Questions: Chain Complexes and the Boundary Operator

Question 1

In a simplicial chain complex, d_1 maps each oriented edge [v_0, v_1] to v_1 - v_0. What is d_1([a,b] + [b,c] + [c,a])?

• A. a + b + c
• B. (b - a) + (c - b) + (a - c) = 0
• C. a - b + c
• D. [a,b,c]

Answer: (B)

Applying d_1 to each edge: d_1([a,b]) = b - a, d_1([b,c]) = c - b, d_1([c,a]) = a - c. Summing these gives (b - a) + (c - b) + (a - c) = 0. This is not a coincidence: the chain [a,b] + [b,c] + [c,a] is a cycle — it forms a closed loop. Its boundary vanishes because it has no endpoints. This is exactly the kind of chain that lives in ker(d_1), and whether it is also in im(d_2) determines if it bounds a filled triangle or represents a genuine 1-dimensional hole.

Question 2

Why is the condition d_{n-1} ∘ d_n = 0 (the boundary of a boundary is zero) essential for defining homology?

• A. It ensures the chain groups are finitely generated
• B. It guarantees that im(d_n) ⊆ ker(d_{n-1}), so the quotient ker(d_{n-1})/im(d_n) is well-defined
• C. It means all boundary operators are isomorphisms
• D. It forces the chain complex to be exact

Answer: (B)

Homology is defined as H_n = ker(d_n)/im(d_{n+1}). For this quotient to make sense, im(d_{n+1}) must be a subgroup of ker(d_n) — that is, every boundary must be a cycle. This is precisely what d_n ∘ d_{n+1} = 0 guarantees: if c is in im(d_{n+1}), say c = d_{n+1}(x), then d_n(c) = d_n(d_{n+1}(x)) = 0, so c is in ker(d_n). Without this condition, boundaries might not be cycles, and the quotient would be undefined.

Question 3

The boundary operator d_2 applied to the oriented 2-simplex [v_0, v_1, v_2] gives [v_1, v_2] - [v_0, v_2] + [v_0, v_1], which equals the oriented boundary traversed counterclockwise.

• T. True
• F. False

Answer: True

The boundary of an oriented n-simplex [v_0, ..., v_n] is the alternating sum of its (n-1)-dimensional faces: d_n([v_0, ..., v_n]) = sum_{i=0}^{n} (-1)^i [v_0, ..., v̂_i, ..., v_n], where v̂_i means omitting vertex v_i. For d_2([v_0, v_1, v_2]): omit v_0 to get [v_1, v_2], omit v_1 to get -[v_0, v_2], omit v_2 to get [v_0, v_1]. This gives the three edges oriented consistently around the triangle, which is the geometric boundary traversed counterclockwise.

Question 4

Explain intuitively why the boundary of a boundary is zero, using the example of a 2-simplex (triangle).

Model answer: The boundary of a triangle [v_0, v_1, v_2] is the cycle of its three edges. Each edge is a 1-simplex whose boundary is the difference of its endpoints. When we take the boundary of the boundary, each vertex appears exactly twice with opposite signs (once as the head of one edge and once as the tail of the adjacent edge), so everything cancels to zero. Geometrically: the boundary of a region is a closed curve, and a closed curve has no endpoints.

This cancellation is not specific to triangles — it is a general combinatorial identity. In the alternating sum formula, applying d twice produces a double sum where each term appears twice with opposite signs due to the sign convention. The geometric intuition is that boundaries are always closed: the boundary of a surface is a closed curve (no endpoints), the boundary of a solid is a closed surface (no boundary edges), and so on. The algebraic condition d ∘ d = 0 encodes this geometric fact.