Chain Complexes

This page follows on from the discussion of delta complexes here.

We now want to represent this topology in algebra terms.

homology face maps

We start by defining two functions on δ (functions from a function to a subset) which define subsets of the codomain and domain respectivly:


Example 1 : Circle

A circle (not filled in) is topologically equivalent to an empty triangle.

chain complex

So our faces are:

  • points: a, b and c
  • edges: ab, -ac and bc


We can calculate cycles and boundaries as follows:


We can now relate them using the boundary function. This boundary function removes each element in turn and alternates the sign.

B0 = Im δ1 = a-b, a-c, b-c

So the boundary of the edges is the verticies.

B1 = Im δ2 = 0

The boundary of 0 is 0.

Higher order boundaries are all zero.


The cycles are given by the 'kernel' of δn which are the 'chains' which map to zero

Z0 = Ker δ0 = a,b,c

This is given by the kernel of δ0 which are the vertices which map to zero, and they all map to zero because δ0 maps everything to zero.

Z1 = Ker δ1 = ab->bc->ca , bc->ca->ab , ca->ab->bc

We want the kernel of δ1 which are the edges which map to zero, here 'zero' can be interpreted as the identity mapping (a mapping from an element back to itself). So the cycles are the loops around the circle starting at any vertex and ending on that vertex.

Z2 = Ker δ2 =

Example 2 : Disk

A disk (filled in circle) is topologically equivalent to a solid triangle.

chain complex

So our faces are:

  • points: a, b and c
  • edges: ab, -ac and bc
  • triangle: abc

We can now relate them using the boundary function:

This boundary function removes each element in turn and alternates the sign.

So if we apply boundary twice we get:

δ(δabc) = δ(ab -ac + bc)
= (a-b) + (c-a) + (b-c)
= 0

So, in this case, applying boundary twice gives zero.

In fact applying delta twice always gives zero: homology chain


In the earlier examples we just looked at the delta operator on single triangles.

Here we look at a chain of triangles:

This gives the boundary of the shape.

chain complex

As a Free Abelian Group

Now we allow integer multiples and sums of these. If i,jk... are integers then we have:

δ( i ab - j ac + k bc) = i (a-b) - j (a-c) + k (b-c)
= (i-j) a + (k-i) b + (j-k) c

This gives a free abelian group isomorphic to Z.

Chain Complex as a Sequence of Abelian Groups or Modules


Zero dimensional chain
= free abelian group on vertices.

Elements of C0 are intergral linear combinations of vertices such as: 2x+7y+5z


One dimensional chain
= free abelian group on directed edges.

Elements of C1 are intergral linear combinations of edges such as: 2xy+7xz+5yz
These don't necessarily need to be a cycle.

We now treat δ as a group homomorphism which extends *.


Being a group homomorphism means respecting the group structure of the two groups, that is,

It also respects the extra structure that groups are required to be continuous.

Algebraic meaning of cycle: δ(X) = 0

We find cycles by finding null space of augmented matrix.

Co-Chain Complex

The difference between co-chain complex and a chain complex is that, in a co-chain complex the boundary operator increases the dimension rather than decreases it.

Next Steps

For information about how this can be implemented in computer code here.

Or to continue on the theoretical route see how this is used in homolgy here.

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Correspondence about this page

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flag flag flag flag flag flag Mathematics++: Selected Topics Beyond the Basic Courses (Student Mathematical Library) Kantor, Ida.


  1. Measure
  2. High Dimensional Geometry
  3. Fourier Analysis
  4. Representations of Finite Groups
  5. Polynomials
  6. Topology

Chapter 6 - Topology. Contains a relatively gentle introduction to homology.


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