If the complex numbers are considered to be the even part of the Clifford algebra C ℓ2,0 (R ), which is spanned by the bases {1, e 1 , e 2 , i =e 1 e 2 } where e 1 2 =e 2 2 =1 and (e 1 e 2 )2 = -1, then the Cauchy–Riemann operator can be identified as e 1 times the Dirac operator (the vector derivative using the Clifford product):
∂
∂
x
+
i
∂
∂
y
=
e
1
(
e
1
∂
∂
x
+
e
2
∂
∂
y
)
=
e
1
∇
{\displaystyle {\partial \over \partial x}+i{\partial \over \partial y}=e_{1}\left(e_{1}{\partial \over \partial x}+e_{2}{\partial \over \partial y}\right)=e_{1}\nabla }
The Cauchy–Riemann equations can therefore be identified with a vector-field equation:
(
∂
∂
x
+
i
∂
∂
y
)
(
u
+
i
v
)
=
0
⇔
(
e
1
∂
∂
x
+
e
2
∂
∂
y
)
(
e
1
u
−
e
2
v
)
=
0
{\displaystyle \left({\partial \over \partial x}+i{\partial \over \partial y}\right)(u+iv)=0\Leftrightarrow \left(e_{1}{\partial \over \partial x}+e_{2}{\partial \over \partial y}\right)(e_{1}u-e_{2}v)=0}
⇔
∇
f
¯
=
0
,
w
h
e
r
e
f
¯
=
[
u
−
v
]
{\displaystyle \Leftrightarrow \nabla {\bar {f}}=0,\qquad \mathrm {where} \qquad {\bar {f}}={\begin{bmatrix}u\\-v\end{bmatrix}}}
Taking even and odd parts of the Clifford product gives:
∇
⋅
f
¯
=
0
a
n
d
∇
∧
f
¯
=
0
{\displaystyle \nabla \cdot {\bar {f}}=0\qquad \mathrm {and} \qquad \nabla \wedge {\bar {f}}=0}
cf #Physical interpretation above.
In turn, this can be related to a (real) scalar potential function,
f
¯
=
∇
ϕ
{\displaystyle {\bar {f}}=\nabla \phi }
so that
∇
2
ϕ
=
0
{\displaystyle \nabla ^{2}\phi =0}
-v come from?
edit
Clifford algebra isn't commutative, so
∇
2
=
∇
∇
,
b
u
t
∇
2
≠
C
^
C
^
;
C
^
C
^
=
e
1
∇
e
1
∇
≠
e
1
2
∇
2
{\displaystyle \nabla ^{2}=\nabla \nabla ,\qquad \mathrm {but} \qquad \nabla ^{2}\;{\neq }\;{\hat {C}}{\hat {C}};\qquad {\hat {C}}{\hat {C}}\;=\;e_{1}\nabla e_{1}\nabla \;\neq \;e_{1}^{2}\nabla ^{2}}
Instead,
∇
2
=
C
^
C
^
†
=
4
∂
∂
z
†
∂
∂
z
=
(
∂
∂
x
+
i
∂
∂
y
)
(
∂
∂
x
−
i
∂
∂
y
)
{\displaystyle \nabla ^{2}\;=\;{\hat {C}}{\hat {C}}^{\dagger }\;=\;4{\partial \over \partial z^{\dagger }}{\partial \over \partial z}\;=\;\left({\partial \over \partial x}+i{\partial \over \partial y}\right)\left({\partial \over \partial x}-i{\partial \over \partial y}\right)}
Therefore
u
+
i
v
=
(
∂
∂
x
−
i
∂
∂
y
)
ϕ
{\displaystyle u+iv=\left({\partial \over \partial x}-i{\partial \over \partial y}\right)\phi }
i.e. (equating imaginary parts):
i
v
=
−
i
∂
∂
y
ϕ
{\displaystyle iv=-i{\partial \over \partial y}\phi }
so it has to be
−
v
=
∂
ϕ
∂
y
{\displaystyle -v={\partial \phi \over \partial y}}
