Cones of weighted quasi-metrics, weighted quasi-hypermetrics and of oriented cuts

0.25 MB·English

by  Michel Deza

Cones of weighted quasi-metrics, weighted
quasi-hypermetrics and of oriented cuts
2
1
0 M.Deza V.Grishukhin E.Deza
2

n
a
J Abstract

5
We show that the cone of weighted n-point quasi-metrics WQMet , the cone of
n
]
G weighted quasi-hypermetrics WHypn and the cone of oriented cuts OCutn are projec-
tions along an extreme ray of the metric cone Met , of the hypermetric cone Hyp
M n+1 n+1
and of the cut cone Cut , respectively. This projection is such that if one knows all
n+1
.
h faces of an original cone then one knows all faces of the projected cone.
t
a
m
1 Introduction
[

1
v Oriented (or directed) distances are encountered very often, for example, these are one-way
9
transport routes, rivers with quick flow and so on.
9
0 The notions of directed distances, quasi-metrics and oriented cuts are generalizations of
1
the notions of distances, metrics and cuts, respectively (see, for example, [DL97]), which are
.
1 central objects in Graph Theory and Combinatorial Optimization.
0
2 Quasi-metrics are used in Semantics of Computations (see, for example, [Se97]) and in
1 computational geometry (see, for example, [AACMP97]). Oriented distances have been used
:
v already by Hausdorff in 1914, see [Ha14].
i
X In [CMM06], authors give an example of directed metric derived from a metric as follows.
r Let d be a metric on a set V 0 , where 0 is a distinguished point. Then a quasi-metric q
a ∪{ }
on the set V is given as
q = d +d d .
ij ij i0 j0
−
This quasi-metric belongs to a special important subclass of quasi-metrics, namely, to a
class of weighted quasi-metrics. We show (cf. also Lemma 1 (ii) in[DDV11]) thatanyweighted
quasi-metric is obtained by a slight generalization of this method.
Allsemi-metricsonasetofcardinalitynformametric coneMet . Therearetwoimportant
n
sub-cones of Met , namely, the cone Hyp of hypermetrics, and the cone Cut of ℓ -metrics.
n n n 1
These three cones form the following nested family Cut Hyp Met , see [DL97].
n n n
⊂ ⊂
We introduce a space Q , called a space of weighted quasi-metrics and define in it a cone
n
WQMet . Elements of this cone satisfy triangle and non-negativity inequalities. Among
n
extreme rays of the cone WQMet there are rays spanned by ocut vectors, i.e., incidence
n
vectors of oriented cuts.
1
We define in the space Q a cone OCut as the cone hull of ocut vectors. Elements of the
n n
cone OCut are weighted quasi-ℓ-metrics.
n
Let semi-metrics in the cone Met be defined on the set V 0 . The cut cone Cut
n+1 n+1
∪{ }
(or the cone of ℓ -metrics on this set is a cone hull of cut semi-metrics δ(S) for all S V 0 .
1
⊂ ∪{ }
The cut semi-metrics δ(S) areextreme rays of all thethree cones Met , Hyp andCut .
n+1 n+1 n+1
In particular, δ( 0 ) = δ(V) is an extreme ray of these three cones.
{ }
In this paper, it is shown that the cones WQMet and OCut are projections of the
n n
corresponding cones Met and Cut along the extreme ray δ(V). We define a cone
n+1 n+1
WQHyp of weighted quasi-hypermetrics as projection along δ(V) of the cone Hyp . So,
n n+1
we obtain a nested family OCut WQHyp WQMet .
n n n
⊂ ⊂
The cones of weighted quasi-metrics, oriented cuts and other related generalizations of
metrics are studied in [DD10] and [DDV11]. The polytope of oriented cuts was considered in
[AM11].
2 Spaces RE and REO
Let V be a set of cardinality V = n. Let E and EO be sets of all unordered (ij) and ordered
ij pairsof elements i,j V. C| o|nsider two Euclidean spaces RE andREOofvectors d RE and
g REO with coordina∈tes d and g , where (ij) E and ij EO, respectively. O∈bviously,
(ij) ij
∈ ∈ ∈
dimensions of the spaces RE and REO are E = n(n−1) and EO = n(n 1), respectively.
Denote by (d,t) = d ts scala|r|produc2t of vect|ors|d,t R−E. Similarly, (f,g) =
(ij)∈E (ij) (ij) ∈
f g is the scaPlar product of vectors f,g REO.
ij∈EO ij ij ∈
P Let e : (ij) E and e : ij EO beorthonormalbasesofRE andREO,respectively.
(ij) ij
{ ∈ } { ∈ }
Then, for f RE and q REO, we have
∈ ∈
(e ,f) = f and (e ,q) = q .
(ij) (ij) ij ij
For f REO, define f∗ REO as follows
∈ ∈
f∗ = f for all ij EO.
ij ji ∈
Each vector g REO can be decompose into symmetric gs and antisymmetric ga parts as
∈
follows:
1 1
gs = (g +g∗), ga = (g g∗), g = gs +ga.
2 2 −
Call a vector g symmetric if g∗ = g, and antisymmetric if g∗ = g. Let REO and REO be
subspaces of the corresponding vectors. Note that the spaces R−EO and REsO are muatually
s a
orthogonal. In fact, for p REO and f REO, we have
∈ s ∈ a
(p,f) = p f = (p f +p f ) = (p f p f ) = 0.
ij ij ij ij ji ji ij ij ij ij
−
ijX∈EO (iXj)∈E (iXj)∈E
Hence
REO = REO REO,
s ⊕ a
2
where is the direct sum.
Obv⊕iously, there is an isomorphism ϕ between the spaces RE and REO. Let d RE have
s ∈
coordinates d . Then
(ij)
dO = ϕ(d) REO , such that dO = dO = d .
∈ s ij ji (ij)
In particular,
ϕ(e ) = e +e .
(ij) ij ji
The map ϕ is invertible. In fact, for q REO, we have ϕ−1(q) = d RE, such that
∈ s ∈
d = q = q . The isomorphism ϕ will be useful in what follows.
(ij) ij ji
3 Space of weights Qw
n
One can consider the sets E and EO as sets of edges (ij) and arcs ij of an unordered and
ordered complete graphs K and KO on the vertex set V, respectively. The graph KO has
n n n
two arcs ij and ji between each pair of vertices i,j V.
It is convenient to consider vectors g REO as fu∈nctions on the set of arcs EO of the graph
KO. So, the decomposition REO = REO ∈REO is a decomposition of the space of all functions
n s ⊕ a
on arcs in EO onto the spaces of symmetric and antisymmetric functions.
Besides, there is an important direct decomposition of the space REO of antisymmetric
a
functions into two subspaces. In the Theory of Electric Networks, these spaces are called
spaces of tensions and flows (see also [Aig79]).
The tension space relates to potentials (or weights) w given on vertices i V of the graph
i
∈
KO. The corresponding antisymmetric function gw is determined as
n
gw = w w .
ij i − j
It is called tension on the arc ij. Obviously, gw = w w = gw. Denote by Qw the subspace
ji j− i − ij n
of REO generated by all tensions on arcs ij EO. We call Qw a space of weights.
∈ n
Each tension function gw is represented as weighted sum of elementary potential functions
q(k), for k V, as follows:
∈
gw = w q(k),
k
Xk∈V
where
q(k) = (e e ), for all k V, (1)
kj jk
− ∈
X
j∈V−{k}
are basic functions that generate the space of weights Qw. Hence, the values of the basic
n
functions q(k) on arcs are as follows:
1, if i = k
q (k) =  1, if j = k (2)
ij
−
 0, otherwise.

3
We obtain
gw = w q (k) = w w .
ij k ij i − j
Xk∈V
It is easy to verify that
q2(k) = (q(k),q(k)) = 2(n 1), (q(k),q(l)) = 2 for all k,l V,k = l, q(k) = 0.
− − ∈ 6
Xk∈V
Hence, there are only n 1 independent functions q(k) that generate the space Qw.
The weighted quasi-−metrics lie in the space REO Qw that we denote as Qn . Direct
complements of Qw in REO and Q in REO is a spasce Q⊕c ofncircuits (or flows). n
n a n n
4 Space of circuits Qc
n
The space of circuits (or space of flows) is generated by characteristic vectors of oriented
circuits in the graph KO. Arcs of KO are ordered pairs ij of vertices i,j V. The arc ij is
n n ∈
oriented from the vertex i to the vertex j. Recall that KO has both the arcs ij and ji for each
n
pair of vertices i,j V.
∈
Let G K be a subgraph with a set of edges E(G ) E. We relate to the graph G
s n s s
⊂ ⊂
a directed graph G KO with the arc set EO(G) EO as follows. An arc ij belongs to G,
⊂ n ⊂
i.e., ij EO(G), if and only if (ij) = (ji) E(G). This definition implies that in this case,
∈ ∈
the arc ji belongs to G also, i.e., ji EO(G).
∈
Let C be a circuit in the graph K . The circuit C is determined by a sequence of distinct
s n s
vertices i V, where 1 k p, and p is the length of C . The edges of C are unordered
k s s
∈ ≤ ≤
pairs (i ,i ), where indices are taken modulo p. By above definition, an oriented bicircuit C
k k+1
of the graph KO relates to the circuit C . Arcs of C are ordered pairs i i and i i , where
n s k k+1 k+1 k
indices are taken modulo p. Take an orientation of C. Denote by C the opposite circuit
−
with opposite orientation. Denote an arc of C direct or opposite if its direction coincides with
or is opposite to the given orientation of C, respectively. Let C+ and C− be subcircuits of C
consisting of direct and opposite arcs, respectively.
The following vector fC is the characteristic vector of the bicircuit C:
1, if ij C+,
∈
fC =  1. if ij C−,
ij − ∈
 0, otherwise.

Note that f−C = (fC)∗ = fC, and fC REO.
− ∈ a
Denote by Qc the space linearly generated by circuit vectors fC for all bicircuits C of the
n
graph KO. It is well known that characteristic vectors of fundamental circuits form a basis of
n
Qc. Fundamental circuits are defined as follows.
n
Let T be a spanning tree of the graph K . Since T is spanning, its vertex set V(T) is the
n
set of all vertices of K , i.e., V(T) = V. Let E(T) E be the set of edges of T. Then any
n
⊂
edge e = (ij) E(T) closes a unique path in T between vertices i and j into a circuit Ce. This
6∈ s
circuit Ce is called fundamental. Call corresponding oriented bicircuit Ce also fundamental.
s
4
There are E E(T) = n(n−1) (n 1) fundamental circuits. Hence
| − | 2 − −
n(n 1)
dimQc = − (n 1), and dimQ +dimQc = n(n 1) = dimREO.
n 2 − − n n −
This implies that Qc is an orthogonal complement of Qw in RO and Q in REO, i.e.
n n a n
REO = Qw Qc and REO = Q Qc = REsO Qw Qc.
a n ⊕ n n ⊕ n ⊕ n ⊕ n
5 Cut and ocut vector set-functions
The space Q is generated also by vectors of oriented cuts, which we define in this section.
n
Each subset S V determines cuts of the graphs K and KO that are subsets of edges
⊂ n n
and arcs of these graphs.
A cut(S) E is a subset of edges (ij) of K such that (ij) cut(S) if and only if
n
⊂ ∈
i,j S = 1.
|{ }∩ |
A cutO(S) EO is a subset of arcs ij of KO such that ij cutO(S) if and only if
⊂ n ∈
i,j S = 1. So, if ij cutO(S), then ji cutO(S) also.
|{ }∩ | ∈ ∈
An oriented cut is a subset ocut(S) EO of arcs ij of KO such that ij ocut(S) if and
⊂ n ∈
only if i S and j S.
We re∈late to th6∈ese three types of cuts characteristic vectors δ(S) RE, δO(S) REO,
q(S) REO and c(S) REO as follows. ∈ ∈ s
∈ a ∈
For cut(S), we set
1, if i,j S = 1
δ(S) = e , such that δ (S) = |{ }∩ |
(ij) (ij) (cid:26) 0, otherwise,
X
i∈S,j∈S
where S = V S. For cutO(S), we set
−
δO(S) = ϕ(δ(S)) = (e +e ) and q(S) = (e e ).
ij ji ij ji
−
X X
i∈S,j∈S i∈S,j∈S
Hence,
1, if i S,j S
1, if i,j S = 1 ∈ 6∈
δO(S) = |{ }∩ | and q (S) =  1, if j S,i S
ij (cid:26) 0, otherwise. ij − ∈ 6∈
 0, otherwise.

Note that, for one-element sets S = k , the function q( k ) is q(k) of section 2. It is easy to
{ } { }
see that
(δO(S),q(T)) = 0 for any S,T V.
⊂
For the oriented cut ocut(S), we set
c(S) = e .
ij
X
i∈S,j∈S
5
Hence,
1, if i S,j S
c (S) = ∈ 6∈
ij (cid:26) 0, otherwise.
Obviously, it holds c( ) = c(V) = 0, where 0 REO is a vector whose all coordinates are
∅ ∈
equal zero. We have
1
c∗(S) = c(S), c(S)+c(S) = δO(S), c(S) c(S) = q(S) and c(S) = (δO(S)+q(S)). (3)
− 2
Besides, we have
1 1
cs(S) = δO(S), ca(S) = q(S).
2 2
Recall that a set-function f(S) on all S V, is called submodular if, for any S,T V,
⊂ ⊂
the following submodular inequality holds:
f(S)+f(T) (f(S T)+f(S T)) 0.
− ∩ ∪ ≥
It is well known that the vector set-function δ RE is submodular (see, for example, [Aig79]).
The above isomorphism ϕ of the spaces RE ∈and REO implies that the vector set-function
s
δO = ϕ(δ) REO is submodular also.
∈ s
Aset-function f(S) iscalled modularif, for anyS,T V, the above submodular inequality
⊂
holds as equality. This equality is called modular equality. It is well known (and can be easily
verified) that antisymmetric vector set-function fa(S) is modular for any oriented graph G.
Hence, our antisymmetric vector set-function q(S) REO for the oriented complete graph KO
∈ a n
is modular also.
Note that the set of all submodular set-functions on a set V forms a cone in the space
R2V. Therefore, the last equality in (3) implies that the vector set-function c(S) REO is
∈
submodular.
Themodularityoftheantisymmetric vectorset-functionq(S)isimportantforwhatfollows.
Itiswell-known(see, forexample, [Bir67])(anditcanbeeasilyverifiedusingmodularequality)
that a modular set-function m(S) is completely determined by its values on the empty set and
on all one-element sets. Hence, a modular set-function m(S) has the following form
m(S) = m + m ,
0 i
Xi∈S
where m = m( ) and m = m( i ) m( ). For brevity, we set f( i ) = f(i) for any
0 i
∅ { } − ∅ { }
set-function f(S). Since q( ) = q(V) = 0, we have
∅
q(S) = q(k), S V, and q(V) = q(k) = 0. (4)
⊂
Xk∈S Xk∈V
Using equations (3) and (4), we obtain
1
c(S) = (δO(S)+ q(k)). (5)
2
Xk∈S
6
Now we show that ocut vectors c(S) for all S V linearly generate the space Q REO.
n
⊂ ⊆
The space generated by c(S) consists of the following vectors
c = α c(S), where α R.
S S
∈
SX⊂V
Recall that c(S) = 1(δO(S)+q(S)). Hence, we have
2
1 1 1 1
c = α (δO(S)+q(S)) = α δO(S)+ α q(S) = (dO +q),
S S S
2 2 2 2
SX⊂V SX⊂V SX⊂V
where dO = ϕ(d) for d = α δ(S). For a vector q, we have
S⊂V S
P
q = α q(S) = α q(k) = w q(k), where w = α .
S S k k S
SX⊂V SX⊂V Xk∈S Xk∈V VX⊃S∋k
Since q = w q (k) = w w , we have
ij k∈V k ij i − j
P
1
c = (dO +w w ). (6)
ij 2 ij i − j
It is well-known (see, for example, [DL97]) that the cut vectors δ(S) RE for all S V
linearly generate the full space RE. Hence, the vectors δO(S) REO, for∈all S V, line⊂arly
generate the full space REO. ∈ s ⊂
s
According to (2), antisymmetric parts of ocut vectors c(S) generate the space Qw. This
n
implies that the space Q = REO Qw is generated by c(S) for all S V.
n s ⊕ n ⊂
6 Properties of the space Q
n
Let x Q and let fC be the characteristic vector of a bicircuit C. Since fC is orthogonal to
n
∈
Q , we have (x,fC) = fCx = 0. This equality implies that each point x Q satisfies
n ij∈C ij ij ∈ n
the following equalitiesP
x = x
ij ij
ijX∈C+ ijX∈C−
for any bicircuit C.
Let K K be a spanning star of K consisting of all n 1 edges incident to a vertex
1,n−1 n n
⊂ −
of K . Let this vertex be 1. Each edge of K K has the form (ij), where i = 1 = j. The
n n 1,n−1
− 6 6
edge (ij) closes a fundamental triangle with edges (1i),(1j),(ij). The corresponding bitriangle
T(1ij) generates the equality
x +x +x = x +x +x .
1i ij j1 i1 1j ji
These equalities are the case k = 3 of k-cyclic symmetry considered in [DD10]. They were
derived by another way in [AM11]. They correspond to fundamental bi-triangles T(1ij), for
7
all i,j V 1 , and are all n(n−1) (n 1) independent equalities determining the space,
∈ −{ } 2 − −
where the Q lies.
n
Abovecoordinatesx ofavector x Q aregivenintheorthonormalbasis e : ij EO .
ij n ij
∈ { ∈ }
But, for what follows, it is more convenient to consider vectors q Q in another basis. Recall
n
thatREO = ϕ(RE). Let, for(ij) E,ϕ(e ) = e +e REO be∈basicvectorsofthesubspace
s ∈ (ij) ij ji ∈ s
REO Q . Let q(i) Qw, i V, be basic vectors of the space Qw Q . Then, for q Q ,
s ⊂ n ∈ n ∈ n ⊂ n ∈ n
we set
q = qs +qa, where qs = q ϕ(e ), qa = w q(i).
(ij) (ij) i
(iXj)∈E Xi∈V
Now, we obtain an important expression for the scalar product (g,q) of vectors g,q Q .
n
∈
Recall that (ϕ(e ),q(k)) = ((e + e ),q(k)) = 0 for all (ij) E and all k V. Hence
(ij) ij ji
∈ ∈
(gs,qa) = (ga,qs) = 0, and we have
(g,q) = (gs,qs)+(ga,qa).
Besides, we have
((e +e ),(e +e )) = 0 if (ij) = (kl), (e +e )2 = 2,
ij ji kl lk ij ji
6
and (see Section 3)
(q(i),q(j)) = 2 if i = j, (q(i))2 = 2(n 1).
− 6 −
Let v , i V, be weights of the vector g. Then we have
i
∈
(g,q) = 2 g q +2(n 1) v w 2 v w .
(ij) (ij) i i i j
− −
(iXj)∈E Xi∈V i6=Xj∈V
For the last sum, we have
v w = ( v )( w ) v w .
i j i i i i
−
i6=Xj∈V Xi∈V Xi∈V Xi∈V
Sinceweightsaredefineduptoanadditivescalar, wecanchooseweightsv suchthat v =
i i∈V i
0. Then the last sum in the product (g,q) is equal to − i∈V viwi. Finally we obPtain that
the sum of antisymmetric parts is equal to 2n i∈V viwi.PSo, for the product of two vectors
g,q Qn we have the following expression P
∈
(g,q) = (gs,qs)+(ga,qa) = 2( g q +n v w ) if v = 0 or w = 0.
(ij) (ij) i i i i
(iXj)∈E Xi∈V Xi∈V Xi∈V
In what follows, we consider inequalities (g,q) 0. We can delete the multiple 2, and rewrite
≥
such inequality as follows
g q +n v w 0, (7)
(ij) (ij) i i
≥
(iXj)∈E Xi∈V
where v = 0.
i∈V i
P
8
Below we consider some cones in the space Q . Since the space Q is orthogonal to the
n n
space of circuits Qc, each facet vector of a cone in Q is defined up to a vector of the space
n n
Qc. Of course each vector g′ REO can be decomposed as g′ = g + gc, where g Q and
n ∈ ∈ n
gc Qc. Call the vector g Q canonical representative of the vector g′. Usually we will
∈ n ∈ n
use canonical facet vectors. But sometimes not canonical representatives of a facet vector are
useful.
Cones Con that will be considered are invariant under the operation q q∗, defined in
→
Section 2. In other words, Con∗ = Con. This operation changes signs of weights:
q = q +w w q +w w = q w +w .
ij (ij) i j (ij) j i (ij) i j
− → − −
Let (g,q) 0 be an inequality determining a facet F of a cone Con Q . Since Con =
n
≥ ⊂
Con∗, the coneConhas, together with thefacet F, also a facet F∗. The facet F∗ isdetermined
by the inequality (g∗,q) 0.
≥
7 Projections of cones Con
n+1
Recall that Q = REO Qw, REO = ϕ(RE) and dimQ = n(n+1) 1. Let 0 V be an
n s ⊕ n s n 2 − 6∈
additional point. Then the set of unordered pairs (ij) for i,j V 0 is E E , where
0
∈ ∪ { } ∪
E = (0i) : i V . Obviously, RE∪E0 = RE RE0 and dimRE∪E0 = n(n+1).
0 { ∈ } ⊕ 2
The space RE∪E0 contains the following three important cones: the cone Met of semi-
n+1
metrics, the cone Hyp of hyper-semi-metrics and the cone Cut of ℓ -semi-metrics, all
n+1 n+1 1
on the set V 0 . Denote by Con any of these cones.
n+1
∪{ }
Recall that asemi-metric d = d is calledmetric ifd = 0 forall (ij) E. Forbrevity
(ij) (ij)
{ } 6 ∈
sake, in what follows, we call elements of the cones Con simply metrics (or hypermetrics,
n+1
ℓ -metrics), assuming that they can be semi-metrics.
1
Note that if d Con is a metric on the set V 0 , then a restriction dV of d on the
n+1
set V is a point of∈the cone Con = Con RE of m∪e{tri}cs on the set V. In other words, we
n n+1
∩
can suppose that Con Con .
n n+1
⊂
TheconesMet ,Hyp andCut containthecutvectorsδ(S)thatspanextremerays
n+1 n+1 n+1
for all S V 0 . Denote by l the extreme ray spanned by the cut vector δ(V) = δ( 0 ).
0
⊂ ∪{ } { }
Consider a projection π(RE∪E0) of the space RE∪E0 along the ray l onto a subspace of RE∪E0
0
that is orthogonal to δ(V). This projection is such that π(RE) = RE and π(RE∪E0) =
RE π(RE0).
⊕
Note that δ(V) RE0, since, by Section 5, δ(V) = e . For simplicity sake, set
∈ i∈V (0i)
P
e = δ( 0 ) = δ(V) = e .
0 (0i)
{ }
Xi∈V
Recall that the vector e spans the extreme ray l . Obviously, the space RE is orthogonal to
0 0
l , and therefore, π(RE) = RE.
0
Let x RE. We decompose this point as follows
∈
x = xV +x0,
9
where xV = x e RE and x0 = x e RE0. We define a map π as
(ij)∈E (ij) (ij) ∈ i∈V (0i) (0i) ∈
follows: P P
1
π(e ) = e for (ij) E, and π(e ) = e e for i V.
(ij) (ij) ∈ (0i) (0i) − n 0 ∈
So, we have
1
π(x) = π(xV)+π(x0) = x e + x (e e ). (8)
(ij) (ij) (0i) (0i) − n 0
(iXj)∈E Xi∈V
Note that the projection π transforms the positive orthant of the space RE0 onto the whole
space π(RE0).
Now we describe how faces of a cone in the space RE∪E0 are projected along one of its
extreme rays.
Let l be an extreme ray and F be a face of a cone in RE∪E0. Let π be the projection along
l. Let dimF be dimension of the face F. Then the following equality holds
dimπ(F) = dimF dim(F l). (9)
− ∩
Let g RE∪E0 be a facet vector of a facet G, and e be a vector spanning the line l. Then
∈
dim(G l) = 1 if (g,e) = 0, and dim(G l) = 0 if (g,e) = 0.
∩ ∩ 6
Theorem 1 Let G be a face of the cone π(Con ). Then G = π(F), where F is a face of
n+1
Con such that there is a facet of Con , containing F and the extreme ray l spanned by
n+1 n+1 0
e = δ(V).
0
In particular, G is a facet of π(Con ) if and only if G = π(F), where F is a facet of
n+1
Con containing the extreme ray l . Similarly, l′ is an extreme ray of π(Con ) if and only
n+1 0 n+1
if l′ = π(l), where l is an extreme ray of Con lying in a facet of Con that contains l .
n+1 n+1 0
Proof. Let be a set of all facets of the cone Con . Then π(F) is a covering of
n+1 f∈F
F ∪
the projection π(Con ). By (9), in this covering, if l F , then π(F) is a facet of
n+1 0
⊂ ∈ F
π(Con ). If l F, then there is a one-to-one correspondence between points of F and
n+1 0
6⊂
π(F). Hence, dimπ(F) = n, and π(F) cannot be a facet of π(Con ), since π(F) fills an
n+1
n-dimensional part of the cone π(Con ).
n+1
If F′ is a face of Con , then π(F′) is a face of the above covering. If F′ belongs only
n+1
to facets F such that l F, then π(F′) lies inside of π(Con ). In this case, it is not
0 n+1
∈ F 6⊂
a face of π(Con ). This implies that π(F′) is a face of π(Con ) if and only if F′ F,
n+1 n+1
⊂
where F is a facet of Con such that l F. Suppose that dimension of F′ is n 1, and
n+1 0
⊂ −
l F′. Then dimπ(F′) = n 1. If F′ is contained in a facet F of Con such that l F,
0 n+1 0
6⊂ − ⊂
then π(F′) = π(F). Hence, π(F′) is a facet of the cone π(Con ) that coincides with the
n+1
facet π(F).
Now, the assertions of Theorem about facets and extreme rays of π(Con ) follow. (cid:3)
n+1
Theorem 1 describes all faces of the cone π(Con ) if one knows all faces of the cone
n+1
Con .
n+1
10