On an $(M,g)$ compact n-dimensional Riemannian manifold the Yamabe operator $Y = 4(n-1)/(n-2)L + s$, where $L$ is the Laplacian and $s$ is the scalar curvature is conformally covariant, which for me just means that transforms in a simple way under conformal transformations, $g \to f g$ then $Y \to f^\alpha Y f^\beta$ with some (known) $\alpha$ and $\beta$ exponents, here $f$ is a smooth positive function.

Let's assume $n=4$. Once a spin structure is chosen the Dirac operator also transforms convariantly under a conformal transformation and for its square we have $D^2 = L + F + s/4$ where $F$ stands for a term coming from the curvature of the underlying gauge field  and $s$ is again the scalar curvature. 

Now what I'm confused about is that if I set $n=4$ in the Yamabe operator, it will be proportional to $L + s/6$ and the square of the Dirac operator with a gauge field with zero curvature we get $L + s/4$. Is the mismatch between $1/4$ and $1/6$ okay? Or am I overlooking some trivial factors coming from different conventions?
 
I guess both shouldn't have the same nice conformal properties...