-div is adjoint to d
edit
The claim is made that −div is adjoint to d :
∫
M
d
f
(
X
)
ω
=
−
∫
M
f
div
X
ω
{\displaystyle \int _{M}df(X)\;\omega =-\int _{M}f\,\operatorname {div} X\;\omega }
Proof of the above statement:
∫
M
(
f
d
i
v
(
X
)
+
X
(
f
)
)
ω
=
∫
M
(
f
L
X
+
L
X
(
f
)
)
ω
{\displaystyle \int _{M}(f\mathrm {div} (X)+X(f))\omega =\int _{M}(f{\mathcal {L}}_{X}+{\mathcal {L}}_{X}(f))\omega }
=
∫
M
L
X
f
ω
=
∫
M
d
ι
X
f
ω
=
∫
∂
M
ι
X
f
ω
{\displaystyle =\int _{M}{\mathcal {L}}_{X}f\omega =\int _{M}\mathrm {d} \iota _{X}f\omega =\int _{\partial M}\iota _{X}f\omega }
If f has compact support , then the last integral vanishes, and we have the desired result.
Laplace-de Rham operator
edit
One may prove that the Laplace-de Rahm operator is equivalent to the definition of the Laplace-Beltrami operator, when acting on a scalar function f . This proof reads as:
Δ
f
=
d
δ
f
+
δ
d
f
=
δ
d
f
=
δ
∂
i
f
d
x
i
{\displaystyle \Delta f=\mathrm {d} \delta f+\delta \,\mathrm {d} f=\delta \,\mathrm {d} f=\delta \,\partial _{i}f\,\mathrm {d} x^{i}}
=
−
∗
d
∗
∂
i
f
d
x
i
=
−
∗
d
(
ε
i
J
|
g
|
∂
i
f
d
x
J
)
{\displaystyle =-*\mathrm {d} {*\partial _{i}f\,\mathrm {d} x^{i}}=-*\mathrm {d} (\varepsilon _{iJ}{\sqrt {|g|}}\partial ^{i}f\,\mathrm {d} x^{J})}
=
−
∗
ε
i
J
∂
j
(
|
g
|
∂
i
f
)
d
x
j
d
x
J
=
−
∗
1
|
g
|
∂
i
(
|
g
|
∂
i
f
)
v
o
l
n
{\displaystyle =-*\varepsilon _{iJ}\,\partial _{j}({\sqrt {|g|}}\partial ^{i}f)\,\mathrm {d} x^{j}\,\mathrm {d} x^{J}=-*{\frac {1}{\sqrt {|g|}}}\,\partial _{i}({\sqrt {|g|}}\,\partial ^{i}f)\mathrm {vol} _{n}}
=
−
1
|
g
|
∂
i
(
|
g
|
∂
i
f
)
,
{\displaystyle =-{\frac {1}{\sqrt {|g|}}}\,\partial _{i}({\sqrt {|g|}}\,\partial ^{i}f),}
where ω is the volume form and ε is the completely antisymmetric Levi-Civita symbol . Note that in the above, the italic lower-case index i is a single index, whereas the upper-case Roman J stands for all of the remaining (n -1) indices. Notice that the Laplace-de Rham operator is actually minus the Laplace-Beltrami operator; this minus sign follows from the conventional definition of the properties of the codifferential . Unfortunately, Δ is used to denote both; reader beware.
Given scalar functions f and h , and a real number a , the Laplacian has the property:
Δ
(
f
h
)
=
f
Δ
h
+
2
∂
i
f
∂
i
h
+
h
Δ
f
.
{\displaystyle \Delta (fh)=f\,\Delta h+2\partial _{i}f\,\partial ^{i}h+h\,\Delta f.}
Δ
(
f
h
)
=
δ
d
f
h
=
δ
(
f
d
h
+
h
d
f
)
=
∗
d
(
f
∗
d
h
)
+
∗
d
(
h
∗
d
f
)
{\displaystyle \Delta (fh)=\delta \,\mathrm {d} fh=\delta (f\,\mathrm {d} h+h\,\mathrm {d} f)=*\mathrm {d} (f{*\mathrm {d} h})+*\mathrm {d} (h{*\mathrm {d} f})\;}
=
∗
(
f
d
∗
d
h
+
d
f
∧
∗
d
h
+
d
h
∧
∗
d
f
+
h
d
∗
d
f
)
{\displaystyle =*(f\,\mathrm {d} *\mathrm {d} h+\mathrm {d} f\wedge *\mathrm {d} h+\mathrm {d} h\wedge *\mathrm {d} f+h\,\mathrm {d} *\mathrm {d} f)}
=
f
∗
d
∗
d
h
+
∗
(
d
f
∧
∗
d
h
+
d
h
∧
∗
d
f
)
+
h
∗
d
∗
d
f
{\displaystyle =f*\mathrm {d} *\mathrm {d} h+*(\mathrm {d} f\wedge *\mathrm {d} h+\mathrm {d} h\wedge *\mathrm {d} f)+h*\mathrm {d} *\mathrm {d} f}
=
f
Δ
h
{\displaystyle =f\,\Delta h}
+
∗
(
∂
i
f
d
x
i
∧
ε
j
J
|
g
|
∂
j
h
d
x
J
+
∂
i
h
d
x
i
∧
ε
j
J
|
g
|
∂
j
f
d
x
J
)
{\displaystyle +*(\partial _{i}f\,\mathrm {d} x^{i}\wedge \varepsilon _{jJ}{\sqrt {|g|}}\partial ^{j}h\,\mathrm {d} x^{J}+\partial _{i}h\,\mathrm {d} x^{i}\wedge \varepsilon _{jJ}{\sqrt {|g|}}\partial ^{j}f\,\mathrm {d} x^{J})}
+
h
Δ
f
{\displaystyle +h\,\Delta f}
=
f
Δ
h
+
(
∂
i
f
∂
i
h
+
∂
i
h
∂
i
f
)
∗
v
o
l
n
+
h
Δ
f
{\displaystyle =f\,\Delta h+(\partial _{i}f\,\partial ^{i}h+\partial _{i}h\,\partial ^{i}f){*\mathrm {vol} _{n}}+h\,\Delta f}
=
f
Δ
h
+
2
∂
i
f
∂
i
h
+
h
Δ
f
{\displaystyle =f\,\Delta h+2\partial _{i}f\,\partial ^{i}h+h\,\Delta f}
where f and h are scalar functions.
