ker di/ Im di−1, i ∈ Z

1. Hodge Decomposition Theorem (Finite Dimensional Case)

A cochain complex of vector spaces (or simply a cochain complex) over a field k is a sequence of k-vector spaces {Cⁱ: i ∈ Z} together with a sequence of k-linear maps {dⁱ : Cⁱ→ Cⁱ⁺¹: i ∈ Z} such that dⁱ◦ dⁱ⁻¹= 0 for any i ∈ Z. A cochain complex over k is denoted by C^• = (Cⁱ, dⁱ)_i∈Z. Since dⁱ◦ dⁱ⁻¹= 0 for all i ∈ Z, Im dⁱ⁻¹is a vector subspace of ker dⁱ. The i-th cohomology of a cochain complex C^• is defined to be the quotient vector space

Hⁱ(C^•) = ker dⁱ/ Im dⁱ⁻¹, i ∈ Z.

We denote H^∗(C^•) =L

i∈ZHⁱ(C^•) and d =L

i∈Zdⁱ.

In this note, all the vector spaces are assumed to be over the real field k = R or over the complex field k = C.

Definition 1.1. A (Riemannian or Hermitian) metric h^C on a cochain complex C is a sequence of functions {hⁱ: Cⁱ× Cⁱ→ k} such that hⁱ is an inner product on Cⁱ for each i ∈ Z.

If h^C = {hⁱ} is a metric on C, the inner product hⁱ is also denoted by h·, ·i_Ci for each i ∈ Z.

Let h^C be a metric on C. For each i ∈ Z, we define a linear map (dⁱ⁺¹)^∗: Cⁱ⁺¹ → Cⁱby hdⁱx, yi_Ci+1 = hx, (dⁱ⁺¹)^∗yi_Ci,

for any x ∈ Cⁱ and for any y ∈ Cⁱ⁺¹. We define a linear operator ∆ⁱon Cⁱ by

∆ⁱ= dⁱ⁻¹◦ (dⁱ)^∗+ (dⁱ⁺¹)^∗◦ dⁱ. We call ∆ⁱ the Laplace operator on C with respect to h^C.

Lemma 1.1. Let C be a cochain complex of vector spaces and h^C be a metric on C. Then ker ∆ⁱ = ker dⁱ∩ ker(dⁱ)^∗.

Proof. If x ∈ ker dⁱ ∩ ker(dⁱ)^∗, then dⁱx = (dⁱ)^∗x = 0 which implies that ∆ⁱx = 0. Therefore x ∈ ker ∆ⁱ. This proves that

ker ∆ⁱ ⊇ ker dⁱ∩ ker(dⁱ)^∗. Suppose x ∈ ker ∆ⁱ. Then ∆ⁱx = 0. Hence

h∆ⁱx, xi_Ci= hdⁱ⁻¹◦ (dⁱ)^∗x + (dⁱ⁺¹)^∗◦ dⁱx, xi_Ci

= h(dⁱ)^∗x, (dⁱ)^∗xi_Ci+ hdⁱx, dⁱxi_Ci

= k(dⁱ)^∗xk²_Ci+ kdⁱxk²_Ci

= 0.

Since k(dⁱ)^∗xk²_C

i ≥ 0 and kdⁱxk²_C

i ≥ 0, we find that k(dⁱ)^∗xk²_C

i = kdⁱxk²_C

i = 0

which implies that (dⁱ)^∗x = dⁱx = 0. Therefore x ∈ ker dⁱ∩ ker(dⁱ)^∗. This proves that ker ∆ⁱ ⊆ ker dⁱ∩ ker(dⁱ)^∗.

We complete the proof of our assertion. 

Theorem 1.1. (Hodge Decomposition Theorem) Let C be a cochain complex of finite dimensional vector spaces (i.e. each Cⁱ is finite dimensional) and h^C be a metric on C. For each i ∈ Z, we have the orthogonal decomposition of vector spaces

Cⁱ= ker ∆ⁱ⊕ Im dⁱ⁻¹⊕ Im(dⁱ⁺¹)^∗.

1

2

Proof. Let us prove the orthogonality of these subspaces of Cⁱ.

Let x ∈ ker ∆ⁱ and y ∈ Im dⁱ⁻¹. Since x ∈ ker ∆ⁱ, dⁱx = (dⁱ)^∗x = 0. Since y ∈ Im dⁱ⁻¹, we write y = dⁱ⁻¹z for z ∈ Cⁱ⁻¹. Therefore

hx, yi_Ci = hx, dⁱ⁻¹zi_Ci= h(dⁱ)^∗x, zi_Ci−1 = h0, zi_Ci−1 = 0.

We prove that ker ∆ⁱ ⊥ Im dⁱ⁻¹. Assume that y⁰ ∈ Im(dⁱ⁺¹)^∗. We write y⁰ = (dⁱ⁺¹)^∗z⁰ for some z⁰∈ Cⁱ⁺¹. Then

hx, y⁰i_Ci= hx, (dⁱ⁺¹)^∗z⁰i_Ci = hdⁱx, z⁰i_Ci+1= h0, z⁰i_Ci+1= 0.

We prove that ker ∆ⁱ⊥ Im(dⁱ⁺¹)^∗. Let us compute hy, y⁰i_Ci:

hy, y⁰i_Ci = hdⁱ⁻¹z, (dⁱ⁺¹)^∗z⁰i_Ci = hdⁱ◦ dⁱ⁻¹z, z⁰i_Ci+1 = h0, z⁰i_Ci+1 = 0 by dⁱ◦ dⁱ⁻¹= 0. We prove that Im dⁱ⁻¹⊥ Im(dⁱ⁺¹)^∗. We obtain that

ker ∆ⁱ⊕ Im dⁱ⁻¹⊕ Im(dⁱ⁺¹)^∗⊆ Cⁱ.

Let us prove the inverse inclusion. It follows from the definition that ∆ⁱ: Cⁱ→ Cⁱ is a self-adjoint operator. By spectral theorem, Cⁱ can be decomposed into orthogonal direct sum of eigenspaces of

∆ⁱ, i.e.

Cⁱ=M

λ

E_∆i(λ) = ker ∆ⁱ⊕M

λ6=0

E_∆i(λ), where E_∆i(λ) = ker(∆ⁱ− λI). For each x ∈ Cⁱ, we can write x = x₀+P

λ6=0x_λ (this is a finite sum because Cⁱ is finite dimensional) where x0 ∈ ker ∆ⁱ and xλ ∈ E_∆i(λ) with λ 6= 0. Then

∆ⁱx =P

λλxλ∈L

λ6=0E_∆i(λ). In other words, Im ∆ⁱ⊆L

λ6=0E_∆i(λ). For each x = xλ∈ E_∆i(λ),

∆ⁱx = λx which implies that x = ∆ⁱ(x/λ) ∈ Im ∆ⁱ. We find that E_∆i(λ) ⊆ Im ∆ⁱ for any λ 6= 0.

Since Im ∆ⁱ is a vector subspace of Cⁱ,L

λ6=0E_∆i(λ) ⊆ Im ∆ⁱ. We prove the equation Im ∆ⁱ=M

λ6=0

E_∆i(λ).

We find that Cⁱ= ker ∆ⁱ⊕ Im ∆ⁱ. For each x ∈ Cⁱ, let us write x = x1+ x2where x1∈ ker ∆ⁱ and x2∈ Im ∆ⁱ. By definition, choose y ∈ Cⁱ so that

x₂= ∆ⁱy = dⁱ⁻¹((dⁱ)^∗y) + (dⁱ⁺¹)^∗(dⁱy) ∈ Im dⁱ⁻¹⊕ Im(dⁱ⁺¹)^∗. This implies that x ∈ ker ∆ⁱ⊕ Im dⁱ⁻¹⊕ Im(dⁱ⁺¹)^∗ for any x ∈ Cⁱ. We find that

Cⁱ⊆ ker ∆ⁱ⊕ Im dⁱ⁻¹⊕ Im(dⁱ⁺¹)^∗.

We complete the proof of our assertion. 

Corollary 1.1. Let C be a cochain complex of finite dimensional vector spaces and h^Cbe a metric on C. Then there is a linear isomorphism of vector spaces

Hⁱ(C) ∼= ker ∆ⁱ for any i ∈ Z.

Proof. Since ker ∆ⁱ = ker dⁱ∩ ker(dⁱ)^∗, ker ∆ⁱ ⊆ ker dⁱ. Since dⁱ◦ dⁱ⁻¹ = 0, Im dⁱ⁻¹ ⊆ ker dⁱ. We find that ker ∆ⁱ⊕ Im dⁱ⁻¹⊆ ker dⁱ. Let us prove that ker dⁱ⊆ ker ∆ⁱ⊕ Im dⁱ⁻¹. Let x ∈ ker dⁱ. By the Hodge decomposition, we can write x = x1+ x2+ x3 for x1∈ ker ∆ⁱ and for x2∈ Im dⁱ⁻¹ and x3∈ Im(dⁱ⁺¹)^∗. Choose z ∈ Cⁱ⁺¹ so that x3= (dⁱ⁺¹)^∗z. Since x ∈ ker dⁱ, dⁱx3= 0. Now,

kx3k²_Ci = hx3, x3i_Ci= hx3, (dⁱ⁺¹)^∗zi_Ci = hdⁱx3, zi_Ci+1= h0, zi_Ci+1 = 0.

This proves that x₃ = 0. Therefore x ∈ ker ∆ⁱ ⊕ Im dⁱ⁻¹ for any x ∈ ker dⁱ and hence ker dⁱ ⊆ ker ∆ⁱ⊕ Im dⁱ⁻¹. We conclude that ker dⁱ = ker ∆ⁱ⊕ Im dⁱ⁻¹. By linear algebra, any (orthogonal) direct sum decomposition U = V ⊕ W of vector spaces induces a linear isomorphism U/W ∼= V. We find that

Hⁱ(C) = ker dⁱ/ Im dⁱ⁻¹∼= ker ∆ⁱ.



3

The orthogonal decomposition Cⁱ= ker ∆ⁱ⊕ Im ∆ⁱ can also be proved using the following basic facts in linear algebra.

Lemma 1.2. Let A : V → W be a linear map between inner product spaces. Then (1) Im A^⊥= ker A^∗, and

(2) (Im A^∗)^⊥= ker A.

Proof. Let z ∈ Im A^⊥.Then hy, ziW = 0 for any y ∈ Im A. For any x ∈ V, hAx, ziW = hx, A^∗ziV = 0.

This implies that A^∗z = 0. In other words, z ∈ ker A^∗. This proves that Im A^⊥ ⊆ ker A^∗. Suppose z ∈ ker A^∗. Then A^∗z = 0. For any y ∈ Im A, we can write y = Ax and hence

hy, ziW = hAx, ziW = hx, A^∗ziV = hx, 0iV = 0.

We see that z ∈ Im A^⊥. Therefore ker A^∗⊆ Im A^⊥. We conclude (1).

 This lemma implies the following result.

Corollary 1.2. Let A : V → V be a self-adjoint linear operator on a finite dimensional inner product space. Then we have the following orthogonal decomposition:

V = ker A ⊕ Im A.

Proof. Since A is self-adjoint, A = A^∗. Hence Im A^⊥ = ker A which implies that ker A ∩ Im A = {0}.

Hence the sum of vector spaces ker A + Im A is a direct sum ker A ⊕ Im A. Since ker A ⊕ Im A is a vector subspace of V and by the rank-nullity lemma,

dim V = dim ker A + dim Im A = dim(ker A ⊕ Im A),

we see that V = ker A ⊕ Im A. (Here we use the fact that if W is a vector subspace of a finite dimensional vector space V so that dim V = dim W, then W = V.)