= Gamma-matrix vector sandwich
{title2=$\gamma_{ab}\gamma^m\gamma^{ab}=-(D-1)(D-4)\gamma^m$}
For <gamma matrices> with $\{\gamma^a,\gamma^b\}=2\eta^{ab}$ and $\gamma^{ab}=[\gamma^a,\gamma^b]/2$, contracted <Clifford algebra> gives $\gamma_a\gamma^m\gamma^a=(2-D)\gamma^m$. Expanding the two antisymmetrized factors gives $\gamma_a\gamma_b\gamma^m\gamma^a\gamma^b-D\gamma^m$. The five-gamma product equals $[2D-(D-2)^2]\gamma^m$, by moving the middle contracted gamma through and using the three-gamma contraction twice. Subtraction yields the displayed identity. A convention without the factor one-half in each bivector multiplies the result by four.
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