Dirac equation without $i$
Whether there is an $i$ $$ (i\gamma^{\mu}D_{\mu}-m)\psi=0 $$ or no $i$ $$ (\gamma^{\mu}D_{\mu}-m)\psi=0 $$ in the Dirac equation is determined by the metric:
- For $(\gamma^0)^2 = I$, there is $i$
- For $(\gamma^0)^2 = -I$, there is no $i$
In Witten's paper, the metric is (-, +, +), therefore no $i$.
Added note:
Since the Hermitian of $\gamma^\mu$ is defined as: $$ (\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0, $$ $\gamma^0$ is non-Hermitian in the $(\gamma^0)^2 = -I$ case: $$ (\gamma^0)^\dagger = \gamma^0\gamma^0\gamma^0 = -\gamma^0. $$
One more note:
The choice (in 4D case) between west coast metric (+, -, -, -) and east coast metric (-, +, +, +) is usually regarded as a matter of convention or personal taste. However, one should note that the respective Clifford algebra Cl(1,3) and Cl(3, 1) are not isomorphic to each other. Instead:
- Cl(1,3) is isomorphic to $M_2(H)$, 2*2 matrices of quaternions.
- Cl(3,1) is isomorphic to $M_4(R)$, 4*4 matrices of real numbers.
Does the different isomorphism have bearing on the physics? John Baez says yes (see the link in the OP). But I would like to know your take.