18.2. THE EXTERIOR DERIVATIVE 477

Definition 18.2.1 Let ω = ∑I ascending aI (x)dxI . Then

dω ≡ ∑I ascending

daI (x)∧dxI

It is clear that d is linear.

It doesn’t matter whether ω is written in terms of ascending indices. The same formulaholds with no change.

Lemma 18.2.2 Let ω = ∑I aI (x)dxI then dω = ∑I daI (x)∧dxI

Proof: Denote by Î the ascending indices. Then if σ I (I) is ascending,

ω = ∑Î

∑{I}={Î}

aI (x)dxI = ∑Î

 ∑{I}={Î}

aI (x)sgn(σ I (I))

dxÎ

Then since d is linear, dω = ∑Î

(∑{I}={Î} daI (x)sgn(σ I (I))

)∧dxÎ . Then since ∧ is also

linear,

= ∑Î

 ∑{I}={Î}

daI (x)sgn(σ I (I))∧dxÎ

= ∑

Î

 ∑{I}={Î}

daI (x)∧dxI

= ∑I

daI (x)∧dxI ■

Next is a product rule. First note that it follows right away from the definition and theproduct rule from beginning calculus that

d ( f g) = d ( f )g+gd (g)

Lemma 18.2.3 Let α,β be in Ωk and Ωl respectively. Then d (α ∧β ) = dα ∧ β +

(−1)kα ∧dβ . Also d2 = 0.

Proof: Let α = ∑I aI (x)dxI ,β = ∑J bJ (x)dxJ . Then α∧β = ∑I,J aIbJdxI∧dxJ andso d (α ∧β ) equals, thanks to the above lemma,

∑I,J

d (aIbJ)∧dxI ∧dxJ = ∑I,J

[d (aI (x))bJ (x)+aI (x)d (bJ (x))]∧dxI ∧dxJ

= ∑I,J

d (aI (x))bJ (x)∧dxI ∧dxJ +∑I,J

aI (x)d (bJ (x))∧dxI ∧dxJ

From the definition of the wedge product and Lemma 18.1.2,

= dα (x)∧β +∑I,J

aI (x)d (bJ (x))∧dxI ∧dxJ

Now we will interchange the 1 form d (bJ (x)) and the k form dxI . From Theorem 18.1.3

= dα (x)∧β (x)+(−1)k∑I,J

aI (x)dxI ∧d (bJ (x))∧dxJ