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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...

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The discrepancy comes from the fact that the square of the Dirac operator is in general not conformally covariant. There are ways to modify powers of the Dirac operator to get a conformally covariant operator (see, for example, this article of Fischmann). However, they involve adding lower-order curvature correction terms which, in your case, will fix the discrepancy in the constant in front of the scalar curvature.

For the first point, note that on an $n$-dimensional manifold, the conformal transformation formula for the Dirac operator is (ignoring how one must identify spin bundles)

$$ D_{e^{2u}g}\psi = e^{-\frac{n+1}{2}u} D_g \left( e^{\frac{n-1}{2}u}\psi \right) . $$

Already from this formula — specifically, the fact that the two exponential factors are not multiplicative inverses — you can see that it shouldn’t be the case that the composition is conformally covariant.

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