Introduction to finite element exterior calculus

Ragnar Winther

CMA, University of OsloNorway

Why finite element exterior calculus?

Recall the de Rham complex on the form:

R →H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0

and the corresponding elasticity complex

T →H1(Ω; R3)ε−→ H(J,Ω; S)

J−→ H(div,Ω; S)div−−→ L2(Ω; R3) −→ 0.

Here J is the second order operator

Jτ = curl(curl τ)T

mapping symmetric matrix fields into symmetric matrix fields, andT is the six dimensional space of rigid motions.Reference: Arnold, Falk, Winther, Mixed finite elements for linearelasticity with weakly imposed symmetry, Math.Comp. 2007.

Why finite element exterior calculus?

Recall the de Rham complex on the form:

R →H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0

and the corresponding elasticity complex

T →H1(Ω; R3)ε−→ H(J,Ω; S)

J−→ H(div,Ω; S)div−−→ L2(Ω; R3) −→ 0.

Here J is the second order operator

Jτ = curl(curl τ)T

mapping symmetric matrix fields into symmetric matrix fields, andT is the six dimensional space of rigid motions.

Reference: Arnold, Falk, Winther, Mixed finite elements for linearelasticity with weakly imposed symmetry, Math.Comp. 2007.

Why finite element exterior calculus?

Recall the de Rham complex on the form:

R →H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0

and the corresponding elasticity complex

T →H1(Ω; R3)ε−→ H(J,Ω; S)

J−→ H(div,Ω; S)div−−→ L2(Ω; R3) −→ 0.

Here J is the second order operator

Jτ = curl(curl τ)T

mapping symmetric matrix fields into symmetric matrix fields, andT is the six dimensional space of rigid motions.Reference: Arnold, Falk, Winther, Mixed finite elements for linearelasticity with weakly imposed symmetry, Math.Comp. 2007.

Mapping of the domain

Consider a map φ : R3 → R3 mapping a domain Ω to Ω′.

Ω’Ωϕ

Desired commuting diagram:

R →H1(Ω′)grad−−→ H(curl,Ω′)

curl−−→ H(div,Ω′)div−−→ L2(Ω′) −→ 0yF 1

φ

yF cφ

yF dφ

yF 0φ

R → H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0

Mapping of the domain

Consider a map φ : R3 → R3 mapping a domain Ω to Ω′.

Ω’Ωϕ

Desired commuting diagram:

R →H1(Ω′)grad−−→ H(curl,Ω′)

curl−−→ H(div,Ω′)div−−→ L2(Ω′) −→ 0yF 1

φ

yF cφ

yF dφ

yF 0φ

R → H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0

Piola transforms

The proper “Piola transforms” are given by

(F 1φu)(x) = u(φ(x))

(F cφu)(x) = (Dφ(x))T u(φ(x)

(F dφ u)(x) = (det Dφ(x))(Dφ(x))−1u(φ(x))

(F 0φu)(x) = (det Dφ(x))u(φ(x))

Alternative de Rham complex

We replace the de Rham complex

R →H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0

by an equivalent complex

R →HΛ0(Ω)d−→ HΛ1(Ω)

d−→ HΛ2(Ω)d−→ L2Λ3(Ω) −→ 0

In fact, the second complex can be defined for any dimension n.

Exterior calculus and differential forms

Instead of consider scalar fields and vector fields, i.e., functions ofthe form

u : Ω→ R or u : Ω→ R3

where Ω ⊂ R3, we consider instead functions of the form

u : Ω→ AltkRn

Here Ω is allowed to be a subset of Rn.

For any vector space V , the space AltkV is the space ofalternating k-linear maps on V

Exterior calculus and differential forms

Instead of consider scalar fields and vector fields, i.e., functions ofthe form

u : Ω→ R or u : Ω→ R3

where Ω ⊂ R3, we consider instead functions of the form

u : Ω→ AltkRn

Here Ω is allowed to be a subset of Rn.

For any vector space V , the space AltkV is the space ofalternating k-linear maps on V

The space AltkV

Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.

Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.As a consequence AltkV is a vector space with

dim AltkV =

(n

k

).

Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.

The space AltkV

Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.

Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.

As a consequence AltkV is a vector space with

dim AltkV =

(n

k

).

Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.

The space AltkV

Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.

Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.As a consequence AltkV is a vector space with

dim AltkV =

(n

k

).

Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.

The space AltkV

Assume dim V = n and that v1, v2, . . . vn is a basis. If ω ∈ AltkVand j1, j2, . . . , jk are integers in 1, 2, . . . , n thenω(vj1 , vj2 , . . . , vjk ) ∈ R.

Furthermore, ω is linear in each argument, and if two neighboringvectors change position then the value of ω change sign.Therefore, it is enough to specify ω(vj1 , vj2 , . . . , vjk ) for anincreasing sequence ji.As a consequence AltkV is a vector space with

dim AltkV =

(n

k

).

Note that dim AltkV = dim Altn−kV . The canonical map betweenthese spaces is the Hodge star operator.

Interior and exterior products

If ω ∈ AltkV and v ∈ V then the interior product, or thecontraction of ω by v , ωyv ∈ Altk−1V is given by

ωyv(v1, v2, . . . vk−1) = ω(v , v1, v2, . . . vk−1)

If ω ∈ AltjV and η ∈ AltkV then the exterior product, or wedge

product, ω ∧ η ∈ Altj+k is given by

(ω ∧ η)(v1, . . . , vj+k)

=∑σ

(signσ)ω(vσ(1), . . . , vσ(j))η(vσ(j+1), . . . , vσ(j+k)), vi ∈ V ,

where the sum is over all permutations σ of 1, 2, . . . , j + k suchthat σ(1) < ... < σ(j) and σ(j + 1) < ... < σ(j + k).

Interior and exterior products

If ω ∈ AltkV and v ∈ V then the interior product, or thecontraction of ω by v , ωyv ∈ Altk−1V is given by

ωyv(v1, v2, . . . vk−1) = ω(v , v1, v2, . . . vk−1)

If ω ∈ AltjV and η ∈ AltkV then the exterior product, or wedge

product, ω ∧ η ∈ Altj+k is given by

(ω ∧ η)(v1, . . . , vj+k)

=∑σ

(signσ)ω(vσ(1), . . . , vσ(j))η(vσ(j+1), . . . , vσ(j+k)), vi ∈ V ,

where the sum is over all permutations σ of 1, 2, . . . , j + k suchthat σ(1) < ... < σ(j) and σ(j + 1) < ... < σ(j + k).

Basis for AltkRn

We let dxi be the natural dual basis in Alt1Rn given by

dxi (ej) = δi ,j

A basis for Alt2Rn is given by

dxi ∧ dxj , i < j ,

while

dxj1 ∧ dxj2 ∧ · · · ∧ dxjk , where ji is increasing

is a basis for AltkRn.

Basis for AltkRn

We let dxi be the natural dual basis in Alt1Rn given by

dxi (ej) = δi ,j

A basis for Alt2Rn is given by

dxi ∧ dxj , i < j ,

while

dxj1 ∧ dxj2 ∧ · · · ∧ dxjk , where ji is increasing

is a basis for AltkRn.

Basis for AltkRn

We let dxi be the natural dual basis in Alt1Rn given by

dxi (ej) = δi ,j

A basis for Alt2Rn is given by

dxi ∧ dxj , i < j ,

while

dxj1 ∧ dxj2 ∧ · · · ∧ dxjk , where ji is increasing

is a basis for AltkRn.

Any ω ∈ AltkRn can be represented uniquely on the form∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

where the sum is over all incresing maps σ mapping 1, 2, . . . , kinto 1, 2, . . . , n, and uσ ∈ R.

Furthermore,

dxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)(v1, . . . vk) = det(dxσ(i)vj)

Any ω ∈ AltkRn can be represented uniquely on the form∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

where the sum is over all incresing maps σ mapping 1, 2, . . . , kinto 1, 2, . . . , n, and uσ ∈ R.

Furthermore,

dxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)(v1, . . . vk) = det(dxσ(i)vj)

Correspondences

Altk and vector fields:

Alt0 R3 = R c ↔ c

Alt1 R3 ∼=−→ R3 u1 dx1 + u2 dx2 + u3 dx3 ↔ u

Alt2 R3 ∼=−→ R3 u3 dx1 ∧ dx2 − u2 dx1 ∧ dx3 + u1 dx2 ∧ dx3 ↔ u

Alt3 R3 ∼=−→ R c ↔ c dx1 ∧ dx2 ∧ dx3

Exterior product:

∧ : Alt1 R3 × Alt1 R3 → Alt2 R3 × : R3 × R3 → R3

∧ : Alt1 R3 × Alt2 R3 → Alt3 R3 · : R3 × R3 → R

Correspondences

Altk and vector fields:

Alt0 R3 = R c ↔ c

Alt1 R3 ∼=−→ R3 u1 dx1 + u2 dx2 + u3 dx3 ↔ u

Alt2 R3 ∼=−→ R3 u3 dx1 ∧ dx2 − u2 dx1 ∧ dx3 + u1 dx2 ∧ dx3 ↔ u

Alt3 R3 ∼=−→ R c ↔ c dx1 ∧ dx2 ∧ dx3

Exterior product:

∧ : Alt1 R3 × Alt1 R3 → Alt2 R3 × : R3 × R3 → R3

∧ : Alt1 R3 × Alt2 R3 → Alt3 R3 · : R3 × R3 → R

Differential forms

Differential forms are functions of the form

ω : Ω→ Altk ≡ Altk Rn.

We let Λk = Λk(Ω) = C∞(Ω; Altk). For such smooth differentiableforms the exterior derivative, d : Λk → Λk+1 is defined by

dωx(v1, v2, . . . , vk+1) =k+1∑j=1

(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),

ω ∈ Λk , v1, . . . , vk+1 ∈ Rn

Using the fact that ∂vi∂vj = ∂vj∂vi it follows that d d = 0.

Differential forms

Differential forms are functions of the form

ω : Ω→ Altk ≡ Altk Rn.

We let Λk = Λk(Ω) = C∞(Ω; Altk). For such smooth differentiableforms the exterior derivative, d : Λk → Λk+1 is defined by

dωx(v1, v2, . . . , vk+1) =k+1∑j=1

(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),

ω ∈ Λk , v1, . . . , vk+1 ∈ Rn

Using the fact that ∂vi∂vj = ∂vj∂vi it follows that d d = 0.

Differential forms

Differential forms are functions of the form

ω : Ω→ Altk ≡ Altk Rn.

We let Λk = Λk(Ω) = C∞(Ω; Altk). For such smooth differentiableforms the exterior derivative, d : Λk → Λk+1 is defined by

dωx(v1, v2, . . . , vk+1) =k+1∑j=1

(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),

ω ∈ Λk , v1, . . . , vk+1 ∈ Rn

Using the fact that ∂vi∂vj = ∂vj∂vi it follows that d d = 0.

Check of d d = 0

dωx(v1, v2, . . . , vk+1) =k+1∑j=1

(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),

ω ∈ Λk , v1, . . . , vk+1 ∈ Rn

Start with a zero form ω.

dωx(v) = ∂vωx

(d d)ω(v1, v2) = ∂v1dω(v2)− ∂v2dω(v1) = 0.

Check of d d = 0

dωx(v1, v2, . . . , vk+1) =k+1∑j=1

(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),

ω ∈ Λk , v1, . . . , vk+1 ∈ Rn

Start with a zero form ω.

dωx(v) = ∂vωx

(d d)ω(v1, v2) = ∂v1dω(v2)− ∂v2dω(v1) = 0.

Check of d d = 0

dωx(v1, v2, . . . , vk+1) =k+1∑j=1

(−1)j+1∂vjωx(v1, . . . , vj , . . . , vk+1),

ω ∈ Λk , v1, . . . , vk+1 ∈ Rn

Start with a zero form ω.

dωx(v) = ∂vωx

(d d)ω(v1, v2) = ∂v1dω(v2)− ∂v2dω(v1) = 0.

Representation by a basis

If ω is given on the form

ωx =∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k),

where each uσ is a scalar field on Ω then

dωx =∑σ

n∑j=1

∂uσ∂xj

dxj ∧ dxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

The de Rham complex and differential forms

Using differential forms the de Rham complex (smooth version)can be written as

R →Λ0 d−→ Λ1 d−→ · · · d−→ Λn −→ 0.

Mapping of the domain

Ω’Ωϕ

The corresponding pull backs, φ∗, which maps Λk(Ω′) to Λk(Ω), isgiven by

(φ∗ω)x(v1, v2, . . . , vk) = ωφ(x)(Dφx(v1),Dφx(v2), . . . ,Dφx(vk)),

where Dφx is the derivative of φ at x mapping TxΩ into Tφ(x)Ω′.This expression should be compared with the expressions we hadfor the Piola maps before.

Mapping of the domain

Ω’Ωϕ

The corresponding pull backs, φ∗, which maps Λk(Ω′) to Λk(Ω), isgiven by

(φ∗ω)x(v1, v2, . . . , vk) = ωφ(x)(Dφx(v1),Dφx(v2), . . . ,Dφx(vk)),

where Dφx is the derivative of φ at x mapping TxΩ into Tφ(x)Ω′.

This expression should be compared with the expressions we hadfor the Piola maps before.

Mapping of the domain

Ω’Ωϕ

The corresponding pull backs, φ∗, which maps Λk(Ω′) to Λk(Ω), isgiven by

(φ∗ω)x(v1, v2, . . . , vk) = ωφ(x)(Dφx(v1),Dφx(v2), . . . ,Dφx(vk)),

where Dφx is the derivative of φ at x mapping TxΩ into Tφ(x)Ω′.This expression should be compared with the expressions we hadfor the Piola maps before.

Commuting diagram

R →Λ0(Ω′)d−→ Λ1(Ω′)

d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)

d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.

So the pullback commutes with the exterior derivative, i.e.,

φ∗(dω) = d(φ∗ω), ω ∈ Λk(Ω′),

and distributes with respect to the wedge product:

φ∗(ω ∧ η) = φ∗ω ∧ φ∗η.

Note this also gives us a way to define the exterior derivative on amanifold.Key identity: (ψ φ)∗ = φ∗ ψ∗

Commuting diagram

R →Λ0(Ω′)d−→ Λ1(Ω′)

d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)

d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.

So the pullback commutes with the exterior derivative, i.e.,

φ∗(dω) = d(φ∗ω), ω ∈ Λk(Ω′),

and distributes with respect to the wedge product:

φ∗(ω ∧ η) = φ∗ω ∧ φ∗η.

Note this also gives us a way to define the exterior derivative on amanifold.Key identity: (ψ φ)∗ = φ∗ ψ∗

Commuting diagram

R →Λ0(Ω′)d−→ Λ1(Ω′)

d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)

d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.

So the pullback commutes with the exterior derivative, i.e.,

φ∗(dω) = d(φ∗ω), ω ∈ Λk(Ω′),

and distributes with respect to the wedge product:

φ∗(ω ∧ η) = φ∗ω ∧ φ∗η.

Note this also gives us a way to define the exterior derivative on amanifold.

Key identity: (ψ φ)∗ = φ∗ ψ∗

Commuting diagram

R →Λ0(Ω′)d−→ Λ1(Ω′)

d−→ · · · d−→ Λn(Ω′) −→ 0.yφ∗ yφ∗ yφ∗R → Λ0(Ω)

d−→ Λ1(Ω)d−→ · · · d−→ Λn(Ω) −→ 0.

So the pullback commutes with the exterior derivative, i.e.,

φ∗(dω) = d(φ∗ω), ω ∈ Λk(Ω′),

and distributes with respect to the wedge product:

φ∗(ω ∧ η) = φ∗ω ∧ φ∗η.

Note this also gives us a way to define the exterior derivative on amanifold.Key identity: (ψ φ)∗ = φ∗ ψ∗

Exterior derivative and wedge product

By combining the wedge product with the exterior derivative weobtain a Leibniz rule of the form

d(ω ∧ η) = dω ∧ η + (−1)jω ∧ dη, ω ∈ Λj , η ∈ Λk .

The pullback of the inclusion ∂Ω → Ω is the trace map, Tr. Thevanishing of Trω for ω ∈ Λk(Ω) does not imply that ωx vanishesfor x ∈ ∂Ω, only that it vanishes when applied to k-tuples oftangent vectors to ∂Ω.

Since Tr is a pull back we will have

Tr d = d∂Ω Tr

Exterior derivative and wedge product

By combining the wedge product with the exterior derivative weobtain a Leibniz rule of the form

d(ω ∧ η) = dω ∧ η + (−1)jω ∧ dη, ω ∈ Λj , η ∈ Λk .

The pullback of the inclusion ∂Ω → Ω is the trace map, Tr. Thevanishing of Trω for ω ∈ Λk(Ω) does not imply that ωx vanishesfor x ∈ ∂Ω, only that it vanishes when applied to k-tuples oftangent vectors to ∂Ω.

Since Tr is a pull back we will have

Tr d = d∂Ω Tr

Exterior derivative and wedge product

By combining the wedge product with the exterior derivative weobtain a Leibniz rule of the form

d(ω ∧ η) = dω ∧ η + (−1)jω ∧ dη, ω ∈ Λj , η ∈ Λk .

The pullback of the inclusion ∂Ω → Ω is the trace map, Tr. Thevanishing of Trω for ω ∈ Λk(Ω) does not imply that ωx vanishesfor x ∈ ∂Ω, only that it vanishes when applied to k-tuples oftangent vectors to ∂Ω.

Since Tr is a pull back we will have

Tr d = d∂Ω Tr

Integration of differential forms

If ω ∈ Λn(Ω), Ω ⊂ Rn, with the representation

ωx = uxdx1 ∧ dx2 ∧ . . . ∧ dxn,

then we can define ∫Ωω =

∫Ω

u dx

If φ : Ω→ Ω′ then we have∫Ωφ∗ω =

∫Ω′ω

Note that this gives us a way to define the integral of an n-form ona manifold.

Integration of differential forms

If ω ∈ Λn(Ω), Ω ⊂ Rn, with the representation

ωx = uxdx1 ∧ dx2 ∧ . . . ∧ dxn,

then we can define ∫Ωω =

∫Ω

u dx

If φ : Ω→ Ω′ then we have∫Ωφ∗ω =

∫Ω′ω

Note that this gives us a way to define the integral of an n-form ona manifold.

Integration of differential forms

If ω ∈ Λn(Ω), Ω ⊂ Rn, with the representation

ωx = uxdx1 ∧ dx2 ∧ . . . ∧ dxn,

then we can define ∫Ωω =

∫Ω

u dx

If φ : Ω→ Ω′ then we have∫Ωφ∗ω =

∫Ω′ω

Note that this gives us a way to define the integral of an n-form ona manifold.

Stokes theorem

With the three concepts exterior derivative, trace operator, andintegral we can formulate Stokes theorem as∫

Ωdω =

∫∂Ω

Trω, ω ∈ Λn−1(Ω)

Furthermore, by combining this with the Leibniz rule we obtain thefollowing integration by parts identity∫

Ωdω ∧ η = (−1)k−1

∫Ωω ∧ dη +

∫∂Ω

Trω ∧ Tr η

where ω ∈ Λk , η ∈ Λn−k−1.

Stokes theorem

With the three concepts exterior derivative, trace operator, andintegral we can formulate Stokes theorem as∫

Ωdω =

∫∂Ω

Trω, ω ∈ Λn−1(Ω)

Furthermore, by combining this with the Leibniz rule we obtain thefollowing integration by parts identity∫

Ωdω ∧ η = (−1)k−1

∫Ωω ∧ dη +

∫∂Ω

Trω ∧ Tr η

where ω ∈ Λk , η ∈ Λn−k−1.

L2 differential forms

If ω is a k form given by

ω =∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

then ω is defined to be in ∈ L2Λk(Ω) if each of the coefficients uσare in L2.

Furthernmore, we let

HΛk(Ω) = ω ∈ L2Λk(Ω) | dω ∈ L2Λk+1(Ω)

We then ontain the L2 de Rham complex:

0→ HΛ0(Ω)d−→ HΛ1(Ω)

d−→ · · · d−→ HΛn(Ω)→ 0

L2 differential forms

If ω is a k form given by

ω =∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

then ω is defined to be in ∈ L2Λk(Ω) if each of the coefficients uσare in L2.

Furthernmore, we let

HΛk(Ω) = ω ∈ L2Λk(Ω) | dω ∈ L2Λk+1(Ω)

We then ontain the L2 de Rham complex:

0→ HΛ0(Ω)d−→ HΛ1(Ω)

d−→ · · · d−→ HΛn(Ω)→ 0

L2 differential forms

If ω is a k form given by

ω =∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

then ω is defined to be in ∈ L2Λk(Ω) if each of the coefficients uσare in L2.

Furthernmore, we let

HΛk(Ω) = ω ∈ L2Λk(Ω) | dω ∈ L2Λk+1(Ω)

We then ontain the L2 de Rham complex:

0→ HΛ0(Ω)d−→ HΛ1(Ω)

d−→ · · · d−→ HΛn(Ω)→ 0

Polynomial de Rham complex

We define

Pr Λk = ω ∈ Λk |ω(v1, . . . vk) ∈ Pr , ∀v1, . . . vk

Alternatively, Pr Λk consists of all∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

where the coefficients uσ are in Pr .The polynomial de Rham complex

0→ Pr Λ0 d−→ Pr−1Λ1 d−→ · · · d−→ Pr−nΛn → 0

is exact.

Polynomial de Rham complex

We define

Pr Λk = ω ∈ Λk |ω(v1, . . . vk) ∈ Pr , ∀v1, . . . vk

Alternatively, Pr Λk consists of all∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

where the coefficients uσ are in Pr .

The polynomial de Rham complex

0→ Pr Λ0 d−→ Pr−1Λ1 d−→ · · · d−→ Pr−nΛn → 0

is exact.

Polynomial de Rham complex

We define

Pr Λk = ω ∈ Λk |ω(v1, . . . vk) ∈ Pr , ∀v1, . . . vk

Alternatively, Pr Λk consists of all∑σ

uσdxσ(1) ∧ dxσ(2) ∧ . . . ∧ dxσ(k)

where the coefficients uσ are in Pr .The polynomial de Rham complex

0→ Pr Λ0 d−→ Pr−1Λ1 d−→ · · · d−→ Pr−nΛn → 0

is exact.

The Kozul complex

The Koszul differential κ of a k-form ω is the (k − 1)-form given by

(κω)x(v1, . . . , vk−1) = ωxyx(v1, . . . , vk−1)

= ωx

(x , v1, . . . , vk−1

),

For each r , κ maps Pr−1Λk to Pr Λk−1, and the Koszul complex

0→ Pr−nΛn κ−→ Pr−n+1Λn−1 κ−→ · · · κ−→ Pr Λ0 −→ R→ 0,

is exact.

The Kozul complex

The Koszul differential κ of a k-form ω is the (k − 1)-form given by

(κω)x(v1, . . . , vk−1) = ωxyx(v1, . . . , vk−1)

= ωx

(x , v1, . . . , vk−1

),

For each r , κ maps Pr−1Λk to Pr Λk−1,

and the Koszul complex

0→ Pr−nΛn κ−→ Pr−n+1Λn−1 κ−→ · · · κ−→ Pr Λ0 −→ R→ 0,

is exact.

The Kozul complex

The Koszul differential κ of a k-form ω is the (k − 1)-form given by

(κω)x(v1, . . . , vk−1) = ωxyx(v1, . . . , vk−1)

= ωx

(x , v1, . . . , vk−1

),

For each r , κ maps Pr−1Λk to Pr Λk−1, and the Koszul complex

0→ Pr−nΛn κ−→ Pr−n+1Λn−1 κ−→ · · · κ−→ Pr Λ0 −→ R→ 0,

is exact.

The spaces P−r Λk

The operators d and κ are related by the homotopy relation

(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,

where Hr denotes the homogeneous polynomials of degree r .

As a consequence we obtain the identity

Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1

We then define the space P−r Λk by

P−r Λk = Pr−1Λk + κPr−1Λk+1.

We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,

0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0

is an exact complex, and the space P−r Λk is affine invariant.

The spaces P−r Λk

The operators d and κ are related by the homotopy relation

(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,

where Hr denotes the homogeneous polynomials of degree r .As a consequence we obtain the identity

Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1

We then define the space P−r Λk by

P−r Λk = Pr−1Λk + κPr−1Λk+1.

We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,

0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0

is an exact complex, and the space P−r Λk is affine invariant.

The spaces P−r Λk

The operators d and κ are related by the homotopy relation

(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,

where Hr denotes the homogeneous polynomials of degree r .As a consequence we obtain the identity

Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1

We then define the space P−r Λk by

P−r Λk = Pr−1Λk + κPr−1Λk+1.

We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,

0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0

is an exact complex, and the space P−r Λk is affine invariant.

The spaces P−r Λk

The operators d and κ are related by the homotopy relation

(dκ+ κd)ω = (r + k)ω, ω ∈ Hr Λk ,

where Hr denotes the homogeneous polynomials of degree r .As a consequence we obtain the identity

Pr Λk = Pr−1Λk + κHr−1Λk+1 + dHr+1Λk−1

We then define the space P−r Λk by

P−r Λk = Pr−1Λk + κPr−1Λk+1.

We note that P−r Λ0 = Pr Λ0 and P−r Λn = Pr−1Λn. Furthermore,

0→ P−r Λ0 d−→ P−r Λ1 d−→ · · · d−→ P−r Λn → 0

is an exact complex, and the space P−r Λk is affine invariant.

Characterization of affine invariant polynomial spaces

TheoremAssume that the polynomial space QΛk , satisfiesPr−1Λk ( QΛk ( Pr Λk . Then

QΛk = P−r Λk or QΛk = Pr−1Λk + dPr+1Λk−1

The four exact sequences ending with Pr Λ3(T ) in 3D

0→ Pr+1Λ0 d−→ P−r+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0

0→ Pr+2Λ0 d−→ Pr+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0

0→ Pr+2Λ0 d−→ P−r+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0

0→ Pr+3Λ0 d−→ Pr+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0

In general there are 2n−1 exact sequences composed of the familiesof P and P− spaces.

The four exact sequences ending with Pr Λ3(T ) in 3D

0→ Pr+1Λ0 d−→ P−r+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0

0→ Pr+2Λ0 d−→ Pr+1Λ1 d−→ P−r+1Λ2 d−→ Pr Λ3 → 0

0→ Pr+2Λ0 d−→ P−r+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0

0→ Pr+3Λ0 d−→ Pr+2Λ1 d−→ Pr+1Λ2 d−→ Pr Λ3 → 0

In general there are 2n−1 exact sequences composed of the familiesof P and P− spaces.

The four sequences ending with P0Λ3(T ) in 3D

0→ grad−−→ curl−−→ div−−→ → 0

0→ grad−−→ curl−−→ div−−→ → 0

0→ grad−−→ curl−−→ div−−→ → 0

0→ grad−−→ curl−−→ div−−→ → 0

Piecewise smooth differential formsIt is a consequence of Stokes theorem that a piecewise smoothk–form ω, with respect to a simplicial mesh Th of Ω, is in HΛk(Ω)if and only if the trace of ω, Trω, is continuous on the interfaces.

Here Trω is defined by restricting thespatial variable x to the interface, andby applying ω only to tangent vectors ofthe interface.

Degrees of freedom

To obtain finite element differential forms—not just pw polynomials—we

need degrees of freedom, i.e., a decomposition of the dual spaces

(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces

associated to subsimplices f of T .

DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate

ω 7→∫

fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )

DOF for P−r Λk(T ):

ω 7→∫

fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )

Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They

are subspaces of HΛk(Ω).

Degrees of freedom

To obtain finite element differential forms—not just pw polynomials—we

need degrees of freedom, i.e., a decomposition of the dual spaces

(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces

associated to subsimplices f of T .

DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate

ω 7→∫

fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )

DOF for P−r Λk(T ):

ω 7→∫

fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )

Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They

are subspaces of HΛk(Ω).

Degrees of freedom

To obtain finite element differential forms—not just pw polynomials—we

need degrees of freedom, i.e., a decomposition of the dual spaces

(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces

associated to subsimplices f of T .

DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate

ω 7→∫

fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )

DOF for P−r Λk(T ):

ω 7→∫

fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )

Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They

are subspaces of HΛk(Ω).

Degrees of freedom

To obtain finite element differential forms—not just pw polynomials—we

need degrees of freedom, i.e., a decomposition of the dual spaces

(Pr Λk(T ))∗ and (P−r Λk(T ))∗ (with T a simplex), into subspaces

associated to subsimplices f of T .

DOF for Pr Λk(T ): to a subsimplex f of dimension d we associate

ω 7→∫

fTrf ω ∧ η, η ∈ P−r+k−dΛd−k(f )

DOF for P−r Λk(T ):

ω 7→∫

fTrf ω ∧ η, η ∈ Pr+k−d−1Λd−k(f )

Given a triangulation T , we can then define Pr Λk(T ), P−r Λk(T ). They

are subspaces of HΛk(Ω).

De Rham cohomology

0 → Λ0(Ω)d−→ Λ1(Ω)

d−→ Λ2(Ω) → 0

0 → C∞(Ω)grad−−→ C∞(Ω; R2)

rot−→ C∞(Ω) → 0

Cohomology

The de Rham complex

HΛk−1(Ω)dk−1

−−−→ HΛk(Ω)dk

−→ HΛk+1(Ω)

is called exact if for all k,

Zk := ker dk = range dk−1 =: Bk .

In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat

〈q, µ〉 = 0 µ ∈ Bk .

This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .

Cohomology

The de Rham complex

HΛk−1(Ω)dk−1

−−−→ HΛk(Ω)dk

−→ HΛk+1(Ω)

is called exact if for all k,

Zk := ker dk = range dk−1 =: Bk .

In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.

The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat

〈q, µ〉 = 0 µ ∈ Bk .

This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .

Cohomology

The de Rham complex

HΛk−1(Ω)dk−1

−−−→ HΛk(Ω)dk

−→ HΛk+1(Ω)

is called exact if for all k,

Zk := ker dk = range dk−1 =: Bk .

In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat

〈q, µ〉 = 0 µ ∈ Bk .

This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .

Cohomology

The de Rham complex

HΛk−1(Ω)dk−1

−−−→ HΛk(Ω)dk

−→ HΛk+1(Ω)

is called exact if for all k,

Zk := ker dk = range dk−1 =: Bk .

In general, Bk ⊂ Zk and we assume throughout that the kthcohomology group Zk/Bk is finite dimensional.The space of harmonic k-forms, Hk , consists of all q ∈ Zk suchthat

〈q, µ〉 = 0 µ ∈ Bk .

This leads to the Hodge decompositionHΛk(Ω) = Zk ⊕ Zk⊥ = Bk ⊕ Hk ⊕ Zk⊥. Note that Hk ∼= Zk/Bk .

Hodge Laplace problem

Λk−1(Ω)dk−1

−−−→ Λk(Ω)dk

−→ Λk+1(Ω)

Formally: Given f ∈ Λk , find u ∈ Λk such that

(dk−1δk−1 + δkdk)u = f .

Here δk is a formal adjoint of dk .

The following mixed formulation is always well-posed: Givenf ∈ L2Λk(Ω), find σ ∈ HΛk−1, u ∈ HΛk and p ∈ Hk such that

〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1

〈dσ, v〉+ 〈du, dv〉+ 〈p, v〉 = 〈f , v〉 ∀v ∈ HΛk

〈u, q〉 = 0 ∀q ∈ Hk

Hodge Laplace problem

Λk−1(Ω)dk−1

−−−→ Λk(Ω)dk

−→ Λk+1(Ω)

Formally: Given f ∈ Λk , find u ∈ Λk such that

(dk−1δk−1 + δkdk)u = f .

Here δk is a formal adjoint of dk .

The following mixed formulation is always well-posed: Givenf ∈ L2Λk(Ω), find σ ∈ HΛk−1, u ∈ HΛk and p ∈ Hk such that

〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1

〈dσ, v〉+ 〈du, dv〉+ 〈p, v〉 = 〈f , v〉 ∀v ∈ HΛk

〈u, q〉 = 0 ∀q ∈ Hk

Hodge Laplace problem

Λk−1(Ω)dk−1

−−−→ Λk(Ω)dk

−→ Λk+1(Ω)

Formally: Given f ∈ Λk , find u ∈ Λk such that

(dk−1δk−1 + δkdk)u = f .

Here δk is a formal adjoint of dk .

The following mixed formulation is always well-posed: Givenf ∈ L2Λk(Ω), find σ ∈ HΛk−1, u ∈ HΛk and p ∈ Hk such that

〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1

〈dσ, v〉+ 〈du, dv〉+ 〈p, v〉 = 〈f , v〉 ∀v ∈ HΛk

〈u, q〉 = 0 ∀q ∈ Hk

Hodge LaplacianWell-posedness of the Hodge Laplace problem follows from theHodge decomposition and Poincare’s inequality:

‖ω‖L2 ≤ c ‖dω‖L2 , ω ∈ (Zk)⊥.

Special cases:Recall that if dim Hk = 0: Find σ ∈ HΛk−1, and u ∈ HΛk

〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1

〈dσ, v〉+ 〈du, dv〉 = 〈f , v〉 ∀v ∈ HΛk

I k = 0: ordinary LaplacianI k = n: mixed LaplacianI k = 1, n = 3: σ = − div u, gradσ + curl curl u = fI k = 2, n = 3: σ = curl u, curl σ − grad div u = f ,

Hodge LaplacianWell-posedness of the Hodge Laplace problem follows from theHodge decomposition and Poincare’s inequality:

‖ω‖L2 ≤ c ‖dω‖L2 , ω ∈ (Zk)⊥.

Special cases:Recall that if dim Hk = 0: Find σ ∈ HΛk−1, and u ∈ HΛk

〈σ, τ〉 − 〈dτ, u〉 = 0 ∀τ ∈ HΛk−1

〈dσ, v〉+ 〈du, dv〉 = 〈f , v〉 ∀v ∈ HΛk

I k = 0: ordinary LaplacianI k = n: mixed LaplacianI k = 1, n = 3: σ = − div u, gradσ + curl curl u = fI k = 2, n = 3: σ = curl u, curl σ − grad div u = f ,

Discretizations of the de Rham complex in threedimensions

R →H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0yI1

h

yIch

yIdh

yI0h

R → H1h

grad−−→ Hh(curl)curl−−→ Hh(div)

div−−→ L2h −→ 0.

The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with grad curl anddiv, but the are not bounded on the spaces H1, H(curl) or H(div).

Discretizations of the de Rham complex in threedimensions

R →H1(Ω)grad−−→ H(curl,Ω)

curl−−→ H(div,Ω)div−−→ L2(Ω) −→ 0yI1

h

yIch

yIdh

yI0h

R → H1h

grad−−→ Hh(curl)curl−−→ Hh(div)

div−−→ L2h −→ 0.

The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with grad curl anddiv, but the are not bounded on the spaces H1, H(curl) or H(div).

Discretizations of the de Rham complex in n dimensions

R →HΛ0(Ω)d−→ HΛ1(Ω)

d−→ · · · d−→ HΛn(Ω) −→ 0.yIh yIh yIhR → Λ0

hd−→ Λ1

hd−→ · · · d−→ Λn

h −→ 0.

The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with d , but the arenot bounded on the HΛ.

Discretizations of the de Rham complex in n dimensions

R →HΛ0(Ω)d−→ HΛ1(Ω)

d−→ · · · d−→ HΛn(Ω) −→ 0.yIh yIh yIhR → Λ0

hd−→ Λ1

hd−→ · · · d−→ Λn

h −→ 0.

The operators Ih are the canonical interpolation operators definedfrom the degrees of freedom. They commute with d , but the arenot bounded on the HΛ.

Discretization, Abstract setting

· · · −→ Λk−1 dk−1

−−→ Λk −→ · · ·

x∪ x∪· · · −→ Λk−1

hdk−1

−−→ Λkh −→ · · ·

Complex of Hilbert spaces with dk bounded and closed range.

Hodge decomposition and Poincare inequality follow.

For discretization, construct a finite dimensional subcomplex.

Define Hkh = (Bk

h)⊥ ∩ Zkh .

Discrete Hodge decomposition follows: Λkh = Bk

h ⊕ Hkh ⊕ (Zk

h)⊥

Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk

h , Hkh

When is it stable?

Discretization, Abstract setting

· · · −→ Λk−1 dk−1

−−→ Λk −→ · · ·

x∪ x∪· · · −→ Λk−1

hdk−1

−−→ Λkh −→ · · ·

Complex of Hilbert spaces with dk bounded and closed range.

Hodge decomposition and Poincare inequality follow.

For discretization, construct a finite dimensional subcomplex.

Define Hkh = (Bk

h)⊥ ∩ Zkh .

Discrete Hodge decomposition follows: Λkh = Bk

h ⊕ Hkh ⊕ (Zk

h)⊥

Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk

h , Hkh

When is it stable?

Discretization, Abstract setting

· · · −→ Λk−1 dk−1

−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1

hdk−1

−−→ Λkh −→ · · ·

Complex of Hilbert spaces with dk bounded and closed range.

Hodge decomposition and Poincare inequality follow.

For discretization, construct a finite dimensional subcomplex.

Define Hkh = (Bk

h)⊥ ∩ Zkh .

Discrete Hodge decomposition follows: Λkh = Bk

h ⊕ Hkh ⊕ (Zk

h)⊥

Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk

h , Hkh

When is it stable?

Discretization, Abstract setting

· · · −→ Λk−1 dk−1

−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1

hdk−1

−−→ Λkh −→ · · ·

Complex of Hilbert spaces with dk bounded and closed range.

Hodge decomposition and Poincare inequality follow.

For discretization, construct a finite dimensional subcomplex.

Define Hkh = (Bk

h)⊥ ∩ Zkh .

Discrete Hodge decomposition follows: Λkh = Bk

h ⊕ Hkh ⊕ (Zk

h)⊥

Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk

h , Hkh

When is it stable?

Discretization, Abstract setting

· · · −→ Λk−1 dk−1

−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1

hdk−1

−−→ Λkh −→ · · ·

Complex of Hilbert spaces with dk bounded and closed range.

Hodge decomposition and Poincare inequality follow.

For discretization, construct a finite dimensional subcomplex.

Define Hkh = (Bk

h)⊥ ∩ Zkh .

Discrete Hodge decomposition follows: Λkh = Bk

h ⊕ Hkh ⊕ (Zk

h)⊥

Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk

h , Hkh

When is it stable?

Discretization, Abstract setting

· · · −→ Λk−1 dk−1

−−→ Λk −→ · · ·x∪ x∪· · · −→ Λk−1

hdk−1

−−→ Λkh −→ · · ·

Complex of Hilbert spaces with dk bounded and closed range.

Hodge decomposition and Poincare inequality follow.

For discretization, construct a finite dimensional subcomplex.

Define Hkh = (Bk

h)⊥ ∩ Zkh .

Discrete Hodge decomposition follows: Λkh = Bk

h ⊕ Hkh ⊕ (Zk

h)⊥

Galerkin’s method: Λk−1, Λk , Hk −→ Λk−1h , Λk

h , Hkh

When is it stable?

Bounded cochain projections

Key property: Suppose that there exists a bounded cochainprojection.

· · · −→ Λk−1 dk−1

−−−→ Λk −→ · · ·yπk−1h

yπkh

· · · −→ Λk−1h

dk−1

−−−→ Λkh −→ · · ·

I πkh uniformly bounded

I ‖πkhω − ω‖ → 0.

I πkh a projection

I πkh dk−1 = dk−1πk−1

h

Theorem

I If ‖v − πkhv‖ < ‖v‖ ∀v ∈ Hk , then the induced map on

cohomology is an isomorphism.

I gap(Hk ,Hk

h

)≤ sup

v∈Hk ,‖v‖=1

‖v − πkhv‖

I The discrete Poincare inequality holds uniformly in h.

I Galerkin’s method is stable and convergent.

Bounded cochain projections

Key property: Suppose that there exists a bounded cochainprojection.

· · · −→ Λk−1 dk−1

−−−→ Λk −→ · · ·yπk−1h

yπkh

· · · −→ Λk−1h

dk−1

−−−→ Λkh −→ · · ·

I πkh uniformly bounded

I ‖πkhω − ω‖ → 0.

I πkh a projection

I πkh dk−1 = dk−1πk−1

h

Theorem

I If ‖v − πkhv‖ < ‖v‖ ∀v ∈ Hk , then the induced map on

cohomology is an isomorphism.

I gap(Hk ,Hk

h

)≤ sup

v∈Hk ,‖v‖=1

‖v − πkhv‖

I The discrete Poincare inequality holds uniformly in h.

I Galerkin’s method is stable and convergent.

Proof of discrete Poincare inequality

Theorem: There is a positive constant c , independent of h, suchthat

‖ω‖ ≤ c‖dω‖, ω ∈ Zk⊥h .

In fact, c ≤ cPcπ, where cP is continuous Poincare constant, andcπ bounds ‖πh‖.

Proof: Given ω ∈ Zk⊥h , define η ∈ Zk⊥ ⊂ HΛk(Ω) by dη = dω. By

the Poincare inequality, ‖η‖ ≤ c‖dω‖, so it is enough to show that‖ω‖ ≤ c‖η‖. Now, ω − πhη ∈ Λk

h and d(ω − πhη) = 0, soω − πhη ∈ Zk

h . Therefore

‖ω‖2 = 〈ω, πhη〉+ 〈ω, ω − πhη〉 = 〈ω, πhη〉 ≤ ‖ω‖‖πhη‖,

whence ‖ω‖ ≤ ‖πhη‖. The result follows from the uniformboundedness of πh.

Proof of discrete Poincare inequality

Theorem: There is a positive constant c , independent of h, suchthat

‖ω‖ ≤ c‖dω‖, ω ∈ Zk⊥h .

In fact, c ≤ cPcπ, where cP is continuous Poincare constant, andcπ bounds ‖πh‖.Proof: Given ω ∈ Zk⊥

h , define η ∈ Zk⊥ ⊂ HΛk(Ω) by dη = dω. Bythe Poincare inequality, ‖η‖ ≤ c‖dω‖, so it is enough to show that‖ω‖ ≤ c‖η‖. Now, ω − πhη ∈ Λk

h and d(ω − πhη) = 0, soω − πhη ∈ Zk

h . Therefore

‖ω‖2 = 〈ω, πhη〉+ 〈ω, ω − πhη〉 = 〈ω, πhη〉 ≤ ‖ω‖‖πhη‖,

whence ‖ω‖ ≤ ‖πhη‖. The result follows from the uniformboundedness of πh.

Construction of bounded cochain projections

The canonical projections, Ih, determined by the degrees offreedom, commute with d . But they are not bounded on HΛk .

If we apply the three operations:

I extend (E )

I regularize (R)

I canonical projection (Ih)

we get a map Qkh : HΛk(Ω)→ Λk

h which is bounded and commuteswith d . But it is not a projection.

However the composition

πkh = (Qk

h |Λkh)−1 Qk

h

can be shown to be a bounded cochain projection.

Construction of bounded cochain projections

The canonical projections, Ih, determined by the degrees offreedom, commute with d . But they are not bounded on HΛk .

If we apply the three operations:

I extend (E )

I regularize (R)

I canonical projection (Ih)

we get a map Qkh : HΛk(Ω)→ Λk

h which is bounded and commuteswith d . But it is not a projection.

However the composition

πkh = (Qk

h |Λkh)−1 Qk

h

can be shown to be a bounded cochain projection.

Construction of bounded cochain projections

The canonical projections, Ih, determined by the degrees offreedom, commute with d . But they are not bounded on HΛk .

If we apply the three operations:

I extend (E )

I regularize (R)

I canonical projection (Ih)

we get a map Qkh : HΛk(Ω)→ Λk

h which is bounded and commuteswith d . But it is not a projection.

However the composition

πkh = (Qk

h |Λkh)−1 Qk

h

can be shown to be a bounded cochain projection.

Bounded projections, definitionsWe combine extension and smoothing into an operatorRε

h : L2Λk(Ω)→ C Λk(Ω) given by

(Rεhω)x =

∫B1

ρ(y)(Eω)x+εhy dy .

Assume that the mesh is quasi–uniform:

Lemma

I For each ε > 0 sufficiently small there is a constant c(ε) suchthat for all h

‖IhRεh‖L(L2Λk (Ω),L2Λk (Ω)) ≤ c(ε).

I There is constant c, independent of ε and h such that

‖I − IhRεh‖L(L2Λk

h ,L2Λk

h) ≤ cε.

Bounded projections, definitionsWe combine extension and smoothing into an operatorRε

h : L2Λk(Ω)→ C Λk(Ω) given by

(Rεhω)x =

∫B1

ρ(y)(Eω)x+εhy dy .

Assume that the mesh is quasi–uniform:

Lemma

I For each ε > 0 sufficiently small there is a constant c(ε) suchthat for all h

‖IhRεh‖L(L2Λk (Ω),L2Λk (Ω)) ≤ c(ε).

I There is constant c, independent of ε and h such that

‖I − IhRεh‖L(L2Λk

h ,L2Λk

h) ≤ cε.

Scaling and compactness

Bounded cochain projections

The operator πh = πkh defined by

πh = (IhRεh|Λk

h)−1 IhRε

h,

where ε is taken small, but not too small, has all the desiredproperties, i.e.

I πkh is bounded in L(L2Λk(Ω), L2Λk(Ω)) andL(HΛk(Ω),HΛk(Ω))

I πkh commutes with the exterior derivative

I πkh is a projection

Removing the assumption of a quasi–uniform mesh

If the family of meshes is shape regular, but not necessarilyquasi–uniform, we introduce a piecewise linear and uniformlyLipschitz continuous function gh(x) to represent the local meshsize.

For each vertex x we let gh(x) be the average of the values ofdiamT for the simplices meeting x , and we defineΦεy

h (x) = x + εgh(x)y .

Then we replace the operator

(Rεhω)x =

∫B1

ρ(y)(Eω)x+εhy dy

by

(Rεhω)x =

∫B1

ρ(y)((Φεyh )∗Eω)x dy .

Removing the assumption of a quasi–uniform mesh

If the family of meshes is shape regular, but not necessarilyquasi–uniform, we introduce a piecewise linear and uniformlyLipschitz continuous function gh(x) to represent the local meshsize.

For each vertex x we let gh(x) be the average of the values ofdiamT for the simplices meeting x , and we defineΦεy

h (x) = x + εgh(x)y .

Then we replace the operator

(Rεhω)x =

∫B1

ρ(y)(Eω)x+εhy dy

by

(Rεhω)x =

∫B1

ρ(y)((Φεyh )∗Eω)x dy .

Removing the assumption of a quasi–uniform mesh

If the family of meshes is shape regular, but not necessarilyquasi–uniform, we introduce a piecewise linear and uniformlyLipschitz continuous function gh(x) to represent the local meshsize.

For each vertex x we let gh(x) be the average of the values ofdiamT for the simplices meeting x , and we defineΦεy

h (x) = x + εgh(x)y .

Then we replace the operator

(Rεhω)x =

∫B1

ρ(y)(Eω)x+εhy dy

by

(Rεhω)x =

∫B1

ρ(y)((Φεyh )∗Eω)x dy .

References

The development of FEEC leans heavily on earlier results takenfrom

I Whitney, Bossavit, Raviart and Thomas, Nedelec, Hiptmair,...

as well as on the theory of finite elements in general.

The presentation here is mostly based on

I D.N. Arnold, R.S. Falk, R. Winther, Finite element exteriorcalculus, homological techniques, and applications, ActaNumerica 2006.

I D.N. Arnold, R.S. Falk and R. Winther Finite element exteriorcalculus: From Hodge theory to numerical stability Bulletin ofthe Amer. Math. Soc. 47 (2010), 281-354.

References

The development of FEEC leans heavily on earlier results takenfrom

I Whitney, Bossavit, Raviart and Thomas, Nedelec, Hiptmair,...

as well as on the theory of finite elements in general.The presentation here is mostly based on

I D.N. Arnold, R.S. Falk, R. Winther, Finite element exteriorcalculus, homological techniques, and applications, ActaNumerica 2006.

I D.N. Arnold, R.S. Falk and R. Winther Finite element exteriorcalculus: From Hodge theory to numerical stability Bulletin ofthe Amer. Math. Soc. 47 (2010), 281-354.