User:Jheald/sandbox/GA/Cauchy-Riemann add

To possibly add to Cauchy–Riemann equations, under :

Clifford analysis
If the complex numbers are considered to be the even part of the Clifford algebra Cℓ2,0(R), which is spanned by the bases {1, e1, e2, i=e1e2} where e12=e22=1 and (e1e2)2 = -1, then the Cauchy–Riemann operator can be identified as e1 times the Dirac operator (the vector derivative using the Clifford product):
 * $${ \partial \over \partial x } + i { \partial \over \partial y } = e_1 \left( e_1 {\partial \over \partial x } + e_2 { \partial \over \partial y } \right) = e_1 \nabla $$

The Cauchy–Riemann equations can therefore be identified with a vector-field equation:
 * $$\left({ \partial \over \partial x } + i { \partial \over \partial y }\right)(u + i v) = 0 \Leftrightarrow \left( e_1 {\partial \over \partial x } + e_2 { \partial \over \partial y }\right)(e_1 u - e_2 v) = 0$$
 * $$\Leftrightarrow \nabla \bar{f} = 0, \qquad \mathrm{where} \qquad \bar{f} = \begin{bmatrix} u \\ -v \end{bmatrix} $$

Taking even and odd parts of the Clifford product gives:
 * $$\nabla \cdot \bar{f} = 0 \qquad \mathrm{and} \qquad \nabla \wedge \bar{f} = 0$$

cf above.

In turn, this can be related to a (real) scalar potential function,
 * $$\bar{f} = \nabla \phi$$

so that
 * $$\nabla^2 \phi = 0$$

Where does the minus sign in -v come from?
Clifford algebra isn't commutative, so
 * $$\nabla^2 = \nabla\nabla, \qquad \mathrm{ but } \qquad \nabla^2 \; {\neq} \; \hat{C}\hat{C};\qquad \hat{C}\hat{C}\; = \; e_1 \nabla e_1 \nabla \; \neq \; e_1^2 \nabla^2$$

Instead,
 * $$\nabla^2 \; = \; \hat{C}\hat{C}^\dagger \; = \; 4 { \partial \over \partial z^\dagger }{ \partial \over \partial z } \; = \; \left( { \partial \over \partial x } + i { \partial \over \partial y } \right)\left( { \partial \over \partial x } - i { \partial \over \partial y } \right)$$

Therefore
 * $$u + iv = \left( { \partial \over \partial x } - i { \partial \over \partial y } \right) \phi$$

i.e. (equating imaginary parts):


 * $$ iv = -i { \partial \over \partial y } \phi $$

so it has to be
 * $$ -v = { \partial \phi \over \partial y } $$