Quadrics Defined by Skew-Symmetric Matrices

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1 International Journal of Algebra, Vol. 11, 2017, no. 8, HIKAI Ltd, Quadrics Defined by Skew-Symmetric Matrices Joydip Saha 1, Indranath Sengupta 1, and Gaurab Tripathi 2 1 Discipline of Mathematics, IIT Gandhinagar Palaj, Gandhinagar, Gujarat , India Corresponding author 2 Department of Mathematics, Jadavpur University Kolkata, WB , India Copyright c 2017 Joydip Saha, Indranath Sengupta and Gaurab Tripathi. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper we propose a model for computing a minimal free resolution for ideals of the form I 1 (X n Y n ), where X n is an n n skew-symmetric matrix with indeterminate entries x ij and Y n is a generic column matrix with indeterminate entries y j. We verify that the model works for n = 3 and n = 4 and pose some statements as conjectures. Answering the conjectures in affirmative would enable us to compute a minimal free resolution for general n. Mathematics Subject Classification: Primary 13D02; Secondary 13C40, 13P10, 13D07 Keywords: Gröbner basis, Betti numbers, determinantal ideals, skewsymmetric matrix, mapping cone 1. Introduction Let K be a field. Let {x ij ; 1 i m, 1 j n}, {y j ; 1 j n} be indeterminates over K, so that = K[x ij, y j ] denotes the polynomial algebra over K. Let X n denote an n n skew-symmetric matrix such that its entries are the indeterminates ±x ij and 0. We call such a matrix a generic skewsymmetric matrix. Let Y n = (y j ) n 1 be the generic n 1 matrix. It is very hard to compute a graded minimal free resolution of the ideal I 1 (X n Y n ).

2 350 Joydip Saha, Indranath Sengupta and Gaurab Tripathi Ideals of the form I 1 (X n Y n ) has been studied by [2] and they appear in some of our recent works; see [4], [5], [6], [7]. We described its Gröbner bases, primary decompositions and Betti numbers through computational techniques, mostly under the assumption that X n is either a generic or a generic symmetric matrix. It is in deed the case that these ideals are far more difficult to understand when X n is a generic skew-symmetric matrix. In this paper, we present a scheme for computing a graded minimal free resolution of the ideal I 1 (X n Y n ), where X n is a n n generic skew-symmetric matrix and and Y n is a generic n 1 matrix. We show that if we assume the truth of two statements then the scheme works for a general n. These two statements which have been proposed as conjectures appear to be correct as seen from symbolic computation using the computer algebra software Singular [1]. We finally verify the validity of these conjectures for n = 3 and n = 4. We refer to [3] for basic knowledge on the techniques used by us. 2. General Scheme and Conjectures 0 x 12 x x 1n x 12 0 x x 2n y 1 Let X n = x 13 x 23.. and Y n =.. y n x 1n x 2n... 0 Our aim is to find a minimal free resolution of I 1 (X n Y n ). Assuming x ij = x ji, if i > j and x ii = 0, let g ki = Σ i j=1x kj y j. Therefore the generators of I 1 (X n Y n ) are g 1i, g 2i,, g ni. Let (i)n denote the Pffafian of the skew symmetric matrix X n with the i-th row and the i-th column deleted. Lemma 2.1. Assuming x ij = x ji, if i > j and x ii = 0 (i) y n g nn = ( y 1 g 1n + y 2 g 2n + + y n 1 g (n 1)n ). (ii) g k(n 1) g nn = x kn y 1 g 1n +x kn y 2 g 2n + +g nn +x kn y k g kn + +x kn y n g n 1n. (iii) (n)n y n = ( (1)n )g 1n + ( (2)n )g 2n + + (( 1) n 1 (n 1)n )g (n 1)n. Proof. A simple calculation gives the proof. Lemma 2.2. {g 1n, g 2n,, g (n 1)n } forms a regular sequence for n 2. Proof. See part (ii) of Theorem 2.2 in [5]. Notations. (i) Let I n = g 1n, g 2n,, g (n 1)n. By Lemma 2.2, the ideal I n is minimally resolved by the Koszul complex 0 (n 1 2 ) ψ 2n (n 1 1 ) ψ 1n ψ 0n /I n 0; where ψ kn : (n 1 k ) ( n 1 k 1) and k {0, 1, 2,, n 1}. (ii) Let J n = g nn and L n = I n + J n = I 1 (X n Y n ).

3 Quadrics defined by skew-symmetric matrices 351 Computations with Singular give us enough evidence in support of the Conjectures proposed below: Conjecture 1. C n := (I n : J n ) = g 1(n 1), g 2(n 2),, g n 1(n 1), y n, (n)n. If n is even then (n)n = 0 and C n = g 1(n 1), g 2(n 2),, g n 1(n 1), y n, for every n 4. Conjecture 2. If n is odd then (n)n 0. For every n 4, P n := ( g 1(n 1), g 2(n 2),, g n 1(n 1) : (n)n ) = (y 1,, y n 1 ). Assuming the validity of these conjectures we can construct a minimal free resolution for I 1 (X n Y n ) through the following steps. We proceed by induction on n 3. We first compute a resolution of L 3, which is not difficult. For 3 i 1 < n, let a resolution of L i 1 be 0 β 1(i) d 1i β 0(i) where d 0i : /L i 1 is the projection map. d 0i /Li 1 0 A resolution of P n is the Koszul complex, which is the following: 0 (n 1 2 ) φ 2n (n 1 1 ) φ 1n φ 0n /P n 0 where φ kn : (n 1 k ) ( n 1 k 1), for k {0, 1, 2,, n 1}. Case 1. For i < n and i is odd, let T i := L i 1 + (i)i. Using mapping cone we get, (i 1 3 ) (i 1 2 ) φ 2i (i 1 1 ) φ 1i φ 0i /P i 0 δ 3i δ 2i β 3(i 1) β 2(i 1) Therefore a resolution of T i is d 2 i 2 δ 1i β 1(i 1) d 1 i 1 δ 0i = i(i) d0i /L i 1 0 (i 1 1 ) β δ 2(i 1) 2i β 1(i 1) δ 1i /T i 0; [ ] where δ ki : k 1) (i 1 β k(kı 1) ( k 2) i 1 β k(kı 1) and δ ki = φk 1i 0. δ k 1i d ki 1 The resolution of T i obtained above may not be minimal. Assuming that we can extract a minimal free resolution from it by identifying matching of degrees and cancelling them (see the computations for some special values of n in the next section) let T i be the minimal free resolution of T i, whose differentials are δ ki, i.e. T i γ 2i δ 2i γ 1i δ 1i /Ti 0

4 352 Joydip Saha, Indranath Sengupta and Gaurab Tripathi To find the resolution of C i, we need to tensor 0 y i 0 with the complex T i, which gives us ( γ 3i ) ( γ 2i ) η 3i ( γ 2i ) ( γ 1i ) η 2i ( γ 1i ) ( ) η 1i [ ] /C i 0; δki y where η ki = i I. (*) 0 δ k 1i We first rewrite complex(*), which gives us a minimal free resolution of C i. Then, we construct the mapping cone of the following complexes with respect to the following connecting maps: (γ i3+γ i2 ) (γ i2+γ i1 ) η 2i (γ i1+1) η 1i η 0i /C i 0 ξ 3i ξ 2i (i 1 3 ) (i 1 2 ) ψ 2(i) ξ 1i (i 1 1 ) ξ 0i =g (i)i ψ 1(i) ψ0i /I i 0 We create a minimal resolution out after the mapping cone construction by suitable cancellation of matched degrees. Case 2. Let i be even and i < n. Then, (i)i = 0. Therefore the ideal T i = L i 1 + (i)i = L i 1. We proceed in a similar way as Case Computation for n = 3 Let X 3 = 0 x 12 x 13 x 12 0 x 23 and Y 3 = y 1 y 2. We write x 13 x 23 0 y 3 such that and I 3 = g 13, g 23 ; J 3 = g 33. I 1 (X 3 Y 3 ) = g 13, g 23, g 33, g 13 = x 12 y 2 + x 13 y 3 g 23 = x 12 y 1 + x 23 y 3 g 33 = x 13 y 1 x 23 y 2 We claim that (I 3 : J 3 ) = x 12, y 3. We first compute a Gröbner basis of I 3. Let us fix the lexicographic monomial order induced by the ordering among the variables y 1 > y 2 > y 3 > x 12 > x 13 > x 23 on. Then h = s(g 13, g 23 ) = x 13 y 3 y 1 + x 23 y 3 y 2. We have Lt(h) = x 13 y 3 y 1 and that it is not divisible by Lt(g 13 ) and Lt(g 23 ). We therefore take the enlarged set {g 13, g 23, h}. It is clear that gcd(lt(g 13 ), LT(h)) = 1, therefore we need to examine only s(h, g 23 ). Now s(h, g 23 ) = x 13 x 23 y x 12 x 23 y 3 y 2 = x 23 y 3 (g 13 ) 0; therefore the set {g 13, g 23, h} forms a Gröbner basis of I 3. We observe that x 12 g 33 = x 13 g 23 x 23 g 13 and y 3 g 33 = (y 1 g 13 + y 2 g 23 ). Therefore, x 12, y 3 (I 3 : J 3 )

5 Quadrics defined by skew-symmetric matrices 353 Let pg 33 I 3, and let r be the remainder term upon division of p by x 12, y 3. We know that x 12, y 3 (I 3 : J 3 ). Therefore, rg 33 I 3. The set {g 13, g 23, h} is a Gröbner basis for I 3, therefore one of the following must hold: x 12 y 2 Lt(r)(x 13 y 1 ) or x 12 y 1 Lt(r)(x 13 y 1 ) or x 13 y 1 y 3 LT(r)(x 13 y 1 ). This gives us x 12 Lt(r) or y 3 Lt(r), which leads to a contradiction if r 0. Therefore r = 0 and p x 12, y 3, and hence (I 3 : J 3 ) = x 12, y 3. Let L 3 = I 3 + J 3 = g 13, g 23, g 33. A minimal free resolution of L 3 is where 0 2 d 23 3 d 13 d 03 /L3 0 d 13 = ( ) g 13 g 23 g 33, d23 = x 23 y 1 x 13 y 2 x 12 y 3 4. Computation for n = 4 0 x 12 x 13 x 14 y 1 X 4 = x 12 0 x 23 x 24 x 13 x 23 0 x 34 and Y 4 = y 2 y 3 x 14 x 24 x 34 0 y 4 By our notation we have, g 14 = x 12 y 2 + x 13 y 3 + x 14 y 4, g 24 = x 12 y 1 + x 23 y 3 + x 24 y 4, g 34 = x 13 y 1 x 23 y 2 + x 34 y 4, g 44 = x 14 y 1 x 24 y 2 x 34 y 3 and I 4 = g 14, g 24, g 34, J 4 = g 44, L 4 = I 4 + J 4, We claim that, C 4 = (I 4 : J 4 ) = g 13, g 23, g 33, y 4. We first find a Gröbner basis of I 4. Let us fix the lexicographic monomial order induced by y 1 > y 2 > y 3 > y 4 > x 12 > x 13 > x 14 > x 23 > x 24 > x 34 on. Consider the s-polynomials: s(g 14, g 24 ) = y 1 y 3 x 13 + y 1 y 4 x 14 + y 2 y 3 x 23 + y 2 y 4 x 23 = y 3 g 34 + y 1 y 4 x 14 + y 2 y 4 x 24 + y 3 y 4 x 34 s(g 24, g 34 ) = y 2 x 12 x 23 + y 3 x 13 x 23 y 4 x 12 x 34 + y 4 x 13 x 24 = x 23 g 14 + y 4 x 12 x 34 y 4 x 13 x 24 + y 4 x 14 x 23 We have gcd(lt(g 14 ), Lt(g 24 ) = 1, therefore s(g 14, g 24 ) 0. Let us take p 1 = y 1 y 4 x 14 + y 2 y 4 x 24 + y 3 y 4 x 34 and p 2 = y 4 x 12 x 34 y 4 x 13 x 24 + y 4 x 14 x 23 and consider the bigger set {g 14, g 24, g 34, p 1, p 2 }. We now compute p 3 = s(g 14, p 2 ) = y 2 y 4 x 13 x 24 y 2 y 4 x 14 x 23 + y 3 y 4 x 13 x 34 + y 2 4x 14 x 34.

6 354 Joydip Saha, Indranath Sengupta and Gaurab Tripathi It is evident that Lt(p 3 ) is not divisible by any element of the set {Lt(g 14 ), Lt(g 24 ), Lt(g 34 ), Lt(p 1 ), Lt(p 2 )}. Therefore we add p 3 in the list and get the set G = {g 14, g 24, g 34, p 1, p 2, p 3 }. It is now straightforward to check that every s polynomial reduces to zero. Hence G is a a Gröbner basis for the ideal I 4. We now compute a Gröbner basis for the ideal g 13, g 23, g 33. Consider the s-polynomials, s(g 13, g 23 ) = x 13 y 3 y 1 + x 23 y 3 y 2 = y 3 g 33 0 s(g 23, g 33 ) = y 2 x 12 x 23 y 3 x 13 x 23 = x 23 g Also, we have gcd(lt(g 13 ), Lt(g 33 ) = 1. Therefore, the set {g 13, g 23, g 33 } itself is a Gröbner basis. Hence it follows easily that {g 13, g 23, g 33, y 4 } is a Gröbner basis for the ideal g 13, g 23, g 33, y 4. Using proposition 2.1 we obtain {g 13, g 23, g 33, y 4 } (I 4 : J 4 ), Let pg 44 I 4 and assume that r is the remainder upon division of p by {g 14, g 24, g 34, p 1, p 2 }. Suppose that r 0. We have rg 44 I 4. Moreover, Lt(rg 44 ) = Lt(r)x 14 y 1 is divisible by one of the leading terms Lt(g 14 ) = x 12 y 2, Lt(g 24 ) = x 12 y 1, Lt(g 34 ) = x 13 y 1, Lt(p 1 ) = y 1 y 4 x 14, Lt(p 2 ) = y 4 x 12 x 34, Lt(p 3 ) = y 2 y 4 x 13 x 24. If Lt(rg 44 ) is divisible by any one of the leading terms Lt(g 14 ) = x 12 y 2, Lt(p 1 ) = y 1 y 4 x 14, Lt(p 2 ) = y 4 x 12 x 34, Lt(p 3 ) = y 2 y 4 x 13 x 24, then we get a contradiction. If Lt(g 24 ) = x 12 y 1 LT(rg 44 ), then x 12 Lt(r). Let r = x 12 m + l. Therefore, r.g 44 = (x 12 m + l)( x 14 y 1 x 24 y 2 x 34 y 3 ) and after division we get q = ( x 34 x 12 y 3 x 14 x 23 y 3 + x 24 x 13 y 3 )m + lg 44 I 4. We have Lt(q) = x 34 x 12 y 3 m and it must be divisible by one of the leading terms Lt(g 14 ) = x 12 y 2, Lt(g 24 ) = x 12 y 1, Lt(g 34 ) = x 13 y 1, Lt(p 1 ) = y 1 y 4 x 14, Lt(p 2 ) = y 4 x 12 x 34, Lt(p 3 ) = y 2 y 4 x 13 x 24. This implies that Lt(r) must be divisible by one of the leading terms Lt(g 14 ) = x 12 y 2, Lt(g 24 ) = x 12 y 1, Lt(g 34 ) = x 13 y 1, Lt(p 1 ) = y 1 y 4 x 14, Lt(p 2 ) = y 4 x 12 x 34, Lt(p 3 ) = y 2 y 4 x 13 x 24, which is a contradiction. Similarly, if Lt(g 34 ) = x 13 y 1 Lt(rg 44 ), we get a contradiction. Therefore r = 0 and our claim is proved. To find the resolution of C 4, we take the tensor product of the complexes: 0 2 d 23 3 d 13 d 03 /L3 0 and 0 y 4 /y 4 0 and obtain a resolution of C 4 as 0 2 η 34 5 η 24 4 η 14 η 04 /C 4 0 where η 14 = [d 13 y 4 ] = [g 13, g 23, g 33, y 4 ],

7 η 24 = η 34 = Quadrics defined by skew-symmetric matrices 355 [ ] x 23 y 1 y d23 y 4 I 3 = x 13 y 2 0 y d 13 x 12 y y 4, 0 0 g 13 g 23 g 33 y y 4 x 23 y 1 x 13 y 2. x 12 y 3 Using the mapping cone between these complexes we get 0 2 η34 5 η 24 4 η 14 η 04 /C 4 0 where ξ34 ξ 24 ξ 14 ξ 04 =g 44 0 ψ34 3 ψ 24 3 ψ 14 ψ04 /I 4 0 ψ 14 = [g 14, g 24, g 34 ], ψ 24 = ψ 34 = g 34 g 14 g 24 g 24 0 g 34 g 14 g g 24 g 14 ξ 04 = [g 44 ], ξ 14 = g 44 + x 14 y 1 x 24 y 1 x 34 y 1 y 1 x 14 y 2 g 44 + x 24 y 2 x 34 y 2 y 2, x 14 y 3 x 24 y 3 g 44 + x 34 y 3 y 3 ξ 24 = x 34 0 y 2 y 1 0 x y 3 y 2, x 24 0 y 3 0 y 1 ξ 24 = ( 1 0 ). Therefore a non-minimal resolution of L 4 is 0 2 d44 6 d 34 7 d 24 4 d 14 d04 /L4 0, where d 14 = [g 14, g 24, g 34, g 44 ], d24 = ξ 14 ψ 24, η 14 0 [ ] [ ] η24 0 η34 d 34 =, d44 =. ξ 24 ψ 34 ξ 34,

8 356 Joydip Saha, Indranath Sengupta and Gaurab Tripathi Therefore a minimal free resolution of L 4 = I 1 (X 4 Y 4 ) is 0 d 44 5 d 34 7 d 24 4 d 14 d 04 /L4 0 where d 14 = d 14, d 24 = d 24, d 34 = [ η24 ξ 24 ], d 44 = Acknowledgements. Joydip Saha is a research associate supported by the research project EM/2015/ sponsored by the SEB, Government of India. Indranath Sengupta is supported by the research project EM/2015/ sponsored by the SEB, Government of India. Gaurab Tripathi thanks CSI for the Senior esearch Fellowship. eferences [1] W. Decker, G.-M. Greuel, G. Pfister, H. Schönemann, Singular A computer algebra system for polynomial computations, (2016). [2] J. Herzog, Certain Complexes Associated to a Sequence and a Matrix, Manuscipta Math., 12 (1974), [3] I. Peeva, Graded Syzygies, Springer-Verlag London Limited, [4] J. Saha, I. Sengupta, G. Tripathi, Ideals of the form I 1 (XY ), arxiv: [math.ac] [5] J. Saha, I. Sengupta, G. Tripathi, Primary decomposition of certain determinanatal ideals, arxiv: [math.ac] [6] J. Saha, I. Sengupta, G. Tripathi, Betti numbers of certain sum ideals, arxiv: [math.ac] [7] J. Saha, I. Sengupta, G. Tripathi, egular sequences from determinanatal conditions, arxiv: [math.ac] eceived: October 12, 2017; Published: November 21, y 4 y 1 y 2 y 3.