# 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 $$i$$.

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.