# Hodge Duality

Hodge duality

We will next deﬁne the Hodge star operator. We will deﬁneit in a chart rather than abstractly.

•

The Hodge star operator, denoted ? in an n-dimensional manifold is a map from p-forms to (n − p)-forms given by p

|g|

n−p+1ν1

(?ω)µ ···µ

ꢀµ ···µ gµ

≡· · · gµ ν ων ···ν , (17.1) np

11np

1n−p p! where ω is a p-form.

The ? operator acts on forms, not on components.

2

Example: Consider R3 with metric +++, i.e. gµν =diag(1, 1, 1) . Then |g| ≡ g = 1 , gµνdiag(1, 1, 1) . Write the coordinate basis 1-forms as dx, dy, dz . Their components are clearly

(dx)i = δi1 , (dy)i = δi2 , (dz)i = δi3 , (17.2) the δ’s on the right hand sides are Kroenecker deltas. So

(?dx)ij = ꢀijkgkl(dx)l = ꢀijkgklδl1 = ꢀijkgk1

11

⇒?dx = (?dx)ijdxi ∧ dxj = ꢀijkgk1dxi ∧ dxj

2! 2! gk1 = 1 for k = 1 , 0 otherwise

ꢀꢁ

1

?dx =

⇒dx2 ∧ dx3 − dx3 ∧ dx2 = dx2 ∧ dx3 = dy ∧ dz .

2!

(17.3)

Similarly, ?dy = dz ∧ dx ,

?dz = dx ∧ dy .

2

65 66 Chapter 17. Hodge duality

Example: Consider p = 0 (scalar), i.e. a 0-form ω in n dimensions. ppp

(?ω)µ ···µ =

|g|ꢀµ ···µn ω

n11

(?1)µ ···µ =

⇒|g|ꢀµ ···µ

11nn

|g|

⇒ꢀµ ···µ dxµ ∧ · · · ∧ dxµ

1n

(?1) =

1nn!

= dV (17.4)

2

Example: p = n . Then p

For the volume form,

|g|

µnνn

ꢀµ ···µ gµ ν · · · g

(?ω) = ων ···ν .(17.5)

11

nn

11n! p

|g|

1ndV =

ꢀµ ···µ dxµ ∧ · · · ∧ dxµ

1nn! p

(dV )ν ···ν =

|g|ꢀν ···ν

11nn

|g|

µnνn

11

ꢀµ ···µ gµ ν · · · g

(?dV ) = ꢀν ···ν

11nnn!

|g| |g| n!

=n!(det g)−1 =

= sign(g) = (−1)s ,(17.6) n! n! g where s is the number of (−1) in gµν

.

2

So we ﬁnd that

?(?1) = ?dV = (−1)s ,

(17.7)

(17.8) and

?(?dV ) = (−1)s(?1) = (−1)sdV , i.e., (?)2 = (−1)s on 0-forms and n-forms.

In general, on a p-form in an n-dimensional manifold with signature (s, n − s) , it can be shown in the same way that

(?)2 = (−1)p(n−p)+s

.(17.9)

In particular, in four dimensional Minkowski space, s = 1, n = 4 , so

(?)2 = (−1)p(4−p)+1

.(17.10) 67

It is useful to work out the Hodge dual of basis p-forms. Suppose

Ip we have a basis p-form dxI ∧ · · · ∧ dx , where the indices are ar-

1ranged in increasing order Ip · · · I1 . Then its components are p!δµI · · · δµI . So pp

1

1

ꢀ

ꢁ

? dxI ∧ · · · ∧ dxI

1p

ν1···νn−p p

|g| ꢀν ···ν

|g|

0

µpµ0p p! δµ0 · · · δµI0

I1 pgµ µ · · · g

1

1

=ꢀν ···ν

n−pµ1···µp

1p

1p! p

µpIp gµ I · · · g

=.(17.11)

11

n−pµ1···µp

1

We will use this to calculate ?ω ∧ ω .

For a p-form ω , we have

1

ω = ωµ ···µ dxµ ∧ · · · ∧ dxµ

1p

1pp!

X

1p

=(17.12)

ωI ···I dxI ∧ · · · ∧ dxI

1p

Iwhere the sum over I means a sum over all possible index sets I =

I1 · · · Ip , but there is no sum over the indices {I1, · · · , Ip} themselves, in a given index set the Ik are ﬁxed. Using the dual of basis p−forms, and Eq. (13.13), we get

X

Ip

1

?ω =

ωI ···I ? (dxI ∧ · · · ∧ dx )

1p

Ip

X

|g|

µpIp gµ I · · · g

ωI ···I dxν ∧ · · · ∧ dxν

111n−p

=.ꢀν ···ν

n−pµ1···µp

1

1p

(n − p)!

I

(17.13)

The sum over I is a sum over diﬀerent index sets as before, and the Greek indices are summed over as usual. Thus we calculate p

X

|g|

µpIp

11

?ω ∧ ω = ×

ꢀν ···ν gµ I · · · g

ωI ···I

n−pµ1···µp

1

1p

(n − p)!

I,J

ꢀꢁdxν ∧ · · · ∧ dxν

∧ ωJ ···J dxJ ∧ · · · ∧ dxJ

n−p

11p

1pp

X

|g|

µpIp

11

=ꢀν ···ν gµ I · · · g ×

ωI ···I ωJ ···J

n−pµ1···µp

1

11pp

(n − p)!

I,J dxν ∧ · · · ∧ dxν

∧ dxJ ∧ · · · ∧ dxJ

(17.14)

n−p

11p68 Chapter 17. Hodge duality

We see that the set {ν1, · · · , νn−p} cannot have any overlap with the set J = {J1, · · · , Jp}, because of the wedge product. On the other hand, {ν1, · · · , νn−p} cannot have any overlap with {µ1, · · · , µp} because ꢀ is totally antisymmetric in its indices. So the set {µ1, · · · , µp} must have the same elements as the set J = {J1, · · · , Jp} , but they may not be in the same order.

Now consider the case where the basis is orthogonal, i.e. gµν is diagonal. Then gµ I = gI I etc. and we can write kkk k p

X

|g|

IpIp gI I · · · g ×

ωI ···I ωJ ···J

11

?ω ∧ ω =

ꢀν ···ν

n−pI1···Ip

111pp

(n − p)!

I,J dxν ∧ · · · ∧ dxν

∧ dxJ ∧ · · · ∧ dxJ . (17.15)

n−p

11p

We see that in each term of the sum, the indices {I1 · · · Ip} must be the same as {J1 · · · Jp} because both sets are totally antisymmetrized with the indices {ν1 · · · νn−p}.

Since both sets are ordered, it follows that we can replace J by

I, p

X

|g|

IpIp

11

?ω ∧ ω = ×

ꢀν ···ν gI I · · · g

ωI ···I ωI ···I

n−pI1···Ip

111pp

(n − p)!

Idxν ∧ · · · ∧ dxν

∧ dxI ∧ · · · ∧ dxI

n−p

11pp

X

|g|

1p

=ꢀν ···ν

n−pI1···Ip

ωI ···I ωI ···I

×

11p

(n − p)!

Idxν ∧ · · · ∧ dxν

∧ dxI ∧ · · · ∧ dx . (17.16)

Ip

n−p

11

In each term of this sum, the indices {ν1 · · · νn−p} are completely determined, so we can replace them by the corresponding ordered set K = K1 · · · Kn−p , which is completely determined by the set I , so that

Xp

1p

?ω ∧ ω = |g| ×

ꢀK ···K

ωI ···I ωI ···I

n−pI1···Ip

11p

IdxK ∧ · · · ∧ dxK

∧ dxI ∧ · · · ∧ dx(1.7.17)

Ip

n−p

11

The indices on this ꢀ are a permutation of {1, · · · , n} , so ꢀ is ±1.

But this sign is the same as that for the permutation to bring the basis to the order dx1 ∧ · · · ∧ dxn , so the overall sign to get both to 69 the standard order is positive. Thus we get

Xp

1p

?ω ∧ ω = |g|

ωI ···I ωI ···I ꢀ1···n dx1 ∧ · · · ∧ dxn

1p

Ip

1

1p

=

|g| ωµ ···µ ωµ ···µ dx1 ∧ · · · ∧ dxn

1pp!

1

=ωµ ···µ ωµ ···µ (vol) (17.18)

1p

1pp!

If we are in a basis where the metric is not diagonal, it is still symmetric. So we can diagonalize it locally by going to an appropriate basis, or set of coordinates, at each point. In this basis, the components of ω may be ωµ0 ···µ , so we can write

1p

ꢂꢃ

1

00p

ωµ ···µ ωµ ···µ0 (vol0)

(17.19)

1

0

?ω ∧ ω = p

1p!

But both factors are invariant under a change of basis. So we can now change back to our earlier basis, and ﬁnd Eq. (17.18) even when the metric is not diagonal. Note that the metric may not be diagonalizable globally or even in an extended region.