Δ-Complexes & Simplicial Homology

Cycles that do not bound: homology you can compute by hand, one triangle at a time.

The idea

Simplicial homology assigns to a space $X$ a sequence of computable abelian groups that count its holes, one dimension at a time. The groups are the tool that proves two spaces differ: spaces with non-isomorphic homology groups are not homeomorphic, and not even homotopy equivalent.

Suppose the space has been cut into triangles, tetrahedra and their higher analogues, and let $\Delta_{n}(X)$ be the group of formal integer combinations of the $n$-dimensional pieces. Each piece has a boundary, the combination of its faces, and this extends to a homomorphism $\partial_{n} \colon \Delta_{n}(X) \to \Delta_{n-1}(X)$. A combination sent to zero is a cycle: it closes up, as a loop of edges does. Some cycles close up only because they are the rim of something one dimension higher; those are boundaries, and they surround nothing. The remaining cycles each enclose a hole, and every boundary is a cycle, so the cycles that fail to bound form a quotient group.

Definition (Simplicial homology).

The $n$-th simplicial homology group of $X$ is the group of $n$-cycles modulo the $n$-boundaries, $H^{\Delta}_{n}(X) = \ker \partial_{n} \,/\, \operatorname{im} \partial_{n+1}.$

Unlike the fundamental group, every group here is abelian and finitely generated, so the answer is a short list of integers rather than a presentation.

Ways to work on it

• Walkthrough. Delta-complexes, chain groups, the alternating boundary map, and the torus computed from end to end.
• Proof. Why applying the boundary map twice always gives zero, and why homology would not exist otherwise.
• Practice. Read boundaries off a triangulated square, compute homology from explicit chain data, and tell cycles from boundaries.
• Hardest. The Klein bottle, computed from its triangulation end to end.

Not sure where to start? Take the ten-question placement test.