A CHARACTERISATION OF THE Z n Z(δ) LATTICE AND DEFINITE NONUNIMODULAR INTERSECTION FORMS

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1 A CHARACTERISATION OF THE Z n Z(δ) LATTICE AND DEFINITE NONUNIMODULAR INTERSECTION FORMS BRENDAN OWENS AND SAŠO STRLE Abstract We rove a generalisation of Elkies characterisation of the Z n lattice to nonunimodular definite forms (and lattices) Combined with inequalities of Frøyshov and of Ozsváth and Szabó, this gives a simle test of whether a rational homology three-shere may bound a definite four-manifold As an examle we show that small ositive surgeries on torus knots do not bound negative-definite four-manifolds 1 Introduction The intersection airing of a smooth comact four-manifold, ossibly with boundary, is an integral symmetric bilinear form Q X on H (X; Z)/Tors; it is nondegenerate if the boundary of X has first Betti number zero Characteristic covectors for Q X are elements of H (X; Z)/Tors reresented by the first Chern classes of sin c structures In the case that Q X is ositive-definite there are inequalities due to Frøyshov (using Seiberg-Witten theory, see [3]) and Ozsváth and Szabó which give lower bounds on the square of a characteristic covector It is helful in this context to rove existence of characteristic covectors with small square The following result was conjectured by the authors in [10] Theorem 1 Let Q be an integral ositive-definite symmetric bilinear form of rank n and determinant δ Then there exists a characteristic covector ξ for Q with n 1 + 1/δ if δ is odd, ξ n 1 if δ is even; this inequality is strict unless Q = (n 1) 1 δ Moreover, the two sides of the inequality are congruent modulo 4/δ For unimodular forms the first art of the theorem was roved by Elkies [] To rove its generalisation we reinterret the statement in terms of integral lattices In Section we introduce the necessary notions and then study short characteristic Date: June 18, 01 B Owens was suorted in art by NSF grant DMS and by the LSU Council on Research Summer Stiend Program SStrle was suorted in art by the ARRS of the Reublic of Slovenia research rogram No P

2 BRENDAN OWENS AND SAŠO STRLE covectors of some secial tyes of lattices The roof of the first art of the theorem is resented in Section 3 The main idea is to use the theory of linking airings to embed four coies of the lattice into a unimodular lattice and then aly Elkies theorem along with results of Section For unimodular forms, a well-known and useful constraint on the square of a characteristic vector is that ξ is congruent modulo 8 to the signature of the form In Section 4 we give a generalisation of this to nonunimodular forms, which secialises to give the congruence in Theorem 1 In Section 5 we combine Theorem 1 with an inequality of Ozsváth and Szabó [1, Theorem 96] to obtain the following theorem, where d(y, t) denotes the correction term invariant defined in [1] for a sin c structure t on a rational homology threeshere Y : Theorem Let Y be a rational homology shere with H 1 (Y ; Z) = δ If Y bounds a negative-definite four-manifold X, and if either δ is square-free or there is no torsion in H 1 (X; Z), then 1 1/δ if δ is odd, max 4d(Y, t) t Sin c (Y ) 1 if δ is even The inequality is strict unless the intersection form of X is (n 1) 1 δ Moreover, the two sides of the inequality are congruent modulo 4/δ Some alications of Theorem are discussed in [10] As a further alication we consider manifolds obtained as integer surgeries on knots Knowing which ositive surgeries on a knot bound negative-definite manifolds has imlications for existence of fillable contact structures and for the unknotting number of the knot (see eg [7, 11]) In articular we consider torus knots; it is well known that (q 1)-surgery on the torus knot T,q is a lens sace It follows that all large ( q 1) surgeries on torus knots bound both ositive- and negative-definite four-manifolds (see eg [11]) We show in Section 7 that small ositive surgeries on torus knots cannot bound negative-definite four-manifolds In articular, we obtain the following result; for a more recise statement see Corollary 7 Proosition 3 Let < q and 1 n ( 1)(q 1) + Then +n-surgery on T,q cannot bound a negative-definite four-manifold with no torsion in H 1 There has been further rogress on the question of which Dehn surgeries on torus knots bound negative-definite manifolds in the time since this aer was originally written In articular Proosition 3 has been imroved in many cases by Greene [4] A comlete answer to the question will aear in [11] Acknowledgements It is a leasure to thank Noam Elkies who introduced to us the notion of gluing of lattices, which is used in the roof of Proosition 31; John Morgan, who demonstrated a version of Corollary 43 to the first author; Peter

3 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 3 Ozsváth, who exlained to us a roof of Theorem 61; and Ed Swartz, for a helful conversation regarding the money-changing roblem We also thank the referee for a careful reading of the aer We started this roject while we were Britton Postdoctoral Fellows at McMaster University, and some of the work was carried out while the first author was visiting Cornell University Short characteristic covectors in secial cases An integral lattice L of rank n is the free abelian grou Z n along with a nondegenerate symmetric bilinear form Q : Z n Z n Z Let V denote the tensor roduct of L with R; thus V is the vector sace R n to which the form Q extends, and L is a full rank lattice in V The signature σ(l) of the lattice L is the signature of Q We say L is definite if σ(l) = n For convenience denote Q(x, y) by x y and Q(y, y) by y A set of generators v 1,, v n for L forms a basis of V satisfying v i v j Z With resect to such a basis the form Q is reresented by the matrix [v i v j ] n i,j=1 The determinant (or discriminant) of L is the determinant of the form (or the corresonding matrix) We will write simly Z n for the standard ositive-definite unimodular lattice of discriminant 1 (the form on this lattice is the standard inner-roduct form) In what follows we will use the well known fact that for n < 8 this is the only ositive-definite unimodular lattice; this also easily follows from Elkies Theorem (Theorem 1 with δ = 1) and the congruence condition for characteristic vectors The dual lattice L = Hom(L, Z) consists of all vectors x V satisfying x y Z for all y L A characteristic covector for L is an element ξ L with ξ y y (mod ) for all y L We say a lattice is comlex if it admits an automorhism i with i given by multilication by 1 A lattice is quaternionic if it admits an action of the quaternionic grou ±1, ±i, ±j, ±k}, with 1 acting by multilication Note that the rank of a comlex lattice is even, and the rank of a quaternionic lattice is divisible by 4 For any lattice L, let L m denote the direct sum of m coies of L There is a standard way to make L L into a comlex lattice and L 4 into a quaternionic lattice; for examle, the quaternionic structure is given by i : (x, y, z, w) ( y, x, w, z) j : (x, y, z, w) ( z, w, x, y) Let Z(δ) denote the rank one lattice with determinant δ; in articular Z = Z(1) Lemma 1 Let δ N, and let be a rime Let L be an index sublattice of Z n 1 Z(δ) Then L has a characteristic covector ξ with ξ < n 1 unless L = Z n 1 Z( δ)

4 4 BRENDAN OWENS AND SAŠO STRLE Proof We may assume L contains none of the summands of Z n 1 Z(δ); any such summand of L contributes 1 to the right-hand side of the inequality and at most 1 to the left-hand side If n = 1 then clearly L = Z( δ) Now suose n > 1 Let e, e 1,, e n, f} be a basis for Z n 1 Z(δ), where e, e 1,, e n have square 1 and f = δ Then multiles of e give coset reresentatives of L in Z n 1 Z(δ); it follows that a basis for L is given by e, e 1 + s 1 e,, e n + s n e, f + te} Here s i, t are nonzero residues modulo in [1, 1], whose arities we may choose if is odd With resect to this basis, the bilinear form on L has matrix with inverse Q = s 1 s s 3 s n t s s 1 s 1 s s 1 s 3 s 1 s n s 1 t s n s 1 s n s s n s 3 s n 1 + s n s n t t s 1 t s t s 3 t s n t δ + t 1+ s i +t /δ s 1 s s 3 s n s Q 1 = s n t δ t δ δ With resect to the dual basis, an element of the dual lattice L is reresented by an n-tule ξ Z n An n-tule corresonds to a characteristic covector if its comonents have the same arity as the corresonding diagonal elements of Q Suose now that is odd Choose s i to be odd for all i and t δ (mod ) Then ξ = (1, 0,, 0) is a characteristic covector whose square satisfies ξ = 1 + s i + t /δ < n 1, noting that s i, t 1 Finally if = then ξ = (0, 0,, 0, (1+δ) mod ) is a characteristic covector with ξ < n 1 Lemma Let δ N be odd, and let be a rime with = or 1 (mod 4) Let M be an index comlex sublattice of Z n Z(δ) Then M has a characteristic covector ξ with ξ < n unless M = Z n Z(δ),

5 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 5 Proof We may assume M contains none of the summands of Z n Z(δ) Suose first that n = 1 Let e, ie} be a basis for Z(δ) with e = δ Then e, ie + se} is a basis for M, for some s Now i(ie + se) = e + sie M = e s e M, from which it follows that divides 1 + s The bilinear form on L has matrix ( ) ( ) Q = δ sδ s sδ δ(1 + s = δ 1+s ) s = δi, since any ositive-definite unimodular form of rank is diagonalisable Thus in this case M = Z(δ) Now suose n > 1 Let e, e 1,, e n 3, f 1, f } be a basis for Z n Z(δ), where e, e i have square 1 and fi = δ A basis for M is given by e, e i + s i e, f i + t i e}, where s i, t i are nonzero residues modulo which we may choose to be odd integers in [1, 1] The matrix in this basis is s 1 s s 3 s n 3 t 1 t s s 1 s 1 s s 1 s 3 s 1 s n 3 s 1 t 1 s 1 t s s 1 s 1 + s s s 3 s s n 3 s t 1 s t Q =, s n 3 s 1 s n 3 s s n 3 s 3 s n s n 3 s n 3 t 1 s n 3 t t 1 s 1 t 1 s t 1 s 3 t 1 s n 3 t δ + t 1 t 1 t t s 1 t s t s 3 t s n 3 t t 1 t δ + t with inverse Q 1 = 1+ s i + t i /δ s 1 s s n 3 t 1 δ t δ s s s n t 1 1 δ δ t 1 δ δ If = then Q is even, and ξ = 0 is characteristic

6 6 BRENDAN OWENS AND SAŠO STRLE Suose now that is odd Then ξ = (k, l 1,, l n 1 ) is characteristic if k is odd and each l i is even By comleting squares we have that ) ξ = 1 n 3 (k + (ks i l i ) + 1 (kt i l n 3+i ) δ Notice that for odd k < and for aroriate choice of even l i each of the squares in the sum above is less than Thus it suffices to rove the statement for δ = 1 and n = ; the following sublemma comletes the roof 3 Sublemma 3 Let F (k, l 1, l, l 3 ) = k + (ks i l i ), where s i are odd numbers with s i Then there exist k odd, k <, and l i even for which F < Proof Let K =, 4,, }, and let K i = k K l i even with ks i l i < 1 } Note that multilication by s i followed by reduction modulo induces a ermutation of K Since 1 (mod 4) it follows that K i = 1 If there exists k 3 K i then with this choice of k and the corresonding choices for l 1, l, l 3 we find ( ) 1 F < + 3 < Otherwise let K ij denote K i K j Then K i = K i K ij, i<j from which it follows that i<j K ij 1 Thus there exists some k K ij with k +1 With this k and aroriate l i we find ( ) ( ) F < + + < Lemma 4 Let δ N be odd, and let q be a rime with q 3 (mod 4) Let N be an index q quaternionic sublattice of Z 4n 4 Z(δ) 4, with the quotient grou (Z 4n 4 Z(δ) 4 )/N having exonent q Then N has a characteristic covector ξ with ξ < 4n 4 unless N = Z 4n 4 Z(qδ) 4 Proof We may assume N contains none of the summands of Z 4n 4 Z(δ) 4 Suose first that n = 1 Let e, ie, je, ke} be a basis for Z(δ) 4 with e = δ Note that e and ie reresent generators of the quotient Z/q Z/q; otherwise we have e + sie N for

7 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 7 some s, and multilication by i yields s (mod q), but 1 is not a quadratic residue modulo q Now a basis for N is given by qe, qie, je+s 1 e+t 1 ie, ke+s e+t ie} The quaternionic symmetry yields s 1 t, s t 1, 1 + s 1 + s 0 (mod q); it follows that the bilinear form Q on N factors as qδ times a unimodular form Thus N = Z(qδ) 4 We now assume n = and δ = 1 (the roof for any other n, δ follows from this case) Let e, ie, v 1 = je, v = ke, v 3 = f, v 4 = if, v 5 = jf, v 6 = kf} be a basis of unit vectors for Z 8 Then qe, qie, v 1 + s 1 e + t 1 ie,, v 6 + s 6 e + t 6 ie} is a basis for N Invariance of N under multilication by i yields s i+1 t i, t i+1 s i (mod q) for i = 1, 3, 5 Using further quaternionic symmetry it follows that 1 + s 1 + s 0 (mod q) s 5 + s 1 s 3 + s s 4 0 (mod q) s 6 + s s 3 s 1 s 4 0 (mod q), hence s 1, s 0 (mod q) From the assumtion that no basis vector for Z 8 is contained in N it follows that s 3, s 4 cannot both be divisible by q, and that also s 5, s 6 cannot both be divisible by q Combining this with the last two congruences above it follows that at most one of s 3,, s 6 is divisible by q We may choose s i, t i in [1 q, q 1] with s i + t i odd Comutation of Q, Q 1 in the above basis for N now yields that ξ = (k 1, k, l 1,, l 6 ) is characteristic if k 1, k are odd and l i are even, and ) ξ = 1 6 (k q 1 + k + (k 1 s i + k t i l i q) There are now two cases to consider, deending on whether or not one of the s i is zero Existence of a characteristic ξ with ξ < 4 follows in each case from one of the following sublemmas Sublemma 5 Let F=k 1 + k + 6 (k 1 s i + k t i l i q), where s i are odd and t i are even integers in [1 q, q 1] Then there exist k 1, k odd and l i even for which F < 4q Proof Set k = 1 Let K = q, 4 q,, q }, and let K i = k K l i even with ks i + t i l i q q 1 }

8 8 BRENDAN OWENS AND SAŠO STRLE Note that K i q 1 (in fact K i either contains q 1 or q+1 elements, deending on the value of t i ) If there is a k which is in the intersection of four of the K i s, then setting k 1 = k and choosing aroriate l i we obtain ( ) q 1 F 1 + 3q + 4 < 4q Otherwise let K ijk denote the trile intersection of K i, K j, K k, and let K ij denote K i K j k K ijk Then 6 q 1 6 K i = K i K ij K ijk, i<j from which it follows that K ij + K ijk (q 1), and hence i<j i<j<k i<j<k K ijk q 1 i<j<k Thus there exists some k K ijk with k q 1 Setting k 1 = k and choosing corresonding l i again yields ( ) q 1 F 1 + 3q + 4 < 4q Sublemma 6 Let F =k 1 + k + 5 (k 1 s i + k t i l i q) + (k t 6 l 6 q), where s i and t 6 are odd, and t 1,, t 5 are even integers in [1 q, q 1] Then there exist k 1, k odd and l i even for which F < 4q Proof First choose k (and l 6 ) with k, k t 6 l 6 q q 1 Again let K = q, 4 q,, q }, and for each i 5 let K i = k K l i even with ks i + k t i l i q q 1 } ; then K i q 1 If there is a k which is in the intersection of four of the K i s, then for k 1 = k and aroriate l i we obtain ( ) q 1 F q + 6 < 4q

9 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 9 Otherwise (with notation as above) we have q 1 5 K i = 5 K i K ij K ijk, i<j from which it follows that K ij + and hence i<j i<j<k i<j<k K ijk 3 (q 1), K ijk q i<j<k Thus there exists some k K ijk with k 3(q+1) Setting k 4 1 = k and choosing corresonding l i yields ( ) ( ) q 1 3(q + 1) F q < 4q 4 3 Proof of the main theorem The bilinear form Q on L induces a symmetric bilinear Q/Z-valued airing on the finite grou L /L, called the linking airing associated to L Such airings λ on a finite grou G were studied by Wall [19] (see also [1, 5]) He observed that λ slits into an orthogonal sum of airings λ on the -subgrous G of G For any rime let A k (res B k) denote the airing on Z/ k for which k times the square of a generator is a quadratic residue (res nonresidue) modulo If is odd then λ decomoses into an orthogonal direct sum of these two tyes of airings on cyclic subgrous The airing on G may be decomosed into cyclic summands (there are four equivalence classes of airings on Z/ k for k 3), lus two tyes of airings on Z/ k Z/ k ; these are denoted E k, F k For an aroriate choice of generators, in E k each cyclic generator has square 0, while in F k they each have square 1 k Proosition 31 Let L be a lattice of rank n Then L 4 may be embedded as a quaternionic sublattice of a unimodular quaternionic lattice U of rank 4n Proof Consider an orthogonal decomosition of the linking airing on L /L as described above Let x L reresent a generator of a cyclic summand Z/ k with k > 1 Then v = k 1 x has v Z and v L Adjoining v to L yields a lattice L 1 which contains L as an index sublattice (so that det L = det L 1 ) Similarly, if x L reresents a generator of a cyclic summand of E k or F k then v = k 1 x may be

10 10 BRENDAN OWENS AND SAŠO STRLE adjoined Finally if x 1, x reresent generators of two Z/ summands then x 1 + x may be adjoined In this way we get a sequence of embeddings L = L 0 L 1 L m1, where each L i is an index sublattice of L i+1 for some rime Moreover the linking airing of L m1 decomoses into cyclic summands of rime order, and the determinant of L m1 is either odd or twice an odd number Now let M 0 = L m1 L m1 Note that M 0 is a comlex sublattice of L L Let be a rime which is either or is congruent to 1 modulo 4, so that there exists an integer a with a 1 (mod ) Let x L reresent a generator of a Z/ summand of the linking airing of L m1 Then v = (x, ax) L L has v Z and v M 0 Adjoining v to M 0 yields a lattice M 1 ; since iv + av M 0, M 1 is reserved by i Continuing in this way we obtain a sequence of embeddings M 0 M 1 M m, where each M i is an index sublattice of M i+1 for some rime with = or 1 (mod 4) Each M i is a comlex sublattice of L L We may arrange that each M i with i > 0 has odd determinant The resulting linking airing of M m is the orthogonal sum of airings on cyclic grous of rime order congruent to 3 modulo 4, and all the summands aear twice Finally let N 0 = M m M m This is a quaternionic sublattice of (L ) 4 Let q 3 (mod 4) be a rime, and suose that x L generates a Z/q summand of the linking airing of M m There exist integers a, b with a +b 1 (mod q) (take m to be the smallest ositive quadratic nonresidue and choose a, b so that a m 1, b m) Let v 1 = (x, 0, ax, bx) and let v = iv 1 = (0, x, bx, ax) Then v i, v 1 v Z, and qv i N 0 Adjoining v 1, v to N 0 yields a quaternionic sublattice N 1 of (L ) 4 ; note that jv 1 + av 1 bv N 0 We thus obtain a sequence N 0 N 1 N m3 = U, where each N i is an index q sublattice of N i+1 for some rime q 3 (mod 4), N i+1 /N i has exonent q, each N i is quaternionic with odd determinant, and U is unimodular Remark 3 Using the methods of Proosition 31 one may show that L embeds in a unimodular lattice if and only if the rime factors of det L congruent to 3 mod 4 have even exonents These embedding results for L and L 4 (without the quaternionic condition in Proosition 31) also follow from [1, Theorem 17] Alying this to rank 1 lattices one recovers the classical fact that any integer is exressible as the sum of 4 squares, and is the sum of squares if and only if its rime factors congruent to 3 mod 4 have even exonents Proof of Theorem 1 Let L be the lattice with form Q Embed L 4 in a unimodular lattice U of rank 4n as in Proosition 31 By Elkies theorem [] either U = Z 4n or

11 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 11 U has a characteristic covector ξ with ξ 4n 8 In the latter case, restricting to one of the four coies of L yields a covector ξ L with (ξ L ) n, which easily satisfies the statement of the theorem This leaves the case that U = Z 4n Let L 0,, L m1, M 0,, M m, N 0,, N m3 be as in the roof of Proosition 31 By successive alication of Lemma 4, we see that either N 0 = Z 4n 4 Z(δ 1 ) 4 for some δ 1 N, or N 0 has a characteristic covector ξ with ξ < 4n 4 In the latter case, restricting ξ to one of the four coies of L yields (ξ L ) < n 1 If N 0 = Z 4n 4 Z(δ 1 ) 4 then M m = Z n Z(δ 1 ) By Lemma we find that either M 0 = Z n Z(δ ) or M 0 has a characteristic covector ξ with ξ < n In the latter case, restricting ξ to one of the two coies of L embedded in M 0 yields (ξ L ) < n 1 Finally if M 0 = Z n Z(δ ) then L m1 = Z n 1 Z(δ ) Successive alication of Lemma 1 now yields that either L = Z n 1 Z(δ) or L has a characteristic covector ξ with ξ < n 1 A roof of the congruence is given in the next section 4 A congruence condition on characteristic covectors Given a ositive-definite symmetric bilinear form Q of rank n and determinant δ with Q (n 1) 1 δ, one may ask for an otimal uer bound on the square of a shortest characteristic covector ξ The square of a characteristic covector of a unimodular form is congruent to the signature modulo 8 (see for examle [17]) Thus if δ = 1 we have ξ n 8 In this section we give congruence conditions on the square of characteristic covectors of forms of arbitrary determinant Lattices in this section are not assumed to be definite (The results in this section may be known to exerts but we have not found them in the literature We also note that the remainder of the aer is indeendent from this section) If ξ 1, ξ are characteristic covectors of a lattice L of determinant δ then their difference is divisible by in L ; it follows that ξ1 ξ modulo 4 For a lattice with δ odd determinant the congruence holds modulo 8 We will give a formula for this δ congruence class in terms of the signature and linking airing of L For a lattice with determinant δ even or odd we will determine the congruence class of ξ modulo 4 in δ terms of the signature and determinant Lemma 41 Suose δ = r k i i s j=1 q l j j where i, q j are odd rimes, not necessarily distinct Let M be an even lattice of determinant δ with linking airing isomorhic

12 1 BRENDAN OWENS AND SAŠO STRLE to r A k i i s j=1 B l q j Then the signature of M satisfies the congruence j σ(m) k i 1 () (1 i ) + l j 1 () (5 q j ) (mod 8) (For the definition of A k and B k see the beginning of Section 3) Proof Let G(M) denote the Gauss sum 1 δ u M /M e iπu (See [18] for more details on Gauss sums and the Milgram Gauss sum formula) Then G(M) deends only on the linking airing of M and in fact factors as G(M) = G(A k i i ) G(B l q j ) j The factors are comuted in [18, Theorem 39] to be: 1 if k is even, G(A k) = e πi(1 )/8 if k is odd; 1 if l is even, G(B q l) = e πi(5 q)/8 if l is odd (Note that in the notation of [18], A k corresonds to (C (k); ) and B q l corresonds to (C q (l); n q ), where n q is a quadratic nonresidue modulo q) The congruence on the signature of M now follows from the Milgram formula: G(M) = e πiσ(m)/8 Proosition 4 Let L be a lattice of odd determinant δ with linking airing isomorhic to A k i B l r s i q j Let ξ be a characteristic covector for L Then j j=1 ξ σ(l) k i 1 () (1 i ) l j 1 () (5 q j ) (mod 8/δ) Proof Let M be an even lattice with the same linking airing as L (Existence of M is roved by Wall in [19]) Then by [6, Satz 3] L and M are stably equivalent; that is to say, there exist unimodular lattices U 1, U such that L U 1 = M U Let ξ be a characteristic covector for L U 1 Then we have decomositions ξ = ξ L + ξ U1 = ξ M + ξ U

13 Taking squares we find ξ L A CHARACTERISATION OF THE Z n Z(δ) LATTICE 13 ξ M + ξ U ξ U1 σ(u ) σ(u 1 ) σ(l) σ(m) σ(l) (1 i ) (5 q j ) (mod 8/δ), k i 1 () where the last line follows from Lemma 41 l j 1 () Corollary 43 Let L be a lattice of determinant δ N, and let ξ be a characteristic covector for L Then σ(l) 1 + 1/δ if δ is odd ξ (mod 4/δ) σ(l) 1 if δ is even Proof If δ is odd then Proosition 4 shows that the congruence class of ξ modulo 4 deends only on the signature and determinant; the formula then follows by taking δ L to be the lattice with the form r 1 s 1 δ where r + s = n 1 and r s = σ(l) 1 If δ is even then as in the roof of Proosition 31 we find that either L or L Z() embeds as an index k sublattice of a lattice with odd determinant It again follows that the congruence class of ξ modulo 4 deends only on the signature and δ determinant 5 Proof of Theorem We begin by noting the following restatement of Theorem 1 for negative-definite forms: Theorem 51 Let Q be an integral negative-definite symmetric bilinear form of rank n and determinant of absolute value δ Then there exists a characteristic covector ξ for Q with n + 1 1/δ if δ is odd, ξ n + 1 if δ is even; this inequality is strict unless Q = (n 1) 1 δ Moreover, the two sides of the inequality are congruent modulo 4/δ Let Y be a rational homology three-shere and X a smooth negative-definite fourmanifold bounded by Y, with b (X) = n For any Sin c structure t on Y let d(y, t) denote the correction term invariant of Ozsváth and Szabó [1] It is shown in [1, Theorem 96] that for each Sin c structure s Sin c (X), (1) c 1 (s) + n 4d(Y, s Y ) Also from [1, Theorem 1], the two sides of (1) are congruent modulo 8

14 14 BRENDAN OWENS AND SAŠO STRLE The image of c 1 (s) in H (X; Z)/Tors is a characteristic covector for the intersection airing Q X on H (X; Z)/Tors Let δ denote the absolute value of the determinant of Q X and Im(Sin c (X)) the image of the restriction ma from Sin c (X) to Sin c (Y ) Then combining (1) with Theorem 51 yields max 4d(Y, t) t Im(Sin c (X)) 1 1/δ if δ is odd, 1 if δ is even, with strict inequality unless the intersection form of X is (n 1) 1 δ Theorem follows immediately since if either δ is square-free or if there is no torsion in H 1 (X; Z), then the restriction ma from Sin c (X) to Sin c (Y ) is surjective, and H 1 (Y ; Z) = δ Similar reasoning yields the following variant of Theorem : Proosition 5 Let Y be a rational homology shere with H 1 (Y ; Z) = rs, with r square-free If Y bounds a negative-definite four-manifold X, then 1 1/r if r is odd, max 4d(Y, t) t Sin c (Y ) 1 if r is even Remark 53 Suose Y and X are as in the statement of Theorem with δ even If in fact max 4d(Y, t) = 1, t Im(Sin c (X)) then it is not difficult to see that this maximum must be attained at a sin structure 6 Surgeries on L-sace knots Let K be a knot in the three-shere and K (T ) = a 0 + j>0 a j (T j + T j ) its Alexander olynomial Torsion coefficients of K are defined by t i (K) = j>0 ja i +j Note that t i (K) = 0 for i N K, where N K denotes the degree of the Alexander olynomial of K For any n Z denote the n-surgery on K by K n K is called an L-sace knot if for some n > 0, K n is an L-sace Recall from [14] that a rational homology shere is called an L-sace if ĤF (Y, s) = Z for each sin c structure s (so its Heegaard Floer homology grous resemble those of a lens sace) There are a number of conditions coming from Heegaard Floer homology that an L-sace knot K has to satisfy In articular (see [14, Theorem 1]), its Alexander

15 olynomial has the form A CHARACTERISATION OF THE Z n Z(δ) LATTICE 15 () K (T ) = ( 1) k + k ( 1) k j (T n j + T n j ) From this it follows easily that the torsion coefficients t i (K) are given by (3) j=1 t i (K) = n k n k n k j i n k j 1 i n k j t i (K) = n k n k n k j n k j 1 n k j i n k j 1, where j = 0,, k 1 and n j = n j, n 0 = 0 In articular, t i (K) is nonincreasing in i for i 0 The following formula for d-invariants of surgeries on such a knot is based on results in [15] and [14] The roof was outlined to us by Peter Ozsváth Note that for sufficiently large values of n the formula for d(k n, i) is contained in [16, Theorem 1] Theorem 61 Let K S 3 be an L-sace knot and let t i (K) denote its torsion coefficients Then for any n > 0 the d-invariants of the ±n surgery on K are given by d(k n, i) = d(u n, i) t i (K), d(k n, i) = d(u n, i) for i n/, where U n denotes the n surgery on the unknot U Before sketching a roof of the theorem we need to exlain the notation The d-invariants are usually associated to sin c structures on the manifold The set of sin c structures on a three-manifold Y is arameterized by H (Y ; Z) = H 1 (Y ; Z) In the case of interest we have H 1 (K ±n ; Z) = Z/nZ, hence sin c structures can be labelled by the elements of Z/nZ Such a labelling is assumed in the above theorem and described exlicitly as follows Attaching a -handle with framing n to S 3 [0, 1] along K S 3 1} gives a cobordism W n from S 3 to K n Note that H (W n ; Z) is generated by the homology class of the core of the -handle attached to a Seifert surface for K; denote this generator by f n A sin c structure t on K n is labelled by i if it admits an extension s to W n satisfying c 1 (s), f n n i (mod n) If K = U is the unknot, the surgery is the lens sace which bounds the disk bundle over S with Euler number n Using either [1, Proosition 48] or [13, Corollary 15], the d-invariants of U n for n > 0 and i < n are given by (n i ) d(u n, i) = 1 4n 4 Proof of Theorem 61 According to [15, Theorem 41] the Heegaard Floer homology grous HF + (K n, i) for n 0 can be comuted from the knot Floer homology of K The knot comlex C = CF K (S 3, K) is a Z -filtered chain comlex, which is a finitely generated free module over T := Z[U, U 1 ] We write (i, j) for the comonents

16 16 BRENDAN OWENS AND SAŠO STRLE of bidegree, and let C(i, j) denote the subgrou generated by elements with filtration level (i, j) Here U denotes a formal variable in degree that decreases the bidegree by (1, 1) The homology of the comlex C is HF (S 3 ) = T, the homology of the quotient comlex B + := Ci 0} is HF + (S 3 ) = Z[U, U 1 ]/U Z[U] =: T + and the homology of Ci = 0} is ĤF (S3 ) = Z The comlexes Cj 0} and B + are quasi-isomorhic and we fix a chain homotoy equivalence from Cj 0} to B + We now recall the descrition of HF + (K ±n ) (n > 0) in terms of the knot Floer homology of K For s Z, let A + s denote the quotient comlex Ci 0 or j s} Let v s + : A + s B + denote the rojection and h + s : A + s B + the chain ma defined by first rojecting to Cj s}, then alying U s to identify with Cj 0} and finally alying the chain homotoy equivalence to B + For any σ 0, 1,, n 1} let A + σ = s σ+nz A + s and B + σ = s σ+nz B s +, where B s + = B + for all s Z Let D ±n + : A + σ B + σ be a homomorhism that on A + s acts by D ±n(a + s ) = (v s + (a s ), h + s (a s )) B s + B s±n + Then HF + (K ±n, σ) is isomorhic to the homology of the maing cone of D ±n, + and hence there is a short exact sequence (4) 0 coker ( ) D ±n + HF + (K ±n, σ) ker ( ) D ±n + 0, where ( ) D + : H ±n (A + σ ) H (B + σ ) is the induced ma on homology Moreover, the isomorhism from the homology of the maing cone of D ±n + to HF + (K ±n, σ) is a homogeneous ma of degree ±d(u n, σ) When comuting for K n, the grading on B s +, where s = σ + nk, is such that U 0 H (B s + ) has grading kσ + nk(k 1) 1 In case of n-surgery, U 0 H (B + σ+kn ) has grading (k + 1)σ nk(k + 1) In either case A + σ is graded so that D ±n + has degree 1 (See [15] for more details) Suose now that K is an L-sace knot with Alexander olynomial as in () Define δ k := 0 and δl+1 (n δ l := l+1 n l ) + 1 if k l is odd δ l+1 1 if k l is even for l = k 1, k,, k, where as above n l = n l Then as in the roof of [14, Theorem 1] (comare also [15, 51]), Ci = 0} is (u to quasi-isomorhism) equal to the free abelian grou with one generator x l in bidegree (0, n l ) for l = k,, k and the grading of x l is δ l To determine the differentials note that the homology of Ci = 0} is Z in grading 0, so generated by (the homology class of) x k It follows that the differential on Ci = 0} is a collection of isomorhisms C 0,nk l+1 C 0,nk l for l = 1,, k Similarly we see that the differential on Cj = 0} is given by a collection of isomorhisms C nk l+1,0 C nk l,0 for l = 1,, k This, together with U-equivariance, determines the filtered chain homotoy tye of the comlex C (For an examle, see Figure 1)

17 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 17 j i Figure 1 The knot Floer comlex CF K (T 3,5 ) Each bullet reresents a Z, and each arrow is an isomorhism The grous and differentials on the axes are determined by the Alexander olynomial T3,5 (T ) = 1 + (T + T 1 ) (T 3 + T 3 ) + (T 4 + T 4 ), and these in turn determine the entire comlex In order to comute the terms in the short exact sequence (4) we need to understand the mas induced by v + s and h + s on homology For notational convenience we will use the same symbols v + s and h + s for these induced mas in what follows Suose that the homology H of a quotient comlex of C is isomorhic to T + = Z[U, U 1 ]/U Z[U] We say H starts at (i, j) if the element U 0 has a reresentative of bidegree (i, j) With this terminology H (B + s ) starts at (0, n k ) and H (Cj s}) starts at (s, s + n k ) It remains to consider H (A + s ) Note that these grous are also isomorhic to T + For s n k the homology of A + s starts at (0, n k ), so v + s = id and h + s = U s For n k 1 s < n k, H (A + s ) starts at (s n k, s), so v + s = U n k s and

18 18 BRENDAN OWENS AND SAŠO STRLE h + s = U n k For nk s < n k 1 it starts at (n k 1 n k, n k 1 ), hence v s + = U n k n k 1 and h + s = U n k n k 1 +s It is now easy to observe that for s 0 the homology H (A + s ) starts at (t s, t s +n k ) and thus v s + = U ts and h + s = U ts+s, where t s is a torsion coefficient of K (comare with equation (3)) Similarly for s < 0 we obtain v s + = U ts s and h + s = U ts Note that since the sin c structures corresonding to i and i are conjugate (so their d invariants are equal) we may restrict to those in the range 0, 1,, n/ } Consider now K n (n > 0) and choose some σ 0, 1,, n/ } Writing (D n + ) relative to the decomositions H (A + σ ) = l Z H (A + σ+ln ) = l Z T +, H (B + σ ) = l Z H (B + σ+ln ) = l Z T +, we obtain for l > 0: b σ+ln = U t σ+ln aσ+ln + U t σ+(l 1)n+σ+(l 1)n a σ+(l 1)n = U t σ+ln aσ+ln = b σ+ln U t σ+(l 1)n+σ+(l 1)n a σ+(l 1)n b σ = U tσ a σ + U t n σ a σ n = U t n σ a σ n = b σ U tσ a σ b σ ln = U t ln σ+ln σ a σ ln + U t (l+1)n σ aσ (l+1)n = U t (l+1)n σ aσ (l+1)n = b σ ln U t ln σ+ln σ a σ ln It is straightforward to use the equations above to find an element of H (A + σ ) with any chosen value for a σ maing to a given element of H (B + σ ) under (D n + ) It follows that (D n + ) is surjective and its kernel contains one T + summand, isomorhic to H (A + σ ) Since D n + shifts grading by 1, U 0 H (B σ + ) has grading 1, and the comonent of (D n + ) from H (A + σ ) to H (B σ + ) is equal to U tσ, it follows that U 0 T + ker D n + has grading t σ The formula for d(k n, σ) now follows using the degree shift between HF + (K n, σ) and ker(d n + ) Finally consider K n (n > 0) and choose σ 0, 1,, n/ } Writing (D n) + with resect to direct sum decomositions as above yields: b σ+ln = U t σ+ln aσ+ln + U t σ+(l+1)n+σ+(l+1)n a σ+(l+1)n (l 0) = U t σ+ln aσ+ln = b σ+ln U t σ+(l+1)n+σ+(l+1)n a σ+(l+1)n b σ n = U tσ+σ a σ + U t n σ+n σ a σ n = U tσ+σ a σ = b σ n U t n σ+n σ a σ n b σ ln = U t ln σ+ln σ a σ ln + U t (l 1)n σ aσ (l 1)n (l ) = U t (l 1)n σ aσ (l 1)n = b σ ln U t ln σ+ln σ a σ ln The to and bottom equations determine a s (or more recisely U N a s for N >> 0) for all s σ + nz, so the middle equation cannot be fulfilled in general It follows that (D n) + has cokernel isomorhic to H (B σ n) + = T + ; the grading of U 0 in this grou is 0 The formula for d(k n, σ) again follows from the degree shift Combining Theorems and 61 yields

19 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 19 Theorem 6 Let n > 0 and let K be an L-sace knot whose torsion coefficients satisfy (n i) +1 t i (K) > 1 if n is odd 8n 4 (n i) 1 if n is even 8n 4 for 0 i n/ Then for 0 < m n, K m cannot bound a negative-definite fourmanifold with no torsion in H 1 Proof The formulas follow from the above-mentioned theorems To see that obstruction for n-surgery to bound a negative-definite manifold imlies obstruction for all m-surgeries with 0 < m n observe that for fixed i the right-hand side of the inequality is an increasing function of n for n i Note that the surgery coefficient m in Theorem 6 is an integer In [11] we will consider in more detail the question of which surgeries (including Dehn surgeries) on a knot K can be given as the boundary of a negative-definite four-manifold In articular it will follow from results in that aer that Theorem 6 holds as stated with m Q 7 Examle: Surgeries on torus knots In this section we consider torus knots T,q ; we assume < q Right-handed torus knots are L-sace knots since for examle q 1 surgery on T,q yields a lens sace [8] Let N = ( 1)(q 1)/ denote the degree of the Alexander olynomial of T,q The following roosition gives a simle function which aroximates the torsion coefficients of a torus knot Proosition 71 The torsion coefficients of T,q are given by t i = #(a, b) Z 0 a + bq < N i} and they satisfy t i g(n i) for 0 i N, where x g(x) is a iecewise linear continuous function, which equals 0 for x 0 and whose sloe on the interval [(k 1)q, kq] is k/ Proof The (unsymmetrised) Alexander olynomial of K = T,q is K (T ) = (1 T q )(1 T ) (1 T )(1 T q ), which is a olynomial of degree N Writing K as a formal ower series in T we obtain K (T ) = (1 T q )(1 T ) T a T bq a 0 b 0 = (1 T ) a 0 T a b 0 T bq T q (1 T ) a 0 T a b 0 T bq

20 0 BRENDAN OWENS AND SAŠO STRLE Clearly only the terms in the first roduct of the last line contribute to the nonzero coefficients of K A ower T k aears with coefficient +1 in this term whenever k = a + bq for some a, b Z 0, and with coefficient 1 whenever k 1 = a + bq for some a, b Z 0 Since the coefficient a j in K (T ) is the coefficient of T N j in K (T ), we obtain a j = m j m j+1, where Then for i 0 t i = j>0 m j := #(a, b) Z 0 a + bq = N j} ja i+j = j>0 m i+j = #(a, b) Z 0 a + bq < N i} In what follows it is convenient to relace t i by Define s(i) := t N i = #(a, b) Z 0 a + bq < i} ḡ(i) := i/ if i 0 0 if i 0 and s(i) := g(i) ; clearly s(i) ḡ(i) for every i Searating the set aearing in the definition of s(i) into subsets with fixed value of b we obtain s(i) = b 0 #a Z 0 a < i bq} = b 0 s(i bq) b 0 ḡ(i bq) =: g(i) The following corollary describes the range of surgeries on T,q that cannot bound negative-definite manifolds according to Theorem 6 (and using Proosition 71) To obtain the result of Proosition 3 note that all the lower bounds in the corollary allow for m = Note that Lisca and Stisicz have shown that Dehn surgery on T,n+1 with ositive framing r bounds a negative-definite manifold (ossibly with torsion in H 1 ) if and only if r 4n [7] Corollary 7 Let < q and N = ( 1)(q 1)/ If is even and m is less than the minimum of 1 + 4N + 1 α(α + 4β) 4 1 α q + 3 where α = q( ) + and β = q + 1 or if is odd and m is less than the minimum of 1 + 4N + 1 α(α + 4β) 4 1 α q( 3) q + 5 q+

21 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 1 where α = q( 4) + + 3q/ and β = q + 1 and 1 n N + m, then +n-surgery on T,q cannot bound a negative definite four-manifold with no torsion in H 1 Proof Write n = N + m; we may assume n < q 1 It suffices to show that (5) h(x) := (m + x) + 1 8n 1 4 < g(x) for m/ x N, where g is the function aearing in Proosition 71 Since h is convex and g is iecewise linear, it is enough to check the inequality for x = kq with k = 0, 1,, N/q and for x = N Consider first x = kq From the definition of g we obtain g(kq) = qk(k + 1)/ Substituting this into (5) we obtain 4k (q n q ) + 4k(qm nq ) + m + 1 n < 0 Since q n/ > 1/, it suffices to consider only k = 0 and k = N/q For k = 0 the last inequality yields m < 1 + 4N Assume first is even; then N/q = / 1 and substituting this into (5) gives (m + q( )) + 1 < n( + q( )) Writing µ = m + q( ), α = q( ) + and β = q + 1 the last inequality becomes µ αµ + 1 αβ < 0, which imlies µ < α + 1 α(α + 4β) 4 ; this is equivalent to the second condition on m in the statement of the corollary Since the sloe of g on the interval from (/ 1)q to N is 1/, we get g(n) = g((/ 1)q) + (N (/ 1)q)/ = (q + )/8 Subsituting this into (5) gives n n(q + 4) + 1 < 0, which holds if n < q + 4 or m < q + 3 Assume now is odd; then N/q = ( 3)/ and substituting this into (5) gives (m + q( 3)) + 1 < n( + q ( 1)( 3)) Writing µ = m + q( 3), α = q( 4) + + 3q/ and β = q + 1 the last inequality becomes µ αµ + 1 αβ < 0, from which the second condition on m in the statement of the corollary follows

22 BRENDAN OWENS AND SAŠO STRLE Now the sloe of g on the interval from q( 3)/ to N is ( 1)/ and g(n) = (q + 4 (q + )/)/8 Subsituting this into (5) gives n n(q + 6 q + ) + 1 < 0, which holds if n < q + 6 q+ or m < q + 5 q+ References [1] J P Alexander, G C Hamrick & J W Vick Linking forms and mas of odd rime order, Trans Amer Math Soc 1 (1976), [] N D Elkies A characterization of the Z n lattice, Math Res Lett (1995), [3] K A Frøyshov The Seiberg-Witten equations and four-manifolds with boundary, Math Res Lett 3 (1996), [4] J E Greene L-sace surgeries, genus bounds, and the cabling conjecture, arxiv: , 010 [5] A Kawauchi & S Kojima Algebraic classification of linking airings on 3-manifolds, Math Ann 53 (1980), 9 4 [6] M Kneser & D Pue Quadratische Formen und Verschlingungsinvarienten von Knoten, Math Zeitschr 58 (1953), [7] P Lisca & A I Stisicz Ozsváth-Szabó invariants and tight contact structures, I, Geometry and Toology 8 (004), [8] L Moser Elementary surgery along a torus knot, Pacific Journal of Math 38 (1971), [9] B Owens & S Strle Rational homology sheres and the four-ball genus of knots, Advances in Mathematics 00 (006), [10] B Owens & S Strle A characterisation of the n 1 3 form and alications to rational homology sheres, Math Res Lett 13 (006) [11] B Owens & S Strle Dehn surgery and negative-definite manifolds, Selecta Math (NS) (01), DOI: /s [1] P Ozsváth & Z Szabó Absolutely graded Floer homologies and intersection forms for fourmanifolds with boundary, Advances in Mathematics 173 (003), [13] P Ozsváth & Z Szabó On the Floer homology of lumbed three-manifolds, Geometry and Toology 7 (003), 5 54 [14] P Ozsváth & Z Szabó On knot Floer homology and lens sace surgeries, Toology 44 (005), [15] P Ozsváth & Z Szabó Knot Floer homology and integer surgeries, Algebr Geom Tool 8 (008), [16] P Ozsváth & Z Szabó Knot Floer homology and rational surgeries, Algebr Geom Tool 11 (011), 1 68 [17] J-P Serre A course in arithmetic, Sringer, 1973 [18] L R Taylor Relative Rochlin invariants, Toology Al 18 (1984), [19] C T C Wall Quadratic forms on finite grous, and related toics, Toology (1963), 81 98

23 A CHARACTERISATION OF THE Z n Z(δ) LATTICE 3 School of Mathematics and Statistics University of Glasgow Glasgow, G1 8QW, United Kingdom address: brendanowens@glasgowacuk Faculty of Mathematics and Physics University of Ljubljana Jadranska Ljubljana, Slovenia address: sasostrle@fmfuni-ljsi