Symmetric Determinantal Singularities I: The Multiplicity of the Polar Curve
Terence Gaffney Michelle Molino
math.AG


Symmetric Determinantal Singularities]Symmetric Determinantal Singularities I: The Multiplicity of the Polar Curve



T. Gaffney]Terence GaffneyT. Gaffney was partially supported by PVE-CNPq Proc. 401565/2014-9
 T. Gaffney, Department of Mathematics

  Northeastern University

  Boston, MA 02215

M. Molino]Michelle MolinoThis paper contains work from this author's PhD dissertation at Universidade Federal Fluminense and was partially supported by Coordenação de Aperfeiçoamento de Pessoal de Nível Superior - (Financiamento 001) and National Council for Scientific and Technological Development - CNPq 
M. Molino, Department of Mathematics

  Universidade Federal Fluminense

  Niteroi, RJ - Brazil


This paper is the first part of a two part paper which introduces the study of the Whitney Equisingularity of families of Symmetric determinantal singularities. This study reveals how to use the multiplicity of polar curves associated to a generic deformation of a singularity to control the Whitney equisingularity type of these curves.


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    December 30, 2023
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§ INTRODUCTION


In this paper and part II <cit.>, we study the Whitney equisingularity of families of symmetric determinantal varieties. It is part of a long term effort by several researchers to connect invariants of algebraic objects (rings, ideals and modules) associated with singularities of complex spaces to equisingularity conditions. The project took off with work of Bernard Teissier in the 70s. Teissier, in <cit.>, in the case of families of hypersurfaces, with isolated singularities, found integral closure descriptions of equisingularity conditions, Whitney A and B, and controlled these conditions using algebraic invariants, such as multiplicities of ideals. Gaffney, in a series of papers, <cit.>, <cit.>, <cit.>, extended the results of Teissier to families of complete intersection, isolated singularities, hypersurfaces with non-isolated singularities, and then constructed a framework for dealing with isolated singularities in general in <cit.>. The approach of <cit.> is based on pairs of modules M,N, M⊂ N. The choice of M is canonical: it is the module generated by the partial derivatives of the defining equations of X, known as the Jacobian module. The best choice of N was less well understood, as some obvious choices lead to technical difficulties in calculating invariants associated to N. Recent work by Gaffney and Rangachev (<cit.>) supports the following approach adapted to the symmetric case:

In this first part of the paper, we start defining the symmetric determinantal varieties and some of our objects of study, such as, the tangent space and the Jacobian and Normal modules. We extend some of the results from <cit.> to the symmetric case. At the end we give a similiar interpretation, from the rectangular case, of the fiber of the conormal space of the symmetric determinantal variety S_r.

In Section two we calculate the multiplicity of the polar curve of N() where  is a 1-parameter stabilization of the symmetric determinantal variety X. First we analyze the multiplicity for the case where X=F^-1(S_n-1), which is similar to the rectangular case proved by Gaffney and Rangachev in <cit.>. After that we move to case where X=F^-1(S_n-2) and prove a formula to calculate the multiplicity based on an intersection number. For this last case, we give a explicit way to calculate the multiplicity by showing an equivalence between (N) and a modification of X based on the presentation matrix of the singularity. This equivalence then gives a decomposition of the multiplicity of the polar of N as a sum of intersection numbers of generic plane sections with the exceptional fiber of the modification. This means we
can compute the  intersection number of the image of the section with the polar variety of S_r of complementary dimension, as a sum of intersection numbers of modules naturally associated with the singularity; these intersection numbers in turn are the colengths of a collection of ideals.
 
 In section three we compute these intersection numbers, called mixed polars, as the alternating sum of intersections of modules which depend only on the presentation matrix, and give an example of a computation for a family of space surfaces. 



§ SYMMETRIC DETERMINANTAL VARIETIES AND THEIR PROPERTIES


In this section we  prove  results for symmetric determinantal varieties analogous to the general case where the total space is  and Σ_r is the set of elements of   of rank less than or equal to r. We define the normal and Jacobian module of a symmetric determinantal variety and show how to calculate the multiplicity of this pair. At the end, we prove a similar result from <cit.> about the fiber of the conormal space of such varieties. 

Let Σ_r be the set of all n× n matrices of rank less than equal to r. Then, the set of all n× n symmetric matrices of rank less than equal to r is given by

    S_r={A∈| A≤ r}=Σ_r∩.



Now consider an element of S_r as a map in . Given a map

    [        F:       ℂ^q         ⟶          ;                   x         ⟼ (f_ij(x)) ]

a determinantal variety X∈ℂ^ q is the pre-image of S_r∈, with the expected codimension, that is, X=F^-1(S_r) and X= S_r. 

 if we consider F as the identity map on , then  S_r is trivially a determinantal variety. 
 The following properties are fundamental in this work. The proof for each one of them is similar to the ones presented in <cit.> for the general case. 

 Let S_r be a symmetric determinantal variety inside of . Then,


  * S_r is irreducible;

  * The codimension of S_r in the ambient space is (n-r)(n-r+1)/2;

  * The singular set of S_r is exactly S_r-1;

  * The stratification of S_r, given by { S_i\ S_i-1}, is locally analytically trivial and hence it is a Whitney stratification of S_r.



Let G(n-r,n) be a grassmannian given by all the linear subspaces of dimension n-r in ^n, and π_1 and π_2 be the projections 

    × G(n-r,n)[dl]_π_1[dr]^π_2   
           G(n-r,n)



  * Consider the set

    S_r={(A,W)∈× G(n-r,n)|. A|_W=0}.

Projection onto G(n-r,n) exhibits S_r as an algebraic vector bundle over the grassmannian G(n-r,n). This implies that S_r is smooth and connected, which means S_r is irreducible. Clearly π_2 maps S_r properly onto S_r, showing that S_r is an irreducible variety of .


  * The sets S_r and S_r are birationally equivalent, which means

    S_r=S_r.
    
The dimension of the fiber (π_1^-1(W)∩S_r) is the same for all W∈ G(n-r,n). In that case, let us consider the linear space W given by

    W={(x_1,…,x_n)∈^n| x_1=x_2=… x_r=0}.

The set of symmetric matrices A such that . A|_W=0 is given by matrices of the following type

    ([ [ a_11    … a_1r;    ⋮    ⋱    ⋮; a_1r    … a_rr ]                                                  0;                                                  0                                                  0 ]).
  
This set is isomorphic to _s(^r,^r) whose dimension is r(r+1)/2. Thus, S_r is a vector bundle of rank r(r+1)/2. In that case,

    S_r= G(n-r,n)+(π_1^-1(W)∩S_r)=2rn-r^2+r/2.
 
Therefore, the codimension of S_r is

    S_r=n(n+1)/2-S_r=(n-r)(n-r+1)/2.


  * First, let A be a point in S_r\ S_r-1, that is, A is a matrix of rank exactly r. This means that there is only one point W∈ G(n-r,n) such that . A|_W=0. Therefore, S_r-1 contain the singular set of S_r. Now, we need to prove the equality of these sets. On the other hand, if A∈ S_r-1 then the rank of A is less than r and, therefore, there are more than one W satisfying . A|_W=0, meaning that A is a singular point of S_r. 

  * Similiar to the proof for the general case seen in <cit.>, this can be deduced by induction from the observation that any point p∈{ S_i\ S_i-1} has a product of analytic spaces

    (S_r,p)≃(S'_r-i,0)×(^i(2n-i+1)/2,0)

where S'_r-i is the symmetric determinantal variety inside of the ambient space _s(^n-i,^n-i) and i(2n-i+1)/2 is the dimension of S_i.  



One of the ways of studying the equisingularity of families is by using the multiplicity of pairs of modules and their polar curves. For this, we will use two specific modules associated with the symmetric determinantal variety X: the Symmetric Normal module, N(X), and the Jacobian module, JM(X). First, we will define the modules JM(X) and N(X), and then we will show some results that will be useful later.

 We first consider the variety S_r in . The Jacobian module of S_r is a submodule of Ø_S_r,0^p, where p is the number of equations defining S_r, generated by the partial derivatives of these set of equations. Whereas the Symmetric Normal module of S_r is the module given by the first order symmetric infinitesimal deformations of S_r. 

As in the general case (<cit.>), the symmetric determinantal varieties S_r are stable, that is, varieties whose Jacobian module is equal to the module of allowable first order linear infinitesimal deformations, which in this case is the Symmetric Normal module . 

Now, let X=F^-1(S_r) be a symmetric determinantal variety, where F is a map from ^q to , and consider the pullback 

    F^*:Ø_S_r,0^p⟶Ø_X,0^p.

Let δ_i,j denote the symmetric matrix with 1 in the (i,j), (j,i) place, and 0 elsewhere. The Jacobian module of X is the module given by the partial derivatives of the equations defining X. Now, the Normal module of X is defined by taking the minors of A+tδ_i,j composing with F, taking the derivative with respect to t and then evaluating at t=0, which means N(X)=F^*(N(S_r)). Since S_r is stable, 

    N(X)=F^*(N(S_r))=F^*(JM(S_r)).


In the next section we want to calculate the multiplicity of the polar of N() in some specific cases, so we want to describe (N(X)). The description we give will apply equally to (N()). 

Let [T_i,j] be the n by n matrix of indeterminates, where T_i,j=T_j,i. We can think of [T_i,j] as the identity map on . The generators of N(X) are in one to one correspondence with the entries of this matrix by the procedure described above. We also have 


    ((N(X)))≃(Ø_X[T_i,j]/I)

where I is the ideal of relations between the T_i,j under the map which sends them to the i,j generator of (N(X)). We can also see I as the kernel of the map 

    φ:Ø_X[T_i,j]⟶(N(X)).

This will help us to prove the next proposition.

 I contains the entries of the matrix [T_i,j][F].

 We work first with the variety S_r inside of  and M the identity map with coordinates m_ij, then extend to the case of general F. In fact for this first case it is convenient to work first with  Σ_r inside , Thus, we have a map 

    [                           φ:                 Ø_Σ_r[T_i,j]                            ⟶                    (JM(Σ_r));                                                     T_i,j                            ⟼ (∂Δ_1/∂ m_ij,…,∂Δ_q/∂ m_i,j);                              ]

where Δ =(Δ_1,…,Δ_q) is the vector of minors of  of size r+1. We first prove that the entries [T_i,j]^t[M] and [M][T_i,j]^t are in the kernel of φ.

A typical entry of [T_i,j]^t[M] is equal to 

    ∑_t=1^n T_tim_ts

and, therefore,

    φ(∑_t T_tim_ts)=∑_t=1^n m_tsφ(T_ti)=∑_t,q m_ts∂Δ_q/∂ m_ti.
  
Now, let us fix q. Observe if Δ_q is the minor of a submatrix which does not contain m_ti then 

    ∂Δ_q/∂ m_ti=0.

So we may assume Δ_q contains m_ti. Now, ∂Δ_q/∂ m_ti=0 is just the cofactor of m_ti so 

    ∑_t,q m_ts∂Δ_q/∂ m_ti

is just the expansion of Δ_q but with m_ts replacing m_ti. So this is either zero if s≠ i and m_ts is already a part of Δ_q, or Δ_q if s=i, or another minor if s≠ i and m_ts is not part of Δ_q. In any event, all terms with fixed  q are zero. The computation for [M][T_i,j]^t is similar.

Now we pass to the symmetric case. We view S_r as a subset of . Its equations are {Δ_q} as before and {(m_ji-m_ji)}. So we have the map

    [                               φ:                     Ø_S_r[T_i,j]                                ⟶                        (JM(S_r));                                                             T_i,j                                ⟼ (∂Δ/∂ m_ij,∂(m_st-m_ts)/∂ m_i,j);                                  ]

For this case, we claim that the entries of ([T]^t+[T])[M] are in the kernel of φ. Indeed, the condition for a element to be in the kernel of φ falls into 2 parts: the first is v·Δ=0 and the second is that v·(m_ij-m_ji)=0. The computations for general matrices done above, apply to satisfying the first part of the condition. 
Note that if the entries of [M][T_i,j]^t satisfy the first part, then so do the entries of ([M][T_i,j]^t)^t=[T_i,j][M] since [M] is symmetric. So the entries of ([T]^t+[T])[M] satisfy the first part. 

We claim the sum satisfies the second part. The element of ([T]^t[M]+[T][M] in position ij is

    ∑_t T_tim_ti+∑_t T_itm_ti.

Under φ this produces the vector field

    ∑ m_ti∂/∂ m_ti+∑ m_ti∂/∂ m_it.

Applying this to (m_ti-m_it) gives us

    m_ti∂ m_ti/∂ m_ti-m_ti∂ m_it/∂ m_it=0.

If we restrict the matrix to the symmetric case, that is, [T_i,j]^t=[T_i,j] then the entries of the matrix [T_i,j][M] are zero as are the entries of [M][T_i,j].


Passing to the general case, F induces maps, F^* Ø_Σ_r[T_i,j]→Ø_X[T_i,j] and F̂^* ℛ(JM(Σ_r))→ℛ(F^*(JM(Σ_r))).
The commutativity of the induced diagram then implies that the entries of the matrix [T_i,j][F] are in the kernel of the map from Ø_X[T_i,j] to ℛ(F^*(JM(Σ_r))), which implies the result.



 The proposition above allows us to conclude that [T_i,j]F=0 are some of the equations of ((N(X))).
 

In the next results we describe the fiber of the conormal modification of S_r. 
The polar varieties of S_r are intimately connected with our invariant. Since the polar varieties are obtained by intersecting the conormal C(S_r) of S_r with enough generic hyperplanes, then projecting to S_r, whether the polar varieties are empty or not depends on the dimension of the fibers  C(S_r).

Let C={(A,B), A∈ S_r, B∈ℙS_n-r, AB=0}. Note that C is the set of points in S_r×ℙS_n-r which satisfy the equations of the previous proposition. Our next goal is to show that C=C(S_r).

 C_|S_r-S_r-1=C(S_r)_|S_r-S_r-1

 If (A,B)∈ C(S_r)_|S_r-S_r-1, then AB=0 by the previous proposition hence in C_|S_r-S_r-1. So suppose  (A,B)∈ C_|S_r-S_r-1. Since Aand B are symmetric AB=0 implies AB^t=B^tA=0.This implies by <cit.>, that B defines a tangent hyperplane to Σ^r-Σ^r-1⊂ at A.
Since B is symmetric and S_r-S_r-1⊂Σ^r-Σ^r-1 is a smooth embedding, B defines a tangent hyperplane to S_r-S_r-1 at A in .




 C_|S_r-S_r-1=C(S_r), and C(S_r) is an component of C

 Since  C_|S_r-S_r-1=C(S_r)_|S_r-S_r-1,  C_|S_r-S_r-1=C(S_r)_|S_r-S_r-1. But C(S_r)_|S_r-S_r-1=C(S_r). Since S_r is irreducible, so is C(S_r), so it is a component of C. 


  There is a GL(n, ) action on C which preserves the rank of A and B, given by S·(A,B)=(S^tAS, S^-1B(S^-1)^t).

For the next proof, it is convenient to decompose the ℂ^n as ℂ^s⊕ℂ^r-s⊕ℂ^n-r, with s<r<n and to write 
ℂ^s⊕ℂ^r-s as ℂ^r, ℂ^r-s⊕ℂ^n-r as ℂ^n-s. Let I_u denote the identity map on ℂ^u, with u=s, r, n-r, n-s, r-s

 Suppose A=I_s∈ S_r, s<r<n, then C_A=C(S_r)_A.

 _s(ℂ^n-s,ℂ^n-s) is embedded in  by the inclusion of ℂ^n-s in ℂ^n and by extension over 
ℂ^s by 0. With this identification the fiber, C_A, of C over A is ℙ(S_r), S_r⊂_s(ℂ^n-s,ℂ^n-s).
If B∈ℙ(S_r), it suffices to prove B∈ C(S_r)_A with rank of B=n-r. Now we can choose an invertible S such that S^tBS=I_n-r. Now

    0=I_r I_n-r=I_rS^tBS=I_rS^t(B).

Consider the line L in S_r parameterized by I_s+t(SI_r-sS^t). Since (I_s+t(SI_r-sS^t))B=I_sB+t(S(I_r-sS^t)B)=0, (I_r+t(SI_n-rS^t),B) is a line in C passing through A at t=0. Since, for t 0 the points on L have rank r, by the previous lemma, the line in C lies in C(S_r), hence B∈ C(S_r).
 
 C(S_r)_A={(A,B), A∈ S_r, B∈ℙS_n-r, AB=0}.

 By <ref>, C(S_r)_A⊂{(A,B), A∈ S_r, B∈ℙS_n-r, AB=0}. So,  suppose B ∈ℙS_n-r AB=0. Using the group action of <ref> we can move (A,B) to (I_s,B̃) in C, s the rank of A. Then there exists a curve by the previous lemma which lies in C(S_r) passing through (I_s, B̃). By the group action, we can move (I_s, B̃) back to (A,B) and the curve along with it. 


We would like to describe the fiber C(S_r)_A in terms of linear spaces associated with A. 


 Given M∈, C(M) denotes the vectors in ℂ^n^* which annihilate the image of M, and K^*(M) the quotient of ℂ^n^* by those vectors which annihilate the kernel of M. Note that every element of hom(K^*(M),C(M)) has a well defined extension to 
 hom(ℂ^n^*,ℂ^n^*), by the inclusion of C(M) in ℂ^n*), and the extension by 0 over those vectors which annihilate the kernel of M, so we can view hom(K^*(M),C(M)) as a subspace of hom(ℂ^n^*,ℂ^n^*). Denote this embedding by Φ. We define hom_s(K^*(M),C(M)) to be 
 
    Φ^-1(Φ(hom(K^*(M),C(M)))∩).

 
Note that since M is symmetric, then v∈ C(M) if and only if v^*∈mathbb C^n is in K(M). 
 Let X_r(M) be elements of ℙ(hom_s(K^*(M),C(M))) of rank less than or equal to r.
 
   Suppose M∈ S_r. Then 
  C(S_r)_M≈ X_n-r(M). If K(M)∩ M(ℂ^n)=0 then
 
  * C(S_r)_M≃ℙS_n-r⊂ℙhom_s(C(M), C(M)).
 
  * C(S_r)_M≃ℙS_n-r⊂ℙhom_s(K(M), K(M)).
 
  * C(S_r)_M≃ℙS_n-r⊂ℙhom_s(K^*(M), K^*(M)).
 
 
 
 

 Suppose B∈ hom_s(K^*(M),C(M)). Since Φ(B) is  symmetric, the element of  induced by duality has the same matrix as  Φ(B), so we can consider M(Φ(B)) and this is the zero element as the image of Φ(B) lies in C(M). Since B∈ X_n-r(M), the rank of Φ(B)≤ n-r, so Phi(B)∈ C(S_r)_M.

Suppose B∈ and MB=0. Then the row space and the column space of B lie in C(M). Further, 

    ann(K(M)=im M^*⊂ K(B).
 
So, since B is 0 on ann(K(M)), B=Φ(B̃) for some B̃. Since Φ is an embedding, this finishes the first part of the proof.

Now suppose K(M)∩ M(ℂ^n)=0. Suppose v∈ K(M), v 0. Consider v^*∈ℂ^n^*. We claim v^*∉ ann(K(M)). If it were, then v^*∈ im(M^*) which implies v∈ im(M). This implies that the canonical map ϕ: K(M)→ K^*(M) is an embedding. Since these spaces have the same dimension and ϕ is linear, ϕ is an isomorphism. Since C(M)≃ K(M) by the symmetry of M and duality the result follows.







The variety X^ν_r coincides with X_n-r.


First, we know that X_r=ℙ(S_r)⊂ℙ(W), where W=. Since W is finite, then W≃ W^* and 
ℙ(W)≃ℙ(W^*). We identify W and W^* using the bilinear form

    (A,B)=∑_i,ja_ijb_ij.

where A,B∈ W. After this identification we can regard X^ν_r⊂ℙ(W^*) as embedded in ℙ(W). We shall explicitly calculate tangent spaces to S_r at smooth points. 

Case 1: Suppose r=n-1, then S_r is a hypersurface of W. Take A∈ S_r a matrix of rank n-1, that is, A=0. If A is the matrix of cofactors of A, the elements of A are defined as follows

    a_ij=(-1)^i+j A_ij

where A_ij=(a_pq)_p i,q j. The matrix A is symmetric, and the dual cone S_n-1^ν is the closure of the set of matrices A corresponding to all matrices A of rank n-1. If A=n-1, by the equivalence given by the group action, we can consider A as the matrix whose left corner is a (n-1)× (n-1) identity matrix and all other entries 0. In this case A_nn=1 and all the other matrices A_i,j, i≠ n or j≠ n, have determinant 0 because they have one zero column or row. Therefore    A=1. Otherwise, if A<n-1, all of the matrices A_ij have determinant 0. Then A=0. 
Now, consider the map 

    .[                                  Φ: {symmetric matrices
    of rank  n-1 
    }                                   ⟶   {symmetric matrices
    of rank  1 
    };                                                                       A                                   ⟼                                   A ].

This map is surjective. Indeed, let B be a symmetric matrix of rank 1. By the group action, B is equivalent to the matrix whose the first entry is 1 and all the other entries are 0, denoted by I_1. Which means that B can be written as 

    B=M^tI_1M.

Where M is a n× n invertible matrix. It is easy to see that I_1 is the adjoint matrix of the matrix whose lower right corner is a (n-1)× (n-1) identity matrix and all the other entries are 0, denoted by I_n-1. Consider A the matrix given by

    A=((√( M))M^-1)^tI_n-1(√( M))M^-1.

By the properties of adjoint matrices we have 

    Φ(A)   =   Φ(((√( M))M^-1)^t )Φ(I_n-1)Φ((√( M))M^-1)
       =   M^tI_1M
       =   B.

Now, we can rewrite S_n-1^ν just in terms of A instead cofactor matrices A, ie

    S_n-1^ν ={A| A=1}={A| A≤ 1}=S_1.

Since S_n-1 is a conic variety we have

    ℙ(S_n-1^ν)=ℙ(S_n-1)^ν=X_n-1^ν.

Therefore,

    X_n-1^ν=X_1.

General Case: Before we prove the general case for symmetric matrices, let us think about the case where the total space is V= and the determinantal singularity is Σ_r. For any S∈ V, we denote S_IJ the submatrix of S with rows from I and columns from J, where I and J are subsets of {1,2,⋯,n} with r+1 elements. Take A∈Σ_r, it follows from the previous case that 

    B∈ T_AΣ_r (B_IJ,A_IJ)=0

for any r+1-element subsets I,J. Denote by [A_IJ] the n× n given by

    [A_IJ]_pq={[ (-1)^p+q ||a_ij||_i∈ I-{p},j∈ J-{q}                         p∈ I, q∈ J;;                                   0                          otherwise. ].
 
Then the orthogonal complement to T_AΣ_r is spanned, as a vector space, by some of these matrices [A_IJ]. Our next step is to show that this orthogonal complement (T_AΣ_r)^⊥ consists of all matrices of rank less than or equal to n-r. For this, let us consider A=I_rn and E(i,j) be the matrix which is 1 in the e_ij coordinate and 0 everywhere else. If one of the subsets I,J does not contain [r]={1,2,…,r}, the matrix ||a_ij||_i∈ I-{p},j∈ J-{q} has at least two zero columns or rows. Suppose I=[r]∪{i} and J=[r]∪{j}, then

    [A_IJ]_pq={[   (-1)^p+q  p=i, q=j;;          0 otherwise. ].

which means that (T_AΣ_r)^⊥ is generated by the vectors {E(i,j)| r+1≤ i,j≤ n}. Therefore, any matrix in (T_AΣ_r)^⊥ is a linear combination of such E(i,j) and has rank less than or equal to n-r.

Going back to the symmetric case, we have that (T_AS_r)^⊥ is generated by the linear combinations {E(i,j)+E(j,i)| r+1≤ i,j≤ n}. Thus, (T_AS_r)^⊥ is composed by all the symmetric matrices of rank less than or equal to n-r. Since all the matrices of a given rank are equivalent under the action of GL(n)

    S_r^ν ={(T_AΣ_r)^| A=r}=S_n-r.

Therefore,

    X_r^ν=X_n-r.



Now, we are ready to work on the fibers of conormal space of S_r at M. First, let M be a smooth point of S_r, which means that M is a symmetric matrix of rank r. In order to simplify the calculations, consider M=I_rn. If we consider M as a map from ℂ^n to ℂ^n, the kernel K(M) and the cokernel C(M) of M are both isomorphic to ℂ^n-r. The vectors B lying in the tangent space T_MS_r are the vector which

    .d/dt|_t=0(M+tB)=0.


As we have seen in the prove of the last proposition, the symmetric matrices E(i,j)+E(j,i) with i,j∈{r+1,⋯,n} lie in the normal space, which allows us to conclude that the normal bundle to S_r at M is  which is isomorphic to _s(K(M),C(M)). As we know, each hyperplane tangent to S_r at M is associated to a normal vector. Therefore, the fiber of the conormal space of S_r at the smooth point M is equal to (_s(K(M),C(M))).

Now, let M be a singular point of S_r, which means the rank of M is less than r, let us say M=s. Then M is a smooth point of S_s and, therefore, the fiber of C(S_s)_M is equal to (_s(K(M),C(M))). Since C(S_r)⊂ C(S_s) we have that the fiber of C(S_r) at M is contained in (_s(K(M),C(M))). In the next proposition we show what exactly is the fiber of C(S_r) at a singular point M. For that, let S_r-s(M) denote the elements of _s(K(M),C(M)) of rank less than equal to r-s. We denote X_r-s(M) the projectivization of S_r-s(M).


Suppose M is a singular point of S_r of rank s. Then the fiber of the conormal of C(S_r) at M is X_n-r(M).


By the Corollary 2.1.3 seen in <cit.>, the fiber of the conormal space of S_r at M consists of the union of the projective dual cone of the tangent cone at M and the projective dual cones of certain cones which are exceptional, ie

    C(S_r)_M=(ℙ(TC(S_r,M)))^ν⋃_iℙ(π_c(V_i))^ν

where π_c(V_i) are the exceptional cones. In our proof we will compute the dual cones of the tangent cones, then show that there are no exceptional cones.

Fix a matrix M whose rank is s. In order to simplify the calculations, we can suppose that the upper left corner of M is the s× s identity matrix and all other entries are 0. Let H be the set of symmetric matrices whose upper left corner is the s× s identity matrix, the lower right corner is a (n-s)× (n-s) symmetric matrix of indeterminates, and all other entries are 0. Thus, considering M as a regular point of S_s and taking the tangent spaces we know that

    T_MS_s+T_MH=

which means that the family H is a transverse section of S_s at M of complementary dimension. As we know, by the properties of Whitney stratification, S_r will be locally a product, ie 

    S_r∩ U = H∩ U× C

where C is a disk of the same dimension as S_s. By the map given by

    [                                                                 Φ:                                                                                                                                     ⟶                                                    _s(ℂ^n-s,ℂ^n-s);                                                                    ([         A_s× s     B_s× (n-s);     C_(n-s)× s D_(n-s)× (n-s) ])                                                                  ⟼                                                     D_(n-s)× (n-s) ]

we can conclude that H is isomorphic to _s(ℂ^n-s,ℂ^n-s) and S_r is isomorphic to S_r-s(M). Moreover, this isomorphism carries M into the origin. 
By the proposition <ref> and some properties of tangent cones we have

    (ℙ(TC(S_r∩ H,0)))^ν=X_r-s(M)^ν=X_(n-s)-(r-s)(M)=X_n-r(M).


Now, we extend the result from (TC(S_r∩ H,0))^ν in of  _s(ℂ^n-s,ℂ^n-s) to (TC(S_r,M))^ν in . The elements of the dual 

    (ℙ(TC(S_r∩ H,0)))^ν

are hyperplanes that are also elements in _s(^n-s,^n-s). We need to extend such elements to hyperplanes in . This is done by adding zeros to the matrix that defines the hyperplane we were first working with. Thus, the elements of X_n-r(M) give elements of X_n-r. By doing this, we still get a tangent hyperplane. The reason for that is: as we have seen earlier in this section, the normal space consists of the symmetric matrices given by

    {E(i,j)+E(j,i)| r+1≤ i,j≤ n}.
 
This implies that the hyperplanes on  determined by the extensions of X_n-r(M) to X_n-r contains the tangent space T_MS_s.

Now, from the analytic triviality we know that the rank stratification is locally a product. Thus, the tangent cone of S_r is a product along the smaller strata S_s. The tangent hyperplane to such product space are the span of the tangent hyperplanes to the fiber and the tangent space to base. Therefore,

    TC(S_r,M)^ν=X_n-r(M).


Now we have to show that there are no exceptional cones. Using again the fact that S_r is locally a product, let us prove that there are no exceptional cones of S_r∩ H at 0. As we know ℙ(TC(S_r∩ H,0))≃ X_r(M) and X_r(M) is irreducible, which means that ℙ(TC(S_r∩ H,0)) has only one component. Suppose, by contradiction, that there are exceptional cones. Thus, there is a component V_i of E_C(S_r∩ H) such that π_c(V_i) has dimension less than the dimension of the component W_i of E_C(S_r∩ H). If π_c(V_i) is contained in the regular set of X_r(M) then, by the group action, π_c(V_i)=X_r(M). Therefore, π_c(V_i) must lie in the projectivized tangent cone of the singular set of S_r∩ H at 0. Supose X_t(M) is the smallest set that contains π_c(V_i). This implies that there is some N of rank t<r such that 

    N∈π_c(V_i)⊂ X_t(M).
 
Again, by the group action, π_c(V_i) contains all the symmetric matrices of rank t, which means that

    π_c(V_i)=X_t(M).
 

Since every tangent hyperplane to π_c(V_i) is a limit tangent hyperplane to S_r∩ H, take N a symmetric matrix of rank t and H⊂ℙ(K(N)),C(N)) a tangent hyperplane to S_t at N of rank n-t. That means that the linear form L in  associated to H has rank n-t. The extension L of L has rank n-t and kernel equal to (N), whose dimension is t. Consider a sequence of points {(N_i,H_i)} from C(S_r∩ H) converging to (N,H). The hyperplanes H_i have rank n-r and the extensions L_H_i has kernel dimension r. Thus, L would have a kernel dimension r>t, which is impossible. Therefore, S_r∩ H has no exceptional cones at 0 which implies that S_r has no exceptional cones at M. 
    
   


§ MULTIPLICITY OF THE POLAR VARIETIES OF N()


In this section we describe the polar variety of a symmetric determinantal variety and show a formula to calculate the multiplicity of a polar curve of N in a deformation to a stabilization. To be more precise, we calculate the multiplicity of the polar of N() when  is the symmetric determinantal variety given by =F^-1(S_n-1) or =F^-1(S_n-2), which is reduced to an intersection number. At the end, for the case where X=F^-1(S_n-2), as in <cit.>, we give an explicit way to calculate these intersection numbers as an alternate sum of colength of ideals.

The polar variety of codimension l of S_r, denoted by Γ_l(S_r), at the origin, is the germ given by intersecting C(S_r)∩ (S_r× H_h-1-d) with l hyperplanes, then projecting to S_r. Altogether, we need to intersect C(S_r) with h-1-d+l hyperplanes and then project to S_r as below

    .[ π:C(S_r)∩ (S_r× H_h-1-d+l);                          ↓;                        S_r ].
 

Sometimes it is better to work with the dimension of the polar variety instead of the codimension. In these cases, if we want the polar variety Γ^l(S_r) of dimension l we are going to intersect C(S_r) with h-1-l hyperplanes and then project to S_r. That means

    Γ^l(S_r)=π(C(S_r)∩ (S_r× H_h-1-l).



The polar variety Γ^l(S_r) of dimension l is empty for all

    l≤r(r+1)/2-1.



Let c denote the codimension of S_n-r in . By taking M=0 in the theorem <ref> we have

    C(S_r)_0=X_n-r(0)≃ X_n-r.

Thus,

    C(S_r)_0= X_n-r= S_n-r-1=(h-c)-1=(h-1)-c.

Since C(S_r)_0⊂ℙ(), C(S_r)_0=c. Let us consider the map  

    π:C(S_r)∩ (S_r× H_j)⟶ S_r.

Using j hyperplanes we have the polar variety of dimension h-1-j. Now, the dimension of the fiber of π over 0 is given as follows:

    π^-1(0)= p^-1(0)-j=(C(S_r)_0)-j=(h-c)-j-1.

If j≥ h-c, then π^-1(0)=∅, which means that the polar variety Γ^h-1-j(S_r) is empty. Therefore, all the polar varieties of dimension less than equal to

    (h-1)-(h-c)=c-1=(n-(n-r))(n-(n-r)+1)/2-1=r(r+1)/2-1

are empty.


Our focus now is to calculate the multiplicity of the polar of N(), where =F^-1(S_n-1). For that we need to describe the ℛ(N(X)). The description we give will apply equally to ℛ(N()).

Consider the map

    [        F:       ℂ^q         ⟶          ;                   x         ⟼ (f_ij(x)) ]

whose entries are complex analytic functions with f_ij(x)=f_ji(x) and X has expected codimension. In this case, X is a hypersurface of dimension q-1. As we have seen, the normal module N(X) is an ideal generated by n(n+1)/2 polynomials, and, for each x∈ X, its row space is generated by one vector that we call v_x. The projective analytic spectrum of the Rees Algebra is given by, 

    ℛ(N(X))={(x,l) | x∈ X_reg and  l∈ℙ(⟨ v_x⟩)}⊆ X×ℙ^n(n+1)/2.

If x is a smooth point of X, then F(x) has rank n-1. By one of the equations of ℛ(N(X)) we have

    F(x)(l_ij)=0

which means that all the columns of (l_ij) (consequently all the rows) are in F(x). Since F(x)=n-1, the dimension of F(x) is equal to 1; therefore, each column of [T_ij] is a multiple of a fixed vector in F(x), which implies that [T_ij] is a matrix of rank 1. Now, define the set X_F, contained in X×ℙ^n-1×ℙ^n-1, by

    X_F={(x,l_1,l_2) | x∈ X_reg and  (l_1,l_2)∈Δ(ℙ( (F(x))×ℙ( (F(x)))}.
 
where Δ is the diagonal.

 X_F≃ℛ(N(X)) as sets.


Both sets are defined by the closures of points over the smooth set of X. So, let us work on this set. Consider the Segre embedding 

    [                                                                     φ:                                                            ℙ^n-1×ℙ^n-1                                                                      ⟶                                                                ℙ^n^2-1;                                                                                                                     ((S_1,…,S_n),(T_1,…,T_n))                                                                      ⟼ ([ S_1T_1      ⋯ S_1T_n;      ⋮      ⋱      ⋮; S_nT_1      ⋯ S_nT_n ]) ]


We need to prove that (ℛ(N(X)))_x⊆φ((X_F)_x), where x is a smooth point of X. For this, let l=(l_ij), with l_ij=l_ji, be a point in (ℛ(N(X)))_x. By the properties of ℛ(N(X)), for all i we have (l_i1,l_i2,…,l_in)∈ F(x).  Since φ(^n-1×^n-1) is the set of all matrices in ^n^2-1 whose rank is equal to one, l∈φ(^n-1×^n-1) and, therefore, there exists (s,t)∈^n-1×^n-1 such that φ(s,t)=l.

Now, we need to show that (s,t)∈Δ(( (F(x))×( (F(x))). The Segre embedding gives us

    [ (
    		l_11   l_12   …   l_1n
    ⋮   ⋮   ⋱   ⋮
    
    		l_1n   l_2n   …   l_nn)=(
    		s_1t_1   …   s_1t_n
    ⋮   ⋱   ⋮
    
    		s_nt_1   …   s_nt_n
    );                                                                                                                              ⇓;                                                                                   s_i(t_1,…,t_n)=t_i(s_1,….s_n)=(l_i1,…,l_in). ]

This implies that s=t in ( F(x)) and, therefore, l∈φ((X_F)_x).

As we have seen, F(x)=1, and dimension of (ℛ(N(X)))_x is one less than the rank of the Jacobian module of X, which is the expected codimension of X. Thus, (X_F)_x= (ℛ(N(X)))_x=0. Since φ((X_F)_x) is irreducible, (ℛ(N(X)))_x is closed and both sets have the same dimension, (ℛ(N(X)))_x = φ((X_F)_x). 


We will use this result to compute the degree over the base ℂ of the polar variety of dimension 1 of N(𝒳), that is _Γ_q-1(N()), where 𝒳 is the total space of the deformation, and a generic fiber is smooth. Consider the projection map 

    p_1:×ℙ^n-1×ℙ^n-1⟶ℙ^n-1

and let h_1 be the pullback of a hyperplane class of ℙ^n-1 via the projection map, and h be a hyperplane class of the diagonal. Denote the fiber over the origin in  of (N()) by E, and consider q-1 the dimension of X. The degree of Γ_q-1(N()) over  is calculated as the next theorem shows.

 Suppose  is a stabilization of X, with base . The degree of the polar curve Γ_q-1(N()) over  at the origin is

    _Γ_q-1(N())=(2h_1)^q-1·φ^*E.


 By definition, the degree of Γ_q-1(N()) over  at the origin is the degree of the projection to  at the origin of Γ_q-1(N()). The generic rank of N() is one, since it is an ideal, and (N()) has dimension q, with generic fiber dimension 0. The fiber of the exceptional divisor E has dimension one less than (N()), that is, q-1. The polar curve Γ_q-1(N()) is given by intersecting (N()) with q-1 generic hyperplanes of ^n(n+1)/2 and projecting to  by p. The degree is calculated as follows: the intersection (N())∩ H^q-1 is a curve in its ambient space, so the degree of the projection of (N())∩ H^q-1 over  is well-defined. By conservation of number, the degree of the polar variety  Γ_q-1N() is the same as the degree
of E as a projective scheme embedded in ^n(n+1)/2. Thus, _Γ_q-1(N())=|E∩ h^q-1|. Now, consider the diagram 

    _F∩φ^-1(h^q-1) [r] [dr]    (N())∩ h^q-1[d]
 
By proposition <ref>, the map from _F∩φ^-1(h^q-1) to (N())∩ h^q-1 is an isomorphism. Moreover, |E∩ h^q-1| has the same numbers of intersection points as |φ^-1(E)∩φ^-1(h^q-1)|. Thus,

    [ |φ^-1(E)∩φ^-1(h^q-1)|                     =     φ^*(E)·φ^*(h^q-1);                                           =    φ^*(E)·(2h_1)^q-1. ]

Therefore, 

    _Γ_q-1(N())=(2h_1)^q-1·φ^*E.



Now, we are interested in knowing what happens if X has codimension 3. For this case, let X_F be defined as follows:

    X_F={(x,l_1,l_2) | x∈ X_reg and  (l_1,l_2)∈ℙ( (F(x))}⊆ X×ℙ^n-1×ℙ^n-1.


Take (x,l)∈ℛ(N(X)), where l=(l_ij), with l_ij=l_ij. Since F(x)=n-2, we have F(x)=2, which means that the rank of l has to be 1 or 2. For this case, we will not have an isomorphism as before. However, we are still able to exhibit a specific map that will be useful to calculate the degree of Γ (N(𝒳)) over ℂ, as we will show in the next results. For that, consider the group action given by:

    [     G: Gl(n)×      ⟶       ;         (A,M)      ⟼  A^tMA ]
 

 Let G be the group action defined above, and consider the map

    [      Φ:     Σ_1       ⟶     S_2;               A       ⟼ A + A^t ]

Then, Φ is equivariant over the action of G, that is, 

    Φ(M· A)=M·Φ(A)

for all M∈ Gl(n) and A∈Σ_1.

 Let M be an element in Gl(n) and A∈Σ_1. Thus,

    [           Φ(M· A)                 =          Φ(M^tAM);                                   = M^tAM + (M^tAM)^t;                                   =   M^tAM + M^tA^tM;                                   =    M^t(AM + A^tM);                                   =     M^t(A + A^t)M;                                   =          M^tΦ(A)M;                                   =           M·Φ(A). ]



 Φ carries orbits of G in Σ_1 to orbits in S_2.

 Follows directly from lemma <ref>.


 Φ is a 2-1 covering map, whose critical set is S_1, and 

    Φ:(Σ_1\ S_1)⟶(S_2\ S_1)

is a local diffeomorphism.
 
 Let us start by showing that Φ is a 2-1 branched cover. First, it is clear that Φ is at least 2-1 cover since for all matrices A in Σ_1 we have Φ(A)=Φ(A^t). Now, suppose A∈ S_2 has rank 2; then there are only two preimages for A. Indeed, by the group action, we can take A as the matrix

    A=([ 0 1 0 ⋯ 0; 1 0 0 ⋯ 0; 0 0 0 ⋯ 0; ⋮ ⋮ ⋮ ⋱ ⋮; 0 0 0 ⋯ 0 ]).

If B is a preimage of A, then B+B^t=A. Let us write B as follows:

    B=([ b_11   b_12
    
    b_21   b_22                        C;                        D                      B_1 ]).
  
The first thing we can say is that b_ii=0 and b_ij+b_ji=0, except for b_21+b_12=1. Therefore, B_1 is a skew symmetric matrix of rank less than equal to 1 whose diagonal elements are 0. However, the only skew symmetric matrix of rank less than or equal to 1 is the zero matrix, so we claim that B_1 is the such matrix. Indeed, if some b_ij is not equal to 0, then b_ji≠ 0, which means that there exists a 2× 2 minor with these elements on it that has a non-zero determinant. But this is impossible since B has rank less than 2. Thus, the preimage of A is contained in the set

    { B∈Σ_1     B=([ 0   b_12
    
    b_21   0                  C;                  D                  0 ])}.
  
Now, let us analyze b_12 and b_21. By the equation  B+B^t=A and the fact that B has rank less than 2, we have

    {[  b_12b_21         =         0; b_12+b_21         =         1 ]..
 
This means we have only two options for the upper left corner of B, 

    [
    0   0
    
    1   0
    ]   or  [
    0   1
    
    0   0
    ].
  
The last step of this part is to analyze C and D. If b_21=1, then the second column of D and the first row of C are zero; and by B+B^t=A, the second column of D being 0 imples that the second row of C is also zero, and the first row of C being 0 implies that the first column of D is zero. By the same analogy, with b_12=1, we also have C and D as zero matrices. Therefore, the only preimages of A are

    ([ 0 1 0 ⋯ 0; 0 0 0 ⋯ 0; 0 0 0 ⋯ 0; ⋮ ⋮ ⋮ ⋱ ⋮; 0 0 0 ⋯ 0 ])   or  ([ 0 0 0 ⋯ 0; 1 0 0 ⋯ 0; 0 0 0 ⋯ 0; ⋮ ⋮ ⋮ ⋱ ⋮; 0 0 0 ⋯ 0 ]).
   
Note that the above argument shows that Φ:(Σ_1\ S_1)⟶(S_2\ S_1) is surjective and since all symmetric matrices of rank 1 have only one pre-image, S_1 is the singular set. Now, we need to show that p(Σ_1\ S_1) is a local diffeomorphism. For this, let B be the matrix of rank 1 given by

    B=([ 0 0 0 ⋯ 0; 1 0 0 ⋯ 0; 0 0 0 ⋯ 0; ⋮ ⋮ ⋮ ⋱ ⋮; 0 0 0 ⋯ 0 ]).

By the group action, in order to prove the local diffeomorphism it is enough to show that D_BΦ has maximal rank. Since Φ is linear, we have

    [     D_BΦ:    T_BΣ_1         ⟶ T_Φ(B)S_2;                   A         ⟼   A + A^t ]

Now, let us calculate these tangent spaces. As we know, the tangent space of a determinantal variety at a smooth point is equal to 

    T_BΣ_1={C∈| C( B)⊂ B}.

In this case, B is generated by the vectors 

    B=⟨ (0,1,0,…,0),(0,0,1,0,…,0),…,(0,…,0,1)⟩

and B={(0,t,0,…,0)| t∈}. Thus, the matrix C applied to the kernel vectors gives us

    C(0,…,0,1_position  i,0,…,0)=(
    c_1i
    
    c_2i
    ⋮
    
    c_ni)

which means c_1i=c_3i=…=c_ni=0 and c_2i is any complex number, for all i=2,…,n. Therefore, if C is a matrix in T_BΣ_1, then 

    C=(
    			c_11    0    ⋯    0
    
    			c_21    c_22   ⋯    c_2n
    	
    			c_31    0    ⋯    0
    ⋮   ⋮   ⋱   ⋮
    									
    			c_n1    0    ⋯    0
    			).

For the tangent space of S_2 at a smooth point we have that

    T_Φ(B)S_2={D∈| D( (Φ(B)))⊂ (Φ(B))}

where Φ(B)=A. In that case,

    [                                            A                                            = ⟨ (0,0,1,0,…,0),(0,0,0,1,0,…,0),…,(0,…,0,1)⟩;                                            A                                            =                        {(s,t,0,…,0) | s,t∈}. ]

Thus, the matrix D applied to the kernel vectors gives us

    D(0,…,0,1_position  i,0,…,0)=(
    d_1i
    
    d_2i
    ⋮
    
    d_ni).

which means d_3i=d_4i=…=d_ni=0 and d_1i,d_2i are any complex number, for all i=3,…,n. Therefore, if D is a matrix in T_AS_2, then 

    D=(
    			d_11    d_12    d_13   ⋯    d_1n
    
    			d_12    d_22    d_23   ⋯    d_2n
    	
    			d_13    d_23    0    ⋯    0
    ⋮   ⋮   ⋮   ⋱   ⋮
    								
    			d_1n    d_2n    0    ⋯    0
    			).

Finally, let us prove that D_BΦ is surjective. If D is a symmetric matrix in T_AS_2, take C as follows
 
    C=(1/2d_11    0    ⋯    0
    
    			d_12   1/2d_22   ⋯    d_2n
    	
    			d_13    0    ⋯    0
    ⋮   ⋮   ⋱   ⋮
    									
    			d_1n    0    ⋯    0
    			).

Thus,

    D_BΦ(C+C^t)=D.



 Φ:Σ_1⟶ S_2 is surjective.

 By proposition <ref>, Φ:(Σ_1\ S_1)⟶(S_2\ S_1) is 2-1; and since Φ:S_1⟶ S_1 maps A∈ S_1 to 2A∈ S_1, Φ is clearly surjective. 


 X_F is a double cover of ℛ(N(X)).

 Before we prove the general case let us consider the case where F is the identity map, denoted by Id, X=Id^-1(S_n-2)=S_n-2 and

    [                                         X_Id                                            = {(h,l_1,l_2) |  h=n-2  and  (l_1,l_2)∈ℙ( h)};                                      ℛ(N(X))                                            =                                    C(S_n-2). ]

Then, we are going to show that the map Ψ:X_Id⟶ C(S_n-2) is a 2-1 branched cover. From the group actions, it suffices to check that Ψ is a 2-1 branched cover on the fiber of X_Id for the representatives

    h_r=([ I_r   0;   0   0 ])
 
for all r∈{1,2,…,n-2}. The fiber of X_Id at each representative is equal to ^n-r-1×^n-r-1, since H_r=^n-r. Now, for the fiber of C(S_n-2) at h_r we are going to use the results from last section. If r≠ n-2, then, by theorem <ref>, the fiber of C(S_n-2) at h_r is equal to X_2(h_r), which means 

    C(S_n-2)_h_r=(S_2)⊂().

If r=n-2, then C(S_n-2)_h_n-2=(S_2)=(_s(^2,^2)), since h_n-2 is a smooth point. Thus, for both cases we can consider the following map given by

    [                                                                                                                       φ:                                                                                                            ^n-r-1×^n-r-1                                                                                                                        ⟶                                                                                                                       ();                                                                                                                                                                                                                                     ((S_i),(T_j))                                                                                                                        ⟼ (
    												2S_1T_1    ⋯    S_n-rT_1+T_n-rS_1
    ⋮   ⋱   ⋮
    												
    												S_1T_n-r+T_1S_n-r   ⋯    2S_n-rT_n-r) ]

This map φ is the composition of the maps Φ∘σ, where σ is the Segre embedding. Indeed, Im(σ)=(Σ_1), which makes the composition well defined, and φ(s,t)=σ(s,t)+σ(s,t)^t=Φ(σ(s,t))=(S_2). Hence, Ψ is a 2-1 branched cover. The general case follows from the commutative diagram where the vertical arrows are embeddings on fibers.

    X_F  [r] [d]    (N(X)) [d]
    
    X_Id[r]_Ψ    C(S_n-2)



As before, we use this theorem to compute the degree over the base ℂ of the polar variety of dimension 1 of N(𝒳), where 𝒳 is the total space of the deformation, and a generic fiber is smooth. Consider the map

    ×ℙ^n-1×ℙ^n-1[dl]_p_1[dr]^p_2   
    ℙ^n-1      ℙ^n-1

and let h_1 be the pullback of a hyperplane class of ℙ^n-1 via the projection map p_1, h_2 be the pullback of a hyperplane class of ℙ^n-1 via the projection map p_2 and h be a hyperplane class on ×× defined as h_1+h_2. Denote the fiber over the origin in  of (N()) by E, p the projection to , and let d be the dimension of X.

 Suppose  is a stabilization of X, with smooth base . The degree of the polar curve Γ_d(N()) over  at the origin is

    _Γ_d(N())=1/2(h_1+h_2)^d+2·φ^*E.


 The reasoning is similiar to the proof of theorem <ref>. In this case, the normal module N() has n(n+1)/2 generators and generic rank 3, so (N()) has dimension d+3, with generic fiber of dimension 2. The fiber E=p^-1(0) has dimension at most one less than (N()), that is, d+2. The polar curve Γ_d(N()) is given by intersecting (N()) with d+2 generic hyperplanes of ^n(n+1)/2 and projecting to  by p. Then,

    deg_Γ_d(N())=[E]·[h]^d+2.

Now, consider the diagram 

    _F∩φ^-1(h^d+2) [r] [dr]    (N())∩ h^d+2[d]
 
By the theorem <ref>, the map from _F∩φ^-1(h^d+2) to (N())∩ h^d+2 is 2-1, which means that if [E]· [h]^d+2 has k points of intersection, then [φ^-1(E)]·[φ^-1(h^d+2)] has 2k points. Thus,

    [ [φ^-1(E)]·[φ^-1(h^d+2)]                       =       φ^*(E)·φ^*(h^d+2);                                               =   φ^*(E)·(h_1+h_2)^d+2. ]

Therefore, 
 
    _Γ_d(N())=1/2(h_1+h_2)^d+2·φ^*E.



Define Γ_i,j(N()) to be the image of the projection of _F∩ h_1^i∩ h_2^j over . We call these mixed polars of type (i,j) of N(). By the same argument as in the proof of theorem <ref> the degree of the mixed polars is

    _Γ_i,j(N())=h_1^ih_2^j·φ*E.
  
In order to simplify notation, let us write the degree of the mixed polars as h_1^ih_2^j. The degree of Γ_d(N()) over  is

    _Γ_d(N())=1/2∑_i=0^d+2(
    																		d+2
    
    																		i		
    																		)h_1^ih_2^d+2-i.


The reasoning follows by intersection theory and the last theorem.



§ COMPUTING THE DEGREES OF THE MIXED POLARS 


The calculation of the degrees of the mixed polars defined in the last section helps us to calculate the degree of the polar varieties Γ_d(N()). In <cit.>, Gaffney and Rangachev found an algorithm to compute these mixed polars when  is a maximum rank determinantal singularity, by using the colength of some specific ideals. We are not working with the maximal minors here, but we can still use some of the ideas presented in the paper to solve our problem. Our next step is to calculate the degree of the mixed polar Γ_i,j(N()) for any possible dimension d of X and  a stabilization of X.

The mixed polar Γ_i,j(N()), where i≥ j and i+j=d+2, is defined by taking the hyperplane classes α^i=[S_n=…=S_n-i+1=0] and β^j=[T_1=…=T_j=0] and their pullbacks via the projections p_1 and p_2, which are:

    [  h_1^i=× (a_1:…:a_n-i:0:…:0)×^n-1; h_2^j=×^n-1× (0:…:0:b_j+1:…:b_n). ]

The mixed polar defined by projecting _F∩ h_1^ih_2^j onto  is equal to:

    Γ_i,j(N())={ x∈_reg    ∃[ (a_1,…,a_n-i,0,…,0)∈F^t(x);   (0,…,0,b_j+1,…,b_n)∈F(x) ]}.


 The degree of the mixed polar Γ_i,0(N()) is equal to 0.
 
 By definition, the degree of Γ_i,0(N()) is the intersection number h_1^i·φ^*E. The dimension of (F(x)) is equal to 1, which means that any generic fiber of _F is isomorphic to ^1×^1. Take a point (x,l_1,l_2) in h_1^i∩φ^*E, because we do not have a h_2 term, l_2 can be any point in ^1, that is, the intersection in this case cannot be a curve. Therefore, the degree of Γ_i,0(N()) must be 0.


To calculate the degree of the defined mixed polar, we are going to consider sets in , prove results about them and then pull them back to  via F. For this, consider the sets


    A(i,j,n)_l={ h∈    [ (p_l-1∘π_n-i∘ h)                ≤            n-i-l;   (p_l∘ h∘ϕ_n-j)                ≤            n-j-1 ]}

where π_n-i:^n⟶^n-i is the projection on the first n-i factors, p_l:C^k⟶^k-l is the projection on the last k-l factors and ϕ_n-j:^n-j⟶^n is the canonical embedding of ^n-j into ^n on the last n-j factors. 

The first inequality, called row condition, analyzes the behavior of the submatrix of h, of size (n-i-l+1)× n, that appears below:

    < g r a p h i c s >

The row condition holds if this submatrix has rank less than the maximal rank. In order to have a well defined submatrix, that is, at least one row, l must be less than or equal to n-i. Increasing l by 1 has the effect of dropping one more row on the upper side of the matrix while the bottom side remains the same, increasing the codimension by 1. The second inequality, called column condition, analyzes the behavoir of the submatrix of h, of size (n-l)×(n-j), that appears below:

    < g r a p h i c s >

The column condition holds if the matrix has rank less than or equal to n-j-1. If l=j+1, then the column conditions holds for all h∈ since the resulting submatrix has exactly n-j-1 rows. For l∈{j+2,…,n}, what happens is that A(i,j,n)_l⊂ A(i,j,n)_j+1 because the row condition for A(i,j,n)_l always implies the row condition for A(i,j,n)_l-1 and the column conditions for A(i,j,n)_j+2,…,A(i,j,n)_n hold for all h∈. For this reason, we want l to be less than or equal to j+1. Increasing l by 1 has the effect of dropping one more row on the upper side of the matrix while the bottom side remains the same, decreasing codimension by 1. Note that the number of considered columns is the same as we change l. For now on we will always assume j≤ i≤ n-1 and l≤min{j+1,n-i}.

 The row and column conditions for A(i,j,n)_l define determinantal varieties.

 Let us start with the row condition. Since the row condition defines a variety by taking the vanishing of some minors of a matrix, the only thing needed to prove is that it has the right codimension. For this, we are going to use the same ideas A. Conca used in the section 2.3 of <cit.>. Let S^n-i-l(i,j,n)_ l,R be the variety defined by the row condition and h be a matrix in it. Following Conca's notation, we have M=0 and N=0. Since rows R_1,… R_l-1 can vary freely, we can move the block S to obtain the format presented in Conca's paper. Then, we can take Z as the rectangular (n-i-l)× n submatrix:

    < g r a p h i c s >

Then, S is the square symmetric matrix marked as red (left block) and P is the submatrix marked as blue (right block). By Conca's results R_n-i-l(Z) is Cohen-Macaulay and also determinantal. Now, consider the projection map 

    [                    τ:                                           ⟶ _s,(n-i-l)(^n,^n-i-l);                                           h                     ⟼        p_l-1∘π_n-i∘ h ]

where _s,(n-i-l)(^n,^n-i-l) is the set of (n-i-l)× n matrices whose left upper (n-i-l)×(n-i-l) submatrix is symmetric. Since the preimage of R_n-i-l(Z) is equal to S^n-i-l(i,j,n)_ l,R, we are done.

For the column condition, let S^n-j-1(i,j,n)_ l,R be the variety defined by the row condition and h be a matrix in it. Then, Z is the rectangular (n-l)×(n-j) submatrix:

    < g r a p h i c s >

Following Conca's notation, we have P=0, N=0, S is the square symmetric matrix marked as red (bottom block) and M is the submatrix marked as blue (upper block). By Conca's results R_n-j-1(Z) is Cohen-Macaulay and also determinantal. Now, consider the map 

    [                    τ:                                           ⟶ _s,(n-j-1)(^n-j,^n-l);                                           h                     ⟼          p_l∘ h∘ϕ_n-j ]

where _s,(n-j-1)(^n-j,^n-l) is the set of (n-l)×(n-j) matrices whose bottom right (n-j)×(n-j) submatrix is symmetric. Since the preimage of R_n-j-1(Z) is equal to S^n-j-1(i,j,n)_ l,C, we are done.


 A(i,j,n)_l is a determinantal variety, for all l.


Let S^n-i-l_l,R be the set satisfying the row condition for A(i,j,n)_l and S^n-j-1_l,C the set satisfying the column condition. Considering the column condition for 
A(i,j,n)_l and Conca's notation, let _s,(n-j)(n-j,n-l) be the set of (n-j)×(n-l) matrices whose last (n-j)×(n-j) submatrix is symmetric. The proof will be by induction. Let us first consider the set A(i,j,n)_j+1. For this set, the column condition holds for all h, so this set is equal to its row condition, which is determinantal by the last lemma. Now, suppose that A(i,j,n)_l+1 is determinantal and consider C a component of A(i,j,n)_l which is not a component of A(i,j,n)_l+1. Take h a generic element of C such that h does not satisfies the row condition for A(i,j,n)_l+1, ie., (R_l+1,…,R_n-i)=n-i-l, which also means that rows R_l+1,…,R_n-i are linearly independent and R_l depends on them. Take a small neighborhood of h, say U, that does not contain any elements of A(i,j,n)_l+1. The map given by

    [                     F:                      U                      ⟶  _s,(n-j-1)(^n-j,^n-l);                                         a_s,t                      ⟼ {a_s,t| s≥ l+1,t≥ j+1} ]
 
is a submersion. Indeed, the linear independence of rows R_l+1,…,R_n-i and the free variation of the last i rows means that there is no relation on the coordinates of the target besides the symmetry ones. By abuse of notation let us consider that S^n-j-1_l,C is inside of _s,(n-j-1)(^n-j,^n-l). Since F is a submersion, the codimension of F^-1(S^n-j-1_l,C) in S^n-i-l_l,R is the same as the codimension of S^n-j-1_l,C in _s,(n-j-1)(^n-j,^n-l). Therefore, A(i,j,n)_l is determinantal.


Before we state the next proposition, let us define Γ_i,j(S_n-2) as the set given by taking F as the identity map in the definition of the mixed polars, that is,

    Γ_i,j(S_n-2)={ h∈ S_n-2\ S_n-3    ∃[   (a_1,…,a_n-i,0,…,0)∈(h); (0,…,0,b_j+1,…,b_n)∈(h^t) ]}.


The pullback of this set by F is equal to Γ_i,j(N()). For now we are going to use it to prove some results and, at the end, pull them back to finally calculate the degree of our mixed polar.

Γ_i,j(S_n-2)⊂ A(i,j,n)_1.


Let h be in Γ_i,j(S_n-2). Assume that our hyperplanes are chosen so that there are vectors v=(v_1,…,v_n-1,0,…,0), u=(0,…,0,u_j+1,…,u_n) with coordinates in  such that v· h=0 and h· u=0. The first equality says that the rows R_1,…,R_n-i are linearly dependent, which means that, (R_1,…,R_n-i)≤ n-i-1. The second equality says that C_j+1,…,C_n are linearly dependent, that is, (C_j+1,…,C_n)≤ n-j-1. Therefore, h∈ A(i,j,n)_1.


Since i+j+1=d+2+1=q, these sets have the right codimension, meaning that their pullbacks will be curves on ^q× and, therefore, we can calculate their degrees. The idea here is to construct a chain of sets starting with Γ_i,j(S_n-2), whose set of components is denoted by C(i,j,n)_0, and ending with a set which is determinantal. The degree of the mixed polar we want will be given by an alternating sum of the degrees of this sets, but in order to do that we need to understand how the chain ends. For this, we are going to analyze two cases: n-i≤ j and j<n-i. Let us start proving the results we need for the case where n-i≤ j. 

 Suppose n-i≤ j. Then, A(i,j,n)_n-i⊂ A(i,j,n)_n-i-1.


Let h be a matrix in A(i,j,n)_n-i. The row condition for A(i,j,n)_n-i implies that the row R_n-i is zero, which means that R_n-i,R_n-i-1 are linearly dependent. Hence h satisfies the row condition for A(i,j,n)_n-i-1. On the other hand, the column condition for A(i,j,n)_n-i says that the matrix given by the last n-j columns and last i rows of h has less than maximal rank. Since R_n-i is zero, the matrix given by the last n-j columns and last i+1 rows of h has also less than maximal rank, meaning that h is in A(i,j,n)_n-i-1.


 Suppose n-i≤ j and C∈ C(i,j,n)_l is a component of A(i,j,n)_l. If h∈ C is a generic point, then (p_l∘ h∘ϕ_n-j)=n-j-1 and (p_l-1∘π_n-i∘ h)=n-i-l.


Before we start the proof, recall that the points in A(i,j,n)_l are defined by (p_l-1∘π_n-i∘ h) and
(p_l∘ h∘ϕ_n-j) both having less than maximal rank. Now, let h be a generic point of C.



  * (p_l∘ h∘ϕ_n-j)=n-j-1.

Pick a column C_p between columns C_j+1… C_n that is linearly dependent on the others and drop it. This gives us a submatrix of size (n-l)×(n-j-1). We would like to consider square submatrices, so let us drop rows R_l+1… R_j+1. By varying the entries, as small as desired,  of the remaining square matrix we can have, as result, a matrix of rank exactly n-j-1. After making such perturbation, we change the entries of C_p using its relation with the remaining columns and the variations on them; so then we will still have C_p linearly dependent on C_j+1,…,C_p-1,C_p+1,…,C_n. Since n-i≤ j, the row condition for A(i,j,n)_l will not be affected. Therefore, the perturbation h̃ of h lies in A(i,j,n)_l and (p_l∘h̃∘ϕ_n-j)=n-j-1. Since h is a generic point, it satisfies this generic property. 


  * (p_l-1∘π_n-i∘ h)=n-i-l.

A similar argument will be made in here. Pick a row R_p between rows R_l,…, R_n-i that is linearly dependent on the others and drop it. This gives us a submatrix of size (n-i-l)× n. We would like to consider square submatrices, so let us drop columns C_1…,C_l. By varying the entries, as small as desired, of the remaining square matrix we can have, as result, a matrix of rank exactly n-i-l. After making such perturbation, we change the entries of R_p using its relation with the remaining rows and the variations on them; so then we will still have R_p linearly dependent on R_l,…,R_p-1,R_p+1,…,R_n-i. Since n-i≤ j, the column condition for A(i,j,n)_l will not be affected. Therefore, the perturbation h̃ of h lies in A(i,j,n)_l and (p_l-1∘π_n-i∘h̃)=n-i-l. Since h is a generic point, it satisfies this generic property. 



The next result is important for both cases. We will state it here and then use it again in the j<n-i case.

 If 1≤ l<min{j+1,n-i}, then 

    C(i,j,n)_l⊂ C(i,j,n)_l-1∪ C(i,j,n)_l+1.

   
 Let h be an element in A(i,j,n)_l. To say that h satisfies both row and column conditions is the same as saying that p_l-1∘π_n-i∘ h and p_l∘ h∘ϕ_n-j are not submersions, and we are going to use this in our proof. Take C an element of C(i,j,n)_l, that is, C is one of the components of A(i,j,n)_l. Let h be a generic point of C. For the row condition, h has a submatrix of size (n-i-l+1)× n whose rank is equal to n-i-l. Consider the map p_l∘π_n-i∘ h:^n⟶^n-i-l used to define the row condition for A(i,j,n)_l+1. Our proof is divided into two situations: p_l∘π_n-i∘ h being a submersion or not. First, if the map is a submersion, then (p_l∘π_n-i∘ h)=n-i-l, meaning that h is not in A(i,j,n)_l+1 and, therefore, C∉ C(i,j,n)_l+1. Now, we need to show that h satisfies the row and column condition for A(i,j,n)_l-1. The map p_l∘π_n-i∘ h being a submersion means that the rows R_l+1,…,R_n-i are linearly independent. Since, by the row condition of A(i,j,n)_l, R_l+1,…,R_n-i are linearly dependent, we have that R_l is dependent on R_l+1,…,R_n-i. This means that h satisfies the row condition for A(i,j,n)_l-1. On the other hand, R_l being dependent on R_l+1,…,R_n-i means that (p_l-1∘ h∘ϕ_n-j)=(p_l∘ h∘ϕ_n-j)≤ n-j-1.Therefore, h∈ A(i,j,n)_l-1 and there is a Z-open subset of C contained in some component of A(i,j,n)_l-1, which means that C∈ C(i,j,n)_l-1.

Now, if p_l∘π_n-i∘ h is not a submersion, then rows R_l+1,…,R_n-i are linearly dependent, meaning that h satisfies the row condition for A(i,j,n)_l+1. The column condition for A(i,j,n)_l automatically implies the column condition for A(i,j,n)_l+1 since (p_l+1∘ h∘ϕ_n-j)=(p_l∘ h∘ϕ_n-j)≤ n-j-1. Therefore, h∈ A(i,j,n)_l+1 and there is a Z-open subset of C contained in some component of A(i,j,n)_l+1, which means that C∈ C(i,j,n)_l+1.   


 Suppose n-i≤ j and 2≤ l≤ n-i. Then,

    C(i,j,n)_l∩ C(i,j,n)_l-2=∅.
 

 The idea here is to take an element h of some element C of C(i,j,n)_l, then we vary the entries of h continuously so we stay in A(i,j,n)_l and the variation h̃ will not be in A(i,j,n)_l-2. We can also assume that h is not in any other component of A(i,j,n)_l. Let h be a generic point of C. Then, as seen before, (p_l∘ h∘ϕ_n-j)=n-j-1 and let C_p be the column between C_j+1,…,C_n which is linearly dependent on the others. Let A be the square matrix given by the last n-j-1 rows and columns C_j+1…,C_p-1,C_p+1,…,C_n of maximal rank, that is, its determinant is not zero. Now, consider the vector v given by the cofactors { c_(j+1)(j+1),…,c_(j+1)p,…, c_(j+1)n} of row R_j+1. This means that v≠ 0 since c_(j+1)p= A≠ 0. Take, v the complex conjugate of v and vary row R_l-1 by tv. This variation does not affect the row and column conditions for A(i,j,n)_l, meaning that the resulting matrix h̃ is in A(i,j,n)_l but it will not satisfy the column condition for A(i,j,n)_l-2. Indeed, R_l-1 is the first row considered in the column condition for A(i,j,n)_l-2. The variation R_l-1+tv, for t≠ 0, gives us the following result when we consider all the rows starting at column C_j+1:

    [
    R_l-1+tv
    
    R_j+2
    ⋮
    
    R_n
    ]=
    [
    R_l-1
    
    R_j+2
    ⋮
    
    R_n
    ]+
    t||v||^2≠ 0 .

Hence, the submatrix of p_l-2∘ h∘ϕ_n-j consisting of rows R_l-1,R_j+2,…,R_n for slight variations has maximal rank, and, therefore, the column condition of A(i,j,n)_l-2 fails. Thus, C∉ C(i,j,n)_l-2. 


In order to understand how the last results help us to calculate the degree of the mixed polar, let us look first at the sets A(i,j,n)_1, A(i,j,n)_2 and A(i,j,n)_3. Suppose C_21,…,C_2k are the components of A(i,j,n)_2. By the proposition <ref>, some of these components are components of A(i,j,n)_1 and the others are components of A(i,j,n)_3. The proposition <ref> says that A(i,j,n)_1 and A(i,j,n)_3 have no components in common, therefore we can suppose that C_21,…,C_2p are the components of A(i,j,n)_2 contained in A(i,j,n)_1 and that C_2(p+1),…,C_2k are the components contained in A(i,j,n)_3. In an alternating sum of degrees, all components of A(i,j,n)_2 would be canceled. Continuing this argument, we will have all the components being canceled, except for some components of A(i,j,n)_1 and some components of A(i,j,n)_n-i. The proposition <ref> takes care of the components of A(i,j,n)_n-i, since A(i,j,n)_n-i is contained in A(i,j,n)_n-i-1. Therefore, the only components left are those in A(i,j,n)_1, which are all the components of Γ_i,j(S_n-2) by the proposition <ref> and the proposition <ref> for l=2.

Now, let us pull back these sets in order to obtain the degree of our mixed polars. The A(i,j,n)_l are determinantal varieties, we can take hyperplanes generic enough so that A_l(F)=F^-1(A(i,j,n)_l) has the expected codimension i+j+1=q, which means we have curves on ^q×. Since determinantal varieties are Cohen Macaulay, the degrees of all A_l(F) are calculated by taking the colength of the ideal given by the maximal minors. Therefore,

    _(Γ_i,j(N()))=∑_l=1^n-i(-1)^l+1_A_l(F).
 
The degrees of the sets A_l(F) are based on the rows and columns of the map F. However, we do not have our hands on such map, we only know the rows and columns of F and, therefore, we would like to use them to find the degree of the polar variety Γ_i,j(N()). For this, consider the projection A_l(F)=A_l⟶. The degree of t is the colength of (t) in Ø_A_l since (A_l,0) is a Cohen-Macaulay variety. In this case we have

    _Ø_A_l,0/(t)=_Ø_^q×,(0,0)/(t,I_A_l)=_Ø_q/I_A_l.

Therefore, for n-i≤ j, the degree over  of the mixed polars of N() is

    _(Γ_i,j(N()))=∑_l=1^n-i(-1)^l+1 I_A_l.


Now, let us procede to the results necessary to analyze the case where j<n-i. It is important to keep in mind that we will need to use some of the results previously stated.

 Suppose j<n-i. Then, A(i,j,n)_j+1⊂ A(i,j,n)_l.

 First of all, let us remember that l≤min{n-i,j+1}. Since, by hypothesis, j<n-i⇒ j+1≤ n-i, we have l≤ j+1. In that case, let h be a matrix in A(i,j,n)_j+1. By the row condition, the rows R_j+1,…,R_n-i are linearly dependent, that is, (R_j+1,…,R_n-i)≤ n-i-j-1. By adding the rows R_l,…,R_j, the rank will increase at most j-l+1, which means that for all l<j+1 (R_j+1,…,R_n-i)≤ n-i-l making h satisfy the row condition for all A(i,j,n)_l, l<j+1. Now, the fact that R_j+1,…,R_n-i are linearly dependent implies that the columns C_j+1,…,C_n-i are also linearly dependent, which means that h∘Φ_n-j has less than maximal rank, satisfying the column condition for all A(i,j,n)_l, l<j+1. Therefore, h∈ A(i,j,n)_l for all l<j+1.


 Suppose j<n-i and 3≤ l≤ j+1. Then, 

    C(i,j,n)_l∩ C(i,j,n)_l-2=C(i,j,n)_j+1.



The idea is to take an element h of some component C of A(i,j,n)_l that it is not in any component of A(i,j,n)_j+1, then we vary entries of h continuosly so we stay in A(i,j,n)_l and the variation h̃ will not be in A(i,j,n)_l-2. Unfortunately, we cannot proceed as we did in the previous case because unlike the previous case, where there was no relation between column and row condition, here we may have a problem with a overlaping block caused by the fact that j<n-i. Let h be a generic point of C∈ C(i,j,n,)_l, where C is not in C(i,j,n)_j+1. Since p_j∘π_n-i∘ h has maximal rank, let us freely deform h by choosing a basis for the row space of p_l-1∘π_n-i∘ h by supplementing the rows of p_j∘π_n-i∘ h, which are linearly independent, with as many additional rows as necessary. Then, in our deformation of rows R_j+1,…,R_n-i we leave fixed the other rows in the basis, deforming the remaining rows using the relation relating each remaining row to the rows in the basis. What we are doing here is similar to the method of deformation seen in lemma <ref>. This technique ensures that the rank of rows R_l,…,R_n-i stay constant in small deformations, preserving the row condition for A(i,j,n)_l and making the deformations stay inside C. 

The next step is to show that we can deform the lower right (n-j)×(n-j) block so that it has rank n-j-1, and p_l∘ h∘ϕ_n-j has rank n-j-1 as well. Let H be the kernel of the map defined by using the rows among R_l+1,…,R_j of h∘ϕ_n-j which are part of the basis elements for the image of p_l-1∘π_n-i∘ h. Since n-j>j-l, H is non-trivial. We can suppose that (p_j∘ h∘ϕ_n-j)<n-j-1. Then the kernel of p_j∘ h∘ϕ_n-j must intersect H non-trivially. Let l be a line in this intersection. We can make small enough deformations of p_j∘ h∘ϕ_n-j so that its kernel is exactly l. This includes a deformation of p_l∘ h∘ϕ_n-j which has kernel rank 1, and still satisfies the row condition. Since this deformation h is small, h is still in C and p_l∘ h∘ϕ_n-j has rank n-j-1.

Now we can deform row R_l-1 of h∘ϕ_n-j without affecting the row condition for A(i,j,n)_l so that the matrix of p_l-2∘h∘ϕ_n-j has rank n-j and, therefore, cannot be in A(i,j,n)_l-2.


In order to state the final formula for the degree of the mixed polars when j<n-i, it is important to prove that C(i,j,n)_0∩ C(i,j,n)_2=∅. For this, let h be a generic element of a component C of A(i,j,n)_2. This means that (p_1∘π_n-i∘ h)=n-i-2 and (p_2∘ h Φ_n-j)=n-j-1. Without loss of generality we can consider h as follows:

    h=(
    1   0   0   0   …   0   0
    
    0   0   0   0   …   0   0
    
    0   0   1   0   …   0   0
    
    0   0   0   1   …   0   0
    ⋮   ⋮   ⋮   ⋮   ⋱   ⋮   ⋮
    
    0   0   0   0   ⋯   1   0
    
    0   0   0   0   ⋯   0   0
    )

since the rank of the following (n-i-1)× n submatrix is:  

    (
    0   0   0   0   …   0   0   …   0
    
    0   0   1   0   …   0   0   …   0
    
    0   0   0   1   …   0   0   …   0
    ⋮   ⋮   ⋮   ⋮   ⋱   ⋮   ⋮   ⋱   ⋮
    
    0   0   0   0   ⋯   1   0   …   0
    )= n-i-2

and the rank of the following (n-2)×(n-j) submatrix is:  

    (
    1   0   …   0   0
    
    0   1   …   0   0
    ⋮   ⋮   ⋱   ⋮   ⋮
    
    0   0   …   1   0
    
    0   0   ⋯   0   0
    )= n-j-1.

The path h+t(δ_1n+δ_n1) satisfies the row condition for A(i,j,n)_2 because this condition does not include the first and last rows, and also satisfies the column condition for A(i,j,n)_2 because this condition does not include the first and last column either. However, the determinant of the (n-1)×(n-1) submatrix formed by dropping the second column and second row is easily seen as non-zero for t≠ 0. So, the generic point of the path does not lie in S_n-2, although it is in A(i,j,n)_2. Therefore, the component C of A(i,j,n)_2 is not contained in any component of Γ_i,j(S_n-2).


Finally, proceeding as in the previous case we still have an alternating sum of degrees, and some of the terms will be canceled because of proposition <ref>. However, in this case the intersection is not always empty, so we need to be careful about it. Inside of each C(i,j,n)_l, l≠ 0 we have a copy of C(i,j,n)_j+1. If j is odd, then all the copies will cancel each other and, again, we just need to worry about the remaining components of A(i,j,n)_1. But since A(i,j,n)_2 has no components in common with A(i,j,n)_0, the only components left in the alternating sum are the components of the mixed polar in question, wich means that,

    _(Γ_i,j(N()))=∑_l=1^j+1(-1)^l+1 I_A_l.
 
Now, if j is even, then we need to take an extra term _A_j+1(F) into consideration. Therefore,

    _(Γ_i,j(N()))=(∑_l=1^j+1(-1)^l+1 I_A_l)- I_A_j+1.
 

 Consider the map

    [                                                                                       F:                                                                                      ℂ^5                                                                                        ⟶                                                                                _s(^4,^4);                                                                                                                                                             (x_1,x_2,x_3,x_4,x_5)                                                                                        ⟼ ([  x_1  x_2  x_3  x_4;  x_2  x_3  x_4  x_5;  x_3  x_4  x_5  x_1;  x_4  x_5  x_1 2x_2 ]) ]

and the symmetric determinantal variety X=F^-1(S_2) of dimension 2 with smoothing . First, let us calculate Γ_3,1(N()). The set A(3,1,n)_1 is given by


    A_1(F)={ x∈^5×    [                                                      (
    														x_1     x_2     x_3   x_4
    														)≤ 0;                                                                                                                  ; (
    														x_3   x_4   x_5
    
    														x_4   x_5   x_1
    
    														x_5   x_1   2x_2
    														)≤ 2 ]}.

The degree of A_1(F) is the colength of

    I_A_1=(x_1,x_2,x_3,x_4,2x_2x_3x_5-x_3x_1^2+2x_1x_4x_5-2x_2x_4^2-x_5^3)

which can be easily calculated by hand as I_A_1=3. As in the curve case, A_2(F) is empty. In fact, A_1(F) is equal to Γ_3,1(N()). Now, let us calculate Γ_2,2(N()). The set A_1(F) here is given by


    [ A_1(F)={ x∈^5×    [   (
    														x_1   x_2   x_3   x_4
    
    														x_2   x_3   x_4   x_5
    														)≤ 1;                                                                                                ; (
    														x_4   x_5
    
    														x_5   x_1
    
    														x_1   2x_2
    														)≤ 2 ]} ].

The degree of A_1(F) is the colength of

    I_A_1=([  x_1x_3-x_2^2,x_1x_4-x_2x_3,x_1x_5-x_2x_4;   x_2x_4-x_3^2,x_2x_5-x_3x_4,x_3x_5-x_4^2; x_4x_1-x_5^2,2x_4x_2-x_5x_1,2x_5x_2-x_1^2 ]).

According to Singular, I_A_1 =12. 

The set A_2(F) is equal to

    [ A_2(F)={ x∈^5×    [               (
    														x_2   x_3   x_4   x_5
    														)≤ 0;                                                                       ; (
    														x_5   x_1
    
    														x_1   2x_2
    														)≤ 2 ]} ].

The degree of A_2(F) is the colength of

    I_A_2=(x_2,x_3,x_4,x_5,2x_5x_2-x_1^2)

which can be easily calculated by hand as I_A_2=2. Thus,

    _Γ_2,2(N())=12-2=10.

Therefore,

    [                 _Γ_2(N())                         = 4_Γ_3,1(N())+3_Γ_2,2(N());                                                   =                4· 3+3· 10;                                                   =                       42. ]



For more examples see <cit.>. 
 
99
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