15.2 Delaunay Polytopes Related to Faces
229
and
the
pentagonal inequalities (6.1.9). For n = 6,
they
are
defined by
the
triangle inequalities,
the
pentagonal
inequalities,
and
the
inequalities
bibjXij
::;
0 for b =
(2,1,1,
-1, -1,
-1)
and
(-2,
-I,
1, 1, 1,
1)
(up
to
permutation
of
the components).
(The
description
in
the
case n = 6
was
obtained
independently by Baranovskii [1971]
and
Avis [1989].)
(ii)
When
n = 7,
the
hypermetric inequality 2:1:5i<j:57 bibjXij
::;
0 for b =
(3,1,1,
I,
-1,
-2,
-2)
defines a facet
of
HYP7,
but
not
of CUT7. Indeed,
there
are precisely
19
linearly independent
cut
semimetrics satisfying
this
hypermetric
inequality
at
equality.
An
additional hypermetric distance
d satisfying equality
can
be
obtained
in
the
following manner: Consider
the
graph
G
g
shown
in
16.2.4 (labeling its nodes as
1,2,3,4,5,6,7
corresponding to degrees
3,2,2,2,5,1,1)
and
set d'j
:=
2 if
ij
is
an
edge
of
Gg, dij
:=
1 if
ij
is
not
an
edge
in
Gg.
This
distance d together
with
19
cut
semimetrics form a set of
20
linearly independent vectors satisfying
the
hypermetric equality; this shows
that
it defines a facet of HYP7.
In
fact, Baranovskii [1995] describes all
the
facets of
the
cone HYP7'
They
are
the
hypermetric facets for CUT7 (see their list
in
Section 30.6) to-
gether
with
the
facets defined by
the
inequalities
2:1::;i<i:57
bibjXij
S;
0, for
b =
(3,1,1,1,-1,-2,
-2),
(-3,1,1,1,1,
2),
(3,-1,-1,-1,1,-2,2),
and
(-3,1,1,
-1,
-1,2,2)
(up
to
permutation).
(iii)
There
is
an
easy way of constructing new hypermetric facets from given
ones, namely, using
the
so-called 'switching' operation.
This
operation
will be described
in
detail
in
Section 26.3;
we
indicate here how
it
acts
on
hypermetric
inequalities. Given b E
zn
and
a subset A C
Vn
{I,
...
,n},
define
the
vector b
A
E
zn
by bf
:=
-bi
if
i E
.4
and
bf
:=
b
i
if i E
Vn
\ A.
If
2::'=1
bi
= 1
and
b(A.) = 0,
then
the
inequality
bfbfXij
S;
0 is
again a hypermetric inequality.
In
fact,
The
hypermetric inequality 2:1
<i<j<n
bibjXij
S;
0 defines a facet
- - A A
of
HYP
n
~
its
switching
2:1:5i<j:5n
b
i
b
j
Xij
S;
0 defines
(a)
a facet of
HYP
n'
See for
an
example
the
facets cited
in
(ii) above. We briefly sketch
the
proof
for assertion (a).
Consider
the
mapping
T5(A) :
~(;)
-----t
~C)
where Y =
TS(A)(X)
is
defined
by
Yij
1
Xij
if
8(A)ij
= 1
and
Yij
:=
Xij
if
8(A)ij
O.
As
HYP
n
is a polyhedral cone,
we
can suppose
that
HYP
n is defined by
the
hypermet-
ric inequalities
2:1
5i
<j:O;n
bibjXij
S;
0 for b E
13,
where
13
is
a finite
subset
of
{bEznl
b
i
I}. Set
13'
:=
{b
A
I
bE
13,
A
~
V
n
}.
For d E
HYP
n
,
set
where
the
minimum is
taken
over all b E
13'
for which 2:1<i<j<n
bibjd;j
>
0,
setting
ad
1 if there is no such
b.
One
can
easily
verify-that
T6(A)(add)
E