The exterior derivative of a “scalar”, i.e., a function f=f(x¹,x²,...,xⁿ) where the xⁱ’s are coordinates of ℝⁿ, is df=∂f/∂x¹dx¹+∂f/∂x²dx²+...+∂f/∂xⁿdxⁿ.

The exterior derivative of a k-blade f,dx∧dx∧...∧dx is df∧dx∧dx∧...∧dx.

The exterior derivative d may be though of as a differential operator del wedge: ∇∧, where ∇=∂/∂x¹dx¹+∂/∂x_2dx²+...+∂/∂xⁿdxⁿ. Then the square of the exterior derivative is d²=∇∧∇∧=(∇∧∇)∧=0∧=0 because the wedge product is alternating. (If u is a blade and f a scalar (function), then fu≡f∧u, so d(fu)=∇∧(fu)=∇∧(f∧u)=(∇∧f)∧u=df∧u.) Another way to show that d²=0 is that partial derivatives commute and wedge products of 1-forms anti-commute (so when d² is applied to a blade then the distributed parts end up canceling to zero.)

Exterior derivative

Meaning

Noun