**THE CONES ASSOCIATED TO SOME** **TRANSVERSAL POLYMATROIDS**

**Alin S¸tefan**

**Abstract**

In this paper we describe the facets cone associated to transversal
polymatroid presented by*A*=*{{*1*,*2*},{*2*,*3*}, . . . ,{n−*1*, n},{n,*1*}}.*Us-
ing the Danilov-Stanley theorem to characterize the canonicale module,
we deduce that the base ring associated to this polymatroid is Gorenstein
ring. Also, starting from this polymatroid we describe the transversal
polymatroids with Gorenstein base ring in dimension 3 and with the
help*Normaliz* in dimension 4.

**1** **Preliminaries on polyhedral geometry**

An aﬃne space generated by *A⊂*R* ^{n}* is a translation of a linear subspace of
R

^{n}*.*If 0

*=*

*a*

*∈*R

^{n}*,*then

*H*

*will denote the hyperplane of R*

_{a}*through the origin with normal vector*

^{n}*a, that is,*

*H** _{a}*=

*{x∈*R

^{n}*|*

*< x, a >= 0},*

where *<, >* is the usual inner product in R^{n}*.* The two closed half spaces
bounded by*H** _{a}* are:

*H*_{a}^{+}=*{x∈*R^{n}*|* *< x, a >≥*0}*and H*_{a}* ^{−}*=

*{x∈*R

^{n}*|*

*< x, a >≤*0}.

Recall that a*polyhedral cone Q⊂*R* ^{n}* is the intersection of a ﬁnite number
of closed subspaces of the form

*H*

_{a}^{+}

*.*If

*A*=

*{γ*1

*, . . . , γ*

_{r}*}*is a ﬁnite set of points in R

*the*

^{n}*cone*generated by

*A, denoted by*

**R**+

*A,*is deﬁned as

**R**+*A*=*{*^{r}

*i=1*

*a*_{i}*γ*_{i}*|* *a*_{i}*∈*R_{+}*, with*1*≤i≤n}.*

Key Words: Transversal polymatroids, Danilov-Stanley theorem 2000 Mathematical Subject Classiﬁcation: 05C50,13A30,13H10,13D40 Received: March, 2007

139

An important fact is that*Q*is a polyhedral cone inR* ^{n}* if and only if there
exists a ﬁnite set

*A⊂*R

*such that*

^{n}*Q*=

**R**+

*A,*see ([15],theorem 4.1.1.).

**Definition 1.1.** A proper face of a polyhedral cone is a subset*F* *⊂* *Q*such
that there is a supporting hyperplane*H** _{a}* satisfying:

1) *F* =*Q∩H*_{a}*=∅;*

2) *QH** _{a}* and

*Q⊂H*

_{a}^{+}

*.*

**Definition 1.2.** A proper face *F* of a polyhedral cone*Q* *⊂* R* ^{n}* is called a

*f acet*of

*Q*if

*dim(F*) =

*dim(Q)−*1.

**2** **Polymatroids**

Let*K* be an inﬁnite ﬁeld,*n*and*m*be positive integers, [n] =*{1,*2, . . . , n}. A
nonempty ﬁnite set*B* of**N*** ^{n}* is the base set of a discrete polymatroid

*P*if, for all

*u*= (u

_{1}

*, u*

_{2}

*, . . . , u*

*),*

_{n}*v*= (v

_{1}

*, v*

_{2}

*, . . . , v*

*)*

_{n}*∈B,*one has

*u*

_{1}+

*u*

_{2}+

*. . .*+

*u*

*=*

_{n}*v*

_{1}+

*v*

_{2}+

*. . .*+

*v*

*and, for all*

_{n}*i*such that

*u*

_{i}*> v*

_{i}*,*there exists

*j*such that

*u*

_{j}*< v*

*and*

_{j}*u*+

*e*

_{j}*−e*

_{i}*∈B, wheree*

*denotes the*

_{k}*k*

*vector of the standard basis of*

^{th}**N**

*. The notion of discrete polymatroid is a generalization of the classical notion of matroid, see [6] [9] [8] [16]. Associated with the base*

^{n}*B*of a discret polymatroid

*P*one has a

*K−algebraK[B*] - called the base ring of

*P*- deﬁned to be the

*K−subalgebra of the polynomial ring in*

*n*indeterminates

*K[x*

_{1}

*, x*

_{2}

*, . . . , x*

*] generated by the monomials*

_{n}*x*

*with*

^{u}*u∈B. From [9], the*algebra

*K[B] is known to be normal and hence Cohen-Macaulay.*

If*A** _{i}*are some non-empty subsets of [n],for 1

*≤i≤m,A*=

*{A*

_{1}

*, A*

_{2}

*, . . . , A*

_{m}*}*, then the set of the vectors

_{m}*k=1**e*_{i}* _{k}*with

*i*

_{k}*∈A*

_{k}*,*is the base of a polymatroid, called transversal polymatroid presented by

*A.*The base ring of a transversal polymatroid presented by

*A*denoted by

*K[A] is the ring :*

*K[A] :=K[x*_{i}_{1}*x*_{i}_{2}*. . . x*_{i}* _{m}* :

*i*

_{j}*∈A*

_{j}*,*1

*≤j≤m].*

**3** **Some Linear Algebra**

Let*n∈***N**be an integer number,*n≥*3 and let be given the following set with
2n*−*3 points with positive integer coordinates :

*{* *R*_{0,1}(2,1,1, . . . ,1,1,0), R_{0,2}(2,1,1, . . . ,1,0,1), . . . , R_{0,n−2}(2,1,0, . . . ,1,1,1),
*R*_{0,n−1}(2,0,1, . . . ,1,1,1), Q_{0,1}(1,2,1,1, . . . ,1,1,0), Q_{0,2}(1,1,2,1, . . . ,1,1,0),

*. . . , Q*_{0,n−3}(1,1,1,1, . . . ,2,1,0), Q_{0,n−2}(1,1,1,1, . . . ,1,2,0)} ⊂**N**^{n}*.*

We shall denote by*A*_{1}*∈* *M** _{n−1,n}*(R) the matrix with rows the coordinates
of points

*{*

*R*

_{0,1}

*, R*

_{0,2}

*, . . . , R*

_{0,n−1}

*}*and for 2

*≤*

*i≤*

*n−*1,

*A*

_{i}*∈*

*M*

*(R) the matrix with rows the coordinates of the points*

_{n−1,n}*{R*_{0,1}*, . . . , R*_{0,n−i}*, Q*_{0,1}*, Q*_{0,2}*, . . . , Q*_{0,i−1}*},*
that is:

*A*_{1}=

⎛

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎝

2 1 1 1 *. . .* *. . .* 1 1 0
2 1 1 1 *. . .* *. . .* 1 0 1
2 1 1 1 *. . .* *. . .* 0 1 1

*.* *.* *.* *.* *.* *.* *.* *.* *.*

2 1 1 0 *. . .* *. . .* 1 1 1
2 1 0 1 *. . .* *. . .* 1 1 1
2 0 1 1 *. . .* *. . .* 1 1 1

⎞

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎠

and, for 2*≤* *i≤* *n−*1,

*↓*(n*−i)*^{th}*column*

*A** _{i}* =

⎛

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎝

2 1 1 *. . .* 1 1 1 *. . .* 1 1 0
2 1 1 *. . .* 1 1 1 *. . .* 1 0 1
2 1 1 *. . .* 1 1 1 *. . .* 0 1 1
*.* *.* *.* *. . .* *.* *.* *. . . . .* *.* *.*
2 1 1 *. . .* 1 1 0 *. . .* 1 1 1
1 2 1 *. . .* 1 1 1 *. . .* 1 1 0
1 1 2 *. . .* 1 1 1 *. . .* 1 1 0
*.* *.* *.* *. . .* *.* *.* *. . . . .* *.* *.*
1 1 1 *. . .* 2 1 1 *. . .* 1 1 0
1 1 1 *. . .* 1 2 1 *. . .* 1 1 0

⎞

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎠

*←* (n*−i*+ 1)^{th}*row* *.*

Let*T** _{i}*be the linear transformation fromR

*intoR*

^{n}*deﬁned by*

^{n−1}*T*

*(x) =*

_{i}*A*

_{i}*x*for all 1

*≤*

*i≤*

*n−*1.

Let*i, j∈***N**, 1*≤i, j≤n.* We denote by*e** _{i,j}* the matrix in

*M*

*(R) with the entries: 1, for the (i, j)-entry, and 0 for the other entries. We denote by*

_{n−1}*T*

*(a) the matrix*

_{i,j}*T** _{i,j}*(a) =

*I*

*+*

_{n−1}*ae*

_{i,j}*∈M*

*(R).*

_{n−1}By*P** _{i,j}* we denote the matrix in

*M*

*(R) deﬁned by*

_{n−1}*P*

*=*

_{i,j}*I*

_{n−1}*−e*

_{i,i}*−e*

*+*

_{j,j}*e*

*+*

_{i,j}*e*

_{j,i}*.*

**Lemma 3.1.** *a)The set of points* *{* *R*_{0,1}*, . . . , R*_{0,n−i}*, Q*_{0,1}*, Q*_{0,2}*, . . . , Q*_{0,i−1}*},*
*for* 2*≤* *i≤* *n−*1 *and{* *R*_{0,1}*, R*_{0,2}*, . . . , R*_{0,n−1}*}* *are linearly independent.*

*b)For*1*≤* *i≤* *n−*1,*the equations of the hyperplanes generated by the points*
*{* *O, R*_{0,1}*, R*_{0,2}*. . . , R*_{0,n−i}*, Q*_{0,1}*, Q*_{0,2}*, . . . , Q*_{0,i−1}*}* *are :*

*H*_{[i]}:=*−(n−i−*1)
*i*
*j=1*

*x** _{j}*+ (i+ 1)

*n*

*j=i+1*

*x** _{j}* = 0,

*where*[i]*is the set*[i] :=*{1, . . . , i}.*

*Proof.* *a) The set of points are linearly independent if the matrices with rows*
the coordinates of the points have the rank*n−*1.

Using elementary row transformations on the matrix*A*_{1}*,*we have:

*B*_{1}=*U*_{1}*A*_{1}*,*where*U*_{1}*∈* *M** _{n−1}*(R)is given by:

*U*_{1}=

2≤i≤^{n}_{2}

*P*_{i,n−i+1}

*n−1*
*i=2*

*T** _{n−i+1,1}*(−1),

and*c*is the greatest integer*≤c.*So*B*_{1} is :

*B*_{1}=

⎛

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎝

2 1 1 1 *. . .* *. . .* 1 1 0

0 *−1* 0 0 *. . .* *. . .* 0 0 1

0 0 *−1* 0 *. . .* *. . .* 0 0 1

*.* *.* *.* *.* *.* *.* *.* *.* *.*

0 0 0 0 *. . .* *. . .* 0 0 1

0 0 0 0 *. . .* *. . .* *−1* 0 1

0 0 0 0 *. . .* *. . .* 0 *−1* 1

⎞

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎠
*.*

For 2*≤* *i≤* *n−*1,using elementary row transformations on the matrix
*A*_{i}*,*we have: *B** _{i}*=

*U*

_{i}*A*

_{i}*,*where

*U*

_{i}*∈*

*M*

*(R),*

_{n−1}*U** _{i}*= [

*n−2*
*j=i*

(

*i−1*
*k=1*

)P*n−j+k−1,n−j+k*][

*i−1*
*k=2*

(

*n−1*
*j=n−i+k*

*T** _{j,n−i+k−1}*(

*−*1

*k*+ 1))]

*·*

*·(* ^{n−1}

*j=n−i+1*

*T** _{j,1}*(−1
2))(

*n−i*
*j=1*

*T** _{j,1}*(−1)),
and so

*B*

*is :*

_{i}*↓*(i+ 1)^{th}*column*

*B** _{i}*=

⎛

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎝

2 1 1 1 *. . .* 1 1 1 *. . .* 1 1 0

0 ^{3}_{2} ^{1}_{2} ^{1}_{2} *. . .* ^{1}_{2} ^{1}_{2} ^{1}_{2} *. . .* ^{1}_{2} ^{1}_{2} 0
0 0 ^{4}_{3} ^{1}_{3} *. . .* ^{1}_{3} ^{1}_{3} ^{1}_{3} *. . .* ^{1}_{3} ^{1}_{3} 0

*.* *.* *.* *.* *. . .* *.* *.* *. . .* *.* *.*

0 0 0 0 *. . .* _{i−1}^{i}_{i−1}^{1} _{i−1}^{1} *. . .* _{i−1}^{1} _{i−1}^{1} 0
0 0 0 0 *. . .* 0 ^{i+1}_{i}^{1}_{i}*. . .* ^{1}_{i}^{1}* _{i}* 0

0 0 0 0 *. . .* 0 0 *−1* *. . .* 0 0 1

*.* *.* *.* *.* *. . .* *.* *.* *. . . . .* *.* *.*

0 0 0 0 *. . .* 0 0 0 *. . .−*1 0 1

0 0 0 0 *. . .* 0 0 0 *. . .* 0 *−1* 1

⎞

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎠

*←* *i*^{th}*row* *.*

Since the rank of*B** _{i}* is

*n−*1,the rank of

*A*

*is*

_{i}*n−*1,for all 1

*≤*

*i≤*

*n−*1.

*b) The hyperplane generated by the points*

*{* *R*_{0,1}*, . . . , R*_{0,n−i}*, Q*_{0,1}*, Q*_{0,2}*, . . . , Q*_{0,i−1}*}*
has the normal vector the generator of the subspace *Ker(T** _{i}*).

For 1*≤* *i≤* *n−*1,using*a),*we obtain that

*Ker(T** _{i}*) =

*{x∈*R

^{n}*|T*

*(x) = 0}=*

_{i}*{x∈*R

^{n}*|A*

_{i}*x*= 0}=

*{x∈*R

^{n}*|B*

_{i}*x*= 0}, that is

*x** _{n}*=

*x*

*=*

_{n−1}*. . .*=

*x*

*= (i+ 1)α and*

_{i+1}*x** _{i}*=

*x*

*=*

_{i−1}*. . .*=

*x*

_{1}=

*−*(n

*−i−*1)α, where

*α∈*R.

Thus, for 1*≤* *i≤* *n−*1,the equations of the hyperplanes generated by
the points*{R*_{0,1}*, . . . , R*_{0,n−i}*, Q*_{0,1}*, Q*_{0,2}*, . . . , Q*_{0,i−1}*}*are :

*H*_{[i]}:=*−(n−i−*1)
*i*
*j=1*

*x** _{j}*+ (i+ 1)

*n*

*j=i+1*

*x** _{j}*= 0.

For 1*≤k≤n−*1,we deﬁne two types of sets of points:

1)

*{R*_{k,1}*, R*_{k,2}*, . . . , R*_{k,n−1}*}*

is the set of points whose coordinates are the rows of the matrix*A*_{1}*P*_{1k+1};
2)

*{* *Q*_{k,1}*, Q*_{k,2}*, . . . , Q*_{k,n−2}*}*

is the set of points whose coordinates are the rows of the matrix*QM** ^{k}*, where

*M*is the matrix

*M* *∈* *M** _{n}*(R), M =

*n−1*
*i=1*

*P*_{n−i,n−i+1}

and *Q* *∈* *M** _{n−2,n}*(R) is the matrix with rows the coordinates of points

*{Q*

_{1}

*, Q*

_{2}

*, . . . , Q*

_{n−2}*}*.

For every 1*≤i≤n−1,*we shall denote by*ν*_{[i]}the normal of the hyperplane
*H*_{[i]}:

*↓i*^{th}*column*
*ν*_{[i]} =

*−(n−i−*1), *. . .* *,−(n−i−*1), (i+ 1), *. . .* *,*(i+ 1)

*∈*R^{n}*.*
For*i*= 1,we denote by*H** _{{k+1}}* the hyperplane having the normal :

*ν** _{{k+1}}*:=

*ν*

_{[i]}

*P*

_{1,k+1}=

*ν*

_{[1]}

*P*

_{1,k+1}

*,*for all 1

*≤k≤n−*1.

For 2*≤i≤n−*1 and 1*≤k≤n−*1,we denote by *H*_{{σ}*k*(1),σ* ^{k}*(2),...,σ

*(i)}*

^{k}the hyperplane which has the normal :

*ν*_{{σ}*k*(1),σ* ^{k}*(2),...,σ

*(i)}:=*

^{k}*ν*

_{[i]}

*M*

^{k}*,*where

*σ∈S*

*is the product of transpositions :*

_{n}*σ*:=

*n−1*
*i=1*

(i, i+ 1).

**Lemma 3.2.** *a)* *For* 1 *≤k* *≤* *n−*1 *and* 2 *≤* *i* *≤* *n−*1, *the set of points*
*{* *R*_{k,1}*, . . . , R*_{k,n−i}*, Q*_{k,1}*, Q*_{k,2}*, . . . , Q*_{k,i−1}*}* *and* *{* *R*_{k,1}*, R*_{k,2}*, . . . , R*_{k,n−1}*}* *are*
*linearly independent.*

*b)* *For* 1 *≤* *k* *≤* *n−*1 *and* 2 *≤* *i* *≤* *n−*1, *the equation of the hyperplane*
*generated by the points{* *O, R*_{k,1}*, R*_{k,2}*. . . , R*_{k,n−i}*, Q*_{k,1}*, Q*_{k,2}*, . . . , Q*_{k,i−1}*}* *is :*

*H*_{{σ}*k*(1),σ* ^{k}*(2),...,σ

*(i)}:=< ν*

^{k}

_{{σ}*k*(1),σ

*(2),...,σ*

^{k}*(i)}*

^{k}*, x >= 0,*

*where* *O* *is zero point,* *O(0,*0, . . . ,0) *and* *σ∈* *S*_{n}*is the product of transposi-*
*tions:*

*σ*:=

*n−1*
*i=1*

(i, i+ 1).

*For*1*≤k≤n−*1,*the equation of the hyperplane generated by the points*
*{* *O, R*_{k,1}*, R*_{k,2}*. . . , R*_{k,n−1}*}is*

*H** _{{k+1}}*:=< ν

_{{k+1}}*, x >= 0.*

*Proof.* *a) Since, for any 1≤k≤n−*1,the matrix *P*_{1,k+1} , *M** ^{k}* are invertible
and the sets of points

*{* *R*_{0,1}*, . . . , R*_{0,n−i}*, Q*_{0,1}*, Q*_{0,2}*, . . . , Q*_{0,i−1}*},{* *R*_{0,1}*, R*_{0,2}*, . . . , R*_{0,n−1}*}*
are linearly independent then the set of points

*{R*_{k,1}*, . . . , R*_{k,n−i}*, Q*_{k,1}*, Q*_{k,2}*, . . . , Q*_{k,i−1}*},{R*_{k,1}*, R*_{k,2}*, . . . , R*_{k,n−1}*}*
are linearly independent.

*b) Since, for any 1* *≤k* *≤n−*1 and 2*≤* *i* *≤* *n−*1, the matrix*M** ^{k}* are
invertible, then the hyperplane generated by the points

*{O, R*_{k,1}*, . . . , R*_{k,n−i}*, Q*_{k,1}*, . . . , Q*_{k,i−1}*}*

has the normal vector obtained by multiplying the normal vector*ν*_{[k]} on the
right with*M** ^{k}*. For any 1

*≤k*

*≤n−*1,the matrix

*P*

_{1,k+1}is invertible, then the hyperplane generated by the points

*{*

*O, R*

_{k,1}*, R*

_{k,2}*. . . , R*

_{k,n−1}*}*has the normal vector obtained by multiplying on the right the normal vector

*ν*

_{[1]}

with*P*_{1,k+1}.

**Lemma 3.3.** *Any point* *P* *∈* **N**^{n}*,* *n* *≥* 3 *which lies in the hyperplane* *H* :
*x*_{1}+*x*_{2}+*. . .*+*x*_{n}*−n*= 0 *such that its coordinates are in the set{0,*1,2} *and*
*has at least one coordinate equal to* 2 *lies in the hyperplaneH** _{{k}}* = 0,

*for an*

*integer*

*k∈ {1,*2, ..., n}.

*Proof.* Let *k* *∈ {1,*2, ..., n} be the ﬁrst position of ”2” that appears in the
coordinates of a point*P* *∈***N**** ^{n}**. Since the equation of the hyperplane

*H*

*is:*

_{{k}}*H** _{{k}}*=

*k−1*

*i=1*

2x_{i}*−*(n*−*2)x* _{k}*+

*n*

*i=k+1*

2x* _{i}*= 0,
it results that

*−2(n−*2) + 2

*i=1,i=k*

*na** _{i}*=

*−2(n−*2) + 2(n

*−*2) = 0,

where *P* = (a_{1}*, a*_{2}*, . . . , a** _{n}*)

*∈H*with

*a*

_{i}*∈ {0,*1,2}and which has at least one coordinate equal to 2.

**4** **The main result**

First let us ﬁx some notations that will be used throughout the remaining of
this paper. Let*K* be a ﬁeld and *K[x*_{1}*, x*_{2}*, . . . , x** _{n}*] be a polynomial ring with
coeﬃcients in

*K. Let*

*n≥*2 be a positive integer and

*A*be the collection of sets:

*A*=*{{1,*2},*{2,*3}, . . . ,*{n−*1, n},*{n,*1}}.

We denote by*K[A*] the*K−*algebra generated by*x*_{i}_{1}*x*_{i}_{2}*...x*_{i}_{n}*,*with
*i*_{1}*∈ {1,*2}, i2*∈ {2,*3}, . . . , i*n−1**∈ {n−*1, n}, i*n**∈ {1, n}.*

This*K−algebra represents the base ring associated to transversal polymatroid*
presented by*A.*

Given *A* *∈* **N**** ^{n}** ﬁnite, we deﬁne

*C*

*as being the subsemigroup of*

_{A}**N**

*generated by*

^{n}*A*:

*C** _{A}*=

*α∈A*

**N***α,*
thus the*cone*generated by*C** _{A}* is:

**R**+*C** _{A}*=

**R**+

*A*=

*{*

*a*_{i}*γ*_{i}*|* *a*_{i}*∈*R+*, γ*_{i}*∈A}.*

With this notation, we state our main result:

**Theorem 4.1.** *Let* *A*=*{log(x*_{i}_{1}*x*_{i}_{2}*...x*_{i}* _{n}*)

*|i*

_{1}

*∈ {1,*2}, i

_{2}

*∈ {2,*3}, . . . , i

_{n−1}*∈*

*{n−*1, n}, i

_{n}*∈ {1, n}} ⊂*

**N**

^{n}*the exponent set of the generators ofK−algebra*

*K[A]*

*andN*=

*{ν*

_{{k+1}}*, ν*

_{{σ}*k*(1),σ

*(2),...,σ*

^{k}*(i)}*

^{k}*|*0

*≤k≤n−*1,2

*≤i≤n−*1},

*then*

**R**+*C** _{A}*=

*a∈N*

*H*_{a}^{+}*,*

*such thatH*_{a}^{+} *with* *a∈N* *are just the facets of the cone* **R**+*C*_{A}*.*

*Proof.* Since *A* = *{log(x*_{i}_{1}*x*_{i}_{2}*...x*_{i}* _{n}*)

*|*

*i*

_{1}

*∈ {1,*2}, i

_{2}

*∈ {2,*3}

*, . . . , i*

_{n−1}*∈*

*{n−*1, n}, i

_{n}*∈ {1, n}} ⊂*

**N**

**is the exponent set of the generators of**

^{n}*K−algebraK[A], then the set{R*

_{0,1}

*, R*

_{0,2}

*, . . . , R*

_{0,n−2}

*, R*

_{0,n−1}

*, I} ⊂A, where*

*I(1,*1, . . . ,1)

*∈*

**N**

^{n}*.*

**First step.**

We must show that the dimension of the cone**R**+*C** _{A}* is

*dim(*

**R**+

*C*

*) =*

_{A}*n.*

We denote by*A∈M** _{n}*(R) the matrix with rows the coordinates of the points

*{* *R*_{0,1}*, R*_{0,2}*, . . . , R*_{0,n−2}*, R*_{0,n−1}*, I}.*Using elementary row transformations to
the matrix *A,* we have: *B* =*UA, where* *U* *∈* *M** _{n}*(R) is an invertible matrix:

*U* = (

*n−1*
*i=2*

*T** _{n−i+1,1}*(−1)))(T

*n,1*(−1 2))(

2≤i≤^{n}_{2}

*P** _{i,n−i+1}*)(

*n−1*
*i=2*

*T** _{n,n−i+1}*(1
2)),
where

*c*is the greatest integer

*≤c.*

So*B* is:

*B*=

⎛

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎝

2 1 1 1 *. . .* *. . .* 1 1 0

0 *−1* 0 0 *. . .* *. . .* 0 0 1

0 0 *−1* 0 *. . .* *. . .* 0 0 1

*.* *.* *.* *.* *.* *.* *.* *.* *.*

0 0 0 0 *. . .* *. . .* 0 0 1

0 0 0 0 *. . .* *. . .* *−1* 0 1

0 0 0 0 *. . .* *. . .* 0 *−1* 1

0 0 0 0 *. . .* *. . .* 0 0 ^{n}_{2}

⎞

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎠
*.*

Then the dimension of the cone**R**+*C** _{A}* is:

*dim(***R**+*C** _{A}*) =

*rank(A) =*

*rank(B) =*

*n,*since

*det(B) = (*

*−*1)

^{n}*n.*

**Second step.**

We must show that *H*_{a}*∩***R**+*C** _{A}* with

*a*

*∈*

*N*are precisely the facets of the cone

**R**+

*C*

_{A}*.*This is equivalent to show that

**R**+

*C*

_{A}*⊂H*

_{a}^{+}and

*dimH*

_{a}*∩*

**R**+

*C*

*=*

_{A}*n−*1

*∀a∈N.*

The fact that *dimH*_{a}*∩***R**+*C** _{A}* =

*n−*1

*∀a∈N*it is clear, from Lemma 3.1 and Lemma 3.2.

For 1*≤k≤ *^{n}_{2}and 1*≤i*_{1}*< i*_{2}*< . . . < i*_{2k−1}*< i*_{2k}*≤n,*let
*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} =*I*+ (e_{i}_{1}*−e*_{i}_{2}) + (e_{i}_{3}*−e*_{i}_{4}) +*. . .*+ (e_{i}_{2k−1}*−e*_{i}_{2k})
and

*I*_{i}^{}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} =*I*+ (e_{i}_{2}*−e*_{i}_{1}) + (e_{i}_{4}*−e*_{i}_{3}) +*. . .*+ (e_{i}_{2k}*−e*_{i}_{2k−1}),
where *I*=*I(1,*1, . . . ,1)*∈***N*** ^{n}* and

*e*

*is the*

_{i}*i*

*unit vector.*

^{th}We set

*A** ^{}* =

*{I, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*, I*

_{i}

^{}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*|1≤k≤ n*2

and 1*≤i*_{1}*< i*_{2}*< . . . < i*_{2k−1}*< i*_{2k} *≤n}.*

We claim that*A*=*A*^{}*.*
Let

*m*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} =

^{n}_{2}
*s=1*

*m*_{s}*, m*^{}_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} =

^{n}_{2}
*s=1*

*m*^{}_{s}*,*
where

*m** _{s}*=

*x*

_{k}

_{i}2s−2+1*. . . x*_{k}_{i}

2s−1*−2**x*^{2}_{i}_{2s−1}*x*_{k}_{i}

2s−1 +1*. . . x*_{k}_{i}_{2s}_{−2}*x*_{i}_{2s}_{−1}*x*_{i}_{2s}_{+1}*,*
*m*^{}* _{s}*=

*x*

_{k}

_{i}2s−2 +1*. . . x*_{k}_{i}

2s−1*−2**x*_{i}_{2s−1}_{−1}*x*_{i}_{2s−1}_{+1}*x*_{k}_{i}

2s−1 +1*. . . x*_{k}_{i}_{2s}_{−2}*x*^{2}_{i}_{2s}*,*
for all 1*≤* *k, s≤ *^{n}_{2}*, i*_{0} = 0 and*k*_{j}*∈ {j, j*+ 1}, for 1 *≤j* *≤n.* Evidently
log(m_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}),log(m^{}_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k})*∈A.*

Since

log(m_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}) =*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}
and

log(m^{}_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}) =*I*_{i}^{}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*,*

for all 1*≤k≤ *^{n}_{2} and 1*≤i*_{1}*< i*_{2}*< . . . < i*_{2k−1}*< i*_{2k}*≤n,*then*A*^{}*⊂A.*

But the cardinal of*A*is*(A) = 2*^{n}*−*1 and since

^{n}_{2}

*s=1*

*n*
2s

= 2^{n−1}*−*1,

the cardinal of*A** ^{}*is:

*(A** ^{}*) = 1 + 2

^{n}_{2}

*s=1*

*n*
2s

= 2^{n}*−*1.

Thus*A*=*A*^{}*.*

Now we start to prove that**R**+*C*_{A}*⊂H*_{a}^{+} for all*a∈N.*

Note that

*< ν*_{{p+1}}*, I >* =*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I >*=

*n >*0, for any 0

*≤p≤n−*1,1

*≤i≤n−*1.

Let 0*≤p≤n−*1.We claim that:

*< ν*_{{p+1}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>≥*0*and < ν*_{{p+1}}*, I*_{i}^{}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>≥*0,

for any 1*≤k≤ *^{n}_{2}and 1*≤i*_{1}*< i*_{2}*< . . . < i*_{2k−1}*< i*_{2k} *≤n.*

We prove the ﬁrst inequality. The proof of the second inequality will be similar.

We have three possibilities:

1) If *< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{p+1}*>= 0,*then*< ν*_{{p+1}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>= 2n >*0;

2) If *< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{p+1}*>= 1 then< ν*_{{p+1}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>=n >*0;

3) If *< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{p+1}*>= 2 then< ν*_{{p+1}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>= 0.*

Let 0*≤p≤n−*1 and 2*≤i≤n−*1 be ﬁxed.

We claim that:

*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>*

*≥*0 and

*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}

^{}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>*

*≥*0, for any 1

*≤k≤*

^{n}_{2}and 1

*≤i*

_{1}

*< i*

_{2}

*< . . . < i*

_{2k−1}

*< i*

_{2k}

*≤n.*

We prove the ﬁrst inequality. The proof of the second inequalities is analogous.

We have:

*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>=H*

_{{σ}*p*(1),σ

*(2),...,σ*

^{p}*(i)}(I*

^{p}

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}) =

=*−(n−i−*1)
*i*
*s=1*

*< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{σ}*p*(s)*>*+

+(i+ 1)
*n*
*s=i+1*

*< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{σ}*p*(s)*> .*
Let

Γ =*{s|< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{σ}*p*(s)*>= 2,*1*≤s≤i}*

be the set of indices of*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*,*where the coordinates are equal to 2.

If the cardinal of Γ is zero, then there exists at most an index *i*_{2t−1} *∈*
*{σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)} with 1*

^{p}*≤*

*t*

*≤*

^{n}_{2}

*.*Otherwise we have two possi- bilities:

1)There exist at least two indices *i*_{2t−1}*, i*_{2t}_{1}_{−1}*∈ {σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)}, with 1*

^{p}*≤*

*t < t*

_{1}

*≤*

^{n}_{2}and, since

*σ*

*(s) = (p+*

^{p}*s)*

*mod n,*then there exists 1

*≤*

*t*

_{2}

*≤*

^{n}_{2}such that

*i*

_{2t}

_{2}

*∈ {σ*

*(1), σ*

^{p}*(2), . . . , σ*

^{p}*(i)} and thus*

^{p}*< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, e*_{σ}*p*(i2t2)*>= 2,* which it is false.

2) There exist at least two indices *i*_{2t−1}*, i*_{2t}_{1} *∈ {σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)},with 1*

^{p}*≤t, t*

_{1}

*≤*

^{n}_{2}

*.*Then as in the ﬁrst case, we have

*< I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*, e*

_{σ}*p*(i2t1)

*>=*

2,which it is false.

If for any 1*≤k≤ *^{n}_{2}*, i*2k−1*∈ {σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)},then*

^{p}*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>=−(n−i−*1)i+ (i+ 1)(n

*−i) =n >*0.

When there exists just one index *i*_{2t−1} *∈ {σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)} with 1*

^{p}*≤*

*t≤*

^{n}_{2}

*,*then

*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>=*

=*−(n−i−*1)(i*−*1) + (i+ 1)(n*−i*+ 1) = 2n >0.

If the cardinal of Γ,is*(Γ) =t≥*1,then we have two possibilities:

1) If

*{1≤i*_{1}*< i*_{2}*< . . . < i*_{2t−3}*< i*_{2t−2}*< i*_{2t−1}*≤n} ⊂ {σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)}, then we have:*

^{p}*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>=−(n−i−1)(i+1)+(i+1)(n−i−1) = 0.*

2) If

*{1≤i*_{1}*< i*_{2}*< . . . < i*_{2t−1}*< i*_{2t}*≤n} ⊂ {σ** ^{p}*(1), σ

*(2), . . . , σ*

^{p}*(i)}, then we have:*

^{p}*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}

*>=−*(n

*−i−*1)(i)+(i+1)(n

*−i) =n >*0.

Thus we have:

**R**+*C*_{A}*⊂*

*a∈N*

*H*_{a}^{+}
Finally let us prove the converse inclusion.

This is equivalent with the fact that the extremal rays of the cone

*a∈N*

*H*_{a}^{+}
are in**R**+*C*_{A}*.*

Let 1 *≤k≤ *^{n}_{2}*,* 1 *≤i*_{1} *< i*_{2} *< . . . < i*_{2k−1} *< i*_{2k} *≤n.* We consider the
following hyperplanes:

*a)H*_{{[i}_{2s−1}_{]\[j]}} if*j∈ {i*_{2s−2}*, . . . i*_{2s−1}*−*1} and 1*≤s≤k,*
*b)H*_{{[j]\[i}_{2s−1}* _{−1]}}*if

*j∈ {i*

_{2s−1}+ 1, . . . , i

_{2s}

*−*1}and 1

*≤s≤k,*

*c)H*

_{{[i}_{2k−1}]∪([n]\[j])}if

*j∈ {i*

_{2k}

*, . . . n−*1},

*d)H*_{{i}_{2s}* _{}}* for 1

*≤s≤k−*1; where [i] :=

*{1, . . . , i}, i*

_{0}= 0 and [0] =

*∅.*

We claim that the point*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} belongs to these hyperplanes.

*a) Letj* *∈ {i*_{2s−2}*, . . . i*_{2s−1}*−*1}and 1*≤s≤k,*then

*< H*_{{[i}_{2s−1}_{]\[j]}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>=< H*_{{j+1,...,i}_{2s−1}_{}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>=*

=<*−(n−*(i_{2s−1}*−j)−*1)

*t∈{j+1,...,i*2s−1*}*

*x** _{t}*+ (i

_{2s−1}

*−j*+ 1)

*t∈[n]\{j+1,...,i*2s−1*}*

*x*_{t}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>=*

=*−*(n*−i*_{2s−1}+*j−*1)(i_{2s−1}*−j*+ 1)+

+(i_{2s−1}*−j*+ 1)(n*−*(i_{2s−1}*−j) + 1) = 0,*
since

*↓j*+ 1^{th}*↓i*^{th}_{2s−1}
*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} = ( *. . .* *,* 1 *,* *. . .* *,* 1 *,* 2 *,* *. . .* ).

*b) Letj∈ {i*_{2s−1}+ 1, . . . i_{2s}*−*1}and 1*≤s≤k.*Then

*< H*_{{[j]\[i}_{2s−1}_{−1]}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>=*

=< H_{{i}_{2s−1}_{,...,j}}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>=<−(n−*(j*−i*_{2s−1}+ 1)*−*1)

*t∈{i*2s−1*,...,j}*

*x** _{t}*+ (j

*−i*

_{2s−1}+ 1 + 1)

*t∈[n]\{i*2s−1*,...,j}*

*x*_{t}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>=*

=*−(n−*(j*−i*_{2s−1}+ 1)*−*1)(j*−i*_{2s−1}+ 1 + 1)+

+(j*−i*_{2s−1}+ 1 + 1)(n*−*(j*−i*_{2s−1}+ 1 + 1)) = 0,
since

*↓i*^{th}_{2s−1} *↓j*^{th}*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} = ( *. . .* *,* 2 *,* 1 *,* *. . .* *,* 1 *,* *. . .* ).

*c) Letj∈ {i*_{2k}*, . . . n−*1}.Then

*< H*_{{[i}_{2k−1}]∪([n]\[j])}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>=< H*_{{1,...,i}_{2k−1}*,j+1,...,n}**, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} *>=*

=<*−(n−*(i_{2k−1}+*n−j)−*1)

*t∈[i*2k−1]∪([n]\[j])

*x** _{t}*+ (i

_{2k−1}+

*n−j*+ 1)

*t∈{i*2k−1+1,...,j}

*x*_{t}*, I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*>=*

=*−*(j*−i*_{2k−1}*−*1)(i_{2k−1}+n*−j+1)+(i*_{2k−1}+n*−j+1)(j−*(i_{2k−1}+1)+1*−*1) = 0,
since

*↓i*^{th}_{2k−1} *↓i*^{th}_{2k} *↓j*+ 1^{th}

*I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k} = ( *. . . ,* 2, 1, *. . . ,* 1, 0, 1, *. . . ,* 1, *. . . ,* 1).

*d) It is clear from Lemma 3.3.*

Since the number of hyperplanes is
*k*

*s=1*

(i_{2s−1}*−*1*−i*_{2s−2}+1)+

*k*
*s=1*

(i_{2s}*−*1*−*(i_{2s−1}+1)+1)+(n*−*1*−i*_{2k}+1)+k*−*1 =

=
*k*
*s=1*

(i_{2s}*−i*_{2s−2})*−k*+*n−i*_{2k}+*k−*1 =*n−*1,
then

*k*
*s=1*

(

*i*2s−1*−1*
*j=i*2s−2

(H_{{[i}_{2s−1}_{]\[j]})*∩* ^{i}^{2s}^{−1}

*j=i*2s−1+1

(H_{{[j]\[i}_{2s−1}* _{−1]}}*))∩

*n−1*

*j=i*2k

(H_{{[i}_{2k−1}]∪([n]\[j])})*∩*^{k−1}

*s=1*

(H_{{i}_{2s}* _{}}*) =

*OI*

_{i}_{1}

_{i}_{2}

_{...i}_{2k−1}

_{i}_{2k}is an extremal ray of the cone

*a∈N**H*_{a}^{+}. But
*OI*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*∈***R**+*C*_{A}*.*

Thus

*a∈N*

*H*_{a}^{+}=**R**+*C*_{A}*.*

For using bellow, we recall that *K−algebraK[A] is a normal domain ac-*
cording to [9].

**Definition 4.2.** Let*R* be a polynomial ring over a ﬁeld*K*and*F* be a ﬁnite
set of monomials in*R.*A decomposition

*K[F] =*
*∞*

*i=0*

*K[F]*_{i}

of the*K−*vector space*K[F] is an* *admissible grading* if*k[F*] is a positively
graded *K−*algebra with respect to this decomposition and each component
*K[F*]* _{i}* has a ﬁnite

*K−*basis consisting of monomials.

**Theorem 4.3**(Danilov, Stanley)**.** *LetR*=*K[x*_{1}*, . . . , x** _{n}*]

*be a polynomial ring*

*over a fieldKandF*

*be a finite set of monomials inR.IfK[F*]

*is normal, then*

*the canonical module*

*ω*

_{K[F}_{]}

*of*

*K[F*],

*with respect to an arbitrary admissible*

*grading, can be expressed as an ideal of*

*K[F*]

*generated by monomials*

*ω*_{K[F}_{]}= ({x^{a}*|* *a∈***N***A∩ri(***R**+*A)}),*

*where* *A*=*log(F*)*andri(***R**+*A)denotes the relative interior of* **R**+*A.*

**Corollary 4.4.** *The canonical moduleω*_{K[A]}*ofK[A]* *is*
*ω** _{K[A]}*= (x

_{1}

*x*

_{2}

*. . . x*

*)K[A].*

_{n}*Thus the* *K−algebraK[A]* *is a Gorenstein ring.*

*Proof.* Since

*< ν*_{{p+1}}*, I >*=*< ν*_{{σ}*p*(1),σ* ^{p}*(2),...,σ

*(i)}*

^{p}*, I >*=

*n >*0,

for any 0*≤p≤n−1,*1*≤i≤n−1 and since for anyI*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, I*_{i}^{}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}
there exist two hyperplanes*H*_{a}*, H** _{a}* with

*a, a*

^{}*∈N*such that

*< I*_{i}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, H*_{a}*>=< I*_{i}^{}_{1}_{i}_{2}_{...i}_{2k−1}_{i}_{2k}*, H*_{a}*>= 0,*

then*I∈ri(***R**+*C** _{A}*)

*is the only point in relative interior of the cone*

**R**+

*C*

_{A}*.*Thus the canonical module is generated by one generator,

*ω** _{K[A]}*= (x

_{1}

*x*

_{2}

*. . . x*

*)K[A].*

_{n}Therefore the *K−*algebra*K[A*] is a Gorenstein ring.

**Conjecture 4.5.** *Let* *n, m* *∈* **N***, m* *≤* *n, A** _{i}* [n],1

*≤*

*i*

*≤*

*m, and*

*A*=

*{A*

_{1}

*, A*

_{2}

*, . . . , A*

_{m}*}.We denote*

*A*=*{log(x**i*1*x*_{i}_{2}*...x*_{i}* _{n}*)

*|i*

_{1}

*∈ {1,*2}, i2

*∈ {2,*3}, . . . , i

*n−1*

*∈ {n−1, n}, i*

*n*

*∈ {1, n}},*

*N*=

*{ν*

_{{k+1}}*, ν*

_{{σ}*k*(1),σ

*(2),...,σ*

^{k}*(i)}*

^{k}*|*0

*≤k≤n−*1,2

*≤i≤n−*1},

*and*

*A*=*{log(x**i*1*x*_{i}_{2}*...x*_{i}* _{n}*)

*|*

*i*

_{1}

*∈A*

_{1}

*, i*

_{2}

*∈A*

_{2}

*, . . . , i*

_{n−1}*∈A*

_{n−1}*, i*

_{n}*∈A*

_{n}*}.*

*Then the base ring associated to transversal polymatroid presented byA, K[* *A],*
*is a Gorenstein ring if and only if there exists* *N* *⊂N* *such that*

**R**+*C** _{A}*=

*a∈N*

*H*_{a}^{+}

*andH*_{a}^{+} *with* *a∈N* *are just the facets of the cone* **R**+*C*_{A}*.*