The Hodge star operator * : Ωᵏ(M) → Ωⁿ⁻ᵏ(M) on an oriented Riemannian n-manifold satisfies α ∧ *β = g(α, β) dVg for k-forms α, β. On ℝ³ with the standard metric, *dx = dy ∧ dz. What does the Hodge star encode?
AThe Hodge star encodes the metric and orientation — it converts between forms of complementary degree by 'filling in' the remaining dimensions using the metric
BThe Hodge star is the exterior derivative in disguise
CThe Hodge star is defined independently of the metric
DThe Hodge star always squares to the identity: ** = id
The Hodge star uses both the metric (to measure 'perpendicular') and the orientation (to choose sign). It sends a k-form to the (n-k)-form that represents the 'orthogonal complement' using the volume form. In ℝ³: *dx = dy∧dz, *dy = dz∧dx, *dz = dx∧dy, *(dx∧dy) = dz, etc. The star does NOT square to the identity in general: **α = (-1)^{k(n-k)} α on an n-manifold (so **= id only when k(n-k) is even). It is metric-dependent and not related to d.
Question 2 True / False
The Hodge decomposition theorem states that on a compact oriented Riemannian manifold, every k-form ω can be uniquely decomposed as ω = α + dβ + δγ where α is harmonic (Δα = 0), dβ is exact, and δγ is co-exact.
TTrue
FFalse
Answer: True
This is the fundamental theorem of Hodge theory. The three components are L²-orthogonal (with respect to the inner product (α,β) = ∫_M α ∧ *β). Since harmonic forms are both closed and co-closed (dα = 0 and δα = 0), each harmonic k-form represents a unique de Rham cohomology class. The Hodge theorem therefore gives an isomorphism ℋᵏ(M) ≅ Hᵏ(M) — the space of harmonic k-forms is isomorphic to the k-th de Rham cohomology. This reduces topology to solving the PDE Δω = 0.
Question 3 Short Answer
Why does Hodge theory require a Riemannian metric, while de Rham cohomology does not?
Think about your answer, then reveal below.
Model answer: De Rham cohomology uses only the exterior derivative d, which is defined from the smooth structure alone — no metric needed. Hodge theory introduces the codifferential δ = (-1)^{n(k+1)+1} *d* (the formal adjoint of d), which requires the Hodge star, which requires a metric. The Laplacian Δ = dδ + δd and the L² inner product on forms both depend on the metric. The harmonic representatives of cohomology classes therefore change when you change the metric — but the cohomology groups themselves (as abstract vector spaces) are metric-independent, since they are topological invariants.
This is a beautiful interplay: topology (de Rham cohomology) provides the abstract structure, and analysis (Hodge theory, via the metric) provides canonical representatives. Different metrics give different harmonic forms, but the same cohomology groups. The proof of the Hodge theorem uses elliptic PDE theory — regularity and Fredholm theory for the Laplacian — which is deeply metric-dependent.
Question 4 True / False
On a compact Kähler manifold, Hodge theory yields a decomposition of cohomology into (p,q)-types: Hᵏ(M; ℂ) = ⊕_{p+q=k} Hᵖ·ᵍ(M).
TTrue
FFalse
Answer: True
On a Kähler manifold (a complex manifold with a compatible Riemannian metric), the complex structure induces a bigrading on forms: Ωᵏ ⊗ ℂ = ⊕ Ωᵖ·ᵍ. The Kähler condition ensures this bigrading is compatible with the Laplacian, so harmonic forms decompose by (p,q)-type. The Hodge numbers h^{p,q} = dim Hᵖ·ᵍ are important invariants of the complex manifold. This Hodge decomposition is one of the central results in complex algebraic geometry.