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Chapter 3: Simplicial Homology Instructor: Yusu Wang
CSE 5559 Computational Topology: Theory, algorithms, and applications to data analysis Chapter 3: Simplicial Homology Instructor: Yusu Wang
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Section 1: Simplicial Homology
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Chains Given a simplicial complex πΎ, a π-chain is
A formal sum of π-simplices π=β π π π π Under π 2 coefficients: a collection of π-simplices
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Chains Given a simplicial complex πΎ, a π-chain is
A formal sum of π-simplices π=β π π π π Under π 2 coefficients: a collection of π-simplices π-th chain group of πΎ πΆ π πΎ : collection of π-chains with operation + π 1 = π 1 + π 3 ; π 2 = π 1 + π 4 ; β π 1 + π 2 = π 3 + π 4 Under π 2 coefficients, πΆ π πΎ is a vector space What is its dimension (rank)? What is a basis for it?
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Chains and boundary operator
π-th boundary operator π π : πΆ π β πΆ πβ1 π π π = set of (πβ1)-faces of π π π π =β π π π π ( π π ) Hence π π is a homomorphism (map preserving + operation) C_p is a free abelian group, which is an abelian group with a basis
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Chains and boundary operator
π-th boundary operator π π : πΆ π β πΆ πβ1 π π π = set of (πβ1)-faces of π π π π =β π π π π ( π π ) Hence π π is a homomorphism (map preserving + operation) Chain complex C_p is a free abelian group, which is an abelian group with a basis Theorem: π π β π π+1 =0
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Cycles and Boundaries Cycles: π-cycle: a π-chain whose boundary is 0
π-th cycle group π π πΎ =kerβ‘( π π ) What is the relation between π π and πΆ π ?
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Under π 2 coefficients, π΅ π , π π , πΆ π are all vector spaces.
Cycles and Boundaries Cycles: π-cycle: a π-chain whose boundary is 0 π-th cycle group π π πΎ =kerβ‘( π π ) Boundary cycles: π-boundary: a π-cycle which is the boundary of some (π+1)- chain π-th boundary group π΅ π πΎ =Imβ‘( π π+1 ) π π β π π+1 =0β π΅ π β π π β πΆ π Under π 2 coefficients, π΅ π , π π , πΆ π are all vector spaces.
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Cycles and Boundaries Cycles: Boundary cycles:
π-cycle: a π-chain whose boundary is 0 π-th cycle group π π πΎ =kerβ‘( π π ) Boundary cycles: π-boundary: the boundary of some (π+1)-chain π-th boundary group π΅ π πΎ =Imβ‘( π π+1 ) π π β π π+1 =0β π΅ π β π π β πΆ π π π+1 π π
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Homology groups π-th cycle group π π πΎ =kerβ‘( π π )
π-th boundary group π΅ π πΎ =Imβ‘( π π+1 ) π-th homology group π» π πΎ = π π / π΅ π π 1 is homologous to π 2 if π 1 + π 2 β π΅ π , i.e, π 1 + π 2 is a boundary cycle β=[π]β π» π : the family π-cycles homologous to π called a homology class B forms a normal subgroup of Z, so the right- / left- cosets coincides. => thus can define quotient group c_1 \in c_2 + B
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Betti numbers Betti number: π½ π πΎ =ππππ π» π Theorem:
π½ π πΎ =ππππ π π βππππ( π΅ π ) Examples: meaning of π½ 0 , π½ 1 , π½ 2 Rank( π» 0 ) = ?; Rank( π» 1 ) = ?
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More results Theorem: Theorem:
For a compact orientable 2-manifold with genus π, we have π½ 0 =1, π½ 1 =2π, π½ 2 =1 Theorem: Given two simplicial complexes πΎ 1 and πΎ 2 such that πΎ 1 β
| πΎ 2 |, then π» β πΎ 1 β π» β ( πΎ 2 ). Hence different triangulations of the same space have the isomorphic homology groups! Thus homology group is a topological invariant
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Euler characteristics
Given a topological space π its Euler characteristics π π = π=0 β1 π π½ π π Hence Euler characteristics is also independent of the triangulation of a space, and is a topological invariant. Theorem (Euler-PoincarΓ© formula) Given a simplicial complexes πΎ, let π π denote the number of π-simplices in πΎ. Then π πΎ βπ πΎ = π=0 β1 π π π Examples of triangulations of 2-manifolds.
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Section 2: Matrix view and computation
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Boundary Matrix πΎ π = πΌ 1 ,β¦, πΌ π π , πΎ πβ1 = π 1 ,β¦, π π πβ1
πΎ π = πΌ 1 ,β¦, πΌ π π , πΎ πβ1 = π 1 ,β¦, π π πβ1 πΎ π forms a basis for p-th chain group πΆ π n πβ1 Γ π π boundary matrix π΄ π π΄ π π π =1 iff π π β π π representing π π : πΆ π β πΆ πβ1
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Boundary Matrix πΎ π = πΌ 1 ,β¦, πΌ π π , πΎ πβ1 = π 1 ,β¦, π π πβ1
πΎ π = πΌ 1 ,β¦, πΌ π π , πΎ πβ1 = π 1 ,β¦, π π πβ1 πΎ π forms a basis for p-th chain group πΆ π n πβ1 Γ π π boundary matrix π΄ π s.t. π΄ π π π =1 iff π π β π π representing π π : πΆ π β πΆ πβ1 w.r.t. basis πΌ 1 ,β¦, πΌ π π and π 1 ,β¦, π π πβ1 π π π π
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Given a p-chain π= π=1 π π π π πΌ π
Under basis πΎ π , vector representation of π is π = π 1 , π 2 ,β¦, π π π π Boundary π π π is a (πβ1)-chain with vector representation π΄ π π w.r.t basis πΎ πβ1 Example.
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Boundary Matrix Let π π , π§ π , π π denote the rank of πΆ π , π π , and π΅ π π½ π =ππππ π» π
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Boundary Matrix Let π π , π§ π , π π denote the rank of πΆ π , π π , and π΅ π π½ π =ππππ π» π Claim: π π π = π§ π + π πβ1 ; ππ π½ π = π§ π β π π Rank-nullity theorem from linear algebra C is isomorphic to direct sum of B and Z Splitting lemma
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Boundary Matrix Let π π , π§ π , π π denote the rank of πΆ π , π π , and π΅ π π½ π =ππππ π» π Claim: π π π = π§ π + π πβ1 ; ππ π½ π = π§ π β π π Consider π΄ π Each columns of π΄ π corresponds to a boundary cycle Rank of π΄ π gives π π =ππππ( π΅ π ) Why?
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Boundary Matrix Let π π , π§ π , π π denote the rank of πΆ π , π π , and π΅ π π½ π =ππππ π» π Claim: π π π = π§ π + π πβ1 ; ππ π½ π = π§ π β π π Consider π΄ π Each columns of π΄ π corresponds to a boundary cycle Rank of π΄ π gives π π =ππππ( π΅ π ) Note, this gives an π π 3 time algorithm for computing all π½ π βs via Gaussian elimination Can be improved to matrix multiplication time
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Right-reduction algorithm
Starting with bnd matrix π= π΄ π For the π-th column corresponding to p-simplex π π , associate a π-chain Ξ π initialized to π π AddColumn(π, π) πΆπ π π π =πΆπ π π π +πΆπ π π π ; Ξ π = Ξ π + Ξ π
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Properties Lemma: Lemma:
Each reduction (column addition) step maintains the following invariance: After π-th stages, π (π) has the same rank as π΄ π , and π π Ξ π (π) =ππ π π π for any π. Lemma: At the end of the reduction algorithm, each non-zero column has a unique low-ID.
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Reduced form: A matrix π is in reduced form is all non-zero columns are linearly independent. It is in Smith-Normal form if it has the following structure: Lemma: A matrix is in reduced form if each non-zero column has a unique low-ID.
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Properties Theorem: Examples.
Procedure Left-Reduction(π) terminates in π( π π 2 π πβ1 ) time The output matrix π is in reduced form The set of non-zero columns in π form a basis for π΅ πβ1 The set { Ξ π β£ππ π π π =0 } form a basis for π π Examples. This is not the only reduction algorithm!! Any elimination via row/column additions to convert a matrix into a reduced form works!
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