Transcript CLASSICAL INFORMATION THEORY

MA5209 Algebraic Topology
Lecture 5. Simplicial Homology
(18, 29 September, 2, 13, 16, 20, 23 October 2009)
Wayne Lawton
Department of Mathematics
National University of Singapore
S14-04-04, [email protected]
http://math.nus.edu.sg/~matwml
Homology Groups
A differential group is an abelian group C and an
endomorphism  : C  C that satisfies   0.
 is the differential or boundary operator
Z (C )  ker()  { z  C : z  0 } subgroup of cycles
B(C )  im()  { z : z  C } subgroup of boundaries
Remark Clearly
  0  B(C)  Z (C)
Definition H (C )  Z (C ) / B(C ) is the homology
group and its elements are homology classes.
Oriented Simplices
Definition An orientation of a d-dimensional
simplex   v0v1 vd is one of two equivalence
classes of orderings (permutations) of its vertices
where two permutations are equivalent if they
differ by an even permutations. [v (0)v (1) v ( d ) ]
denotes the equivalence class for permutation  .
An oriented simplex is a simplex with an orientation.
Examples [v0v1 ]  [v1v0 ]
[v0v1v2 ]  [v1v2v0 ]  [v2v0v1 ]  [v1v0v2 ]  [v2v1v0 ]  [v0v2v1 ]
Result Two permutations are equivalent iff one
can be obtained from the other using an even
number of transpositions.
Group of q-Chains
Definition Let K be a simplicial complex and let
q  0 and let Cq denote the abelian group
generated by the oriented q  simplices in K
together with the relations  1  q2 whenever
 1 and  2 are oriented simplices having the same
simplices but inequivalent orientations.
Example K  { v0 , v1, v2 , v0v1 }; Cq  0, q {0,1}
C0 (K )  gen{ [v0 ],[v1 ],[v2 ] : no relations}  Z
C1 ( K )  gen{ [v0v1 ],[v1v0 ] : [v1v0 ]  [v0v1 ]}  Z
3
1
Cq (K ) is a free abelian group whose rank equals
the number of q  simplices in K and is
called the group of q  chains in K .
Boundary Operator
Definition The group of chains of a simplicial
complex K is C(K )  C1 (K )  C0 (K )  C1 (K )  C2 (K ) 
Lemma The homomorphisms  q : Cq ( K )  Cq1 (K )
i
given by  q [v0 vq ]   (1) [v0 vˆi vq ], where we
0i q
define [v0 vˆi vq ]  [v0 vi 1vi 1 vq ], are well
defined, satisfy  q 1 q  0, and define a boundary
operator by    1  0  1   2 : C(K )  C( K ).
Proof 1st claim: a direct tedious computation gives
 q [v0 v j 1v j 1v j v j 2 vq ]  q [v0 vq ] whence the
claim follows since every permutation is a product
of transpositions (permutation switching 2 things).
Boundary Operator
Lets do this tedious work since it is necessary !
 q [v0 v j 1v j 1v j v j 2 vq ] 
(1) j [v0 v j 1v j v j 2 vq ] 
i
(

1
)
 [v0 vˆi v j 1v j v j2 vq ]
0i  j 1
 (1) j 1[v0 v j 1v j 1v j 2 vq ] 
 (1) j 1[v0 vˆ j 1 vq ] 
 (1) j [v0 vˆ j vq ] 
i
(

1
)
 [v0 v j 1v j 1v j vˆi vq ]
j  2i q
i
(

1
)
 [v0 vˆi vq ]
0i  j 1
i
(

1
)
 [v0 vˆi vq ]
j  2i q
   (1)i [v0 vˆi vq ]  q [v0 vq ].
0i q
Boundary Operator
Proof 2nd claim: it suffices to show that
 q1 q [v0 vq ] 
i
(

1
)
  q1[v0 vˆi vq ]  0.
0i q
We compute
 q1[v0 vˆi vq ] 
 (1) [v0 vˆ j vˆi vq ] 
j
0 j i 1
 (1) [v vˆ vˆ v ]
j
0
i 1 j q
i
j
q
The result now follows since
i
(

1
)

0i q

j
(

1
)
 [v0 vˆ j vˆi vq ] 
0 j i 1
i j
(

1
)
[v0 vˆi vˆ j vq ] 

0i  j q
i j
(

1
)
[v0 vˆ j vˆi vq ]

0 j i q
i
(

1
)

0i q
j
(

1
)
 [v0 vˆi vˆ j vq ].
i 1 j q
Boundary Operator
Proof 3rd claim: follows since
 
 1  0  1   2  1  0  1   2  
 0 1  1 0   21  3 2  
0 : C ( K )  C ( K ).
Remark Cq (K )  0, q  1 (the trivial group) and
 q  0 : Cq (K )  Cq1 ( K ), q  0.
C (K ) is a graded group and  : C ( K )  C ( K )
is a graded homomorphism of degree  1.
Computation of Homology Groups
Definition H q ( K )  ker( q ) / im( q1 ), q  Z
is the q  th homology group of K .
Result H ( K )  H0 ( K )  H1 ( K )  H3 ( K ) 
2
K

{
v
,
v
};
C

gen
{[
v
],
[
v
]}

Z
, C1  0
Example 1.
0 1
0
0
2
2


ker(0 )  C0 and im(1 )  0  H0 ( K )  Z .

Example 2. K  {v0 , v1 , v2 , v0v1 , v1v2 , v1v2} 

C0  gen{[v0 ],[v2 ],[v2 ]}  Z 3 , C1  gen{[v0v1 ],[v1v2 ],[v2v0 ]}  Z 3.
1[vavb ]  [va ]  [vb ]  Z1 ( K )  ker(1 )  {k[v0v1 ]  k[v1v2 ]  k[v2v0 ] : k  Z}  Z.
and B0 (K )  im(1 )  {k0[v0 ]  k1[v1 ]  k2[v2 ] : k0  k1  k2  0}  Z 2 .
Z0 (K )  ker(0 )  ker(0)  C0  Z 3. B1(K)  im( 2 )  im(0)  0.
H0 (K )  Z0 (K ) / B0 (K )  Z 3/Z2  Z.
H1 ( K )  Z1 ( K ) / B1 ( K )  Z/0  Z .
Computation of Homology Groups
Example 3. K  { properfacesof v0v1v2v3}
C3  0 , 3  0, B2 ( K )  0
C2  { k1[v0v1v2 ]  k2[v0v2v3 ]  k3[v0v3v1 ]  k4[v1v3v2 ]}  Z 4
C1  { k1[v0v1 ]  k2[v0v2 ]  k3[v0v3 ]  k4[v1v2 ]  k5[v1v3 ]  k6[v2v3 ]}  Z 6
 2 : C2  C1 is represented by the following 6 x 4 matrix
 1 0 1 0 
 1 1 0 0 


 0 1 1 0 
M 

 1 0 0  1
 0 0 1 1 


 0 1 0  1
Linear Algebra 
r
Z 2 ( K )  Z where
r  dim(C2 )  rank(M )  4  3  1
and hence
H 2 ( K )  Z 2 ( K ) / B2 ( K )  Z / 0  Z
The sphere has one 3-dim hole
Computation of Homology Groups
Example 4. K  { properfacesof v0v1v2v3} {v0v1v2v3}
C0  Z[v0 ]  Z[v1 ]  Z[v2 ]  Z[v3 ]
C1  Z[v0v1 ]  Z[v0v2 ]  Z[v0v3 ]  Z[v1v2 ]  Z[v1v3 ]  Z[v2v3 ]
C2  Z[v0v1v2 ]  Z[v0v2v3 ]  Z[v0v3v1 ]  Z[v1v3v2 ]
C3  Z[v0v1v2v3 ]
B2 ( K )  Z ([v0v1v2 ]  [v0v2v3 ]  [v0v3v1 ]  [v1v3v2 ])
 2 : C2  C1 represented 6 x 4 matrix M on preceding page
Z2 ( K )  Z ([v0v1v2 ]  [v0v2v3 ]  [v0v3v1 ]  [v1v3v2 ])  B2 ( K )
and hence H 2 ( K )  Z 2 ( K ) / B2 ( K )  0.
The ball does NOT have a 3-dim hole
Smith Normal Form
Definition A commutative ring R is principle if
every ideal in R has the form Rg for some g in R
Examples Z = ring of (rational) integers,
F[z] = ring of polynomials over a field F
Theorem [1] Every m x n matrix A over a PID R equals
A  USV where U  GL(m, R), V  GL(n, R)
mn
S  diag(d1, d2 ,...,ds )  R , di | di 1
Proof and Algorithm: see the very informative article
http://en.wikipedia.org/wiki/Smith_normal_form
which also discusses applications to homology and the
structure theorem for finitely generated modules over pid
1] Smith, H. M. S., On systems of indeterminate equations and congruences, Philos. Trans., 151(1861), 293--326.
http://en.wikipedia.org/wiki/Henry_John_Stephen_Smith
MATLAB Code
MATLAB Code available at
http://mathforum.org/kb/thread.jspa?forumID=80&threadID=257763&messageID=835342#835342
>> help smith
Smith normal form of an integer matrix.
[U,S,V] = smith(A) returns integer matrices U, S, and V such that
A = U*S*V',
S is diagonal and nonnegative, S(i,i) divides S(i+1,i+1) for all i,
det U =+-1, and det V =+-1.
s = smith(A) just returns diag(S).
Uses function ehermite.
[U,S,V] = smith(A);
This function is in some ways analogous to SVD.
Remark The matrix V is the transpose of the matrix
appearing in the previous page!
Example
>> A = [2 4 4;-6 6 12;10 -4 -16]
>> U
U=
A=
2
-6
10
4 4
6 12
-4 -16
>> [U,S,V] = smith(A);
>> S
S=
2
0
0
0
6
0
0
0
12
1
-12
17
0
-3
4
0
-2
3
2
-3
0
2
-2
-1
>> V'
ans =
1
-1
0
>> U*S*V'
ans =
2
-6
10
4 4
6 12
-4 -16
Finitely Generated Abelian Groups
Theorem Every finitely generated abelian group
r where
di
G  Z  Z  Z  Z
d1
Proof If
G
d2
ds
is generated by elements g1 ,..., g m then
h:Z G
h(ei )  gi , i  1,...,m. Since h is surjective
there exists a unique homomorphism
such that
the 1st
| di 1
http://en.wikipedia.org/wiki/Isomorphism_theorem
m
for groups gives
G  Z / ker(h). Clearly ker(h)  AZ where
mn
A  Z is any matrix whose columns generate ker(h).
Let A  USV be the Smith form of A where
m
S  diag(d1, d2 ,...,ds )  Z
m
mn
.
The result follows by
G  Z m / USVZ n  Z m / SZ n  Z m /(d1Z  ds Z  (0)r ), r  m  s
Smith Form in Homology
 q1
q
Consider Cq 1  Cq  Cq 1
mi  dim Ci and choose a fixed bases for Cq1, Cq , Cq1
 q 1 is represented by
mq mq1
mq1 mq
and  q is represented A  Z
Aq 1  Z
q
with Smith form Aq  U q SqVq , Sq  diag(d1 ,...,dt ).
then
Ci  Z
mi
and
Hq  ker(Aq ) / im( Aq1 )  ker(Sq ) / im(Vq Aq1 ) where
( mq  s )mq1
mq  s
s
so we let B  Z
ker(Sq )  (0)  Z
be the lower submatrix of Vq Aq1 and B  UDV be its
( mq s )mq1
Smith form with D  diag(b1 ,...,bt )  Z
r
Then Hq  Zb1  Zbt  Z , r  mq  s  t.
Homology of the Torus
K
v 3 e3
e1
v2 e 2
v1
e 4 f 1 e5 e6 f 3 e7 e8 f 5 e9
f
f
f
e12 6
v4 e10 2 v6 e11 4 v7
e
e13 f 7 e14 e15 f 9 16f e17 f11 e18
10
f
f8 v
8
v5 e19
e21 12
e20 v9
e
e
e
e22 f13 23e f15 25 e26 f17 27
f16
f18
f14 24
v1
e1
e2
v2
v 3 e3
v1
e4
v4
e13
v5
e22
v1
2  Z
2718
, 1  Z
927
all 27 edges oriented
all 18 faces oriented
counterclockwise
Question Compute the matrices for
 2 and 1 and use
them with MATLAB to compute the homology groups of
K.
Suggestion : I attempted to compute the matrix for
 2 on the next page but did not check it for errors, the
matrix should have rank = 18 - dim ker = 17
Homology of the Torus
 
T
2
1
1
2
3
1
4
5
1
1
1
2
1
3
6
7
1
6
11 12
13 14
1
1
1
1
23
24
25
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
27
1
1
1
17
26
1
1
1
15
18
1
1
1
13
16
21 22
1
1
9
14
20
1
8
12
19
1
1
11
18
1
7
10
15 16 17
1
1
1
10
1
1
5
9
1
1
4
8
1
1
1
1
1
Computation of H 0 ( K )
Theorem H0 ( K )  Z r , r # connectedcomponentsof | K |
Proof
v, w  vert( K ) are in the same connected
component of the polyhedron | K | iff there exist a path
p : [0,1]  | K | such that p(0)  v, p(1)  w. The
simplicial approximation theorem implies that this occurs
iff there exist
such that
v  v0 , v1,...,vn  w  vert( K )  Z0 ( K )
vi vi 1  K , i  0,1,...,n 1. Then the chain
z   i 0 [vi vi 1 ] satisfies 1 z   n1 (vi 1  vi )  w  v  B0 ( K )
i 0
n 1
and hence
v  B0 ( K )  w  B0 ( K )  H1 ( K ).
We leave it to the reader to prove the converse.
Why Elements in Z q (K ) are Called Cycles
We consider the case q  1.
Definition A loop in K is a chain z 

n 1
i 0
[vi vi 1 ]  C1 ( K )
where v  v0 , v1 ,...,vn  v0  vert( K )
vi vi 1  K , i  0,1,...,n 1.
n 1
Clearly 1 z   i 0 (vi 1  vi )  vn  v0  0
such that
and hence z  Z1 ( K ) so every loop is a cycle.
Theorem Every element in Z1 ( K ) is a lin. comb. of loops.
Proof Express Z1 ( K )  z 

ij
aij [vi v j ] with aij  0.
then construct a loop w and c  0 such that
z  cw   i j bij [vi v j ] has less coefficients, then use induction.
We suggest that the reader work out a detailed proof.
Normal Subgroups
Definition Let G be a group. A subgroup S  G is
1
normal if for every g  G we have gSg
 S.
Lemma If S is a normal subgroup of G then the set
G / S of left cosets of S is a group (called the quotient
of G by S ) under the binary operation defined by
g1S g2 S   g1g2 S . Moreover, if Si  Gi , i  1, 2
are normal subgroups then every homomorphism
h : G1  G2 that satisfies h S1   S2 induces a
~
homomorphism h : G1 / S1  G2 / S2 defined by
~
h ( g1S )  h ( g1 ) S , g1  G1.
Proof Left to the reader.
Commutator Subgroups
Definition Let G be a group. The commutator subgroup
[G, G] (also called the first derived subgroup and
(1)

denoted by G or G ) of G is the subgroup generated
by the set  aba
1 1

b : a, b  G .
Lemma For every homomorphism
h : G  G,
h([G, G])  [G, G]. Proof Left to the reader.
Corollary [G, G] is a normal subgroup of G .
Proof For each g  G construct
1
hg ( x)  gxg , x  G. Clearly
hg : G  G by
hg is a homomorphism
and hence the lemma implies that for every g  G
g[G, G]g 1  hg ([G, G])  [G, G] so [G, G] is normal.
The Abelianization Functor
Lemma If G is a group then  (G)  G /[G, G] is abelian.
1 1
xyx
y  1 (G) , x, y  (G).
Proof It suffices to show that
Given x, y  G /[G, G] choose a, b  G such that
x  a[G, G] and y  b[G, G]. Then compute
1
1
1 1
xyx y  aba b [G, G]  [G, G]  1 (G)
Question For groups G1 and G2 and a homomorphism
h : G1  G2 define  (h) :  (G1 )   (G2 ) by
 (h)g1[G1 , G1 ]  h( g1 )[G2 , G2 ].

Show that
is a functor from the category of groups
to the category of abelian groups.

Question Prove
is the left adjoint of the inclusion
functor. See http://en.wikipedia.org/wiki/Commutator_subgroup
The Abelianization Functor
Theorem If S is a closed surface with genus g  0 then
 1 ( S )  a1 , b1 ,, a g , bg | a1b1a11b11  a g bg a g1bg1  1
when
S is orientable and
 1 ( S )  a1 , , a g | a12  a g2  1 when S is nonorientable.
Proof http://wapedia.mobi/en/Fundamental_polygon
Theorem H1 (S )    1 ( S ) Proof To be given later.
Corollary H1 ( S )  Z 2 g when
S is orientable and
g 1
H1 (S )  Z  Z2 when S is nonorientable.
Proof The first assertion is left to the reader. Clearly if
S is nonorientable then  1 ( S )  c1 ,, cg | cg2  1 where
c1  a1,, cg 1  ag 1, cg  a1a2 ag whence the 2nd assertion.
Homology from Homotopy
Theorem If K is a simplicial complex whose polyhdron
| K | is connected and v  | K | then H1 ( K )   1 (| K |, v).
Proof Let E ( K , v) be the edge group defined on page 132
in Armstrong’s Basic Topology (BT). Theorem 6.1 in BT
proves (using the simplicial approximation theorem) that
1 (| K |, v)  E( K , v) and hence it suffices to prove that
H1 ( K )   E( K , v). Define  : E( K , v)  H1 ( K ) by
 ( g  {  vv1 vk v})  z()  B1 ( K ), z( L)  [vv1 ]  [v1v2 ]  [vk v].
Clearly equivalent edge loops e give homologous z(e),
so  is well defined.  is surjective since every cycle is
a linear combination of ‘loops’ of the form z(g). Moreover
g  ker  v v  K , i  1,...,m  z ()  
i
1
i
pi

m
i
i
n
[
v

v
pi ]
i 1 i 1

Homology from Homotopy
For each i  1,...,m choose an edge path  i joining
to
i
1 and construct the edge loop
v

m
~
i
i
1
  i 1  i v1 v pi  i
Since

ni
v
L is a
max tree
~
v v  K , i  1,...,m  { }  1 E ( K , v)
i
1
and hence
i
pi
~ 1
{  }  {} E(K , v).
Moreover z ()  

m
i
1
i
pi
n [v  v ]
i 1 i
 implies that
~ 1
z(  )  0. Therefore for every oriented edge (a,b)
~1
that occurs k-times in z(  ) the oriented edge (b,a)
also occurs k-times. Since by Theorem 6.12 in BT
~ 1
E ( K , v)  G( K , L) we have {  }[G( K , L),G( K , L)].
Chain Complexes and Chain Maps
Chain Complex: differential graded abelian group
(C, d ) : 
 Cq1 ( K )  Cq ( K )  Cq1 ( K ) 
d q2
d q1
dq
d q1
Chain Maps: deg 0 homomorphism

(C, d ) 

(C, d ) that gives a commutative diagram:
d q1
dq
 Cq 1 

Cq  Cq 1 
  q 1
 q
  q 1
d q 1
d q
 Cq 1 

Cq  Cq 1 
Theorem Chain complexes & chain maps are a
category. Homology is a functor from it to the cat.
of graded abelian groups&deg-0 homomorphisms
Proof Left to the reader.
Example: Simplicial Homology
Theorem If K, L are simplicial complexes and
s : | K |  | L | is a simplicial map then the maps
q : Cq ( K )  Cq ( L) defined by:
q ([v0v1 vq ])  [s(v0 )s(v1 )s(vq )]
( = 0 if s(v0 ), s(v1 ),, s(vq ) are not distinct)
  C (s)
give a chain map (Csimp ( K ), ) (Csimp ( L), )
Proof Left to the reader.
Corollary Csimp is a functor from the category of
simplicial complexes & simplicial maps to the
category of chain complexes and chain maps.
Corollary H  Csimp is a functor from SIMP to GAB.
sim p
Example: Singular Homology
Definition For n  0,1,... let   (v0v1 vn )  R
be a standard ordered n-simplex and define
i
n1
n
simplicial maps  n :    , i  0,...,n
n
by
n1
v
,
j

i
n 1
j
(this
maps
 (v j ) 
 onto
v j 1 , j  i the i  th face of n )
i
n
For any topological space X let Csing, n ( X ) be the
n
free abelian group generated by maps f :   X
let the boundary operator  n : Csing, n ( X )  Csing, n-1 ( X )
n
i
i
be  n f   i 0 (1) f   n
Theorem Csing is a functor from category TOP to
category CHAIN and H  Csing= sing. hom. functor.
Chain Homotopy
between two chain maps f , g : ( A, d A )  ( B, d B )
is a degree 1 homomorphism h : ( A, d A )  ( B, d B )
 An1 
 An 
 An1 
d A,n1
hn
d A,n
n
hn 1
 Bn1 
 Bn 
 Bn1 
such that   f  g  d B h  hdA
d B ,n1
d B ,n
Lemma f h g, g h k  g h f , f h h k
Theorem f h g  H ( f )  H ( f ) : H ( A)  H ( B)
Proof a  Z n ( A)  f (a)  g (a)  d B h(a)  hdA (a)
1
2
1
1
2
 d B h(a)  Bn ( B)  H ( f )(a)  H ( g )(a)  H n ( B)
Chain Homotopy
f 2 f1 , g 2 g1
(
A
,
d
)



(C, dC ) of chain
Lemma Composites
A
f1 , g1
f 2 , g2
(B, dB ) (C, dC )
homotopic maps ( A, d A ) 
are chain homotopic. This means that
f1  g1  d B h1  h1d A and f 2  g2  dC h2  h2d B
  a chain homotopy h : ( A, d A )  (C, dC )
such that f 2 f1  g2 g1  dC h  hdA .
Proof Let h  f 2h1  h2 g1 : ( A, d A )  (C, dC ). Then
h is a degree 1 homomorphism and dC h  hdA 
dC f 2h1  dC h2 g1  f 2h1d A  h2 g1d A 
f 2d B h1  dC h2 g1  f 2h1d A  h2d B g1 
f 2 ( f1  g1 )  ( f 2  g2 ) g1  f 2 f1  g2 g1.
Chain Homotopy
Definition A chain equivalence between two chain
complexes is a chain map f : ( A, d A )  ( B, d B ) such
that there exists a chain map g : ( B, d B )  ( A, d A )
with g f , f g chain homotopic to 1A , 1B respectively.
Lemma Chain equivalence  same homology.
Definition A chain contraction of a chain complex
(C, dC ) is a chain homotopy between 1C and 0C.
Lemma A contractible chain complex is acyclic
(all its homology groups equal 0).
Theorem A free acyclic chain comp.is contractible.
Proof Zq1free   rq1 : Zq1  Cq  dC rq1  1C
q
q1
 hq  rq (1Cq  rq1dCq ) : Cq  Cq1 satisfies dCq1 hq  hq1dCq  1Cq .
Chain Homotopy
Definition An augmentation of a chain complex
(C, dC ) with Cq  0, q  0 is a chain complex
~
~
~
~
(C , d ~ ) such that Cq  Cq , d q1  d q1 , q  1, C1  Z .
C
Lemma If L is a simplicial complex then C (L ) has
~
a unique augmentation given by d0 ([v])  1, v  vert(L).
Theorem If K  w is a cone (Lecture 2, slide 8)
~
then C (K * w) is acyclic.
~
~
Proof Define a hom. h : C ( K  w)  C ( K  w)
by h([v0 ,...,v p ])  [w, v0 ,...,v p ], h(1)  [w] and
Homework due 20 Oct: compute dC~ h  hdC~  1C~.
Stellar Subdivision
Lemma Barycentric subdivision can be obtained
by a sequence of stellar subdivisions of the form
K    v0v1...vk vk 1...vd  vv0v1...vˆi ...vk vk 1...vd
where v is the barycenter of v0v1...vk vk 1...vd
v2   [v0 , v1 , v2 ]
Step 1
v









Lemma Stellar subdivision
defines a chain map

 : C ( K )  C ( K ) (the subdivision operator)
k
by  ( )   i 0 (1)i [v, v0 , v1 ,...,vˆi ,...,vk , vk 1 ,...,vd ]
Proof (Step 1)  ( )  [v, v1, v2 ]  [v, v0 , v2 ]  [v, v0 , v1 ] 
v0
v1
[v1, v2 ]  [v0 , v2 ]  [v0 , v1 ]  [v0 , v1, v2 ]  [v0 , v1, v2 ].
Stellar Subdivision
Theorem If K  is a stellar subdivision of K ,
s :| K  | | K | is a simplicial approximation of 1|K |
that induces the chain map  s : C( K )  C ( K )

([
v
v

v
])

[
s
(
v
)
s
(
v
)

s
(
v
)],
defined
by
s
,
q
0
1
q
0
1
q
v
s
v
then s   1c( K ) : C(K )  C(K ).
v 
v
Corollary H (s )H (  )  1H (K ).
Proof (Step 1)s  ( )  s [v, v1, v2 ] s [v, v0 , v2 ]  s [v, v0 , v1 ] 
2
0
1
[s(v), s(v1 ), s(v2 )]  [s(v), s(v0 ), s(v2 )]  [s(v), s(v0 ), s(v1 )] 
[v0 , v1, v2 ]  [v0 , v0 , v2 ]  [v0 , v0 , v1 ]  [v0 , v1, v2 ]  1C ( K ) ([v0 , v1, v2 ]).
Theorem H ( )H (s )  1H (K). Proof Subcomplex L
of simplices containing v is a cone and z  Z (K ) 
z  Z (K )  z  s ( z)  Zq (L)  Bq (L)  Bq (K ), q  1.Finish it!
Contiguity Classes
Definition Simplicial maps s , t : | K | | L | are
contiguous if v0 vq  K  s(v0 )s(vq )t (v0 )t (vq )  L
(delete repeated vertices) and in the same
contiguity class if there exist simplicial maps
s  s1 , s2 ,...,sn  t with si and si 1 contiguous.
Lemma If s , t : | K | | L | are simplicial approximations
to a map f : | K | | L | then s and t are contiguous.
(review simplicial approx. on slide 17, lecture 3)
Proof v0 vq  K    f

q
 
St
(
s
(
v
))

L
i
i 0

q
q
St
(
t
(
v
))
L
i
i 0
 s(v0 )s(vq )t (v0 )t (vq )  L.

St
(
v
)

K
i
i 0

Contiguity Classes
Lemma Simplicial maps s , t : | K | | L | that are
in the same contiguity class are homotopic.
Proof Without loss of generality we may assume
that s and t are contiguous.
Proof Define a homotopy H : | K | [0,1] | L |
by H ( x, u)  (1  u)s( x)  ut(s), x | K |, u [0,1].
Remark Let L be a simplicial complex and realize
its polyhedron | L | as a subset of Euclidean space
with norm || || . Let   0 be the Lebesque number
for the open cover of { StL (v) : v  vert( L) } of | L | .
Hence X  | L | and diam ( X )  sup u,vX || u  v ||  
 v  vert( L)  X  StL (v).
Contiguity Classes
Lemma Let f , g : | K | | L | be maps such that
max | f ( x)  g ( x) |   / 3 where   0 is the Lebesgue
number for the open cover { StL (v) : v  vert( L) } of | L | ,
1
1
{
f
(
St
(
v
))

g
(StL (v)) : v  vert( L) } is an open
then
L
cover of | K | and there exists m  1 and a simplicial
m
m
s
:
|
K
|

|
L
|
f
,
g
:
|
K
| | L | .
approximation
to both
Proof The first assertion is obvious. Choose m  1
m
such that w  vert(K )  diam f (StK m (w))  / 3
hence diam  f (StK (v))  g (StK (v))   
m
m
 v  vert(L)  f (StK m (v))  g (StK m (v))  StL (v).
Then construct s : | K m | | L | so that s( w)  v.
Contiguity Classes
Corollary If f , g : | K | | L | are homotopic then
there exist m  1 and simplicial approximations
m
m
f
s : | K | | L | to and t : | K | | L | to g
such that s and t are in the same contiguity class.
Proof Homework Due Tuesday 27 October.
Contiguity Classes
Theorem If s , t : | K | | L | are simplicial maps
that are in the same contiguity class then their
induced chain maps  s ,t : C( K )  C( L)
are chain homotopic.
Proof We may assume that s and t are contiguous.
For   [v], v  vert( K ) define h0 ( )  [s(v)t (v)] and
observe that h( ) is a subset of a simplex (called
the carrier of  ) in L and that h0  h0  s,0  t ,0 .
Assume we have homomorphisms hi : Ci (K )  Ci1 (L)
for 0  i  n  1 so that hi1  hi  s,i t ,i : Ci (K )  Ci (L)
and hi ( ) is a chain in the the carrier of . For an
oriented n-simplex  let z( )  s ( ) t ( )  hn1( )  Cn (K ).
Then z( )  0  (why?)  Cn1 (K )     so let hn ( )  .
Corollary H ( s )  H (t ).
Invariance
Lemma If K and L are simplicial complexes,
f : | K |  | L | is a map with simplicial approx.
m
n
t : | K |  | L | and s : | K |  | L | with n  m,
1 : C( K )  C( K m ) , 2 : C( K m )  C( K n ) are subdivision
chain maps (by composing stellar subdiv. maps),
n
m
s1 : | K |  | K | is a simplicial approx. of 1|K | and
t , s , s chain maps induced by the simp. maps,
then H (t ) H ( 2 ) H ( 1 )  H ( s ) H ( 1 ) : H ( K )  H ( L).
n
n
Proof. Since both ss1 : | K |  | L | and t : | K |  | L |
are simplicial approx. to f : | K |  | L | they are
contiguous so H (s )H (s )  H (t ). Also H (s )H (2 )  1H ( K )
so H (t )H (2 )H (1 )  H (s )H (s )H (2 )H (1 )  H (s )H (1 ).
m
1
1
1
1
m
Invariance
Theorem 1. Every map f : | K |  | L | induces
a homomorphism H ( f ) : H ( K )  H ( L)
by the rule: H ( f )  H ( s ) H ( 1 ) where
s : | K m |  | L | is a simplicial approxroximation to f
m

:
C
(
K
)

C
(
K
) is the subdivision chain map.
and 1
Proof Preceding lemma  the rule is well defined.
Invariance
Theorem 2. The identity 1|K | : | K |  | K | satisfies
f
g
 | M |
H (1|K | )  1H ( K ) and maps | K |  | L | 
satisfy H ( fg )  H ( f ) H ( g ).
n
Proof Let t : | L |  | M | be a simplicial approx.
m
n
to g : | L |  | M | then let s : | K |  | L | be
m
n
a simplicial approx. to f : | K |  | L | .
m
n
Let 1 : C( K )  C( K ), 2 : C( L)  C( L ) be
n
be subdivision chain maps and s1 : | L |  | L | be
a simplicial approx. to 1|L| (a standard simp. map)
and s : C( Ln )  C( L) its induced chain map.
1
Since s1s, ts are simp. approx. to f , gf resp.
H ( g )H ( f )  H (t )H (2 )H (s1 )H (s ) H (1 )  H (t )H (s )H (1 )  H ( gf ).
Invariance
Theorem 3. If f , g : | K |  | K | are homotopic
then H ( f )  H ( g ) : H (| K |)  H (| L |).
Proof Homework due Friday 30 October
Applications
Theorem S  S  m  n.
n
Proof H m (S )  Z  m  0 or m  n
m
n
R

R
 m  n.
Corollary
Proof S m  Rm  
m
n
Theorem (Brouwer-Fixed Point) Every map
m
m
g : B  B , m  2 has a fixed point.
Proof Otherwise there exists a retraction
i
r
r : B m  S m1 so that S m1 

Bm 

S m1
m
m
Since B is contractible H m1 ( B )  0.
However ri  1S m1  H m1 (r ) H m1 (i)  1H ( S m1 )  1Z
m1
and this ccontradiction concludes the proof.