Types of problems Given a combinatorial object what are the - - PDF document
Folie 1/17 Informal Introduction General type of situation (with k points) Combinatorial object geometric realization (can be regarded as a point in IR nk ) IR n (Examples: face lattice of a polytope graph simplicial manifold)
‘geometric realization’ (can be regarded as a point in IRnk ) IRn (Examples: face lattice of a polytope graph simplicial manifold) Realization space: Set of (standard) realizations (⊆
′
IRnk )
(Standard: Fix a base (affine or projective))
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– Given a combinatorial object what are the properties
such as ≠ ∅? (existence), connected? (isotopy problem) – Characterize (up to stable equivalence, …) the subsets
combinatorial objects of the type under consideration!
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Example. A map on the torus
1 2 3 1 1 2 3 1 4 5 4 5 6 8 7 9
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Let f be an injective mapping of the set of vertices of K into
IR3 and P ∈ K a polygon
The image of P is the closed polygon arising from connecting the images of the vertices by straight line segments corresponding to the edges of P. The relative interior of a simple planar polygon (i.e. without self-intersections) is the bounded open planar region bounded by the polygon. A polyhedral embedding of K is an injective mapping f from the set of vertices of K into IR3 , such that
polygon;
are pairwise disjoint.
fa f K ⊆ IR3 denotes the union of these images.
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Let v
v
1 5 1
, , , , , … … w wn
a fixed order. A polyhedral embedding of K is called standard if
f ( )
v1
=F
)
v2 1
=F
)
v3 1
=F
f ( )
v4 1
=F
)
v5 1 1 1
=F
The realization space of K is
), , ( )) f f w wn
n 1 3
… ∈IR f is a polyhedral standard
embedding of K.p
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Definition: A (primary) semialgebraic set P ⊆ IRn is a finite intersection of sets of the form k
x x x ∈ = < IRn f g ( ) , ( ) 0 ,
where f g
x x x x
n n
, , , , ( , , ) ∈ =
I
Q
1 1
… … x
.
Proposition 1. The realization space of an abstract polyhedral complex is a semialgebraic set.
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Theorem: Let n k
, ≥ 0 and G a graph with vertex set
v
v
1 5 1
Let
3
be a semialgebraic set. Then there exists a map M which contains only triangles and quadrangles and which contains G as an induced graph, such that for each subfield
and each standard embedding
3
the following are equivalent: i) f can be extended to a polyhedral embedding
3
. i') f can be extended to an nc-embedding
3
. ii)
1
n
.
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numbers. Then there is a map M which can be polyhedrally embedded in IR3 but not in L3.
time equivalent to the ‘Existential Theory of the Reals’ and thus NP-hard.
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edges
polyhedral complex
map
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Proposition 2. Let
3
be a semialgebraic set. Then there are sets
1 5 1
n
v v
such that for each subfield K ⊆ IR and each standard embedding
3
the following are equivalent: i) f can be extended to an embedding f of the graph
3
f f ( ), ( ), ( ) x y z are collinear
for each x y z
. i') Same as i) such that in addition the image of the graph
ii)
1
n
.
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Definition: A standard handle is a map with boundary which is isomorphic to H, where H:= The quadrangle 6 7 9 8 is removed. Lemma 1: a) For each polyhedral embedding of H the quadrangle
6 7 9 8 is planar and strictly convex.
b) Each embedding of 6 7 9 8 as a strictly convex quadrangle can be extended to a polyhedral embed- ding of H within any arbitrary pyramid having the quadrangle as its base.
6 7 8 9
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Lemma 2: There exists an admissible polyhedral complex K0 with three vertices x y z
, , , such that for each injective mapping f : , , x y z
i) f can be extended to a polyhedral embedding f of K0 ii) f
Moreover:
f f ( ), ( ), ( ) x y z are collinear then for each ε > 0 and
each extension of f to an embedding g of the graph
K3 3
, the extension f can be chosen such that
f (
) ( ( )).
,
K0
3 3
⊆ U g K
ε
v
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Proof of Lemma 2:
1 is composed from 8 standard handles which intersect
quadrangles are
The planarity of these quadrangles for a polyhedral embedding of K
1 (Lemma 1) implies that all the vertices
are coplanar:
not collinear.
e f z b g a d c h y x
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z x y e f b g a d c h z x y e f b g a d c h
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Construction of K0
1, which are disjoint except
for the common vertices x y z
, , .
Since the graph K3 3
, is not planar, at least two of the
three planes are different and thus x y z
, , are collinear.
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We begin with the sets W of vertices, E of edges constructed in the first step and encode each collinearity in
C according to Lemma 2.
Then we add the edges of G and additional vertices ∗ such that the resulting polyhedral complex K is connected.
K is admissible and orientable and all its polygons are
quadrangles. Using Proposition 2 and Lemma 2 we get: pairs of edges
∗
with additional
K has all properties required for M in the
universality theorem, except for the property of being a map.
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Universal extension
Definition: Let L be an abstract polyhedral complex and K
⊆
a subcomplex.
L is called a universal extension of K if for each ε > 0
and each polyhedral embedding f : K → IR3 there exists an extension of f to f : L → IR3 such that i) f is a polyhedral embedding, ii) the image f L
f K
iii) the vertices of L
are in Q
I 3.
Proposition 3: Let K be a connected admissible abstract polyhedral complex. a) Then there exists a universal extension of K to a map
M (with boundary).
b) If K is orientable then there even exists a universal extension of K to a map M (without boundary). Moreover M can be chosen such that all new polygons are triangles.
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