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Article

On Lagrangian Grassmannian Variety and Plücker Matrices

by
Jesús Carrillo-Pacheco
Academia de Matemáticas, Universidad Autónoma de la Ciudad de México, Ciudad de México 09390, Mexico
Mathematics 2024, 12(6), 858; https://doi.org/10.3390/math12060858
Submission received: 18 January 2024 / Revised: 3 March 2024 / Accepted: 4 March 2024 / Published: 14 March 2024
(This article belongs to the Special Issue Advances of Linear and Multilinear Algebra)

Abstract

:
The Plücker matrix B L ( n , E ) of the Lagrangian Grassmannian L ( n , E ) , is determined by the linear envelope L ( n , E ) of the Lagrangian Grassmannian. The linear envelope L ( n , E ) is the intersection of linear relations of Plücker of Lagrangian Grassmannian, defined here. The Plücker matrix B L ( n , E ) is a direct sum of the incidence matrix of the configuration of subsets. These matrices determine the isotropy index r n and r n -atlas which are invariants associated with the symplectic vector space E.

1. Introduction

Given E a symplectic vector space of dimension 2 n , the set L ( n , E ) of all Lagrangian subspaces of E is called the Lagrangian Grassmannian of E. These spaces have a prominent role in symplectic geometry that, in the words of Dusa McDuffSymplectic geometry: is the geometry of closed skew-symmetric form, [ ] thus symplectic geometry is essentially topological in nature  [ ] ”, see [1]. In this article it is shown that the Lagrangian Grassmannian L ( n , E ) has a rich algebraic structure when we assign a coordinate system known as Plücker coordinates. For this we prove the existence of a matrix B L ( n , E ) , whose kernel contains all the Lagrangian subspaces of E. This matrix is built with the minimal family of linear form in ( n E ) * that nullify the Lagrangian Grassmannian under Plücker inclusion. another way of seeing B L ( n , E ) is as a sum of matrices of the family { L r n , L r n 1 , , L 2 } where L r n k , is a ( 0 , 1 ) -matrix, sparce, with r n k -ones in each row and r n ( k + 1 ) -ones in each column where k = 0 , , r n 2 and r n = n + 2 2 is an index that measures the degree of isotropy in E. In this paper, we have:
In Section 3 and Section 4 we construct a family of homogeneous polynomial equations whose solutions parameterize the elements of L ( n , E ) , we call this family of polynomials Plücker relations of Lagrangian Grassmannian. In Section 5 and Section 6 we show that the linear relations of Plücker of the Lagrangian-Grassmannain is the minimal family, up to linear combination, of homogeneous linear polynomials that nullify L ( n , E ) .
In Section 7 calculate the matrix B L ( n , E ) associated with the linear envelope L ( n , E ) of L ( n , E ) , we call this the Plücker matrix of the Lagrangian Grassmannian. We can see that B L ( n , E ) is the incidence matrix of a family of subsets of the set of indices I ( n , 2 n ) and so B L ( n , E ) is a direct sum of submatrices, each belonging to the set { L r n , L r n 1 , , L 2 } . Where L k is a sparce matrix of zeroes and ones with k-ones in each row and k 1 -ones in each column.
Section 8 the isotropy index  r n : = n + 2 2 is studied as an invariant of L ( n , E ) and that, among other things, allows us to compare Lagrangian Grassmannian.
De Concini and Lakshmibai [1981] [2] show that the Lagrangian Grassmannian L ( n , E ) is defined by quadratic relations. These relations are obtained by expressing L ( n , E ) as a linear section of G ( n , E ) , so L ( n , E ) = G ( n , E ) P ( L ( ω n ) ) , where P ( L ( ω n ) ) is the projectivization of a vector space L ω n such that n E L ω n n 2 E , where L ω n is S p 2 n ( F ) -representation of highest weight ω n = h 1 * + + h n * , and where L ( n , E ) P ( L ( ω n ) ) see [3] (pages 182–184), [2,4].
The advantage of our approach is that we have the equations { Q α , β , Π α r s } that define L ( n , E ) as a projective variety and with this we obtain a totally explicit information of L ( n , E ) as a linear section of the Grassmannian G ( n , E ) (see Theorem 6) and in this way, we can give a connection with matrix theory and symplectic geometry which opens a computational horizon in these topics. In [5] we show that the homogeneous linear functionals { Π α r s : α r s I ( n , 2 n ) } also allow us to describe the k-Grassmannian-Isotropic I G ( k , E ) , of a symplectic vector space E of dimension 2 n and we give the Plücker matrix of I G ( k , E ) which is a generalization of B L ( n , E ) . The following bibliography is relevant for this research in [6,7,8,9,10] we can see a few results about L ( n , E ) . See [9,11,12,13,14], where you can see some applications of { Π α r s : α r s I ( n , 2 n ) } .

2. Preliminaries

Using terminology and definitions as given in [15,16], let E be an vector space defined over an arbitrary field F . A symplectic form is a bilinear map
, : E × E F
that satisfies
v , w = w , v f o r a l l v , w E v , v = 0 f o r a l l v E a n d i f v , w = 0 f o r a l l v E w = 0
and it is said to be skew-symmetric nondegenerate. ( E , , ) is called a symplectic vector space, a symplectic vector space E is necessarily of even dimension and there is a basis e 1 , , e n , u 1 , u n of E such that
e i , e j = u i , u j = 0 y e i , u j = δ i j
where δ i j is the Kroneker delta function. A subspace U E is said to be isotropic if u , u = 0 for all u , u U . A subspace L E is said to be Lagrangian subspace if L is isotropic and dim ( L ) = n . The collection of Lagrangian subspace of E, we call it Lagrangian Grassmannian of E or simply Lagrangian Grassmannian we denote it by L ( n , E ) . A subspace W E is a symplectic subspace of E if the symplectic form in E when restricted to W remains symplectic.

Indices

Let m be an integer we denote
[ m ] : = { 1 , 2 , , m }
to the set of the first m integers. Let m and be positive integers such that < m as usual in the literature C m denotes binomial coefficient. If α = ( α 1 , , α ) N then, s u p p { α } : = { α 1 , , α } . If s 1 is a positive integer and Σ N is a non-empty set, we define the sets
C s ( Σ ) : = { α = ( α 1 , , α s ) N s : α 1 < < α s a n d s u p p { α } Σ }
if | Σ | = m then | C s ( Σ ) | = C m ; with this notation if < m we define I ( , m ) = C ( [ m ] ) . So we have
I ( , m ) = { α = ( α 1 , , α s ) : 1 α 1 < < α m } .
We say
α = ( α 1 , , α s ) C s ( Σ )
if there is a permutation σ such that arrange the elements of s u p p { α } in increasing order so we have ( σ ( α 1 ) , , σ ( α s ) ) C s ( Σ ) .
if α and β are elements of C s ( Σ ) then we say that
α = β s u p p { α } = s u p p { β } .
Let α I ( n , 2 n ) , suppose there are i , j s u p p α such that i + j = 2 n + 1 in this case j = 2 n i + 1 and we write this pair as P i = ( i , 2 n i + 1 ) so we define the set
Σ n : = P 1 , , P n
and if α I ( n , 2 n ) such that { i , 2 n i + 1 } s u p p { α } , then we say that P i s u p p { α } and that P i s u p p { α } Σ m , note that | s u p p P β | = 2 | s u p p β | .
If k n is a even integer with notation similar to (2) we define
C k 2 ( Σ n ) : = { P β = ( P β 1 , , P β k 2 ) : P β i Σ n a n d 1 β 1 < < β k 2 n } .
If 1 a 1 < a 2 < < a 2 k 2 n such that a i + a j 2 n + 1 then we define
Σ a 1 , , a 2 k = Σ n { P a 1 , , P a 2 k }
so | Σ a 1 , , a 2 k | = n 2 k .
We define the sets
( a 1 , , a 2 k ) × C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) : = : = { ( a 1 , , a 2 k , P θ ) I ( n 2 , 2 n ) | P θ C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) }
similarly we define ( a 1 , , a 2 k + 1 ) × C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k ) .
Lemma 1.
(A) 
Let n 4 be an even number such that r n = n + 2 2 , then
I ( n 2 , 2 n ) = C n 2 2 ( Σ n ) ( k = 1 r n 2 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 ( a 1 , , a 2 k ) × C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) ) .
(B) 
Let n 5 be an odd and such that r n = n + 1 2 , then
I ( n 2 , 2 n ) = j = 1 n C n 3 2 ( Σ n { P j } ) × { j } ( k = 1 r n 2 1 a 1 < < a 2 k + 1 2 n a i + a j 2 n + 1 ( a 1 , , a 2 k + 1 ) × C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) ) .
Proof. 
(A) Let n 4 even integer and let r n = n + 2 2 , then it is it is enough to prove the inclusion ⊆. If s u p p { α r s } Σ n then it exists P θ C n 2 2 ( Σ n ) such that α r s = P θ C n 2 2 ( Σ n ) . If 1 | s u p p { α r s } Σ n | < n 2 there are P θ C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) and ( a 1 , , a 2 k ) such that 1 a 1 < , < a 2 k 2 n and α r s = ( a 1 , , a 2 k , P θ ) ( a 1 , , a 2 k ) × C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) .
(B) If n 5 odd integer and let r n = n + 1 2 , analogously it is sufficient to show the ⊆.
Let α r s I ( n 3 , 2 n ) then if 1 | s u p p { α r s } Σ n | n 3 2 , if | s u p p { α r s } Σ n | = n 3 2 then α r s = ( P θ , j ) with P θ C n 3 2 ( Σ n { P j } ) .
If 1 | s u p p { α r s } Σ n | < n 3 2 then it exists 1 a 1 < < a 2 k + 1 2 n such that a i + a j 2 n and P θ C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) with α r s = ( a 1 , , a 2 k + 1 , P θ ) ( a 1 , , a 2 k + 1 ) × C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 )
with which the demonstration ends. □
Given a canonical basis B = { e 1 , , e n , u 1 , u n } , see [15], of symplectic vector space E, in this article, we redefine its elements as follows e 2 n : = u 1 , e 2 n 1 : = u 2 , e n + 1 : = u n and we have B = { e 1 , , e 2 n } such that
e i , e j = 1 if j = 2 n i + 1 , 0 otherwise .
For α = ( α 1 , , α n ) I ( n , 2 n ) , write
e α : = e α 1 e α n , e α r s : = e α 1 e ^ α r e ^ α s e α n ,
where e ^ α k means that the corresponding term is omitted.
Denote by n E the n-th exterior power of E, which is generated by { e α : α I ( n , 2 n ) } . For w = α I ( n , 2 n ) p α e α n E , the coefficients p α are the Plücker coordinates of w, see also [17] (p. 42). Writing w n E as w = α I ( n , 2 n ) P α e α , the vector
w ρ = [ p α ] α I ( n , 2 n ) P ( n E )
we call the Plücker vector of w.

3. Contraction Map

In this section, E is a symplectic vector space of dimension 2 n .
Definition 1.
The contraction map is defined as the linear transformation
f : n E n 2 E
f ( w 1 w n ) = 1 r < s n w r , w s ( 1 ) r + s 1 w 1 w ^ r w ^ s w n
where w ^ means that the corresponding term is omitted; see [18] (p. 283).
Lemma 2.
Let f the contraction map, and w = α I ( n , 2 n ) X α e α n E arbitrary element, in coordinates of Plücker, then the contraction map
f ( w ) = α r s I ( n 2 , 2 n ) ( i = 1 n X ( α r s , P i ) ) e α r s
where X α r s P i disappears from the equation i = 1 n X ( α r s , P i ) if | s u p p { α r s P i } | n .
Proof. 
Let w = α I ( n , 2 n ) X α e α n E arbitrary element then
f ( w ) = α I ( n , 2 n ) X α f ( e α ) = α I ( n , 2 n ) X α ( 1 r < s n e α r , e α s 1 r + s 1 e α r s ) = 1 r < s n ( i = 1 n X ( α r s , ( i , 2 n i + 1 ) ) e i , e 2 n i + 1 1 [ i + 2 n i + 1 ] 1 ) e α r s = α r s I ( n 2 , 2 n ) ( i = 1 n X ( α r s , P i ) ) e α r s
where clearly X ( α r s P i ) disappears from the equation if | s u p p { α r s P i } | n .
Let ψ : E E a symplectomorphism that sends the symplectic basis { e i } i = 1 2 n in the symplectic basis { ϵ i } i = 1 2 n such that ψ ( e i ) = ϵ i then let us consider the following linear transformations defined in generators ψ n : n E n E such that ψ n ( e α ) = ϵ α and ψ n 2 : n 2 E n 2 E such that ψ n 2 ( e α r s ) = ϵ α r s .
Lemma 3.
Let f : n E n 2 E the contraction map then the following diagram commutes
Mathematics 12 00858 i001
that is to say f ψ n = ψ n 2 f .

4. Plücker Relations of Lagrangian Grassmannian

For an m-dimensional vector space E, denote by G ( , E ) the set of vector subspaces of dimension of E. The Grassmannian G ( , E ) is a algebraic variety of dimension ( m ) and can be embedded in a projective space P c 1 , where c = m by Plücker embedding. The Plücker embedding is the injective mapping ρ : G ( , E ) P ( E ) given on each W G ( , E ) by choosing a basis w 1 , , w of W and then mapping the vector subspace W G ( , E ) to the tensor w 1 w E . Since choosing a different basis of W changes the tensor w 1 w by a nonzero scalar, this tensor is a well-defined element in the projective space P ( E ) P N 1 , where N = C m = dim F ( E ) . If w = α I ( , m ) P α e α P ( E ) , then w G ( , E ) if and only if for each pair of tuples 1 α 1 < < α 1 m and 1 β 1 < < β + 1 m , the Plücker coordinates of w satisfy the Plücker relations
Q α , β : = i = 1 + 1 ( 1 ) i P α 1 α 1 β i P β 1 , β 2 β i ^ β + 1 = 0 ,
where β i ^ means that the corresponding term is omitted and where α = ( α 1 , , α 1 ) I ( 1 , m ) , β = ( β 1 , , β i , , β + 1 ) I ( + 1 , m ) , see [17] (Section 4) and [19]. Under the inclusion of Plücker the Lagrangian Grassmannian is given by
L ( n , E ) = { w 1 w n G ( n , E ) : w i , w j = 0 for all 1 i < j n }
Lemma 4.
L ( n , E ) = G ( n , E ) ker f .
Proof. 
Using the Definition 1 we clearly have L ( n , E ) G ( n , E ) ker f given that w r , w s = 0 for all 1 r < s 2 n .
Let v 1 v n G ( n , E ) ker f then { w 1 w ^ r w ^ s w n } 1 r < s 2 n is a family of linearly independent vectors in n 2 E more over by hypothesis f ( w 1 w n ) = 0 then w r , w s = 0 for all 1 r < s 2 n and so G ( n , E ) ker f L ( n , E ) . □
The proof of the following lemma is a consequence of the Lemma 2 where the kernel of the contraction map f is characterized as follows
Lemma 5.
Let w = α I ( n , 2 n ) X α e α n E written in Plücker coordinates, then we have
w ker f i = 1 n X α r s P i = 0 , for all α r s I ( n 2 , 2 n ) .
where X α r s P i disappears from the equation if | s u p p { α r s P i } | n .
For all α r s I ( n 2 , 2 n ) we define a homogeneous linear polynomial
Π α r s = i = 1 n c α r s P i X α r s P i ( n E ) *
where
c α r s P i = 1 if | supp { α r s P i } | = n , 0 otherwise ,
Remark 1.
Throughout this article, we write the Equation (16) simply as
Π α r s = i = 1 n X α r s P i
where the addend X α r s P i disappears from the Equation (17) if | s u p p { α r s P i } | n .
Corollary 1.
ker f is independent of the symplectic basis and
ker f = α r s I ( n 2 , 2 n ) ker Π α r s .
Proof. 
From the Lemma 2, we have f ψ n = ψ n 2 f so then ker f ψ n = ker ψ n 2 f = ker f since ψ n 2 is an isomorphism and then ker f = α r s I ( n 2 , 2 n ) ker Π α r s . □
Following [20] for definitions of algebraic set, we have below that ker f , G ( n , E ) and L ( n , E ) are algebraic sets in P ( n E )
ker f = Z Π α r s : α r s I ( n 2 , 2 n ) .
So we have to G ( n , E ) is an algebraic set of
G ( n , E ) = Z Q α , β : α I ( 1 , m ) , β I ( + 1 , m )
see (14).
Theorem 1.
Let E symplectic vector space of dimension 2 n then
L ( n , E ) = Z Q α , β , Π α r s
where Q α , β and Π α r s are as in (14) and (16), respectively.
Proof. 
Of Lemma 4, (20) and (19) we have
L ( n , E ) = G ( n , E ) ker f = Z Q α , β Z Π α r s = Z Q α , β , Π α r s
Definition 2.
To the set of homogeneous polynomials
{ Q α , β , Π α r s }
where α = ( α 1 , , α n 1 ) I ( n 1 , 2 n ) , β = ( β 1 , β i , , β n + 1 ) I ( n + 1 , 2 n ) , we call it relations of Pücker of Lagrangian Grassmannian. To the set of linear homogeneous polynomials
{ Π α r s : α r s I ( n 2 , 2 n ) }
we call it linear relations of Plücker of the Lagrangian Grassmannian.
Example 1.
In the case L ( 2 , E ) we have the relations of Plücker of Lagrangian Grassmannian
Q : = X 12 X 34 X 13 X 24 + X 14 X 23 = 0 Π : = X 14 + X 23 = 0
Example 2.
The linear relations of Plücker of L ( 3 , E ) are
Π 1 = X P 2 , 1 + X P 3 , 1 Π 2 = X P 1 , 2 + X P 3 , 2 Π 3 = X P 1 , 3 + X P 2 , 3 Π 4 = X P 1 , 4 + X P 5 , 4 Π 5 = X P 1 , 5 + X P 3 , 5 Π 6 = X P 2 , 6 + X P 3 , 6 .
Example 3.
In the case L ( 4 , E ) it was shown in [21] (example 4) that linear relations { Π α r s : α r s I ( 2 , 8 ) } consists of 28 homogeneous linear equations in 70 variables.

Ideal of L ( n , E )

The ideal of L ( n , E ) in F [ X α ] α I ( n , 2 n ) is defines by
I ( L ( n , E ) ) = { f F [ X α ] α I ( n , 2 n ) : f ( w ) = 0 f o r a l l w L ( n , E ) } .
Proposition 1.
If the field of definition of the symplectic vector, space E is algebraically closed then
I ( L ( n , E ) ) = Q α , β , Π α r s
so L ( n , E ) is a projective variety.
Proof. 
By the Theorem 1 we have L ( n , E ) = Z Q α , β , Π α r s and by Hilbert’s Nullstellensatz theorem, see [20] (Theorem 1.3 A), the result is fulfilled. □
Let F q be a finite field with q elements, and denote by F ¯ q an algebraic closure of F q . For a vector space E over F q of finite dimension k, let E ¯ = E F q F ¯ q be the corresponding vector space over the algebraically closed field F ¯ q . We will be considering algebraic varieties in the projective space P ( E ¯ ) = P k 1 ( F ¯ q ) . Recall that a projective variety X P k 1 ( F ¯ q ) is defined over the finite field F q if its vanishing ideal can be generated by polynomials with coefficients in F q . If E is a symplectic vector space, of dimension 2 n define over a finite field F q , then the rational points of L ( n , E ) are defined as the set
L ( n , E ) ( F q ) : = Z Q α , β , Π α r s , x ϵ q x ϵ
where α I ( n 1 , 2 n ) , β I ( n + 1 , 2 n ) , α r s I ( n 2 , 2 n ) and ϵ I ( n , 2 n ) more over
| L ( n , E ) ( F q ) | = Π i = 1 n ( 1 + q i )
see [22] (Prop. 2.14).
We define the ideal I F ¯ q F [ x α ] α I ( n , 2 n ) as
I F ¯ q = Q α , β , Π α r s , g ϵ
where α I ( n 1 , 2 n ) , β I ( n + 1 , 2 n ) , α r s I ( n 2 , 2 n ) , ϵ I ( n , 2 n ) y g ϵ = x ϵ q x ϵ .
Lemma 6.
The ideal I F ¯ q is radical
Proof. 
The ideal I F ¯ q is zero-dimensional since the set of solutions to the homogeneous polynomial equations
| S o l { Q α , β , Π α r s , g ϵ } | <
given that (26) implies | S o l { Q α , β , Π α r s , g ϵ } | = | L ( n , E ) ( F q ) | moreover g ϵ = q x ϵ q 1 1 = 1 so g c d ( g ϵ , g ϵ ) = 1 , and by Seindeber’s lemma, ver [23] (Proposition 3.7.15), I F ¯ q is radical. □
Let t : = Z h 1 , h 2 , , h t P ( n E ) a hyperplane of codimension t, we say that L ( n , E ) t is a linear section of the Lagrangian Grassmannian and let
I t : = Q α , β , Π r s , x ϵ q x ϵ , h 1 , h 2 , , h t .
Lemma 7.
Suppose the basis field F q is perfect then the linear section of the Lagrangian Grassmannian satisfies | L ( n , E ) ( F q ) t | = dim F q F [ x α ] α I ( n , 2 n ) I t .
Proof. 
Given the | L ( n , E ) t | | L ( n , E ) | then the ideal I r is zero dimensional, m c d ( g α , g α ) = 1 , thus by Seindeber’s lemma we have that the ideal I t is radical. □

5. Factorable Morphisms

Let 1 < α ¯ k < < 2 n α ¯ k + 1 < 2 n two integers, we define
S : = [ 2 n ] { α ¯ k , 2 n α ¯ k + 1 }
with the notation (2) we define
I α ¯ k ( n 1 , 2 n 2 ) : = C n 1 { S } .
Let
ϕ α ¯ k : = { ( β , α ¯ k ) I ( n , 2 n ) : β I α ¯ k ( n 1 , 2 n 2 ) } .
For e α ¯ k B be a basic vector and let = e α ¯ k E generated by the vector e α ¯ k and
U ( ) = { W L ( n , E ) : W } .
In [16] (Lemma 1.4.38) it shows that there is a one-to-one correspondence between U ( ) and L ( n 1 , / ) , where / is a symplectic vector space of dimension 2 n 2 , generated by the symplectic basis { e 1 + , , e ^ α ¯ k + , , e ^ 2 n α ¯ k + 1 + , , e 2 n + } , recall ^ means that the term was omitted.
As consequence we have U ( ) = e α ¯ k ( L ( n 1 , / ) ) and
U ( ) = L ( n 1 , / ) e α ¯ k
so in Plücker coordinates we have
U ( ) = ( w L ( n , E ) : w = ( β , α ¯ k ) ϕ α ¯ k p ( β , α ¯ k ) e ( β , α ¯ k ) ) .
X β : = e β + * denotes the basis vector of the dual vector space n 1 / * and X β X α ¯ k : = e ( β , α ¯ k ) * the basis vector of dual vector space n E * . Now with this notation we define in generators an injective linear transformation
ξ : ( n 1 ( / ) ) * ( n E ) * X β X β X α ¯ k
with β I α ¯ k ( n 1 , 2 n 2 ) and X β X α ¯ k = X ( β , α ¯ k ) .
Definition 3.
We say that h = α I ( n , 2 n ) A α I ( n , 2 n ) X α I ( n , 2 n ) ( n E ) * is factored if h = h 0 X α ¯ k for some h 0 ( n 1 / ) * where = e α ¯ k and 1 α ¯ k 2 n . We say h satisfies the factoring property if there are at least one coefficient A α ¯ 0 and there are at least one element α ¯ k s u p p { α ¯ } such that 2 n α ¯ k + 1 s u p p { α ¯ } .
Example 4.
The homogeneous linear polynomials given in the Example 2 are factored
Π 1 = ( X P 2 + X P 3 ) X 1 Π 2 = ( X P 1 + X P 3 ) X 2 Π 3 = ( X P 1 + X P 2 ) X 3 Π 4 = ( X P 1 + X P 5 ) X 4 Π 5 = ( X P 1 + X P 3 ) X 5 Π 6 = ( X P 2 + X P 3 ) X 6
Proposition 2.
Let h ( n E ) * such that h ( L ( n , E ) ) = 0 and h = h 0 X α ¯ k is factored. Then h 0 ( L ( n 1 , / ) ) = 0 where = e α ¯ k .
Proof. 
U ( ) = L ( n 1 , / ) e α ¯ L ( n , E ) then h ( U ( ) ) = 0 and
h ( U ( ) ) = h ( L ( n 1 , / ) e α ¯ k ) = ( h 0 X α ¯ k ) ( L ( n 1 , / ) e α ¯ k ) = h 0 ( L ( n 1 , / ) ) · X α ¯ k ( e α ¯ k ) = 0
h 0 ( L ( n 1 , / ) ) = 0 .
We denote by β r s = ( β 1 , , β ^ r , , β ^ s , , β n 1 ) I α ¯ k ( n 3 , 2 n 2 ) , where ^ means that the corresponding term is omitted. We define
Π ( β r s , α ¯ k ) : = i = 1 n X ( β r s , α ¯ k ) P i
an element of in ( n E ) * such that | s u p p { ( β r s , α ¯ k ) P i } | = n .
Lemma 8.
Let β r s I α ¯ k ( n 3 , 2 n 2 ) where 1 r < s 2 n and r , s S see (28) then
U ( ) β r s I α ¯ k ( n 3 , 2 n 2 ) ker Π ( β r s , α ¯ k ) .
Proof. 
Let w U ( ) be from (33) we have w = β I α ¯ k ( n 1 , 2 n 2 ) X ( β , α ¯ k ) e ( β , α ¯ k ) and f ( w ) = 0 for f contraction map, we have
f ( w ) = β I α ¯ k ( n 1 , 2 n 2 ) X ( β , α ¯ k ) f ( e ( β , α ¯ k ) ) =
= β I α ¯ k ( n 1 , 2 n 2 ) X ( β , α ¯ k ) ( 1 r < s n e α r , e α s ( 1 ) r + s 1 e ( β , α ¯ k ) r s )
= 1 r < s n ( 1 ρ 1 < ρ 2 n X ( β r s , α ¯ k , ρ 1 , ρ 2 ) e ρ 1 , e ρ 2 ( 1 ) ρ 1 + ρ 2 1 ) e ( β , α ¯ k ) r s = 0
then
1 ρ 1 < ρ 2 n X ( β r s , α ¯ k , ρ 1 , ρ 2 ) ( 1 ) ρ 1 + ρ 2 1 e ρ 1 , e ρ 2 = 0 .
where β r s I α ¯ k ( n 3 , 2 n 2 ) and as stated before ( β r s , α ¯ k , ρ 1 , ρ 2 ) I ( n , 2 n ) . Now e ρ 1 , e ρ 2 = 1 iff ρ 1 + ρ 2 = 2 n + 1 , note that with this condition we have ( 1 ) ρ 1 + ρ 2 1 = 1 . Renaming ρ 1 = i , we have ρ 2 = 2 n i + 1 . Then
1 ρ 1 < ρ 2 n X ( β r s , α ¯ k ) ρ 1 ρ 2 = i = 1 n X ( β r s α ¯ k ) P i = 0
for all β r s I α ¯ k ( n 3 , 2 n 2 ) , that is U ( ) ker Π ( β r s , α ¯ k ) where
Π ( β r s , α ¯ k ) : = i = 1 n X ( β r s α ¯ k ) P i
for all ( β r s , α ¯ k ) I α ¯ k ( n 3 , 2 n 2 ) × { α ¯ k } .
Remark 2.
Note that
{ Π ( β r s , α ¯ k ) : β r s I α ¯ k ( n 3 , 2 n 2 ) } { Π α r s : α r s I ( n 2 , 2 n ) }
for all 1 r < s 2 n 2 .

6. Linear Envelope of L ( n , E )

In this section, E is a symplectic vector space of dimension 2 n .
H : = h ( n E ) * : h ( L ( n , E ) ) = 0
Remark 3.
Sometimes in (37), it is necessary to distinguish the even case from the odd case so we write H e v e n n ( n E ) * to n even number and H o d d n ( n E ) * for n odd number.
Lemma 9.
H is a nontrivial vector subspace of ( n E ) *
Proof. 
The proof follows from (18) given that L ( n , E ) ker f , so Π α r s : α r s I ( n 2 , 2 n ) is a vector subspace of H . □
Let h H to the set H h : = ker h we call it a hyperplane containing L ( n , E ) .
Definition 4.
The Linear Envelope L ( n , E ) of L ( n , E ) is the smallest linear variety that contains L ( n , E ) in P ( n E ) .
The proof of the following corollary follows directly from the Definition 4.
Corollary 2.
L ( n , E ) = h H H h .
Proposition 3.
Let h = α I ( n , 2 n ) A α X α H and A α ¯ 0 a coefficient different from zero of h then it exists { α ¯ i , α ¯ j } s u p p { α ¯ } such that α ¯ i + α ¯ j = 2 n + 1 .
Proof. 
Suppose that for each { α ¯ i , α ¯ j } s u p p { α ¯ } you have to α ¯ i + α ¯ j 2 n + 1 this means that e α ¯ i , e α ¯ j = 0 then e α ¯ = e α ¯ 1 e α ¯ n L ( n , E ) and so then h ( e α ¯ ) = A α ¯ = 0 which is a contradiction. Then there are { α ¯ i , α ¯ j } s u p p { α ¯ } such that α ¯ i + α ¯ j = 2 n + 1 . □
For each P θ C n 2 2 ( Σ n ) , we define
h P θ = i = 1 n A P θ P i X P θ P i
for each ( a 1 , , a 2 k , P θ ) { ( a 1 , , a 2 k ) } × C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) we define
h ( a 1 , , a 2 k , P θ ) = i = 1 n A ( a 1 , , a 2 k , P θ ) X ( a 1 , , a 2 k , P θ ) .
Similarly
for each 1 j n and ( P θ , j ) C n 3 2 ( Σ n { P j } ) × { j } we define
h P ( θ , j ) = i = 1 n A ( P θ , j , P i ) X ( P θ , j , P i )
for each ( a 1 , , a 2 k + 1 , P θ ) { ( a 1 , , a 2 k + 1 ) } × C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) we define
h ( a 1 , , a 2 k + 1 , P θ ) = i = 1 n A ( a 1 , , a 2 k + 1 , P θ ) X ( a 1 , , a 2 k + 1 , P θ ) .
Corollary 3.
Let E symplectic vector space of dimention 2 n
(i) 
If n 4 even, r n = n + 2 2 and let h H e v e n n then
h = P θ C n 2 2 ( Σ n ) h P θ + k = 1 r n 2 ( a 1 , , a 2 k , P θ ) h ( a 1 , a 2 k , P θ )
with P θ C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) such that 1 a 1 < < a 2 k 2 n and a i + a j 2 n + 1 .
(ii) 
If n 5 odd, r n = n + 1 2 and let h H o d d n then
h = j = 1 n ( P θ , j ) C n 3 2 ( Σ n { P j } ) × { j } h ( P θ , j ) + k = 1 r n 2 ( a 1 , , a 2 k + 1 , P θ ) h ( a 1 , a 2 k + 1 , P θ )
with P θ C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) such that 1 a 1 < < a 2 k + 1 2 n + 1 and a i + a j 2 n + 1 .
Proof. 
The proof follows directly from the Lemma 1 and Proposition 3. □
Lemma 10.
Let E symplectic vector space of dimension 4, and h ( 2 E ) * such that h ( L ( 2 , E ) ) = 0 then h = A Π for A a non-zero constant and Π : = X P 1 + X P 2 .
Proof. 
Clearly by Proposition 3 each h H e v e n 2 it is of the form h = A 14 X 14 + A 23 X 23 it is easy see that e 12 + e 14 e 23 + e 34 L ( 2 , E ) since it satisfies the Equation (24) more over h ( e 12 + e 14 e 23 + e 34 ) = A 14 A 23 = 0 consequently h = A Π where A = A 14 = A 23 .
Theorem 2.
Let E symplectic vector space of dimension 2 n , h ( n E ) * such that h ( L ( n , E ) ) = 0 and h = h 0 X α ¯ k is factored then h Π α r s : α r s I ( n 2 , 2 n ) F .
Proof. 
The proof is by induction on n. If n = 2 it follows from Lemma 10.
We induction hypothesis is, let E symplectic vector space of dimension 2 m with m < n and h 0 ( m E ) * such that h 0 ( L ( m , E ) ) = 0 then h 0 Π β r s : β r s I ( n 3 , 2 n 2 ) .
If h ( n E ) * such that h ( L ( n , E ) ) = 0 and h = h 0 X α ¯ k is factored, then by Lemma 2 h 0 ( L ( n 1 , / ) ) = 0 where = e α ¯ k then by induction step h 0 Π β r s : β r s I α ¯ ( n 3 , 2 n 2 ) so h = h 0 X α ¯ k Π β r s X α ¯ k : β r s I α ¯ ( n 3 , 2 n 2 ) Π α r s : α r s I ( n 2 , 2 n ) . □
Proposition 4.
Let h = α I ( n , 2 n ) A α X α ( n E ) * such that h ( L ( n , E ) ) = 0 and satisfies the factoring property from the Definition 3, then h Π α r s : α r s I ( n 2 , 2 n ) or h = h + h , where h Π α r s : α r s I ( n 2 , 2 n ) and h satisfies that s u p p { α } Σ n for all coefficients A α 0 moreover h ( L ( n , E ) ) = 0 .
Proof. 
The proof is by induction on n. If n = 2 it follows from Lemma 10.
Induction Step): Let E symplectic vector space of dimension 2 m with m < n and h 0 ( m E ) * such that h 0 ( L ( m , E ) ) = 0 then h 0 Π β r s : β r s I ( n 3 , 2 n 2 ) .
Let A α ¯ a non-zero coefficient that satisfies the factoring property, from the Definition 3, then there is α ¯ k s u p p { α ¯ } but 2 n α ¯ k + 1 s u p p { α ¯ } . Now let ϕ α ¯ k = { ( β , α ¯ k ) I ( n , 2 n ) : β I α ¯ k ( n 1 , 2 n 2 ) } as in (30). So if h = Σ α I ( n , 2 n ) A α X α ( n E ) * then h = Σ α ϕ α ¯ k A α X α + Σ α ϕ α ¯ k c A α X α where I ( n , 2 n ) = ϕ α ¯ k ϕ α ¯ k c . We define homogeneous linear polynomials h : = Σ α ϕ α ¯ k A α X α and h : = Σ α ϕ α ¯ k c A α X α such that h , h ( n E ) * and h = h + h so h = ( β , α ¯ k ) ϕ α ¯ k A ( β , α ¯ k ) X ( β , α ¯ k ) and h = α ϕ α ¯ k c A α X α . Note that h 0 , because as we mentioned before α ¯ ϕ α ¯ k , from (33) we have that h ( U ( ) ) = 0 that is h ( U ( ) ) = h ( U ( ) ) , since h ( L ( n , E ) ) = 0 we obtain h ( U ( ) ) = 0 moreover h = h 0 X α ¯ k with h 0 = β I α ¯ k ( n 1 , 2 n 2 ) A β X β ( n 1 / ) * where A β = A ( β , α ¯ k ) then by Proposition 2 we have h 0 ( L ( n 1 , / ) ) = 0 then h = h 0 X α ¯ k Π β r s X α ¯ k : β r s I α ¯ k ( n 3 , 2 n 2 ) } Π α r s : α r s I ( n 2 , 2 n ) . Given that h ( L ( n , E ) ) = ( h h ) ( L ( n , E ) ) = 0 and if h satisfies the factoring property from the Definition 3 continuing recursively in the same way, the process ends in a finite number of steps in h Π α r s : α r s I ( n 2 , 2 n ) or h = h + h , where h Π α r s : α r s I ( n 2 , 2 n ) and h 0 with s u p p { α } Σ n for all coefficients A α 0 and h ( L ( n , E ) ) = 0 . □
Corollary 4.
Let E symplectic vector space of dimension 2 n
(a) If n 5 odd then H o d d n = Π α r s : α r s I ( n 2 , 2 n )
(b) If h H e v e n n then h = h + h such that h Π α r s : α r s I ( n 2 , 2 n ) F and h satisfies that s u p p { α } Σ n for all coefficients A α 0 and h ( L ( n , E ) ) = 0 .
Proof. 
(a) If h H o d d n then by Corollary 3 ( i i )
h = j = 1 n ( P θ , j ) C n 3 2 ( Σ n { P j } ) × { j } h ( P θ , j ) + k = 1 r 2 ( a 1 , , a 2 k + 1 , P θ ) h ( a 1 , a 2 k + 1 , P θ )
with P θ C n 2 ( k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) such that 1 a 1 < < a 2 k + 1 2 n + 1 and a i + a j 2 n + 1 satisfies the factoring property for all non-zero coefficients of h then by Proposition 4 we have h Π α r s : α r s I ( n 2 , 2 n )
(b) If h H e v e n n then by Corollary 3 we have h = h + h where h = P θ C n 2 2 ( Σ n ) h P θ and h = k = 1 r 2 ( a 1 , , a 2 k , P θ ) h ( a 1 , a 2 k , P θ ) with P θ C n 2 ( k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) such that 1 a 1 < < a 2 k + 1 2 n + 1 and a i + a j 2 n + 1 . Clearly h satisfies factoring property of the Definition 3 for all non-zero coefficients of h , so by Proposition 4 we have h Π α r s : α r s I ( n 2 , 2 n ) F and each coefficient A P θ 0 of h satisfies that s u p p { P θ } C n 2 2 ( Σ n ) and h ( L ( n , E ) ) = 0 . □
For P θ : = ( P θ 1 , , P θ n 2 ) C n 2 ( Σ n ) an arbitrary element where 1 θ 1 < < θ n 2 n we say that
θ 1 = min s u p p { P θ }
For Ω C n 2 ( Σ n ) an arbitrary subset then there are a partition of Ω of the form
Ω = j = 1 t Ω k j
where
Ω k j : = { P θ Ω : k j = min s u p p { P θ } }
to j { 1 , , t } and 1 k 1 < < k t n .
h Ω = j = 1 t ( P θ ¯ j Ω k j ( i = 1 r A P k j , P θ 2 j , P θ n 2 2 j P ϵ i j X P k j , P θ 2 j , P θ n 2 2 j P ϵ i j ) ) .
where P θ ¯ j : = P k j , P θ 2 j , P θ n 2 2 j P ϵ i j
Lemma 11.
Let E symplectic vector space of dimention 2 n , Ω C n 2 2 ( Σ n ) non-empty set and h Ω ( n E ) * such that h Ω ( L ( n , E ) ) = 0 then h Ω Π α r s : α r s I ( n 2 , 2 n ) .
Proof. 
Let h Ω = j = 1 t ( P θ ¯ j Ω k j ( i = 1 r A P k j , P θ 2 j , P θ n 2 2 j P ϵ i j X P k j , P θ 2 j , P θ n 2 2 j P ϵ i j ) ) and without loss of generality we can assume that A P k j , P θ 2 j , P θ n 2 2 j P ϵ 1 j 0 for all j { 1 , , t } . Now let Π ¯ : = j = 1 t ( P θ ¯ j Ω k j A P k j , P θ 2 j , P θ n 2 2 j P ϵ 1 j Π P k j , P θ 2 j , P θ n 2 2 j ) Π α r s : α r s I ( n 2 , 2 n ) We define
h ¯ Ω : = h j = 1 t Ω k j Π ¯ = j = 1 t ( P θ ¯ j Ω k j ( i = 2 r ( A P k j , P θ 2 j , P θ n 2 2 j P ϵ i j A P k j , P θ 2 j , P θ n 2 2 j P ϵ 1 j ) X P θ 2 j , P θ n 2 2 j P ϵ i j k j ) X 2 n k j + 1
then for each 1 j t we define
h ( 0 , k j ) : = P θ j Ω k j ( i = 2 r ( A P k j , P θ 2 j , P θ n 2 2 j P ϵ i j A P k j , P θ 2 j , P θ n 2 2 j P ϵ 1 j ) X P θ 2 j , P θ n 2 2 j P ϵ i j k j
such that
h ¯ Ω = j = 1 t ( h ( 0 , k j ) X 2 n k j + 1 )
Let 1 j 0 t a fixed element, j 0 = e j 0 E and L ( n 1 , j 0 / j 0 ) e 2 n k j 0 + 1 L ( n , E ) then
h ¯ Ω ( L ( n 1 , j 0 / j 0 ) e 2 n k j 0 + 1 ) = = ( j = 1 t ( h ( 0 , k j ) X 2 n k j 0 + 1 ) ) ( L ( n 1 , j 0 / j 0 ) e 2 n k j 0 + 1 ) = = h ( 0 , k j 0 ) ( L ( n 1 , j 0 / j 0 ) ) · X 2 n j 0 + 1 ( e 2 n j 0 + 1 ) = = h ( 0 , k j 0 ) ( L ( n 1 , j 0 / j 0 ) ) .
Moreover, h ¯ Ω ( L ( n , E ) ) = 0 implies h ( 0 , k j 0 ) ( L ( n 1 , j 0 / j 0 ) ) = 0 so by Corollary 4 ( a ) given that n 1 is odd number and h ( 0 , k j 0 ) ( n 1 j 0 / j 0 ) * we have h 0 Π β r s : β r s I k j 0 ( n 3 , 2 n 2 ) so
h ( 0 , k j 0 ) X 2 n k j 0 + 1 Π ( β r s , k 2 n j 0 + 1 ) : ( β r s , k 2 n j 0 + 1 ) I ( n 2 , 2 n ) Π α r s : α r s I ( n 2 , 2 n )
then h ¯ Ω = j = 1 t ( h ( 0 , k j ) X 2 n k j + 1 ) Π α r s : α r s I ( n 2 , 2 n ) . □
Corollary 5.
If n 4 even then H e v e n n = Π α r s : α r s I ( n , 2 n ) .
Proof. 
By Corollary 4(b) and by the Lemma 11, we have Π α r s : α r s I ( n , 2 n ) H e v e n n . □
Theorem 3.
Let E be a symplectic vector space of dimension 2 n then
(I) 
H = Π α r s : α r s I ( n , 2 n )
(II) 
L ( n , E ) = P ( ker f ) .
Proof. 
( I ) From the Corollary 4 ( a ) and Corollary 5 we have H = Π α r s : α r s I ( n , 2 n )
( I I ) By (I) above, Corollary 2 and (18) we have
L ( n , E ) = h H H h = α r s I ( n 2 , 2 n ) ker Π α r s = ker f .
Lemma 12.
Let g : n E n 2 E a linear transformation such that g ( L ( n , E ) ) = 0 then g ( ker f ) = 0 .
Proof. 
g ( L ( n , E ) ) = 0 we have L ( n , E ) ker g so ker f ker g that is g ( ker f ) = 0 . □
Corollary 6.
Suppose the contraction map f is surjective and suppose G : n E n 2 E is a surjective linear transformation that vanishes L ( n , E ) then there exists a unique isomorphism such that G = h f .
Proof. 
By Lemma 12 we have ker f ker G , more over ker f = ker G since both have the same dimension because f and G are surjective, then there exists a unique linear isomorphism h that makes the following diagram commute.
Mathematics 12 00858 i002
and so we have to G = h f . □
Corollary 7.
Suppose the contraction map f is surjective then
(i) 
If H is a matrix of order C n 2 2 n × C n 2 n and maximum rank that annuls L ( n , E ) , then H = P B L ( n , E ) , where P is an invertible matrix.
(ii) 
Suppose that there exists R matrix such that L ( n , E ) = G ( n , E ) ker R . Then R = P B L ( n , E ) where P is an invertible matrix.
Proof. 
The proof of (i) follows directly from the Lemma 6. For the (ii) suppose that R = h 1 h 2 h ϵ is a rank matrix ϵ such that L ( n , E ) = G ( n , E ) ker R , then L ( n , E ) ker R and ϵ C n 2 n . If ϵ = C n 2 n the affirmation is followed by the previous clause of this lemma. Now suppose that t < C n 2 n then ker B L ( n , E ) ker R ; this implies that
L ( n , E ) = G ( n , E ) ker B L ( n , E ) G ( n , E ) ker R = L ( n , E ) ,
which is a contradiction and therefore ϵ = C n 2 n . □

7. The Plücker Matrix of the Lagrangian Grassmannian

Definition 5.
Let E symplectic vector space of dimension 2 n defined over an arbitrary field F arbitrary. The matrix
B L ( n , E )
of order C n 2 2 n × C n 2 n associated to the linear equations system { Π α r s : α r s I ( n 2 , 2 n ) } we call the Plücker matrix of the Lagrangian Grassmannian.
The main result of this section is
Theorem 4.
Let E symplectic vector space of dimension 2 n and let r n = n + 2 2 , then there exists a family
A = { L r n , L r n 1 , , L 2 }
of ( 0 , 1 ) -matrices, regular, sparse such that
(A) 
If n 4 be an even integer and 1 k r n 2 , then
B L ( n , E ) = L r n k = 1 r n 2 ( 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 L r n k ( a 1 , , a 2 k ) )
where L r n k ( a 1 , , a 2 k ) is a copy of L r n k , for each 1 k r n 2 .
(B) 
If n 5 be an odd integer then B L ( n , E )
B L ( n , E ) = L r n n k = 1 r n 2 ( 1 a 1 < a 2 < < a 2 k + 1 2 n a i + a j 2 n + 1 L r n k ( a 1 , a 2 , , a 2 k + 1 ) )
where L r n k ( a 1 , a 2 , , a 2 k + 1 ) is a copy of L r n k for each 1 k r n 2 and L r n n = L r n L r n n-times.

7.1. Configuration of Subsets

Following [24] (p. 3) we call X = { x 1 , , x n } an n-set. Now let X 1 , X 2 , , X m be m distinct subsets of the n-set X. We refer to this collection of subsets of an n-set as a configuration of subsets. We set a i j = 1 if x j X i and we set a i j = 0 if x j X i . The resulting ( m × n ) -matrix A = ( a i , j ) , i = 1 , , m , j = 1 , , n of size m by n is the incidence matrix for the configurations of subsets X 1 , X 2 , , X m of the n-set X. The 1 s in row i of A display the elements in the subsets X j and the 1 s in column j display the occurrences of the element x j among the subsets. If a matrix A has all its coefficients equal 0 or 1 is called a ( 0 , 1 ) -matrix. Give a ( 0 , 1 ) -matrix A we say that is regular if the number of 1’s is fixed in each column and has a fixed number of 1’s in each row. If A is not regular we say that is irregular see [10,25,26], for more information. A sparse matrix is a ( 0 , 1 ) -matrix in which most of the elements are zero.

Configuration of Incidence

Let S = { s 1 , s n } an n-set and S 1 , , S m be m subsets of the n-set S and L the m × n incidence matrix, for the configuration of subsets S 1 , , S m . The pair
( S , S i ) i = 1 m
we call configuration of incidence of S. If ( S , S i ) i = 1 m ,where S = { s 1 , s n } , is other configuration of incidence then they are isomorphic if and only if there is a bijection
ψ : S S
ψ ( s i ) = s i
such that ψ ( S i ) = S i for all i = 1 , , m and note L = L where L and L are ( m × n ) -incidence matrices.
Let ( S , S i ) i = 1 m be an incidence configuration, with S an n-set and { a } a set of cardinality 1 then using the Cartesian product we define the Cartesian incidence configuration as follows
a × ( S , S i ) i = 1 m : = ( { a } × S , S ( a , i ) ) i = 1 m
where S ( a , i ) : = { a } × S i .
Lemma 13.
The Cartesian incidence configuration { x } × ( S , S i ) i = 1 m is isomorphic to ( S , S i ) i = 1 m and they have the same incidence matrix.
Proof. 
Since | { x } × ( S , S i ) i = 1 m | = | ( S , S i ) i = 1 m | then the projection mapping
ψ : { x } × ( S , S i ) i = 1 m ( S , S i ) i = 1 m
( x , s ) s
is one–one and clearly ψ | { x } × S i = S i ; thus, the configurations are isomorphic. Moreover, ( x , s ) { x } × S i if and only if s S i and so both configurations have the same incidence matrix. □
Let m 2 even integer, r m = m + 2 2 and Σ m = { P 1 , , P m } as in (6)
( C m 2 ( Σ m ) , S P α ) P α C m 2 2 ( Σ m )
where C m 2 ( Σ m ) and C m 2 m -set and
S P α = { P β C m 2 ( Σ m ) : s u p p { α } s u p p { β } } ,
a configuration of subsets of C m 2 ( Σ m ) .
Remark 4.
Note that S α = { P β C m 2 ( Σ m ) : | s u p p { α } s u p p { β } | = m 2 2 } .

7.2. Properties

Let m 4 even integer, given the incidence configuration (50)
( C m 2 ( Σ m ) , S P α ) P α C m 2 2 ( Σ m )
let r m = m + 2 2 then we have
Lemma 14.
For all P α C m 2 2 ( Σ m ) we have | S P α | = r m .
Proof. 
If P β S P α and | s u p p { α } | = m 2 2 then
| S P α | = | { β I ( m / 2 , m ) : s u p p { β } = s u p p { α } { i } w i t h 1 i m } | = | { i [ m ] : | s u p p { α } { i } | = m / 2 } | = m | s u p p { α } | = m m 2 2 = m + 2 2
Lemma 15.
Let P α and P α ¯ two different elements of C m 2 2 ( Σ m ) then
S P α S P α ¯ i f a n d o n l y i f | s u p p { α } s u p p { α ¯ } | = m 4 2 .
Proof. 
) : Let P β S P α S P α ¯ . Then there are two different positive integers N and M such that s u p p { α } { N } = s u p p { β } = s u p p { α ¯ } { M } , also s u p p { α } { M } = s u p p { β } { N , M } = s u p p { α ¯ } { N } , as a consequence we have to s u p p { β } { N , M } = s u p p { α } s u p p { α ¯ } so | s u p p { P α } s u p p { P α ¯ } | = | s u p p { P β } | | { N , M } | = m 4 2
) : Suppose | s u p p { α } s u p p { α ¯ } | = m 4 2 then there exist M, N distinct positive integers such that { M } = s u p p { α } s u p p { α } s u p p { α ¯ } and { N } = s u p p { α ¯ } s u p p { α } s u p p { α ¯ } and so it exists P β C m 2 ( Σ m ) such that s u p p { β } = ( s u p p { α } s u p p { α ¯ } ) { N , M } then we have P β S P α S P α ¯ and so S P α S P α ¯ . □
Corollary 8.
Let P α and P α ¯ be two different of C m 2 2 ( Σ n ) then | S P α S P α ¯ | 1 .
Proof. 
Be P β S P α S P α ¯ then there are two positive integers N and M different such that s u p p { α } { N } = s u p p { β } = s u p p { α ¯ } { M } , also s u p p { α } { M } = s u p p { β } { N , M } = s u p p { α ¯ } { N } , as a consequence we have to s u p p { β } = ( s u p p { α } s u p p { α ¯ } )   { N , M } clearly ( s u p p { α } s u p p { α ¯ } ) { M } s u p p { α } then for Lemma 15 we have | ( s u p p { α } s u p p { α ¯ } ) { M } | = m 4 2 + 1 = m 2 2 so we have to s u p p { α } = ( s u p p { α } s u p p { α ¯ } ) { M } , analogously we have to s u p p { α ¯ } = ( s u p p { α } s u p p { α ¯ } ) { N } . Suppose there is another P β S P α S P α ¯ then there exist M y N distinct positive integers such that s u p p { β } = s u p p { α } s u p p { α ¯ } { N , M } , as s u p p { α } s u p p { β } then ( s u p p { α } s u p p { α ¯ } ) { M } ( s u p p { α } s u p p { α ¯ } ) { N , M } then M { N , M } . Analogously, s u p p { α ¯ } s u p p { β } , and so we have to N { N , M } so { N , M } = { N , M } then s u p p { β } = s u p p { β } so by (5) we have | S P α S P α ¯ | 1 .
Corollary 9.
If S P α = S P α ¯ then P α = P α ¯ .
Proof. 
Suppose S P α = S P α ¯ and P α P α ¯ then by Corollary 8 we have to | S P α S P α ¯ | 1 ; however, this is not possible since by Lemma 14, | S P α S P α ¯ | = | S P α | = r m but m 4 and so r m 3 which implies that P α = P α ¯ .
We call w P α the intersection count of the S P α
Lemma 16.
w P α = r m 1
Proof. 
Clearly P β S P α if and only if α C m 2 2 { s u p p { β } } , so the number subsets S P α that contain P β is equal to
| C ( m 2 2 ) { s u p p { β } } | = C ( m 2 ) / 2 m / 2 = m / 2 = r m 1
in consequence we have w P α = r m 1
Let us denote by
L r m
the C m 2 2 m × C m 2 m -incidence matrix from ( C m 2 ( Σ m ) , S P α ) P α C m 2 2 ( Σ m )
Proposition 5.
Let 2 m even positive integer and let r m = m + 2 2 then
(a) 
L r m has r m -ones in each row;
(b) 
L r m has r m 1 -ones in each column;
(c) 
every two lines have at most one 1 in common;
(d) 
L r m is sparse.
Proof. 
The case m = 2 generates the matrix L 2 which trivially satisfies all the statements of this statement. So we assume that m 4 for the rest of the proof.
(a): the 1 s in row P α of L r m display the elements in the subset S P α so by Lemma 14 each row has exactly r m ones in each row.
(b): the 1 s in the column P β display the occurrences of the elements of S P α among the subsets, which follows from Lemma 16.
(c): follows directly from Corollary 8.
(d): The density of ones in the matrix is given by r m × C m 2 2 m = ( r m 1 ) × C m 2 m
r m C m 2 m = r m 1 C m 2 2 m
which approaches zero as m approaches infinity. □
Definition 6.
Let n 2 is an integer and r n = n + 2 2 we define the r n -atlas
A : = { L r n , L r n 1 , , L 2 }
of incidence matrices corresponding to the family
C m 2 ( Σ m ) , S P α P α C m 2 2 ( Σ m ) m = 2 n
of incidence configurations.

7.3. Cartesian Configurations

For n 4 even, r n = n + 2 2 , 1 k r n 2 and 1 a 1 < a 2 < < a 2 k 2 n such that a i + a j 2 n + 1 then we define Σ a 1 , , a 2 k = Σ n { P a 1 , , P a 2 k } as in (8). We define an Cartesian incidence configuration as in (49)
( a 1 , , a 2 k ) × C n 2 k 2 ( Σ a 1 , , a 2 k ) , S ( a 1 , , a 2 k , P α ) P α C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k )
where
{ ( a 1 , , a 2 k ) } × C n 2 k 2 ( Σ a 1 , , a 2 k )
is an C n 2 k 2 n 2 k -set and the subsets are
S ( a 1 , , a 2 k , P α ) : = { ( a 1 , , a 2 k , P β ) : s u p p { α } s u p p { β } }
for all P α C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) .
Lemma 17.
For n 4 even, r n = n + 2 2 , 1 k r n 2 and 1 a 1 < a 2 < < a 2 k 2 n such that a i + a j 2 n + 1 then the incidence matrix of (55) is L r n k a 1 , , a 2 k = L r n k an element of A .
Proof. 
As | Σ a 1 , , a 2 k | = n 2 k if we make m = n 2 k and r m = m + 2 2 , clearly 2 n 2 k n 2 and r m = r n k . A simple calculation shows that 2 r m r n 1 , then renumber the elements of Σ a 1 , , a 2 k = { P 1 , , P m } so by the Lemma 13 the incidence configuration
( a 1 , , a 2 k ) × C n 2 k 2 ( Σ a 1 , , a 2 k ) , S ( a 1 , , a 2 k , P α ) P α C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k )
is isomorphic to
C m 2 ( Σ m ) , S P α P α C m 2 2 ( Σ m )
so also for the Lemma 13 both have the same C n ( 2 k + 1 ) 2 n 2 k × C n 2 k 2 n 2 k incidence matrix L r m = L r n k . We denote this matrix by
L r n k a 1 , , a 2 k = L r n k .
For n 5 odd and r n = ( n + 1 ) / 2 and j { 1 , , n } , we define an Cartesian incidence configuration as in (49)
j × C n 1 2 ( Σ n { P j } ) , S ( j , P α ) P α C n 3 2 ( Σ n { P j } )
where
{ j } × C n 1 2 ( Σ n { P j } )
is a C n 2 n -set and its subsets S ( j , P α ) are defined by
S ( j , P α ) = { ( j , P β ) : s u p p { α } s u p p { β } }
with P α C n 3 2 ( Σ n { P j } ) .
For n 5 odd and r n = n + 1 2 , 1 k r n 2 , consider 1 a 1 < a 2 < < a 2 k + 1 2 n such that a i + a j 2 n + 1 we define Σ a 1 , , a 2 k + 1 = Σ n { P a 1 , , P a 2 k + 1 } , as in (8).
We define a Cartesian incidence configuration as in (49)
( a 1 , , a 2 k + 1 ) × C n ( 2 k + 1 ) 2 ( Σ a 1 , , a 2 k + 1 ) , S ( a 1 , , a 2 k + 1 , P α ) P α C n ( 2 k + 3 ) 2 ( Σ a 1 , a 2 k + 1 )
where
{ ( a 1 , , a 2 k + 1 ) } × C n ( 2 k + 1 ) 2 ( Σ a 1 , , a 2 k + 1 )
is C n ( 2 k + 1 ) 2 n ( 2 k + 1 ) -set where the family of subsets is given by
S ( a 1 , , a 2 k + 1 , P α ) = { ( a 1 , , a 2 k + 1 , P β ) : s u p p { α } s u p p { β } }
for all P α C n 2 ( k + 3 ) 2 ( Σ a 1 , , a 2 k ) .
Lemma 18.
(a) 
The incidence matrix (58) is L r n
(b) 
The C n ( 2 k + 3 ) 2 n ( 2 k + 1 ) × C n ( 2 k + 1 ) 2 n ( 2 k + 1 ) incidence matrix (60) is L r m ( a 1 , , a 2 k + 1 ) = L r n k and is an element of A .
Proof. 
For the proof of (a) If we do m = n 1 and define Σ m = Σ n { P j } so on
j × C n 1 2 ( Σ n { P j } ) , S ( j , P α ) ( j , P α ) C n 3 2 ( Σ n { P j } )
C m 2 ( Σ m ) , S P α P α C m 2 2 ( Σ m )
so both have the same incidence matrix L r n = L r m . Given that r m = m + 2 2 = n + 1 2 = r n , we denote the incidence matrix (58) by
L r n { j } = L r m
and part (a) has been proved.
For the proof of (b), clearly | Σ a 1 , , a 2 k + 1 | = n ( 2 k + 1 ) if we do m = ( n 1 ) + 2 k and r m = m + 2 2 , clearly we have that r m = r n k . Now if we rename the elements of Σ a 1 , , a 2 k + 1 = { P 1 , , P m } then
( a 1 , , a 2 k + 1 ) × C n ( 2 k + 1 ) 2 ( Σ a 1 , , a 2 k + 1 ) , S P α P α C n ( 2 k + 3 ) 2 ( Σ a 1 , a 2 k + 1 )
is isomorphic to the incidence configuration
( C m 2 ( Σ m ) , S P α ) P α C m 2 2 ( Σ m )
then L r m ( a 1 , , a 2 k + 1 ) = L r n k note that 2 m m 3 implies that 2 r m r n 1 . □
For all α r s I ( n 2 , 2 n ) making a change in the notation we rewrite (16) as
Π α r s = i = 1 n c α r s P i X α r s P i
where
c α r s P i = 1 if | s u p p { α r s P i } | = n , 0 otherwise ,
For each α r s I ( n , 2 n ) , consider
S α r s = { α r s P i I ( n 2 , 2 n ) × Σ n : | s u p p { α r s P i } | = n } .
Remark 5.
Note that depending on where α r s I ( n 2 , 2 n ) is, we have S α r s is equal to (51), (56), (59) or (61), respectively.

7.4. Function φ

For n 4 , we consider
φ : I ( n 2 , 2 n ) { 0 , 1 } C n 2 n
( α r s ) ( c β ) β I ( n , 2 n ) ; w h e r e
c β = 1 if β S α r s 0 otherwise .
Clearly
B L ( n , E ) = φ ( I ( n 2 , 2 n ) )
up to permutation of rows.
Lemma 19.
The function φ is injective.
Proof. 
Let α r s , α r s I ( n 2 , 2 n ) such that φ ( α r s ) = φ ( α r s ) then ( c β ) β I ( n , 2 n ) = ( c β ) β I ( n , 2 n ) this implies S α r s = S α r s and by Corollary 9 we have α r s = α r s
Corollary 10.
Let n 4 integer even, r n = n + 2 2 then
(a) 
φ ( C n 2 2 ( Σ n ) ) = L r n
(b) 
φ ( ( a 1 , , a 2 k ) × C n 2 k 2 ( Σ a 1 , , a 2 k ) ) = L r n k ( a 1 , , a 2 k )
Let n 5 odd integer, r n = n + 1 2 then
(c) 
φ ( { j } × C n 2 2 ( Σ n { P j } ) ) = L r n j for all j = 1 , , n
(d) 
φ ( ( a 1 , , a 2 k + 1 ) × C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k + 1 ) ) = L r n k ( a 1 , , a 2 k + 1 )
Proof. 
The proof follows directly from the Lemma 19 and Remark 5. □

7.5. Proof of the Theorem 4

(A) by Lemma 1 we have
I ( n 2 , 2 n ) = C n 2 2 ( Σ n ) ( k = 1 r n 2 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 ( a 1 , , a 2 k ) × C n 2 ( k 1 ) 2 ( Σ a 1 , , a 2 k ) ) .
since φ is injective we have a partition in the image, so
φ ( C n 2 2 ( Σ n ) ) ( k = 1 r n 2 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 φ ( ( a 1 , , a 2 k ) × C n 2 ( k 1 ) 2 ( Σ a 1 , , a 2 k ) ) ) .
Associating the corresponding matrix, using the Corollary 10, we have that
B L ( n , E ) = L r n k = 1 r n 2 ( 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 L r n k ( a 1 , , a 2 k ) ) .
Part (B) Proceeding as in part (A) of this proof, we have
j = 1 n ( { j } × C n 2 ( Σ n ) ) ( k = 1 r n 2 1 a 1 < < a 2 k + 1 2 n a i + a j 2 n + 1 ( a 1 , , a 2 k + 1 ) × C n 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) )
we obtain
φ ( I ( n 2 , 2 n ) ) = j = 1 n φ ( { j } × C n 2 2 ( Σ n ) ) k = 1 r n 2 ( 1 a 1 < a 2 < < a 2 k + 1 2 n a i + a j 2 n + 1 φ ( ( a 1 , , a 2 k + 1 ) × C n ( 2 k + 3 ) 2 ( Σ a 1 , , a 2 k + 1 ) ) ) B L ( n , E ) = j = 1 n L r n j k = 1 r n 2 ( 1 a 1 < < a 2 k + 1 a 2 n a i + a j 2 n + 1 L r n k ( a 1 , a 2 , , a 2 k + 1 ) ) .
Example 5.
Let E be a symplectic vector space of dimension 12 and r 6 = 6 + 2 2 = 4 then his r 6 -atlas is given by the triplet of matrices
A = { L 4 , L 3 , L 2 }
where
Mathematics 12 00858 i003
and
L 3 = 1 1 1 0 0 0 1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 .
L 2 is a row matrix filled with zeros and with only two entries equal to 1.
The matrix B L ( 6 , E ) , of size 12 4 × 12 6 , associated to the homogeneous system Π = { Π α r s : α r s I ( 4 , 12 ) } can be given by a block diagonal matrix as follows
B L ( 6 , E ) = L 4 ( 1 α 1 < α 2 12 α 1 + α 2 13 L 3 ( α 1 , α 2 ) ) ( 1 α 1 < α 2 < α 3 < α 4 12 α i + α j 13 L 2 ( α 1 , α 2 , α 3 , α 4 ) ) ,
Mathematics 12 00858 i004
where there are 1 matrix L 4 , 60 submatrices L 3 , and 240 submatrices L 2 .
Example 6.
Let E be a symplectic vector space of dimension 14 and r 7 = 7 + 2 2 = 4 then his 4-atlas is given by
A = { L 4 , L 3 , L 2 }
B L ( 7 , E ) = i = 1 7 L 4 { i } ( 1 a 1 < a 2 < a 3 14 a i + a j 15 L 3 ( a 1 a 2 a 3 ) ) ( 1 a 1 < a 2 < < a 4 < a 5 14 a i + a j 15 L 2 ( a 1 a 2 a 3 a 4 a 5 ) ) ,
where L 4 { i } = L 4 , L 3 ( a 1 a 2 a 3 ) = L 3 and L 2 ( a 1 a 2 a 3 a 4 a 5 ) = L 2 .
Mathematics 12 00858 i005

8. Isotropy Index r n and r n -Atlas

Definition 7.
Let E symplectic vector space of dimension 2 n ; we call r n = n + 2 2 isotropy index of E.
Let B L ( n , E ) as Equation (47) and using notation as in (13) and [ ] α I ( n , 2 n ) T denotes the transposed vector.
Lemma 20.
Let f be the contraction map, then f ( w ) ρ = B L ( n , E ) ( w ρ ) T .
Proof. 
Let w = α I ( n , 2 n ) X α e α n E and ρ Plücker embedding we denote w ρ = [ X α ] α I ( n , 2 n ) , it as in (13), then by Lemma 2 and by Equation (68) the following diagram commutes i.e., ρ f = B L ( n , E ) ρ
Mathematics 12 00858 i006
so we have
( ρ 2 f ) ( w ) = ρ 2 ( α r s I ( n 2 , 2 n ) ( i = 1 n X ( α r s , P i ) ) e α r s ) = i = 1 n X ( α r s , P i ) α r s I ( n 2 , 2 n ) = B L ( n , E ) [ X α ] α I ( n , 2 n ) T = B L ( n , E ) ( w ρ 1 ) T
which proves commutativity so ρ 2 f = B L ( n , E ) ρ 1 .
Corollary 11.
ker f is isomorphic to ker B L ( n , E ) as vector spaces.
Proof. 
From the Lemma 20 we have ρ 1 ( ker f ) ker B L ( n , E ) , and both have the same dimension so ker f ker B L ( n , E ) is an isomorphism of vector spaces. □
The following corollary follows directly from the Lemma 20.
Corollary 12.
Let E symplectic vector spaces of dimension 2 n then
dim ker f = C n 2 n r a n k B L ( n , E )
Moreover, C n 2 n C n 2 2 n dim ker f C n 2 n 1 .
Theorem 5.
Let E symplectic vector space of dimension 2 n defined over an arbitrary field F and r n = n + 2 2 the isotropy index.
Then, the following are equivalent:
(a) C h a r F = 0 or C h a r F r n
(b) r a n k B L ( n , E ) = C n 2 2 n
(c) dim ker f = C n 2 n C n 2 2 n
(d) dim H = C n 2 2 n and { Π α r s : α r s I ( n 2 , 2 n ) } is a base of H .
Proof. 
(a) and (b) are equivalent by [27] (Theorem 6).
(b) and (c) are equivalent by Corollary 11 and by Corollary 12.
(c) and (d) are equivalent by Theorem 3. □
We say that the embedding rank e r ( L ( n , E ) ) of L ( n , E ) is the dimension of the linear envelope L ( n , E ) .
Lemma 21.
Embedding rank of L ( n , E ) is e r ( L ( n , E ) ) = C n 2 n r a n k B L ( n , E ) .
Proof. 
We have that
e r ( L ( n , E ) ) = dim L ( n , E ) = dim k e r f = C n 2 n r a n k B L ( n , E )
Corollary 13.
Let E a symplectic vector space of dimension 2 n , let f the contraction map the isotropy index r n = n + 2 2 are equivalent
(1) 
f is surjective
(2) 
e r ( L ( n , E ) ) = C n 2 n C n 2 2 n
(3) 
r a n k B L ( n , E ) = C n 2 2 n
(4) 
char F = 0 or char F r n
(5) 
r a n k L r n k is maximum for everything 0 k r n 2 .
Proof. 
(1) is equivalent (2), (2) is equivalent (3) given that dim ker f = dim ker B y dim ker f = 2 n n r a n k B .
(3) is equivalent (4) is followed from [27] (Theorem 6) and finally (3) is equivalent (5) is obvious. □
Example 7.
By Corollary 12 dim ker f = 70 r a n k B L ( 4 , 8 ) , the order of B L ( 4 , 8 ) is 28 × 70 and by Theorem 4, L 3 is a submatrix of B L ( 4 , 8 ) and it is also easy to see that
L 3 = 1 1 1 0 0 0 1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 0 2 2 2 .
and
r a n k B L ( 4 , 8 ) = 28 if char F 3 27 if char F = 2
Example 8.
Consider the matrices L 4 and L 3 in a field F 2 . By elementary matrix operations we have
L 4 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
This matrix we denote by ( i d | A 3 3 ) .
As we saw in the Example 7, L 3 in F 2 it is of the form
L 3 1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 0 0 0 0
and we denote by ( I d | A 2 2 ) .
In [18] (Corollary 1.2) can find a more general case of Theorem 5 also see [28] for some examples where n = 2 , 3 , 4 , 5 , 6 , and 7.
Let E be a symplectic vector space of dimension 2 n and let r n = n + 2 2 consider the family of matrices given in (53)
A = { L r n , L r n 1 , , L 2 }
and we call the r n -atlas of L ( n , E ) .
Lemma 22.
Let n integer and let E and E ¯ symplectic vector spaces of dimension 2 n and 2 m , respectively, then
r n = r m i f a n d o n l y i f n = m o r m = n + 1 .
Proof. 
) If both m and n are even integers or if both m and n odd integers then n = m .
Suppose that n is even integer and m = 2 k + 1 is odd integer. If r n = r m then n + 2 2 = 2 k + 1 + 2 2 so n + 2 2 = 2 k + 2 2 what it implies n = 2 k so m = n + 1 .
) If n = m then r n = r m . Now if m = n + 1 then r m = m + 2 2 = n + 1 + 2 2 = n + 2 2 = r n .
Corollary 14.
(a) If E and E ¯ both are symplectic vector spaces of dimension 2 n so they share the same r n -atlas.
(b) Let E symplectic vector space of dimension 2 n and let E ¯ symplectic vector space of dimension 2 ( n + 1 ) then both spaces share the same r n -atlas.
Proof. 
The proof follows directly from the Lemma 22. □
Example 9.
Let E 1 and E 2 two symplectic vector spaces of dimension 2 n defined over a field F and isotropy index r n = n + 2 2 . If char F =0 or char F r then E 1 and E 2 they share
(a) the same isotropy index r n ;
(b) the same r n -atlas { L r n , , L 2 } ;
(c) the same Plücker relations { Q α , β , Π α r s } of the Lagrangian Grassmannian variety.
Example 10.
Consider the symplectic vector spaces R 6 ( R 6 ) * and F 2 6 ( F 2 6 ) * , two symplectic vector space non-symplectomorphisms, them
(a) They share the same isotropy index r 6 = 6 + 2 2 = 4 .
(b) They share the same Plücker relations { Q α , β , Π α r s } of the Lagrangian Grassmannian variety.
However, they do not share the same 4-atlas so we have:
the 4-atlas of R 6 ( R 6 ) * is { L 4 , L 3 , L 2 } see Example 5
but the 4-atlas of F 2 6 ( F 2 6 ) * is { ( i d | A 3 3 ) , ( i d | A 2 2 } , L 2 } see Example 8.

Hypersurfaces in L ( n , E )

The linear sections of the Lagrangian-Grassmannian L ( n , E ) have applications in other fields of mathematics see [14]. Using the notation α r s : = Z Π α r s P ( n E ) we define the following linear varieties.
Definition 8.
Let n 4 even integer then
(a) 
: = α r s C n 2 2 ( Σ n ) α r s
(b) 
C n 2 ( k + 1 ) 2 ( a 1 , , a 2 k ) : = α r s C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k ) α r s
Let n 5 odd integer then
(c) 
n : = α r s j = 1 n { j } × C n 2 2 ( Σ n ) α r s
(d) 
C n 2 ( k + 1 ) 2 ( a 1 , , a 2 k + 1 ) : = α r s C n 2 ( k + 1 ) 2 ( Σ a 1 , , a 2 k + 1 ) α r s
Theorem 6.
Let E be a symplectic vector space of dimension 2 n then L ( n , E ) is intersection of linear sections of the Grassmannian variety and is included in a projective space of a direct sum of matrix kernels
(A) 
If n 4 even integer and let r n = n + 2 2 , then
L ( n , E ) = ( G ( n , E ) ) ( k = 1 r n 2 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 ( G ( n , E ) C n 2 k 2 ( a 1 , , a 2 k ) ) ) P ( ker L r n k = 1 r n 2 ( 1 a 1 < < a 2 k 2 n a i + a j 2 n + 1 ker L r n k ( a 1 , , a 2 k ) ) )
(B) 
If n 5 odd integer and let r n = n + 1 2 , then
L ( n , E ) = ( G ( n , E ) n ) ( k = 1 r n 2 1 a 1 < < a 2 k + 1 2 n a i + a j 2 n + 1 ( G ( n , E ) C n 2 k 2 ( a 1 , , a 2 k + 1 ) ) ) P ( ker L r n n k = 1 r n 2 ( 1 a 1 < a 2 < < a 2 k + 1 2 n a i + a j 2 n + 1 ker L r n k ( a 1 , a 2 , , a 2 k + 1 ) ) ) .

Funding

The APC was funded by research project 2024 del Colegio de Ciencia y Tecnología of the Universidad Autónoma de la Ciudad de México.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The author does not declare a conflict of interest.

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Carrillo-Pacheco J. On Lagrangian Grassmannian Variety and Plücker Matrices. Mathematics. 2024; 12(6):858. https://doi.org/10.3390/math12060858

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Carrillo-Pacheco, Jesús. 2024. "On Lagrangian Grassmannian Variety and Plücker Matrices" Mathematics 12, no. 6: 858. https://doi.org/10.3390/math12060858

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