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P: 161
 Quote by ismaili I reconsidered this recently. I think for my first question, the reason that reduces the independent components $$2m^2$$ into $$m^2$$ is the (anti-)hermitian properties of the gamma matrices, not the Clifford algebra. Then, the antisymmetrized products of gamma matrices form a basis for the algebra, hence, by matching independent components $$m^2$$ and the number of basis $$2^d$$, we see that $$m = 2^{d/2}$$ for even dimension $$d$$. The solution to the second equation is due to the role of $$\gamma_5$$. That's why only in odd dimension, those basis are related by Levi-Civita tensor, and the number of basis is reduced to $$2^{(d-1)/2}$$ for odd $$d$$.
For your information.
I found this reference which deals with Clifford algebra in a mathematical rigorous way.
http://arxiv.org/abs/hep-th/9811101
But at least those $$B,C$$ matrices are not suddenly popped.