversitat 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 twodimensional cones (corresponding to twodimensional 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 DAADfellowship. 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 \barvalue" 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 barvalue 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 S`nE` 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) ICalgebra. Then, the Bmodule 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 Bmodule Q all maps d 0 : B! Q meeting the properties similar to (i) factorize through d by a uniquely determined Blinear 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 IClinear map IC[S]! Z n Z IC[ Z n ] ; x s 7! s x s is a ICderivation: 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 ICvector 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 onedimensional vector space concentrated in the singular point. 3 Twodimensional, cyclic quotient singularities (3.1) Twodimensional, 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 twodimensional 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 Zbasis 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 IRbasis of IR, and, for an s Z, the condition \s S" is equivalent to having nonnegative 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 nonnegative 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 barvalues. 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 nonnegative 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 (barvalue 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 barvalue 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 barvalue 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 barvalue ` 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 twodimensional 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 ith 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 ith 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 ith 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), SpringerVerlag,
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