ISEGORIA / MATH ENCYCLOPEDIA
Algebraic topology: detecting holes
Build complexes, distinguish cycles from holes, and vary scale.
Before you begin: Sets, graphs, and elementary linear algebra
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Build a complex
Build a square one simplex at a time: four vertices, the four sides, a diagonal, then the two triangles. Beside the picture are the boundary matrices over the two-element field. Watch each new edge either join two components, raising rank ∂₁ and lowering β₀, or close a cycle, raising β₁; each face turns one cycle into a boundary.
Worked example. After the four sides, V = 4, E = 4 and rank ∂₁ = 3, so β₀ = 1 and β₁ = 1. The diagonal makes β₁ = 2, and the two triangles bring it back to 0.
Watch out. A drawn loop is not necessarily a homological hole: it may bound a filled surface.
Why does filling a face never change the number of components?
β₀ = V − rank ∂₁ does not involve faces at all: a face's boundary already lies in one component.
2. Holes across scales
Put a disk of diameter ε around each point, join two points when their disks touch, and fill every triangle whose three edges are present. As ε grows, components merge and loops are born and then filled. The barcode records each feature from the ε where it appears to the ε where it dies, computed by reducing the boundary matrix over the two-element field. Drag the points to reshape the cloud.
Worked example. For eight evenly spaced points on the unit circle, adjacent points join at ε ≈ 0.765, giving an octagon with β₁ = 1. The loop survives until ε ≈ 1.848, when triangles through every third point fill it.
Watch out. Only the two-skeleton is displayed. It suffices for H₀ and H₁, but not for higher-dimensional homology.
Why can a larger threshold destroy a hole?
New triangles can fill a cycle that was not a boundary before, so the class dies in H₁.
3. Euler characteristic
Choose a triangulated surface and count vertices, edges and faces. The alternating sum V − E + F equals the alternating sum of Betti numbers, computed here from the boundary matrices. Press Play to place the triangles one at a time: the two sums agree at every stage, not only for the finished surface.
Worked example. A tetrahedron surface has 4 − 6 + 4 = 2 with β = (1, 0, 1). The 3 by 3 torus has 9 − 27 + 18 = 0 with β = (1, 2, 1).
Watch out. The sphere here is the surface of the tetrahedron, not the solid. Betti numbers are computed over the two-element field; χ is the same over any field.
Why does a disk have Euler characteristic one?
It has one component and no positive-dimensional homology.