BALANCE DEL COMBUSTIBLE CONSUMIDO (GAS PROPANO)

Definition 11.13. Let (G, l) = (V, E, l) be anN-labelled graph. We say that

(G, l) iscircuit-coprimeif for every circuitC⊂G, gcd{l(e)|eis an edge ofC}= 1.

Example 11.14. In fig. 3 the_N-labelled graph (a) is circuit-coprime, whereas the_N-labelled graph (b) is not, as it contains a loop labelled by 3 in addition to a circuit labelled by 6,10 and 10.

Lemma 11.15. Let (G, l) = (V, E, l) be an _N-labelled graph. Denote by r its nullity. The Smith normal form of the matrix M(G,l) has diagonal entries

d1=d2=. . .=dr= 1 if and only if(G, l)is circuit-coprime.

Proof. Assume first that (G, l) is not circuit-coprime. LetCbe a circuit whose labels have greatest common divisorD6= 1. Pick an edgeeofC. The subgraph

C\e is a tree; letT be a spanning tree of Gcontaining it. Then eis a link forT, andCis its associated fundamental circuit. The lfc-matrixN(G,l)has a

2 3 6 1 10 15 (a) 2 3 6 3 10 10 (b)

Figure 3: A circuit-coprimeN-labelled graph (a) and anN-labelled graph that

is not circuit-coprime (b).

by D. Then the linear mapf:Zn → Zr defined by N(G,l) is not surjective;

hence the linear map associated to the Smith normal form of N(G,l) is not

surjective either. Therefore, some (necessarily non-zero) diagonal entry of the Smith normal form of N(G,l) is different from ±1. As previously remarked,

the Smith normal forms ofM(G,l)andN(G,l)have the same non-zero diagonal

entries, hencedr6=±1.

Conversely, assume thatGis circuit-coprime. After fixing some spanning tree

T, consider the lfc-matrix N(G,l). We only need to prove that the diagonal

entries of the Smith normal form ofN(G,l)are all 1, which amounts to proving

that the greatest common divisor dof the minors of orderrof the lfc-matrix

N(G,l)is 1.

As we have seen in section 11.2, we have the relation

N(G,l)=NG·L.

Let N0 be a maximal square submatrix of N(G,l). Then N0 corresponds to

r edges ofG, which we denote ei1, ei2, . . . , eir. Let N

00 _{be the corresponding}

square submatrix of NG. We have the relation

detN0= r Y j=1 l(eij) detN 00

By [TS92], Theorem 6.15, all minors ofNG are either 1,0 or−1, hence detN00

is either 1,0 or −1. Moreover, by [TS92], Theorem 6.10, a square submatrix of order rofNG has determinant±1 if and only if the correspondingr edges

are the complement of a spanning tree. Hence detN0 =±Qr

i=1l(eij) if the edges ei1, ei2, . . . , eir form the complement of a spanning tree ofG, otherwise detN0 = 0. We claim that

Letpbe a prime number and denote byEpthe set of edgeseofGwhose label

l(e) is divisible byp. Because (G, l) is circuit-coprime,Ep contains no circuit;

henceEpis contained in some spanning treeT ofG. There are exactlyredges,

e1, e2, . . . , er, that do not belong to T. These give a square r×r submatrix

of N(G,l) whose determinant isQ

r

i=1l(ei)6≡0 (mod p), sincee1, . . . , er 6∈Ep.

Hence p- d. It follows that d= 1; since di|di+1 for alli = 1, . . . , r−1 and

dr|d, we obtain the result.

Proposition 11.16. Let (G, l) = (V, E, l) be anN-labelled graph. The group

homomorphism :H →He is an isomorphism if and only if (G, l) is circuit- coprime.

Proof. We already know that : H → He is injective by lemma 11.12. It is

surjective if and only if for every vertex-labelling α∈ZVe_{, there exists} e ϕ∈Ce

such thateδ(ϕ_e) +αis in the image of the extension-by-zero map:ZV →ZVe,

i.e. eδ(ϕ_e) +αis supported on the set of old vertices. We may of course assume

that αbelongs to the canonical basis ofZVe_{. That is,α=χ}

v for some vertex

v ofGe, where

χv(w) =

(

1 ifw=v

0 ifw6=v.

Ifvis an old vertex ofGe,χvis an extension by zero of a vertex-labelling onG,

so we may assume thatvis a new vertex. Thenvbelongs to some pathP ⊂Ge

associated to some edgee∈E. Denote byw0, w1, . . . , wl(e)the vertices of the

pathP, so thatw0andwl(e)are old vertices, and the numbering of the indices

follows the order of the vertices on the path. For every i = 1, . . . , l(e), let

αi=χwi−χw0 ∈Z e

V _{be the vertex-labelling that has value 1 atw}

i, value−1

at w0, and value 0 everywhere else. Then it is easy to check that the images

αi of theαi in He satisfykα1 =αk for all k= 1, . . . , l(e). Hence, if α1 is in

the image of:H →He, so are all the αi for 1≤i≤l(e). This shows that we

can takevto be equal tow1; henceχv=χw1 takes value 1 on a new vertex v

adjacent to an old vertex, and value zero at all other vertices.

We ask whether an element ϕ_e ∈ Ceexists such that δe(ϕ_e) +χw1 is supported

only on the old vertices. In other words,eδ(ϕ_e) must be zero on all new vertices

except for the vertexw1, where it has to take the value−1. This is equivalent

to asking that, for every new vertex z, adjacent to vertices z1, z2,

( e ϕ(z)−ϕ_e(z1) =ϕ_e(z2)−ϕ_e(z) ifz6=w1 (ϕ_e(z1)−ϕe(z)) + (ϕe(z2)−ϕe(z)) =−1 ifz=w1 (31) holds.

We claim that such aϕ_eexists if and only if there exists a vertex-labellingϕof the graph G, such that, for every edgee∈E with endpointsv0, v1,

(

ϕ(v1)−ϕ(v0)≡0 mod l(e) ife6=e

ϕ(v1)−ϕ(v0)≡1 mod l(e) ife=e, v0=w0, v1=wl(e)

(32) where we have identified the old verticesw0, wl(e)with the corresponding ver-

tices inG. Indeed, givenϕ_eone obtainsϕsimply by restriction to old vertices. Conversely, given a ϕas in (37), ϕ_e is obtained as follows: for an edgee6=e, we define ϕ_eon the corresponding path {z0=v0, z1, z2, . . . , zl(e)=v1}by:

∀k= 0,1, . . . , l(e), ϕ_e(zk) =

kϕ(v1) + (l(e)−k)ϕ(v0)

l(e) .

On the path{w0=v0, w1, . . . , wl(e)=v1} corresponding to the edgee, we set

instead e ϕ(wk) = (kϕ(v1)+(l(e)−k)(ϕ(v0)+1) l(e) ifk∈ {1,2, . . . , l(e)} e ϕ(v0) ifk= 0;

which establishes the claim.

If the graph Gis a tree it is clear that such a ϕ can be found. If there are circuits in G, the existence of a solution ϕ depends of course on the labels of the circuits. Fix an orientation on G, so that we have source and target functions s, t:E → V, and so that s(e) = w0, t(e) = wl(e). Assume that a

vertex-labelling ϕof Gsatisfying the conditions (37) exists. In particular we have thatϕ(t(e))−ϕ(s(e))≡1 modl(e). For every edgee∈Elet

x(e) :=

(ϕ(t(e))−ϕ(s(e))

l(e) ife6=e

ϕ(t(e))−ϕ(s(e))−1

l(e) ife=e

Let C⊂Gbe a circuit consisting of vertices v0, v1, . . . , vs=v0 connected by

edgese0, e1, e2, . . . , es=e0, so thateiconnectsviandvi+1for everyi∈Z/sZ.

Notice that the increasing numbering gives an orientation toC. We have (ϕ(vs)−ϕ(vs−1)) + (ϕ(vs−1)−ϕ(vs−2)) +. . .+ (ϕ(v1)−ϕ(vs)) = 0. Setting ai= ( 1 ift(ei) =vi+1, s(ei) =vi −1 ift(ei) =vi, s(ei) =vi+1 (33)

for every i∈_Z/s_Z, we obtain

X

if the edge edoes not belong to the circuitC, whereas if e∈Cwe have

X

aixeil(ei) =

(

−1 if the orientations ofC andeagree; 1 if the orientations ofC andedo not agree; LetC1, . . . , Cmbe the circuits ofG. Choose an orientation for each circuit, so

that we can form the labelled circuit matrix M(G,l) associated toG. We see

that the vectorx= (x1, . . . , xn) is a solution of

M(G,l)x=b(e) where b(e) = (b1, . . . , bm) with bi=                0 ife6∈Ci;

−1 ife∈Ci and the orientation ofeagrees with the

orientation ofCi;

1 ife∈Ci and the orientation ofedoes not agree with the

orientation ofCi.

Conversely, a solution x ∈ Zn to the system M(G,l)x = b(e) yields a vertex

labelling ϕ as in (37). We conclude that the map: H →He is surjective if

and only if for every edgee∈E, there is a solution x∈_Zn _to

M(G,l)x=b(e).

After having chosen a spanning tree T and formed the lfc-matrixN(G,l), this

is in turn equivalent to the mapZn →Zr defined byN(G,l) being surjective.

Indeed, the set{b(e)|eis a link ofT}is a basis forZr. Now,N(G,l)is surjective

if and only if its Smith normal form (or equivalently the one of M(G,l)) has

only 1’s on the diagonal. By lemma 11.15, we conclude.