# Simplifying a seemingly simple gamma matrix identity

When studying from the book by Wise and Manohar, Heavy Quark Physics (pg 102), I came across a seemingly simple identity that I am not able to prove. It's likely an easy problem but I can't for the life of me figure out why.

First they define the transverse covariant derivative,

$$D_\perp \equiv D^\mu - D \cdot v v ^\mu$$

where $v^\mu$ is some four-vector that squares to 1 and then they write down the identity,

$$\require{cancel} \cancel{D_{\perp}}\cancel{D_{\perp}} = D_\perp^2 +\frac{1}{2}[\gamma_\mu,\gamma_\nu] D_\perp^\mu D_\perp^\nu$$

I would have naively thought that we would still have the usual, $D_\perp^2= \cancel{D_{\perp}}\cancel{D_{\perp}}$. Why is this not the case?

It is not the case because the gamma matrices do not commute and neither do the covariant derivatives. We can always write a 2-index tensor as the sum of its antisymmetric and its symmetric part, $$D^\mu D^\nu = \frac{1}{2}(D^\mu D^\nu + D^\nu D^\mu) + \frac{1}{2}(D^\mu D^\nu - D^\nu D^\mu).$$ The latter part vanishes precisely when the curvature (= field strength) vanishes. And we also have similarly for the gamma matrices $$\gamma^\mu\gamma^\nu = \frac{1}{2}( \gamma^\mu\gamma^\nu + \gamma^\mu\gamma^\nu ) + \frac{1}{2}( \gamma^\mu\gamma^\nu - \gamma^\mu\gamma^\nu ) = g^{\mu\nu} + \frac{1}{2}[\gamma^\mu, \gamma^\nu].$$ Since an antisymmetric tensor contracted with a symmetric tensor gives 0, $$\gamma^\mu\gamma^\nu D_\mu D_\nu = g^{\mu\nu}\frac{1}{2}(D_\mu D_\nu + D_\nu D_\mu) + \frac{1}{2}[\gamma^\mu, \gamma^\nu] D_\mu D_\nu = D^\mu D_\mu + \frac{1}{2} [\gamma^\mu, \gamma^\nu] D_\mu D_\nu.$$ The one half in the first term is cancelled by the metric being symmetric.