Complete case
We follow these
conventions, with
and
non-rational non-simplicial a priori. We use the standard inner product.
By locally we mean for each compartment.
Conventions and notations
Let
be
-compartmentalized
polytopes,
-compartmentalized
convex bodies.
is a full dimensional strictly convex cone.
Summarizing basic properties
To see the proofs, overlay the link, read, and then click the
link.
We systematically study the links between :
- (c) compartmentalized I :
,
- (c*) co-compartmentalized :
.
- (max-c) max-compartmentalized II :
,
and the following properties
In the following cases,
being complete :
(*) any
,
(+)
positive,
(<)
acute-angled.
(△) P simplicial
In particular via the followings operations :
- (⊞) Minkowski sums,
- (∩) intersection,
- (⊍) convex hull,
- (∨) polarization.
Study of the fans properties
:
(+) ⇐ (<) ⇒ (△)
(+) ⇏ (△)
(+) is stable by : subfaces star tropical modification
(<) is stable by : subfaces star tropical
modification
(△) is stable by : subfaces star tropical
modification
Case (*)
- Case (c) :
- Case (max-c)
- ⇒ (c) ❓
- (⊞) → (max-c)
- (∩)
- (⊍) → (max-c)
- (∨)
Case (+)
Lemmas
- Lemma
“Characterization of acute-angled cones via orthogonal projections”
1
-
For
a cone, the followings are equivalent.
- the orthogonal projection of
on
where
is a facet of
is included in
,
- the angle between two facets is at most
,
- Proof
-
- ⇒ 1. because you take a facet
and a facet
of
.
Then
is a support hyperplane of
.
Indeed, this is obvious if you take the cone
.
- ⇒ 2. quite clear: take two adjacent facets
whose intersection is some
.
Then, projecting along
,
we get the result.
- Lemma
“Faces of acute-angled cones are acute-angled” 2
-
(<) stable by subfaces
- Proof
(🔴)
-
It suffices to check it for facets. Assume there is a facet
diamond
with
.
The lattice above
is the one of a polygon. Take two consecutive facets
apart from
containing
.
Let
be the orthogonal onto
.
Let us prove that
.
We have
hence
We just have to prove that the three inner products are nonnegative, and
not all zeroes. For the two last ones, the non-negativity comes from
point 2, and they cannot be both zero, since otherwise
would be a codimension 1 cone containing
and parallel to
,
that is would be included in
which is clearly impossible. For the first scalar product,
is nonnegative on
,
hence on
.
So
(the dual of
in its own tangent space). Hence,
.
- Lemma
“Acute-angled cones are simplicial” 3
-
(<) ⇒ (△)
- Proof
-
By induction, all the facets are simplicial cones. Moreover, by
induction, all proper star fans are simplicial. Hence, the link of
in
is a simple and simplicial polytope. Hence it is a simplex, or of
dimension 2. It remains to deal with the case
.
This follows from the fact that the sum of the angles of a
-gon
on
is greater than
.
Hence, if all the angles are at most
,
then
.
- Lemma
“Acute-angled cones are positive” 4
-
(<) ⇒ (+)
- Proof
-
Take
and
in two rays
in the same cone. By Lemma “Acute-angled cones are
simplicial” 3, both rays belong to the same 2-cone, which is
acute-angled by @lem:acute-subface. Hence
.
- Lemma
“Compartmentalized structures are co-compartmentalized”
5
-
- ⇒ (c*)
- Proof
-
Let
and consider
.
Let
be a facet of
which is not in a facet of
.
If
,
then there exists
such that
.
Hence
,
for some
in the interior of
,
does not belong to
,
so
is not compartmentalized.
Hence, the outer normal vectors to the facets of
are either in
,
or in a ray of
.
Now,
So
verifies the condition of
.
(Counter-)examples
- Example “A
compartmentalized structure not containing zero” 1
-

- Example
“A positive fan which is not acute-angled” 2
-
Take the fan of orthants in
.
Blow-up the positive orthant in the barycenter. span of
,
,
.
- Example
“A compartmentalized structure which is not max-compartmentalized”
3
-
Take the fan in Example “A positive fan which is not
acute-angled” 2 with the polytope
.
- Example
“Two compartmentalized structures whose Minkowski sum and convex hulls
are not compartmentalized” 4
-
Take the polytope in Example “A compartmentalized structure
which is not max-compartmentalized” 3 and another one which
is
.
Then both their sums and their convex hull are not compartmentalized.
Archive
Basic
properties (of complete compartmentalized structures obsolete)
- Example
5
-
Wrong:
.
- Proof
-

- Lemma
6 (A cone always intersects its dual)
-
Let
be a cone. Then
is a full-dimensional cone.
- Proof
-
Let
a unit vector in
maximizing the minimum of
.
Then
is in the interior, otherwise we can add an normal inner vector and we
get something better.
is strictly convex. If it is
,
then we are done. Otherwise, take
in the interior of
.
Then
is strictly better than
.
Indeed, the maxima are all obtained in the same open half-space of
(they are in
and then look at the segment between them and
).
- Lemma
7
-
Let
be a cone. Then the following are equivalent
- the orthogonal projection of
on
where
is a facet of
is included in
,
- the angle between two facets is at most
,
- any face verifies 1.,
- any star fan verifies 2.,
- star fan around rays verifies 2.,
and they strictly imply
- for any
,
,
-
is simplicial.
- Proof
-
- ⇒ 1. because you take
and a facet
of
.
Then
is a support hyperplane of
.
Indeed, size is obvious if you take the cone
.
- ⇒ 2. quite clear: take two adjacent facets
whose intersection is some
.
Then, projecting along
,
we get the result.
- ⇒ 1. obvious.
- ⇔ 4. ⇔ 5., star fans preserve facets angles,
- ⇏ 2. take the span of
,
,
,
a. is true since it is true on rays, but the projection of
on the opposite facet is equal to
.
- ⇏ 2. take the span of
and
.
- ❗1. & 2. ⇒ 3. It suffices to check it for facets. Assume there
is a facet
diamond
with
.
The lattice above
is the one of a polygon. Take two consecutive facets
apart from
containing
.
Let
be the orthogonal onto
.
Let us prove that
.
We have
hence
We just have to prove that the three inner products are nonnegative, and
not all zeroes. For the two last ones, the non-negativity comes from
point 2, and they cannot be both zero, since otherwise
would be a codimension 1 cone containing
and parallel to
,
that is would be included in
which is clearly impossible. For the first scalar product,
is nonnegative on
,
hence on
.
So
(the dual of
in its own tangent space). Hence,
.
- ❗2. ⇒ b. By induction, all the facets are simplicial cones.
Moreover, by induction, all proper star fans are simplicial. Hence, the
link of
in
is a simple and simplicial polytope. Hence it is a simplex, or of
dimension 2. It remains to deal with the case
.
This follows from the fact that the sum of the angles of a
-gon
on
is greater than
.
Hence, if all the angles are at most
,
then
.
Here is an alternative proof for the dimension 3 case. Take the dual
cone
.
Take a vector
in the interior of
such that
.
Such a vector exists by Lemma 6 (A cone always intersects its
dual). Point 2. is equivalent to requiring
whenever
.
Project orthogonally the
on
.
Let us call the projection map
.
Then
(⋆) since both vectors lies in the open half-space
.
If
,
then clearly (⋆) cannot holds. Hence, the projection of
induces a complete fan of dimension two with only strictly obtuse
angles. This can only have three facets.
There seems to be no much acute-angled fans. In article,
Theorem C state that no complete fan strictly acute-angled of dimension
5 of higher exists. Maybe a reference?
We say that a cone
is acute-angled if any of the previous statement holds.
We say that a fan is acute-angled if its cones are.
A cone is positive if
.
- Example
6 (Positive cone not in orthant)
-
Take the cone
spanned by the permutations of
.
Then
for any
but
cannot be mapped orthogonally inside the positive orthant. Indeed, take
an orthant cone
.
Then three orthogonal vectors
in
.
Then ❓
❓same question when simplicial
- Example
7 (A compartmentalized structure locally but not globally
convex)
-
Wrong: Any body
-compartmentalized
whose compartments are convex is convex.
- Proof
-
Take
.
It is anti-blocking and locally convex but not globally.
- Lemma
8
-
For each
,
.
- Proof
-
Otherwise, let
be the argmax on
and
be the global argmax. Then,
belongs to a cone
for
small enough.
- Question
1
-
Are compartmentalized polytopes stable by Minkowski sum?
- Answer
-
Question
2
Question
3
Question
4