11.2. DERIVATIVES OF DIFFERENTIAL FORMS 277
Next, we derive formulas for the derivatives of 1 and 2-forms.
Example 11.3. For a 1-form ω = F
1
dx + F
2
dy + F
3
dz , we find dω by calculating d(F
1
dx),
d(F
2
dy), and d(F
3
dz ) separately. For instance, by definition:
d(F
1
dx)= dF
1
∧ dx
=
(
∂F
1
∂x
dx +
∂F
1
∂y
dy +
∂F
1
∂z
dz
)
∧ dx
=
∂F
1
∂x
dx ∧ dx +
∂F
1
∂y
dy ∧ dx +
∂F
1
∂z
dz ∧ dx
=0 -
∂F
1
∂y
dx ∧ dy +
∂F
1
∂z
dz ∧ dx
= -
∂F
1
∂y
dx ∧ dy +
∂F
1
∂z
dz ∧ dx.
Similarly:
d(F
2
dy)=
∂F
2
∂x
dx ∧ dy -
∂F
2
∂z
dy ∧ dz
and d(F
3
dz )= -
∂F
3
∂x
dz ∧ dx +
∂F
3
∂y
dy ∧ dz.
Adding these calculations gives:
d(F
1
dx + F
2
dy + F
3
dz )=
(
∂F
3
∂y
-
∂F
2
∂z
)
dy ∧ dz
+
(
∂F
1
∂z
-
∂F
3
∂x
)
dz ∧ dx +
(
∂F
2
∂x
-
∂F
1
∂y
)
dx ∧ dy,
a formula that is reproduced in row 2(b) of Table 11.2 below.
Example 11.4. For the derivative of a 2-form η = F
1
dy ∧ dz + F
2
dz ∧ dx + F
3
dx ∧ dy, we compute,
for example:
d(F
2
dz ∧ dx)= dF
2
∧ dz ∧ dx
=
(
∂F
2
∂x
dx +
∂F
2
∂y
dy +
∂F
2
∂z
dz
)
∧ dz ∧ dx
=
∂F
2
∂x
dx ∧ dz ∧ dx +
∂F
2
∂y
dy ∧ dz ∧ dx +
∂F
2
∂z
dz ∧ dz ∧ dx
= -
∂F
2
∂x
dx ∧ dx ∧ dz -
∂F
2
∂y
dy ∧ dx ∧ dz +0
=0+
∂F
2
∂y
dx ∧ dy ∧ dz
=
∂F
2
∂y
dx ∧ dy ∧ dz.
Likewise:
d(F
1
dy ∧ dz )=
∂F
1
∂x
dx ∧ dy ∧ dz and d(F
3
dx ∧ dy)=
∂F
3
∂z
dx ∧ dy ∧ dz.