Simplicial homology

A key concept in defining simplicial homology is the notion of an orientation of a simplex. By definition, an orientation of a k-simplex is given by an ordering of the vertices, written as (v₀,...,v_k), with the rule that two orderings define the same orientation if and only if they differ by an even permutation. Thus every simplex has exactly two orientations, and switching the order of two vertices changes an orientation to the opposite orientation. For example, choosing an orientation of a 1-simplex amounts to choosing one of the two possible directions, and choosing an orientation of a 2-simplex amounts to choosing what "counterclockwise" should mean.

Let S be a simplicial complex. A simplicial k-chain is a finite formal sum

${\displaystyle \sum _{i=1}^{N}c_{i}\sigma _{i},\,}$

where each c_i is an integer and σ_i is an oriented k-simplex. In this definition, we declare that each oriented simplex is equal to the negative of the simplex with the opposite orientation. For example,

${\displaystyle (v_{0},v_{1})=-(v_{1},v_{0}).}$

The group of k-chains on S is written C_k. This is a free abelian group which has a basis in one-to-one correspondence with the set of k-simplices in S. To define a basis explicitly, one has to choose an orientation of each simplex. One standard way to do this is to choose an ordering of all the vertices and give each simplex the orientation corresponding to the induced ordering of its vertices.

Let σ = (v₀,...,v_k) be an oriented k-simplex, viewed as a basis element of C_k. The boundary operator

${\displaystyle \partial _{k}:C_{k}\rightarrow C_{k-1}}$

is the homomorphism defined by:

${\displaystyle \partial _{k}(\sigma )=\sum _{i=0}^{k}(-1)^{i}(v_{0},\dots ,{\widehat {v_{i}}},\dots ,v_{k}),}$

where the oriented simplex

${\displaystyle (v_{0},\dots ,{\widehat {v_{i}}},\dots ,v_{k})}$

is the i^th face of σ, obtained by deleting its i^th vertex.

In C_k, elements of the subgroup

${\displaystyle Z_{k}:=\ker \partial _{k}}$

are referred to as cycles, and the subgroup

${\displaystyle B_{k}:=\operatorname {im} \partial _{k+1}}$

is said to consist of boundaries.

Because ${\displaystyle (-1)^{i+j-1}(v_{0},\dots ,{\widehat {v_{i}}},\dots ,{\widehat {\widehat {v_{j}}}},\dots ,v_{k})=-(-1)^{i+j}(v_{0},\dots ,{\widehat {\widehat {v_{i}}}},\dots ,{\widehat {v_{j}}},\dots ,v_{k})}$, where ${\displaystyle {\widehat {\widehat {v_{x}}}}}$ is the second face removed, ${\displaystyle \partial ^{2}=0}$. In geometric terms, this says that the boundary of anything has no boundary. Equivalently, the abelian groups

${\displaystyle (C_{k},\partial _{k})}$

form a chain complex. Another equivalent statement is that B_k is contained in Z_k.

As an example, consider a tetrahedron with vertices oriented as w,x,y,z. By definition, its boundary is given by: xyz - wyz + wxz - wxy. The boundary of the boundary is given by: (yz-xz+xy)-(yz-wz+wy)+(xz-wz+wx)-(xy-wy+wx) = 0.

A simplicial complex with 2 1-holes

The k^th homology group H_k of S is defined to be the quotient abelian group

${\displaystyle H_{k}(S)=Z_{k}/B_{k}\,.}$

It follows that the homology group H_k(S) is nonzero exactly when there are k-cycles on S which are not boundaries. In a sense, this means that there are k-dimensional holes in the complex. For example, consider the complex S obtained by gluing two triangles (with no interior) along one edge, shown in the image. The edges of each triangle can be oriented so as to form a cycle. These two cycles are by construction not boundaries (since every 2-chain is zero). One can compute that the homology group H₁(S) is isomorphic to Z², with a basis given by the two cycles mentioned. This makes precise the informal idea that S has two "1-dimensional holes".

Holes can be of different dimensions. The rank of the k^th homology group, the number

${\displaystyle \beta _{k}=\operatorname {rank} (H_{k}(S))\,}$

is called the k^th Betti number of S. It gives a measure of the number of k-dimensional holes in S.

Let S be a triangle (without its interior), viewed as a simplicial complex. Thus S has three vertices, which we call v₀, v₁, v₂, and three edges, which are 1-dimensional simplices. To compute the homology groups of S, we start by describing the chain groups C_k:

• C₀ is isomorphic to Z³ with basis (v₀), (v₁), (v₂),
• C₁ is isomorphic to Z³ with a basis given by the oriented 1-simplices (v₀, v₁), (v₀, v₂), and (v₁, v₂).
• C₂ is the trivial group, since there is no simplex like ${\displaystyle (v_{0},v_{1},v_{2})}$ because the triangle has been supposed without its interior. So are the chain groups in other dimensions.

The boundary homomorphism ∂: C₁ → C₀ is given by:

${\displaystyle \partial (v_{0},v_{1})=(v_{1})-(v_{0})}$

${\displaystyle \partial (v_{0},v_{2})=(v_{2})-(v_{0})}$

${\displaystyle \partial (v_{1},v_{2})=(v_{2})-(v_{1})}$

Since C₋₁ = 0, every 0-chain is a cycle (i.e. Z₀ = C₀); moreover, the group B₀ of the 0-boundaries is generated by the three elements on the right of these equations, creating a two-dimensional subgroup of C₀. So the 0th homology group H₀(S) = Z₀/B₀ is isomorphic to Z, with a basis given (for example) by the image of the 0-cycle (v₀). Indeed, all three vertices become equal in the quotient group; this expresses the fact that S is connected.

Next, the group of 1-cycles is the kernel of the homomorphism ∂ above, which is isomorphic to Z, with a basis given (for example) by (v₀,v₁) − (v₀,v₂) + (v₁,v₂). (A picture reveals that this 1-cycle goes around the triangle in one of the two possible directions.) Since C₂ = 0, the group of 1-boundaries is zero, and so the 1st homology group H₁(S) is isomorphic to Z/0 ≅ Z. This makes precise the idea that the triangle has one 1-dimensional hole.

Next, since by definition there are no 2-cycles, C₂ = 0 (the trivial group). Therefore the 2nd homology group H₂(S) is zero. The same is true for H_i(S) for all i not equal to 0 or 1. Therefore, the homological connectivity of the triangle is 0 (it is the largest k for which the reduced homology groups up to k are trivial).

Let S be a tetrahedron (without its interior), viewed as a simplicial complex. Thus S has four 0-dimensional vertices, six 1-dimensional edges, and four 2-dimensional faces. The construction of the homology groups of a tetrahedron is described in detail here.[3] It turns out that H₀(S) is isomorphic to Z, H₂(S) is isomorphic to Z too, and all other groups are trivial.Therefore, the homological connectivity of the tetrahedron is 0.

If the tetrahedron contains its interior, then H₂(S) is trivial too.

In general, if S is a d-dimensional simplex, the following holds:

• If S is considered without its interior, then H₀(S) = Z and H_d−1(S) = Z and all other homologies are trivial;
• If S is considered with its interior, then H₀(S) = Z and all other homologies are trivial.