Noun
exterior derivative (plural exterior derivatives)
(calculus) A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0.
The exterior derivative of a “scalar”, i.e., a function
f
=
f
(
x
1
,
x
2
,
.
.
.
,
x
n
)
{\displaystyle f=f(x^{1},x^{2},...,x^{n})}
where the
x
i
{\displaystyle x^{i}}
’s are coordinates of
R
n
{\displaystyle \mathbb {R} ^{n}}
, is
d
f
=
∂
f
∂
x
1
d
x
1
+
∂
f
∂
x
2
d
x
2
+
.
.
.
+
∂
f
∂
x
n
d
x
n
{\displaystyle df={\partial f \over \partial x^{1}}dx^{1}+{\partial f \over \partial x^{2}}dx^{2}+...+{\partial f \over \partial x^{n}}dx^{n}}
.
The exterior derivative of a k-blade
f
d
x
i
1
∧
d
x
i
2
∧
.
.
.
∧
d
x
i
k
{\displaystyle f\,dx^{i_{1}}\wedge dx^{i_{2}}\wedge ...\wedge dx^{i_{k}}}
is
d
f
∧
d
x
i
1
∧
d
x
i
2
∧
.
.
.
∧
d
x
i
k
{\displaystyle df\wedge dx^{i_{1}}\wedge dx^{i_{2}}\wedge ...\wedge dx^{i_{k}}}
.
The exterior derivative
d
{\displaystyle d}
may be though of as a differential operator del wedge:
∇
∧
{\displaystyle \nabla \wedge }
, where
∇
=
∂
∂
x
1
d
x
1
+
∂
∂
x
2
d
x
2
+
.
.
.
+
∂
∂
x
n
d
x
n
{\displaystyle \nabla ={\partial \over \partial x^{1}}dx^{1}+{\partial \over \partial x_{2}}dx^{2}+...+{\partial \over \partial x^{n}}dx^{n}}
. Then the square of the exterior derivative is
d
2
=
∇
∧
∇
∧
=
(
∇
∧
∇
)
∧
=
0
∧
=
0
{\displaystyle d^{2}=\nabla \wedge \nabla \wedge =(\nabla \wedge \nabla )\wedge =0\wedge =0}
because the wedge product is alternating. (If u is a blade and f a scalar (function), then
f
u
≡
f
∧
u
{\displaystyle fu\equiv f\wedge u}
, so
d
(
f
u
)
=
∇
∧
(
f
u
)
=
∇
∧
(
f
∧
u
)
=
(
∇
∧
f
)
∧
u
=
d
f
∧
u
{\displaystyle d(fu)=\nabla \wedge (fu)=\nabla \wedge (f\wedge u)=(\nabla \wedge f)\wedge u=df\wedge u}
.) Another way to show that
d
2
=
0
{\displaystyle d^{2}=0}
is that partial derivatives commute and wedge products of 1-forms anti-commute (so when
d
2
{\displaystyle d^{2}}
is applied to a blade then the distributed parts end up canceling to zero.)