2
$\begingroup$

I am sure that is very well-known question and see on this site several similar questions but I would like to specify the answer

1) I know that in $(2+1)$-dimensions one can construct $\gamma$-matrices as in $(1+1)$ dimensions

2) But I see in several books/papers that in $(2+1)$ one can also take $\gamma$-matrices in $(3+1)$ and just remove $\gamma^3$

I do not understand why the second way is possible. Can anybody explain it?

$\endgroup$
4
$\begingroup$

Gamma matrices don't have a unique representation. They only requirement is that they satisfy the axiom of the Clifford algebra $$ \{\gamma^\mu,\gamma^\nu\} = 2 \,\eta^{\mu\nu}\,. \tag{1} $$ The usual choice is to take the representation which has the smallest dimension (for obvious reasons). For space-time dimension $d$ the matrices would be $n\times n$ where $n = 2^{\lfloor d/2\rfloor}$.

The construction 1) would be the one with smallest dimension and thus the preferable one. The construction 2) has bigger dimension but we trivially see that it satisfies the Clifford algebra as it is inherited from the $(3+1)$ dimensional one. Indeed any subset of $d'$ $\gamma$ matrices of a Clifford algebra in dimension $d$ satisfies his own Clifford algebra in $d'$ dimensions.

The proof is trivial, if $(1)$ holds for all $\mu,\nu \in 1,\ldots,d$, then it holds for $\mu,\nu \in 1,\ldots,d'<d$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.