Документ взят из кэша поисковой машины. Адрес оригинального документа : http://higeom.math.msu.su/people/taras/TTC/slides/lu.pdf
Дата изменения: Fri Jun 2 20:49:02 2006
Дата индексирования: Sun Apr 10 01:25:14 2016
Кодировка:
Home Page

Top ology on Graphs
Title Page Contents

Zhi Lu Ё Institute of Mathematics, Fudan University, Shanghai.

Page 1 of 22

Osaka, 2006
Go Back

Full Screen

Close

Quit


Home Page

§1. Ob jective
Title Page Contents

· Graphs = · · · = Geometric Ob jects

· Two basic problems
Page 2 of 22

- - - Under what condition, is a geometric ob ject a closed manifold? - - - Can any closed manifold be geometrically realizable by the above way?

Go Back

Full Screen

Close

Quit


Home Page

§2. Background · GKM theory--by Goresky, Kottwitz and MacPherson in 1998 (see [Invent. Math. 131, 25-83]).
GKM manifolds T
k

Title Page

Contents

M

2n

A unique regular GKM graphs

M

of valency n
Page 3 of 22

A GKM manifold is a T -manifold M with · |M T | < + · M having a T k -invariant almost complex structure · for p M T , the weights of the isotropy representation of T k on TpM being pairwise linearly independent.

k

2n

Go Back

Full Screen

Close

Quit


Home Page

§3. Coloring graphs and faces
Title Page

Let G = (Z2)k . Given a G-manifold M with |M G| < with properties as follows: a natural map : EM - Hom(G, Z2) e - such that

regular graph M

Contents

Page 4 of 22

A) for each p VM , (Ep) spans Hom(G, Z2) B) for each e = pq EM and (Ep), the number of times which and + (e) o ccur in (Ep) is the same as that in (Eq ).

Go Back

Full Screen

Close

Quit


Home Page

-- Abstract definition Let G = (Z2)k . We shall work on H (B G; Z2) = Z2[a1, ..., ak ] ( Hom(G, Z2)).

Title Page

H 1(B G; Z2) =

Contents

n: a connected regular graph of valency n with n k and no loops. If there is a map : E - H 1(B (Z2)k ; Z2) - {0} s. t. (1) for each vertex p V, the image (Ep) spans H 1(B (Z2)k ; Z2), and (2) for each edge e = pq E, (x )
xEp -Ee y Eq -Ee
Full Screen

Page 5 of 22

Go Back

(y )

mod (e),

then the pair (, ) is called a coloring graph of typ e (k , n).
Close

Quit


Home Page

-- Examples
a3

Title Page

(, 1 ) is a coloring graph
a1 a2

Contents

a1

a2 a3

1 : E - H 1 (B (Z2 )3 ; Z2 ) where H (B (Z2 )3 ; Z2 ) = Z2 [a1 , a2 , a3 ].

Page 6 of 22

a3

(, 2 ) is not a coloring graph
a1 a1 + a2

Go Back

a1

a2 a3

a1 a2 a1 (a1 + a2 ) mod a

3

Full Screen

Close

Quit


Home Page

--Faces (, ): a coloring graph of type (k , n). : a connected -valent subgraph of where 0 n. If ( , | ) satisfies a) for any two vertices p1, p2 of , ((E | )p1 ) and ((E | )p2 ) span the same subspace of H 1(B G; Z2); b) for each edge e = pq E | , (x )
x(E | )p -(E | )e y (E | )q -(E | )e

Title Page

Contents

Page 7 of 22

(y ) mod (e)
Go Back

then ( , | ) is an -face of (, ).
Full Screen Close

Quit


Home Page

Example

Title Page

Contents

a2 a1 a2 a3 a1

is a 2-face
a1 + a2

a1 + a2

a2 + a3

a1 + a3

a coloring graph (, )
a1 a3

Page 8 of 22

Go Back

is not a 2-face
Full Screen

a1 + a2

a2 + a3
Close

Quit


Home Page

Assumption--Case: valency n of = rank k of G = (Z2) (, ): a coloring graph of type (n, n) with connected. F(,): the set of all faces of (, ). -- An application for the n-connectedness of a graph.

k
Title Page

Contents

Theorem (Whitney) A graph with at least n + 1 vertices is n-connected if and only if every subgraph of , obtained by omitting from any n - 1 or fewer vertices and the edges incident to them, is connected. Theorem (Z. Lu and M. Masuda). Suppose that (, ) is a Ё coloring graph of type (n, n) with connected. If the intersection of any two faces of dimension 2 in F(,) is either connected or empty, then is n-connected.

Page 9 of 22

Go Back

Full Screen

Close

Quit


Home Page

Example
a a
2 1

Title Page

a a2 + a

3

a
3

3

a

2

Contents

3

a1 + a

2

a1 + a

a2 + a3 a1 + a3 a1 + a a
1

2

a a
2

1

a a
2

1

a

3

a

3
Page 10 of 22

a1 + a

2

a1 + a2 a a
1

a1 + a

3

a1 + a a
1

3
Go Back

a2 + a
2

3

a2 + a
3

3

a
3

3

a1 + a2

a1 + a

a

3

a a2 + a
3

2

a1 + a

3

a1 + a2

Full Screen

a2 + a

Close

Quit


Home Page

§4. Geometric realization (, ):a coloring graph of type (n, n)=F(,)=|F(,)| Example 1.
four 2-faces:
a a a1 + a
2 1

Title Page

Contents

1

a

1

a1 + a a
3

a1 + a
3

2

a1 + a a
3

3

a1 + a

2

a1 + a

3

a a
2

2

Page 11 of 22

a2 + a

3

a

2

a

a2 + a
3

3
Go Back

(, ):a coloring graph of type (3, 3)

a2 + a

3

The geometric realization |F(,)| = S

2

Full Screen

Close

Quit


Home Page

Example 2.
a1 a3 a1 a2 a3 a2 a1 a2 a3 a1 a3 a2 a3 a3

Title Page

Contents

(, ):a coloring graph of type (3, 3).
Page 12 of 22

a1 a2

The geometric realization |F(,)| = RP

2

Go Back

a2
Full Screen

a1
Close

Quit


Home Page

Generally, Fact. F(,) forms a simplicial poset of rank n with respect to reversed inclusion with (, ) as smal lest element. |F(,)| is a pseudo manifold. poset means partially ordered set A poset P is simplicial if it contains a smallest element ^ and for each a P the segment [^, a] is a boolean 0 0 algebra (i.e., the face poset of a simplex with empty set as the smallest element).

Title Page

Contents

Page 13 of 22

Go Back

Full Screen

Close

Quit


Home Page

P : a simplicial poset a simplicial cell complex KP in the following way: for each a = ^ in P , one 0 such that its face poset all obtained geometrical to the ordered relation in complex as desired. obtains a geometrical simplex is [^, a], and then one glues 0 simplices together according P , so that one can get a cell

Title Page

Contents

Page 14 of 22

By |P | one denotes the underlying space of this cell complex, and one calls |P | the geometric realization of P .

Go Back

Full Screen

Close

Quit


Home Page

Basic problems: (I). Under what condition, is the geometric realization |F(,)| a closed topological manifold? (II). For any closed topological manifold M n, is there a coloring graph (, ) of type (n + 1, n + 1) such that M n |F(,)|?

Title Page

Contents

Page 15 of 22

Go Back

Full Screen

Close

Quit


Home Page

Basic problem (I) (, ): a coloring graph of type (n, n) with connected. The case n = 1: |F(,)| S
0

Title Page

Contents

The case n = 2: it is easy to see that for any coloring graph (, ) of type (2, 2), the geometric realization |F(,)| is always a circle. The case n = 3: Fact.|F(,)| is a closed surface S .

Page 16 of 22

Go Back

Full Screen

Close

Quit


Home Page

Generally, if n > 3, the geometric realization |F(,)| is not a closed topological manifold. For example, see the following coloring graph (, ) of type (4, 4).
a
2

Title Page

Contents

a a a
3 1

3

a

1

a

2

a

4

a

4

a

4

a

4

Page 17 of 22

a a a
2 3

1

a a
1

2

a

3
Go Back

(|F(,)|) = 5 - 12 + 16 - 8 = 1 = 0 so |F(,)| is not a closed topological 3-manifold.

Full Screen

Close

Quit


Home Page

The case n = 4. Write v = |V| and e = |E| so 2v = e. f : the number of all 2-faces in F(,) f3: the number of all 3-faces in F(,) Theorem. Let n = 4. |F(,)| is a closed connected topological 3-manifold f = f3 + v .

Title Page

Contents

Page 18 of 22

Problem: for n > 4, to give a sufficient (and necessary) condition that |F(,)| is a closed connected topological manifold.

Go Back

Full Screen

Close

Quit


Home Page

Basic problem (I I) M n:n-dim closed connected topological manifold 1-dim case: M S . S1 is realizable by any coloring graph of type (2, 2).
1 1

Title Page

Contents

Page 19 of 22

2-dim case: Prop. Any closed surface can be realized by some coloring graph of type (3, 3).

Go Back

Full Screen

Close

Quit


Home Page

3-dim case: Conjecture: Any closed 3-manifold M 3 is geometrical ly realizable by a coloring graph (, ) of type (4, 4), i.e., M 3 |F(,)|. 4-dim case: It is well known that there exist closed topological 4-manifolds that don't admit any triangulation. closed topological 4-manifolds that cannot be realized by any coloring graph of type (5, 5).

Title Page

Contents

Page 20 of 22

Go Back

Full Screen

Close

Quit


Home Page

Prop osition. Let M n be a closed manifold. If M n admits a simplicial cel l decomposition with at least n + 2 vertices, then M n can be geometrical ly realizable by a coloring graph.

Title Page

Contents

Page 21 of 22

Go Back

Full Screen

Close

Quit


Home Page

Restatement Prop osition. Suppose that is a 3-valent graph and is at least 2-connected. Then is planar if and only if admits a coloring of type (3, 3) such that |F(,)| S 2.

Title Page

Contents

Page 22 of 22

Go Back

Full Screen

Close

Quit