Singularities of dual varieties and ϕ dimension of Nakayama algebras. by Emre Sen

Transcription

1 Singularities of dual varieties and ϕ dimension of Nakayama algebras by Emre Sen B.S. in Mathematics, Bilkent University M.S. in Mathematics, Bilkent University A dissertation submitted to The Faculty of the College of Science of Northeastern University in partial fulfillment of the requirements for the degree of Doctor of Philosophy April 18, 2018 Dissertation directed by Gordana Todorov Professor of Mathematics Jerzy Weyman Professor of Mathematics 1

2 Dedication Dedicated to my father Cihat Şen,

3 Acknowledgments I would like to begin with expressing my deep gratitude to my supervisor Professor Gordana Todorov. I could not thank her enough for her endless patience, unconditional support and precious guidance. Without her, this thesis could not be done. I am grateful to Prof. Jerzy Weyman and Prof. Kiyoshi Igusa for helpful and enjoyable conversations. I am thankful to Prof. Zelevinsky and my former advisor Prof. Klyachko, for accepting me to Northeastern and suggesting me Northeastern, respectively. I would like to thank to the committee members Prof. Venkatramani Lakshmibai and Prof. Anthony Iarrobino for reading this thesis and valuable comments. I want to thank to all my friends. Last but not least, I am grateful to my family for their support and confidence throughout my education. 3

4 Abstract of Dissertation In the first part of this thesis we study the singular locus of projectively dual varieties of certain Segre-Plücker embeddings. We give a classification of the irreducible components of the singular locus of several representation classes including dual Grassmannian varieties. Basically, they admit two types of singularities: cusp type and node type which are degeneracies of a certain Hessian matrix, and the closure of the set of tangent planes having more than one critical point, respectively. In the second part, we study ϕ function which is generalization of the notion of projective dimension. We analyze the behavior of the ϕ functions for cyclic Nakayama algebras of infinite global dimension. We prove that it is always an even number. Also we give a sharp upper bound for ϕ dimension in terms of number of monomial relations which describes the algebra. 4

5 Table of Contents dedication 2 Acknowledgements 3 Abstract of Dissertation 4 Table of Contents 5 1 Introduction 6 Chapter 2: The Dual Grassmannian Cusp Type Singularities The Generic Node Component Special node components Decomposition of singular locus Symmetric Matrices with diagonal lacunae and skew symmetric blocks List of Matrices Chapter 3: Analysis of k C N C M The Cusp Component The Node Component The Determinant of The Hessian Matrix Chapter 4: Other Formats The Cusp Component The Node Component The Hessian Matrices Conclusion Chapter 5: The ϕ dimension for Nakayama Algebras Introduction Set up and Notation The ϕ-dimension Even ϕ-dimension Upper Bound For ϕ-dimension

6 Chapter 1 Introduction The main results that I present in the thesis are in two separate areas: algebraic geometry, in particular projective duality in representations of algebraic groups, and representation theory of artin algebras with emphasis on homological properties. Projective duality is one of the fundamental and historical notions in algebraic geometry. Briefly it is the correspondence between lines and hyperplanes in projective space. This relation can be extended to the nonlinear subvarieties of projective space. Namely, for a given irreducible projective variety X, the closure of the set of all hyperplanes containing tangents to X is the projectively dual variety X. An important class of dual varieties - introduced by I. Gelfand, M. Kapranov and A. Zelevinsky [GKZ94] - is hyperdeterminants which are the defining equations of projective duals of Segre embeddings. Also they are the analogs of determinants for multidimensional matrices. The first treatment of the subject was initiated by Cayley and then Schläfli, but it was not studied for 150 years. In the early 90 s the subject 6

7 was revitalized by I. Gelfand, M. Kapranov and A. Zelevinsky. The first problem we consider is: Problem Consider the Segre-Plücker embedding : X = P ( k1 ) ( kr ) ( k1 ) k r C N 1... P C N r P C N 1 C N r (1.0.1) where N i 2k i for 1 i r. Describe the singular locus of the dual variety X provided that it is a hypersurface, and classify its components. We point out that, if all k i = 1, then the Segre-Plücker embedding reduces to the Segre embedding and the dual is a hyperdeterminant hypersurface. J. Weyman and A. Zelevinsky worked on singularities of hyperdeterminants [WZ96]; they classified the irreducible components of the singular locus of hyperdeterminants which are cusp and node components. Our results generalize the work of J. Weyman and A.Zelevinsky, we extend it to the tensor products of exterior power representations. Another connection occurs in the case of i = 1, T. Maeda [Mae01] and F. Holweck [Hol11] studied dual Grassmannians i.e. duals of Plücker embeddings. In addition, we also solve a linear algebra problem of independent interest which is about polynomial factors of determinants of a certain type of matrices. Let V be a vector space, V be its dual. The set of one dimensional subspaces of V is called projectivization of V and is denoted by P (V ). For each point in P (V ) we can associate a hyperplane. After regarding those hyperplanes as points, dual projective space P (V ) = P (V ) is obtained. In the case of V = C N+1, the usual projective space P N = P ( C N+1) is obtained. One can extend this to nonlinear subvarieties of projective 7

8 spaces. Definition Let X P N be a projective variety. The dual variety X ( P N) is the closure of the set of all tangent hyperplanes to X. To analyze singularities, an important tool is the Hessian matrix which is the matrix of second partial derivatives of a form F in local coordinates and evaluated at a point p, i.e. H (F ) p = 2 F p (1.0.2) Definition The cusp type singularity is a subvariety of X such that the the determinant of a Hessian matrix vanishes. Formally: X cusp := {F p X s.t PT p X F and det H (F ) = 0} Definition T he node type singularity is a subvariety of X which is the set of forms such that the form F is tangent to X at least two distinct points. Formally: X node := {F p, q X such that PT p X, PT q X F } We list here some of our main results regarding the singular locus of the dual varieties. Theorem The singular locus of the dual Grassmannian k C N (i.e. r = 1 in problem 1.0.1) is of codimension 1 in X for N 2k, k 4 or N 9, k = 3. For these cases the singular locus X sing has two irreducible components X node and X cusp both having codimension two in the ambient space. For the cases k = 3 and N = 6, 7, 8, the codimensions of X cusp in X are 4, 3, 2 respectively and X node is a subvariety of X cusp. Regarding the cusp component we have: 8

9 Theorem Let X be the dual variety given in problem with conditions r = 2 and k 2 = 1. We have: 1 -) If N 1 2k 1 with k 1 3 and if N 2 4, X cusp is of codimension one in X. 2 -) If k 1 (N 1 k 1 ) + 1 = N 2, then X cusp is of codimension 2 in X Regarding the node component we have: Theorem Let X be the dual variety given in problem with conditions r = 2 and k 2 = 1. We have: 1 -) If k 1 (N 1 k 1 ) + 1 = N 2, then X node is of codimension 1 in X and it has two components. 2 -) If X is one of the following 2 C 4 C 2, 2 C 4 C 3, 2 C 4 C 4, 2 C 5 C 3, 2 C 5 C 4, 2 C 6 C 2, then X node is contained in X cusp. 3 -) Aside of these, X node is always codimension one in X. Theorem The node type singularity of k 1 C N 1 kr C N r is of codimension one if k j (N j k j ) < i j k i (N i k i ). If for some j, k j (N j k j ) = i j k i (N i k i ), then X node is of codimension one and contains X cusp. The second part of the thesis is about homological properties of finite dimensional algebras. The finitistic dimension conjecture asserts that the supremum of finite projective dimensions of modules is finite. K. Igusa and G. Todorov introduced the ϕ function [IT05] to prove the finitistic dimension conjecture for some classes of artin algebras. 9

10 Let Λ be an artin algebra. In order to define ϕ(m) for each module M we need the following set-up: let K 0 be the abelian group generated by all symbols [X], where X is a finitely generated Λ-module, modulo the relations: i) [C] = [A] + [B] if C A B ii) [P ] = 0 if P is projective. Then K 0 is the free abelian group generated by the isomorphism classes of indecomposable finitely generated nonprojective Λ-modules. For any finitely generated Λ-module M let L [M] := [ΩM] where ΩM is the first syzygy of M. Since Ω commutes with direct sums and takes projective modules to zero this gives a homomorphism L : K 0 K 0. For every finitely generated Λ-module M let addm denote the subgroup of K 0 generated by all the indecomposable summands of M, which is a free abelian group since it is a subgroup of the free abelian group K 0. Definition For a given Λ module M, let ϕ (M) be the minimal integer t such that rank (L t addm ) = rank (L t+j addm ) for all j 1. The ϕ dimension of Λ is the supremum of the possible values of the ϕ function i.e. ϕ dim(λ) := sup{ϕ(m) M modλ}. Remark Let ϕ be the function defined above. Then: (1) ϕ(m) is finite for each module M. (2) Suppose the projective dimension p dim(m) <. Then ϕ(m) = p dim(m). (3) Suppose M is indecomposable with p dim(m) =. Then ϕ(m) = 0. By this remark, ϕ(m) is generalization of notion of projective dimension. 10

11 Λ is a Nakayama algebra if both indecomposable injective modules and indecomposable projective modules are uniserial. We present the properties of the ϕ dimension of Nakayama algebras and the proof of the following theorem in the last chapter: Theorem Let Λ be cyclic Nakayama algebra of infinite global dimension. Then ϕ dim Λ is an even number. 11

12 Chapter 2 The Dual Grassmannian Now we consider the case that X is the Grassmannian G(k, C N ) which is the set of all k-dimensional vector subspaces of C N. Consider the Pücker embedding: G(k, C N ) P ( k C N) (2.0.1) which takes a k dimensional subspace L C N to the one dimensional subspace k L k C N. If we choose the coordinate matrix K of size k N: x 1 1 x 1 2 x 1 N x 2 1 x 2 2 x 2 N K = x k 1 x k 2 x k N (2.0.2) then, minors of this matrix give us local coordinates in the ambient space, which are called Plücker coordinates. They are subject to quadratic relations which are called Plücker relations. 12

13 For a given index set I = (i 1,..., i k ), η I is the minor of coordinate matrix where columns are indexed by I. To describe the dual Grassmannian, we can use the form: F (A, x) = a i1...i k η i1...i k (2.0.3) 1 i 1 <i 2 <...<i k N The form F belongs to the dual Grassmannian G(k, C N ) if and only if F and its partial derivatives with respect to local coordinates are zero at some nonzero point x G(k, C N ), i.e. the system of equations F (A, x) = 0, F (A,x) x j i = 0, for all i,j has a nontrivial solution for some x. Definition We call such a point x a critical point of the form Remark Let A be the space of arrays i.e coefficients of the form F = F (A, x) for G ( k, C N). Then projection onto the first factor of the incidence variety: Z := {(A, x) A X F (A, x) X } (2.0.4) gives points of the dual Grassmannian. Remark There is a natural action of G = GL (V ) on V. There is an induced action of G on k V and A. Explicitly: F (A, g. (v 1... v k )) = F (A, gv 1... gv k ) = F (A g, v 1... v k ) (2.0.5) for g G. To analyze singularities, an important tool is the Hessian matrix which is matrix of 13

14 the second partial derivatives in the local coordinates and evaluated at point p, i.e. H (F ) p = 2 F p (2.0.6) We analyze its algebraic properties in section 2.5. We prove the following in section 2.1: Theorem For k 3 and N 9 or k 4 and N 2k, X cusp is an irreducible hypersurface in X. We will show the following in section 2.2: Theorem For k 3 and N 9 or k 4 and N 2k, X node is an irreducible hypersurface in X. It is well known that under the SL N action the space 3 C N for N = 6, 7, 8 has finitely many orbits [KW12], [KW13], [Hol11]. For those, we show that X sing = X cusp by simply checking the orbit representatives in the section 2.4. For the other cases we show that X cusp X node at theorem Remark Since k C N = N k (C N ), it is enough to work with N 2k in general. The dual Grassmannian is a hypersurface if k 3 and N 2k. We give the proof in section 2.5. If it is not specified, we use codimension with respect to the ambient space A. So the dual Grassmannian X is codimension one subvariety in A. Now we present the main theorem by putting together all the above results: Theorem (MAIN). The singular locus of the dual Grassmannian k C N is of codimension 2 in A for N 2k, k 4 or N 9 and k = 3. For these cases singular locus X sing 14

15 has two components X node and X cusp which are irreducible, both having codimension two in the ambient space. For the cases k = 3 and N = 6, 7, 8, the codimensions of X cusp in X are 4, 3, 2 respectively and X node is a subvariety of X cusp. Remark A weaker versions of the main theorem also appears in the literature: [Hol11], [Mae01] The dual Grassmannian G(k, C N ), k 3 is normal if and only if k = 3 and N = 6, 7, Cusp Type Singularities In this section we study the cusp component. Let Y be the span of k linearly independent vectors, i.e. v 1,..., v k {0} Y. Let x 0 be the point which is the span of basis vectors e 1,..., e k which is equivalent to saying x 0 is the point with coordinates x j i = δ i,j, which means the first k k submatrix of K given in is the identity matrix, and the rest is zero. It will be convenient for us to dehomogenize the multilinear form F (A, x) = a i1...i k η i1...i k corresponding to A A by setting all x j i = δ i,j where i, j k. More precisely, we consider the subset E = { x X x j i = 1 for i = j, 0 otherwise for the first k k submatrix}. So the image of E under the Plücker map is the affine chart near the point [1 : 0 :... 0]. We introduce new notation for A A. a 12...k is denoted by a. a 12...t...k is denoted by a t p where t is the value > k and p is the position of t and it is done for other terms similarly. For example we will write a quadratic term a 1,2,...,t,t,...,k as a tt pp of t and t are p and p respectively. where the positions 15

16 Under the new notation, F can be written as: F (A, x) = a + p,t a t px p t + a tt pp xp t x p t,t,p,p t +... (2.1.1) Remark By abuse of notation we do not add signs, it is clear that a tt pp = at t pp Let (x) be the set of arrays A in A having x as a critical point i.e. (x) = {A F (A, x) X }. In the view of dehomogenized form it becomes (x) = { A A a = a t p = 0 t, p }. Let H = H (F (A, x 0 )) = H (A) be the Hessian matrix of the form F at x 0 or the Hessian matrix of A at x 0. Entries of H are a tt pp the column index. In this set up the cusp variety takes the form: where (t, p) is the row index and (t, p ) is 0 cusp = { A ( x 0) det H (A) = 0 }. (2.1.2) and cusp = 0 cusp.g. Proposition Cusp variety is irreducible. Proof. It is enough to show that 0 cusp is irreducible. This follows from a study of the polynomial factors of the determinant of the Hessian matrix. Proofs are given at the last section. Here, we just mention that, the determinant is irreducible for all k = 3, N 8 or k 4, N 2k. For k = 3, N = 6, 7, it is the cube and the square of an irreducible polynomial which implies irreducibility. Instead of working with the group G = GL (V ), we use E G where g E has the following shape: I k k 0 K I (N k) (N k) (2.1.3) 16

17 where the N k block is the transposition of the coordinate matrix K described in E. By abuse of notation the group and the space have the same letter E. E acts simply transitively by translations on the space E. One can compute the action of E on A via partial derivatives of the form F. Theorem The tangent space of the cusp variety at the point x 0 is of codimension 2 in the tangent space of all arrays. Proof. Let Z = {(A, x) A E A.x (x 0 )}. By definiton of (x 0 ) and the group action of E on A the tangent space T (A,x 0 ) Z is given by equations: da = 0, da t p + t,p a tt pp dxp t = 0 (2.1.4) where the differentials da, da t p,... and dx p t are the usual coordinates in T A A and T x 0E respectively. Now consider the variety Z cusp { (A, x) A E A.x 0 cusp } = { (A, x) Z } det H (A.x) = 0 (2.1.5) (2.1.6) The tangent space of this variety at (A, x 0 ) is given by above equations together with d det (H (A.x)) = 0 (2.1.7) Let Ĉpp tt be the cofactor of the Hessian matrix corresponding the entry a tt pp. Then the above equation evaluated at x 0 has the following expression: Ĉ pp tt da tt pp + a tt t pp p = 0 (2.1.8) dxp t t,t,p,p t,p The tangent space T A cusp at a generic point A of 0 cusp is the image of T (A,x 0 ) Z cusp under the map pr 1 induced by the first projection. In other words T A cusp is defined 17

18 by all the linear equations between the coordinates da, da t p that are consequences of equations and 2.1.7,that is, by the linear combinations of the equations not containing the terms dx p t. One of them is da = 0. We can choose A as described in subsection so that the Hessian matrix H(A) is corank one. This implies equations produces exactly one relation between da t p which can be written as c p t da t p = 0 (2.1.9) t,p where c p t are the coefficients of the linear dependency between rows of H (A). We will show that it is the only relation by showing that equation does not produce any additional relations. To do this, it is enough to produce A A satisfying: t,t,p,p,t,p Ĉ pp tt c p t a tt t pp p 0 (2.1.10) Since C has corank 1, the adjoint has rank 1, we can choose any nonzero row of adc as the vector of coefficients c p t and then c p t described in 2.5.4, there exists Ĉp p + t t 0. Since a tt t pp p = Ĉpp tt. By the existence of matrices show that its coefficient is nonzero. The coefficient is Ĉpp tt Ĉ p p t t 0. is arbitrary it is enough to 2.2 The Generic Node Component Let I f and I l be the ordered sets {1, 2,..., k} and {N k + 1,..., N} respectively. Recall that we chose N 2k, which implies I f I l =. We study the following variety: 18

19 Definition The generic node variety 0 node ( ) is the set of forms which are tangent to X at I f and I l i.e: { ai1...ik η i1...ik a i1...ik = 0, when If {i 1... i k } k 1 or I l {i 1... i k } k 1} for all possible indices {i 1... i k }. Its Zariski closure is denoted by node ( ). Theorem For all k 3 and N 2k, except k = 3, N = 6, 7, 8, the generic node component node ( ) of the dual Grassmannian G(k, C N ) is of codimension 2 in T A A, given by equations: da I f = 0, da I l = 0 (2.2.1) Proof. Similar to the definition of x 0, let x be the span of basis vectors e N k+1,..., e N. Let w G be the element which sends points around the image of x 0 to x. Consider the variety: Z 2 = {(A, x, y) A E we (A, pr(x)) Z, (A, pr(y)) Z} (2.2.2) Recall that H(A, x 0 ), H(A, x ) are the Hessian of A at x 0 and x respectively. We use the same notation for coefficients requiring I f introduced at section 2.1. For the coefficients requires I l we put w. For instance we use a N k+1,...,n = w a, a N k+1...t...n is denoted by w a t p where t is the value < N k + 1 and p is the position of t and it is done for the other terms similarly. The tangent space T Z (A,x 0,x ) 2 is given by equations together with da = 0, da t p + t,p a tt pp dxp t = 0 (2.2.3) d w a = 0, d w a t p + t,p w a tt pp dxp t = 0 (2.2.4) 19

20 The tangent space at a generic A (x 0 ) (x ) is given by da = d w a = 0 if and only if the other equations in and do not produce additional relations between da t p, d w a t p. If both H(A, x 0 ), H(A, x ) are invertible, then it holds. Assume that k 5. There is no common entries of H(A, x 0 ), H(A, x ) and those entries are arbitrary, so the determinant is nonzero. In this case the generic node component is of codimension 2 in A. If k = 4, there are 36 nonzero common entries of H(A, x 0 ) and H(A, x ). For N = 8 such matrices exist. We can embed it into the Hessian matrices with larger N. In more details: common entries of H(A, x 0 ) and H(A, x ) are of the form a i,j,i,j such that i, j I f, i, j I l. After row and column operations, they satisfy ( H(A, x 0 ) αβ )γθ = ( ) H(A, x ) γθ where inner subscripts give position of skew symmetric block in the whole Hessian, outher subscripts are inner position in the block, overline means complementary indices i.e. αβ = {1, 2, 3, 4} {α, β}. The most complicated case occurs when k = 3, since we have more conditions on αβ the entries of H(A, x 0 ) and H(A, x ) as in the following: H(A, x 0 ) = satisfying: 0 A 12 A 13 A 12 0 A 23, H(A, x ) = A 13 A B 12 B 13 B 12 0 B 23 B 13 B 23 0 (2.2.5) For each A ij (respectively, B ij ) the last (respectively, the first) 3 3 block is zero matrix. 20

21 The first row of B 12 (respectively, B 13, B 23 ) is the transpose of (N 3)th (respectively, (N 4)th, (N 5)th) column of A 23 The second row of B 12 (respectively,b 13, B 23 ) is the transpose of (N 3)th (respectively, (N 4)th, (N 5)th) column of A 13 The third row of B 12 (respectively, B 13, B 23 ) is the transpose of (N 3)th (respectively, (N 4)th, (N 5)th) column of A 23 Such a tuple of matrices exists for N = 9, 10, 11. Then we embed them into larger matrices described in the section 2.5 G (3, C 6 ) has no generic node component, since the skew symmetric blocks of its Hessian matrix is 3 3, by the first item in the above list, it is the zero matrix. Generic node components for G (3, C 7 ) and G (3, C 8 ) lie in the cusp component, since the determinant of the Hessian matrices of those are zero. 2.3 Special node components We follow the notation in the previous section. Recall that for generic node type, there is no intersection of the sets I f and I l. Now we consider the case with possible intersections. Let J be an ordered set such that J I f I l, J = k (2.3.1) The complement of J in I f I l is denoted by J, i.e. J J = and J J = I f I l. Let x (J) Y be the point with coordinate matrix K of size k N such that K i,j(i) = 1 21

22 otherwise zero where J(i) means ith element of the ordered set J. In particular if J = I f, it is the point x 0. Recall that (x) is the set of arrays a A such that x is the critical point of A. (x) is a vector subspace in A and it has codimension k (N k)+1. We use star of a multiindex (i 1,..., i k ) which is the set of all multiindices that differs from (i 1,..., i k ) in at most one position. Let P I f. Replacement r(p ) is the multiindex in which we replace the θth position of I f with the θth position of J. We have : (x (J)) = {A A a i1...i k = 0, (i 1... i k ) in the star of J} (2.3.2) By using this we get: node (J) = ( (x (I f )) (x (J))).G (2.3.3) where J I l, and bar denotes Zariski closure. Indeed node component node described in the definition has the decomposition according to sets J, i.e. node = J node(j). In the case of J = I l, we call it the generic node component node ( ) just to keep the analogy with [WZ96]. Observe that the star of multiindices I f and I l do not meet each other. Therefore, ( x ( I f)) ( x ( I l)) has codimension 2 (1 + k (N k)) in A as a vector subspace. Using the action of the group G, we see that (x) (y) has codimension 2 (1 + k (N k)) whenever (x) (y) node ( ). By the same argument we conclude that ( x ( I f)) (x (J)) has codimension 2 (1 + k (N k)), whenever I f J k 3. If I f J k 2, and J I f = {α, α }, I f J = {t, t }, star of multiindices has 4 22

23 intersections. They are: a α t, a α t aα t a α t (2.3.4) Hence in this case codimension is 2 (k (N k) 1) in A. If I f J k 1, node (J) is not a component of node by the next proposition. Proposition For G ( k, C N), if J I f = k 1 then node (J) cusp. [ Proof. Hessian matrix associated to I f is given by blocks B λ iλ j ij ], 1 i, j k. Let J be the set with I f J = α and J I f = β. So intersection of J and I f is of size k 1. With respect to J, a λ β β for all possible λ β is zero, since they are linear terms. Moreover a βj αj are zero in the Hessian matrix, since they are linear terms of J. For all possible j, a βj αj forms zero column and row in the Hessian matrix, so its determinant is zero. This implies node (J) cusp if J I f = k Construction For every J and every T 0 we define the point x (J, T ) as follows: Let K be k N matrix with the first k k block is the identity matrix. Entries of the last k k block are either T, T 1 or zero described as: Put T into the position K rs, r, s are row and column indices with r I f J, s I l J is the replacement. For positions I l J, put T 1. Fill the remaining positions with zeros. We give here some examples for points parametrized by J, T. 23

24 Example For G (4, C 8 ), given J = {1, 6, 7, 8} T T K = T T 1 Replacement is between 1 and 5. For G (4, C 10 ), given J = (2, 3, 8, 9) T T K = T T 1 0 (2.3.5) (2.3.6) Replacement is 1 8, 2 7, 3 10, 4 9. For G (5, C 14 ), given J = (2, 4, 12, 13, 15), last square block: 0 T T K = T T T (2.3.7) Lemma If I f J k 3 then lim ( x ( I f)) (x (J, T )) = ( x ( I f)) (x (J)), T 0 the limit taken in the Grassmannian of subspaces of codimension 2 (k (N k) + 1). Proof. We will exhibit a system of linear forms φ i,t, (i = 1,..., 2k (N k) + 2) on space A defining the subspace ( x ( I f)) (x (J, T )) and having the following property: 24

25 Each φ i,t is a polynomial function of T and the linear forms φ i,0 are linearly independent. The above property implies existence of a limit operation, and limit is given by evaluation of forms at T = 0. The first k (N k) + 1 forms φ i,t will be independent of T, these are the degree zero and degree one forms defining ( x ( I f)). For the remaining forms we take forms associated to x (J, T ) and its partial derivatives. Explicitly: ( ) F (A, x (J, T )) = T J P J P a r(p ) P I f (2.3.8) where J is the complement of J, r (P ) is replacement. Then to make it polynomial we multiply it with suitable power of T, explicitly T If J. Example G (4, C 10 ), J = (2, 3, 8, 9), I f = (1, 2, 3, 4), I l = (7, 8, 9, 10), J = (1, 4, 7, 10). Replacement is given by 1 8, 2 7, 3 10, 4 9. We give a few of them: r (P ) J P J P r ({1}) = 8, 2, 3, r ({2}) = 1, 7, 3, r ({3}) = 1, 2, 10, r ({4}) = 1, 2, 3, r ({1, 2}) = 8, 7, 3, r ({1, 3}) = 8, 2, 10, r ({1, 2, 3}) = 8, 7, 10, The limit of this form is a r(j). 25

26 All other forms F (A, x (J, T )) x j i (2.3.9) are treated in a similar way for possible indices i, j. They are all possible 1 j k and 1 i N excluding: (a) i = j I f J, i I f (b) i T l J, j I f J and replacement of j is i, i I l. In the limit we obtain a i1,...,i k for all (i 1,..., i k ) in the star of J. Those forms define (x (J)), we are done. Lemma If I f J = k 2, J I f = {α, α }, I f J = {t, t } lim T 0 ( x ( I f)) (x (J, T )) = ( x ( I f)) (x (J)), exists and is equal to the subspace in ( x ( I f)) (x (J)) given by four additional equations: j I f J j I f J where r(j) is the replacement of j. a r(j)α) jt = j I f J a r(j)α) jt = j I f J a r(j)α jt = 0 a r(j)α jt = 0 Proof. We prove it in a similar way to the previous lemma but with a slight modification. When we normalize the form by multiplying a suitable power of T, we get : A x t α F (A, x (J, T )) 26

27 Let ϕ T be the form whose first two leading terms are given by: + T a + a α t j I f J a r(j)α jt Since the forms a and a α t are degree zero and degree one terms with respect to If which are already zero, we replace the form by ( ) ϕ T = T 1 ϕ T T a a α t Now we have ϕ 0 = j I f J a r(j)α jt (2.3.10) We get the other equations by just differentiating the form at x t α, x t α and xt α. Proposition For G ( k, C N), if J I f k 3 then node (J) node ( ). Theorem (a) Suppose that I f J = k 1. Then node (J) cusp (b) Suppose that I f J k 2 for all k, N except (k, N) = (3, 6), (3, 7), (3, 8), (3, 9). Then node (J) node ( ) We already proved part a. Part b, if the cardinality intersection I f J is smaller than k 2, it follows from lemma In the equality case we have the following arguments: To show that node (J) lies in the node ( ), it suffices to show the existence of a dense Zariski open set U ( x ( I l)) (x (J)) satisfying the conditions below: For every array A U, there exists a g G such that A.g ( x ( I l)) (x (J)) (2.3.11) 27

28 A.g satisfies the additional equations in the lemma. We need to find g G such that g fixes every coordinate except the basis indexed by I l J. And it sends elements of I l J to c j i e i. With this A.g satisfies the above conditions, and the number of unknowns (i.e. c j i ) is more than 4, so there exist a solution with nonzero coefficients. Hence it is Zariski open and dense. 2.4 Decomposition of singular locus In this section we prove that X sing = X node X cusp. To show this, we will modify the arguments of [WZ96], which follows from the works [Kat73] and [Dim86]. Let sm = sing be the set of smooth points of and O be the complement of node cusp in. The following result is due to N. Katz [Kat73]: Proposition ([WZ96], Proposition 6.1). Suppose the projection pr 1 : Z is generically unramifired. Then this projection is birational. Furthermore O consists of smooth points of and is the biggest open set in for which the projection pr 1 : pr 1 1 (O) O is an isomorphism. By this proposition O is a subvariety of sm, which is equivalent to: sing node cusp (2.4.1) Now we need to show the reverse inclusion. We will deduce it by the following two propositions: Proposition ([WZ96] Lemma 6.2). The variety of arrays A having infinitely many critical points in X is of codimension 2 in. 28

29 This result is due to A. Dimca [Dim86]: Proposition ([WZ96], Proposition 6.3.). Suppose an array A has finitely many critical points in X. Let H A be the hyperplane in P( k C N ) defined by A. Then the multiplicity of at A is equal to mult A ( ) = µ(x H A, x) (2.4.2) x X(A) where the sum is over all critical points of A in X and µ(x H A, x) is the Milnor number of X H A at x. In particular µ(x H A, x) = 1 if and only if the Hessian of A at x is nondegenerate. We see that for all k C N except 3 C 6, 3 C 7, 3 C 8, all of the irreducible components of node and cusp are of codimension one in. By proposition 2.4.2, the generic point A of each of them has finitely many critical points in X. By the multiplicity of at A is 2 which implies A is a singular point of. Therefore node cusp sing. We combine it with to get: sing = node cusp (2.4.3) For the remaining cases, since node cusp, we get sing = cusp. Also one can check this by just looking at their orbits [Hol11], [KW12],[KW13]. Theorem Let X be dual Grassmannian. Then we have: X sing = X node X cusp (2.4.4) Theorem The cusp and the generic node components are different for the dual Grassmannians except k = 3, N = 6, 7, 8. 29

30 Proof. To prove node ( ) cusp in A, we modify arguments used in [WZ96]. We will construct conormal bundles of cusp and node component, and show that their generic fibers are different. We can identify the space of arrays A with k (C ) N. Let e 1, e 2,..., e N be a basis of (C ) N. Now we can make a pairing between da i1...i k and e i1 e ik. We deduce the following: For a generic A (x 0 ) (x ), the conormal space T A, node A is the two dimensional subspace in A spanned by e 1 e k and e N k+1 e N. By the group action, the generic fiber of the bundle T node A is spanned by two vectors u 1 u k and v 1 v k where all u i, v j are linearly independent. By similar arguments the generic fiber of the conormal bundle T cusp A is a two dimensional vector space of the form W = k w 1 w j 1 (C ) N w j+1 w k (2.4.5) j=1 where w j (C ) N and nonzero for all j = 1,..., k. For any two alternating decomposable tensors u 1 u k and v 1 v k contained in W, there is an index j such that u j is proportional to v j. Hence W cannot contain a generic fiber of T node A. This implies that node cusp. 2.5 Symmetric Matrices with diagonal lacunae and skew symmetric blocks We consider the following type of matrices: for given k, N, construct a k (N k) k (N k) matrix 30

31 which has a k k skew symmetric blocks of size (N k) (N k). They have the following shape: 0 A 12 A 1k A 12 0 A 2k A k 1,k A 1k A 2k A k 1,k 0 (2.5.1) where each A ij and 0 are skew symmetric blocks of size (N k) (N k). The aim is to find irreducible factors of the determinant of that kind of matrices. H k,n denotes the matrix as above. Theorem Let H k,n be the Hessian matrix of the dual Grassmanian evaluated at x 0. Then we have the following: the determinant of H k,n is an irreducible homogeneous polynomial for all N 2k, k 3 except k = 3, N = 6, 7. In these cases we have: (1) det H 3,6 is the cube of degree 3 irreducible polynomial. (2) det H 3,7 is the square of degree 6 irreducible polynomial. Remark The dual variety is hypersurface if and only if there exists an invertible Hessian matrix [GKZ94]. So, by the theorem 2.5.1, for all N 2k, k 3, the dual Grassmannian is a hypersurface. To prove theorem 2.5.1, first we study the case k = det H 3,N Since k = 3, we have the following 3 3 block matrix 31

32 0 A 12 A 13 A 12 0 A 23. A 13 A 23 0 To analyse the determinant we use Young diagrams. Consider the direct sum decomposition of C N to C 3 C N 3. Entries of the Hessian matrix can be obtained from the quadratic part of 3 C N which is C 3 2 C N 3. The decomposition of 3 C N is 3 C 3 2 C 3 C N 3 C 3 2 C N 3 3 C N 3, and entries of H 3,N comes from the underlined part. We call the first and the second summands the zero and linear terms respectively. The existence of an invariant of degree d is equivalent to the existence of rectangular ( diagrams of size d in the decomposition of Sym d C 3 2 C N 3). We apply Cauchy s formula to get: ( Sym d ) ) C 2 3 C N 3 = S λ C ( 2 3 S λ C N 3 λ d (2.5.2) where λ = (λ 1 λ 2 λ 3 ) is Young diagram with height ht (λ) 3. Otherwise, the representation S λ C 3 = 0. Now we obtain arithmetic conditions, since we are looking at rectangular partitions λ appearing above: 3 d, (N 3) 2d, d 3 (N 3) The last condition arose since we are looking for possible polynomial factors of the determinant and the degree of it is 3 (N 3). If N 3 is a prime number greater than 3, det H 3,N is irreducible, since d becomes the product of two primes. This implies det H 3,8 and det H 3,10 are irreducible polynomials since those determinants are unique 32

33 invariants of degrees 15 and 21 respectively. We obtain the factors of exceptional cases and irreducibility of det H 3,9 by Maple computation. Before analysis of the other cases, we point out that by the canonical isomorphism G ( k, C N) ( = G N k, (C N ) ), we get H k,n = H N k,n. We give a proof in coordinates: Proposition (Duality). The Hessian matrix of type H k,n can be obtained by row and column changes on H N k,k. Proof. For the matrix H 3,N, arrange rows and columns in this order: [1, N 2, 2N 5 2, N 1, 2N 4... N 3, 2N 6, 3N 9]. It gives the block structure of H N k,n. For general k and N, the row and the column changes below gives the block structure of H N k,n [1, N k + 1, 2 (N k) , (k 1) (N k) + 1] [2, N k + 2, 2 (N k) , (k 1) (N k) + 2] [N k, 2 (N k),..., k (N k)] Let S be a skew symmetric matrix of size k k. We can rewrite it in terms of smaller blocks i.e. S 1 U S = (2.5.3) U t S 2 where the blocks on diagonal S i are skew symmetric of size k 1, k 2 with 2 k 1, k 2 and k 1 + k 2 = k and U is the k 1 k 2 rectangular block with arbitrary entries, U t is the 33

34 k 2 k 1 block which is the transpose of U multiplied by 1. We want to make a similar block decomposition to H 3,N. The shape of H 3,N is 0 A B A 0 C. B C 0 where A, B, C are skew symmetric matrices of size (N 3) (N 3), 0 are the sero matrices matrix of size (N 3) (N 3). We can divide A, B, C into blocks described in (2.5.3) above. Now H 3,N becomes 0 0 A 1 A B 1 B 0 0 A t A 2 B t B 2 A 1 A 0 0 C 1 C A t A C t C 2 B 1 B C 1 C 0 0 B t B 2 C t C with sizes A 1, B 1, C 1 = k 1 k 1, A 2, B 2, C 2 = k 2 k 2, k 1 + k 2 = N 3, k 1 k 2. After row and column operations (2, 4, 5, 3) applied to blocks we get: 34

35 0 A 1 B 1 0 A B A 1 0 C 1 A 0 C B 1 C 1 0 B C 0 0 A t B t 0 A 2 B 2 A t 0 C t A 2 0 C 2 B t C t 0 B 2 C 2 0 The unshaded 3 3 blocks are in the form of H 3,k1 +3 and H 3,k2 +3. If we call the upper right shaded block U, the remaining block is the transpose of U. After setting A, B, C to zero we have H 3,N = H 3,k1 +3 H 3,k2 +3, overline denotes the matrix with the zero blocks. We call this method specialization with zero and it allows us to embed smaller matrices into larger ones. Lemma If N 11, det H 3,N is an irreducible polynomial. Proof. Consider the embeddings: H 3,6 H 3,N 3 H 3,N (2.5.4) H 3,7 H 3,N 4 H 3,N (2.5.5) For small N s we give the possible polynomial factors of these embeddings, then we derive a contradiction. The basic tool is the following lemma which we do not know 35

36 any reference of it: Lemma Let P be a homogeneous multivariate polynomial. Suppose that by setting some variables to zero we get P = P 1 P 2, deg P 1 deg P 2 and P 1, P 2 are irreducible homogeneous polynomials. Similarly, suppose that by setting some variables not the same as above we get P = Q 1 Q 2, deg Q 1 deg Q 2 and Q 1, Q 2 are irreducible homogeneous polynomials. If deg P 1 > deg Q 1 and deg P 1 deg Q 2, then P has to be an irreducible polynomial. Proof. Assume to the contrary that P has nonconstant polynomial factors. Specialization with zero cannot increase the degrees of those irreducible factors. Thus, the polynomial factors assumed in the lemma at specializations with zero can only appear if P is irreducible. By the lemma 2.5.5, it is enough to analyze the degrees of possible polynomial factors of the determinant. If they are not compatible we conclude that it has to be an irreducible polynomial. First we analyze H 3,11. We can embed H 3,6 and H 3,8 into H 3,11 and H 3,7 H 3,7 as in Degrees of irreducible factors are and which are not compatible by lemma 2.5.5, hence det H 3,11 is irreducible polynomial of degree 24. Now we can use proof by induction. We have showed that det H 3,8, det H 3,9, det H 3,10,det H 3,11 are irreducible. Assume that for all n < N det H 3,n is irreducible, and they have degree 3 (n 3). If we apply embedding 2.5.4, we get possible factors of det H 3,N as (N 6) and (N 7) and they are not compatible by lemma This forces that det H 3,N is irreducible. 36

37 2.5.2 det H 4,N Now we focus on H 4,N, N 8. The matrix we are dealing with is: 0 A 12 A 13 A 14 A 12 0 A 23 A 24 A 13 A 23 0 A 34 A 14 A 24 A 34 0 (2.5.6) Observe that if we set A 34 to be the zero matrix in (2.5.6), det H 4,N is the square of the irreducible determinant. Together with lemma this implies that det H 4,N has at most two factors of the same degree. Lemma If N 8, det H 4,N is irreducible Proof. We have two assumptions. After assuming those, the result holds. By computer computation we obtain that, det H 4,8 and det H 4,9 are irreducible. Now we modify the method we used for H 3,N. For each A ij of matrix 2.5.6, we apply This allows us to embed smaller Hessian matrices into larger ones. It is enough to use the following embeddings. H 4,6 H 4,N 2 H 4,N (2.5.7) H 4,7 H 4,N 3 H 4,N (2.5.8) For N = 10, degrees of possible factors are and By 2.5.5, they are compatible only det H 4,10 is irreducible. 37

38 Now we can use induction. Suppose that for all n < N, det H 4,n is irreducible. By and 2.5.8, degrees of possible factors are (N 6) and (N 7). They are compatible if det H 4,N is irreducible det H k,n, k 5 First we analyze the case k = 5. We obtain irreducibility of det H 5,10 by computer computation. In H 5,11, we can embed H 5,8 H 5,8 and H 3,9 H 2,8 by using duality and setting A i,5, i 4 to zero respectively. Degrees of possible factors are and , by lemma we obtain irreducibility of det H 5,11. In the case of H 5,12, we can use Young diagrams. A degree d invariant exists if and only if there exists rectangular diagrams satisfying 5 3d and 7 2d. This implies d = 35 which is the degree of determinant. For the remaining cases we can use induction together with the embeddings: H 5,8 H 5,N 3 H 5,N (2.5.9) H 5,9 H 5,N 4 H 5,N (2.5.10) In general we can use embeddings H k,k+2 H k,n 2 H k,n (2.5.11) H k,k+3 H k,n 3 H k,n (2.5.12) H k,k+4 H k,n 4 H k,n (2.5.13) and induction to show that det H k,n is irreducible depending on whether k is odd or even. If k is odd, det H k,k+2 is identically zero. By these the only open case is H 7,14, 38

39 we obtain irreducibility of its determinant by computer computation Rank Lemmas We need to show existence of Hessian matrices H k,n satisfying the below conditions: (i) It has corank 1. (ii) Each row block of size k k (N k) is of full rank. Observe that, if there exist a matrix A satisfying the above, then A A satisfy it when A is of full rank. That s why it is enough to produce those matrices for G(3, 8), G(3, 9), G(3, 10), G(3, 11), G(4, 8), G(4, 9, ), G(5, 10). Since we can use the following embeddings inductively: (1) 3 C 6 3 C a 3 C a+3, the first matrix is generic. (2) 4 C 6 4 C a 4 C a+2, the first matrix is generic. (3) 5 C 8 5 C a 5 C a+3, a 8, the first matrix is generic. (4) in general: 3 C N k+3 k 3 C N 3 k C N, the second matrix is generic. 2.6 List of Matrices Here we give the list of matrices which satisfies conditions in part C 9 39

40 0 A 1,2 A 1,3 A 1,2 0 A 2,3 A 1,3 A 2,3 0 A 12 = , A 13 = A 23 = C 10 40

41 A 12 = , A 13 = A 23 = C 11 41

42 A 12 = , A 13 = A 23 = C 8 42

43 0 A 1,2 A 1,3 A 1,4 A 1,2 0 A 2,3 A 2,4 A 1,3 A 2,3 0 A 3,4 A 1,4 A 2,4 A 3,4 0 A 12 = , A 13 = , A 14 = A 23 = , A 24 = , A 34 = C 9 43

44 A 12 = , A 13 = , A 14 = A 23 = , A 24 = , A 34 = C 10 0 A 1,2 A 1,3 A 1,4 A 1,5 A 1,2 0 A 2,3 A 2,4 A 2,5 A 1,3 A 2,3 0 A 3,4 A 3,5 A 1,4 A 2,4 A 3,4 0 A 4,5 A 1,5 A 2,5 A 3,5 A 4,5 0 44

45 A 12 = , A 13 = , A 14 = A 15 = , A 23 = , A 24 = A 25 = , A 34 = , A 35 =

46 A 45 =

47 Chapter 3 Analysis of k C N C M Consider the following Segre embedding: ( k X = P C N) P ( ( k ) C M) P C N C M (3.0.1) Let η i1...i k be the minor of the coordinate matrix: x 1 1 x 1 2 x 1 N x 2 1 x 2 2 x 2 N x k 1 x k 2 x k N indexed by (i 1... i k ). For simplicity we choose coordinates y 1,..., y m for C M. A form F in local coordinates is given by : F (A, z) = a i1...i k :jη i1...i k.y j (3.0.2) i 1,...,i k,j Again we have the criteria for the points of X F X F (A, p) = F x j i (p) = F y t (p) = 0 for all possible indices i, j, t (3.0.3) Definition We call such a point z, the critical point of the form or A A. 47

48 Remark Let A be the space of arrays i.e. the coefficients of the form F = F (A, z) for X described above. Projection onto the first factor of the incidence variety Z = {(A, z) A X z is a critical point of A} (3.0.4) is the points of the dual variety. There is a natural action of G = SL ( C N) SL ( C M) on X which induces the action on F, F (A, g.z) = F (A.g, z) where g = (g 1, g 2 ) G. Let Y be the product of the span of k linearly independent vectors with another vector i.e. v 1,..., v k w 0 Y. Let z 0 be the point which is the product of the span of the basis vectors e 1,..., e k with f 1, which is equivalent to say z 0 is the point with coordinates x j i = δ i,j and y t = δ 1,t. 3.1 The Cusp Component We will work on affine chart given by z 0, so we dehomogenize the form F = F (A, z) = i 1,...,i k,j a i 1...i k :jη i1...i k.y j corresponding to A A by setting all x j i = δ i,j where i, j k and y 1 = 1. More precisely, we consider the subset E of Y, which is the preimage of point [1 : 0 :... : 0]. For simplicity we introduce a new notation for A A. We have: a 12...k;1 = a a 12...t...k;s = a t p;s we keep track of indices coming from the Plücker coordinates similar to the dual Grassmannian case. We keep the additional index as it is. For example a 12...k;t is denoted by a ;t. 48

49 Now the form F becomes: F (A, z) = a + p,t a t p;1x p t + t a ;t y t + a tt pp xp t x p t,t,p,p t + t,t,p a t p;t xp t y t +... (3.1.1) Remark For the first part of index, we assume sign permutation, i.e. a tt pp ;u = a t t p p;u. Let (z) be the set of arrays A A having z as a critical point: (z) = {A F (A, z) X }. (3.1.2) We can rewrite it (z) = { A A a = a t p;1 = a ;u = 0, p, t, u } (3.1.3) Let H (F (A, z 0 )) be the Hessian matrix of form F. Entries of H are quadratic coefficients either a tt pp or at p;u. Now, 0 cusp = {A (z 0 ) det H (A) = 0}, and cusp = 0 cusp.g. If the determinant of the Hessian matrix is irreducible or it has only one irreducible factor, then immediately the cusp component is irreducible. If there are factors, we analyze that case later. We work with the group of affine translations E in SL ( C N) SL ( C M), its elements look like: I k k 0 K 1 I (N k) (N k) 1 0 K 2 I (M 1) (M 1) (3.1.4) The action of G on A can be computed by partial derivatives of the form in the local coordinates. 49

50 Definition If M = k (N k) + 1, we call the format boundary. Otherwise we call it interior. Remark If M is greater than k (N k) + 1, the dual variety is not a hypersurface. Theorem If the format is not boundary, and the determinant of the Hessian matrix is irreducible, then the tangent space of the cusp variety is of codimension 2 in the tangent space of all arrays Irreducibility of the cusp component Since the determinant of the Hessian matrices have factors for some formats, to show that they are irreducible varieties we need another approach C N+2 C 2 We can identify 2 C N+2 C 2 with 2 slices of skew symmetric matrices, and N is even. Front slice S 1 and rear slice S 2 are: S 1 = b, S 2 = 0 0 c 0 A b t c t B with: (I) S 1 and S 2 are (N + 2) (N + 2) skew symmetric matrices. (II) A and B are N N skew symmetric matrices. (III) b and c are 1 N row vectors. 50

51 The action of (g, h) SL(C N ) SL(C 2 ) on [S 1 S 2 ] is given by: (g, h).[s 1 S 2 ] = [gs 1 g t h 11 + gs 2 g t h 21 gs 1 g t h 12 + gs 2 g t h 22 ] (3.1.5) If we specify g and h as: 1 0 x g = 0 1 y, h = 1 z (3.1.6) I Front slice(skew symmetric) is: 0 xay t + ( xc t + by t + xby t )z xa + (b + xb)z S 1 = 0 ya + (c + yb)z A + Bz Rear slice(skew symmetric) is: 0 xc t + by t + xby t b + xb S 2 = 0 c + yb B (3.1.7) (3.1.8) Consider two irreducible components: = { A (z 0 ) pf(a) = 0 } (3.1.9) = { A (z 0 ) U = 0 } (3.1.10) where U is the polynomial det A b divided by the Pfaffian of A. c t 0 The following lemma proves irreducibility of the cusp component. Proposition The set of arrays A such that A.g for some g E is dense in. Therefore cusp =.E. 51

52 Proof. [S 1 S 2 ] lies in the (x 0 ) if : (i) xay t + ( xc t + by t + xby t )z = 0 (ii) xa + (b + xb)z = 0 (iii) ya + (c + yb)z = 0 Now, we assume that A + Bz is invertible for some z 0. If we substitute ii, iii into i, we get: x(a + Bz) 1 y t = 0 (3.1.11) Recall that the inverse of a skew symmetric matrix is again skew symmetric. If x or y is zero, then follows immediately. If both of them are nonzero, we can apply the base change to send x, y to (1, 0,..., 0). The equation gives zero if the cofactor corresponding to entry (1, 1) is zero. Since the matrix is skew symmetric, the determinant of the principal minor is identically zero. Hence, A.g 0 cusp but not in, since A + Bz is invertible. Therefore A.g. To complete proof we need: Lemma Every pair (A, B) of N N skew symmetric matrices such that the Pfaffian of A is zero lies in the closure of the set {(A, B) pf(a) = 0, det(a + Bz) 0 for some z C} It is enough to show that det(a + Bz) is not identically zero. Choose B as direct 0 1 sums of J =. Then, in the expansion of the determinant, the coefficient of 1 0 the highest degree monomial z is det B which is nonzero. 52