Dual invariants, the cup product, and cohomology rings
Cohomology flips homology around: instead of drawing loops, you put numbers on edges. The payoff is a way to multiply, and that product tells apart spaces homology cannot.
On this torus (a square with opposite edges glued), use cycle mode to click edges into a loop and cocycle mode to mark edges with 1. The pairing 〈φ, c〉 counts the loop's marked edges, mod 2.
Cohomology is the algebraic dual of homology: instead of building cycles from chains, we assign values to chains and ask when those assignments are consistent. A cocycle is a function on chains that vanishes on boundaries — it “counts intersections” with a dual object.
Why bother with this dual viewpoint? Because cohomology carries strictly morestructure than homology. The cup product turns H*(X) into a graded ring, and this ring structure can distinguish spaces that homology alone cannot.
The punchline: two spaces can have identical homology groups yet completely different cohomology rings. The cup product detects how loops and surfaces are “linked” inside a space.
Key insight: The Kronecker pairing 〈φ, c〉 is the bridge between cohomology and homology. A cocycle φ “evaluates” a cycle c by summing φ(e) over all edges e of c. This is the algebraic version of counting intersection points.
The torus and S¹ ∨ S¹ ∨ S² have the same homology groups. Show α, then β, then their cup product, and watch it come out nonzero on the torus and zero on the other space.
| H&sup0; | ℤ |
| H¹ | ℤ ⊕ ℤ |
| H² | ℤ |
| α ∪ β | ≠ 0 (generates H²) |
| H&sup0; | ℤ |
| H¹ | ℤ ⊕ ℤ |
| H² | ℤ |
| α ∪ β | = 0 (trivial!) |
The cup product α ∪ β combines two cocycles into a higher-degree cocycle. On the torus T², the two 1-cocycles α and β (dual to the horizontal and vertical loops) have a nontrivial cup product that generates H². But on S¹ ∨ S¹ ∨ S² — which has the exact same homology groups — the cup product is zero!
Key insight: The cup product detects “linking” — on the torus, the two generating loops intersect transversally in a single point, so α ∪ β ≠ 0. In the wedge sum, the loops live in separate circles and never interact, so the cup product vanishes. This is the fundamental example showing that cohomology rings are strictly stronger than homology groups.
Select a space to see its multiplication table; nonzero products glow rose. Click a cell for its geometric meaning, and use compare mode to set two spaces side by side.
The full cohomology ring H*(X; ℤ) organizes all cup products into a multiplication table.
Key insight: The cohomology ring is a graded-commutative ring: α ∪ β = (-1)&sup(deg α)(deg β) β ∪ α. For odd-degree classes this means 2(α ∪ α) = 0, so the square vanishes whenever the cohomology has no 2-torsion — which is why the torus has an exterior algebra. For even-degree classes like the hyperplane class of ℂP², self-cup-products can be nontrivial — giving a polynomial ring truncated at the top dimension.
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