Differential Forms

Wedge product

dx∧dy =
1
2
(dx⊗dy - dy⊗dx)		 (1)

Properties:

dx∧dx = 0
dx'∧dy' = Jacobian(x',y';x,y)dx∧dy
(2)

A differential form of rank p or a p-form is defined by:

A =
1
p!
Ai1….ipdxi1∧….dxip		 (3)

We have

ap∧bq = ( - 1)pqbq∧ap (4)

Exterior derivative

dAp =
1
p!
Ai1….ip,jdxj∧dxi1∧….dxip		 (5)

It follows that d is nilpotent:d2A = 0

Leibnitz's rule:

d(ap∧bq) = d ap∧bq + ( - 1)pap∧dbq (6)

Hodge'star:

⋆(d xi1∧….d xip) =
1
(n - p)!
εi1….ipip + 1….ind xip + 1∧….d xin		 (7)

We have that:

⋆⋆ωp = ( - 1)p(n - p)ωp (8)

Proof:

w =
1
p!
wi1….ipd xi1∧….d xip
⋆w =
1
p!
1
(n - p)!
wi1….ipεi1….ipip + 1….ind xip + 1∧….d xin
⋆⋆w =
1
p!
1
(n - p)!
wi1….ipεi1….ipip + 1….in⋆(d xip + 1∧….d xin) =
1
p!
1
(n - p)!
wi1….ipεi1….ipip + 1….in
1
p!
εip + 1…inj1…jpd xj1∧d xjp =

( - 1)p(n - p)
1
p!
wj1….jpd xj1∧…d xjp = ( - 1)p(n - p)w

We have used the identity:

εi1….ipip + 1….inεip + 1…inj1…jp = ( - 1)p(n - p) εi1….ipip + 1….inεj1….jpip + 1….in =
( - 1)p(n - p)(n - p)!(δi1j1δi2j2….δipjp + …)(p! terms)

Induction on p:

p = 1, εi1….ipip + 1….inεj1….jpip + 1….in = Aδi1j1
Put i1 = j1 and sum
We get:
A =
n!
n
= (n - 1)!
p = 2, εi1….ipip + 1….inεj1….jpip + 1….in = A(δi1j1δi2j2 - δi1j2δi2j1)
∑(i1 = j1,i2 = j2) gives
n! = A(n2 - n)
A = (n - 2)!
Suppose that for p = a, εi1….iaia + 1….inεj1….jaia + 1….in = (n - a)!∑σϵSasgn(σ)δi1σ(j1)….δiaσ(ja)

Define

(ap,bp) = ∫Map∧⋆bp (9)

We have that:

ap∧⋆bp =
1
p!
ap i1….ip
1
p!
bp j1…jp
1
(n - p)!
εj1….jpjp + 1….jnd xi1∧….d xip∧d xjp + 1 …∧d xjn =
1
p!2(n - p)!
ap i1….ipbp j1…jpεj1….jpjp + 1….jnεi1….jnd x1…∧d xn =
1
p!
ap i1….ipbp i1…ipd x1…∧d xn =

bp∧⋆ap		 (10)

Therefore:

(ap,bp) =
1
p!
∫Map i1….ipbp i1…ipd x1…∧d xn		 (11)

From equation (10) is easy to see that (a,b) defines an inner product of (real) p-forms.

In particular:

(ap,bp) = (bp,ap) (12)

Adjoint of the exterior derivative:

(ap,d bp - 1) = (δap,bp - 1) (13)

We start from the identity:

∫Md(bp - 1∧⋆ap) = 0 = ∫M(d bp - 1∧⋆ap + ( - 1)p - 1bp - 1∧d⋆ap)

That is:

(ap,d bp - 1) = (d bp - 1,ap) = ( - 1)p( - 1)(n - p + 1)(p - 1)(bp - 1,⋆d⋆ap)
p + n p - n - (p - 1)2
Notice that p - (p - 1)2 = - (p - 1)(p - 2) + 1 is odd
( - 1)p( - 1)(n - p + 1)(p - 1) = ( - 1)np + n + 1

So,

δ = ( - 1)n p + n + 1⋆d⋆ (14)

From (13) we get:

(δ2ap + 1, bp - 1) = (ap + 1,d2bp - 1) = 0 , for all ap + 1

Then,

δ2 = 0 (15)