 Solution

Take $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$ and ${\gamma }^{ab}={\gamma }^{[a}{\gamma }^{b]}=[{\gamma }^a,{\gamma }^b]/2$, with unit-weight antisymmetrization. The Clifford algebra first givesExpand the two antisymmetrized gamma matrices using ${\gamma }_{ab}={\gamma }_a{\gamma }_b-{\eta }_{ab}$:Anticommuting ${\gamma }^m$ past the next contracted factor givesSubtracting the remaining $D{\gamma }^m$ gives the gamma-matrix vector sandwichIn particular it vanishes in four dimensions. The result is independent of the choice of Lorentz signature when all indices are contracted consistently. Defining the bivector without its factor $1/2$ would multiply $\alpha$ by four.